October 2015

edit Vol. XXIII

No. 10

cErN discovers new properties of

October 2015

higgs boson, the fundamental particle

Corporate Office : Plot 99, Sector 44 Institutional area, Gurgaon -122 003 (Hr). tel : 0124-4951200 e-mail : [email protected] website : www.mtg.in

H

iggs boson, the fundamental particle was discovered by cerN, three years back, in collaboration with the scientists of other laboratories.

Regd. Office 406, taj Apartment, Near Safdarjung Hospital, ring road, New Delhi - 110029.

Managing Editor Editor

: :

the Atlas and cMS collaborators of cerN have obtained the sharpest picture of the Higgs boson for the first time. Now they are on the way to study its properties, production, decay and how its interacts with other particles.

Mahabir Singh Anil Ahlawat (be, MbA)

contents Physics Musing (Problem Set-27)

rial

8

JEE Workouts

11

Core Concept

15

Thought Provoking Problems

20

Ace Your Way CBSE XI

23

Series 2 JEE Accelerated Learning Series

31

Brain Map

46

Bihar CECE

59

All the measured properties of this particle are in agreement with the standard model. the present studies have the best precision for performing measurements for these particles. A lot of work is yet to be done in the field of more familiar particles. Krane had been able to measure the radius and the error of determination for half a dozen well-known fundamental particles such as electron, proton, etc. these values were theoretically verified by our scientists. this story of cerN has proved that with international cooperation, discoveries are faster and their confirmation is also better when more than one lab is attached to the working teams of the bigger laboratories.

Solved Paper 2015 Exam Prep

66

You Ask We Answer

73

Physics Musing (Solution Set-26)

74

Ace Your Way CBSE XII

75

Series 5 Crossword

85

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Physics for you | OctOber ‘15

7

P

PHYSICS

MUSING

hysics Musing was started in August 2013 issue of Physics For You with the suggestion of Shri Mahabir Singh. The aim of Physics Musing is to augment the chances of bright students preparing for JEE (Main and Advanced) / AIIMS / Other PMTs with additional study material. In every issue of Physics For You, 10 challenging problems are proposed in various topics of JEE (Main and Advanced) / various PMTs. The detailed solutions of these problems will be published in next issue of Physics For You. The readers who have solved five or more problems may send their detailed solutions with their names and complete address. The names of those who send atleast five correct solutions will be published in the next issue. We hope that our readers will enrich their problem solving skills through “Physics Musing” and stand in better stead while facing the competitive exams.

Set 27 single oPtion correct tyPe

1. In a Young’s double-slit experiment the slit separation is 0.5 mm and the screen is 0.5 m from the slit. For a monochromatic light of wavelength 500 nm the distance of third maxima from the second minima on the other side is (a) 2.75 mm (b) 2.5 mm (c) 22.5 mm (d) 2.25 mm 2. A particle of mass m is projected from the surface of earth with a speed v0 (v0 < escape velocity). The speed of particle at height h = R (radius of earth) is (a)

gR

(b)

v02 − 2 gR

(c)

v02 − gR

(d)

v02 + 2 gR

3. A uniform magnetic field of induction B is confined to a cylindrical region of radius R. The magnetic field is increasing at a constant rate of dB (tesla/second). An electron of charge e, placed at dt the point P on the periphery of the field experiences an acceleration

8

Physics for you | OctOber ‘15

(a) 1 eR dB toward left 2 m dt (b)

1 eR dB toward right 2 m dt

eR dB toward left m dt (d) zero (c)

4. After one second the velocity of a projectile makes an angle of 45° with the horizontal. After another one second it is travelling horizontally. The magnitude of its initial velocity and angle of projection are (Take g = 10 m s–2) (a) 14.62 m s–1, 60° (b) 14.62 m s–1, tan–1(2) (c) 22.36 m s–1, tan–1(2) (d) 22.36 m s–1, 60° 5. A simple pendulum is suspended from the ceiling of a lift. When the lift is at rest its time period is T. With what acceleration should the lift be accelerated T

upwards in order to reduce its period to ? (g is the 2 acceleration due to gravity) (a) 4g (b) g (c) 2g (d) 3g

6. A block released from rest from the top of a smooth inclined plane of angle q1 reaches the bottom in t1. The same block released from rest from the top of another smooth inclined plane of angle q2, reaches the bottom in time t2. If the two inclined planes have the same height, the relation between t1 and t2 is 1/2

 sin q1  t (a) 2 =  t1  sin q2  (c)

t2 sin q1 = t1 sin q2

t (b) 2 = 1 t1 (d)

7. The magnetic flux f through a stationary loop of wire having a resistance R varies with time as f = at2 + bt (a and b are positive constants). The average emf and the total charge flowing in the loop in the time interval t = 0 to t = t respectively are (a) at + b,

2

at + bt R

(b) at + b,

2

at + bt 2R

at2 + bt at + b at2 + bt , (d) 2(at + b), 2R 2 R 8. Assuming all the surfaces to be frictionless, find the magnitude of net acceleration of smaller block m with respect to ground (c)

10

Physics for you | OctOber ‘15

2 5mg (5m + M )

(b)

(c)

7 5mg (5m + M )

(d) none of these



t2 sin2 q1 = t1 sin2 q2

2mg (5m + M )

(a)



9. Vectors A and B include an angle q between them.     If ( A + B) and ( A − B) respectively subtend angles a  and b with A , then (tana + tanb) is AB sin q 2 AB sin q (a) (b) A2 + B2 cos2 q

(c)

2

A2 sin2 q 2

2

A + B cos q

A2 − B2 cos2 q

(d)

B2 sin2 q

A2 − B2 cos2 q

10. Resultant of two vectors having same magnitude forms an angle with any of the vectors. If the magnitude of second vector is reduced to half of initial magnitude without changing the angle between the direction of new resultant vector and first vector is also reduced to half, then the angle between the two vectors is (a) 120° (b) 60° (c) 90° (d) 45° nn

integer tyPe Questions class-Xii 1. A monochromatic light of wave length 500 Å is incident on two identical slits separated by a distance of 5 × 10–4 m. The interference pattern is seen on a screen placed at a distance of 1 m from the plane of slits. A thin glass plate of thickness 1.5 × 10–6 m and refractive index 1.5 is placed between one of the slits and the screen. Find the intensity at the center of the slit now. 2. A totally reflecting, small plane mirror placed horizontally, faces a parallel beam of light as shown in the figure. The mass of the mirror is 20 g. Assume that there is no absorption in the lens and that 30% of the light emitted by the source goes through the lens. The power of the source needed to support the weight of the mirror is x × 108 W. Find x. (Take g = 10 m s–2) 3. A certain radioactive material can undergo three different types of decay, each with a different decay constant l, 2l, and 3l. Then, the effective decay constant leff is equal to nl. What is the value of n? 4. A coil is connected to an alternating emf of voltage 24 V and of frequency 50 Hz. The reading on the ammeter connected to the coil in series is 10 mA. If a 1 mF capacitor is connected to the coil in series the ammeter shows 10 mA again. What would be the approximate reading on a dc ammeter (in A) if the coil was connected to a 180 V dc voltage supply? (Take p2 = 10) 5. A metal disc of radius 25 cm rotates with a constant angular velocity 130 rad s–1 about its axis. Find the potential difference in nV between the centre and

rim of the disc if the external magnetic field is absent. 6. Figure shows a potentiometer circuit for determining the internal resistance of a cell. When switch S is open, the balance point is found to be at 76.3 cm of the wire. When switch S is closed and the value of R is 4.0 W, the balance point shift to 60.0 cm. Find the internal resistance of cell C′.

7. The focal length of a thin biconvex lens is 20 cm. When an object is moved from a distance of 25 cm in front of it to 50 cm, the magnification of its m image changes from m25 to m50. The ratio 25 is m50 8. An a-particle and a proton are accelerated from rest by a potential difference of 100 V. After this, their de Broglie wavelengths are la and lp respectively. lp The ratio , to the nearest integer, is la 9. When two identical batteries of internal resistance 1 W each are connected in series across a resistor R, the rate of heat produced in R is J1. When the same batteries are connected in parallel across R, the rate is J2. If J1 = 2.25J2 then the value of R in W is Physics for you | OctOber ‘15

11

10. To determine the half life of a radioactive element, dN (t ) versus t. a student plots a graph of ln dt dN (t ) is the rate of radioactive decay at time t. Here dt If the number of radioactive nuclei of this element decreases by a factor of p after 4.16 years, the value of p is

11. At time t = 0, a battery of 10 V is connected across points A and B in the given circuit. If the capacitors have no charge initially, at what time (in seconds) does the voltage across them become 4 V? [Take : ln5 = 1.6, ln3 = 1.1]

PQ = 3x, PR = 4x and QR = 5x. If the magnitude of  m I  the magnetic field at P due to this loop is k  0  ,  48 px  find the value of k. solutions

1. (0) : Take a point at a distance y from the centre of the slit. Path difference between waves reaching this point, dy Dx = − (m − 1)t D At the centre of slit y = 0 \ Dx = –(m – 1)t 2p 2p Dx = − (m − 1)t Phase difference φ = l l p φ Intensity, I = 4 I0 cos2   = 4 I0 cos2  − (m − 1)t   l  2 −6   = 4 I0 cos2  − p(1.5 − 1)1.5 × 10   5000 × 10−10  3p = 4 I0 cos2   = 0  2 

2

12. Image of an object approaching a convex mirror of radius of curvature 20 m along its optical 50 25 m axis is observed to move from m to 7 3 in 30 seconds. What is the speed of the object in km per hour ? 13. A large glass slab (m = 5/3) of thickness 8 cm is placed over a point source of light on a plane surface. It is seen that light emerges out of the top surface of the slab from a circular area of radius R cm. What is the value of R ? 14. Three charges q1 = 3 mC, q2 = –3 mC and q3 are kept at the vertices of a triangle as shown in the figure. If the net force acting on q1  is F, the charge q3 is then 2

 1 given as 1 +  mC.  n Find the value of n. 15. A steady current I goes through a wire loop

PQR having shape of a right angle triangle with

12

Physics for you | OctOber ‘15

(1) : Let n photons (each of frequency f) per second are emitted from source. Then power of source is P = nhu But only 30% of the photons go towards mirrors. Then force exerted on mirror is 30  h 3 nhu 3 P F = 2  = n = 5c 100  l 5 c This force should be equal to weight of mirror, so 3P = 20 × 10−3 × g 5c 5 × 3 × 108 × 20 × 10−3 × 10 or P = = 1 × 108 W 3 \ x=1

3. (6) : Effective decay constant will be the sum of all different decay constants So, leff = l l + 2l + 3l = 6l, hence n = 6. 2 2 4. (1) : Z = (wL) + R =

24 10 × 10−3

1   R2 + (wL)2 = R2 +  wL −   wC 

2

1 (wL) = −wL + wC 1 1 L= 2 = 2w C 2 × 100 p × 100 p × 10−6

= 5 H; (2400)2 = (500p)2 + R2

7. (6) : Magnification, m =

According to cartesian sign conventions 20 =−4 m25 = 20 − 25

R = (2400)2 − (5p × 100)2 = 10 (24)2 − 25p2 = 10 × 326 ≅ 180, I = 1 A 5. (3) : Centripetal force required for circular motion of electron is generated by radial electric field caused by the redistribution of the electron in the disc. 2 F = eE = mrw2 or E = mrw e From dV = –E dr, mw2 r dr or dV = − e V2 −mw2 R r dr or ∫ dV = V1 e ∫0 or V1 − V2 =

2 2

−31

2

m50 = 8.

m −2 20 12 = \ 25 = = 6 20 − 50 3 m50 2 h

(3) : de Broglie wavelength l =

h

l=

\

h

= 3.0 × 10–9 V = 3.0 nV 6. (1) :

, la =

2m p q pV

lp

=

la

=

2mK

[ kinetic energy, K = qV ]

2mqV

lp =

2

(9.1 × 10 )(130) (0.25) mw R = 2e (2)(1.6 × 10−19 )

f f +u

h 2m p q pV ma qa mpq p

=

×

h 2ma qaV 2ma qaV h

(4m p )(2q p ) mpq p

= 8 =2 2 ≈3

9. (4) :

Let e be the emf of the cell C′ and r its internal resistance. Let l = AJ be the balance length when switch S is open. When a resistance R is introduced by closing the switch a current begins of flow through the cell C′ and resistance R. The potential difference between the terminals of the cells falls and the balance length decreases to l′ = AJ′. The terminal resistance of the cell is given by E −V r= I where V is the terminal voltage of C′ and I is the current in the circuit involving C′ and R. V Also I = . R E \ r =  − 1 R V 

But E = l . V l′ \

′  r = R  l − l  = 4.0 ×  76.3 − 60.0   1 W  60.0   l′ 

In series,

2

 2e  R Rate of heat produced in R is J1 =   R + 2 

In parallel Rate of heat produced in R is 2

2

 2e   e  R J2 = R =  1  2R + 1   R +  2 \

2

2

 2R + 1   2e   2 R + 1  = × =    J 2  R + 2   2e   R+2  J1

2

According to given problem J1 = 2.25J2 Physics for you | OctOber ‘15

13

2

2R + 1  2R + 1  or 1.5 = \ 2.25 =   R+2  R+2  or 1.5R + 3 = 2R + 1 ⇒ 0.5R = 2 2 \ R= =4W 0. 5 10. (8) : According to radioactive decay N = N0e–lt dN = − lN 0e − lt dt dN = − lN 0 e − lt dt dN = lN 0 e − lt dt Taking natural logarithms of both sides of above equation, we get dN = ln(lN 0 ) − lt ln dt dN = − lt + ln(lN 0 ) dt Comparing the above equation with equation of a straight line i.e. y = mx + c, we get 3−4 From graph, slope = − l = 6−4 1 or l = year −1 2 0.693 Half life T1/2 = = 2 × 0.693 years =1.386 years l 4.16 years is approximately 3 half lives Nuclei will decay by a factor of 23 = 8 \ p=8 ln

11. (2) : The equivalent resistance of the two parallel resistors is (2 MW) (2 MW) R= = 1 MW (2 MW) + (2 MW) The equivalent capacitance of the two parallel capacitors is C = 2 mF + 2 mF = 4 mF The corresponding equivalent diagram is as shown in the figure.

Time constant of the circuit, t = RC = (1 MW)(4 mF) =4s –t/t Since V(t) = V0(1 – e ) Here, V(t) = 4 V, V0 = 10 V \ 4 = 10(1 – e–t/t) 6 3 10e–t/t = 6 or e −t /t = = 10 5 −t = ln 3 − ln 5 t t or = ln 5 − ln 3 = 1.6 − 1.1 = 0.5 t or t = (0.5)(4 s) = 2 s 12. (3) : Focal length of a convex mirror,

R 20 = m = 10 m 2 2 For first object 25 v1 = + m, f = + 10 m 3 1 1 1 Using mirror formula + = v u f 1 1 1 1 1 3 + = \ or = − (25 / 3) u1 10 u1 10 25 u1 = – 50 m For second object 50 v2 = + m, f = + 10 m 7 1 1 1 \ + = v2 u2 f 1 1 1 1 1 7 or + = = − (50 / 7) u2 10 u2 10 50 f =

u2 = – 25 m

13. (6) : sin θc =

From figure, sin θc = 14

Physics for you | OctOber ‘15

25 m s −1 30 25 18 –1 = × km h −1 = 3 km h 30 5

Speed of the object =

1 3 = m 5

R

R2 + 82

...(i)

...(ii) ....Contd. on page no. 84

Centre of Mass The centre of mass (COM) for any given system of masses represents that particular point for the entire system whose motion is not dependent upon the point of application of force but rather what matters is how much force has been applied and in which direction the force has been applied, and not where it has been applied. To understand this better, let us consider three cases of a uniform rod being hit at three different points. F C

vc

C 

F

vc

C



1. For discrete point masses

Let m1, m2, .... mn, be point masses with position vectors    r1 , r2 , .... rn then the position vector of COM is,     Σmi ri m r + m2r2 + ..... + mnrn =  rCOM = 1 1 Σmi m1 + m2 + ..... + mn  \ rCOM = x COM i^ + y COM j^ + z COM k^ Σmi xi Σmi y i Σmi z i , y COM = ,z = Σmi Σmi COM Σmi where (xi , yi , zi) are the x, y and z co-ordinates of the ith mass. Note : They are co-ordinates and hence + or – has to be taken care of. where x COM =

2. For continuous mass distribution vc

F

The diagram clearly shows, that the motion of different points of the rod is different in all the three cases but there, at point C (in this case the geometrical centre) whose motion is identical in all the three cases. This point C is the COM. Hence, one more definition of COM is also prevalent amongst most books which says it is one such point for the entire system of given masses where the entire mass may be assumed to be concentrated. Mathematically, the location of COM is calculated as shown :

Choose an elemental mass dm, the position vector of  whose COM is r , then the COM of the entire system becomes  dm r  ∫ rCOM = ∫ dm Now, with the formula being known, let us calculate the location of COM for some standard configuration.

Contributed By: Bishwajit Barnwal, Aakash Institute, Kolkata

Physics for you | OctOber ‘15

15

COM of a two point mass system

Let us take COM to be somewhere in between as shown and from there we have taken x and y axis for simplification of calculation. m x + m2 x2 m1 (−r1 ) + m2 (+r2 ) \ xCOM = 0 = 1 1 = m1 + m2 m1 + m2 ⇒ m1r1 = m2r2 Hence greater the mass, lesser its distance from COM. Thus COM is shifted more towards heavier mass. also, r1 + r2 = r  m1   m1  ...(i) r1 =  r , r2 =  r   m1 + m2   m1 + m2  Clearly, if the masses been equal, the COM would be located midway on the Line of Symmetry (LOS). Hence if we can find any LOS for the given system then draw it and the COM will lie on this LOS. If there are two or more LOS, then their point of interaction gives COM. Q1. Locate the COM of the system shown in the figure.

Q2. From a uniform disc of radius 2R, a disc of radius R is cut out touching the periphery as shown. Locate the COM.

Soln.: The LOS is drawn as below

Let us fill in the portion which was cut-out, now the COM of 2 is known as well as combined 1 + 2 is known. Let us assume the COM of the given figure (part 1 in our diagram) at a distance x from the combined COM. For 1 + 2 combined system

Soln.: Seeing this as three separate uniform rods, their respective COM will be at their geometric centre and mass being proportional to length, we can see the same figure as shown below

m x + m2 x2 xCOM = 1 1 =0 m1 + m2 ⇒ m1x1 + m2x2 = 0 But since the disc is uniform, m ∝ A (area) \ A1x1 + A2x2 = 0 ⇒ p[(2R)2 – R2] (–x) + (pR2)(R) = 0 ⇒ x = R/3 COM of non-uniform rod with linear mass density linearly dependent on distance measured from one end A

x

dx

B

(x) = (ax + b)kg m–1

Given : The linear mass density of rod AB of length l is proportional to distance measured from one end A. To locate COM, we choose an elemental mass dm where, dm = ldx = (ax + b)dx 16

Physics for you | OctOber ‘15

l

xCOM =

∫ xdm = 0 ∫ dm l (ax + b)dx ∫ 0

al 3 bl 2 + 2al 2 + 3bl = 32 2 = 3(al + 2b) al + bl 2 2al 2 + 3bl ...(ii) \ xCOM = 3(al + 2b) Note : If the linear mass density is given to be proportional to distance being measured from one end, l(x) = ax i.e., b = 0 2l \ xCOM = 3 COM of a thin uniform semicircular ring

COM R

d

=

Rsin = y



LOS p

\ yCOM =

m dm = R ·Rd = md 

R

∫

∫ dmy = 0 ∫ dm

2R p A quarter ring is identical from y-axis as well as x-axis hence,  2R  2R \ OC = 2   xCOM = yCOM =  p  p COM of a uniform semicircular disc Let us divide the entire disc, into small triangular strips as below : \ yCOM =

2 ∫ (ax + bx)dx

Since, each thin triangular sheet is uniform, their COM will lie at centroid, i.e., at 2R from common vertex. 3 Joining all the point masses, we have a ring of radius 2R r= 3 2r 2  2R  4 R ...(iv) \ yCOM = =   = p p  3  3p COM of a thin uniform hollow hemisphere

m dθ ⋅ R sin θ p m

p

R 2R sin θdθ = ∫ p p

...(iii)

0

Q3. Locate the COM of a thin uniform quarter ring. C 45° COM 45° O LOS

OC = ?

Soln.:Let us add the other quarter so that it becomes a half ring as shown in the figure below

We divide the hemisphere into thin rings.  m  \ dm =  (2 p R sin θ)(Rdθ)  2 pR2  \ yCOM =

∫ dmy ∫ dm p /2

∫

m sin θdθ ⋅ R cos θ

= 0 =

R 2

p /2

∫ 0

m sin 2θ dθ =

R 2

...(v)

COM of a uniform solid hemisphere

We divide the hemisphere, into thin hemispherical shells. Physics for you | OctOber ‘15

17

  \ If vCOM = 0 initially and Fext = 0  ⇒ DrCOM = 0    m1Dr1 + m2 Dr2 + ........... + mn Drn = 0

COM r

y= r 2

dr

Q4. A man of mass m stands on a rough horizontal plank of mass M whose lower surface is smooth.

R

3m 2  M  (2 pr 2dr ) = r dr 2 3 3 R pR   3  r The COM of this element is at . 2 dm =

R

\ yCOM =

∫ dmy ∫ dm

3m 2 r dr ⋅ 2

∫ R3 ⋅ r

=0

m

=

3R 8

...(vi)

If the man walks a distance l over plank, find the distance moved by plank. Soln.: Even though we understand that the plank would recoil towards left due to friction force of man’s foot, we assume the plank to be shifting towards right for simplification of calculation.

There are certain standard results which needs to be memorized. Motion of COM

 Σmi ri  rCOM = Σmi \ Velocity of COM,

 dri   Σmi dr  dt = Σmi vi vCOM = COM = Σmi Σmi dt    m1v1 + m2v2 + ....... + mnvn  \ vCOM = m1 + m2 + ...... + mn

\ Linear momentum of COM is,      pCOM = MVCOM = m1v1 + m2v2 + ...... + mnvn Where M = m1 + m2 + ..... + mn     dpCOM dv dv dv \ = m1 1 + m2 2 + ........ + mn n dt dt  dt  dt = m1a1 + m2a2 + ............ + mnan    = F1 + F2 + .......... + Fn which is the vector sum of all the forces acting on the combined system which is the rate of change of momentum of COM. This proves that for COM, the net force is important and not the point of application of force. This gives us one important conclusion, if there is no net external force in any direction for combined system, the linear momentum of the system in this direction cannot be changed. And specifically speaking, if the COM was initially at rest and due to internal forces, objects start moving, then they move such that COM does not move. 18

Physics for you | OctOber ‘15

On the combined system, Fhorizontal = 0 ⇒ DxCOM = 0 ⇒ m1Dx1 + m2Dx2 = 0 ⇒ m(x + l) + Mx = 0  ml  ⇒ x=−  m + M  \ Negative sign signifies that plank moves opposite to ml . assumed direction by a distance | x | = m+ M This result gives us the answer without calculations in few questions as shown here. m

m R M

x

M

Smooth

m

R mR  |x| = m+M

l



M

M x

Smooth

 m lsin

mlsin  |x| = m+M

Acceleration of COM

 dvi   Σ m i dv  dt = Σmi ai aCOM = COM = Σmi Σmi dt    m1a1 + m2a2 + ......... + mnan  \ aCOM = m1 + m2 + ..... + mn =

Q5.

m1

m2

Q6. F1

m1

k

m2

F2

Feff

m2

x2

1 ⇒ Feff (x1 +x2 ) = k(x1 +x2 )2 2 2(m2 F1 + m1F2 ) 2F ⇒ xmax = x1 + x2 = eff = k (m1 + m2 )k

Smooth

m x + m2 x2 1 \ SCOM = aCOMt 2 = 1 1 m1 + m2 2 1 2 ⇒ Ft − m1x1 = m2 x2 2 1 F 2  m1  t − x1 ⇒ x2 = 2 m2  m2 

m2

At maximum elongation v1 = v2 = 0 \ Using work energy theorem, 1 Feff x1 + Feff x2 = DPE = k(x1 +x2 )2 2

m1 F

The spring is in its natural length and a constant force starts acting on m1 as shown. In a time t, m1 displaces by x1 then find the displacement of m2? F Soln.: aCOM = = constant m1 + m2

m1

x1

vector sum of all the forces total mass k

Feff

Q7.

For (m2 > m1), find : (i) aCOM (ii) Net force acting on the combined system. Soln.:

Smooth

Find the maximum elongation in spring if the spring is initially in its natural length. (Assume F2 > F1) F2 − F1 towards right. Soln.: aCOM = m1 + m2

If we analyse the motion from COM, we need to include a pseudo force for both, as shown below

 F − F  m F + m1F2 \ F1eff = F1 + m1  2 1  = 2 1 m1 + m2  m1 + m2   F − F  m2 F1 + m1F2 F2eff = F2 − m2  2 1  = m1 + m2  m1 + m2  F1eff = F2eff = Feff With respect to COM, equal and opposite forces act on the two objects.

Clearly,  m − m1  a2 = a1 =  2  g =a  m1 + m2  m (−a ) + m2 (a2 ) aCOM = 1 1 m1 + m2 2

 m − m1  = 2  g  m1 + m2  Fnet = (m1 + m2 ) aCOM =

(m2 − m1 )2 g (m1 + m2 ) nn

Physics for you | OctOber ‘15

19

By : Prof. Rajinder Singh Randhawa*

ROTATIONAL MOTION 1. Determine the moment of inertia about the x-axis of the solid spherical segment of mass M. O

X

R/2 R/2

2. A body consisting of a cone and hemisphere of radius r fixed on the same base rests on a table, the hemisphere being in contact with the table. Find the maximum height of the cone, so that the combined body may stand upright. 3. Consider the motion of the cylinder of mass m and radius r moving up a plane inclined at angle q to the horizontal. To begin with, cylinder rotates with angular velocity wo about its axis when gently placed on the plane so that its initial translational velocity is zero. Find the distance moved by cylinder before sliding stops. o 

along the smooth horizontal surface by a horizontal force F = 60 N. Determine the angle q for translation. What is the accompanying acceleration?

4. As shown in figure, a homogeneous slider bar with a mass of 4 kg and a length of 500 mm being pushed

m

0m

F

50



5. The mass of block A is twice the mass of block B. Find the acceleration of A in terms of the gravitational acceleration. Neglect the mass of the pulleys shown in figure.

6. A boy is pushing a ring of mass 2 kg and radius 0.5 m with a stick as Ground shown in figure. The stick applies a force of 2 N on the ring and rolls it without slipping with an acceleration of 0.3 m s–2. The coefficient of friction between the ground and the ring is large enough that rolling always occurs and coefficient of friction between the stick and the ring is x/10. Find the value of x?

*Randhawa Institute of Physics, S.C.O. 208, First Fl., Sector-36D & S.C.O. 38, Second Fl., Sector-20C, Chandigarh, Ph. 09814527699

20

Physics for you | OctOber ‘15

y

SOLUTIONS

1. The solid sphere is generated by revolving circle about the diameter. Mass of spherical segment, M = Density × Volume of segment Consider a small disc of radius y and thickness dx as shown in figure. Mass of small disc, dm = r·py2·dx ...(i) M M M r= = R = R V 2 2 2 ∫ py dx ∫ p(R − x )dx R /2

r=

M 3

5p R /24

=

R /2

24 M

...(ii)

5pR3

y

A R

y x

C

O x

Moment of inertia of disc about x-axis 1 1 dI = dm ⋅ y 2 = rpy 4dx 2 2

I=

rp 2

R

∫

R

∫

R /2

y 4 dx =

O

r

x

Weight of cone, W2 = rg p r 2h. 3 Distance of centre of gravity of cone from bottom, y2 = r + h/4. Centre of gravity of combined body W y + W2 y2 y= 1 1 W1 + W2 2p 5r 1  h rg r 3 × + rg pr 2h  r +   3 8 3 4 y= 2p 3 1 2 rg r + rg pr h 3 3 For stable equilibrium, y ≤ r 2 p 3 5r rg 2  h  r × + pr h  r +   3 8 3 4 ≤r 2p 3 1 r + rg pr 2h rg 3 3 \ h2 ≤ 3r2 ⇒ h ≤ 1.732 r rg

dx B x2 + y 2 = R 2

1 \ I = ∫ dI = rp 2

h

rp 2

R

∫

(using (i))

(R2 − x 2 )2 dx

R /2

3. The cylinder initially slides on the plane. Surface of the cylinder slides downward (backward) with respect to the plane; hence, the frictional force acts upwards (forward) as shown in figure. The equation of translational motion is mmgcosq – mgsinq = ma

(R 4 − 2R2 x 2 + x 4 )dx

R /2

R 24 pM  4 2R2 x 3 x 5  = +  R x − (using (ii)) 3 5  10 pR3  R /2 53 2 Solving, we get, I = MR 200

2. The composite body is symmetrical about y axis therefore its centre of gravity will lie on this axis. Now consider two parts of the body, hemisphere and cone. Let the bottom of the hemisphere be the axis of reference as shown in figure. 2p 3 Weight of hemisphere, W1 = rg r . 3 Distance of centre of gravity of hemisphere from 5 bottom y1 = r 8

or a = g(mcosq – sinq) Velocity of cylinder after time t is v = at = g (mcosq – sinq)t Frictional force produces a torque. Hence, t = (mmgcosq)r = Ia

...(i) ...(ii)

mmg cos q.r 2mg cos q = ...(iii) 1 2 r mr 2 Angular velocity decreases with angular retardation a with time t, \ w = wo – at ...(iv) or a =

Physics for you | OctOber ‘15

21

At the moment sliding stops and pure rolling begins, we have v = wr, i.e. rolling begins at time t = T. v w = = wo − aT r g (m cos q − sin q)T = wo − aT ⇒ r 2mg cos qT g (m cos q − sin q)T = ⇒ wo − r r wo r or T = g (3m cos q − sin q) The distance moved by the cylinder before sliding stops is 1 S1 = aT 2 2   wor 1 = g (m cos q − sin q) ×   2  g (3m cos q − sin q)  \ S1 =

2

w2o r 2 (m cos q − sin q)

Equation of motion for body A is 2mg – T = 2ma2 ...(i) Equation of motion for body B is 2T – mg = ma1 ...(ii) Length of the rope L = y1 + (y1 – l) + (y2 – l) + 2 × (half of circumference) or 2y1 + y2 = constant Differentiating twice w.r.t. t, we get, d 2 y2 + = 0 ⇒ 2a1 + a2 = 0 dt 2 dt 2 Acceleration of body A = 2 × Acceleration of body B From equation (i), 2mg – T = 2ma2 = 4ma1 ...(iii) Solving (ii) and (iii), we get a1 = g/3, a2 = 2a1 = 2g/3 2

2g 3 6. There is no slipping between ring and ground. Hence f2 is not maximum. \

2 g (3m cos q − sin q)2

4. Free body diagram of slider bar is shown in figure.

d 2 y1

Acceleration of body A =

a

Angular acceleration, a = 0 and linear acceleration = a Applying the equation SFH = ma ⇒ 60 = 4a \ a = 15 m s–2 ...(i) SFV = 0 ⇒ RA = 4g = 39.24 N ...(ii) Taking moment about G, we get, 60 × 0.25sinq – RA × 0.25cosq = 0 39.24 ⇒ tan q = = 0.654 60 q = tan–1 (0.654) ⇒ q = 33.18°

5. The free body diagram of two masses are shown in figure. y1

a1

l

T T

B

T T 2T

T T A 2m 2mg

mg

22

Physics for you | OctOber ‘15

y2 a2

 N mg 1 N2

f1 f2

But there is slipping between ring and stick. Therefore, f1 is maximum. 1 Now, I = mR2 = 2(0.5)2 = kg m2 2 N1 – f2 = ma or N1 – f2 = 2 × 0.3 = 0.6 N Rt R a=Ra= = R( f2 − f1 ) = R2 ( f2 − f1 ) / I I I

...(i)

or 0.3 = (0.5)2 (f2 – f1)/I or 0.3 =

(0.5)2 ( f2 − f1 ) 1/ 2

or f2 – f1 = 0.6 N

...(ii)

N1 = 2 N

...(iii)

x Further, f1 = mN1 =   N1  10  Solving these four equations, we get, x = 4

...(iv) nn

Series 22 Series

CHAPTERWISE UNIT TEST : Laws of Motion| Work, Energy and Power| System of Particles and Rotational Motion GENERAL INSTRUCTIONS (i) (ii) (iii) (iv) (v) (vi) (vii)

All questions are compulsory. Q. no. 1 to 5 are very short answer questions and carry 1 mark each. Q. no. 6 to 10 are short answer questions and carry 2 marks each. Q. no. 11 to 17 are also short answer questions and carry 3 marks each. Q. no. 18 is a value based question and carries 4 marks. Q. no. 19 and 20 are long answer questions and carry 5 marks each. Use log tables if necessary, use of calculators is not allowed.

1. Why are porcelain objects wrapped in paper or 2. 3. 4. 5. 6.

7. 8.

9. 10.

straw before packing for transportation? A light body and a heavy body have the same kinetic energy. Which one will have the greater momentum? Find the radius of gyration of a hollow sphere rotating about its diameter. Why are rockets given a conical shape? About 4 × 1010 kg of matter is converted into energy in the Sun each second. What is the power output of the Sun? A rigid spherical body is spinning around an axis without any external torque. Due to change in temperature, the volume increases by 1%. What will be the percentage change in angular velocity? What is the tension in rod of length L and mass M at a distance y from F1 when the rod is acted on by two unequal forces F1 and F2 (F2 < F1) at its ends? A sphere of mass m moving with a velocity u hits another stationary sphere of same mass. If e is the coefficient of restitution, what is the ratio of the velocities of two spheres after the collision? A particle is projected making an angle of 45° with horizontal having kinetic energy K. What is the kinetic energy at highest point ? A sphere of radius r is rolling without sliding. What is the ratio of rotational kinetic energy and total kinetic energy associated with the sphere?

OR

An automobile engine develops 100 hp when rotating at a speed of 1800 rpm. Find the torque acting. 11. An object of weight W hangs from a rope that is tied to other ropes that are fastened to the ceiling as shown in figure. The upper ropes make angles q and φ with the horizontal. Find the tensions T1, T2 and T3 in the three ropes. 



T1

T2 T3 W

12. A ball A moving with a velocity of 9 m s–1 strikes an

identical stationary ball B such that after collision the direction of each ball makes an angle of 30° with the original line of motion. Find the speeds of the two balls after the collision. Is the kinetic energy conserved in the collision process? 13. A solid cylinder rolls up an inclined plane of angle of inclination 30°. At the bottom of the inclined plane, the centre of mass of the cylinder has a speed of 5 m s–1. (a) How far will the cylinder go up the plane? (b) How long will it take to return to the bottom ? Physics for you | OctOber ‘15

23

14. Explain why

15.

16.

17.

18.

24

(a) a horse cannot pull a cart and run in empty space. (b) passengers are thrown forward from their seats when a speeding bus stops suddenly. (c) It is easier to pull a lawn mower than to push it. A student holds two dumbbells without stretched arms while standing on a turntable. He is given a push until he is rotating at a rate of 0.5 rps. Then the student pulls the dumbbells towards his chest. What is the new rate of rotation? Assume the dumbbells are originally 60 cm from his axis of rotation and are pulled into 10 cm from the axis of rotation. The mass of the dumbbells is such that the student and dumbbells have equal angular momentum when at 60 cm distance. A locomotive of mass m starts moving so that its velocity varies according to the law v = k s , where k is constant, and s is the distance covered. Find the total work performed by all the forces which are acting on the locomotive during the first t second after the beginning of motion. OR Two discs of moment of inertia I1 and I2 about their respective axes (normal to the disc and passing through the centre) and rotating with angular speeds w1 and w2 are brought into contact face to face with their axes of rotation coincident. (a) What is the angular speed of the two-disc system? (b) Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy? (Take w1 ≠ w2). If 28 × 1023 molecules of a gas strike a surface of area 14 cm2 normally per second with velocity of 500 m s–1 and rebound in the opposite direction with the same speed, find the pressure exerted by the gas on the surface if mass of each molecule is 5 × 10–23 g. Sam went to shopping mall to purchase certain goods .There he noticed an old lady struggling with her shopping. Immediately he showed her the lift and explained to her how it carries the load from one floor to the next. Even then the old lady was not convinced. Then Sam took her in the lift and showed her how to operate it. That old lady was very happy. (a) What values does Sam possess? (b) An elevator can carry a maximum load of 1800 kg in moving up with a constant speed of 2 m s–1. The frictional force opposing the motion is 4000 N. Determine the minimum power delivered by the motor to the elevator in watts as well as in horse power. Physics for you | OctOber ‘15

19. A bob of mass m is suspended by a light string of

length L. It is imparted a horizontal velocity v0 at the lowest point A such that it completes a semicircular trajectory in the vertical plane with the string becoming slack only on reaching the topmost point, C as shown in figure. Obtain an expression for (a) v0 ; (b) the speeds at points B and C; (c) the ratio of the kinetic energies at B and C (KB/KC). Comment on the nature of the trajectory of the bob after it reaches the point C.

OR Prove the result that the velocity v of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height h is given by 2 gh v2 = (1 + k 2 / R2 ) using dynamical consideration (i.e., by consideration of forces and torques). Note k is the radius of gyration of the body about its symmetry axis, and R is the radius of the body. The body starts from rest at the top of the plane. 20. A particle of mass m slides from the top of the surface of sphere of radius r. It loses contact and strikes the ground. At what distance below the top, the particle will lose contact with the surface? At what distance from the initial position the particle strikes the ground? OR A small mass slides down an inclined plane of inclination q with the horizontal. The coefficient of friction, µ = µ0x, where x is the distance through which the mass slides down and µ0 is a constant. Find the distance covered by the mass before it stops. What is the maximum speed over this distance? solutions 1. When porcelain objects are wrapped in paper or

straw, the time of impact between themselves is very much increased during jerk while transportation. Dp , as Dt increases, F decreases. Dt Hence, force on the porcelains is reduced during transportation, and saves them from breakage. Since F =

p2 , \ p = 2mK . 2. Kinetic energy, K = 2m Since K is same for both bodies, p ∝ m , i.e., the heavier body has more momentum than the lighter body. 3. The moment of inertia of a hollow sphere about its 2 diameter = MR2 ..(i) 3 If K is the radius of gyration, then moment of inertia of a body = MK2 ...(ii) From (i) and (ii) 2 2 MR2 = MK 2 \ K = R 3 3 4. Rockets are given a conical shape because conical

shape of rockets minimize the atmospheric friction. It also helps to maintain its direction. 5. Energy liberated per second by the conversion of 4 × 1010 kg of matter, i.e., E = mc2 =[(4 × 1010) (3 × 108)2] J = 3.6 × 1027 J Power output of the Sun = energy (work) liberated per second = 3.6 × 1027 J s–1 = 3.6 × 1027 W 6. Angular momentum of spherical body, 2  2 Q I = 5 MR 

2 L = I w = MR2 w 5 Also, volumeV =

4 3 πR or 3

2  3V  \ L= M  5  4π 

2 /3

 3V  R2 =    4π 

2 /3

w

Q L and M do not change with temperature \ w ∝ V −2/3

7.

dw 2 dV × 100% = − × 100% w 3 V So, the angular velocity decreases by 0.67%. F1 T

A y B

L

F2

Refer to figure, acceleration of the rod along F1, (F − F ) a= 1 2 M Mass of the part (say AB) of length y, M m= y L If T is the tension in AB, then M F1 − T = ma =  L

y  (F − F ) y  1 2 = (F1 − F2 )  M L

or T = F1 − (F1 − F2 )

y y  y = F1 1 −  + F2   L  L L

8. Here, u1 = u, u2 = 0

\ e=

v2 − v1 v2 − v1 = u1 − u2 u−0

or v2 – v1 = e u By the law of conservation of momentum, mu + m × 0 = mv1 + mv2 or v1 + v2 = u Adding (i) and (ii), 2v2 = u + eu = u (1 + e) u(1 + e) or v2 = 2 Again, from (ii), u(1 + e) u(1 − e) v1 = u − v2 = u − = 2 2 Divide eqn (iii) by eqn (iv)

...(i) ...(ii)

...(iii)

...(iv)

v2 1 + e = . v1 1 − e 1 2 Velocity at the highest point, u' u = horizontal component of u = u cos 45° = 2 Hence KE at the highest point, 1 K ′ = mu ′ 2 2

9. Initial kinetic energy, K = mu2

2

1  u  1 1  K K′ = m =  mu2  = .   2 2  2 2 2 KE at highest point will be half of the initial kinetic energy. 1 10. Translational kinetic energy, ET = mv 2 2 1 and Rotational kinetic enegy ER = I w2 2 Physics for you | OctOber ‘15

25

1 1 Total energy, E = ET + ER = mv 2 + I w2 2 2 1 2 1 2 2 v2 = mv + × mr × 2 2 2 5 r 1 1 7 = mv 2 + mv 2 = mv 2 2 5 10 1 2 mv ER 2 \ = 5 = . 7 E 7 mv 2 10 OR Here, N = 1800 rpm, P = 100 hp = 100 × 746 W Angular velocity, 2 π × 1800 w = 2π N = = 60 π rad s −1 60 Now power = tw or 100 × 746 = t × 60 π 746 × 100 = 395.9 N m 60 π 11. The free body diagram for the forces acting at O, is shown here. ⇒ t=

From ∑ Fx = 0, T2 cos φ − T1cosq = 0 From ∑ Fy = 0, T1 sin q + T2 sin φ − W = 0 From eqns. (i) and (ii),  T cos q  T1 sin q +  1 sin φ = W  cos φ  or T1 (sin q + cos q tan φ) =W W or T1 = (sin q + cos q tan φ) Similarly, T2 =

W (sin φ + cos φ tan q)

and T3 = W 26

Physics for you | OctOber ‘15

...(i) ...(ii)

12.

Initial momentum of A and B along X-axis = m × 9 + m × 0 = 9m ...(i) Final momentum of A and B after collision along X-axis = mv1 cos 30° + mv2 cos 30° m 3 (v1 + v2 ) 2 From eqns. (i) and (ii), =

...(ii)

m 3 (v1 + v2 ) = 9m 2 or (v1 + v2 ) = 6 3 ...(iii) Applying the law of conservation of momentum along Y-axis, mv1 sin 30° = mv2 sin 30° or v1 = v2 ...(iv) From eqns. (iii) and (iv), v1 = v2 = 3 3 m s −1 1 81 Initial KE = m(9)2 = m = 40.5 m 2 2 1 2 1 2 1 Final KE = mv1 + mv2 = m(v12 + v22 ) 2 2 2 1 2 2 = m[(3 3 ) + (3 3 ) ] = 27m 2 KE lost in the collision 40.5 m – 27 m = 13.5 m So, kinetic energy is not conserved. 1 2 (a) As the cylinder goes up the plane, it acquires potential energy at the expense of its KE of translational and rotational motion. Let the cylinder go up the plane upto height h. Then according to principle of conservation of energy, we have, 1 2 1 2 mv + I w = mgh 2 2

13. Here q = 30°, v = 5 m s–1, I = mr 2

or

1 2 1  1 2  v2 mv +  mr  2 = mgh r 2 2 2

[... v =rw]

3 2 mv = mgh 4 3v 2 3 × (5)2 \ h= = = 1.913 m 4g 4 × 9. 8 Suppose that the cylinder covers a distance S along the incline in reaching height h on the plane. Then, h h 1.913 sin q = or S = = = 3.826 m S sin q sin 30° (b) For a cylinder rolling down an inclined plane, linear acceleration. 2 2 a = g sin q = × 9.8 × sin 30° = 3.27m s −2 3 3 1 1 Now S = v0t + at 2 = 0 + at 2 {Q v0 = 0} 2 2 2S \ t= = 1.53 s a or

14. (a) While trying to pull a cart, a horse pushes the

ground backwards with a certain force at an angle. The ground offers an equal reaction in the opposite direction, on the feet of the horse. The forward component of this reaction is responsible for motion of the cart. In empty space, there is no reaction and hence a horse cannot pull the cart and run. (b) This is due to inertia of motion. When the speeding bus stops suddenly, lower part of the bodies in contact with the seats stop. The upper part of the bodies of the passengers tend to maintain the uniform motion. Hence the passengers are thrown forward. (c) While pulling a lawn mower, force is applied upwards along the handle. The vertical component of this force is upwards and reduces the effective weight of the mower, as shown in figure (i). While pushing a lawn mower, force is applied downward along the handle. The vertical component of this force is downwards and increases the effective weight of the mower, as shown in figure (ii). As the effective weight is lesser in case of pulling than in case of pushing, therefore, pulling is easier than pushing.

15. As L = mvr = mr2w,

initial angular momentum of both the dumbbells of total mass m, Ld1 = mr12w. If Ls1 is the initial angular momentum of the student, then initial angular momentum of the system, L1 = Ls 1 + Ld 1 = Ls 1 + mr12 w1 ...(i) Final angular momentum of the system,

L2 = Ls2 + Ld2 = Ls2 + mr22 w2 ...(ii) From the law of conservation of angular momentum L1 = L2 ...(iii) From eqns. (i), (ii) and (iii), ...(iv) Ls2 + mr22 w2 = Ls 1 + mr12 w1 Since the angular momentum of the student is proportional to his rate of spin, w ...(v) Ls 2 = 2 Ls 1 w1 From eqns. (iv) and (v),  w2  2 2  w Ls 1  + mr2 w2 = Ls1 + mr1 w1 1 or

 w2  2 2 2 2  w  mr1 w1 + mr2 w2 = mr1 w1 + mr1 w1 1 = 2mr12 w1

(Q Ls 1 = Ld 1 = mr12 w1 )

or

 2r 2   2(0.6)2  w2 =  2 1 2  w1 =  (0.5) 2 2  (0.6) + (0.1)   r1 + r2  = 0.97 rps

16. Given: v = k s

k 1 dv k ds k × k s = k2 = = v = 2 dt 2 s dt 2 s 2 s Force on the locomotive, F =m Again,

dv 1 2 = mk dt 2

dv k 2 = dt 2

or

dv =

k2 dt 2

k2 t +c 2 where c is the constant of integration. Suppose v = 0 at t = 0. Then, c = 0 Integrating, v =

k2 t or 2 k 2t or ds = dt 2

\ v=

ds k 2t = 2 dt

Physics for you | OctOber ‘15

27

k 2t 2 + c′ 4 k 2t 2 Suppose s = 0 at t = 0. Then, c′ = 0 \ s = 4 Work done = Fs 1 k 2t 2 mk 4t 2 = mk 2 × = 2 4 8 OR (a) Total initial angular momentum of the discs, L1 = I1w1 + I2w2 Moment of inertia of the two disc system = (I1 + I2) If w is the angular speed of the combined system, then, final angular momentum of the system, L2 = (I1 + I2)w As no external torque is acting, therefore, according to principle of conservation of angular momentum, we have, (I1 + I2)w = I1w1 + I2w2 I w +I w or w = 1 1 2 2 I1 + I2 1 1 (b) Initial KE of two discs, E1 = I1w12 + I 2 w22 2 2 1 2 Final KE of the system, E2 = (I1 + I 2 )w 2 1 1 1 2 2 \ E1 − E2 = I1w1 + I 2 w2 − (I1 + I 2 )w2 2 2 2 Putting the value of w from part (a) and solving, we get, Integrating, s =

E1 − E2 =

I1 I 2 (w1 − w2 )2 2(I1 + I2 )

Rate of change of momentum 140 kg m s-1

= 140 kg m s-2 1s But rate of change of momentum is equal to the applied force = 140 kg m s–2 = 140 N By Newton's third law of motion this must also be the magnitude of the force exerted by the molecules on the surface. \ Force exerted by molecules on surface = 140 N Area of surface = 14 cm2 = 14 × 10–4 m2 force 140 Pressure on surface = = area 14 × 10−4 =

= 105 N m–2 18. (a) Sam is sympathetic and also has the attitude of helping others. He has patience. (b) The downward force on the elevator is F= mg + f = (1800 × 10) + 4000 = 22000 N. The motor must supply enough power to balance this force. Hence P = F.v = 22000 × 2 = 44000 W = 59 hp (1hp = 746 W) 19. (a) Two external forces act on the bob are gravity and tension (T) in the string. At the lowest point A, the potential energy of the system can be taken zero. So at point A,

which is a positive quantity

Hence the rotational KE of the combined system is less than the sum of the initial rotational KE of the two discs. Note that there is loss of KE in the process. This loss of energy is due to dissipation of energy due to frictional contact of the two discs. It may be noted that angular momentum is conserved in the process as torque due to friction is only an internal torque. 17. Let the direction in which the molecules rebound after striking the surface be taken as positive. Momentum of each molecule after striking the surface =mv2 = 5 × 10–26 kg × 500 m s–1 Momentum of each molecule before striking the surface = mv1 = 5 × 10–26 kg × (– 500 m s–1) 28 × 1023 molecules strike the surface per second. Change in momentum of the molecules striking the surface in 1 second = 28 × 1023 [(5 × 10–26 × 500) – 5 × 10–26 (– 500)] kg m s–1 = 28 × 1023 × 5 × 10–26 × 1000 kg m s–1 = 140 kg m s–1 28

\

Physics for you | OctOber ‘15

Total mechanical energy = kinetic energy 1 E = mv02 ...(i) 2 If TA is the tension in the string at point A, then

mv02 ...(ii) L At the highest point C, the string slackens, so the tension TC becomes zero. If vC is the speed at point C, then by conservation of energy, 1 E = K + U or E = mvC2 + 2 mgL ...(iii) 2 mvC2 Also, mg = ...(iv) L 2 or mvC = mgL ...(v) TA − mg =

Using (v) in (iii), 1 5 E = mgL + 2mgL = mgL ...(vi) 2 2 From equations (i) and (vi), we get m 5 ...(vii) mgL = v02 or v0 = 5 gL 2 2 (b) From equation (iv), we have vC = gL The total energy at B is 1 E = mv B2 + mgL ...(viii) 2 From equations (i) and (viii), we get 1 2 1 mv B + mgL = mv02 2 2 1 2 1 mv B + mgL = m × 5 gL [using (vii)] 2 2 v B = 3 gL (c) The ratio of kinetic energies at B and C is 1 2 K B 2 mv B 3 = = KC 1 2 1 mvC 2 OR Consider a body of mass M and radius R rolling down a plane inclined at an angle q with the horizontal, as shown in figure. It is only due to friction at the line of contact that body can roll without slipping. The centre of mass of the body moves in a straight line parallel to the inclined plane. The external forces on the body are (i) The weight Mg acting vertically downwards. (ii) The normal reaction N of the inclined plane. (iii) The force of friction acting up the inclined plane.

Let a be the downward acceleration of the body. The equation of motion for the body can be written as N – mg cos q = 0 F = ma = mg sin q – f As the force of friction f provides the necessary torque for rolling, so a t = f × R = I a = mk2   R

or

f =m

k2

a R2 where k is the radius of gyration of the body about its axis of rotation. Clearly k2 ma = mg sin q – m a R2 g sin q or a = (1 + k 2 / R2 ) Let h be height of the inclined plane and s the distance travelled by the body down the plane. The velocity v attained by the body at the bottom of the inclined plane can be obtained as follow: v2 – u2 = 2as

or

v2 – 02 = 2

or

v2 =

or

v=

g sin q

(1 + k 2 / R2 ) 2 gh 2

1+ k / R

2

s  h  Q s = sin q  

2 gh

(1 + k 2 / R2 ) 20. The particle of mass m is initially at A. At the point P where the particle loses contact with the surface of the sphere, normal reaction is zero and the only force acting on the particle is its weight mg acting vertically downwards, as shown in figure.

The radial component (mg cos q) of the weight provides the necessary centripetal force, mv 2 mg cos q = r or v 2 = rg cos q ...(i) OQ (r − h) As cos q = = , OP r v2 = (r – h)g ...(ii) Since the velocity v has been acquired by the particle after falling through a height h, Physics for you | OctOber ‘15

29

...(iii)

v = 2 gh From eqns. (ii) and (iii), 2gh = (r – h)g r \ h= 3 From eqns. (iii) and (iv),

...(iv)

2 gr ...(v) v= 3 If x is the horizontal and y is the vertical distance covered by the particle as it hits the ground at C after time t, x = (v cos q)t ...(vi) 1 2 and y = (v sin q)t + gt ...(vii) 2 From eqns. (vi) and (vii), y = x tan q +

gx 2

...(viii)

2v 2 cos2 q

r 5 It is clear that, y = 2r − h = 2r − = r , 3 3 r −h r −r /3 2 cos q = = = , r r 3 and tan q = sec2 q − 1 = (3 / 2)2 − 1 =

5 2

From eqn. (viii),  5 gx 2 5 r =x + 3  2  2(2 gr / 3)(2 / 3)2 or

27 x 2 + (8 5r )x − (80 / 3)r 2 = 0

i.e., x =

−8 5r + 320r 2 + 2880r 2 2 × 27 (neglecting negative root)

−17.9r + 56.6r 38.7r = = 0.72r 54 54 If s is the distance from the initial position where the particle strikes the ground at C, s = BC = BD + DC = rsin q + x or

x=

 5 = r   + 0.72r = 0.75r + 0.72r ≈ 1.5r  3  OR

Let us consider the mass after it has slide down a distance x [as shown in figure]. 30

Physics for you | OctOber ‘15

If a is the acceleration at this instant, then ma = mg sin q – f = mg sin q– µ0x mg cos q [Q f = µR = (µ0x)mg cos q] or a = g(sin q – µ0x cos q) ...(i) dv  dv   dx   dv  = = v, dt  dx   dt   dx  v dv = a dx = g (sin q – µ0 xcos q) dx

As a =

or ∫ v v dv = ( g sin q) ∫ x dx − (µ0 g cos q)∫ x x dx 0 0 0 2 2 x  v or = ( g sin q)x − µ0 g cos q   2  2  or v2 = (2g sin q)x – (µ0g cos q)x2 ...(ii) If the mass comes to rest after covering a distance s, then if x = s, v = 0. From eqn. (ii), (2g sin q)s – (µ0g cos q)s2 = 0 or s[2g sin q – (µ0g cos q)s] = 0 As s ≠ 0, 2g sin q – (µ0g cos q)s = 0 2 g sin q 2 or s = = ...(iii) tan q µ0 g cos q µ0 When v = vmax , dv d a= = (v ) = 0 dt dt max If x = x0, for a = 0, from eqn. (i) g(sin q – µ x0 cos q) = 0 or µ x0 cos q = sin q tan q or x0 = ...(iv) µ0  tan q  Putting v = vmax and x = x0 =  in eqn. (ii),  µ  0

2 vmax

or

 tan q   tan q  = (2 g sin q)  − (µ0 g cos q)    µ   µ  0

2 vmax =

=

g 2g sin q tan q − sin q tan q µ0 µ0

2

0

g sin q tan q µ0

or vmax =

g sin q tan q µ0 nn

Class XI

ACCELERATED LEARNING SERIES

Unit

4

Thermodynamics | Kinetic Theory of Gases | Oscillations and Waves thermodynamics

It is the branch of physics which deals with the study of transformation of heat energy into other forms of energy and vice versa. In thermodynamics, the collection of objects on which attention is being focused is called the system, while everything else in the environment is called the surroundings. thermal equilibrium

If microscopic variables such as pressure, temperature, volume, mass, composition etc., which characterize a system, do not change with time, the system is said to be in thermodynamic equilibrium. Zeroth Law of thermodynamics

If two systems A and B are each in thermal equilibrium with a third system C, then A and B will be in thermal equilibrium with each other. heat

Heat is the energy transferred between a system and its environment because of a temperature difference that exists between them. It is a scalar quantity. Its SI unit is joule. It is path dependent. Work

The work done by the system is W = ∫ dW =

Vf

∫

Vi

During the change in volume, the pressure and temperature of the system also change. It is a scalar quantity. Its SI unit is joule. It is path dependent. In P-V diagram (indicator diagram) the area under P-V curve represents work done. internal energy

Internal energy of a system is the energy possessed by the system due to molecular motion and molecular configuration. Change in internal energy is path independent and depends only on the initial and final states of the system. The internal energy of an ideal gas is only dependent upon temperature whereas that of a real gas it is a function of both temperature and volume. thermodynamic variables

These are macroscopic physical quantities like pressure (P), volume (V) and temperature (T) etc. which are used to describe the state of the system. The relation between these variables is called equation of state. Heat and work are not thermodynamic variables. Thermodynamic variables are of two kinds, extensive and intensive. Internal energy U, volume V, total mass M are extensive variables. Pressure P, temperature T and density r are intensive variables. first Law of thermodynamics

PdV

Here Vi and Vf referred to the change in volume from state i to state f.

First law of thermodynamics is simply a restatement of principle of conservation of energy. Imagine a gaseous system for which pressure, volume and temperature are related by certain equation of state and is undertaken through a process. If DQ, DU and DW represent the heat Physics for you | OctOber ‘15

31

given to the system, change in internal energy and work done by the system respectively. During the process the first law of thermodynamic states that DQ = DU + DW Sign Convention Heat absorbed by the system is taken as positive while the heat lost by the system is taken as negative. Work done by the system is taken as positive while work done on a system is taken as negative. The increase in internal energy of the system is taken as positive while decrease in internal energy is taken as negative.

dP P =− g  dV V In an adiabatic expansion temperature of the gas will fall (T2 < T1) while in adiabatic compression temperature of the gas will rise (T2 > T1).

isothermal Process

reversible and irreversible Processes

For such a process temperature remains constant throughout the process. Hence PV = constant as T is constant. Work is done at the same rate as heat is supplied, hence there is no increase of internal energy. Due to this, DU = 0 and thus DQ = DW. Work done during an isothermal process, V  P  W = mRT ln  2  = mRT ln  1   V1   P2 

The slope of isothermal curve on a P-V diagram at any point on the curve is given by dP −P = dV V KEY POINT In an isothermal expansion the gas absorbs heat and does work and in an isothermal compression work is done on the gas by the environment and heat is released. adiabatic Process

An adiabatic process is one that occurs so rapidly or occurs in a system that is so well insulated that no transfer of energy as heat occurs between the system and its environment. Equation of adiabatic process, PVg = constant TVg – 1 = constant

second Law of thermodynamics

Kelvin-Planck statement : No process is possible whose sole result is the absorption of heat from a reservoir and the complete conversion of the heat into work. Clausius statement : No process is possible whose sole result is the transfer of heat from a colder object to a hotter object. Reversible process : A reversible process is one which can be retraced in the opposite direction. A quasi-static isothermal expansion of an ideal gas in a cylinder fitted with a frictionless movable piston is an example of a reversible process. Irreversible process : An irreversible process is one which cannot be retraced back in the opposite direction. All spontaneous processes of nature are irreversible processes. e.g. transfer of heat from a hot body to a cold body, diffusion of gases, etc. are all irreversible processes.

SELF CHECK

1. An ideal gas goes V c through a reversible b cycle a → b → c → d has the V-T diagram shown d a below. Process d → a T and b → c are adiabatic. The corresponding P-V diagram for the process is (all figures are schematic and not drawn to scale) P

(a)

C  where, g =  P  is called adiabatic exponent.  CV  Work done during an adiabatic process W=

(PV (T − T ) 1 1 − P2V2 ) = mR 1 2 (g − 1) g −1

The slope of adiabatic curve on a P-V diagram at any point on the curve is given by 32

Physics for you | OctOber ‘15

P

b

a d

(b)

c

c

d

b

a

V

V P

(c)

(d)

c

d

a

b V

(JEE Main 2015)

2. One mole of diatomic ideal gas undergoes a cyclic process ABC as shown in figure. The process BC is adiabatic. The temperatures at A, B and C are 400 K, 800 K and 600 K respectively. B

800 K

P A

400 K

600 K C V

Choose the correct statement. (a) The change in internal energy in the process BC is – 500 R. (b) The change in internal energy in whole cyclic process is 250 R. (c) The change in internal energy in the process CA is 700 R. (d) The change in internal energy in the process AB is – 350 R. (JEE Main 2014) 3. The above p-v diagram represents the thermodynamic cycle of an engine, operating with an ideal monoatomic gas. The amount of heat, extracted from the source in a single cycle is p

2p0 p0

(a) 4p0v0  13  (c)   p0v 0 2 carnot engine

v0

(b) p0v0

2v0

v

 11  (d)   p0v 0 2 (JEE Main 2013)

Carnot engine is a reversible heat engine operating between two temperatures T1 (source) and T2 (sink). Carnot cycle : Carnot engine works in series of operations. The operations consist of an isothermal expansion and then adiabatic expansion. Further operations are isothermal compression and adiabatic compression so that the working substance is back at the initial state at the end of each cycle. This cycle of operations is called Carnot cycle. The efficiency of a Carnot engine is given by

T2 T1 The efficiency of Carnot engine depends on the temperature of source (T1) and temperature of the sink (T2) , but does not depend upon the nature of the working substance. Carnot theorem : No heat engine operating between two given temperatures can be more efficient than a Carnot engine operating between the same two temperatures. h = 1−

SELF CHECK

4. A Carnot engine, whose efficiency is 40%, takes in heat from a source maintained at a temperature of 500 K. It is desired to have an engine of efficiency 60%. Then, the intake temperature for the same exhaust (sink) temperature must be (a) 1200 K (b) 750 K (c) 600 K (d) efficiency of Carnot engine cannot be made larger than 50% (JEE Main 2012) kinetic theory of gases equation of ideal gas or Perfect gas

The relation between pressure P, volume V and absolute temperature T of a gas is given by PV = nRT = kBNT Where n is the number of moles of the gas and R is the universal gas constant, its value, R = 8.31 J mol–1 K–1 and kB is Boltzmann’s constant. R kB = NA Here, NA is Avogadro’s constant or Avogadro’s number. kB = 1.38 × 10–23 J K–1 and NA = 6.022 × 1023 per mol. While deriving the ideal gas equation, the following two assumptions are used : (i) The size of the gas molecules is negligibly small. (ii) There is no force of attraction amongst the molecules of the gas. ideal gas Laws

Boyle’s law : It states that at constant temperature, the volume of a given mass of a gas is inversely proportional to the pressure. 1 or PV = constant. V∝ P Physics for you | OctOber ‘15

33

Charles’ law : It states that at constant pressure, volume of a given mass of gas is directly proportional to its absolute temperature. V ∝ T or V/T = constant. Gay Lussac’s law or Regnault’s law : It states that at constant volume, pressure of the given mass of a gas is directly proportional to its absolute temperature. P ∝ T or P/T = constant. Avogadro’s law : It states that at same temperature and pressure, equal volumes of all the gases contain equal number of molecules. N1 = N2, if P, V and T are the same. Graham’s law of diffusion : It states that at constant pressure and temperature, the rate of diffusion of a gas is inversely proportional to the square root of its density, 1 Rate of diffusion ∝ , if P and T are constant. r Dalton’s law of partial pressure : It states that the total pressure exerted by a mixture of non-reactive ideal gases is equal to the sum of the partial pressure which each would exert, if it alone occupied the same volume at the given temperature. P = P1 + P2 + P3 + ..... kinetic theory of gases

The kinetic theory of gases correlates the macroscopic properties of gases e.g., pressure, temperature etc. to the microscopic properties of gas molecules e.g. speed, momentum, kinetic energy of molecules etc. The kinetic theory of gases is based on the following assumptions : • The intermolecular forces are negligible during a collision. • The volume of the molecules themselves can be neglected compared with the volume occupied by the gas. • The time occupied by a collision is negligible compared with the time spent by a molecule between collisions and between collisions a molecule moves with uniform velocity. ideal gas

An ideal gas or a perfect gas is that gas which strictly obeys the gas laws. Following are the characteristics of the ideal gas : • The size of the molecule of a gas is zero, i.e., each molecule of a gas is a point mass with no dimensions. • There is no force of attraction or repulsion amongst the molecules of the gas. Equation of an ideal gas : PV = µRT = kBNT 34

Physics for you | OctOber ‘15

where µ is the number of moles and N is the number of molecules, R is the universal gas constant and kB is the Boltzmann constant. R kB = NA NA is the Avogadro’s number. m N Here, m = = M NA where m is the mass of the gas containing N molecules, M is the molar mass. rRT Another form of an ideal gas equation P = M where r is the density of the gas. kinetic theory of an ideal gas

According to kinetic theory of an ideal gas pressure exerted by an ideal gas is given by 1 P = mn v 2 3 where n is the number density (number of molecules per unit volume), m is the mass of molecule and v 2 is the mean square speed. KEY POINT • The internal energy of an ideal gas is equal to its

internal kinetic energy, which depends only on the temperature of the ideal gas. • Average kinetic energy per molecule of a gas depends only on its absolute temperature. It is independent of pressure, volume and nature of the ideal gas. Kinetic Interpretation of Temperature Average kinetic energy per molecule of a gas is

1 3 E = mv 2 = k BT 2 2 Average kinetic energy per molecule of a gas depends only on its absolute temperature. It is independent of pressure, volume and nature of the ideal gas. At a given temperature, the average kinetic energy per molecule of all the gases is same. Average kinetic energy per mole of a gas is 3 E = RT 2 Relation between pressure and kinetic energy of gas 2 PV = E 3 Absolute zero of temperature : Absolute zero of temperature may be defined as that temperature at which the velocity of the gas molecules become zero.

Also, all molecular motion ceases at absolute zero of temperature. The value of absolute zero for an ideal gas is —273°C. This definition is true only in the case of an ideal gas. different type of speed of gas molecules Root mean square speed, v rms =

3RT 3k BT = M m

Average speed, v = 8RT = 8k BT pM pm Most probable speed, v mp =

2RT 2k BT = M m

v rms > v > v mp degree of freedom

The number of degrees of freedom of a gas molecule is given by f = 3N – K where N = Number of atoms in the molecule K = Number of independent relations between atoms of a molecule Relation between g and degree of freedom (f) 2 g =1+ f Law of equipartition of energy

It states that if a system is in equilibrium at absolute temperature, T, the total energy is distributed equally in different energy modes of absorption, the energy in each mode being equal to (1/2) kBT. Each translational and rotational degree of freedom corresponding to one energy mode of absorption and has energy (1/2)kBT. Each vibrational frequency has two modes of energy (kinetic and potential) with corresponding energy equal to 1 2 × k BT = k BT . 2 specific heat capacity

Monatomic Gases The molecule of a monatomic gas has only three translational degrees of freedom. The total internal energy of a mole of such a gas is given by 3 3 U = k BT × NA = RT 2 2 The molar specific heat at constant volume, CV is dU 3 CV = = R dT 2

The molar specific heat at constant pressure, CP is 5 CP = R 2 C 5 Ratio of specific heats, g = P = CV 3 Diatomic Gases A diatomic gas molecule treated as a rigid rotator like a dumbbell has 5 degrees of freedom: 3 translational and 2 rotational. The total internal energy of a mole of such a gas is given by 5 5 U = k BT × NA = RT 2 2 The molar specific heats are given by 5 CV (rigid diatomic) = R 2 7 C P (rigid diatomic) = R 2 7 g (rigid diatomic) = 5 If the diatomic molecule is not rigid but has in addition a vibrational mode, then 7 5  U =  k BT + k BT  NA = RT 2 2  7 9 9 CV = R, C P = R, g = 2 2 7 Polyatomic Gases In general, a polyatomic gas molecule has 3 translational, 3 rotational degrees of freedom and a certain number (f ) of vibrational mode. The total internal energy of a mole of such a gas is given by 3 3  U =  k BT + k BT + f k BT  NA = (3 + f )RT 2 2  The molar specific heats are given by CV = (3 + f )R, CP = (4 + f )R (4 + f ) g= (3 + f ) Note : CP – CV = R. This relation is known as Mayer’s relation. It is true for any ideal gas whether monatomic, di or polyatomic. mean free Path

It is the average distance covered by a molecule between two successive collisions and is given by 1 l= 2npd 2 where n is the number density and d the diameter of the molecule. Physics for you | OctOber ‘15

35

Mean free path related to the temperature (T) and pressure (P) as k BT l= 2 pd 2P KEY POINT

• The square root of the absolute temperature of an

ideal gas is directly proportional to the root mean square velocity of its molecules. • The mean free path is of great significance in understanding transport phenomena like diffusion, viscosity, and thermal conduction.

SELF CHECK

5. Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as V q, where V is the volume of C   the gas. The value of q is g = p  Cv  (a) g +1 2

(b) g −1 2

(c) 3g + 5 6

(d) 3g − 5 6

(JEE Main 2015)

7. Using equipartition of energy, the specific heat (in J kg–1 K–1) of aluminium at room temperature can be estimated to be (atomic weight of aluminium = 27) (a) 25 (b) 410 (c) 925 (d) 1850 (JEE Main 2015) osciLLations Periodic and oscillatory motions

Periodic motion : Any motion which repeats itself after regular interval of time is called periodic motion. The time interval after the motion is repeated is called time period or period of motion. A periodic motion can be either rectilinear or closed or open curvilinear. Physics for you | OctOber ‘15

Periodic function

A function f (t) is said to be periodic, if f (t) = f (t + T) where T is the time period of the periodic function. Any periodic function can be expressed as a superposition of sine and cosine functions of different time periods with suitable coefficients sinwt, coswt and sinwt + coswt are the periodic 2p –wt . e and log(wt) are non functions with a period w periodic functions. simple harmonic motion (shm)

6. In an ideal gas at temperature T, the average force that a molecule applies on the walls of a closed container depends on T as T q. A good estimate for q is 1 (a) 2 (b) 1 (c) (d) 1 2 4 (JEE Main 2015)

36

In case of periodic motion, force is always directed towards a fixed point which may or may not be on the path of motion. Oscillatory motion : A periodic motion in which body moves to and fro (or back and forth) along the same path about a fixed point (called mean or equilibrium position) is called oscillatory or vibratory motion. Oscillatory or vibratory motion is a constrained periodic motion between two fixed limits (called extreme positions) on either side of mean position.

It is a kind of periodic motion, in which a particle moves to and fro (or up and down) about a mean position under a restoring force, which is always directed towards the mean position and whose magnitude at any instant is directly proportional to the displacement of the particle from the mean position at that instant. i.e., F ∝ –x or F = –kx where k is known as the force constant. The negative sign shows that restoring force F is always directed towards the mean position. The SI unit of k is N m–1 and its dimensional formula is [ML0T–2]. Examples of simple harmonic motion : • Motion of bob of a simple pendulum. • Motion of a block connected to spring. KEY POINT • Every oscillatory motion is periodic but every

periodic motion need not be oscillatory. e.g., circular motion is a periodic, but it is not oscillatory. • Every periodic motion is not simple harmonic motion. Only that periodic motion governed by the force law F = –kx is simple harmonic. Geometrical interpretation of simple harmonic motion : Simple harmonic motion is the projection of a uniform circular motion on a diameter of the reference circle.

simple harmonic motion are of two types :

Linear simple harmonic motion : When a particle moves to and fro (or up and down) about a mean position along a straight line, then its motion is called linear simple harmonic motion. e.g., motion of a block connected to spring. Angular simple harmonic motion : When a system oscillates angularly with respect to a fixed axis, then its motion is called angular simple harmonic motion. e.g., motion of bob of a simple pendulum. Comparison between linear SHM and angular SHM Linear SHM Angular SHM 1. Restoring force Restoring torque F = –kx t = – Cq where k is the restoring where C is the restoring force constant torque constant i.e., restoring force per i.e., restoring torque per unit displacement. unit twist 2. Acceleration k a=− x m

Angular acceleration α=−

C q I

where m is the mass of a where I is the moment of body inertia of a body 3. The differential equation The differential equation of angular SHM is of linear SHM is 2 d x d 2q 2 + w = 0 + w 2q = 0 x dt 2 dt 2 k C where w 2 = where w 2 = m I displacement in simple harmonic motion

The displacement of a particle in SHM at any instant t from its mean position is given by x = Acos(wt + f) ...(i) y = Asin(wt + f) ...(ii) where A is the amplitude (maximum displacement on either side of mean position) of motion, the argument (wt + f) is the phase of the motion, f is the initial phase or phase constant (or phase angle) and w is the angular frequency. If s is the span of a particle executing SHM, its amplitude s A= 2

SELF CHECK

8. x and y displacements of a particle are given as x(t) = a sinwt and y(t) = a sin2wt. Its trajectory will look like

y

y

(a)

(b)

x

y

(c)

x

y

x

(d)

x

(JEE Main 2015) Velocity in simple harmonic motion

Velocity of a particle in SHM is given by dx v= = − wA sin(wt + f) dt v = w A2 − x 2 In SHM, velocity is maximum at the mean position and minimum at the extreme positions. The maximum value of velocity is called velocity amplitude in SHM and is given by vm = Aw. The direction of velocity of a particle in SHM is either towards or away from the mean position and the velocity varies simple harmonically with the same frequency as that of the displacement. In SHM, the graph between velocity and displacement v2 x2 is an ellipse as 2 2 + 2 = 1 Aw A –1 If w = 1 rad s , then the graph will be a circle. acceleration in simple harmonic motion

Acceleration of a particle in SHM is given by dv d 2x a= = = −w 2 A cos(wt + f) dt dt 2 a = –w2x In SHM, acceleration is proportional to the displacement and is always directed towards the mean position acceleration is maximum at the extreme positions and minimum at the mean position. The maximum value of acceleration is called acceleration amplitude in SHM and is given by am = w2A. In SHM, the acceleration varies simple harmonically with the same frequency as that of the displacement. The graph between acceleration and displacement is a straight line which passes through the origin and has slope (–w2). Physics for you | OctOber ‘15

37

Phase relationship Between displacement, Velocity and acceleration in simple harmonic motion

The displacement of a particle executing SHM is given by x = Acos(wt + f) dx Velocity, v = = − wA sin(wt + f) dt p  = wA cos (wt + f) +   2 dv Acceleration, a = = − w 2 A cos(wt + f) dt = w 2 A cos[(wt + f) + p] From above, we get that Phase of displacement = (wt + f) p Phase of velocity = (wt + f) +   2 Phase of acceleration = (wt + f + p) Thus, we conclude that, the velocity in SHM is leading the displacement by a phase p/2 radian. The acceleration in SHM is leading the displacement by a phase p radian. The acceleration in simple harmonic motion is leading the velocity by a phase p/2 radian. energy in simple harmonic motion

The kinetic energy of a particle in SHM is given by 1 1 Kinetic energy K = mv 2 = mw 2 A2 sin 2(wt + f) 2 2 1 = mw 2(A2 − x 2) 2 Kinetic energy of a particle executing SHM is periodic with period T/2. It is zero at extreme positions and maximum at mean position. The potential energy of a particle in SHM is given by 1 1 U = kx 2 = kA2 cos 2(wt + f) 2 2 1 = mw 2 A2 cos 2(wt + f) 2 Potential energy of a particle executing SHM is periodic with period T/2. It is zero at the mean position and maximum at the extreme positions. Total energy of a particle in SHM is given by E=K+U 1 1 = mw 2 A2 sin 2(wt + f) + mw 2 A2 cos 2(wt + f) 2 2 1 = mw 2 A2 2 In SHM, total energy remains constant at all instants and at all displacements. It depends upon the mass, amplitude and frequency of vibration of the particle. 38

Physics for you | OctOber ‘15

In SHM, at the mean position total energy is in the form of its kinetic energy and at the extreme positions, total energy is in the form of its potential energy. The average value of kinetic energy or potential energy of a particle in SHM in a time of one complete oscillation, 1 < K > = < U > = mw 2 A2. 4 But the average value of total energy of a particle in SHM in one complete oscillation, 1 < E > = mw 2 A2. 2 If frequency of oscillation in SHM is u, then frequency of oscillation of KE = frequency of oscillation of PE = 2u. In SHM, the frequency of oscillation of total energy is zero. Graphical variation of energy of SHM E

KE

TE PE

�A �A 2

O

A A 2

x

the spring-mass system

Let us find out the time period of a spring-mass system oscillating on a smooth horizontal surface as shown in the figure below.

At the equilibrium position the spring in relaxed. When the block is displaced through a distance x towards right, it experiences a net restoring force F = –kx towards left. The negative sign shows that the restoring force is always opposite to the displacement. That is, when x is positive, F is negative, the force is directed to the left. When x is negative, F is positive, the force always tends to restore the block to its equilibrium position x = 0. F = –kx Applying Newton’s second law, d 2x F = m 2 = −kx ...(i) dt

d 2x  k  + x =0 dt 2  m  Comparing equation (i) with, F = ma or

a= we get,

d 2x = −w 2x 2 dt

m k or T = 2p k m Note : The time period is independent of the amplitude. For a given spring constant, the period increases with the mass of the block that means more massive block oscillates more slowly. For a given block, the period decreases as k increases. A stiffer spring produces quicker oscillations. Note : Gravity does not influence the time period of the spring-mass system, it merely changes the equilibrium position. w2 =

Time period of a simple pendulum is independent of mass, shape and material of bob and it is also independent of the amplitude of oscillation provided it is small. Time period of a simple pendulum depends on L as T ∝ L , so the graph between T and L will be a parabola while between T2 and L will be a straight line. Time period of a simple pendulum depends on

(

T = 2p

1

1 1  g +   L Re 

series and Parallel combination of springs

where Re is the radius of the earth. Special cases : If L > > Re, then

Series combination of springs When two springs are joined in series, the equivalent stiffness of the combination may be obtained as

6.4 × 106 Re = 2p = 84.6 min g 9. 8 If L = Re , then

1 1 1 = + k k1 k2

Parallel combination of springs When two springs are joined in parallel, the equivalent stiffness of the combination is given by k = k1 + k2

T = 2p

T = 2p

6.4 × 106 Re = 2p = 60 min 2g 2 × 9. 8

The simple pendulum having time period of 2 s is called second pendulum. If a simple pendulum is suspended in a lift and lift is accelerating downwards with an acceleration a, then its time period is given by L g −a If a simple pendulum is suspended in a lift and lift is accelerating upwards with an acceleration a, then its time period is given by T = 2p

T = 2p

simple Pendulum

The time period of a simple pendulum is given by T = 2p

L . g

where L is the length of the pendulum and g is the acceleration due to gravity.

)

acceleration due to gravity as T ∝ 1/ g . With increase in g, T will decrease or vice versa. If the length of a simple pendulum is comparable with the radius of earth (Re), then time period T is given by

L g +a

If a simple pendulum is suspended in a lift and the lift is moving upwards or downwards with constant velocity v, then its time period is given by T = 2p

L g

If a simple pendulum is suspended in a lift and lift is freely falling with acceleration g, then its time period is given by Physics for you | OctOber ‘15

39

E

L =∞ g−g

T = 2p

(c)

If a simple pendulum is suspended in a carriage which is accelerating horizontally with an acceleration a, then its time period is given by T = 2p

L 2

2

( g +a )

If a simple pendulum is suspended from the roof of a trolley which is moving down an inclined plane of inclination q, then the time period is given by T = 2p

L g cos q

If a simple pendulum whose bob is of density r oscillates in a non-viscous liquid of density s(s < r), then its time period is given by T = 2p

L  s 1 − r  g

SELF CHECK

9. A pendulum made of a uniform wire of cross sectional area A has time period T. When an additional mass M is added to its bob, the time period changes to TM. If the Young’s modulus of 1 the material of the wire is Y then is equal to Y (g = gravitational acceleration) 2 2   (a) 1 −  TM   A (b) 1 −  T   A     T   Mg   TM   Mg (c)

 T 2  A   T  2  Mg M (d) − 1     M  − 1   Mg  A   T    T 

(JEE Main 2015) 10. For a simple pendulum, a graph is plotted between its kinetic energy (KE) and potential energy (PE) against its displacement d. Which one of the following represents these correctly? (graphs are schematic and not drawn to scale) E

KE

(a)

E d

(b)

PE

40

Physics for you | OctOber ‘15

PE KE

d

E

KE PE

(d) d

PE KE

d

(JEE Main 2015)

undamped and damped oscillations

Undamped oscillations : When a system oscillates with a constant amplitude which does not change with time, its oscillations are called undamped oscillations. The energy of the system executing undamped oscillations remains constant and is independent of time. The dissipative forces (i.e., frictional or viscous forces) are not present in the system executing undamped oscillations. Damped oscillations : When a system oscillates with a decreasing amplitude with time, its oscillations are called damped oscillations. The energy of the system executing damped oscillations will go on decreasing with time but the oscillations of the system remain periodic. The dissipative forces or damping forces are active in the oscillating system which are generally the frictional or viscous forces. The damping force is given by Fd = –bv, where, v is the velocity of the oscillator and b is damping constant. Negative sign shows that damping force acts opposite to the velocity at every moment. The SI unit of b is kg s–1. The differential equation of damped harmonic oscillator is given by d 2x dx m 2 + b + kx = 0 dt dt The displacement of the damped oscillator at any instant t is given by x = Ae −bt /2m cos(w′t + f) where w′ is the angular frequency of the damped oscillator is given by k b2 − m 4m2 If the damping constant b is small, then w′ ≈ w, where w is the angular frequency of the undamped oscillator. The mechanical energy E of the damped oscillator at any instant t is given by 1 E = kA2e −bt /m 2 Maintained oscillations : Due to the damping forces, the amplitude of oscillator will go on decreasing with w′ =

time. If we can feed the energy to the damped oscillatory system at the same rate at which it is dissipating the energy, then the amplitude of such oscillations would become constant. Such oscillations are called maintained oscillations. forced oscillations and resonance

Free oscillations : When a system oscillates with its own natural frequency, without the help of any external periodic force, its oscillations are called free oscillations. Forced or driven oscillations : When a system oscillates with the help of an external periodic force, other than its own natural angular frequency, its oscillations are called forced or driven oscillations. The differential equation of forced damped harmonic oscillator is given by d 2x dx + b + kx = F0 cos w dt 2 dt dt where wd is the angular frequency of the external force. The displacement of the forced damped harmonic oscillator at any instant t is given by x = Acos(wdt + f) F0 where A = 2 2 {m (w − w 2d )2 + w 2db 2}1/2 m

=

F0

( )

2  wb  m (w 2 − w d2 )2 + d  m  

1/2

−v 0 and tanf = wd x0 where w is the natural angular frequency of the oscillator, x0 and v0 are the displacement and velocity of the oscillator at time t = 0, when the periodic force is applied. Special cases : Case I. For small damping driven frequency is far from natural frequency wd2b2 or > , i.e., d > 10 v 10 20 If a person standing between two parallel hills fires a gun and hears the first echo after t1 s, the second echo after t2 s, and v is the velocity of sound, then the distance between the two hills is given by s1 + s2 = (vt1/2) + (vt2/2) = [v(t1 + t2)/2] KEY POINT • The concept of rarer and denser media for a wave

is through the velocity of propagation and not density. Lesser the velocity, denser is the medium and vice versa. Interference of waves : When two waves of same frequency or wavelength having constant phase difference travelling with same speed in the same direction superpose on each other, they give rise to an effect called interference of waves. Condition for constructive interference Phase difference φ = 2nπ where n = 0, 1, 2, ... l Path difference δ = × f = nl where n = 0, 1, 2,.. 2p Condition for destructive interference Phase difference φ = (2n + 1)π where n = 0, 1, 2, ... l  1 Path difference δ = × f = n +  l . 2p  2 where n = 0, 1, 2 .... KEY POINT • The phenomenon of interference is based on

conservation of energy.

stationary Waves

When two waves of same frequency, wavelength and amplitude travel in opposite directions at same speed, their superposition gives rise to a new type of waves known as stationary waves or standing waves. Energy does not propagate in this type of wave hence, it is named as stationary wave. Stationary waves are of two types • Longitudinal stationary waves • Transverse stationary waves Longitudinal stationary waves : It is produced in organ pipe and resonance tube.

Transverse stationary waves : It is produced in stretched string and sonometer. Equation of a stationary wave is given by y = (2Asinkx)coswt Stationary waves are characterised by nodes and antinodes. Nodes are the points for which the amplitude is minimum whereas antinodes are the points for which the amplitude is maximum. In a stationary wave nodes and antinodes are formed alternately and distance between them is l/4. At antinodes, displacement and velocity is maximum. At nodes, displacement and velocity is zero. Distance between two consecutive nodes or antinodes is l/2. Distance between a node and adjoining antinode is l/4. Vibrations in a stretched string of Length l fixed at Both ends

Speed of waves in a stretched string is given by T v= m where T is the tension of the string, m is the mass per unit length of the string. Fundamental mode or first mode, l1 = 2L Fundamental frequency v v 1 T u1 = = = l1 2L 2L m This frequency is called first harmonic. Second mode, l2 = L Frequency u2 =

v v = = 2u1 l2 L

This frequency is called second harmonic or first overtone. Third mode, l = 2L 3 3 Frequency v 3v u3 = = = 3u1 l 3 2L This frequency is called third harmonic or second overtone.

For the nth mode, l = 2L n n th Frequency of n mode

un =

v nv n T where n = 1, 2, 3, .... = = nu1 = l n 2L 2L m

th th This frequency is called n harmonic or (n – 1) overtone. p T Note : In general, u p = , 2L m where p = number of loops. Laws of vibrating stretched string : The fundamental frequency of a stretched string is given 1 T by u = . 2L m

1 when T and m are constants. L • Law of tension, u ∝ T when L and µ are constants 1 • Law of mass, u ∝ when L and T are constants m Note : If r is the density of the material of the string and D is the diameter of string, then mass per unit length, •

Law of length, u ∝

pD 2r . 4 1 4T 1 T = . \ u= 2L pD 2r LD pr m=

Laws of vibration of stretched string can be verified experimentally by using a sonometer. melde’s experiment

In longitudinal mode, the prongs of the tuning fork vibrate in a direction parallel to the length of string. In the longitudinal mode when fork completes one vibration, the string completes only half the vibration. So, the frequency of the string is one half of that of the fork. uL =

p T L m

In transverse mode, the prongs of the tuning fork vibrate in a direction perpendicular to the length of the string. In the transverse mode of vibration when fork completes one vibration, the string also completes one vibration. So, the frequency of the string is equal to the frequency of the fork. uT =

u p T = L. 2L m 2

The number of loops in the transverse mode is twice that in the longitudinal mode. Physics for you | OctOber ‘15

45

James Clerk Maxwell made great strides in helping to understand electromagnetism and

BRAIN

produced a unified model of electromagnetism. His research in kinematics and electricity laid

ELECTROMAGNETIC WAVE

foundation for modern quantum mechanics and special relativity.

James Clerk Maxwell

Displacement current It is the current which is produced when electric field and hence electric flux changes with time.

Ampere’s circuital law The line integral of magnetic field around any closed path in vacuum is equal to m0 times the total current passing through that closed path.

Properties of electromagnetic waves

Do not carry any charge. Do not deflected by electric and magnetic field. l Travel with speed of light in vacuum. l Frequency does not change when it goes from one medium to another, but its wavelength changes. l Transverse in nature. l Do not require any material medium for propagation l

l

ELECTROMAGNETIC WAVE

This law implies the fact that not only a conduction current but a displacement current, associated with a changing electric field, also produces a magnetic field

An electromagnetic wave is a wave radiated by an accelerated charge as coupled electric and magnetic field oscillating perpendicular to each other and to the direction of propagation of wave. Y

E

B

E

Maxwell’s four equations

B

X B

E

B

E

Direction of propagation

Magnitude of

and

are related as

Speed of an electromagnetic wave in free space is given by

Electromagnetic spectrum The orderly distribution of e l e c t r o m a g n e t i c w av e s i n accordance with their wavelength or frequency into distinct groups. l

l

Modified Ampere’s law

Radiowaves Used in radio communication .

l

Microwaves Radar communication Analysis of fine details of molecular and atomic structure.

l l

Infrared Useful for elucidating molecular structure. Us e f u l f o r h a z e photography.

l

l

Visible light Detected by stimulating nerve endings of human retina. Can cause chemical reaction.

l

l

Radio

Microwave

Infrared

Visible

102 to 10–1

10–1 to 10–3

10–3 to 10–6

8 ´ 10–7 to 4 ´10–7

Ultraviolet Can cause many chemical reactions, e.g., the tanning of the human skin. Ionize atoms in atmosphere, resulting in the ionosphere.

l l l l

Ultraviolet 3.5 ´ 10–7 to 1.5 ´10–7

X-rays Penetrate matter (e.g., radiography) Ionize gases Cause fluorescence Cause photoelectric emission from metals

X-rays 10–8 to 10–11

l l

Gamma rays In the treatment of cancer and tumours To produce nuclear reaction

Gamma rays 10–10 to 10–14

Vibrations of A closed organ Pipe

In a closed organ pipe, one end is closed and other end is open. In a closed organ pipe, the closed end is always a node while the open end is always an antinode. Fundamental mode, or first mode λ1 = 4L where L is the length of the pipe. Fundamental frequency υ1 =

v v = λ 1 4L

where v is the speed of sound in air. This frequency is called first harmonic. Second mode, 4L 3 Frequency, λ2 =

υ2 =

v 3v = = 3υ1 λ 2 4L

This frequency is called third harmonic or 1st overtone. Third mode, λ = 4L 3 5 Frequency, v 5v = = 5υ1 λ 3 4L This frequency is called fifth harmonic or second overtone. υ1 : υ2 : υ3 = 1 : 3 : 5 Only odd harmonics are present. υ3 =

For nth mode, λ n = Frequency, υn =

4L (2n − 1)

v v(2n − 1) = = (2n − 1)υ1 λn 4L

where n = 1, 2, 3, .... This frequency is called (2n – 1)th harmonic or (n – 1)th overtone. Vibration of An open organ Pipe

In an open organ pipe, both ends are open. In an open organ pipe, at both ends there will be antinodes. 48

Physics for you | OctOber ‘15

Fundamental mode or first mode, λ1 = 2L where L is the length of the pipe. Fundamental frequency v v υ1 = = λ1 2L where v is the speed of sound in air. This frequency is called first harmonic. Second mode, λ2 = L Frequency, v v υ2 = = = 2υ1 λ2 L This frequency is called second harmonic or first overtone. 2L Third mode, λ 3 = 3 v 3v Frequency, υ3 = = = 3υ1 λ 3 2L This frequency is called third harmonic or second overtone. υ 1 : υ 2 : υ3 = 1 : 2 : 3 Hence in open pipe all harmonics are present, whereas in a closed pipe only odd harmonics are present. For nth mode, λ n = 2L n v nv Frequency, υn = = = nυ1 , where n = 1, 2, ... λ n 2L

This frequency is called nth harmonic or (n – 1)th overtone. The fundamental frequency of an open organ pipe is twice that of a closed organ pipe of the same length. If an open pipe of length L is half submerged in water, it will become a closed organ pipe of length half that of open pipe as shown in figures (a) and (b). So its frequency will become v v vC = = = vO . 4(L / 2) 2L i.e., equal to that of open pipe, i.e., frequency will remain unchanged.

End correction

The antinode at the open end of a pipe is not formed exactly at the open end but a little outside. This is called the end correction. This is denoted by e and is given by e = 0.6r where r is the radius of the pipe. If L is the length of pipe then for closed pipe L is replaced by L + e while for open pipe L is replaced by L + 2e. Due to the end correction the fundamental frequency of a closed organ pipe is given by v v υC = = 4[L + e] 4[L + 0.6r] Due to the end correction, the fundamental frequency of an open pipe is given by v v υO = = 2[L + 2e] 2[L + 1.2r] Speed of sound in air at room temperature using resonance tube is given by v = 2υ(L2 – L1) where, υ = frequency of the tuning fork L1 = first resonance length L2 = second resonance length End correction e =

L2 − 3L1 2

SELF CHECK

13. A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m s–1. (a) 4 (b) 12 (c) 8 (d) 6 (JEE Main 2014) Beats

When two waves of nearly equal (but not exactly equal) frequencies travelling with same speed in the same direction superpose on each other, they give rise to beats. Beat frequency : It is defined as number of beats heard per second. Beat frequency = no. of beats/sec = (υ1 – υ2) = difference in frequencies.

Applications of the Phenomenon of Beats • The phenomenon of beats is used to determine frequency of a tuning fork. • The phenomenon of beats is used in tuning of musical instruments. • The phenomenon of beats is used in detecting the presence of dangerous gases in mines. • The phenomenon of beats is used in radio reception in many ways. Tuning fork is a source of sound of single frequency and frequency of a tuning fork of arm length L and thickness d in the direction of vibration is given by  d Y Y d υ =  2 v = 2  v = r  r L L    where Y is the Young’s modulus and r is the density of the material of the tuning fork. KEY POINT • Loading a tuning fork with wax decreases its

frequency while filing a tuning fork increases its frequency.

Doppler’s Effect

When a source of sound or an observer or both are in relative motion, there is an apparent change in the frequency of sound as heard by the observer. This phenomenon is known as Doppler’s effect. According to Doppler’s effect the apparent frequency heard by the observer is given by  v ± vo  υ′ = υ    v  vs  where vs, vo and v are the speed of source, observer and sound relative to air. The upper sign on vs (or vo) is used when source (observer) moves towards the observer (source) while lower sign is used when it moves away. If the wind blows with speed vw in the direction of sound, v is replaced by v + vw in the above equation. If the wind blows with speed vw in a direction opposite to that of sound, v is replaced by v – vw in the above equation. Physics for you | OctOber ‘15

49

When a source is revolving in a circle and observer is stationary outside, as shown in the figure. vυ At A, υ max = v − vs vυ At C, υ min = v + vs Beat frequency = υmax – υmin. There is no Doppler effect at B and D. When an observer is revolving in a circle with a stationary source outside, as shown in the figure.

(v + vo )υ v (v − vo )υ At C, υ min = v Beat frequency = υmax – υmin. There is no Doppler effect at B and D. If source and observer both are stationary i.e. vs = vo = 0 then υ′ = υ. Similarly, if source and observer both are moving in the same direction with same speed, At A, υ max =

H

i.e. vs = vo, then υ′ = υ. Thus it is clear that if there is no relative motion between the source and the observer then there is no Doppler effect.

SELF CHECK 14. A train is moving on a straight track with speed 20 m s–1. It is blowing its whistle at the frequency of 1000 Hz. The percentage change in the frequency heard by a person standing near the track as the train passes him is (speed of sound = 320 m s–1) close to (a) 18% (b) 24% (c) 6% (d) 12% (JEE Main 2015) –1 15. A bat moving at 10 m s towards a wall sends a sound signal of 8000 Hz towards it. On reflection it hears a sound of frequency f. The value of f in Hz is close to (speed of sound = 320 m s–1) (a) 8258 (b) 8516 (c) 8000 (d) 8424 (JEE Main 2015) AnswEr kEys (sElf chEck) 1. (a) 6. (b) 11. (d)

Physics for you | OctOber ‘15

3. (c) 8. (c) 13. (d)

4. (b) 9. (c) 14. (d)

5. (a) 10. (d) 15. (b) nn

PerPetual Motion Machine

ave you ever heard of a perpetual motion machine? A perpetual motion machine is a hypothetical device that would produce useful energy out of nothing. This is generally accepted as being impossible, according to laws of physics. In particular, perpetual motion machines would violate either the first or the second law of thermodynamics. A perpetual motion machine that violates the first law of thermodynamics is called a machine of the first kind. In general, the first law says that you can never get something for nothing. This means that without energy input, there can be no change in internal energy, and without a change in internal energy, there can be no work output. Machines of the first kind typically use no fuel or make their own fuel faster than they use it. If this type of machine appears to work, look for some hidden source of energy. A machine of the second kind does not attempt to make energy out of nothing. Instead, it tries to extract either random molecular motion into useful work or useful energy from some degraded source, such as outgoing radiant energy. The second law of thermodynamics says this cannot happen any more than rocks can roll uphill on their own. This just does not happen.

50

2. (a) 7. (c) 12. (a)

1. A gas is suddenly compressed to 1/4th of its original volume. Find the rise in temperature, the original temperature being 27°C and g = 1.5. (a) 200 K (b) 300 K (c) 500 K (d) 600 K 2. The bottom of a dip on a road has a radius R. A rickshaw of mass M, left a little away from the bottom, oscillates about this dip. Then the expression for the time period of oscillation is R2 (a) T = 2p g

g (b) T = 2p R

R (d) T = 2p gR g 3. A cyclic process ABCA is shown in V – T diagram, is performed with a constant mass of an ideal gas. Which of the following graphs in figure represents the corresponding process on a P – V diagram. (c) T = 2p

C

A

O A

T

B

(a) P

C

A

(b) P

B C

V

V

A

A B

(c) P

C V

(d) P

6. Two identical containers A and B with frictionless pistons contain the same ideal gas at the same temperature and the same volume V. The mass of gas in A is mA and that in B is mB. The gases in each cylinder are now allowed to expand isothermally to the same final volume 2V. The changes in pressure in A and B are found to be DP and 1.5 DP respectively. Then (a) 4 mA = 9 mB (b) 2 mA = 3 mA (c) 3 mA = 2 mB (d) 9 mA = 4 mB 7. A person carrying a whistle emitting continuously a note of 272 Hz is running towards a reflecting surface with a speed of 18 km h–1. The speed of sound in air is 345 m s–1. The number of beats heard by him is (a) 4 (b) 6 (c) 8 (d) 3

B

V

5. Two particles move parallel to x-axis about the origin with the same amplitude and frequency. At a certain instant, they are found at distance r/3 from the origin on opposite sides but their velocities are found to be in the same direction. The phase difference between the two particles is (a) cos–1(2/3) (b) cos–1 (4/9) –1 (c) cos (7/9) (d) cos–1 (5/9)

B

C V

4. Two uniform wires are vibrating simultaneously in their fundamental notes. The tensions, lengths, diameters and the densities of the two wires are in the ratio 8 : 1, 36 : 35, 4 : 1 and 1 : 2 respectively. If the note of the higher pitch has a frequency 360 Hz, the number of beats produced per second is (a) 5 (b) 15 (c) 10 (d) 20

8. Two balloons are filled, one with pure He gas and other by air, respectively. If the pressure and temperature of these balloons are same, then the number of molecules per unit volume is (a) more in the He filled balloon (b) same in both balloons (c) more in air filled balloon (d) in the ratio of 1 : 4 9. Equations of a stationary wave and a travelling wave are y1 = asin kx coswt and y2 = asin (wt – kx). The phase p 3p difference between two points x1 = and x 2 = 3k 2k are f1 and f2 respectively for the two waves. The ratio f1/f2 is (a) 1 (c)

3 4

5 6 6 (d) 7

(b)

Physics for you | OctOber ‘15

51

10. One kg of coal generates 0.7 × 104 kcal of heat during combustion. Assuming that 2% of this energy is utilized for mechanical purpose, find the mass of coal required per hour for running a 10 metric hp engine. (1 metric hp = 735.5 W, 1 cal = 4.186 J) (a) 35.2 kg (b) 45.2 kg (c) 73.5 kg (d) 63.5 kg 11. An ideal gas has a specific heat at a constant pressure, CP = (5/2)R. The gas is kept in a closed vessel of volume 0.0083 m3 at a temperature of 300 K and a pressure of 1.6 × 106 N m–2. If 2.49 × 104 J amount of heat energy is supplied to the gas. Find the final pressure of the gas. (a) 2.4 × 106 N m–2 (b) 3.6 × 106 N m–2 (c) 4.2 × 106 N m–2 (d) 8.3 × 106 N m–2 12. An aeroplane is going towards east at a speed of 510 km h–1 at a height of 2000 m. At a certain instant, the sound of the plane heard by an observer on ground appears to come from a point vertically above him. Where is the plane at this instant? Speed of the sound in air = 340 m s–1. (a) 425 m (b) 530 m (c) 833 m (d) 920 m 13. The displacement y of a wave travelling in x-direction is given by p  y = 10 −4 sin  600t − 2x +  3 where x and y are in metres and t is in second. The speed of wave motion in m s–1 is (a) 300 (b) 600 (c) 1200 (d) 200 14. Starting from the origin, a body oscillates simple harmonically with a period of 2 s. After what time will its kinetic energy be 75% of the total energy? 1 1 1 1 s (b) s (a) (c) s (d) s 6 3 4 12 15. The displacement of a particle varies with time as x = 12sinwt – 16sin3wt (in cm). If its motion is simple harmonic then its maximum acceleration is 2 (a) 12w2 (b) 192 w (c) 36 w2 (d) 144 w2

18. A certain simple harmonic vibrator of mass 0.1 kg has a total energy of 10 J. Its displacement from the mean position is 1 cm when it has equal kinetic and potential energies. The amplitude A and frequency u of vibration of the vibrator are 500 (a) A = 2 cm, u = Hz p 1000 (b) A = 2 cm, u = Hz p 1 500 (c) A = cm, u = Hz p 2 1 1000 cm, u = Hz (d) A = p 2 19. A wave travelling along the x-axis is described by the equation y(x, t) = 0.005 cos(ax – bt). If the wavelength and the time period of the wave are 0.08 m and 2.0 s respectively, then a and b in appropriate units are p (a) a = 12.50 p, b = 2. 0 (b) a = 25.00 p, b = p 0.08 2. 0 (c) a = , b= p p 0.04 1. 0 , b= (d) a = p p 20. A Carnot engine whose sink is at 300 K has an efficiency of 40%. By how much should the temperature of source be increased so as to increase its efficiency by 50% of original efficiency? (a) 380 K (b) 275 K (c) 325 K (d) 250 K 21. An air column in a pipe which is closed at one end, will be in resonance with the vibrating body of frequency 166 Hz, if the length of the air column is (a) 0.5 m (b) 1.0 m (c) 1.5 m (d) 2.0 m

16. A gaseous mixture consists of 16 g of helium and 16 g of oxygen. The ratio CP /CV of the mixture is (a) 1.4 (b) 1.54 (c) 1.59 (d) 1.62

22. A cylinder containing one gram mole of the gas was compressed adiabatically until its temperature rises from 27°C to 97°C. The heat produced in the gas (g = 1.5) is (a) 250.6 cal (b) 276.7 cal (c) 298.5 cal (d) 320 cal

17. A gas expands with temperature according to the relation V = KT2/3. Calculate work done when the temperature changes by 60 K? (a) 10R (b) 30R (c) 40R (d) 20R

23. Heat is supplied to a diatomic gas at constant pressure. The ratio of DQ : DU : DW is (a) 5 : 3 : 2 (b) 7 : 5 : 2 (c) 2 : 3 : 5 (d) 2 : 5 : 7

52

Physics for you | OctOber ‘15

24. Suppose ideal gas equation follows VP3= constant. Initial temperature and volume of gas are T and V respectively. If gas expands to 27V, then its temperature will become (a) T (b) 9T T (c) 27T (d) 9 25. Mean free path of a gas molecule is (a) inversely proportional to number of molecules per unit volume (b) inversely proportional to diameter of the molecule (c) directly proportional to the square root of the absolute temperature (d) directly proportional to the molecular mass 26. The amplitude of a damped oscillator becomes half in one minute. The amplitude after 4 minutes will 1 times the original, then x is be x (a) 32 (b) 24 4 (c) 3 (d) 23 27. The efficiency of Carnot’s heat engine is 0.5 when the temperature of the source is T1 and that of sink is T2. The efficiency of another Carnot’s heat engine is also 0.5. The temperature of source and sink of the second engine are respectively T (a) 2T1, 2T2 (b) 2T1, 2 2 (c) T1 + 5, T2 – 5 (d) T1 + 10, T2 – 10 28. What is the change in internal energy of one mole of a gas, when volume changes from V to 2V at constant pressure P? R (b) PV (g − 1) gPV PV (c) (g − 1) (d) (g − 1) where g is the ratio of specific heats of the gas at constant pressure to that at constant volume. (a)

29. A source of sound of frequency 600 Hz is placed inside water. The speed of sound in water is 1500 m s–1 and in air it is 300 m s–1. The frequency of sound recorded by an observer who is standing in air is (a) 200 Hz (b) 3000 Hz (c) 120 Hz (d) 600 Hz

30. Match the Column I with Column II. Column I

Column II

A

Molar specific heat of (i) helium gas at constant volume (when R is gas constant)

3R

B

Molar specific heat (ii) of oxygen at constant volume Molar specific heat (iii) of carbondioxide at constant volume

7 R 2

C

D

3 R 2

Molar specific heat of (iv) hydrogen at constant pressure (a) (b) (c) (d)

5 R 2

A – (i), B – (ii), C – (iii), D – (iv) A – (ii), B – (iii), C – (iv), D – (i) A – (iii), B – (iv), C – (i), D – (ii) A – (iv), B – (i), C – (ii), D – (iii) SolutionS

1. (b) : Let the initial volume, V1 = V Final volume, V2 = V/4 Here, initial temperature, T1 = 27°C = (273 + 27) K = 300 K and g = 1.5 Let T2 be the temperature after compression. Since the change in adiabatic, T1V1g–1 = T2V2g–1 g −1

1. 5 − 1

V   V  T2 = T1  1  = 300  V /4  V2  = (300 K)(4)1/2 = (300 K) × 2 = 600 K Rise in temperature = T2 – T1 = 600 K – 300 K = 300 K

or

2. (c) : Let R be the radius of the dip and C be its centre. As shown in figure, let the rickshaw of mass M be at A at any instant. C

 R Road

Mg sin  A

O

 Mg

Mg

cos



This case is similar to that of a simple pendulum. Physics for you | OctOber ‘15

53

The force that produces oscillations in the rickshaw is F = Mg sin q. If q is small and measured in rad, sin q ≈ q and F ≈ Mg q Displacement of the rickshaw, i.e., OA = Rq. Force constant, force F Mg q Mg k= = = = displacement OA Rq R Inertia factor = M inertia factor Time period, T = 2p spring factor or

T = 2p

M Mg / R

or T = 2p

3. (a) : From A to B, V ∝ T or

R g

V = constant T

PV = R = constant \ P is constant T (A′ B′ is a straight line parallel to volume axis) From B to C, volume V is constant (B′ C′ is a straight line parallel to pressure axis). From C to A, temperature T is constant. \ PV = constant (Boyle’s Law) 1 Thus, C′ A′ is a curve such that P ∝ . V T 8 l 36 d1 4 ρ1 1 = ; = 4. (c) : Here, 1 = , 1 = , T2 1 l2 35 d2 1 ρ2 2 As

u2 = 360 Hz, u2 = ? l d T2 ρ1 u Now, 2 = 1 1 u1 l2 d2 ρ2 T1 u2 36 4 1 1 36 = × × = u1 35 1 8 2 35 So, u2 > u1. When u2 = 360 Hz, u1 = 350 Hz Number of beats per second = u2 – u1 = 360 – 350 = 10 \

5. (c) : Let f be the phase difference between two waves. Then x1 = r sin wt ...(i) and x2 = r sin (wt + f) ...(ii) r 1 Given that = r sin wt or sin wt = ...(iii) 3 3 r and − = r sin (wt + f) or sin (wt + f) = − 1 / 3 3 or sin wt cos f + cos wt sin f = –1/3 1 1 or cos f + (1 − 1 / 9) sin f = − 3 3 7 On solving, we get cos f = –1 or 9 and f = p or cos–1 (7/9) 54

Physics for you | OctOber ‘15

Differentiating (i) and (ii), we get v1 = r w cos wt and v2 = rw cos (wt + f) If we put f = p, we find that v1 and v2 are of opposite signs. So, f = p is not true as per question. Hence f = cos–1(7/9). 6. (c) : As the expansion of gases is isothermal, 1 therefore, T = constant and P ∝ . V Volume of gas is increasing, pressure would decrease. In container A DP = (PA)i – (PA)f n RT nART nART ...(i) = A − = V 2V 2V In container B 1.5 DP = (PB)i – (PB)f n RT nBRT nBRT ...(ii) = B − = V 2V 2V From (i) and (ii) DP 2 m nA = = \ n ∝m As n = M nB 1.5 DP 3 mA nA 2 = = or 3 mA = 2 mB mB n B 3 7. (c) : Original frequency of whistle, u = 272 Hz. Apparent frequency of whistle striking the wall (listener) when person (source) is moving towards the wall with velocity vs = 18 km h–1 = 5 m s–1 is v 345 …(i) u′ = ×u= × 272 v − vs 345 − 5 On reflection, wall acts as source and person act as listener moving towards the source. Therefore, v + vL (345 + 5) 345 u′′ = × u′ = × × 272 = 280 v 345 340 \ No. of beats per second = u′′ – u = 280 – 272 = 8 8. (b) : Assuming the balloons have the same volume, as PV = nRT, if P, V and T are the same, n, the number of moles present will be the same, whether it is He or air. Hence, number of molecules per unit volume will be same in both the balloons. 9. (d) : Equation of a stationary wave is y1 = asinkx coswt and equation of progressive wave is y2 = asin(wt – kx) = a(sinwt coskx – coswt sinkx) p 3p At x1 = and x 2 = 3k 2k sin kx1 or sin kx2 is zero.

\ neither x1 nor x2 is node. 3p p 7p Dx = x 2 − x1 = − = 2k 3k 6k 7p 2p p As Dx = , therefore, > Dx > . 6k k k l 2p But = l. So, l > Dx > 2 . k In case of a stationary wave, phase difference between any two points is either zero or p. 7p 7 \ f1 = p and f 2 = kDx = k = p 6k 6 f1 p 6 = = f2 7 7 p 6 10. (b) : Power of the engine = 10 metric hp = (10 × 735.5) W = 7355 W Work required to run the engine for one hour = 7355 × 3600 J Amount of heat required 7355 × 3600 J = = 6325 kcal 4.186 Jcal −1 \

Since 2% (i.e., 2/100) of the heat of combustion of coal is used for mechanical purpose and one kg of coal generates 0.7 × 104 kcal, mass of coal required to run the engine for one hour 6325 kcal 100 = × = 45.2 kg 4 −1 2 0.7 × 10 kcal kg (As 2/100 = output/input, input = output × 100/2) 11. (b) : Given, P = 1.6 × 106 N m–2, V = 0.0083 m3, T = 300 K According to gas equation, PV = nRT 6 PV (1.6 × 10 )(0.0083) 16 or n = = = RT 8.3 × 300 3 5 3 R−R= R As CP – CV = R, CV = CP – R = 2 2 Let dT be the rise in temperature when 2.49 × 104 J of heat energy is supplied to the gas at constant volume. So, 2.49 × 104 = nCVdT 2.49 × 10 4 2.49 × 10 4 J or dT = K = 375 K = 8 × 8. 3 (16 / 3) × (3 / 2)R

()

Final temperature of the gas = T + dT = (300 + 375)K = 675 K Since the gas is heated at constant volume,

()

P T = P′ T ′

6 −2 PT ′ (1.6 × 10 N m )(675 K) = T 300 K = 3.6 × 106 N m–2 12. (c) : Let O be the observer and A be the position of plane vertically above him. So, AO = 2000 m. The sound waves that reach the observer are emitted when the plane is at A. During the time sound waves travel from A to O (with velocity v), the plane moves eastwards through a distance AB with vp B A velocity vp, AO AB \ = v vp 2000 m v vp or AB = AO × v

or

P′ =

But vp = 510 km h–1 O = 510 × (5/18) m s–1 = (425/3) m s–1, and v = 340 m s–1 Thus,  (425 / 3) ms −1  2500 = AB = (2000 m)  m = 833 m −1  3   340 ms p  −4 13. (a) : Here, y = 10 sin  600t − 2x +  3 Compare it with the standard equation of a travelling wave y = A sin  2pt − 2px + f   T  l 2p 2p p = 600,T = = s T 600 300 2p 2p = +2, l = =pm l 2 l p v= = = 300 m s −1 T p / 300 75 75 1 2 2 E= × ma w 100 100 2 1 2 2 75 1 2 2 \ ma w cos 2 wt = × ma w 2 100 2 3 3 p or cos 2 wt = or cos wt = = cos 4 2 6 p p p 1 or wt = or t = = = s 6 6w 6 × (2p / 2) 6

14. (b) : Here, K.E. =

15. (c) : Given, x = 12 sin wt – 16 sin3wt = 12 sinwt – 4(4 sin3wt) [Q sin3q = 3sinq – 4 sin3q or 4 sin 3q = 3sinq – sin3q] \ x = 12 sinwt – 4(3sinwt – sin 3 wt) Physics for you | OctOber ‘15

55

x = 4 sin 3wt dx velocity,v = = 4 × 3w cos 3wt dt dv acceleration, a = = 12w × 3w(− sin 3wt ) dt = – 36w2 sin3wt Maximum acceleration = 36 w2

or

Frequency, u =

1 K.E. = 5 = × 0.1[A2 − (1 × 10 −2)2](106) 2 or A2 − 10 −4 = 10 −4 or A2 = 2 × 10 −4 or A = 2 × 10 −2 m = 2 cm 19. (b) : The wave travelling along x-axis is given by y(x, t) = 0.005 cos(ax – bt). 2p Therefore a = k = . As l = 0.08 m. l 2p p p \ a= = ⇒ a = × 100.00 = 25.00 p. 0.08 0.04 4 2p 2p w= = b. As T = 2.0 s, b = ⇒ b=p T 2. 0 \ a = 25.00 p, b = p

16 =4 4 16 1 For 16 g of oxygen, n2 = = 32 2 For mixture of gases, n1CV1 + n2CV2 f where CV = R CV = 2 n1 + n2

16. (d) : For 16 g of helium, n1 =

n1C P1 + n2C P2 f  where C P =  + 1 R n1 + n2 2  For helium, f = 3, n1 = 4 CP =

For oxygen, f = 5, n2 =

1 2

5  1 7   4 × R +  × R 2   2 2  47 C P  \ = = 1.62 = CV  3   1 5  299 4 × R + × R  2   2 2  17. (c) : dW = PdV =

RT dV V

...(i)

2 As V = KT \ dV =K T −1/3 dT 3 2 −1/3 K T dT dV 2 dT \ = 3 2/3 = V 3 T KT 2/3

T2

From (i), W = ∫ RT T1

dV = V

T2

2 dT T

∫ RT 3

T1

2 2 R (T2 − T1) = R × 60 = 40R 3 3 18. (a) : For a displacement x, the kinetic and potential energies are, 1 2 2 1 K.E. = m(A2 − x 2)w 2 and P.E. = mx w 2 2 10 Each of these = = 5 J when x = 1 cm 2 1 2 2 1 \ mx w = × 0.1 × (1 × 10 −2)2 w 2 = 5 2 2 10 2 or w = = 106 0.1 × 10 −4 or w = 103 = 1000 rad s–1 2p p T= = s w 500 W=

56

Physics for you | OctOber ‘15

1 500 −1 500 = s = Hz p p T

20. (d) : Efficiency of a Carnot engine, T2 40 3 T = 1− h = 1− = h = 1 − 2 or T1 100 5 T1 5 5 \ T1 = × T2 = × 300 = 500 K 3 3 Increase in efficiency = 50% of 40% = 20% \ New efficiency, h′ = 40% + 20% = 60% T2 60 2 \ = 1− = 100 5 ′ T 1

(where T′1 is the increased source temperature) 5 5 \ T1′ = × T2 = × 300 = 750 K 2 2 Increase in temperature of source = T1′ – T1 = 750 – 500 = 250 K

21. (a) : When air is blown at the open end of a closed pipe a longitudinal wave travels in the air of the pipe from closed end to open end. When l is wavelength, l is the length of pipe, u is the frequency of note emitted and v is the velocity of sound in air, then v u = , (fundamental note) l Given, u = 166 Hz, v = 332 m s–1 332 \ l= =2m 166 But, l = 4l l 2 \ l = = =0.5 m 4 4 22. (b) : Here, initial temperature, T1 = 27°C = 273 + 27 = 300 K Final temperature, T2 = 97° C = 273 + 97 = 370 K

Physics for you | OctOber ‘15

57

When a gas is compressed adiabatically, work done on the gas is given by R W= (T − T ) 8.3 × (370 − 300) (1 − g) 2 1 = 1 − 1. 5 2 or W = –11.62 × 10 J \ Heat produced, W 11.62 × 102 H= = = 276.7 cal J 4. 2 5 7 23. (b) : For a diatomic gas, CV = R, C P = R 2 2 7  DQ = nC P DT = n  R  DT 2  5  DU = nCV DT = n  R  DT 2  DW = DQ − DU = nR DT \ DQ : DU : DW =

7 5 : : 1 or 7 : 5 : 2 2 2

24. (b) : Given: VP 3 = constant = k or P = According to ideal gas equation, PV = nRT K 1/3 \ V = nRT (say (k) = K) 1/3 V nRT or V 2/3 = K Hence,  V1   V  2  V   27V   

2/3

=

2/3

=

(k)1/3 (V )1/3

T1 T2

T1 or T2 = 9 T1 = 9T T2

1 25. (a) : Mean free path, l = 2 pd 2n where, n = Number of molecules per unit volume d = Diameter of the molecule As PV = kBNT k BT N P \ l= n= = V k BT 2 pd 2P where, kB = Boltzmann constant P = Pressure of gas T = Absolute temperature of the gas 26. (b) : Amplitude of damped oscillation is A = A0 e–bt/2m A As, A = 0 , t = 1 minute, 2 58

Physics for you | OctOber ‘15

A0 1 −b/2m −b × 1/2m = A0 e ⇒ =e ⇒ e b/2m = 2 2 2 A0 When, A = , t = 4 minutes x b/2m A0 −b × 4/2m {Q e = 2} \ = A0 e x ⇒ x = e 4b/2m = (e b/2m)4 = (2)4 27. (a) : Efficiency of Carnot’s heat engine, T h =1− 2 T1 T For first Carnot’s heat engine, 0.5 = 1 − 2 T1 T′ For second heat engine, 0.5 = 1 − 2 T1′ Efficiency remains the same when both T1 and T2 are increased by same factor. Therefore, the temperature of source and sink of second engine are T1′ = 2T1, T2′ = 2T2 \

28. (c) : According to ideal gas equation PV = nRT When pressure is constant PdV = nRdT P(2V – V) = nRdT PV = nRdT ...(i) CP As g = CV C − CV \ g −1= P CV C P − CV or CV = g −1 R = (Q C P − CV = R) g −1 nRdT DU = nCV dT = g −1 PV DU = (Using (i)) g −1 29. (d) : Frequency of sound wave, does not change during refraction. Frequency of wave depends upon the vibrator which produces the wave. It does not depend on medium in which it is propogated. The wavelength and velocity of the wave undergo change during refraction. \ The observer records the frequency as 600 Hz. 30. (c) 

SECOND STAGE

BIHAR CECE SOLVED PAPER 2015

1. A ball of mass M1 collides elastically and head on with another ball of mass M2 initially at rest. In which of the following cases the transfer of momentum will be maximum? (a) M1 > M2 (b) M1 = M2 (c) M1 < M2 (d) Data is insufficient to predict it. 2. Two particles of equal masses have velocity   v1 = 2i m/s and v2 = 2 j m/s . The first particle  has an acceleration a1 = (3i + 3j) m/s2 while the acceleration of the other particle is zero. The centre of mass of the two particles moves in a (a) circle (b) parabola (c) ellipse (d) straight line 3. A body is kept on a horizontal disc of radius 2 m at a distance of 1 m from the centre. The coefficient of friction between the body and the surface of the disc is 0.4. The speed of rotation of the disc at which the body starts slipping is (g = 10 m/s2) (a) 4 rad/s (b) 2 rad/s (c) 0.4 rad/s (d) 2 rad/s 4. A wheel has moment of inertia 5 kgm2. If 105 J of work is done in producing rotational kinetic energy, then the wheel attains an angular speed equal to (a) 200 rad/s (b) 20 rad/s (c) 100 rad/s (d) 10 rad/s 5. A balloon has 5 g of air. A small hole is pierced into it. The air escapes at a uniform rate with a velocity of 4 cm/s. If the balloon shrinks completely in 2.5 s, then the average force acting on the balloon is (a) 2 dyne (b) 2 N (c) 8 dyne (d) 8 N 6. A body weighs 700 g on the surface of the earth. How much will it weigh on the surface of a planet 1 whose mass is and radius is half that of earth? 7 (a) 200 g (b) 400 g (c) 350 g (d) 50 g

7. Three uniform spheres, each having mass m and radius r, are kept in such a way that each touches the other two. The magnitude of the gravitational force on any sphere due to the other two is (a)

Gm2

(b)

Gm2

r2 4r 2 2 Gm Gm2 (c) 2 2 (d) 3 4r 4r 2 8. Which of the following statements is not correct for the decrease in the value of acceleration due to gravity? (a) As we go down from the surface of the earth towards its centre. (b) As we go up from the surface of the earth. (c) As we go from equator to the poles on the surface of the earth. (d) As the rotational velocity of the earth is increased. 9. There are four point masses m each on the corners of a square of side length l. About one of its diagonals, the moment of inertia of the system is (a) 2ml2 (b) ml2 (c) 4ml2 (d) 6ml2 10. Two cylinders A and B of radii r and 2r are soldered coaxially. The free end of A is clamped and the free end of B is twisted by an angle q. The twist produced at the junction is 16 17 q q (c) q (a) (b) (d) zero 17 16 11. A thick rope of density r and length L is hung from a rigid support. The increase in length of the rope due to its own weight is (Y is Young’s modulus) rL2 g rL2 g rL2 g rLg (a) (b) (c) (d) 2Y 4Y Y Y 12. The excess pressure inside the first soap bubble is three times that inside the second bubble. Then the ratio of the volumes of the first and second bubbles is (a) 1 : 3 (b) 3 : 1 (c) 1 : 27 (d) 27 : 1

Contributed by : Paradise Institute, Patna Physics for you | OctOber ‘15

59

13. A long capillary tube of radius 0.2 mm is placed vertically inside a beaker of water. The surface tension of water is 7.0 × 10–2 N/m. Water rises into the capillary tube upto a height of 5 cm. The angle of contact between the glass and water is (g = 10 m/s2) −1  5  −1  4  (a) cos   (b) cos   5 7 4 −1  2  − 1 (c) cos   (d) cos   7 7 14. A metal plate of area 500 cm2 is kept on a horizontal surface with a layer of oil of thickness 0.5 mm between them. The horizontal force required to drag the plate with a velocity of 2 cm/s is (coefficient of viscosity = 0.9 kg/m s) (a) 180 N (b) 18 N (c) 0.018 N (d) 1.8 N 15. In motors, more viscous oil is used in summer than in winter due to (a) the rise in temperature in summer, the viscosity of oil decreases. (b) the rise in temperature in summer, the viscosity of oil increases. (c) the decrease in surface tension of oil. (d) the increase in surface tension of oil. 16. Two vessels A and B are identical. A has 1 g hydrogen at 0°C and B has 1 g oxygen at 0°C. Vessel A contains x molecules and B contains y molecules. The average kinetic energy per molecule in A is n times the average kinetic energy per molecule in B. The value of n is (a) 16 (b) 8 (c) 32 (d) 1 17. An ideal gas having f degrees of freedom is isobarically heated. The ratio of the work done by it to the change in its internal energy will be f 2 2 f −2 (a) (b) (c) (d) 2 f −2 f 2 18. Pressure remaining constant, at what temperature will the r.m.s. velocity of a gas be half of its value at 0°C? (a) 0°C (b) 32°C (c) –273°C (d) –204°C 19. A diatomic gas undergoes same change of temperature by two different processes (i) at constant volume and (ii) at constant pressure. The heat supplied in the two cases will be in the ratio of (a) 1 : 1 (b) 3 : 5 (c) 5 : 7 (d) 7 : 5 20. An ideal gas A and a real gas B have their volumes increased from V to 2V under isothermal conditions. The increase in internal energy (a) will be same in both A and B 60

Physics for you | OctOber ‘15

(b) will be zero in both the cases (c) of B will be more than that of A (d) of A will be more than that of B 21. For a certain gas g = 1.5, then for this gas (a) CV = 3R (b) CP = 3R (c) CP = 5R (d) CV = 5R

22. Wien’s constant is 2892 × 10–6 S.I. unit and the value of lm for moon is 14.46 micron. The surface temperature of moon is (a) 200 K (b) 2000 K (c) 20 K (d) 200°C 23. The temperature of equal masses of three different liquids A, B and C are 12°C, 19°C and 28°C respectively. The temperature when A and B are mixed is 16°C, when B and C are mixed is 23°C. What is the temperature when A and C are mixed? (a) 26.02°C (b) 22.60°C (c) 20.26°C (d) 21.62°C 24. Which of the following constants is not related to radiation? (a) Solar constant (b) Boltzmann’s constant (c) Stefan’s constant (d) Wien’s constant 25. Air conditioners are good example of (a) conduction (b) convection (c) radiation (d) both conduction and radiation 26. A refrigerator works between 2°C and 27°C. To keep the temperature of the refrigerated space constant, 660 calories of heat are to be removed every second. The power required is (a) 60 watt (b) 55 watt (c) 252 watt (d) 231 watt 27. Two bodies A and B are placed in an evacuated vessel maintained at a temperature of 27°C. The temperature of A is 327°C and that of B is 227°C. The ratio of heat loss from A and B is about (a) 9 : 4 (b) 6 : 5 (c) 36 : 25 (d) 3 : 2 28. For any given scale X, the ice point is 40° and the steam point is 120°. For another scale Y, the ice point and steam point are –30° and 130° respectively. If X reads 50°, then Y would read (a) –5° (b) –8° (c) –10° (d) –12° 29. 56 tuning forks are so arranged in series that each fork gives 4 beats per second with the previous one. The frequency of the last fork is three times that of the first. The frequency of the first fork is (a) 52 Hz (b) 56 Hz (c) 60 Hz (d) 110 Hz

30. The velocity of sound is greatest in (a) steel (b) ammonia (c) air (d) water 31. The equation of a wave is y = 60 cos (1800t – 6x), where y is in microns, t in seconds and x in metres. The ratio of maximum particle velocity to the wave velocity of wave propagation is (a) 3.6 (b) 3.6 × 10–6 –11 (c) 3.6 × 10 (d) 3.6 × 10–4 32. Two boys stand close to a long straight metal pipe, at some distance from each other. One boy fires a gun and the other hears two explosions with a time interval of 1 s between them. If the velocity of sound in metal is 3630 m/s and in air is 330 m/s, then the distance between the two boys is (a) 36.3 m (b) 363 m (c) 72.6 m (d) 726 m 33. The lengths of two organ pipes open at both ends are L and L + d. If they are sounded together, then the beat frequency will be 2Vd Vd (a) (b) L(L + d ) L(L + d ) 2L(L + d ) Vd (c) (d) Vd 2L(L + d ) 34. Two particles P and Q describe simple harmonic motions of same amplitude a, frequency u along the same straight line. The maximum distance between the two particles is a 2 . The initial phase difference between the particles is (a) zero (b) 45° (c) 60° (d) 90° 35. A car with a horn of frequency 620 Hz travels towards a large wall with a speed of 20 m/s. Velocity of sound is 330 m/s. The frequency of echo of sound of horn as heard by the driver is (a) 700 Hz (b) 660 Hz (c) 620 Hz (d) 550 Hz 36. When the length of a simple pendulum is decreased by 600 mm, the period of oscillation is halved. The original length of the pendulum was (a) 800 mm (b) 1000 mm (c) 1200 mm (d) 2400 mm 37. A particle of mass 0.2 kg moves with simple harmonic motion of amplitude 2 cm. If the total energy of the particle is 4 × 10–5 J, then the time period of the motion is

3p seconds 2 p (c) p seconds (d) seconds 2 38. Which of the following cannot represent a travelling wave? (a) y = f(x – vt) (b) y = ym sin k(x + vt) (c) y = (x – vt)/(x + vt) (a) 2p seconds

(b)

(d) y = Ae–b(x – vt)

2

39. Which one of the following emits sound of higher pitch? (a) Lion (b) Man (c) Donkey (d) Mosquito 40. It is possible to recognize a person by hearing his voice even if he is hidden behind a solid wall. This is due to the fact that his voice (a) has a definite pitch (b) has a definite quality (c) has a definite capacity (d) can penetrate the wall 41. The region surrounding a stationary electric dipole has (a) electric field only (b) magnetic field only (c) both electric and magnetic fields (d) neither electric nor magnetic field 42. Five balls numbered 1 to 5 are suspended using separate threads. Pairs (1, 2), (2, 4) and (4, 1) show electrostatic attraction while pairs (2, 3), (4, 5) show repulsion. Therefore, ball 1 must be (a) neutral (b) metallic (c) positively charged (d) negatively charged 43. Two spherical conductors of radii 4 m and 5 m are charged to same potential. If s1 and s2 be the respective values of the surface density of charge on the two conductors, then the ratio s1/s2 is 25 5 16 4 (a) (b) (c) (d) 16 4 25 5 10 44. Electric charge of × 10−9 C are placed at each 3 of the four corners of a square of side 0.08 m. The potential at the intersection of diagonals is (a) 900 V

(b) 900 2 V

(c) 150 2 V

(d) 1500 2 V Physics for you | OctOber ‘15

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45. A condenser having a capacity of 6 mF is charged to 100 V and is joined to an uncharged condenser of 14 mF. The ratio of charges after connection and the potential on 6 mF and 14 mF condensers respectively will be 6 14 (a) and 50 V (b) and 30 V 14 6 6 14 (c) and 30 V (d) and 0 V 14 6 46. The effective capacitance between points A and B will be 2 F 12 F 28 mF (a) 9 2 F B (b) 5 mF A (c) 4 mF 2 F (d) 18 mF 47. The radius of earth is 6400 km. Its capacitance will be (a) zero (b) 7.1 × 10–4 F –4 (c) 6.4 × 10 F (d) 6.4 × 106 F 48. Eight drops of mercury of equal radii and possessing equal charges combine to form a big drop. The capacitance of the bigger drop as compared to capacitance of each individual drop is (a) 16 times (b) 8 times (c) 2 times (d) 32 times 49. A voltmeter reads 6 V at full scale deflection and is graded as 3000 W/V. What resistance should be connected in series with it so that it reads 12 V at full-scale deflection? (a) 1.8 × 104 W (b) 3.6 × 104 W 4 (c) 5.4 × 10 W (d) 7.2 × 104 W 50. In the given circuit, with steady current, the potential drop across the capacitors must be (a) V R V V (b) C V 2 2V V 2V 2R (c) (d) 3 3 51. Eddy currents are produced in a material when it is (a) heated (b) placed in a time varying magnetic field (c) placed in an electric field (d) placed in a uniform magnetic field 52. In the circuit, the voltmeter resistance is 10000 W and the ammeter resistance is 2 W. The voltmeter 62

Physics for you | OctOber ‘15

reads 12 V and ammeter reads 0.1 A. The value of R is (a) 118 W R A (b) 122 W V (c) 10022 W (d) 10018 W 53. A magnetic dipole of moment 0.72 Am2 is placed horizontally with the north pole pointing south. The magnetic field of earth is 18 mT. The neutral point is at (a) 0.1 m on axial line (b) 0.2 m on equatorial line (c) 0.2 m on axial line (d) 0.16 m on equatorial line 54. A magnetizing field of 5000 A/m produces a magnetic flux of 5 × 10–5 Wb in an iron rod of 0.5 cm2 area of cross-section. The permeability of the rod is (in Wb A–1 m–1) (a) 4 × 10–6 (b) 3 × 10–5 –4 (c) 2 × 10 (d) 1 × 10–3 55. A uniformly charged spherical metal ball of mass M, radius R and charge q is rotated with an angular velocity w about one of its diameter. The ratio of its magnetic dipole moment to the angular momentum is q q R (a) (b) 2M 2Mw qw q (c) (d) 2 MR 2M 56. Flux f (in Weber) in a closed circuit of resistance 10 W varies with time t (in seconds) according to the equation f = 6t2 – 5t + 1 The magnitude of the induced current in the circuit at t = 0.25 s is (a) 0.2 A (b) 0.6 A (c) 0.8 A (d) 1.2 A 57. A transformer has an efficiency of 80%. It works at 4 kilowatt and 100 volts. If the secondary voltage is 240 volts, then the current in the secondary coil is (a) 1.333 A (b) 4 A (c) 13.33 A (d) 40 A 58. When inductance of 1 henry is connected to 200 volts 50 Hz source, the maximum value of the current is (a) 0.1 A (b) 0.9 A (c) 1 A (d) 9 A 59. If in a moving coil galvanometer, a current i produces a deflection q, then

(a) i ∝ tan q (c) i ∝ q2

(b) i ∝ q (d) i ∝ q

60. An electric heater consumes 1 kilowatt power when connected across a 100 volt D.C. supply. If this heater is to be used with 200 V, 50 Hz A.C. supply, the value of inductance to be connected in series with it is (a) 5.5 H (b) 0.55 H (c) 0.055 H (d) 1.1 H 61. In an oscillating LC circuit the maximum charge on the capacitor is Q. When the energy is shared equally between the electric and the magnetic fields, then the charge on the capacitor is Q (a) Q (b) 2 Q Q (c) (d) 3 2 62. A 16 mF capacitor is charged to 20 volts. The battery is then disconnected and a pure 40 mH coil is connected across the capacitor so that LC oscillations are set up. The maximum current is the coil is (a) 0.2 A (b) 40 mA (c) 2 A (d) 0.4 A 63. Which of the following statements about electromagnetic waves is/are true? 1. Electromagnetic waves travel at the same speed in any medium. 2. All electromagnetic waves can ionize living cells. 3. All electromagnetic waves are transverse waves. (a) 3 only (b) 1 and 3 only (c) 2 and 3 only (d) 1, 2 and 3 64. Which of the following electromagnetic waves will cause heating? 1. Radio waves 2. Infrared radiation 3. X-rays (a) 2 only (b) 1 and 2 only (c) 2 and 3 only (d) 1, 2 and 3 65. A linearly polarized electromagnetic wave given as E = E0 i cos(kz − wt ) is incident normally on a perfectly reflecting infinite wall at z = a. Assuming that the material of the wall is optically inactive, the reflected wave will be given as (a) Er = E0 i sin(kz − wt )

(b) Er = − E0 i cos(kz − wt ) (c) E = − E i cos(kz + wt ) r

0

(d) Er = E0 i cos(kz + wt )

66. Which of the following statements is incorrect? (a) The magnification produced by a convex mirror is always less than one. (b) A virtual, erect, same sized image can be obtained using a plane mirror. (c) A virtual, erect, magnified image can be formed using a concave mirror. (d) A real, inverted, same sized image can be formed using a convex mirror. 67. A concave mirror of focal length f (in air) is 4  immersed in water  m =  . The focal length of  3 mirror in water will be 3 7 4 f (c) f f (a) (b) (d) f 4 3 3 68. A diamond sparkles because of its (a) hardness (b) emission of light by the diamond (c) absorption of light by the diamond (d) high refractive index. 69. By placing a convex lens of focal length equal to 15.0 cm between an object and a screen separated by a distance of 75.0 cm, the sizes of the images 2 obtained are 6.0 cm and cm . The size of the 3 object must be (a) 2.0 cm (b) 4.0 cm (c) 3.0 cm (d) 1.5 cm 70. A glass cube is placed on a white paper having spots of red, blue, yellow and green colour. Then, the one that appears least raised is (a) blue (b) red (c) yellow (d) green 71. The refractive index of air is 1.0003. The thickness of air column which has one more wavelength of yellow light (l = 600 nm), than the same thickness of vacuum is (a) 0.02 mm (b) 0.2 mm (c) 2 mm (d) 2 cm 72. Two convex lenses of focal lengths f1 and f2 are separated co-axially by a distance d. The power of the combination will be zero if (a) d = (f1 + f2) (b) d = (f1 – f2) (f − f ) (c) d = f1 f2 (d) d = 1 2 2 Physics for you | OctOber ‘15

63

73. The time of coherence is of the order of (a) 10–2 s (b) 10–4 s –6 (c) 10 s (d) 10–8 s 74. The first minimum of a single slit diffraction pattern is observed at angle 2° with a light of wavelength 698 nm. The width of this slit is (a) 2 mm (b) 0.2 mm (c) 0.02 mm (d) 0.002 mm 75. When a thin sheet of transparent material of thickness 4 × 10–3 mm is placed in the path of one of the interfering beams in the Young’s double slit experiment, it is found that the central bright fringe shifts through a distance equal to four fringes. Wavelength of light used is 5893 Å. The refractive index of transparent material is (a) 1.5893 (b) 1.2946 (c) 1.884 (d) 1.9853 76. The intensity at a point where the path difference l is (l = wavelength of light) is I. If I0 is the 6 maximum intensity, then I/I0 is equal to (a) 3 /2 (b) 1/2 (c) 3/4

(d) 1/ 2

77. A ray of light from air is incident on the surface of glass with angle of incidence equal to the angle of polarization. It suffers a deviation of 22° on entering glass. The angle of polarization is (a) 22° (b) 56° (c) 34° (d) 68° 78. The photoelectric work function of a surface is 2.2 eV. The maximum kinetic energy of photoelectrons emitted when light of wavelength 6200 Å is incident on the surface is (a) 0.4 eV (b) 1.2 eV (c) 1.6 eV (d) photoelectrons are not emitted 79. What potential must be applied on an electron microscope so that it may produce an electron of wavelength 1 Å? (a) 50 V (b) 120 V (c) 150 V (d) 200 V 80. The de Broglie wavelength of thermal neutrons at 27°C will be (a) 1.77 Å (b) 1.77 mm (c) 1.77 cm (d) 1.77 m 81. If E and P are the energy and the momentum of a photon respectively, then on reducing the wavelength of photon 64

Physics for you | OctOber ‘15

(a) (b) (c) (d)

both P and E will decrease both P and E will increase P will increase but E will decrease P will decrease but E will increase

82. Half-life of a radioactive substance is 20 minutes. The time interval between 20% and 80% decay will be (a) 20 minutes (b) 30 minutes (c) 40 minutes (d) 60 minutes 83. If the aluminium nucleus 13Al27 has nuclear radius of about 3.6 Fermi, then the tellurium nucleus 125 will have nuclear radius nearly as 52Te (a) 3.6 Fermi (b) 6.0 Fermi (c) 8.9 Fermi (d) 16.7 Fermi 84. Consider the fusion reaction : 2 2 2 4 1 1 1H + 1H + 1H 2He + 0n + 1H The atomic masses are 1H2 = 2.0141 amu, 1H1 = 1.00783 amu, 2He4 = 4.0026 amu 0n1 = 1.008665 amu. In producing 8.2 × 1013 J of energy by this reaction, the mass of 1H2 fuel required is (a) 1.433 × 10–1 kg (b) 2.4 × 10–1 kg –1 (c) 1.433 × 10 g (d) 2.4 × 10–1 g 85. A positronium atom undergoes a transition from n = 4 to n = 2. The energy of the photon emitted in this process is (a) 1.275 eV (b) 2.55 eV (c) 3.4 eV (d) 3.825 eV 86. The most suitable energy of neutrons which will produce nuclear fission in a reactor is (a) 200 MeV (b) 2 MeV (c) 40 eV (d) 0.04 eV 87. A p-n junction diode is connected to a battery of e.m.f. 5.5 V and external resistance 5.1 kW. The barrier potential in the diode is 0.4 V. The current in the circuit is (a) 1.08 mA (b) 0.08 mA (c) 1 mA 5.5 V 5.1 k (d) 1 A 88. Generally, the base of a transistor has a thickness of the order of (a) 10–6 m (b) 10–3 m (c) 0.1 mm (d) 1 cm 89. The circuit diagram shown performs the logic funtion of A (a) XOR gate Y B (b) AND gate (c) NAND gate (d) OR gate

90. The equation of an FM signal is e = 6 sin(106t + 4 sin103t). Then, the modulating frequency is (a) 1000 Hz (b) 500 Hz (c) 300 Hz (d) 159 Hz 91. The dimensions of solar constant are (a) [MLT–2] (b) [M0L0T0] (c) [ML0T–3] (d) [M0LT–3] 92. If x = (a – b), the maximum percentage error in the measurement of x will be ∆b   ∆a − ×100 (a)   a − b a − b  ∆b   ∆a + ×100 (b)   a − b a − b   ∆a ∆b  (c)  +  ×100  a b   ∆a ∆b  (d)  −  ×100  a b  93. A body of mass m kg is rotating in a vertical circle at the end of a string of length r metre. The difference in the kinetic energy at the top and bottom of the circle is 1 (a) mgr (b) mgr (c) 2mgr (d) 4mgr 2 94. A long spring is stretched by 2 cm, its potential energy is U. If the spring is stretched by 10 cm, the potential energy stored in it will be U U (a) (b) (c) 5U (d) 25U 25 5 95. Two balls of different masses ma and mb are dropped from two different heights a and b. The ratio of times taken by the two to drop through these distances is (a) a : b (b) b : a (c) a : b (d) a2 : b2 96. A particle moves along x-axis obeying the equation x = t(t – 1)(t – 2), where x (in metres) is the position of the particle at any time t (in seconds). The displacement when the velocity of the particle is zero, is 5 5 2 2 m, m m, m (b) − (a) − 3 3 3 3 3 3 3 3 (c) –3 m, 3 m (d) –5 m, 5 m 97. A projectile is given an initial velocity of i + 2 j . The cartesian equation of its path is (g = 10 m/s2) (a) y = 2x – 5x2 (b) y = x – 5x2 2 (c) 4y = 2x – 5x (d) y = 2x – 25x2

98. A heavy uniform chain lies on a horizontal table. If the coefficient of friction between the chain and the table is 0.25, then the maximum fraction of the length of the chain that can hang over one edge of the table is (a) 20% (b) 25% (c) 30% (d) 40% 99. A body of mass M is resting on a rough horizontal plane surface, the coefficient of friction being equal to m. At t = 0, a horizontal force F = F0t starts acting on it, where F0 is a constant. The time T at which the motion starts is (a)

mMg F0

(b)

mF Mg

(d)

Mg mF0

F F0 100. Force acting on a particle moving in a straight line K varies with the velocity of the particle v as F = . v where K is a constant. The work done by this force in time t is 2Kt K (a) 2 t (b) 2 v v (c)

(c) Kt 1. 6. 11. 16. 21. 26. 31. 36. 41. 46. 51. 56. 61. 66. 71. 76. 81. 86. 91. 96.

(b) (b) (b) (d) (b) (c) (d) (a) (a) (b) (b) (a) (b) (d) (c) (c) (b) (d) (c) (a)

(d) 2Kt 2. 7. 12. 17. 22. 27. 32. 37. 42. 47. 52. 57. 62. 67. 72. 77. 82. 87. 92. 97.

(d) (d) (c) (c) (a) (a) (b) (a) (a) (b) (a) (c) (d) (c) (a) (b) (c) (c) (b) (a)

Answer Keys 3. (d) 4. 8. (c) 9. 13. (b) 14. 18. (d) 19. 23. (c) 24. 28. (c) 29. 33. (d) 34. 38. (c) 39. 43. (c) 44. 48. (c) 49. 53. (c) 54. 58. (b) 59. 63. (a) 64. 68. (d) 69. 73. (d) 74. 78. (d) 79. 83. (b) 84. 88. (a) 89. 93. (c) 94. 98. (a) 99.

(a) (b) (d) (c) (b) (d) (d) (d) (d) (a) (c) (b) (d) (a) (c) (c) (b) (b) (d) (a)

5. 10. 15. 20. 25. 30. 35. 40. 45. 50. 55. 60. 65. 70. 75. 80. 85. 90. 95. 100.

(c) (a) (a) (c) (b) (a) (a) (b) (c) (c) (d) (c) (d) (b) (a) (a) (a) (d) (c) (c)

nn Physics for you | OctOber ‘15

65

chapterwise McQ’s for practice

Useful for All National and State Level Medical/Engg. Entrance Exams vg

Gravitation

1. Three identical bodies of mass M are located at the vertices of an equilateral triangle of side L. They revolve under the effect of mutual gravitational force in a circular orbit, circumscribing the triangle while preserving the equilateral triangle. Their orbital velocity is 3GM (a) GM (b) 2L L 2GM 3GM (d) 3L L 2. An earth satellite is moved from one stable circular orbit to a farther stable circular orbit. Which one of the following quantities increase? (a) Linear orbital speed (b) Gravitational force (c) Centripetal acceleration (d) Gravitational potential energy (c)

3. Two satellites S1 and S2 are revolving around a planet in coplanar and concentric circular orbits of radii R1 and R2 in the same direction respectively. Their respective periods of revolution are 1 h and 8 h. If R1 = 104 km, their relative speed when they are closest, in km h–1 is p (a) × 104 (b) p × 104 2 (c) 2 p × 104 (d) 4 p × 104 4. Select the proper graph between the gravitational potential (Vg) due to hollow sphere and distance (r) from its centre. vg

vg

(a) r

66

r

(b) vg

Physics for you | OctOber ‘15

R

(c)

vg

r

R

(d)

r

–vg

5. The mass of the moon is (1/8) of the earth but the gravitational pull is (1/6) of the earth. It is due to the fact that (a) moon is the satellite of the earth. (b) the radius of the earth is (8/6) of the moon. (c) the radius of the earth is ( 8 / 6 ) of the moon. (d) the radius of the moon is (6/8) of the earth. 6. The orbit of geostationary satellite is circular, the time period of satellite depends on which of the following factors? (i) mass of the satellite (ii) mass of the earth (iii) radius of the orbit (iv) height of the satellite from the surface of the earth (a) (i) only (b) (i) and (ii) (c) (i), (ii) and (iii) (d) (ii), (iii) and (iv) 7. R and r are the radii of the earth and moon respectively, re and rm are densities of earth and moon respectively. The ratio of the acceleration due to gravity on the surfaces of the earth and moon is r re R re (a) r r (b) R r m m R rm r rm (c) (d) r r R re e 8. Three uniform spheres of mass M and radius R each are kept in such a way that each touches the other two. The magnitude of gravitational force on any of the spheres due to other two is 3 GM 2 3 GM 2 (b) (a) 2 R2 2 R2 (c)

3GM 2 R2

(d)

3 GM 2 4 R2

9. A geostationary satellite is orbiting the earth at a height of 5R above the surface of the earth, R being the radius of the earth. The time period of another satellite in hours at a height of 2R from the surface of the earth is 6 (a) 5 (b) 10 (c) 6 2 (d) 2 10. A satellite is orbiting around the earth with kinetic energy K. What will happen if the satellite’s kinetic energy is made 2K? (a) Radius of the orbit is doubled (b) Radius of the orbit is halved (c) Period of revolution is doubled (d) Satellite escapes away 11. The height at which the weight of a body becomes  1  th  16  , its weight on the surface of earth is (R is the radius of the earth) (a) 5R (b) 15R (c) 3R (d) 4R 12. A satellite is moving in a circular orbit at a certain height above the earth’s surface. It takes 5.26 × 103 s to complete one revolution with a centripetal acceleration equal to 9.32 m s–2. The height of the satellite orbit above the earth’s surface is (Radius of earth = 6.37 × 106 m) (a) 70 km (b) 160 km (c) 190 km (d) 220 km 13. A synchronous satellite goes around the earth once in every 24 h. What is the radius of orbit of the synchronous satellite in terms of the earth’s radius? (Given mass of the earth, Me = 5.98 × 1024 kg, radius of the earth, Re = 6.37 × 106 m, Universal constant of gravitation, G = 6.67 × 10–11 N m2 kg–2). (a) 2.4 Re (b) 3.6 Re (c) 4.8 Re (d) 6.6 Re 14. Two satellites of earth, S1 and S2 are moving in the same orbit. The mass of S1 is four times the mass of S2. Which one of the following statements is true? (a) The potential energies of earth and satellite in the two cases are equal. (b) S1 and S2 are moving with the same speed. (c) The kinetic energies of the two satellites are equal. (d) The time period of S1 is four times that of S2. 15. The potential energy of gravitational interaction of a point mass m and a thin uniform rod of mass M and length l, if they are located along a straight line at distance a from each other is GMm  a + l  ln  (a) U =  a  l

a + l  (c) U = − GMm ln   a  l GMm (d) U = − a mechanical ProPerties of solids

16. Two rods of different materials having coefficients of thermal expansion a1 and a2 and Young’s modulli Y1 and Y2 respectively are fixed between two rigid massive walls. The rods are heated such that they undergo the same increase in temperature. There is no bending of the rods. If a1 : a2 = 2 : 3 the thermal stresses developed in the two rods are equal provided Y1 : Y2 is equal to (a) 2 : 3 (b) 1 : 1 (c) 3 : 2 (d) 4 : 9 17. When a wire of length 10 m is subjected to a force of 100 N along its length, the lateral strain produced is 0.01 × 10–3. The Poisson’s ratio was found to be 0.4. If the area of cross-section of wire is 0.025 m2, its Young’s modulus is (a) 1.6 × 108 N m–2 (b) 2.5 × 1010 N m–2 (c) 1.25 × 1011 N m–2 (d) 16 × 109 N m–2 18. Two wires of the same material and length but diameter in the ratio 1 : 2 are stretched by the same force. The ratio of potential energy per unit volume for the two wires when stretched will be (a) 1 : 1 (b) 2 : 1 (c) 4 : 1 (d) 16 : 1 19. A wooden wheel of radius R is made of two semicircular parts (see figure). R The two parts are held together by a ring made of a metal strip of cross sectional area A and length L. L is slightly less than 2pR. To fit the ring on the wheel, it is heated so that its temperature rises by Dq and it just steps over the wheel. As it cools down to surrounding temperature, it presses the semicircular parts together. If the coefficient of linear expansion of the metal is a, and its Young’s modulus is Y, the force that one part of the wheel applies on the other part is (a) 2pAYaDq (b) AYaDq (c) pAYaDq (d) 2AYaDq 20. A rod of length L and negligible mass is suspended at its two ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in the figure. The cross-sectional areas of wires A and B are 1 mm2 and 2 mm2 respectively. (YAl = 70 × 109 N m–2 and Ysteel = 200 × 109 N m–2)

1  1 (b) U = GMm  −  a a + l  Physics for you | OctOber ‘15

67

To have equal stress in both the wires, mass m should be suspended at a distance of 1 (a) L from the wire A 3 1 (b) L from the wire B 2 2 (c) L from the wire B 3 2 (d) L from the wire A 3 21. A wire is suspended from the ceiling and stretched under the action of a weight F suspended from its other end. The force exerted by the ceiling on it is equal and opposite to the weight. (a) Tensile stress at any cross section A of the wire is F/A. (b) Tensile stress at any cross section is zero. (c) Tensile stress at any cross section A of the wire is 2F/A. (d) Tension at any cross section A of the wire is 2F. 22. A steel rod of length 1 m and area of cross-section 1 cm2 is heated from 0°C to 200°C, without being allowed to extend or bend. The tension produced in the rod is (Given : Young’s modulus of steel = 2 × 1011 N m–2 and coefficient of linear expansion of steel = 10–5 °C–1) (a) 4 × 103 N (b) 4 × 104 N (c) 4 × 105 N (d) 4 × 106 N 23. A mild steel wire of length 2L and cross-sectional area A is stretched, well within elastic limit, horizontally between two pillars as shown in the figure. A mass M is suspended from the midpoint of the wire. Strain in the wire is x x2 x (a) (b) (c) 2 (d) x 2 L L 2L 2L 24. A spring is stretched by applying a load to its free end. The strain produced in the spring is (a) volumetric (b) shear (c) longitudinal and shear (d) longitudinal. 25. A thick rope of rubber of density 1.5 × 103 kg m–3 and Young’s modulus 5 × 106 N m–2, 8 m in length is hung from the ceiling of a room. The increase in its length due to its own weight is (Take g = 10 m s–2) (a) 9.6 × 10–2 m (b) 9.6 × 10–5 m –7 (c) 9.6 × 10 m (d) 9.6 m 26. The radii and Young’s moduli of two uniform wires A and B are in the ratio 2 : 1 and 1 : 2 respectively. Both wires are subjected to the same longitudinal force. If the increase in length of the wire A is one percent, the percentage increase in length of the 68

Physics for you | OctOber ‘15

wire B is (a) 1.0

(b) 1.5

(c) 2.0

(d) 3.0

27. If the ratio of lengths, radii and Steel Young’s modulii of steel and m brass wires in the figure are a, b, c respectively. Then the corresponding Brass ratio of increase in their lengths 2m would be 2 2ac 3a 3c (a) (b) (c) (d) 2a c 2 2 b b 2b c 2ab2 28. If the shear modulus of a wire material is 5.9 ×1011 dyne cm–2 then the potential energy of a wire of 4 × 10–3 cm in diameter and 5 cm long twisted through an angle of 10′, is (a) 1.253 × 10–12 J (b) 2 × 10–12 J –12 (c) 1.00 × 10 J (d) 0.8 × 10–12 J 29. The Poisson’s ratio of the material is 0.5. If a force is applied to a wire of this material, there is a decrease in the cross-sectional area by 4%. The percentage increase in its length is (a) 1% (b) 2% (c) 2.5% (d) 4% 30. The stress-strain graph for a metal wire is as shown in the figure. In the graph, the region in which Hooke’s law is obeyed, the ultimate strength and fracture points are represented by (a) OA, C, D (b) OB, D, E (c) OA, D, E (d) OB, C, D solutions

1. (a) :

C

v

L

O

F2 M A

M

F 30° v

F1

L v

N L

B M

From figure, O be the centre of equilateral triangle ABC. AN L/2 L = = . Let OA = r = cos 30° 3 /2 3 The gravitational force on mass M at A due to mass at B is, GMM F1 = along AB. L2 The gravitational force on mass M at A due to mass at C is, GMM along AC. F2 = L2

Here, F1 = F2 =

GM 2

L2 These forces F1 and F2 are inclined at angle 60°, their resultant F is along AO, which is given by F = [F12 + F22 + 2F1 F2 cosq]1/2 1/2

GM 2  1 GM 2 3GM 2 = 2 1 + 1 + 2 ×  = 2 3= 2 L  L L2 As this force is providing the required centripetal force, so Mv 2 3GM 2 = r L2 GM On solving, we get v = L GM 2. (d) : Linear orbital speed, v = r GM m Gravitational force on satellite, F = r2 v2 Centripetal acceleration, ac = r GM m Gravitational PE, U = − r As r increases gravitational potential energy becomes less negative, i.e., gravitational potential energy increases, whereas speed, force and acceleration will decrease. 3. (b) : Here,

T22 T12

=

R23 R13

2/3

2/3

gM = gE

2 RM GM

=

RE2

2 8 RM

GM E 6. (d) : Orbital velocity, v0 = R + h E 2 p(RE + h) 2 p(RE + h)3/2 Time period, T = = v0 (GM E )1/2 Thus, the time period of satellite is independent of mass of satellite but depends on mass of the earth, radius of the orbit (RE + h), height of the satellite from the surface of the earth. 7. (a) : Acceleration due to gravity on the surface of the earth is GM ge = 2 R where M and R be mass and radius of the earth Acceleration due to gravity on the surface of the moon is Gm gm = 2 r where m and r be mass and radius of the moon. M 4  re = or M = pR3re 4 3 3 pR 3 m 4 and rm = or m = pr 3rm 4 3 3 pr 3 \

T  8 or R2 = R1  2  = 104   = 4 × 104 km 1  T1  Velocity of satellite S1, 2 pR1 2 p × 104 v1 = = = 2 p × 104 km h −1 T1 1 Velocity of satellite S2, 2 pR2 2 p × 4 × 104 v2 = = = p × 104 km h −1 T2 8 Relative velocity = v1 – v2 = 2p × 104 – p × 104 = p × 104 km h–1 4. (c) G( M E / 8) 5. (c) : Here,

From (i) and (ii), we get RE2 1 8 R = or RE = 2 6 M 8 RM 6

2 3 2 g e  M   r   R   re   r   R   re  =    =    = g m  m   R   r   rm   R   r   rm 

8. (d) : The situation is as shown in the figure. A 2R B

…(i)

RE2 where the subscripts M and E are for the moon and the earth respectively. g 1 …(ii) Given, M = gE 6

2R FCA 60° 2R FCB C

Gravitational force on sphere C due to sphere A is GM × M GM 2 FCA = = along CA (2R)2 4 R2 Gravitational force on sphere C due to sphere B is GM × M GM 2 FCB = = along CB (2R)2 4 R2 These two forces are equal in magnitude and inclined at an angle 60°. \ The total gravitational force on sphere C due to other two spheres is 2 2 Ftotal = FCA + FCB + 2FCA FCB cos 60° Physics for you | OctOber ‘15

69

2

2

 GM 2   GM 2   GM 2   GM 2   1  2 =  + +         4 R2   4 R2   4 R2   4 R2   2  =

3GM 2 4 R2

T2  r2  = T1  r1 

3/2

 R + 2R  =  R + 5R 

3/2

3 =  6

3/2

=

1

23/2

T2 1 24 24 = 3/2 or T2 = 3/2 = = 6 2 hours 24 2 2 2 2 10. (d) : Kinetic energy of the satellite orbiting the 1 earth is K = mvo2 , where vo is the orbital velocity. 2 If v′o is the new velocity when kinetic energy of the satellite becomes double, then 1 1 1 mvo′2 = 2K = 2 × mvo2 = m( 2vo )2 2 2 2 \ v o′ = 2 vo It is the escape velocity of the satellite from the earth. Therefore the satellite will escape away when its kinetic energy is made 2K. 11. (c) : Acceleration due to gravity at a height h from the surface of earth is g gh = 2 ....(i)  h 1 +  R where g is the acceleration due to gravity at the surface of earth and R is the radius of earth. Multiplying by m (mass of the body) on both sides in (i), we get mg mg h = 2  h 1 +  R \ Weight of a body at height h, Wh = mgh Weight of the body at the surface of earth, W = mg 1 According to question,Wh = W 16 1 1 \ = 16  h  2 1 +  R 2

h  h or 1 +  = 16 or 1 + = 4  R R h or = 3 or h = 3R R 70

(R + h)3 GM

(R + h)3 T 2 = 2 GM 4p Centripetal acceleration, a =

...(i)

or

9. (c) : According to Kepler’s third law T 2 ∝ r3 \

12. (b) : As, T = 2 p

Physics for you | OctOber ‘15

(R + h)2 1 = GM a

or

GM

(R + h)2 ...(ii)

Divide (i) by (ii), we get

2

 5.26 × 103  ( R + h) = 2 × a =   × 9.32  2p 4p R + h = 6.53 × 106 m h = 6.53 × 106 m – 6.37 × 106 m = 0.16 × 106 m = 160 × 103 m = 160 km 13. (d) : Time period of satellite T2

T = 2p Also, g = ⇒

gRe2 = T2 =

R3 GMe

GMe

Re2 GMe

...(i)

2 3

4p R

(Using (i)) gRe2 Substituting the given values in above equation, we get 4 × (3.14)2 R3 (24 × 60 × 60)2 = 9.8Re2 or R = 6.6Re 14. (b) : Both, orbital speed of satellite (vo = GMe / r ) and time period of revolution of satellite, 1/2  4 p2 r 3  T=  are independent of mass of satellite.  GMe  Therefore orbital speed and time period of revolution of both the satellites are same. GMe m The kinetic energy of a satellite, K = and 2r GMe m potential energy of a satellite, U = − both r depend on mass of satellite. \

15. (c) :

l dx

a x

m

M l M Mass of element of length dx, dm = dx l The gravitational potential energy between this element and point mass is Mass per unit length of rod =

Gmdm dU = − =− x \ U =−

GmM l

a+l

∫ a

M  Gm  dx   l  x

dx x

GmM  a + l  or U = − ln   a  l F/A 16. (c) : As, Y = since the rods are fixed between Dl / l two rigid massive walls, their lengths are equal. 1 So Y ∝ Dl a lDT a2 3 Y Dl Hence, 1 = 2 = 2 = = Y2 Dl1 a1lDT a1 2 Lateral strain 17. (a) : Poisson’s ratio = Longitudinal strain Lateral strain Longitudinal strain = Poisson’s ratio 0.01 × 10−3 = ...(i) 0.4 Normal stress Young’s modulus, Y = Longitudinal strain F or Y = (Using (i))  0.01 × 10−3  A×  0.4   100 × 0.4 = = 1.6 × 108 N m −2 0.025 × 0.01 × 10−3 18. (d) : As the two wires are of the same material, therefore their Young moduli are same. DL F L F L or 1 = 4 \ Y= 2× = × 2 DL1 p(2r ) DL2 DL2 pr Potential energy per unit volume of the wire is 1 u = × stress × strain 2 DL 1 F DL 1 F × 2 \ u1 = × 1 and u2 = 2 2 2 p(2r ) L 2 pr L \

2

DL u1  2r  =   × 1 = 4 × 4 = 16 u2  r  DL2

19. (d) : If Dq is the increase in temperature, then increase in length L of metal ring is DL DL = aLDq or = aDq L Let T be the tension developed in the ring, then T

T L YADL × or T = = YAaDq A DL L Refer to figure shown here, if F is the force that one part of the wheel applies on the other part, then F = 2T = 2AYaDq Y=

20. (d) :

Let mass m be suspended at a distance x from the wire A. Let TA and TB be tensions in the wire A (steel) and wire B (aluminium) respectively. T \ Stress in wire A = A AA TB AB For equal stress in both the wires, TA T = B AA AB Stress in wire B =

2 TA AA 1 mm 1 ...(i) = = = 2 2 TB AB 2 mm As the system is in equilibrium, taking moments about C, we get TAx = TB (L – x) L − x TA = x TB L− x 1 (Using (i)) = x 2 21. (a) : When a wire is suspended from the ceiling and stretched under the action of a weight (F) suspended from its other end, the force exerted by the ceiling on it is equal and opposite to the weight. However, the tension at any cross-section A of the wire is just F and not 2F. Hence, tensile stress which is equal to the tension per unit area is equal to F/A. 22. (b) : Let DL be increase in the length of the rod due to increase in temperature of the rod. Then DL = LaDT where, a is the coefficient of the linear expansion, DT is the rise in temperature and L is the length of the rod. DL \ = a DT ...(i) L Physics for you | OctOber ‘15

71

Let the compressive tension of the rod be T and A be cross-section area. Then T/A Y= DL / L DL \ T =Y A = Y × a DT × A (Using (i)) L 11 –2 Here, Y = 2 × 10 N m a = 10–5°C–1 DT = 200°C – 0°C = 200°C L = 1 m, A = 1 cm2 = 1 × 10–4 m2 \ T = 2 × 1011 N m–2 × 10–5°C–1 × 200°C × 1 × 10–4 m2 = 4 × 104 N 23. (a) :

Increase in length DL = (PR + RQ) – PQ = 2PR – PQ 1/2  x2  2 2 1/2 DL = 2 (L + x ) − 2 L = 2 L 1 + 2  − 2L L    1 x2  = 2 L 1 +  − 2 L (Using binomial theorem) 2 L2   x2 = L

DL x2 \ Strain = = 2 2L 2L 24. (c)

L (LArg ) Fl 2 25. (a) : Using, Dl = = YA YA weight of wire acts at its centre of mass, l i.e., l = 2 L2 Arg L2rg \ Dl = = 2YA 2Y 2 8 × 1.5 × 103 × 10 = = 9.6 × 10−2 m 6 2 × 5 × 10 26. (c) : Young’s modulus, Y = DL F or = 2 L pr Y For the same force, \

72

DLB  DL  = 2 A  LB  LA  DLB   DL × 100 = 2  A × 100  = 2% LB   LA Fl ADl Fl DlS FS lS AB YB or Dl = ; = × × × AY DlB FB lB AS YS

27. (b) : Young’s modulus, Y =

or

DlS  3 mg  1 1 3a = ×a× 2 × = 2  DlB  2 mg  b c 2b c

28. (a) : To twist wire through the angle dq, it is necessary to do the work dW = tdq 10 p p and q = 10′ = × = rad 60 180 1080 q

W = ∫ t dq = ∫ 0

W=

q

0

ηpr 4 q dq ηpr 4 q2 = 2l 4l

5.9 × 10 × 10−5 × p (2 × 10−5 )4 p2 11

10−4 × 4 × 5 × 10−2 × (1080)2

W = 1.253 × 10–12 J 29. (d) : Given, Poisson’s ratio = 0.5. It shows that the density of material is constant. Therefore, the change in volume of the wire is zero. Thus V = A × l = a constant logV = logA + logl Dl DA =− l A Dl × 100 = −(−4) = 4% or % increase in length = l 30. (c) : In the region OA, the graph is linear showing that stress is proportional to the strain. Thus, in this region Hooke’s law is obeyed. \

DV DA Dl = + = 0 or V A l

FL

pr 2 DL

DLB 2 2 LB  r  Y   2   1  = A   A =   =2 DLA  rB   YB   1   2  LA Physics for you | OctOber ‘15

The point D on the graph is known as ultimate tensile strength. The point E on the graph is known as fracture point. nn

Y U ASK

WE ANSWER

Do you have a question that you just can’t get answered? Use the vast expertise of our mtg team to get to the bottom of the question. From the serious to the silly, the controversial to the trivial, the team will tackle the questions, easy and tough. The best questions and their solutions will be printed in this column each month.

Q1. When comets enter the Earth’s atmosphere, they burn up. But when space shuttles re-enter they do not. Why? – Neetu Verma (UP) Ans. Meteorites and comets do not generally land unharmed at the surface of the Earth. In spite of their random and uncontrolled entry, many a time, pieces of meteorites do escape complete annihilation during their trip through the atmosphere. The space shuttle, on the other hand, is brought into the atmosphere along a controlled and precisely calculated glide path to reduce heating due to friction. In addition, ablative coatings that dissipate heat through high temperature sublimation protect the critical surfaces of the shuttle. These coatings might be made of a ceramic or composite material. Q2. Why does water condense on the inside surface of the car windshield during the monsoon?

Ans. The determination could be made by noticing the order of the colors in the spectrum relative to the direction of the original beam of white light. For a prism, in which the separation of light is a result of dispersion, the violet light will be refracted more than the red light, so the order of the spectrum will be from red, closest to the original direction, to violet. For a diffraction grating, the angle of diffraction increases with wavelength. Thus, the spectrum from the diffraction grating will have colors in the order of violet, closest to the original direction, to red. Q4. Suction cups can be used to hold objects onto surfaces. Why don’t astronauts use suction cups to hold onto the outside surface of the space shuttle? – Arjun Sharma (Haryana)

Ans. A suction cup works because air is pushed out from under the cup when it is pressed against a surface. When the cup is released, it tends to spring back a bit, causing the trapped air under the cup to expand. This expansion causes a reduced pressure inside the cup. Thus, the pressure difference between the atmospheric pressure on the outside of the cup against the surface. For astronaut in orbit around the Earth, almost no air exists outside the surface of the spacecraft. Thus, if a suction cup were to be pressed against the pressure outside surface of the spacecraft, the pressure difference needed to press the cup to the surface is not present. So suction cups do not get stick to the surface of space shuttle.  solution of sePtember 2015 crossword

– Arun Nair (Kerala)

Ans. The inside of a car whose windows are rolled up can get quite humid because of the moisture in the exhaled breath of passengers. Also, the temperature inside the car is higher than the outside and, sometimes, because of heat leaking into the cabin from the engine compartment. The rainwater cools the windshield and the windows of the car. On touching these glass surfaces, the inside air cools below the dew point and condenses, which often blocks the driver’s view. In modern cars, the windshield is often equipped with embedded elements (at the rear) and hot air blower (at the front). If your car has neither, try opening the windows just a little, to increase the exchange of air, this should help equalise the temperature and/or moisture content and should clear the condensation. Q3. White light enters through an opening in an opaque box, exits through an opening on the other side of the box, and a spectrum of colors appears on the wall. How would one determine whether the box contains a prism or a diffraction grating? – Rashmi Agarwal (Delhi)

 

  

WinnErS (September 2015) Ajay Chauhan (Bihar) Dibyakanti Kumar (WB) Solution Senders (August 2015) Sumit Sharma (Delhi) Sahil Gupta (WB) neha Gupta (UP) Physics for you | OctOber ‘15

73

a



= 2 p  ∫ Ax 3dx + ∫ Bx 4 dx 

0

Solution Set-26

1. (c): Heat current, Q = Q1 + Q2

λ × 2 A(T1 − T2 ) λ1A(T1 − T2 ) λ2 A(T1 − T2 ) = + d d d λ1 + λ2 ⇒ λ= 2 2 (a) : Thermal expansion of Cu = lacT F l Elastic contraction of Cu = A Yc F l Net expansion of Cu = lacT − A Yc F Either steel plate is subjected to the tensile force 2 from the sides of the Cu plate. F l Net expansion of steel = la sT + × 2 A Ys The plus sign is on account of the fact that there is elastic expansion, of steel plates. Since the plates suffer same net expansion F l F l \ lacT − = la sT + A Yc 2 A Ys 2 AYcYs (ac − a s )T Solving, we get F = 2Ys + Yc 9M 3. (b) : Mass per unit area of disc = p R2 Mass of removed portion of disc 2

9M × p R = M 2  3 pR Moment of inertia of removed portion about an axis passing through centre of disc and 3 R/ perpendicular to the plane of disc, 2R/3 using theorem of parallel axis is O R 2 2 M  R 1 I1 = + M  2R  = MR 2  3 2  3 2 When portion of disc would not have been removed, then the moment of inertia of 9 complete disc about the given axis is I 2 = MR 2 2 So moment of inertia of the disc with removed portion, about the given axis is 9 1 I = I2 – I1 = MR 2 − MR 2 = 4MR 2 2 2 4. (c) : Consider an elemental ring of thickness dx at a distance x from its centre. Area of this ring is 2pxdx Moment of inertia of ring, dI = (A + Bx)2pxdx x2 \ Moment of inertia of disc, =

a

a

0

0

I = ∫ dI = ∫ ( A + Bx )2 px 3dx 74

Physics for you | OctOber ‘15

  Aa 4 Ba5  On solving, we get, I = 2 p  +  5   4 5. (b) : Frequency of fundamental note in an open v organ pipe u = 2L Frequency just audible to a person of normal hearing = 20 Hz 340 340 ;L= ( v = 340 m s −1 ) = 8.5 m \ 20 = 2L 40 1 1 1 6. (b) : As u ∝ , u ∝ , u ∝ L D ρ \

2L 2D uA LB DB ρ ρ = ⋅ B = A ⋅ A ⋅ =4 uB LA DA ρA LA DA ρ

So, we get uA = 4uB

log(umax / umin ) log 2 Here, umax = 20000 Hz and umin = 20 Hz log(20000 / 20) N= = 10 log 2 8. (b) : Frequency of nth harmonic or (n – 1)th overtone nv must be just equal to or less than 20000 Hz 2L 7. (c): Number of octaves =

20000 × 2 × L 20000 × 2 × 0.15 nv = = 18 = 20000 \ n = v 330 2L

i.e., 17th overtone dq V dq dR V = ⇒ ⋅ = dt R dR dt R dq  5  V ⇒ × = dR  60  R 40 dR dR dq = 12V ⇒ q = 12V ∫ = 12V ln2 = 120 ln2 C R R 20 NE 10. (b) : Here, I = R + Nr Let n cells are reversed N I (N − 2n)E N ⇒ n= = ⇒ 3= 3 3 R + Nr (N − 2n) 9. (b) : I =

nn Solution Senders of Physics Musing sET-26 1. Sayan Rakshit 2. Arnab Pauli (WB) 3. Pratim Majumdar (Gujarat) 4. Debdutta Bose sET-25 1. Abinav Mittal (Haryana) 2. Sakshi Saxena (Delhi)

Series 5 CHAPTERWISE PRACTICE PAPER : Dual Nature of Radiation and Matter | Atoms and Nuclei Time Allowed : 3 hours

Maximum Marks : 70 GENERAL INSTRUCTIONS

(i) All questions are compulsory. There are 26 questions in all. (ii) This question paper has five sections: Section A, Section B, Section C, Section D and Section E. (iii) Section A contains five questions of one mark each, Section B contains five questions of two marks each, Section C contains twelve questions of three marks each, Section D contains one value based question of four marks and Section E contains three questions of five marks each. (iv) There is no overall choice. However, an internal choice has been provided in one question of two marks, one question of three marks and all the three questions of five marks weightage. You have to attempt only one of the choices in such questions.

section-A

1. It is harder to remove a free electron from Cu than from Na. Which metal has greater work function? Which has higher threshold frequency ? 2. A radioactive isotope of silver has half-life of 20 minutes. What fraction of the original mass would remain after one hour? 3. Write a typical nuclear reaction in which a large amount of energy is released in the process of nuclear fission. 4. The energy of an electron in the nth orbit of H-atom is given by En = –13.6/n2 eV. Find the energy required to take an electron from ground state to the second excited state. 5. Define atomic mass unit. Write its energy equivalent. section-b

6. Explain with an example, whether the neutron to proton ratio increases or decreases during (i) alpha decay and (ii) beta decay.

7. The value of ground state energy of hydrogen atom is –13.6 eV. (i) What does the negative sign signify? (ii) How much energy is required to take an electron in this atom from the ground state to the first excited state? 8. An electromagnetic wave of wavelength l is incident on a photosensitive surface of negligible work function. If the photoelectrons emitted from this surface have the de Broglie wavelength l1, prove  2mc  2 that l =   l1 .  h  9. Which level of the doubly ionized lithium (Li++) has the same energy as the ground state energy of the hydrogen atom? Compare the orbital radii of the two levels. 10. Red light however it is bright, cannot produce the emission of electrons from a clean zinc surface, but even weak ultraviolet radiation can do so; why? OR If 200 MeV energy is released in the fission of a single nucleus 235 92U, how many fissions must occur per second to produce a power of 1 kW? Physics for you | OctOber ‘15

75

section-c

11. Show that the speed of an electron in the innermost orbit of H-atom is 1/137 times the speed of light in vacuum. 12. The deuteron is bound by nuclear forces just as H-atom is made up of p and e bound by electrostatic forces. If we consider the force between neutron and proton in deuteron as given in the form of a Coulomb potential but with an effective charge e′: 1 e′2 F= 4 pε0 r Estimate the value of (e′/e) given that the binding energy of a deuteron is 2.2 MeV. 13. (a) A monoenergetic electron beam with electron speed of 5.20 × 106 m s–1 is subjected to a magnetic field of 1.30 × 10–4 T, normal to the beam velocity. What is the radius of the circle traced by the beam, given e/m for electron equals 1.76 × 1011 C kg–1. (b) Is the formula you employ in (a) valid for calculating radius of the path of a 20 MeV electron beam? If not, in what way is it modified? 14. Find the height of potential barrier for a head-on collision of two deuterons. The effective radius of deuteron can be taken to be 2 fm. Note that height of potential barrier is given by the Coulomb repulsion between two deuterons when they just touch each other. 15. (a) For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.40 × 10–10m? (b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter having an 3 average kinetic energy of kT at 300 K. 2 16. An electron and a proton are accelerated through the same potential. Which one of the two has (i) greater value of de Broglie wavelength associated with it and (ii) less momentum? Justify your answer. OR (i) What characteristic property of nuclear force explains the constancy of binding energy per nucleon (BE/A) in the range of mass number A lying 30 < A < 170? (ii) Show that the density of nucleus over a wide range of nuclei is constant independent of mass number A. 76

Physics for you | OctOber ‘15

17. A source of light of frequency u > u0 is placed at 2 m from the cathode of a photocell. The stopping potential is found to be V0. If the distance of the light source is halved, state with reason what changes occur in (i) stopping potential (ii) photoelectric current, and (iii) maximum velocity of photoelectrons emitted. 18. Why are de Broglie waves associated with a moving football not visible? The wavelength, l, of a photon and the de Broglie wavelength of an electron have the same value. Show that the energy of the photon 2lmc times the kinetic energy of the electron, is h where m, c and h have their usual meanings. 19. Write Einstein’s photoelectric equation. State clearly how this equation is obtained using the photon picture of electromagnetic radiation. Write the three salient features observed in photoelectric effect which can be explained using this equation. 20. Deuteron is a bound state of a neutron and a proton with a binding energy B = 2.2 MeV. A g-ray of energy E is aimed at a deuteron nucleus to try to break it into a (neutron + proton) such that the n and p move in the direction of the incident g-ray. If E = B, show that this cannot happen. Hence, find how much bigger than B must E be for such a process to happen. 21. (a) Using Bohr’s second postulate of quantization of orbital angular momentum show that the circumference of the electron in the nth orbital state in hydrogen atom is n times the de Broglie wavelength associated with it. (b) The electron in hydrogen atom is initially in the third excited state. What is the maximum number of spectral lines which can be emitted when it finally moves to the ground state? 22. Find the energy equivalent of one atomic mass unit, first in joule and then in MeV. Using this, express the mass defect of 168O in MeV/c2. section-D

23. Mr. Raju a daily wages worker got affected by cancer. On knowing about it all his coworkers started avoiding him, fearing that it was contagious. Mr. Raju felt very depressed. Mr. Rahul a close friend immediately took Mr. Raju to radiologist who examined him and said it was the begining stage of cancer and it can be easily cured and he also

certified that it is not a comunicable disease. (i) What moral values did Mr. Rahul exhibit? (ii) A radioactive substance X has a half life of 140 days. Initially it is 8 g. Find the time for this substance X when it is reduced to 1 g. section-e

24. State the basic postulates of Bohr’s theory of atomic spectra. Hence obtain an expression for radius of orbit and the energy of orbital electron in hydrogen atom. OR What do you mean by binding energy? Draw the graph to show variation of binding energy per nucleon with mass number. Explain the graph. 25. Describe Davisson and Germer experiment to establish the wave nature of electrons. Draw a labelled diagram of the apparatus used. OR (a) Describe in detail alpha particle scattering experiment along with diagrams? (b) How can you estimate the size of nuclei? 26. (i) Compare the properties of a-particles b-particles and g-particles. (ii) Distinguish between nuclear fission and fusion. In a fusion reaction 2 3 4 1 H + 1 H → 2 He + n Find the amount of energy (in MeV) released.

( ) = 2.014102 u m ( ) = 3.016049 u m ( 24 H ) = 4.002603 u 2

(Given m 3 1H

2 1H

mn = 1.00867 u; 1u = 931.5 MeV/c .)

OR State and explain the laws of radioactive disintegration. Hence define disintegration constant and half life period. Establish relation between them. solutions

1. Since we need more energy to remove an electron from Cu than an electron from Na, Cu (4.65 eV) has greater work function (f0) than Na (2.75 eV). Since f0 = hu0, u0 = f0/h; u0 threshold frequency (u0) for Cu is more than that for Na. 2. Number of half-lives, n=

Total time of disintegration Half- life

=

60 min =3 20 min

3

N 1 1 =  = . N0  2  8

\ 3.

Nuclear fission reaction is 235 1 144 89 1 92 U + 0n → 56Ba + 36 Kr + 3(0 n) + 200 MeV

4. In ground state n = 1, so E1 = – 13.6 eV In second excited state n = 3, so E3 = −

13.6 32

= − 1.51 eV

= –1.51 eV Required energy is E = E3 – E1 = – 1.51 + 13.6 or E = 12.09 eV

5. 1 atomic mass unit is defined as 1/12th of the mass of the 126 C atom. 1 amu = 1.66 × 10–27 kg. In terms of energy, 1 amu = 931.5 MeV. 6. (i) 92U238 → 90Th234 + 2He4 Neutron to proton ratio before a-decay 238 − 92 146 = = 92 92 Neutron to proton ratio after a-decay 234 − 90 144 = = 90 90 As

144 146 > 90 92

Thus, the neutron to proton ratio increases in an a-decay. (ii)

83 Bi

210

 → 84 Po210 +

−1e

0

+u

Neutron to proton ratio before b-decay =

210 − 83 127 = 83 83

Neutron to proton ratio after b-decay 210 − 84 126 = = 84 84 126 127 As < 84 83 Thus, the neutron to proton ratio decreases in b-decay. 7. (i) The negative sign signifies that the electron is bound to the nucleus and the force is attractive. So, energy has to be supplied to remove the electron from the nucleus. (ii) Energy of hydrogen atom in nth state is 13.6 En = − 2 where, n = 1, 2, 3, …… n In the ground state, n = 1 Physics for you | OctOber ‘15

77

E1 = −

13.6

= −13.6 eV

12 In the first excited state, n = 2 13.6 E2 = − 2 = −3.4 eV 2 \ Required energy = E2 – E1 = (–3.4) – (–13.6) = 10.2 eV 8.

As the work function of metal can be neglected, so KE of the emitted electron = energy of X-ray photon, p2 hc 2mhc 1 2 = or p = mv = hu or 2m l l 2 h h hl = = p 2mc 2mhc / l hl l12 = 2mc  2mc  2 l=  l1  h 

Also, l1 = or or

9. As En = –13.6 Z2 n2

Z2 n2

, for the same energy

= constant, i.e.,

Z12 n12

=

Z22 n22

For hydrogen (in the ground state), Z1 = 1, n1 = 1. For Li++, Z2 = 3 12

Thus, or

12

=

(3)2 n22

n2 = 3

Further, as r = r∝

∈0 n2 h2 p mZe

n r1 n12 / Z1 , = Z r2 n22 / Z2 2

(\ 3rd level of Li++ will have the same energy)

2

 ∈ h2   0 2 = constant   pme 

2

r1 (1) / 1 1 r ⇒ 2 =3 = = r2 (3)2 / 3 3 r1 where the subscripts 1 and 2 stand for hydrogen and Li++ respectively. or

10. The photoemission of electrons does not depend on the intensity but it depends on the frequency and hence on the energy of photon of incident light. If the energy of photon is greater than the work function, the photoemission of electrons results however weak the incident radiation may 78

Physics for you | OctOber ‘15

be. The energy of photon of red light is less than the work function of zinc, so red light cannot emit photoelectrons. The energy of photon of ultraviolet light is greater than the work function of zinc, so ultraviolet light can emit photoelectrons. OR Energy released in the fission of a single nucleus of 235 92U = 200 MeV = 200 (1.6 × 10–13 J) = 3.2 × 10–11 J Energy required per second = Power × Time = 1 kW × 1 s = 1000 J If n is the number of fissions required per second to produce an energy of 1000 J, 1000J n= = 3.125 ×1013 3.2 × 10−11 J 11. Speed of an electron in the innermost (n = 1) orbit of H-atom is 2p ke 2 2p ke 2 .c v= = h ch =

2p × 9 × 109 × (1.6 × 10−19 )2

3 × 108 × 6.63 × 10−34 1 c= c 137 12. We know that binding energy of hydrogen atom in ground state, me 4 …(i) E = 2 2 = 13.6 e V 8 ∈0 h Replacing e by e′ and m by m′, reduced mass of neutron – proton, M × M M 1836m m′ = = = = 918 m M+M 2 2 M = mass of neutron/proton \ Binding energy, E ′ =

918 me , 4

8 ∈20 h2 Dividing (ii) by (i), we get 4

= 2.2 MeV …(ii)

 e ′  2.2 × 106 918   = 13.6 e e′ = (176.21)1/ 4 = 3.64 e 13. (a) Here, v = 5.20 × 106 m s–1, B = 1.30 × 10–4 T e Specific charge, = 1.76 × 1011 C kg −1 , θ = 90° m

The normal magnetic field provides necessary centripetal force to the electron beam so that it can follow a circular path. Thus Force on an electron = Centripetal force due to magnetic field on an electron mv v mv 2 or r = = or evB sin 90° = eB (e / m)B r 5.20 × 106 = m 1.76 × 1011 × 1.30 × 10−4 = 0.227 m = 22.7 cm. (b) The formula for radius of circular path is not valid at very high energies because such high energy electrons have velocities comparable to the speed of light. In such situation we use relativistic formula for mass of electron. m0 m= , where m0 is rest mass v2 1− 2 c m0  v  mv Radius, r = =   eB v 2  eB  1− 2 c 14. For head on collision, distance between centres of two deuterons = d = 2 × radius –15

d = 4 fm = 4 × 10

m

charge of each deuteron, e = 1.6 × 10–19 C Potential energy 9 × 109 (1.6 × 10−19 )2 e2 J U= = 4 pε0 d 4 × 10−15 9 × 1.6 × 1.6 × 10−14 = keV = 360 keV 4 × 1.6 × 10−16 As P.E. = 2 × K.E. of each deuteron = 360 keV 360 \ K.E. of each deuteron = = 180 keV 2 h 15. (a) de Broglie wavelength, l = 2mE \ Kinetic energy (E) of neutron, (6.63 × 10−34 )2 h2 E= = 2ml2 2 × 1.677 × 10−27 × (1.40 × 10−10 )2 = 6.686 × 10–21 J = (b) 80

6.686 × 10−21 1.6 × 10

E=

−19

= 4.174 × 10–2 eV.

3 kT 2

Physics for you | OctOber ‘15

where k = Boltzman constant \

l=

2 mE

=

h 3 mkT 6.63 × 10 −34

=

=

h

3 × 1.677 × 10 −27 × 1.38 × 10 −23 × 300 6.63 × 10 −10 20.8

=

m

6.63 × 10 −10 m 4.56

= 1.45 × 10–10 m = 0.145 nm. 16. When a charged particle of charge, q and mass m is

accelerated under a potential difference V, let v be the velocity acquired by the particle. Then 1 2 qV = mv or mv = 2qVm 2 h h 1 = (i) l = or l ∝ mv 2qVm qm q m \ le = p p = e × 1837me > 1 lp qe me e × me So, le > lp, i.e., greater value of de Broglie wavelength is associated with electron as compared to proton. (ii) Momentum of particle, p = mv = \ p∝ Hence,

2qVm

qm pe qe me me e = = × u0. Distance of source from cathode of photocell = 2 m. Stopping potential = V0 If distance of light source from cathode of photocell is halved, intensity of light increases. (i) Stopping potential will remain the same as it is independent of the intensity of the source of light. (ii) Photoelectric current is directly proportional to the intensity of incident light. So it will increase. (iii) Maximum velocity of photoelectrons, emitted remains unchanged, because it depends upon the frequency of incident light, which is not changing. h 18. de Broglie wavelength, l = mv Due to large mass of a football, the de Broglie wavelength associated with the moving football is very small, so its wave character is not visible. de Broglie wavelength of an electron is h h l = or p = p l p2 Kinetic energy of the electron, K = 2m h2 \ K= ...(i) 2ml2 hc Energy of the photon, E = hu = ( l is same) l ...(ii) Divide (ii) by (i), we get E hc 2ml2 2lmc = × 2 = K l h h or Energy of the photon  2lmc  =  kinetic energy of the electron  l 

19. Einstein’s photoelectric equation Kmax = hu – f0 According to Planck’s quantum theory, light radiations consist of small packets of energy called quanta. One quantum of light radiation is called a photon. Energy of a photon, E = hu where h is Planck’s constant and u is the frequency of light. When a photon falls on a photosensitive metal surface, the energy of the photon (= hu) is spent in two ways : (i) A part of the energy of the photon is used in liberating the electron from the metal surface which is equal to the work function f0 of the metal. (ii) The rest of the energy of the photon is used in imparting the maximum kinetic energy Kmax to the emitted photoelectron. If vmax is the maximum velocity of the emitted photoelectron and m is its mass, then Maximum kinetic energy of the photoelectron, 1 2 K max = mvmax 2 1 2 \ hu = f0 + mvmax or K max = hu − f0 2 This equation is called Einstein’s photoelectric equation. The salient features observed in photoelectric effects are : (i) Kmax is independent of the intensity of incident radiation. (ii) Kmax depends upon the frequency of incident radiation. (iii) Below the threshold frequency no emission of photoelectrons takes place, no matter how intense the incident light is. 20. Applying principle of conservation of energy, 2 pn2 p p E – B = Kn + Kp = …(i) + 2m 2m From law of conservation of momentum, pn + pp = E/c when E = B, from equation (i), pn = pp = 0 \ Process cannot take place. For process to take place, let E be slightly bigger than B so that E = B + l, (l 0, q2 < 0, the charge q3 should be positive  to get the net force F .

15. (7) :

Using the concept of area of triangle 1 1 × PD × 5x = × 3x × 4 x 2 2 12 x \ PD = 5 QD = (PQ)2 − (PD)2 = 9 x 2 −

144 x 2 9 x = 25 5

9 x 16 x = 5 5 Magnetic field at P due to current elements PQ and PR is zero as the point P is on the conductor. Therefore, magnetic field at P due to current element QR is µ I B = 0 (sin φ1 + sin φ2 ) 4 πPD and DR = 5x −

tan θ = \

F2

, F1 =

Kq1q2

16 F1 3 q3 16 = × 4 9 q2

\ q3 =

, F2 =

Kq1q3 9

2

81  9  27 27 × q2 × (3) = =   µC 64  8  64 64 2

 1 = 1 +  µC  8 Hence n = 8

84

 1   K = 4 πε   0

Physics for you | OctOber ‘15

B= = \

µ0 I × 5  (9 x / 5) (16 x / 5)  +   4 π × 12 x  3x 4x 

µ0 I 5  3 4  7 µ0 I  + = 48 πx  5 5  48 πx k=7 

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across

3. A type of magnetic tape designed for computer storage and backup. (3) 6. Large groups of wind turbines. (4, 5) 8. Electricity produced in a power plant by fission of uranium atom. (7) 9. Prefix for 10–24. (5) 12. Electronic circuit having two stable state. (8) 15. The most common type of battery. (3, 4) 17. The measure of gravity pulling on mass. (6) 18. A variable that gives your location relative to an origin. (8) 22. A device for measuring hardness of matter. (11) 26. The variable parameter, by means of which information is conveyed through an electronic circuit or system. (6) 28. A device capable of making the performance of mechanical work easier. (7) 29. The current atomic model. (5)

Cut Here

2

1

3

4 5 6

7

8 9 10 12

11 13

15

14 16

17

19

18

20 22

21 23

24 25

26

27 28

Down 1. A ray that does not propagate in a plane that contain both the object point and optical axis. (4, 3) 2. A K–meson. (4) 3. Used on a bicycle to generate electricity. (6) 4. Streams of positive ions produced in a discharge tube by boring holes in the cathode. (5, 4) 5. The point on the celestial sphere that lies directly above an observer. (6) 7. A material which darkens under electron bombardment and recovers on heating. (9) 10. A set of quantum number that uniquely identify the type of particle it is. (6) 11. Solid carbon dioxide, used as a refrigerant. (3, 3) 13. An a.c. component superimposed on a d.c. (6) 14. A form of differential manometer with two sealed bulbs attached to the limbs. (10)

29

16. Weight is measured in the same unit as this physical quantity. (5) 19. Familiar form of renewable energy. (3) 20. A glass mirror or metal mirror for producing interference fringes in overlapping beams. (5, 6) 21. Sources of energy that is formed from the remains of living organisms that were buried millions of years ago. (6, 5) 23. The galaxy in which earth is located. (5, 3) 24. A mineral crystal that exhibit its dichroism. (10) 25. Having a low degree of reverberation with little or no reflection of sound. (8) 27. An electronic circuit with a single output and one or more inputs. (4)

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Physics for you | OctOber ‘15

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Physics for you | OctOber ‘15