February 8, 2017 | Author: Phan Hồ Nghĩa | Category: N/A
Volume 24 Managing Editor Mahabir Singh Editor Anil Ahlawat (BE, MBA)
No. 10
October 2016
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CONTENTS
Class 11 NEET | JEE Essentials
8
Ace Your Way CBSE
25
MPP-4
36
Brain Map
46
Class 12 NEET | JEE Essentials
40
Brain Map
47
Ace Your Way CBSE
56
Exam Prep
65
Core Concept
72
MPP-4
76
Competition Edge Physics Musing Problem Set 39
80
You Ask, We Answer
82
Physics Musing Solution Set 38
83
Crossword
85
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PHYSICS FOR YOU | OCTOBER ‘16
7
CENTRE OF MASS OF A RIGID BODY •
Ideally a rigid body is a body with a perfectly definite and unchanged shape. The distances between all pairs of particles of such a body do not change. e.g., fan, pen, stone etc. For a system of particles, centre of mass is an imaginary point at which its total mass is supposed to be concentrated. If co-ordinates of particles of masses m1, m2, ...... are (x1, y1, z1), (x2, y2, z2), ...... then position vector of their center of mass is RCM = xCM i + yCM j + zCM k
•
•
•
m1 (x1 i + y1 j + z1 k) + m2 (x2 i + y2 j + z2 k) =
=
8
+ m3 (x3 i + y3 j + z3 k) + ...... m1 + m2 + m3 + ......
•
(m1x1 + m2 x2 + ......)i + (m1 y1 + m2 y2 .....) j
•
+ (m1z1 + m2 z2 + .....)k m1 + m2 + m3 + ......
⎛ m x + m2 x2 + ............ ⎞ 1 = ∑ mi xi xCM = ⎜ 1 1 ⎝ m1 + m2 + .......... ⎟⎠ M ⎛ m y + m2 y2 + ............ ⎞ 1 yCM = ⎜ 1 1 = ∑ mi yi ⎝ m1 + m2 + .................... ⎟⎠ M ⎛ m z + m2 z2 + ............ ⎞ 1 zCM = ⎜ 1 1 = ∑ mi zi ⎝ m1 + m2 + .................... ⎟⎠ M PHYSICS FOR YOU | OCTOBER ‘16
• •
•
If the system has continuous distribution of mass, treating the mass element dm at position r as a point mass and replacing summation by integration. 1 RCM = ∫ r dm M 1 1 1 and xCM = ∫ xdm, yCM = ∫ ydm, zCM = ∫ zdm M M M It may be inside or outside of the body. Its position depends on the shape of the body. For a given shape it depends on the distribution of mass within the body and is closer to massive part. For symmetrical bodies having homogeneous distribution of mass it coincides with centre of symmetry of geometrical centre. If we know the centre of mass of parts of the system and their masses, we can get the combined centre of mass by treating the parts as point particles placed at their respective centre of masses.
•
•
It is independent of the co-ordinate system, e.g., the centre of mass of a ring is at its centre whatever be the co-ordinate system. If the origin of co-ordinate system is at centre of mass, i.e., RCM = 0 1 Then by definition ∑ mi ri = 0 ⇒ ∑ mi ri = 0 M The sum of the moments of the masses of a system about its centre of mass is always zero.
Hemispherical shell R yc = , xc = 0 2
Solid hemisphere yc =
For a system of two point masses m1r1 = m2r2
3R , xc = 0 8
Circular cone (solid)
yc =
h , x =0 4 c
The centre of mass lies closer to the heavier mass. •
yc =
Centre of mass of symmetric body
Rectangular plate (By symmetry) b xc = , 2 L yc = 2
Semi-circular ring 2R yc = , xc = 0 π
Semi-circular disc yc =
4R , xc = 0 3π
PHYSICS FOR YOU | OCTOBER ‘16
MOTION OF CENTRE OF MASS •
For a system of particles, position of centre of mass m r + m2r2 + m3r3 + ....... is RCM = 1 1 m1 + m2 + m3 + ......
•
Velocity of CM m v + m2v2 + ...... vCM = 1 1 m1 + m2 + ......
⎛ dr ⎞ =v⎟ ⎜⎝∵ ⎠ dt
Acceleration of CM m a + m2a2 + .... aCM = 1 1 m1 + m2 + ....
dv ⎞ ⎛ ⎜⎝∵ a = ⎟⎠ dt
•
•
•
Linear momentum of a system of particles is equal to the product of mass of the system with the velocity of its centre of mass. If no external force acts on a system the velocity of its centre of mass remains constant, i.e., velocity of centre of mass is unaffected by internal forces. d(MvCM ) From Newton’s second law Fext = dt If Fext = 0 then vCM = constant.
PROBLEM SOLVING TRICKS •
10
h , x =0 3 c
Triangular plate (By qualitative argument) At the centroid, h yc = 3 xc = 0
Circular cone (hollow)
(For centre of mass problems) Make full use of the symmetry of the object, be it point, line, or plane.
•
•
If the object can be divided into several parts, treat each of these parts as a particle, located at its own center of mass. Choose the axis wisely : If given system is a group of particles, choose one of the particles as origin. If system is a body with a line of symmetry, consider it as x-axis. The choice of origin is completely arbitrary ; the location of the centre of mass is same regardless of the origin from which it is measured.
PURE ROTATIONAL MOTION •
•
• • •
•
A body is said to be in pure rotational motion if the perpendicular distance of each particle remains constant from a fixed line or point and do not move parallel to the line, and that line is known as axis of rotation. s Angular displacement θ = r Where s = length of arc traced by the particle. r = distance of particle from the axis of rotation. Angular velocity ω =
n
I = ∑ mi ri 2 i =1
•
•
•
dθ dt
dω dt All the parameters θ, ω and α are same for all the particles. Axis of rotation is perpendicular to the plane of rotation of particles. If α = constant, then where ω0 = initial angular speed ω = ω0 + αt
The moment of inertia of a rigid body about a given axis is the sum of the product of the masses of its constituent particles and the square of their respective distances from the axis of rotation.
Moment of inertia of a body about an axis not only depend upon the mass of the body but also upon the distribution of its mass about the axis of rotation. Greater is the part of the mass of the body away from the axis of rotation, greater is the moment of inertia of the body about that axis. The radius of gyration K of a body about an axis of rotation is defined as the root mean square distances of the particles from the axis of rotation and its square when multiplied with the mass of the body gives moment of inertia of the body about the axis.
Angular acceleration α =
θ = ω0t +
•
•
1 2 αt 2
t = time interval
K=
•
= root mean square distance of a particles from axis OZ. Two Important Theorems on Moment of Inertia :
ω2 = ω20 + 2αθ These equations are known as equations of rotational motion. 1 1 Total kinetic energy = m1v21 + m2v22 + .......... 2 2 1 = (m1r12 + m2r22 + ........) ω2 2 1 = Iω2; where I = Moment of inertia = m1r 12 + m2r 22 + .... 2 ω = angular speed of body.
12
The property of a body by virtue of which it opposes any change in its state of rotation about an axis is known as moment of inertia. PHYSICS FOR YOU | OCTOBER ‘16
Perpendicular axes theorem (Only applicable to plane lamina (that means for 2-D objects only)): Iz = Ix + Iy
(Object is in x-y plane)
Where Iz = moment of inertia of the body about z-axis. Ix = moment of inertia of the body about x-axis. Iy = moment of inertia of the body about y-axis. Iy = Ix + Iz (Object is in x-z plane) Ix = Iy + Iz (Object is in y-z plane)
MOMENT OF INERTIA •
r12 + r22 + ..... + rn2 n
Parallel axes theorem (Applicable to any type of object) :
PHYSICS FOR YOU | OCTOBER ‘16
13
IAB = Icm + Md2 Where Icm = Moment of inertia of the object about an axis passing through centre of mass and parallel to axis AB •
IAB = Moment of inertia of the object about axis AB M = Total mass of object d = Perpendicular distance between axis about which moment of inertia is to be calculated and the one passing through the centre of mass.
Moment of inertia and radius of gyration of some regular bodies about specific axis Body
Axis of rotation
Moment of inertia (I)
Radius of gyration (K)
about an axis passing through centre and perpendicular to its plane
MR2
R
1 MR2 2
R
3 MR2 2
3 R 2
about a tangent perpendicular to its plane
2MR2
R 2
about an axis passing through centre and perpendicular to its plane
1 MR2 2
R
Uniform circular ring about a diameter of mass M and radius R about a tangent in its own plane
Uniform circular disc of mass M and radius R
Solid sphere of radius R and mass M
about a diameter about a tangent in its own plane
1 MR2 4 5 MR2 4
2
2
R/2 5
R 2
about a tangent perpendicular to its own plane
3 MR2 2
3 R 2
about its diameter
2 MR2 5
2 R 5
about a tangential axis
7 MR2 5
7 R 5
about its diameter
2 MR2 3
2 R 3
about a tangential axis
5 MR2 3
5 R 3
about its own axis
1 MR2 2
R
Hollow sphere of radius R and mass M
2
⎡ l 2 R2 ⎤ about an axis passing through centre of mass and + ⎥ M ⎢ Solid cylinder of length perpendicular to its own axis ⎢⎣12 4 ⎥⎦ l, radius R and mass M
l 2 R2 + 12 4
⎡ l 2 R2 ⎤ M⎢ + ⎥ 4 ⎦⎥ ⎢⎣ 3
l 2 R2 + 3 4
about the diameter of one of the faces of cylinder
14
PHYSICS FOR YOU | OCTOBER ‘16
MR2
about its own axis
Hollow cylinder of ⎛ R2 l 2 ⎞ about an axis passing through its centre of mass and mass M and radius R + ⎟ M⎜ perpendicular to its own axis ⎝ 2 12 ⎠
R R2 l 2 + 2 12
about an axis through centre of mass and perpendicular to the rod
ML2 12
L
about an axis through one end and perpendicular to rod
ML2 3
L
Thin rod of length L
12
3
⎡ l 2 + b2 ⎤ Rectangular lamina of about an axis passing through its centre of mass and M⎢ ⎥ length l and breadth b perpendicular to plane ⎢⎣ 12 ⎥⎦
l 2 + b2 12
Uniform cone of radius about an axis passing through its centre of mass and R and height h joining its vertex to centre of base
R
Parallelopiped of length l, breadth b, height h about its central axis and mass M
⎛ l 2 + b2 ⎞ M⎜ ⎟ ⎝ 12 ⎠
TORQUE • •
point τ = r × F , where F = force applied P = point of application of force Q = point about which we want to calculate the torque. r = position vector of the point of application of force from the point about which we want to determine the torque.
•
3 10
l 2 + b2 12
F = force applied on the body If forces F1 and F2 are applied on the body to rotate it in anti-clockwise direction and F3 makes body to rotate in clockwise direction. Then, τresultant = F1r1 + F2r2 – F3r3 (in anti-clockwise direction)
It represents the capability of a force to produce change in the rotational motion of the body. Torque of force F about a
τ = rF sinθ = r⊥F = rF⊥ Where θ = angle between the direction of force and the position vector of P with respect to Q. r⊥ = perpendicular distance of line of action of force from point Q. F⊥ = force arm Torque acting on the body about the axis of rotation, τ = r × F r = position vector of the point of application of force about the axis of rotation
3 MR2 10
τnet = τ1 + τ2 + τ3 + ......... Torque produced by a force about an axis can be zero if force vector (i) is parallel to the axis of rotation. (ii) passes through the axis of rotation.
•
If net external torque acting on the body is zero, then the body is said to be in rotational equilibrium. i.e., ∑ τ ext = 0.
•
Relation between τ and α (angular acceleration): τ=I α PHYSICS FOR YOU | OCTOBER ‘16
15
•
•
Torque is also change in angular momentum dL τ= dt
•
Work done by a torque and power of a torque : If a torque τ applied on a body rotates it through an angle Δθ, the work done by the torque is ΔW = τΔθ or work done = torque × angular displacement Power of a torque is given as ΔW τΔθ P= = = τω Δt Δt
axis, then angular momentum of rigid body about this axis will be given by L = Iω Angular momentum of a rigid body in rotation plus translation about a general axis
i.e., Power of a torque = torque × angular velocity
ANGULAR MOMENTUM •
Angular momentum of a particle in motion about a fixed point (O) Suppose a particle A has a linear momentum p = mv as shown in the figure.
Angular momentum of particle A about point O will be, L = r × p = r × (mv ) = m(r × v ) Magnitude of L is L = mvr sin θ = mvr⊥ where θ is the angle between r and p .
Direction of L will be given by right hand screw rule. As shown in figure, direction of L is perpendicular to paper inwards. If the particle passes through point O, r⊥ = 0. Therefore, angular momentum is zero. Angular momentum of a rigid body in pure rotation about axis of rotation
ROTATIONAL PLUS TRANSLATIONAL MOTION OF A RIGID BODY •
•
There will be two terms (a) Icω (b) mvcr⊥ = mvr⊥ From right hand screw rule, we can see that Icω and mvr⊥ both terms are perpendicular to the paper in inward direction. Hence, they are added. or LTotal = Icω + mvr⊥ In a problem, if these two terms are in opposite directions then they will be subtracted. Law of conservation of angular momentum : If no external torque acts on a system, total angular momentum of the system remains unchanged. In the absence of any external torque, L = Iω = constant 2π 2π or I1ω1 = I2ω2 or I1 ⋅ = I2 ⋅ T1 T2
•
A complex motion of rotation plus translation can be simplified by considering The translational motion of centre of mass of the rigid body. Rotational motion about the centre of mass. Now, velocity of point (90 – ) P is the vector sum of two terms v and rω. = Here v is common for all points, while rω is different for different points, as r is different. vp = v 2 + (rω)2 + 2(v )(rω)cos(90° − θ)
16
If a rigid body is in pure rotation about a fixed
PHYSICS FOR YOU | OCTOBER ‘16
= v 2 + r 2 ω2 + 2vr ω sin θ •
Acceleration of point P is the vector sum of three terms.
•
a an = rω2
(acceleration of CM) (acting towards centre O)
dω dt Again, here a is common for all points, while an and at are different. at = rα = r
(acting tangentially)
MOTION OF A SPHERICAL BODY ON ROUGH INCLINED SURFACE •
tan θ
mR2 I If μ = 0, body will slip downwards (only translational motion) with an acceleration, a1 = g sin θ If μ > μmin , body will roll down without slipping with an acceleration, (rotation + translation both) 1+
Minimum value of coefficient of friction required for pure rolling, μmin =
•
Accelerated pure rolling : If v and ω are not constant then, a = Rα is an additional condition for pure rolling on horizontal ground, which takes place in the presence of some external forces. Here friction plays very important role. Magnitude and direction of friction is so adjusted that equation a = Rα is satisfied. If friction is insufficient for satisfying the equation a = Rα, slipping (either forward or backward) will occur and kinetic friction will act.
Uniform pure rolling : In which v and ω remain constant. Condition of pure rolling is v = Rω. In this case bottommost point of the spherical body is at rest. It has no slipping with its contact point on ground. Because ground point is also at rest. If v > Rω, then net velocity of point P is in the direction of v. This is called forward slipping. If v < Rω, then net velocity of point P is in opposite direction of v. This is called backward slipping.
•
If a spherical body is rolling over a plank, condition for no slipping between spherical body and plank is, v – Rω = v0
•
a2 =
g sin θ I 1+ mR2
PHYSICS FOR YOU | OCTOBER ‘16
17
When μ > μmin force of friction will act upwards. Magnitude of this force is, mg sin θ f = mR2 1+ I If μ < μmin , body will roll downwards with forward slipping. Maximum friction will act in this case. The acceleration of body is a3 = g sin θ – μg cos θ Comparison of pure rolling and pure sliding on a inclined plane
Angular velocity of rigid body is, p × r⊥ ω= I (p × r⊥) = angular impulse = change in angular momentum = Iω.
•
Physical quantities
Pure rolling g sin θ
Acceleration
1+ Velocity
gsinθ
K2 r2
2 gh 1+
Time taken to cover the distance s, where K is the radius of gyration.
Pure sliding
K
2gh
2
2s g sin θ
ANGULAR IMPULSE •
Linear impulse when multiplied by perpendicular distance gives angular impulse. Angular impulse is also equal to change in angular momentum.
Translational motion
Rotational motion about a fixed axis
Displacement x
Angular displacement θ
Velocity v = dx/dt
Angular velocity ω = dθ/dt
Acceleration a = dv/dt
Angular acceleration α = dω/dt
Mass M
Moment of inertia I
Force F = Ma
Torque τ = Iα
Work dW = Fds
Work dW = τdθ
Kinetic energy of a Kinetic energy of a rotational translational motion motion KR = Iω2/2 KT = Mv2/2
r2
⎛ K2 ⎞ 2s ⎜1 + ⎟ ⎝ r2 ⎠ g sin θ
ANALOGY BETWEEN TRANSLATIONAL MOTION AND ROTATIONAL MOTION
Power P = Fv
Power P = τω
Linear momentum p = Mv
Angular momentum L = Iω
Equations of translational motion (i) v = u + at
Equations of rotational motion ω = ω0 + αt 1 θ = ω0t + αt 2 2
1 2 (ii) s = ut + at 2 (iii) v2 – u2 = 2as a (iv) snth = u + (2n − 1) 2
ω2 – ω02 = 2αθ θnth = ω0 +
α (2n − 1) 2
•
18
A rigid body is kept over a smooth table. It is hit at a point by a linear impulse p at a perpendicular distance r⊥ from C as shown. Since it is hit at some perpendicular distance from C its motion is rotational plus translational. Velocity of centre of p mass will be given by, v = m PHYSICS FOR YOU | OCTOBER ‘16
ANSWER KEY
MPP-4 CLASS XII 1. 6. 11. 16. 21. 26.
(a) (b) (b) (d) (b,c) (5)
2. 7. 12. 17. 22. 27.
(c) (d) (a) (a) (a,b,d) (a)
3. 8. 13. 18. 23. 28.
(c) (a) (c) (b) (a,d ) (a)
4. 9. 14. 19. 24. 29.
(d) (a) (d) (d) (1) (c)
5. 10. 15. 20. 25. 30.
(c) (c) (c) (a,b,c) (6) (b)
1. The centre of mass of a non-uniform rod of length
(a) 6 mv 02
(b) 12 mv 02
Kx 2 , where K is a L constant and x is the distance from one end is
(c) 4 mv 02
(d) 8 mv 02
L whose mass per unit length λ =
3L L K 3K (b) (c) (d) 4 8 L L 2. A uniform rod AB of mass m and length 2a is falling freely without rotation under gravity with AB horizontal. Suddenly the end A is fixed when the speed of the rod is v. The angular speed with which the rod begins to rotate is 4v v v 3v (a) (b) (c) (d) 3a 2a 3a 4a 3. A semicircular lamina of mass m, radius r and centre at C is shown in the figure. Its centre of mass is at a distance x from C. Its moment of inertia about an axis through its centre of mass and perpendicular to its plane is 1 2 1 2 (a) mr (b) mr 2 4 1 2 1 2 (c) mr + mx 2 (d) mr − mx 2 2 2 4. Three identical rings, each of mass M and radius R are arranged as shown in figure. The moment of inertia of the arrangement about YY′ is (a) 1 MR2 2 (b) MR2 (c) 5 MR2 2 7 (d) MR2 2 (a)
5. A ring of mass m and radius R has three particles attached to the ring as shown in the figure. The centre of the ring has a speed v0. If the system is in pure rolling then the kinetic energy of the system is
6. A solid sphere and a solid cylinder of same mass are rolled down on two inclined planes of heights h1 and h2 respectively. If at the bottom of the plane the two objects have same linear velocities, then the ratio of h1 : h2 is (a) 2 : 3 (b) 7 : 5 (c) 14 : 15 (d) 15 : 14 7. A horizontal turn table in the form of a disc of radius r carries a gun at G and rotates with angular velocity ω0 about a vertical axis passing through the centre O. The increase in angular velocity of the system if the gun fires a bullet of mass m with a tangential velocity v with respect to the gun is (moment of inertia of gun + table about O is I0) 2 mvr mvr (b) (a) I0 I + mr 2 0
mvr v (d) 2I 0 2r 8. Three identical rods, each of length l, are joined to form an equilateral triangle. Its radius of gyration about an axis passing through a corner and perpendicular to the plane of the triangle is (c)
(a)
l 2
(b)
3 l 2
(c)
l 2
(d)
l 3
9. A uniform disc of radius R lies in x-y plane with its centre at origin. Its moment of inertia about the axis x = 2R and y = 0 is equal to the moment of inertia about the axis y = d and z = 0. Where d is equal to 15 4 17 R R R (c) 3R (b) (d) 2 3 2 10. A constant power is supplied to a rotating disc. Angular velocity (ω) of disc varies with number of rotations (n) made by the disc as (a)
(a) ω ∝ n1/3 (c) ω ∝ n2/3
(b) ω ∝ n3/2 (d) ω ∝ n2 PHYSICS FOR YOU | OCTOBER ‘16
19
11. A disc is rolling on the inclined plane. What is the ratio of its rotational KE to the total KE? (a) 1 : 3 (b) 3 : 1 (c) 1 : 2 (d) 2 : 1 12. A boat of 90 kg is floating in still water. A boy of mass 30 kg walks from the stern to the bow. The length of the boat is 3 m. Find the distance through which the boat will move. (a) 0.75 m (b) 0.90 m (c) 1.0 m (d) 1.5 m 13. A disk and a sphere of same radius but different masses roll off on two inclined planes of the same altitude and length. Which one of the two objects gets to the bottom of the plane first? (a) Both reach at the same time (b) Depends on their masses (c) Disk (d) Sphere [NEET Phase I 2016] 14. From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre? (a) 11 MR2/32 (b) 9 MR2/32 2 (c) 15 MR /32 (d) 13 MR2/32 [NEET Phase I 2016] 15. A uniform circular disc of radius 50 cm at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of 2.0 rad s–2. Its net acceleration in m s–2 at the end of 2.0 s is approximately (a) 6.0 (b) 3.0 (c) 8.0 (d) 7.0 [NEET Phase I 2016] 16. An automobile moves on a road with a speed of 54 km h–1. The radius of its wheels is 0.45 m and the moment of inertia of the wheel about its axis of rotation is 3 kg m2. If the vehicle is brought to rest in 15 s, the magnitude of average torque transmitted by its brakes to the wheel is (a) 10.86 kg m2 s–2 (b) 2.86 kg m2 s–2 (c) 6.66 kg m2 s–2 (d) 8.58 kg m2 s–2 [AIPMT 2015] 17. In the figure shown ABC is a uniform wire. If centre of mass of wire lies vertically BC below point A, then is AB close to (a) 1.85 (b) 1.5 (c) 1.37 (d) 3 [JEE Main Online 2016] 20
PHYSICS FOR YOU | OCTOBER ‘16
18. A roller is made by joining together two cones at their vertices O. It is kept on two rails AB and CD which are placed asymmetrically (see figure), with its axis perpendicular to CD and its centre O at the centre of line joining AB and CD (see figure). It is given a light push so that it starts rolling with its centre O moving parallel to CD in the direction shown. As it moves, the roller will tend to (a) turn left (b) turn right (c) go straight (d) turn left and right alternately [JEE Main Offline 2016] 19. A cubical block of side 30 cm is moving with velocity 2 m s–1 on a smooth horizontal surface. The surface has a bump at a point O as shown in figure. The angular velocity (in rad/s) of the block immediately after it hits the bump, is (a) 13.3 (b) 5.0 (c) 9.4 (d) 6.7 [JEE Main Online 2016] 20. Concrete mixture is made by mixing cement, stone and sand in a rotating cylindrical drum. If the drum rotates too fast, the ingredients remain stuck to the wall of the drum and proper mixing of ingredients does not take place. The maximum rotational speed of the drum in revolutions per minute (rpm) to ensure proper mixing is close to (Take the radius of the drum to be 1.25 m and its axle to be horizontal) (a) 27.0 (b) 0.4 (c) 1.3 (d) 8.0 [JEE Main Online 2016] SOLUTIONS 1. (a) :
2
Mass of the element PQ is dm = λdx = Kx . dx L 4 ⎛L ⎞ 3 L Kx L ⎜4⎟ dx ⎝ ⎠ 3L ∫0 x dm = ∫0 L x = = = CM L 2 ∴ 3 ∫0 dm ∫ L Kx dx ⎜⎛ L ⎟⎞ 4 0 L ⎝3⎠
2. (d) :
Angular momentum about A will be conserved, i.e., Li = Lf m(2a)2 or mva = Iω or mva = ω 3 3v ∴ ω= 4a 3. (d) : We know, IC = mr2/2 Using parallel axes theorem, IC = ICM + mx2 ∴ ICM = IC – mx2 = mr2/2 – mx2 4. (d) : Moment of inertia of ring I about YY′ =
3 MR2 2
3 2 Moment of inertia of ring II about YY′ = MR 2 1 Moment of inertia of ring III about YY′ = MR2 2 Moment of inertia of the system about YY′ 3 3 1 7 = MR2 + MR2 + MR2 = MR2 2 2 2 2 5. (a) : As we know, KE = KEtran. + KErot. 2
1 1 ⎛v ⎞ KE = m(v02 ) + (mR2 ) ⎜ 0 ⎟ ⎝R⎠ 2 2 1 1 + (3m)( 2v0 )2 + (m)(2v0 )2 = 6mv2 0 2 2 6. (c) : In case of pure rolling of a solid sphere, KR 2 = KT 5 ∴
Where KR = Rotational kinetic energy, KT = Translational kinetic energy. Then at the bottom of the plane, KR + KT = mgh1 or
5 KT = mgh1 7
For solid cylinder,
KR KT
⎡ KR 2 ⎤ ⎢as K = 5 ⎥ ⎣ ⎦ T 1 = 2
As, KR + KT = mgh2 2 KT = mgh2 3 At the bottom, both have the same linear velocities, i.e., they have the same translational kinetic energies ∴
h1 14 5 2 mgh1 = mgh2 ∴ = h2 15 7 3 7. (a) : Given, I0 is the moment of inertia of table and gun and m is the mass of bullet. Initial angular momentum of system about centre Li = (I0 + mr2) ω0 ...(i) Let ω be the angular velocity of table after the bullet is fired. Final angular momentum Lf = I0ω + mr2ω – mvr or Lf = I0ω – m (v – rω) r ...(ii) where (v – rω) is absolute velocity of bullet to the right. From eqs. (i) and (ii), we get mvr (ω − ω0 ) = I0 + mr 2 This is the increase in angular velocity. 8. (c) : Moment of inertia of all the three rods is ∴
2 ⎛ ⎛ 3l ⎞ ⎞⎟ 3 2 Ml 2 Ml 2 ⎜ Ml 2 I= + +⎜ + M⎜ ⎟ ⎟ = Ml ⎝ 2 ⎠ ⎠ 2 3 3 ⎝ 12 As I = MK2 3 l So, 3MK 2 = Ml 2 ⇒ K = 2 2 9. (b) : An axis passing through x = 2R, y = 0 is in ⊗ direction as shown in figure. Moment of inertia about this axis will be 1 9 I1 = mR2 + m(2R)2 = mR2 ...(i) 2 2 Axis passing through y = d, z = 0 is shown by dotted line in figure. Moment of inertia about this axis will be 1 I2 = mR2 + md 2 ...(ii) 4 By equations (i) and (ii), we get
1 9 17 mR2 + md 2 = mR2 or d = R 4 2 2 10. (a) : We have P = τ.ω P ⎛ dω ⎞ or I ⎜ ω ⎟ ω = P or ω2dω = dθ I ⎝ dθ ⎠ On integration, we find that ω ∝ (θ)1/3 or ω ∝ (n)1/3 PHYSICS FOR YOU | OCTOBER ‘16
21
1 1 11. (a) : Etot = Etran + Erot = mv 2 + I ω2 2 2 1 1 1 v2 = mv 2 + × mr 2 × 2 2 2 r2 1 1 3 = mv 2 + mv 2 = mv 2 2 4 4 1 2 Erot 4 mv 1 = = = 1: 3 Etot 3 mv 2 3 4 12. (a) : As shown in figure, let C1, C2 and C be the centres of mass of the boy, boat and the system (boy and boat) respectively. Let x1 and x2 be the distances of C1 and C2 from the shore. Then the centre of mass will be at a distance, 30 x1 + 90 x2 xCM = 30 + 90
Since g is constant and l, R and sin θ are same for both ∴ td
ts
3 5 15 ⇒ td > ts × = 2 7 14 Hence, the sphere gets to the bottom first. M 14. (d) : Mass per unit area of disc = 2 Mass of removed portion of disc, πR =
30[x1 − (3 − d )] + 90(x2 + d ) 30 + 90 As x′CM = xCM ′ = xCM
30(x1 − 3 + d ) + 90(x2 + d ) 30 x1 + 90 x2 = 120 120 or –90 + 30 d + 90 d = 0 or d = 0.75 m 13. (d) : Time taken by the body to reach the bottom when it rolls down on an inclined plane without slipping is given by
t=
22
2 ⎞ ⎛ 2l ⎜ 1 + k ⎟ ⎝ R2 ⎠ g sinθ
PHYSICS FOR YOU | OCTOBER ‘16
2
M ⎛R⎞ × π⎜ ⎟ = 2 4 ⎝2⎠ πR Moment of inertia of removed portion about an axis passing through centre of disc O and perpendicular to the plane of disc, 2 IO ′ = IO′ + M′d M′ =
=
As the boy moves from the stern to the bow, the boat moves backward through a distance d so that position of the centre of mass of the system remains unchanged.
k2 R2 1 + d2 1+ R = 2R2 ⎛∵k = R , k = 2 R ⎞ = ⎜ d ⎟ s 5 ⎠ 2R2 ⎝ 2 ks2 1+ 1+ 2 5R 2 R
M
1 M ⎛ R ⎞2 M ⎛ R ⎞2 × ×⎜ ⎟ + ×⎜ ⎟ 2 4 ⎝2⎠ 4 ⎝2⎠
2 2 2 = MR + MR = 3MR 32 16 32 When portion of disc would not have been removed, the moment of inertia of complete disc about centre O is 1 IO = MR2 2 So, moment of inertia of the disc with removed portion is
I = IO – IO ′ =
1 3MR2 13MR2 MR2 − = 2 32 32
15. (c) : Given, r = 50 cm = 0.5 m, α = 2.0 rad s–2, ω0 = 0 At the end of 2 s, Tangential acceleration, at = rα = 0.5 × 2 = 1 m s–2 Radial acceleration, ar = ω2r = (ω0 + αt)2r = (0 + 2 × 2)2 × 0.5 = 8 m s–2 ∴ Net acceleration, a=
at2 + ar2 = 12 + 82 = 65 ≈ 8 m s −2
16. (c) : Here, Speed of the automobile, 5 v = 54 km h −1 = 54 × m s −1 = 15 m s −1 18
Radius of the wheel of the automobile, R = 0.45 m Moment of inertia of the wheel about its axis of rotation, I = 3 kg m2 Time in which the vehicle brought to rest, t = 15 s The initial angular speed of the wheel is −1
1500 100 v 15 m s rad s −1 = rad s −1 = = 45 3 R 0.45 m and its final angular speed is ωf = 0 (as the vehicle comes to rest) ∴ The angular retardation of the wheel is 100 ωf − ωi 0 − 3 100 α= = =− rad s −2 t 15 45 The magnitude of required torque is ωi =
q 2 p2 + 2 p 2 4 ⇒ p2 + pq = q2 + p = 2 ( p + q) 2 1+
q = p
q = p ∴
18. (a)
2
q2
q 1 1 ⎛q⎞ + ⇒⎜ ⎟ − − =0 2 2 p 2 ⎝ p⎠ p
⎛ 1⎞ −(−1) ± (−1)2 − 4(1) ⎜ − ⎟ ⎝ 2⎠ 2 ×1
Possible value of
=
1± 3 2
q 1+ 3 = = 1.366 ≈ 1.37 p 2
19. (b) : Since no external torque acts on the system, therefore total angular momentum of the system about point O remains constant. 24
PHYSICS FOR YOU | OCTOBER ‘16
After hitting , Lf = Iω mva a ∴ mv = I ω or ω = 2I 2 Here, I = moment of inertia of cube about its edge 2
⎛ 2a ⎞ 2 2 2 a2 = m + m⎜ ⎟ = ma + ma = 2ma ⎜ ⎟ 6 6 2 3 ⎝ 2 ⎠ mva × 3 3v 3×2 ∴ ω= = = = 5 rad s −1 2 4 a 4 × 0 .3 2 × 2ma 20. (a) : Radius of the drum, R = 1.25 m For just one complete rotation, speed of the drum at top position, Angular velocity of the drum, ω =
17. (c) : Let AB = p BC = q λ = linear mass density of the rod According to question, centre of mass of the rod lies vertically below point A. ⎛q⎞ ⎛ p⎞ (λq) ⎜ ⎟ + (λp) ⎜ ⎟ cos 60° ⎝2⎠ ⎝2⎠ ∴ XCM = p cos 60° = λ( p + q)
⇒
a 2
v = Rg
⎛ 100 ⎞ τ = I | α | = (3 kg m2 ) ⎜ rad s −2 ⎟ ⎝ 45 ⎠ 20 = kg m2s −2 = 6.66 kg m2s −2 3
⇒
Before hitting, Li = mv
ω=
v g = R R
10 60 10 rad s −1 = rpm = 27 rpm 1.25 2π 1.25
CLASS XI Series 4
Gravitation and Mechanical Properties of Solids
Time Allowed : 3 hours Maximum Marks : 70
GENERAL INSTRUCTIONS (i)
All questions are compulsory.
(ii)
Q. no. 1 to 5 are very short answer questions and carry 1 mark each.
(iii) Q. no. 6 to 10 are short answer questions and carry 2 marks each. (iv) Q. no. 11 to 22 are also short answer questions and carry 3 marks each. (v)
Q. no. 23 is a value based question and carries 4 marks.
(vi) Q. no. 24 to 26 are long answer questions and carry 5 marks each. (vii) Use log tables if necessary, use of calculators is not allowed.
SECTION-A 1. What is the Young’s modulus for a perfect rigid body? 2. Why are space rockets usually launched from west to east in the equatorial plane? 3. Why do spring balances show wrong readings after they have been used for a long time? 4. The shear modulus of a material is always considerably smaller than the Young modulus for it. What does it signify? 5. If the earth were hollow, but still had the same mass and radius, would your weight be different? SECTION-B 6. A particle is projected upward from the surface of the earth (radius R) with kinetic energy equal to half the minimum value needed for it to escape. To which height does it rise above the surface of the earth? 7. Discuss the variation of acceleration due to gravity with the altitude.
8. Define Poisson's ratio. Write an expression for it. What is the significance of negative sign in this expression? 9. The breaking stress for aluminium is 7.5 × 107 N m–2. Find the greatest length of aluminium wire that can hang vertically without breaking. Density of aluminium is 2.7 × 103 kg m–3. OR How is the knowledge of elasticity useful in selecting metal ropes used in cranes for lifting heavy loads? 10. How will you weigh the sun, that is estimate its mass? The mean orbital radius of the earth around the sun is 1.5 × 108 km. SECTION-C 11. A rocket is fired from the earth towards the sun. At what distance from the earth’s centre is the gravitational force on the rocket zero? Mass of the sun = 2 × 1030 kg, mass of the earth = 6 × 1024 kg. Neglect the effect of other planets etc. (orbital radius = 1.5 × 1011 m). PHYSICS FOR YOU | OCTOBER ‘16
25
12. State and explain Kepler's laws of planetary motion. Name the physical quantities which remain constant during the planetary motion. 13. Compute the bulk modulus of water from the following data : Initial volume = 100.0 litre, pressure increase = 100.0 atm (1 atm = 1.013 × 105 Pa), final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large. 14. Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 cm and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column. The Young’s modulus of steel is 2.0 × 1011 Pa. 15. A structural steel rod has a radius of 10 mm and a length of 1 m. A 100 kN force F stretches it along its length. Calculate (a) the stress, (b) elongation, and (c) strain on the rod. Given that the Young's modulus, Y, of the structural steel is 2.0 × 1011 N m–2. 16. Define the term stress. Give its units and dimensions. Describe the different types of stress. 17. Two identical heavy spheres are separated by a distance 10 times their radius. Will an object placed at the mid point of the line joining their centres be in stable equilibrium or unstable equilibrium? Give reason for your answer. 18. A mass m is placed at P a distance h along the normal through the centre O of a thin circular ring of mass M and radius r as shown in the figure. If the mass is removed further away such that OP becomes 2h, by what factor the force of gravitation will decrease, if h = r ? 19. Frictional force increases the velocity of a satellite. Discuss. OR A star like the sun has several bodies moving around it at different distances. Consider that all of them are moving in circular orbits. Let r be the distance of the body from the centre of the star and let its linear velocity be v, angular velocity ω, kinetic energy K, gravitational potential energy U, total energy E and angular momentum L. As the radius r of the orbit increases, determine which of the above quantities increase and which ones decrease. 20. If the normal density of sea water is 1.00 g cm–3, what will be its density at a depth of 3 km? Given 26
PHYSICS FOR YOU | OCTOBER ‘16
compressibility of water = 0.0005 per atmosphere, 1 atmospheric pressure = 106 dyne cm–2, g = 980 cm s–2. 21. How will the value of g be affected if (i) the rotation of the earth stops (ii) the rotational speed of the earth is doubled (iii) the rotational speed of the earth is increased to seventeen times its present value? 22. The planet mars has two moons, phobos and deimos. (i) phobos has a period of 7 h, 39 min and an orbital radius of 9.4 × 103 km. Calculate the mass of mars. (ii) Assume that earth and mars move in circular orbits around the sun, with the martian orbit being 1.52 times the orbital radius of the earth, what is the length of the martian year in days? SECTION-D 23. Sohan is a student of class XI and reads an article on astronomy in a magazine. Astronauts spend weeks and months in orbiting spacecrafts and space stations. Although gravity acts on them, the astronauts experience long durations of zero gravity due to centripetal motion. On earth, gravity provides the force that causes our muscles and bones to develop to the proper strength so that we may function in our environment. After reading the article some questions arise in his mind. He goes to his physics teacher to know the answer. Answer the following questions based on the information. (a) What happens to muscles and bones in a zero gravity environment? (b) How is our circulatory system affected? (c) Do astronauts lose blood in zero gravity environment. (d) What inference do you draw from the above discussion? SECTION-E 24. What is escape velocity? Obtain an expression for the escape velocity on the earth in different form. Why is it that there is no atmosphere on the moon? Explain. OR
Define orbital velocity of a satellite. Derive expressions for the orbital velocity of a satellite. Show that the escape velocity of a body from the earth's surface is 2 times its velocity in a circular orbit just above the earth's surface. 25. What is meant by gravitational potential energy of a body? What is the zero level of potential energy? Derive an expression for the gravitational potential
energy of a body of mass m located at distance r from the centre of the earth. OR Explain what happens when the load on a metal wire suspended from a rigid support is gradually increased. Illustrate your answer with a suitable stress-strain graph. 26. (a) A steel wire with a circular cross section has a radius of 0.1 cm. The wire is 10 m long when measured lying horizontal, and hangs from a hook on the wall. A mass of 25 kg is hung from the free end of the wire. Assuming the wire to be uniform and lateral strains R. Suppose at any instant the body is at point A, such that OA = x The gravitational force of attraction on the body at A is
GMm ⎡ 1⎤ ⎡1 1 ⎤ = GMm ⎢ − ⎥ = −GMm ⎢ − ⎥ = − r ⎣ x ⎦∞ ⎣r ∞ ⎦
GM R+h
or
GMm
(i) The initial part OA of the graph is a straight line indicating that stress is proportional to strain. Upto the point A, Hooke's law is obeyed. The point A is called the proportional limit. In this region, the wire is perfectly elastic. (ii) After the point A, the stress is not proportional to strain and a curved portion AB is obtained. However, if the load is removed at any point between O and B, the curve is retraced along BAO and the wire attains its original length. The portion
OB of the graph is called elastic region and the point B is called elastic limit or yield point. The stress corresponding to the yield point is called yield strength (Sy). (iii) Beyond the point B, the strain increases more rapidly than stress. If the load is removed at any point C, the wire does not come back to its original length but traces dashed line CE. Even on reducing the stress to zero, a residual strain equal to OE is left in the wire. The material is said to have acquired a permanent set. The fact that the stress-strain curve is not retraced on reversing the strain is called elastic hysteresis. (iv) If the load is increased beyond the point C, there is large increase in the strain or the length of the wire. In this region, the constrictions (called necks and waists) develop at few points along the length of the wire and the wire ultimately breaks at the point D, called the fracture point. In the region between B and D, the length of wire goes on increasing even without any addition of load. This region is called plastic region and the material is said to undergo plastic deformation. The stress corresponding to the breaking point is called ultimate strength or tensile strength of the material. 26. (a) Let AB = L, Mass of wire = m,
mg = weight of wire Wire is fixed at point A and a mass M is hanged at point B.
(i) Suppose ΔL1 is the extension in the wire of length L due to its mass. (mg )(L / 2) mgL = ... (i) YA 2YA (ii) Suppose ΔL2 is the extension in the wire due to hanged mass M. (Mg )L MgL ∴ ΔL2 = = ... (ii) YA YA Hence, total extension in the wire, mgL MgL + ΔL = ΔL1 + ΔL2 = 2YA YA gL ⎛ m ⎞ +M⎟ ⇒ ΔL = ... (iii) YA ⎜⎝ 2 ⎠ Given, r = 0.1 cm = 10–3 m, L = 10 m M = 25 kg, Y = 2 × 1011 N m–2, ρ = 7860 kg m–3 A = πr2 = 3.14 × 10–6 m2 m = (πr2L)ρ = 3.14 × 10–6 × 10 × 7860 ≈ 0.25 kg Put these values in eqn. (iii),
Then, ΔL1 =
ΔL =
10 × 10
(0.125 + 25) 2 × 10 × 3.14 × 10−6 = 4 × 10–3 m (b) Yield strength = 2.5 × 108 N m–2 Yield force = Yield strength × A = 2.5 × 108 × 3.14 × 10–6 = 785 N At the yield point, (m + Mmax)g = 785 Here, Mmaxg = maximum weight sustain by the wire 0.25 + Mmax = 78.5 ⇒ Mmax = 78.25 kg 11
OR
Here, F = Fsinθ, F|| = Fcosθ F produces tensile stress and F|| produces shear stress, on the plane aa′. Let A′ = area of plane aa′ A A ∴ sin θ = ⇒ A′ = A′ sin θ Tensile stress on the plane aa′ is F F sin θ F 2 = ⊥= = sin θ A′ A / sin θ A Shearing stress on the plane aa′ F|| F cos θ F sin θ cos θ F (2 sin θ cos θ) = = = = A′ A / sin θ A 2A F sin 2θ = 2A (a) For maximum tensile stress, sin2θ = 1 ⇒ θ = 90° (b) For maximum shearing stress, sin2θ = 1 ⇒ 2θ = 90° ⇒ θ = 45°
PHYSICS FOR YOU | OCTOBER ‘16
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Class XI
T
his specially designed column enables students to self analyse their extent of understanding of specified chapters. Give yourself four marks for correct answer and deduct one mark for wrong answer. Self check table given at the end will help you to check your readiness.
Rotational Motion | Gravitation Total Marks : 120
NEET / AIIMS / PMTs Only One Option Correct Type
1. If R is the average radius of earth, ω is its angular velocity about its axis and g is the gravitational acceleration on the surface of earth then the cube of the radius of orbit of a geostationary satellite will be equal to Rg R 2 ω2 R2 g R2 g (a) (b) (c) 2 (d) g ω ω2 ω 2. A man stands at one end of a boat which is stationary in water. Neglect water resistance. The man now moves to the other end of the boat and again becomes stationary. The centre of mass of the man plus boat system will remain stationary with respect to water (a) in all cases (b) only when the man is stationary initially and finally (c) only if the man moves without acceleration on the boat (d) only if the man and the boat have equal masses 3. Weights of 1 g, 2 g, ...., 100 g are suspended from the 1 cm, 2 cm, ...., 100 cm marks respectively of a light metre scale. Where should it be supported for the system to be in equilibrium? (a) 55 cm mark (b) 60 cm mark (c) 66 cm mark (d) 72 cm mark 4. The gravitational force between two bodies is 1 ⎛ 1 ⎞ directly proportional to not ⎟ , where R is R ⎜⎝ R2 ⎠ 36
PHYSICS FOR YOU | OCTOBER ‘16
Time Taken : 60 min
the distance between the bodies. Then the orbital speed for this force in circular orbit is proportional to 1 1 (a) (b) R2 (c) R (d) 2 R R0 5. The imaginary angular velocity of the earth for which the effective acceleration due to gravity at the equator shall be zero is equal to (Take g = 10 m s–2 and Re = 6400 km) (a) 1.25 × 10–3 rad s–1 (b) 2.50 × 10–3 rad s–1 (c) 3.75 × 10–3 rad s–1 (d) 5.0 × 10–3 rad s–1 6. A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is K. If radius of the ball be R, then the fraction of total energy associated with its rotation will be (a)
(c)
K 2 + R2 R2 K2 K 2 + R2
(b)
(d)
K2 R2 R2 K 2 + R2
7. A uniform rod of mass m and length l makes a constant angle θ with an axis of rotation which passes through one end of the rod. Its moment of inertia about this axis is (a)
ml 2 3
(c)
ml 2 2 sin θ 3
ml 2 sin θ 3 ml 2 cos2 θ (d) 3
(b)
8. A particle of mass m is placed at the centre of a uniform spherical shell of mass 3m and radius R. The gravitational potential on the surface of the shell is 3Gm Gm (a) − (b) − R R 4Gm 2Gm (d) − (c) − R R 9. A rope is wound round a hollow cylinder of mass 3 kg and radius 40 cm. If the rope is pulled with a force of 30 N, angular acceleration of the cylinder will be (a) 10 rad s–2 (b) 15 rad s–2 –2 (c) 20 rad s (d) 25 rad s–2
(b) If both assertion and reason are true but reason is not the correct explanation of assertion (c) If assertion is true but reason is false (d) If both assertion and reason are false.
10. Two point masses A and B having masses in the ratio 4 : 3 are separated by a distance of 1 m. When another point mass C of mass M is placed in between A and B, the force between A and
15. Assertion : A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling motion). Reason : For perfect rolling motion, work done against friction is zero.
rd
⎛1⎞ C is ⎜ ⎟ of the force between B and C. Then the ⎝3⎠ distance of C from A is 1 2 1 2 m (d) m m (b) m (c) (a) 4 7 3 3 11. The diameter of a flywheel is 1 m. It has a mass of 20 kg. It is rotating about its axis with a speed of 120 rotations per minute. Its angular momentum in kg m2 s–1 is (a) 13.4 (b) 31.4 (c) 41.4 (d) 43.4 12. Two bodies of masses m1 and m2 are initially at rest and infinite distance apart from each other. Now, they are allowed to move towards each other under mutual gravitational attraction. Their relative velocity of approach at a distance r between them is (a)
2G(m1 + m2 ) r
(b)
(c)
G(m1 + m2 ) r
(d)
2G m1m2 (m1 + m2 )r G m1m2 (m1 + m2 )r
Assertion & Reason Type
Directions : In the following questions, a statement of assertion is followed by a statement of reason. Mark the correct choice as : (a) If both assertion and reason are true and reason is the correct explanation of assertion
13. Assertion : The time period of revolution of a satellite close to surface of earth is smaller than that revolving far from surface of earth. Reason : The square of time period of revolution of a satellite is directly proportional to cube of its orbital radius. 14. Assertion : Position of centre of mass is independent of the reference frame. Reason : Centre of mass is same as centre of gravity.
JEE MAIN / JEE ADVANCED / PETs Only One Option Correct Type
16. A uniform hollow sphere has internal radius a and external radius b. Taking the potential at infinity be zero, the ratio of the gravitational potential at a point on the outer surface to that on the inner surface is (a) (c)
2(b3 − a3 ) 3b(b2 − a2 ) 3b (b2 − a2 ) 2 (b3 − a3 )
(b) (d)
(b3 − a3 ) 3b(b2 − a2 ) a (b2 − a2 )
(b3 − a3 ) 17. A carpet of mass M, made of inextensible, material is rolled along its length in the form of a cylinder of radius R and is kept on a rough floor. The carpet starts unrolling without sliding on the floor when a negligibly small push is given to it. Calculate the horizontal velocity of the axis of the cylindrical part of the carpet when its radius reduces to R/2. 3 14 21 Rg (b) 3Rg (c) Rg Rg (d) 2 3 2 18. A circular plate of uniform thickness has a diameter of 56 cm. A circular portion of diameter 42 cm is removed from one edge of the plate as shown in the figure. Find the position of the centre of mass of the remaining portion. (a)
PHYSICS FOR YOU | OCTOBER ‘16
37
(a) 9 cm to right of centre of bigger circle (b) 2.5 cm to left of centre of bigger circle (c) 9 cm to left of centre of bigger circle (d) 4.2 cm to right of centre of bigger circle 19. A cockroach, mass m, runs counterclockwise around the rim of a lazy Susan (a circular dish mounted on a vertical axle) of radius R and rotational inertia I with frictionless bearings. The cockroach’s speed (relative to the earth) is v, whereas the lazy Susan turns clockwise with angular speed ω. The cockroach finds a bread crumb on the rim and, of course, stops. What will be the angular speed of the lazy Susan after the cockroach stops? mvR − I ω I ω − mvR (a) (b) 2 I + mR mR2 ω(2m + 1) (c) ω (d) m More than One Options Correct Type
20. A light thread with a body of mass m tied to its end is wound on a uniform solid cylinder of mass M and radius R as shown in the figure. At a moment t = 0 the system is set in motion. Assuming the friction in the axle of the cylinder to be negligible, at any time t (a) the angular velocity of the cylinder is gt ⎛ M⎞ ⎜⎝1 + ⎟ R 2m ⎠ (b) the angular velocity of the cylinder is gt ⎛ m ⎞ ⎜⎝1 + ⎟ R 2M ⎠ (c) the kinetic energy of the whole system is mg 2t 2 ⎛ M⎞ ⎜⎝1 + ⎟ 2 2m ⎠ (d) the kinetic energy of the whole system is m2 g 2t 2 2
⎛ 2m ⎞ ⎜⎝1 + ⎟ M⎠
21. A rocket is accelerated to speed of 2 gR near earth’s surface and then coast upward. (R is radius of the earth and g is acceleration due to gravity.) 38
PHYSICS FOR YOU | OCTOBER ‘16
(a) Rocket will orbit the earth. (b) Rocket will escape from the earth. (c) Speed of the rocket at very far from the earth is 2gR . (d) Speed of the rocket in the orbit is gR / 2 . 22. A thin uniform rod of mass m and length l is free to rotate about its upper end. When it is at rest, it receives an impulse J at its lowest point, normal to its length. Immediately after impact, (a) the angular momentum of the rod is Jl (b) the angular velocity of the rod is 3J/ml (c) the kinetic energy of the rod is 3J2/2m (d) the linear velocity of the midpoint of the rod is 3J/m. 23. A sphere of uniform density ρ has within it a spherical cavity whose centre is at distance a from the centre of the sphere. Then (a) The gravitational field within the cavity is nonuniform. (b) The gravitational field within the cavity is uniform. (c) The magnitude of gravitational field within the 4πGρa2 . 3r (d) The magnitude of gravitational field within the 4πGρa cavity is . 3 cavity is
Integer Answer Type
24. A bullet of mass m moving with velocity v0 (−k ) strikes the bottom of a stationary vertical uniform ring of same mass m and radius R = 1 m. The ring lies in xy plane with its topmost point hinged on the ceiling. The ring can rotate about x-axis. There is no friction between the hinge and the ring. The bullet gets embedded in the ring immediately after
collision. Find the angular velocity of the system (in rad s–1) just after collision. (Take v0 = 11 m s–1]
Column I (A) Angular momentum (P)
25. The gravitational potential energy of a satellite revolving around the earth in circular orbit is –4 MJ. Find the additional energy (in MJ) that should be given to the satellite so that it escapes from the gravitational field of earth. Assume earth’s gravitational force to be the only gravitational force on the satellite and no atmospheric resistance. 26. A solid disc is rolling with slipping on a level surface at constant speed of 5 m s–1. If the disc rolls up a 28° ramp, how far (in m) along the ramp will it move before it stops? (Given sin 28° = 0.4695) Comprehension Type
A rod of mass M and length L is suspended by a frictionless hinge at the point O as shown in figure. A bullet of mass m moving with velocity v in a horizontal direction strikes the end of the rod and gets embedded in it.
Column II Increases
(B) Area of earth covered (Q) Decreases by satellite signal (C) Potential energy (R) Becomes double (D) Kinetic energy (S) Becomes half (a) (b) (c) (d)
A S P P Q
B Q, S P S P
C Q P, S Q, S P, S
D S Q, S P R
30. In case of pure rolling of a rigid body of radius R on a stationary horizontal surface with an angular velocity ω and with v0 as the velocity of its centre of mass, match the entries in column I with those given in column II. Column I
27. The angular momentum of the system, about O before collision is (a) mvL (b) MvL 1 1 (c) mvL (d) MvL 2 2 28. The angular velocity acquired by the rod just after the collision is mv 2mv (a) (b) (3m + M )L ML 3mv 3mv (c) (d) L (3m + M )L Matrix Match Type
29. A satellite is in a circular equatorial orbit of radius 7000 km around the earth. If it is transferred to a circular orbit of double the radius then match the entries of column I with those given in column II.
Column II
(A) Distance moved by the CM of the body in one full rotation while slipping forward (B) The speed of the bottommost point on its circumference (C) The speed of the topmost point on its circumference (D) The speed of a point on the circumference at angle 90° with line joining the bottom-most point and the centre of the body (a) (b) (c) (d)
A S Q R P
B Q P S P
C R Q Q R
(P)
2 v0
(Q)
2v0
(R)
> 2πR
(S)
zero
D P R P S
Keys are published in this issue. Search now! ☺
Check your score! If your score is > 90%
EXCELLENT WORK !
You are well prepared to take the challenge of final exam. You can score good in the final exam. You need to score more next time
No. of questions attempted
……
90-75%
GOOD WORK !
No. of questions correct
……
74-60%
SATISFACTORY !
Marks scored in percentage
……
< 60%
NOT SATISFACTORY! Revise thoroughly and strengthen your concepts.
PHYSICS FOR YOU | OCTOBER ‘16
39
Maximize your chance of success, and high rank in NEET, JEE (Main and Advanced) by reading this column. This specially designed column is updated year after year by a panel of highly qualified teaching experts well-tuned to the requirements of these EntranceTests.
•
•
•
• •
•
Faraday’s Law of Induction First law : Whenever there is change in magnetic flux with respect to time for a coil or circuit, an emf induced in it and remains in it till change in flux takes place. Second law : The magnitude of the induced emf is directly proportional to the rate of change of flux through the coil. Mathematically, the induced emf is given by dφ ε=− B dt The negative sign indicates the direction of ε and hence the direction of current in the closed loop. Induced emf does not depend on nature of the coil i.e. resistance. Magnitude of induced emf is directly proportional to the relative speed of coil and magnet system. Lenz’s Law This law states that the direction of induced current in the coil is in such a way that it always opposes the cause by which it is produced.
dφ where negative sign indicates the Lenz’s law. dt Direction of induced emf (ε)
ε = (−) •
•
Some basic induced parameters in a circuit −Δφ f Average induced emf εav = Δt f Instantaneous induced emf
f
40
PHYSICS FOR YOU | OCTOBER ‘16
⎛ −Δφ ⎞ ε = lim ⎜ ⎟ = − dφ Δt →0 ⎝ Δt ⎠ dt Induced current flow at this instant in the ε 1 ⎛ dφ ⎞ closed circuit I = = ⎜ ⎟ R R ⎝ dt ⎠
f
In time interval dt, induced charge dq = Idt =
f
Induced heat H = ∫ I 2 Rdt = ∫
f
t
t 2
0
0
ε dt R
dφ R
• •
If any two out of v , B and dl become parallel or antiparallel, ε will become zero. Some special cases for motional emf : f
Induced electric field : A time varying magnetic field (dB/dt) always produces induced electric field in all space surrounding it. Induced emf ε = ∫ Ein ⋅ dl As ε = −
dφ dt
so ε = ∫ Ein ⋅ dl = −
f
dφ dt
If θ = 90° i.e. v , B and l are perpendicular to each other. Induced emf ε = Blv
Rotating straight conductor : emf induced in small element dx dε = (v × B) ⋅ dx = −vBdx
l
l
0
0
1 Net emf ε = − ∫ vBdx = − Bω∫ xdx = − Bωl 2 2
•
– It is non conservative and non electrostatic in nature. – Its field lines are concentric circular closed curves. Induced emf may exist in an open circuit, but there is no induced current and induced charge in the open circuit.
f
f
•
•
Motional Electromotive Force When a conducting rod moves in an external magnetic field in such a way it cuts the field. So all the free electrons in the rod transfer from one end to another end (C to D) and emf is induced in the rod. This emf is known as motional emf.
emf induced within dl dε = (v × B) ⋅ dl Net emf across the rod, ε = ∫ (v × B) ⋅ dl = vBl sin θ Induced electric field in the rod, Ein = −(v × B) = (B × v )
•
Negative sign indicates that end C will be at higher potential. A rotating conducting disc : Induced emf between centre C and circumference D is 1 ε = BωR2 2 A conductor of arbitrary shape : Induced emf in this conductor = Induced emf in straight conductor connected between C and D = Blv sin θ
Effect of Motional emf Developed in a Circuit For a given circuit, if the metal rod moves with uniform velocity v by an external agent then all the induced parameters are possible in the circuit.
•
Induced emf in the circuit ε = Bvl
•
Current flows through the circuit I =
ε Bvl = R R
PHYSICS FOR YOU | OCTOBER ‘16
41
•
Retarding opposing force exerted on metal rod by action of induced current Fm = I (l × B) ; Fm = BIl where θ = 90°
B 2l 2v R External mechanical force required for uniform velocity of metal rod For constant velocity resultant force on metal f rod must zero and for that Fext = Fm B 2l 2v f Fext = Fm = R f If (B, l, R) → constant ⇒ Fext ∝ v For uniform motion of metal rod, mechanical power delivered by external source is given as Pmech = Pext = Fext ⋅ v = Fext v
•
When current through a coil changes with respect to time dI dB dφ f → → ⇒ electromagnetic induction link dt dt dt dφ dI = −L , f Nφ = LI or, −N dt dt dφ where −N called self induced emf of coil εs. dt dI εs = −L dt
•
Self inductance always opposes the change of current in a electric circuit so it is also called inertia of the electric circuit. For a real inductor
Fm =
•
•
Pext = Pmech =
•
B 2l 2v 2 R
If (B, l, R) → constant ⇒ Pmech ∝ v2 Thermal power developed across resistor
f
•
2
•
⎛ Bvl ⎞ B l v Pth = I 2 R = ⎜ ⎟ R= ⎝ R ⎠ R f
•
•
22 2
It is clear that Pth = Pmech which is consistent with the principle of conservation of energy.
Self Induction When current through a coil changes with respect to the time then magnetic flux linked with the coil also changes with respect to time. Due to this an emf and a current induced in the coil. According to Lenz’s law induced current opposes the change in magnetic flux. This phenomenon is called self induction and a factor by virtue the coil shows opposition for change in magnetic flux called self inductance of coil. When current through a coil is constant,
I → B → φ → constant ⇒ No electromagnetic induction Total flux of coil (Nφ) ∝ current through the coil f f
Nφ ∝ I Nφ = LI N φ NBA φtotal , L= = = I I I where L is self inductance of the coil. 42
PHYSICS FOR YOU | OCTOBER ‘16
N
I + – V K
Rn
Role of L : To oppose the change in current. If current becomes constant then there is no role of L. Coefficient of self inductance of planar circular coil μ N 2 πR Lc = 0 2
•
Coefficient of self inductance of solenoid μ N 2A Ls = 0 = μ0n2 Al = μ0n2V l Here, V = volume of solenoid = Al A = area of cross section of frame of solenoid
•
Mutual Induction Whenever current passing through a primary coil or circuit changes with respect to time then magnetic flux in a neighbouring secondary coil or circuit will also changes with respect to time. According to Lenz’s law for opposition of flux change an emf and a current induced in the neighbouring coil or circuit. This phenomenon is called as mutual induction.
•
When current through primary coil is constant, f Total flux linked into secondary coil is directly proportional to the current flowing through the primary coil. f
•
N2 φ2 ∝ I1, N2 φ2 = MI1
N 2φ2 N 2 B1A2 (φT )s = = , I1 I1 Ip where M is mutual inductance of circuits. When current through primary coil changes with respect to time,
f
f
f
•
M=
dI1 dB dφ dφ → 1→ 1→ 2 dt dt dt dt ⇒ Electromagnetic induction link N2φ2 = MI1 or, −N 2
dφ2 dI = −M 1 , dt dt
Current growth in an LR circuit
Emf equation : V = IR + L
•
Current at any instant : When key is closed the current in circuit increases exponentially with respect to time. The current in circuit at any instant t is given by I = I0(1 – e–t/τ) Just after the closing of key, inductance behaves like open circuit and current in circuit is zero.
•
⎛ dI ⎞ εm = − M ⎜ 1 ⎟ ⎝ dt ⎠ •
Mutual inductance of two concentric and coplanar coils μ N N πr 2 Mc c = 0 1 2 2 1 2 2r1
•
Mutual inductance of two co-axial solenoids
•
μ NN A Ms s = 0 1 2 12 l Combination of Inductances Two inductors of inductances L1 and L2 are connected in series and are kept apart so that their mutual inductance is negligible, then the equivalent inductance of the combination is given by LS = L1 + L2
•
dI dt
•
dφ ⎞ ⎛ where ⎜ −N 2 2 ⎟ called total mutual induced emf dt ⎠ ⎝ of secondary coil εm.
Two inductors of inductances L1 and L2 are connected in parallel and are kept far apart so that mutual inductance between them is negligible, then their equivalent inductance is given by LL 1 1 1 = + or L = 1 2 L L1 L2 L1 + L2
Two inductors of self-inductance L1 and L2 are connected in series and they have mutual inductance M, then the equivalent inductance of the combination is given by
•
Some time after closing of the key inductance behaves like short circuit and current in circuit is constant.
f
I0 =
V (maximum or peak value of current) R
Peak value of current in circuit does not depend on self inductance of coil. Time constant : It is a time in which current increases up to 63% or 0.63 times of peak current L value. τ = R
f
•
L = L1 + L2 ± 2M The plus sign occurs if windings in the two coils are in the same sense, while minus sign occurs if windings are in opposite sense. PHYSICS FOR YOU | OCTOBER ‘16
43
•
Current decay in an LR circuit dI Emf equation : IR + L = 0 dt
•
Average value or mean value f The mean value of ac over any half cycle is that value of dc which would send same amount of charge through a circuit as is sent by the ac through same circuit in the same time. Average value of current for half cycle f T /2
= •
Current at any instant : Once current acquires its final maximum steady value, if suddenly required switching positions (S1 and S2) are interchanged then current starts decreasing exponentially with respect to time. The current in the circuit at any instant t is given by
•
44
PHYSICS FOR YOU | OCTOBER ‘16
dt
T /2
Iav •
∫ = 0
I0 sin ωt dt T /2
∫0
dt
Irms =
•
•
2I0 2I [− cos ωt ]T0 /2 = 0 ωT π
T
T 2
•
=
Root mean square (rms) value f It is that value of dc which would produce same heat in a given resistance in a given time as is done by the alternating current when passed through the same resistance for the same time.
∫0 I dt T ∫0 dt
=
∫0 (I0 sin ωt ) T ∫0 dt
2
dt
I 1 T ⎡1 − cos 2ωt ⎤ dt = 0 ∫ ⎢ ⎥ 0 2 T ⎣ ⎦ 2
rms value = virtual value = apparent value Phase : I = I0 sin (ωt ± φ) f Initial phase = φ (it does not change with time) f Instantaneous phase = ωt (it changes with time) Phase difference : V = V0 sin (ωt + φ1), I = I0 sin (ωt + φ2) f Phase difference of I with respect to V φ = φ2 – φ1 f Phase difference of V with respect to I φ = φ1 – φ2 V leads I or I lags V → It means, V reaches maximum before I. Let if V = V0 sin ωt then I = I0 sin (ωt – φ) and if V = V0 sin (ωt + φ) then I = I0 sin ωt f
•
∫
Average value of I = I0 sin ωt over the positive half cycle :
= I0
Fundamental of Alternating Current Voltage or current is said to be alternating if it changes continuously magnitude and periodic in direction with the time. It can be represented by a sine curve or cosine curve. I = I0 sin ωt or I = I0 cos ωt I = Instantaneous value of current at time t I0 = Amplitude or peak value 2π ω = Angular frequency (rad s–1) = = 2πυ T T = Time period, υ = Frequency
Idt
0 T /2 0
f
I = I0 e–t/τ
V f Just after opening of key t = 0 ⇒ I = I0 = R f Some time after opening of key t → ∞ ⇒ I0 → 0 Time constant (τ) : It is a time in which current decreases up to 37% or 0.37 times of peak current L value τ = R
∫
•
•
V lags I or I leads V → It means, V reaches maximum after I Let if V = V0 sin ωt then I = I0 sin (ωt + φ) and if V = V0 sin (ωt – φ) then I = I0 sin ωt Components L, C and R in ac circuit separately
R
Term
L
C
Circuit
Supply voltage
V = V0 sin ωt
V = V0 sin ωt
V = V0 sin ωt
Current
I = I0 sin ωt
π⎞ ⎛ I = I0 sin ⎜ ωt − ⎟ 2⎠ ⎝
π⎞ ⎛ I = I0 sin ⎜ ωt + ⎟ 2⎠ ⎝
Peak current
V I0 = 0 R
V I0 = 0 ωL
I0 =
Impedance (Ω)
V0 =R I0
V0 = ωL = X L I0
V0 1 = = XL I 0 ωC
XL = Inductive reactance π + (V leads I) 2
XC = Capacitive reactance π − (V lags I) 2
1 XL Inductive susceptance L passes dc easily (because XL = 0) while gives a high impedance for the ac of high frequency (XL ∝ υ)
1 XC Capacitive susceptance
V V Z = 0 = rms I0 Irms
R = Resistance Phase difference and zero Phasor diagram (V and I are in same phase)
Variation of Z with υ
V0 1 / ωC
R does not depend on υ
G, SL, SC (mho, seiman)
G=
1 = conductance R
Behaviour of device Same in ac and dc in dc and ac
Ohm’s law
VR = IR
SL =
VL = IXL
SC =
C blocks dc (because XC = ∞) while provides an easy path for the ac of high frequency 1⎞ ⎛ ⎜ XC ∝ ⎟ υ⎠ ⎝ VC = IXC
Combination of R – L, R – C and L – C in an ac circuit
Term
R–L
R–C
L–C
I is same in R and L
I is same in R and C
I is same in L and C
Circuit
PHYSICS FOR YOU | OCTOBER ‘16
45
Phasor diagram
V Phase difference between V and I
2
= VR2
+ VL2
π⎞ ⎛ V leads I ⎜ 0 to ⎟ 2⎠ ⎝
V 2 = VR2 + VC2 ⎞ ⎛ π V lags I ⎜ − to 0 ⎟ ⎝ 2 ⎠
V = VL – VC
(VL > VC)
V = VC – VL
(VC > VL)
⎞ ⎛ π V lags I ⎜ − , if XC > X L ⎟ ⎝ 2 ⎠ ⎞ ⎛ π V leads I ⎜ + , if X L > XC ⎟ ⎝ 2 ⎠
Impedance
Z = R2 + X L2
Z = R2 + XC2
Z = X L − XC Z
XC
Z
XL
At very low υ
Z
R ( X L → 0)
Z
XC
At very high υ
Z
XL
Z
R
•
Series LCR Circuit Circuit diagram
•
I is same for R, L and C. Phasor diagram
•
•
•
•
f
If VL > VC then
f
If VC > VL then
f
48
applied voltage. For resonance both L and C must be present in circuit. At resonance, f XL = XC , VL = VC f φ = 0 (V and I are in same phase) V f Zmin = R, I max = R 1 Resonance frequency : ... XL = XC ⇒ ωL = ωC 1 1 ω= or, υr = LC 2π LC Variation of Z with υ If υ < υr then XL < XC , circuit is capacitive, φ f (negative). f At υ = υr , XL = XC, circuit is resistive, φ = zero. If υ > υr then XL > XC circuit is inductive, φ f (positive). As υ increases, Z first decreases then increases,
V = VR2 + (VL − VC )2 , Z = R2 + ( X L − XC )2 Impedance triangle X − XC VL − VC tanφ = L = R VR
•
( XC → 0)
•
Variation of I with υ
V = IZ, VR = IR, VL = IXL , VC = IXC
Resonance in Series LCR Circuit A circuit is said to be resonant when the natural frequency of the circuit is equal to frequency of the PHYSICS FOR YOU | OCTOBER ‘16
as υ increases, I first increases then decreases.
•
At resonance impedance of the series resonant circuit is minimum so it is called acceptor circuit as it most readily accepts that current out of many currents whose frequency is equal to its natural frequency. In radio or TV tuning we receive the desired station by making the frequency of the circuit equal to that of the desired station.
•
Band width Δυ = υ2 – υ1
•
Quality factor (Q) : Q-factor of ac circuit basically gives an idea about stored energy and lost energy. maximum energy stored per cycle Q = 2π maximum energy lost per cycle f It represents the sharpness of resonance. f It is unitless and dimensionless quantity.
( XL )r
Q=
f
υr 1 L υr = = R C Δυ band width Sharpness ∝ quality factor R decrease ⇒ Q increases ⇒ sharpness increases
R
=
( XC )r 2πυr L = R R
f
=
•
•
Power in ac circuit Let V = V0 sin ωt and I = I0 sin (ωt – φ) Instantaneous power P = V0 sin ωt . I0 sin (ωt – φ) = V0 I0 sin ωt (sin ωt cos φ – sinφ cos ωt) Average power
• •
• •
•
• •
• •
• •
•
π⎞ ⎛ circuit ⎜∵φ = ⎟ . This current is refered as wattless ⎝ 2⎠ current. Power is dissipated only in resistor even circuit has RL, RC or LCR combination.
Resistance of an ideal coil is zero. LC Oscillation The oscillation of energy between capacitor (electric field energy) and inductor (magnetic field energy) is called LC-oscillation. 1 Frequency of oscillation υ = 2π LC If charge varies sinusoidally with time t as q = q0 cos ωt then current varies periodically with t π⎞ dq ⎛ as I = = q0ω cos ⎜ ωt + ⎟ 2⎠ dt ⎝ If initial charge on the capacitor is q0 then electrical
1 q02 2C If the capacitor is fully discharged, then total electrical energy is stored in the inductor in the form of magnetic energy. 1 U B = LI02 where I0 = Maximum current 2 Transformer A transformer is an electrical device which is used for changing alternating voltages. It is based on the phenomenon of mutual induction. If it is assumed that there is no loss of energy in the transformer then the power input = the power output, and since P = I × V then IPVP = ISVS Although some energy is always lost, this is a good approximation, since a well designed transformer may have an efficiency of more than 95%. I P VS N S = = I S VP N P A transformer affects the voltage and current. We have energy stored in capacitor is U E =
•
T
1 (V I sin2 ωt cos φ − V0 I0 sin ωt cos ωt sin φ)dt T∫ 0 0 0 V I cos φ = 0 0 ⇒ < P > = Vrms Irms cos φ 2 rms power Prms = Vrms Irms Average power Power factor (cos φ) = rms power R cos φ = Z Power dissipation is maximum in resistive circuit or at resonance in a LCR series circuit. No power is dissipated in purely inductive or capacitive circuit even a current is flowing in the
=
•
Choke Coil Circuit with a choke coil is a series L – R circuit. If resistance of choke coil = r (very small). Current in V the circuit, I = with Z = (R + r )2 + (ωL)2 Z It has a high inductance and negligible resistance coil. It is used to control current in ac circuit at negligible power loss. V r r ≈ →0 ∴ cos φ = = 2 2 2 Z ω L r +ω L
•
•
•
•
PHYSICS FOR YOU | OCTOBER ‘16
49
⎛N VS = ⎜ S ⎝ NP •
•
•
•
⎞ ⎛ NP ⎟VP and I S = ⎜ N ⎠ ⎝ S
⎞ ⎟ IP ⎠
Now, if NS > NP, the voltage is stepped up (VS > VP). This type of arrangement is called a stepup transformer. If NS < NP, we have a step-down transformer. In this case VS < VP and IS > IP. The voltage is stepped down and the current is increased. Displacement Current Maxwell assumed that a current also flows in the gap between the two plates of a capacitor, during the process of charging, known as displacement current ID. This displacement current originates due to time varying electric field between the plates of capacitor and is given by dφ I D = ε0 ε dt where φε is the electric flux linked with the space between the two plates of the capacitor. Using the concept of displacement current ID, Ampere’s circuital law can be modified as
∫ B ⋅ dl = μ0 (IC + I D ) •
f f
•
•
•
50
•
∫ B ⋅ dA = 0 (Gauss’s law for magnetism) dφ ∫ E ⋅ dl = − dtB (Faraday’s law)
•
•
•
dφ E dt (Ampere-Maxwell law)
Velocity of electromagnetic waves in free space is given by, 1 c= = 3 × 108 m s −1 μ0 ε0 The instantaneous magnitude of the electric and magnetic field vectors in electromagnetic wave are PHYSICS FOR YOU | OCTOBER ‘16
•
με μ 0 ε0
The energy is equally shared between electric field and magnetic field vectors of electromagnetic wave. Therefore the energy density of the electric field, 1 uE = ε0 E 2 and 2 1 B2 . the energy density of magnetic field, uB = 2 μ0 Average energy density of the electric field, 1 < uE > = ε0 E02 and average energy density of the 4 B2 1 magnetic field < uB > = 0 = ε0 E02 . 4μ 0 4 Average energy density of electromagnetic wave is =
∫ B ⋅ dl = μ0 IC + μ0ε0
Electromagnetic Waves An electromagnetic wave is a wave radiated by an accelerated charge which propagates through space as coupled electric and magnetic fields, oscillating perpendicular to each other and to the direction of propagation of the wave.
In a medium of refractive index n, the velocity v of an electromagnetic wave is given by 1 c 1 1 , Also v = v= = ⋅ n n μ0 ε0 με So that n =
•
Maxwell’s equations: q f ∫ E ⋅ dA = ε0 (Gauss’s law for electricity) f
related as |E| = c or E = Bc |B|
B2 1 ε0 E02 = 0 . 2 2μ0
Intensity of electromagnetic wave is defined as energy crossing per unit area per unit time perpendicular to the directions of propagation of electromagnetic wave. The intensity I is given by the relation 1 I = < u > c = ε0 E02c 2 The electromagnetic wave also carries linear momentum with it. The linear momentum carried by the portion of wave having energy U is given by p = U/c.
•
If the electromagnetic wave incident on a material surface is completely absorbed, it delivers energy U and momentum p = U/c to the surface.
•
If the incident wave is totally reflected from the surface, the momentum delivered to the surface is U/c – (–U/c) = 2U/c. It follows that the electromagnetic wave incident on a surface exerts a force on the surface.
1. Two coils have a mutual inductance 0.005 H. The current changes in the first coil according to equation I = I0 sin ωt, where I0 = 10 A and ω = 100 π rad s–1. The maximum value of emf in the second coil is (in V) (a) 2π (b) 5π (c) 6π (d) 12π 2. An express train takes 16 hours to cover the distance of 960 km between Patna and Gaziabad. The rails are separated by 130 cm and the vertical component of the earth’s magnetic field is 4.0 × 10–5 T. If the leakage resistance between the rails is 100 Ω, the retarding force on the train due to the magnetic field will be (a) 5 × 10–10 N (b) 8 × 10–10 N –5 (c) 15 × 10 N (d) 5 × 10–5 N 3. An inductor of inductance L = 400 H and resistors of resistances R1 = 2 Ω and R2 = 2 Ω are connected to a battery of emf 12 V as shown in figure. The internal resistance of the battery is negligible. The switch S is closed at t = 0. The potential drop across L as a function of time is 12 –3t (a) 6e–5t V (b) e V t (c) 6(1 – e–t/0.2) V (d) 12e–5t V 4. A uniform magnetic field B exists in a direction perpendicular to the plane of a square frame made of copper wire. The wire has a diameter of 2 mm and a total length of 40 cm. The magnetic field changes with time at a steady rate dB/dt = 0.02 T s–1. What will be the current induced in the frame? (Resistivity of copper = 1.7 × 10–8 Ω m) (a) 0.1 A
(b) 0.2 A
(c) 0.3 A
(d) 0.4 A
5. In a uniform magnetic field of induction B, a wire in the form of semicircle of radius r rotates about the diameter of the circle with angular frequency ω. The axis of rotation is perpendicular to the field. If the total resistance of the circuit is R, then mean power generated per period of rotation is
Bπr 2 ω (Bπr 2 ω)2 (b) 2R 8R 2 (Bπr ω) (Bπr ω2 )2 (c) (d) 2R 8R 6. If a resistance of 100 Ω, an inductance of 0.5 H and a capacitance of 10 × 10–6 F are connected in series through 50 Hz ac supply, the impedance will be (a)
(a) 1.87 Ω (c) 18.7 Ω
(b) 101.3 Ω (d) 189.7 Ω
7. Two circular loops of equal radii are placed coaxially at some separation. The first loop is cut and a battery is inserted in between to drive a current in it. The current changes slightly because of the variation in resistance with temperature. During this period, the two loops (a) attract each other (b) repel each other (c) do not exert any force on each other (d) attract or repel each other depending on the sense of the current. 8. The instantaneous values of alternating current and voltage in a circuit are given as 1 1 I= sin (100πt) A and ε = sin (100πt + π/3) V 2 2 The average power in watts consumed in the circuit is 1 3 1 1 (b) (c) (d) 4 4 2 8 9. An electromagnetic wave of frequency 3 MHz passes from vacuum into a dielectric medium with permittivity εr = 4. Then (a) the wavelength and frequency both remain unchanged (b) the wavelength is doubled and the frequency remains unchanged (c) the wavelength is doubled and the frequency becomes half (d) the wavelength is halved and the frequency remains unchanged. (a)
PHYSICS FOR YOU | OCTOBER ‘16
51
10. The r.m.s. value of the electric field of the light coming from the sun is 720 N C–1. The average total energy density of the electromagnetic wave is (a) 3.3 × 10–3 J m–3 (b) 4.58 × 10–6 J m–3 (c) 6.37 × 10–9 J m–3 (d) 81.35 × 10–12 J m–3 11. The amplitude of the electric field in a parallel beam of light of intensity 2.0 W m–2 is (a) 38.8 N C–1 (b) 49.5 N C–1 –1 (c) 32.7 N C (d) 35.5 N C–1 12. A free electron is placed in the path of a plane electromagnetic wave. The electron will start moving (a) along the electric field (b) along the magnetic field (c) along the direction of propagation of the wave (d) in a plane containing the magnetic field and the direction of propagation. 13. An inductor 20 mH, a capacitor 50 μF and a resistor 40 Ω are connected in series across a source of emf V = 10 sin 340t. The power loss in ac circuit is (a) 0.76 W (b) 0.89 W (c) 0.51 W (d) 0.67 W [NEET Phase I 2016] 14. A small signal voltage V(t) = V0 sinωt is applied across an ideal capacitor C (a) Current I(t) is in phase with voltage V(t) (b) Current I(t) leads voltage V(t) by 180° (c) Current I(t), lags voltage V(t) by 90° (d) Over a full cycle the capacitor C does not consume any energy from the voltage source. [NEET Phase I 2016] 15. A long solenoid has 1000 turns. When a current of 4 A flows through it, the magnetic flux linked with each turn of the solenoid is 4 × 10–3 Wb. The self-inductance of the solenoid is (a) 2 H (b) 1 H (c) 4 H (d) 3 H [NEET Phase I 2016] 16. Out of the following options which one can be used to produce a propagating electromagnetic wave? (a) A chargeless particle (b) An accelerating charge (c) A charge moving at constant velocity (d) A stationary charge [NEET Phase I 2016] 17. An arc lamp requires a direct current of 10 A at 80 V to function. If it is connected to a 220 V (r.m.s.), 52
PHYSICS FOR YOU | OCTOBER ‘16
50 Hz ac supply, the series inductor needed for it to work is close to (a) 80 H (b) 0.08 H (c) 0.044 H (d) 0.065 H [JEE Main Offline 2016] 18. A series LR circuit is connected to a voltage source with V(t) = V0 sinωt. After very large time, current ⎛ ⎝
I(t) behaves as ⎜ t0 >>
L⎞ R ⎠⎟
(a)
(b)
(c)
(d) [JEE Main Online 2016]
19. A conducting metal circular-wire-loop of radius r is placed perpendicular to a magnetic field which −
t
varies with time as B = B0e τ , where B0 and τ are constants, at time t = 0. If the resistance of the loop is R then the heat generated in the loop after a long time (t → ∞) is (a)
π 2r 4 B04 2 τR
(b)
π 2r 4 B02 2 τR
(c)
π 2r 4 B02 R τ
(d)
π 2r 4 B02 τR
[JEE Main Online 2016] 20. Microwave oven acts on the principle of (a) giving rotational energy to water molecules (b) giving translational energy to water molecules (c) giving vibrational energy to water molecules (d) transferring electrons from lower to higher energy levels in water molecule. [JEE Main Online 2016] SOLUTIONS d dI 1. (b) : As |ε| = M = M (I0 sin ωt) = MI0 ω cos ωt, dt dt εmax = MI0ω = (0.005)(10)(100π) V = 5π V 2. (a) : As the train moves in a magnetic field, a motional emf ε = vBl is produced across its width. Here B is the component of the magnetic field in a direction perpendicular to the plane of the motion, i.e., the vertical component.
960 km = 16.67 m s −1 16 h Thus, ε = (16.67 m s–1) (4.0 × 10–5 T)(1.3 m) = 8.6 × 10–4 V The leakage current is I = ε/R and the retarding force is The speed of the train is v =
F = IlB =
8.6 × 10−4 V × 1.3 m × 4.0 × 10−5 T 100 Ω
= 4.47 × 10–10 N 5 × 10−10 N 3. (d) : If I1 is the current through R1 and I2 is the ε current through L and R2, then I1 = and R1 I2 = I0(1 – e–t/), L 400 × 10 −3 = 0.2 s = Where τ = R2 2 ε 12 =6A and I0 = = R2 2 Thus, I2 = 6(1 – e–t/0.2) Potential drop across L, i.e., ε – R2I2 = 12 V – 2 × 6(1 – e–t/0.2) V = (12e–5t) V 4. (a) : Here, total length l = 40 cm = 40 × 10–2 m, Resistivity = 1.7 × 10–8 Ω m The area A of the loop ⎛ 40 cm ⎞ ⎛ 40 cm ⎞ 2 =⎜ ⎟⎜ ⎟ = 0.01 m ⎝ 4 ⎠⎝ 4 ⎠ If the magnetic field at an instant is B, the flux through the frame at that instant will be φ = BA. As the area remains constant, the magnitude of the emf induced will be dφ dB ε= =A dt dt = (0.01 m) (0.02 T s–1) = 2 × 10–4 V The resistance of the loop is (1.7 × 10−8 Ω m)(40 × 10−2 m) 3.14 × 1 × 10−6 m2
= 2.16 × 10−3 Ω
Hence, the current induced in the loop will be I=
2 × 10−4 V −3
2.16 × 10 Ω
= 9.3 × 10−2 A ≈ 0.1 A
⎛1 2⎞ 5. (b) : As φB = BA cos ωt = B ⎜ πr ⎟ cos ωt, ⎝2 ⎠ dφ B 1 ε=– = Bπr2ω sin ωt dt 2
Instantaneous power, P = T
Pav =
∫0
Pdt T
=
ε2 (Bπr 2 ω)2 = sin2 ωt R 4R
(Bπr 2 ω)2 (T / 2) 4R T T (as ∫0 sin2 ωt = T / 2)
(Bπr 2 ω)2 8R 6. (d) : As XL = 2πυL = 2π (50)(0.5) = 157.1 Ω, 1 1 = Ω = 318.4 Ω XC = 2πυC 2π(50)(10 −5 ) =
|XL – XC| = 161.3 Ω Z = R2 + ( X L − XC )2 = (100)2 + (161.3)2 Ω = 189.7 Ω 7. (a) 8. (d) : As εrms = I0
ε0
(1 / 2 )
=
2 (1 / 2 )
2
=
1 V, 2
1 A, 2 2 2 1 and cosφ = cos π/3 = 2 ⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ 1 Pav = εrmsIrms cosφ = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = W ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠ 8 Irms =
=
=
9. (d) : Frequency remains unchanged with change of medium. c 1 / ε0 μ0 = εr μr Refractive index, n = = v 1 / εμ Since μr is very close to 1, n = εr = 4 = 2 λ λ Thus, λmedium = = n 2 1 1 2 10. (b) : uav = ε0 E02 = ε0 ( 2 Erms )2 = ε0 Erms 2 2 E (as Erms = 0 ) 2 = (8.85 × 10–12)(720)2 = 4.58 × 10–6 J m–3 11. (a) : The intensity of a plane electromagnetic wave is 1 I = uav c = ε0E02c 2 2I 2 × 2. 0 or E0 = = ε0 c 8.85 × 10 −12 × 3 × 108 = 38.8 N C –1 PHYSICS FOR YOU | OCTOBER ‘16
53
12. (a) 13. (c) : Here, L = 20 mH = 20 × 10–3 H, C = 50 μF = 50 × 10–6 F R = 40 Ω, V = 10 sin 340t = V0 sinωt ω = 340 rad s–1, V0 = 10 V XL = ωL = 340 × 20 × 10–3 = 6.8 Ω 1 1 104 XC = = = = 58.82 Ω ωC 340 × 50 × 10−6 34 × 5
Arc lamp will glow if I = 10 A, ∴
15. (b) : Here, N = 1000 , I = 4 A , φ0 = 4 × 10–3 Wb Total flux linked with the solenoid, φ = Nφ0 = 1000 × 4 × 10–3 Wb = 4 Wb Since, φ = LI ∴ Self-inductance of solenoid, φ 4 Wb L= = =1 H I 4A 16. (b) : An accelerating charge is used to produce oscillating electric and magnetic fields, hence the electromagnetic wave. 17. (d) : For a dc source I = 10 A, V = 80 V Resistance of the arc lamp, V 80 R= = =8 Ω I 10 For an ac source, εrms = 220 V υ = 50 Hz ω = 2π × 50 = 100 π rad s–1
54
PHYSICS FOR YOU | OCTOBER ‘16
εrms
ε or R2 + ω2 L2 = ⎜⎛ rms ⎟⎞ 2 2 2 ⎝ I ⎠ R +ω L
⎛ 220 ⎞ or 82 + (100 π)2 L2 = ⎜ ⎝ 10 ⎟⎠ 2
2
2
2
22 − 8 or L2 = (100 π)2
2 2 Z = R2 + (XC − X L )2 = (40) + (58.82 − 6.8)
= (40)2 + (52.02)2 = 65.62 Ω The peak current in the circuit is V 10 40 ⎞ R I0 = 0 = A , cos φ = = ⎜⎛ ⎟ Z 65.62 Z ⎝ 65.62 ⎠ Power loss in ac circuit, 1 = Vrms Irms cos φ = V0 I0 cos φ 2 1 10 40 = × 10 × × = 0.46 W 2 65.62 65.62 14. (d) : When an ideal capacitor is connected with an ac voltage source, current leads voltage by 90°. Since, energy stored in capacitor during charging is spent in maintaining charge on the capacitor during discharging. Hence over a full cycle, the capacitor does not consume any energy from the voltage source.
I=
∴
L=
30 × 14 = 0.065 H 100 π
18. (d) : Current in LR circuit is I=
π⎞ ⎛ sin ⎜ ωt − ⎟ 2⎠ ⎝ R2 + ω2 L2 V0
i.e., it is sinusoidal in nature. −
t
19. (b) : Here, B = B0e τ Area of the circular loop, A = πr2 Flux linked with the loop at any time, t, −
t
φ = BA = πr 2 B0e τ t
dφ 1 − Emf induced in the loop, ε = − = πr 2 B0 e τ dt τ Net heat generated in the loop ∞ 2
∞
0
0
2t
π2r 4 B02 − τ ε = ∫ dt = ∫ e dt R τ2 R =
π2r 4 B02
=
−π2r 4 B02
τ2 R
2τ2 R
×
1 2t ×⎡ − τ ⎢ ⎛ 2 ⎞ ⎣e ⎜− τ ⎟ ⎝ ⎠ × τ(0 − 1) =
∞
⎤ ⎥⎦ 0
π2r 4 B02 2τR
20. (a)
ANSWER KEY
MPP-4 CLASS XI 1. 6. 11. 16. 21. 26.
(d) (c) (b) (a) (b,c) (4)
2. 7. 12. 17. 22. 27.
(a) (c) (a) (d) (a,b,c) (a)
3. 8. 13. 18. 23. 28.
(c) (c) (a) (c) (b,d) (d)
4. 9. 14. 19. 24. 29.
(d) (d) (c) (a) (4) (b)
5. 10. 15. 20. 25. 30.
(a) (a) (b) (a,c) (2) (c)
CLASS XII Series 5
Optics
Time Allowed : 3 hours Maximum Marks : 70
GENERAL INSTRUCTIONS (i)
All questions are compulsory.
(ii)
Q. no. 1 to 5 are very short answer questions and carry 1 mark each.
(iii) Q. no. 6 to 10 are short answer questions and carry 2 marks each. (iv) Q. no. 11 to 22 are also short answer questions and carry 3 marks each. (v)
Q. no. 23 is a value based question and carries 4 marks.
(vi) Q. no. 24 to 26 are long answer questions and carry 5 marks each. (vii) Use log tables if necessary, use of calculators is not allowed.
SECTION-A 1. For what angle of incidence, the lateral shift produced by a parallel sided glass slab is maximum? 2. Light of wavelength 6000 Å in air enters a medium of refractive index 1.5. What will be its frequency in the medium? 3. No interference pattern is detected when two coherent sources are infinitely close to one another. Why? 4. Differentiate a ray and a wavefront. 5. Why are mirrors used in searchlights (or car headlights) parabolic in shape and not concave spherical? SECTION-B
6. A light ray travels from medium 1 of refractive index μ1 to medium 2 of refractive index μ2, where μ2 < μ1. Write an expression for critical angle of incidence. 7. A convex lens made of material of refractive index μ2 56
PHYSICS FOR YOU | OCTOBER ‘16
Previous Years Analysis 2016 2015 2014 Delhi AI Delhi AI Delhi AI VSA SA-I SA-II VBQ LA
– 1
1 1
1 1
2 _
2 _
2 _
1
1
1
1 1
1 1
1 1
2
2 _
2 _
1
1
– 1
is held in a reference medium of refractive index μl. Trace the path of a parallel beam of light passing through the lens when (a) μl = μ2 , (b) μ1 < μ2 and (c) μl > μ2 . 8. What is the focal length of a combination of a convex lens of focal length 30 cm and a concave lens of focal length 20 cm ? Is the system a converging or diverging lens? Ignore thickness of the lenses. 9. In a double-slit interference experiment, the two coherent beams, have slightly different intensities I and (I + δI) where δI lLi and EH > ELi (b) lH = lLi and EH < ELi (c) lH = lLi and EH > ELi (d) lH < lLi and EH < ELi 5. The ratio of the longest and shortest wavelengths in Brackett series of hydrogen spectra is 9 4 25 17 (a) (b) (c) (d) 5 3 9 6 6. The electric potential between a proton and an ⎛r⎞ electron is given by V = V0 ln ⎜ ⎟ , where r0 is a ⎝ r0 ⎠
constant. Assuming Bohr’s model to be applicable, write variation of rn with n, n being the principal quantum number. 1 (a) rn ∝ n (b) rn ∝ n 1 (c) rn ∝ n2 (d) rn ∝ 2 n 7. A hydrogen atom is excited up to 9th level. The total number of possible spectral lines emitted by the hydrogen atom is (a) 36 (b) 35 (c) 37 (d) 38 8. The radius of the hydrogen atom in its ground state is a0. The radius of a muonic hydrogen atom in which the electron is replaced by an identically charged muon with mass 207 times that of an electron, is aμ equal to a0 a0 (c) (d) a0 207 (a) 207a0 (b) 207 207 9. Which energy state of doubly ionised lithium has the same energy as that of the ground state of hydrogen ? (Given Z for lithium = 3) (a) 4 (b) 3 (c) 2 (d) 1 10. When the electron in hydrogen atom is excited from the 4th stationary orbit to the 5th stationary orbit, the change in the angular momentum of the electron is (Planck’s constant, h = 6.63 × 10–34 J s) (a) 4.16 × 10–34 J s (b) 3.32 × 10–34 J s (c) 1.05 × 10–34 J s (d) 2.08 × 10–34 J s 257 11. If the atom 100 Fm follows the Bohr model and the 257 Fm is N times the Bohr radius of fifth orbit of 100 radius, then the value of N is (a) 100 (b) 200 (c) 4 (d) 1/4 PHYSICS FOR YOU | OCTOBER ‘16
65
12. The angular momentum of electron in 3d orbital of an atom is ⎛ k ⎞ (a) 2 ⎜ ⎟ (b) 3 ⎜⎛ k ⎞⎟ ⎝ 2π ⎠ ⎝ 2π ⎠ ⎛ k ⎞ 6⎜ ⎟ ⎝ 2π ⎠
⎛ k ⎞ 12 ⎜ ⎟ ⎝ 2π ⎠ 13. Taking the Bohr radius as a0 = 53 pm, the radius of Li++ ion in its ground state, on the basis of Bohr’s model, will be about (a) 53 pm (b) 27 pm (c) 18 pm (d) 13 pm (c)
(d)
14. An alpha particle of energy 5 MeV is scattered through 180° by a fixed uranium nucleus. The distance of closest approach is of the order of (a) 1 Å (b) 10–10 cm –12 (c) 10 cm (d) 10–15 cm 15. The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 Å. The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom is (a) 1215 Å (b) 1640 Å (c) 2430 Å (d) 4687 Å NUCLEI
16. In the following reaction, the energy released is 1 4 4 1H → 2 He + 2e+ + Energy Given
1 : Mass of 1H = 1.007825 u 4 Mass of 2 He = 4.002603 u +
Mass of e = 0.000548 u
(a) 12.33 MeV (c) 25.7 MeV
(b) 24.67 MeV (d) 49.34 MeV
17. A radioactive isotope A with a half life of 1.25 × 1010 years decays into B which is stable. A sample of rock from a planet is found to contain both A and B present in the ratio 1 : 16. The age of the rock is (a) 9.6 × 1010 years (b) 4.2 × 1010 years (c) 5 × 1010 years (d) 1.95 × 1010 years 18. A heavy nucleus at rest breaks into two fragments which fly off with velocities in the ratio 3 : 1. The ratio of radii of the fragments is (a) 1 : 31/3 (b) 31/3 : 4 (c) 4 : 1
(d) 2 : 1
19. A radioactive nucleus emits 3α-particles and 5β-particles. The ratio of number of neutrons to that of protons will be 66
PHYSICS FOR YOU | OCTOBER ‘16
A−Z A − Z − 12 (b) Z −1 Z −6 A − Z − 11 (d) (c) A − Z − 11 Z −1 Z −6 20. A radioactive sample has half life of 5 days. To decay from 8 microcurie to 1 microcurie, the number of days taken will be (a) 40 (b) 25 (c) 15 (d) 10 21. A free neutron decays spontaneously into (a) a proton, an electron and anti-neutrino (b) a proton, an electron and a neutrino (c) a proton and electron (d) a proton, and electron, a neutrino and an anti-neutrino. 22. The radius of germanium (Ge) nuclide is measured 9 to be twice the radius of 4Be. The number of nucleons in Ge are (a) 72 (b) 73 (c) 74 (d) 75 (a)
23. Fusion reaction takes place at high temperature because (a) nuclei break up at high temperature (b) atoms get ionised at high temperature (c) kinetic energy is high enough to overcome the coulomb repulsion between nuclei (d) molecules break up at high temperature. 24. To generate power of 3.2 MW, the number of fissions of 235U per minute is (Energy released per fission = 200 MeV, 1 eV = 1.6 × 10–19 J) (a) 6 × 1018 (b) 6 × 1017 17 (c) 10 (d) 6 × 1016 232
25. The 90Th atom has successive alpha and beta decays 208 to the end product 82Pb. The numbers of alpha and beta particles emitted in the process respectively are (a) 4, 6 (b) 4, 4 (c) 6, 2 (d) 6, 4 26. An element X decays first by positron emission and then two α-particles are emitted in successive radioactive decay. If the product nucleus has mass number 227 and atomic number 89, the mass number and atomic number of element X are (a) (273, 93) (b) (235, 94) (c) (238, 93) (d) (237, 92) 27. A radioactive isotope has a decay constant λ and a molar mass M. Taking the Avogadro constant to be L, what is the activity of a sample of mass m of this isotope?
λmL M mλ mL (c) (d) ML λM 28. In a nuclear reactor, moderators slow down the neutrons which come out in a fission process. The moderator used have light nuclei. Heavy nuclei will not serve the purpose because (a) they will break up (b) elastic collision of neutrons with heavy nuclei will not slow them down (c) the net weight of the reactor would be unbearably high (d) substances with heavy nuclei do not occur in liquid or gaseous state at room temperature. (a) λmML
(b)
29. The half life of radioactive radon is 3.8 days. The time at the end of which (1/20)th of radon sample will remain undecayed (given log10 2 = 0.30103) is (a) 3.8 days (b) 16.5 days (c) 33 days (d) 76 days 30. A radioactive sample consists of two distinct species having equal number of atoms initially. The mean life time of one species is τ and that of the other is 5τ. The decay products in both cases are stable. A plot is made of the total number of radioactive nuclei as function of time. Which of the following figures best represents the form of this plot?
(a)
(d)
SOLUTIONS 1. (b) 2. (d) :
E3 – E1 = 2E – E = E From figure, E2 – E1 = E2 – E3 + E3 – E1
Tn =
4ε02 h3n3 me 4
Tn ∝ n3 3
∴
T2 ⎛ 2 ⎞ 8 =⎜ ⎟ = T1 ⎝ 1 ⎠ 1
4. (b) : In second excited state, n = 3, ⎛ h ⎞ So, lH = lLi = 3 ⎜ ⎟ ⎝ 2π ⎠ 2 while E ∝ Z and ZH = 1, ZLi = 3 So, ELi = 9 EH or EH < ELi 5. (a) : For Brackett series, 1 1⎤ ⎡1 = R⎢ − ⎥ 2 λ ⎣ 4 n2 ⎦ where n = 5, 6, 7, 8, ........ For longest wavelength, n = 5 1 ⎡1 1⎤ ∴ = R⎢ − ⎥ λ Longest ⎣ 42 52 ⎦ .... (i)
.... (ii)
⎝ r0 ⎠
∴
hc =E λ
3. (c) : In Bohr model of hydrogen atom, the period of revolution of electron in nth orbit is given as
6. (a) : Given : V = V0 ln ⎛⎜ r ⎞⎟
(Using (i))
9 ⎡1 1⎤ = R⎢ − ⎥ = R ⎣16 25 ⎦ 400 For shortest wavelength, n = ∞ 1 1 ⎤ R ⎡1 ∴ = R⎢ − ⎥= 2 λShortest ⎣4 ∞ 2 ⎦ 16 Dividing (ii) by (i), we get λ Longest R 400 25 = × = λShortest 16 9R 9
(b)
(c)
4 = E − 2E + 2E − E 3 2 1 =− E+E= E 3 3 hc 1 hc = λ′ 3 λ λ′ = 3λ
...(i)
Potential energy, U = eV dU ⎛r⎞ ⎛r ⎞ 1 or U = eV0 ln ⎜ ⎟ ∴ = eV0 ⎜ 0 ⎟ ⎝ r ⎠ r0 dr ⎝ r0 ⎠ eV dU Force, F = − =− 0 dr r or
F =
=
eV0 r
eV0 r PHYSICS FOR YOU | OCTOBER ‘16
67
This force provides the necessary centripetal force. ∴
mv 2 r
=
eV0
or v =
r
eV0
m nh By Bohr’s postulate, mvr = 2π nh or v = 2πmr
...(i)
...(ii)
From equations (i) and (ii), we get eV0 nh = m 2πmr
or ∴
⎡ h ⎣⎢ 2π
r=⎢
or r = 1 meV0
nh m × 2πm eV0
⎤ ⎥×n ⎥⎦
rn ∝ n 7. (a) : Number of spectral lines emitted is n(n − 1) N= 2 9(9 − 1) 9 × 8 = = 36 Here n = 9 ∴ N = 2 2 h 2 ε0 a = 8. (b) : 0 πme 2 h 2 ε0 aμ = π(207m)e 2 Dividing (ii) by (i), we get aμ a 1 = or aμ = 0 a0 207 207
.... (i) .... (ii)
CZ 2
, where C is a constant. n2 For ground state of hydrogen atom, Z = 1, n = 1 ∴
C(1)2
=C (1)2 For nth state of Li2+ ion (Z = 3) En =
E1 =
C(3)2
=
9C
n n2 As En = E1 9C ∴ = C or n2 = 9 or n = 3 n2 10. (c) : According to Bohr’s quantisation condition nh Ln = 2π 4h 5h ∴ For n = 4, L4 = and for n = 5, L5 = 2π 2π ∴ Change in angular momentum when an electron is excited from n = 4 to n = 5 is 68
2
PHYSICS FOR YOU | OCTOBER ‘16
n2 a , where n is the orbit number. Z 0 For 257 100 Fm, Z = 100 25 1 ⎛ rs ⎞ a0 = a0 ⇒ r5 = ⎜⎝∵ a = N ⎟⎠ 100 4 0 r5 1 =N = ∴ a0 4
11. (d) : As rn =
12. (c) : The angular momentum is given by ⎛ k ⎞ L = l(l + 1) ⎜ ⎟ ⎝ 2π ⎠ For 3d orbital, l = 2 ⎛ k ⎞ ⎛ k ⎞ ∴ L = 2(3) ⎜ ⎟ = 6 ⎜ ⎟ ⎝ 2π ⎠ ⎝ 2π ⎠ 13. (c) : Radius of Li++ ion in its ground state, i.e.,
9. (b) : The energy of nth state of a hydrogen like atom is given as En =
5h 4h − 2π 2π h 6.63 × 10−34 J-s = 1.05 × 10–34 J s = = 2π 2 × 3.14 ΔL = L5 − L4 =
r0 =
ε0 h 2
πmZe 2 53 pm ε h2 a ≈ 18 pm As Bohr radius, a0 = 0 2 ⇒ r0 = 0 = πme
Z
3
(∵ for Li, Z = 3)
14. (c) : Gain in potential energy of α-particle = Loss in kinetic energy of α-particle 1 Ze × 2e 5 × 1.6 × 10−13 = ⋅ 4 πε0 r0 2 1 2Ze ⋅ r0 = 4 πε0 5 × 1.6 × 10−13 =
9 × 109 × 2 × 92 × (1.6 × 10−19 )2
5 × 1.6 × 10−13 = 5.3 × 10–14 m = 5.3 × 10–12 cm ≈ 10–12 cm 15. (a) : ∴
1 1⎤ ⎡1 = RZ 2 ⎢ − ⎥ 2 2 λ ⎣ n1 n2 ⎦ 1 ⎡1 1⎤ 5 = R(1)2 ⎢ − ⎥ = R λ1 ⎣ 22 32 ⎦ 36
1 1 ⎞ 3 ⎛ 1 = R(2)2 ⎜ − ⎟ = R 2 λ2 42 ⎠ 4 ⎝2 ∴ λ2 = 5 × 4 = 5 λ1 36 3 27 5 5 or λ2 = λ1 = × 6561 Å = 1215 Å 27 27 16. (c) : The given nuclear reaction is 1 4 4 1H → 2 He + 2e+ + Energy The energy released during the process is 1 4 Q = [4m(1H) – m( 2He) – 2(me+)]c2 = [4 × 1.007825 – 4.002603 – 2 × 0.000548]u × c2 = [4.0313 – 4.002603 – 0.001096]u × c2 = (0.027601 u)c2 = (0.027601)(931.5) MeV = 25.7 MeV 17. (c) : According to Rutherford and Soddy law for radioactive decay, Number of atoms remained undecayed after time t is N = N0e–λt N N e λt = 0 or λt = ln 0 N N 1 N0 t = ln λ N T1/2 N 0 1.25 × 1010 16 ln t= ln = ln 2 N ln 2 1 10 1.25 × 10 × 4 × ln 2 = 5 × 1010 years = ln 2 18. (a) : As the heavy nucleus at rest breaks, therefore according to law of conservation of momentum, we get m1v1 + m2v2 = 0 and
v1 m2 3 = = v2 m1 1 As nuclear density is same, 4 3 m1 ρ 3 πR1 R13 ∴ = = 3 4 m2 ρ πR23 R2 3 R13 m1 1 = = or R23 m2 3
or
....(i)
(Using (i))
∴ R1 : R2 = 1 : 31/3 19. (d) : During the emission of α-particle, the mass number and atomic number decreases by four and two respectively. During the emission of β-particle the mass number remains the same while the atomic number increases by 1. 3α 5β −12) ( A −12) ⎯⎯→ ( A (Z −6)Y ⎯⎯→ (Z −1)Y ′ Number of neutrons A − 12 − (Z − 1) ∴ = Number of protons Z −1 A − Z − 11 = Z −1 20. (c) : Here, Half life, T1/2 = 8 days Initial activity, R0 = 8 microcurie Final activity, R = 1 microcurie A ZX
n
R ⎛1⎞ = ⎜ ⎟ where n is the number of half lives As R0 ⎝ 2 ⎠ n
3
n
⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎜ ⎟ = ⎜ ⎟ or ⎜ ⎟ = ⎜ ⎟ ⎝2⎠ ⎝2⎠ ⎝8⎠ ⎝2⎠ or n = 3 t As n = or t = nT1/2 = (3)(5 days) = 15 days T1/2 ∴
21. (a) : A free neutron is unstable. It decays spontaneously into a proton, an electron and antineutrino. – n → p + e– + υ neutron
proton
electron
anti-neutrino
22. (a) : Nuclear radii, R = R0(A)1/3 (where R0 = 1.2 fm) or R ∝ (A)1/3 ∴ or
RBe (9)1/3 = RGe ( A)1/3 RBe (9)1/3 = 2RBe ( A)1/3
∴ (A)1/3 = 2 × (9)1/3 or A = 23 × 9 = 72 The number of nucleons in Ge is 72. PHYSICS FOR YOU | OCTOBER ‘16
69
23. (c) : For fusion to take place, high temperature is needed because at high temperature, the kinetic energy becomes large enough to overcome the coulomb repulsion between nuclei. 24. (a) : Here, power = 3.2 MW = 3.2 × 106 W Energy released per fission = 200 MeV = 200 × 106 eV = 200 × 106 × 1.6 × 10–19 J Number of fissions per minute 3.2 × 106 × 60 = = 6 × 1018 200 × 106 × 1.6 × 10−19 25. (d) : Let number of α particles emitted be x and number of β-particles emitted be y. Difference in mass number = 4x = 232 – 208 = 24 or x = 6 Difference in atomic number = 2x – y = 90 – 82 = 8 ⇒ 12 – y = 8 or y = 4 26. (b) : Let A and Z be mass number and atomic number of element X. β+
and M is the molar mass. λmL ⎛m⎞ ⇒ N = ⎜ ⎟ L or A = M ⎝M⎠ 28. (b) : During an elastic collision between two particles, the maximum kinetic energy is transferred from one particle to the other when they have the same mass. Consequently, a neutron loses all of its kinetic energy when it collides head-on with a proton, in analogy with the collision between a moving billiard ball and a stationary one. For this reason, materials which are abundant in hydrogen such as paraffin and water, are good moderators for neutrons. N 29. (b) : Here, T1/2 = 3.8 days, N = 0 20 ∴ Disintegration constant, 0.693 0.693 day −1 λ= = T1/2 3.8 As N = N0e–λt
2α
⎯⎯→ Z −A1Y ⎯⎯→ ZA−−58Y ′ Since the product nucleus has mass number 227 and atomic number 89. ∴ A – 8 = 227 or A = 235 and Z – 5 = 89 or Z = 94 A ZX
27. (b) : A = λN N = n⋅L where n is the number of moles and L is the Avogadro number. m Now, n = where m is the mass of the isotope, M
N0 = N 0e −λt or e λt = 20 20 Taking natural logarithm, λt = loge 20 or λt = 2.303 × log10 20 ∴
2.303 × 1.3010 2.303 × 1.3010 × 3.8 = days λ 0.693 =16.43 days 16.5 days or
t=
30. (d)
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70
PHYSICS FOR YOU | OCTOBER ‘16
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rnt by around all over It is a lesson lea —they wobble les bb bu g in tists had fun blow . But now scien never last long ey th le d bb an bu e y ac the pl mobilise a tin managed to im lp he d ul co at th in France have hrough rprising breakt in water in a su ood clots. pushed doctors treat bl ill naturally be les in a liquid w bb bu edes in y im all ch rm Ar No ed by omenon describ d in a se er m im upwards, a phen y all t, wholly or parti jec ob ht of ny “A eig C. w 250B ual to the up by a force eq ician at m he at m fluid, is buoyed e th by the object,” ed lac stop sp to di ay id the flu d found a w now, no-one ha til s un er d ch An ar . se te re ro , w However m happening. eate cr d ul co this process fro ey th d Université foun h a tiny at Aix-Marseille electricity throug g in nn ru by es cy of the en qu microbubbl fre changing the By . er at w bubble in de electro could make the discovered they . electricity they ctrode nce from the ele ould stay a set dista h the water, it w ug ro th g in ris ly w slo ctrode, of ele ad So inste ey moved the sition. And if th po té ed rsi fix ive a Un in y ille sta x-Marse t with it. The Ai ea ilis ob m im the bubble wen d ul co monstrated they is as if the researchers de water electrolys m fro d te ea cr ally push it rm no microbubble ld t force that wou an oy bu ’ es Physics ed Archim ican Institute of exist. The Amer ’t uld dn di co … ds ar on w en up phenom and surprising dustry or in ar cle said: “This new nu e th ne, tions in medici lead to applica logy.” no ch te n io at ul micromanip
Courtesy : The Times of India
PHYSICS FOR YOU | OCTOBER ‘16
71
INDUCED ELECTRIC FIELD
This article expects a brief knowledge of induction;. i.e., when the magnetic flux through a closed conducting loop is changed, a current is induced within the loop such that it opposes the change that has produced it. According to Faraday's law, the induced emf in the loop is, dφ ε=− B dt where φB is magnetic flux through the loop. ∴ Induced current, dφ − B ε I= = dt R resistance of the loop Now, the question that should pop up in our head is how does current appear; i.e., what actually makes the free electrons of the loop push along the length of the wire? In current electricity chapter, you must have learnt that the terminals of the battery creates a potential difference across the length of the conductor which creates an electric field and hence the free electrons get pushed due to which current appears. So, it was the conservative electric field there since they were produced by static charges at the ends of the battery terminals. But how do we explain the generation of current without battery? An interesting observation found in nature is that a changing magnetic field leads to an induction of electric field in the surrounding and vice versa. So in time varying magnetic field, current is generated due to induced electric field which is very different from the conservative electric field. How do we know that its electric field only which is making the charges push along the length of conductor and not magnetic forces? Had it been magnetic forces, they would have been perpendicular to velocity [F = q(v × B)] and hence
perpendicular to length of wire too and hence charges could not have moved. Now, how do we find the strength of the induced electric field? We go back to the definition of emf of a battery which was the work done per unit charge by the nonconservative forces developed within the battery in moving the charges from one terminal to the other. Similarly, for induced emf, it will be the work done per unit charge by induced electric field in taking the charge through a closed loop once. Induced emf can be found out from Faraday's law as – dφ ε=− B dt where φB is the magnetic flux through the loop. Hence combining the two definitions, ε=
Winduced electric field q
∫ qEi ⋅ dl
dφ B q dt dφ ⇒ ∫ Ei ⋅ dl = − B dt where Ei is the induced electric field's value at an elemental length dl of the loop. To avoid the confusion of positive and negative in the above result, it would be convenient to use this law in two parts. 1. Calculation of magnitude using – dφ ∫ E ⋅ dl = dtB 2. Calculation of direction of induced electric field using Lenz's law, according to which if a conducting loop is placed in a region of time varying magnetic field (TVMF), current will be induced in it such that the induced magnetic field due to induced current is in opposite direction with respect to inducing ⇒
=−
Contributed By: Bishwajit Barnwal, Aakash Institute, Kolkata
72
PHYSICS FOR YOU | OCTOBER ‘16
magnetic field if the strength of inducing field increases with respect to time and in same direction if the strength decreases with respect to time. The direction of induced current would give us the direction of induced electric field. We will restrict our discussion to a cylindrical region of TVMF. Let us consider a cylindrical region of radius R which has a magnetic field uniform in space (along the length of cylinder) but variable in time and changing at the dB rate . dt We would be interested in finding out the induced electric field (Ei ) for internal as well as external points. At whichever point Ei is to be found out, we imagine that if we had placed a conducting loop in this cylindrical region symmetrically and passing through the given point, there would have been an induced current certainly and this induced current would give us the direction of Ei . dB >0 dt For internal points ( r < R): Current would have induced in anticlockwise sense, so would be the direction of Ei . Hence, note that Ei is forming closed circular loop. ∴ ∫ Ei ⋅ dl = ∫ Eidl cos θ
Let us assume
= Ei ∫ dl cos 0° = Ei (2πr) dB dφ B = πr 2 dt dt r ⎞ ⎛ dB ⎞ ⇒ Ei = ⎛⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎠ ⎝ dt ⎠
∴ Ei (2πr) =
For external points (r ≥ R): dB ∴ Ei (2πr) = πR2 dt ⎛ 2 ⎞ dB ⇒ Ei = ⎜ R ⎟ ⎝ 2r ⎠ dt If the strength of Ei is plotted against r : (i) Ei ∝ r for r < R 1 (ii) Ei ∝ for r ≥ R r 74
PHYSICS FOR YOU | OCTOBER ‘16
Some of the important noteworthy points regarding Ei are These are non-conservative in nature and hence work done by such field in a closed path is non-zero. In fact, work done in closed path = charge × emf of loop. Induced electric field lines form closed loops. Potential energy cannot be associated with them. This comes not due to static charges but rather due to TVMF. Hence if a charge particle is placed at rest in a TVMF, it will experience a non-zero force due to induced electric field. Now, let us see some applications of it : Q1. A charge particle of charge q, mass m is placed at a distance 2R from the centre of a cylindrical region of radius R with TVMF where magnetic field varies ^ as B = (4t 2 − 2t + 6) k where t in sec gives B in tesla. Find the acceleration of the charge at t = 2 s. (Take R = 1 m) Sol.: dB = 8t − 2 dt ∴ At t = 2 s dB = 14 T s −1 dt which is greater than zero, so induced electric field must try to decrease the strength of inducing magnetic field, so must be circular loops in anticlockwise sense. 7 R2 dB 1 = (14) = N C −1 2 2(2R) dt 4 F qE 7q ∴ a= = i = m m 2m ∴ Ei =
Q2. A charge q, mass m is placed in a cylindrical region of uniform magnetic field of strength B and of radius R. It is attached to one end of a relaxed spring of stiffness
constant k as shown. If the magnetic field is suddenly switched off, find the maximum elongation or compression in the spring. Sol.: Switching off the magnetic field is decreasing the magnetic flux, to oppose which electric field would be induced in anticlockwise sense for the short duration in which the magnetic field changed its value from non-zero to zero value due to which the particle will experience an impulsive force and gain a kinetic energy using which the spring gets deformed. Clearly to increase the flux electric field gets induced in clockwise direction, hence spring will be elongated. The momentum gained by charge in switching duration Δp = mv − 0 = ∫ F ⋅ dt = ∫ (qEi ) dt = ∫ q ⋅
a dB dt 2 dt
qa q dB = aB ∫ 2 2 qaB ∴ v= 2m 1 2 1 ∴ kxmax = mv 2 2 2 =
m ⎛ qaB ⎞ ⇒ xmax = ⎟ ⎜ k ⎝ 2m ⎠
Q3. A conducting rod of length l is placed as a chord in a cylindrical region of TVMF of radius R where dB > 0. Find the potential difference across the dt ends of the rod. Sol.: Isn't there a contradiction in what has been taught in theory and numerical since we learnt that potential term cannot be associated with induced electric field. No, there isn't! Induced electric field pushes the free electrons of the conductor from one end to the other which creates a separation of charges across the ends of the rod and these separated charges create conservative electric field along the length of the rod in opposite direction to the component of induced electric field. We consider an arbitrary point on the rod at a radial distance r. The component of Ei along the length of rod,
E|| = Eicosθ ⎛ r dB ⎞ ⎛ y ⎞ =⎜ ⎝ 2 dt ⎟⎠ ⎜⎝ r ⎟⎠
y ⎞ ⎛ dB ⎞ = ⎜⎛ ⎟ ⎜ ⎟ ⎝ 2 ⎠ ⎝ dt ⎠
= 2– 2
2
Do you notice something interesting here? The value is independent of r ! ∴ This E|| pushed the free electrons towards left. ∴ At equilibrium, if the drift of electron stops, conservative field strength, y dB Ec = E|| = 2 dt ∴ Potential difference across the ends y dB l= ΔV = Ecl = 2 dt
l2 4 dB l 2 dt
R2 −
Q4. Consider a rectangular conducting loop PQRS placed in a cylindrical region of radius R of dB TVMF where >0 dt is known. Find the induced emf across the ends QR. Sol.: Let us observe the distribution of induced electric field in the region once– There cannot be an induced emf in the section PS since Ei is perpendicular to length [Ei ⋅ dl = 0]. In the remaining 4 sections, PQ, QT, TR and RS, the induced emf will be identical as can be seen from symmetry. Hence, the induced emf of the loop = 4 times in QT ∴ ε=
dφ B πR2 dB = = 4εQT 2 dt dt
∴ εQT =
πR2 dB 8 dt
∴ εQR = 2εQT =
PHYSICS FOR YOU | OCTOBER ‘16
πR2 dB 4 dt 75
Class XII
T
his specially designed column enables students to self analyse their extent of understanding of specified chapters. Give yourself four marks for correct answer and deduct one mark for wrong answer. Self check table given at the end will help you to check your readiness.
Magnetic Effect of Current and Magnetism Total Marks : 120
Time Taken : 60 min NEET / AIIMS / PMTs
Only One Option Correct Type
1. Consider a long, straight wire of cross-sectional area A carrying a current I. Let there be n free electrons per unit volume. An observer places himself on a trolley moving in the direction opposite to the I current with a speed v = and separated from πAe the wire by a distance r. The magnetic field seen by the observer is very nearly μ0 I 2μ 0 I μ I (a) 0 (b) zero (c) (d) πr πr 2 πr 2. Two short magnets of equal dipole moments M are fastened perpendicularly at their centres as shown in the figure. The magnitude of the magnetic field at a distance d from the centre on the bisector of the right angle is μ 0 2M μ M (a) 0 3 (b) 4 π d3 4π d μ 0 2 2M μ0 2 M (c) 4 π (d) 3 4 π d3 d 3. An electron accelerated through a potential difference V passes through a uniform transverse magnetic field and experiences a force F. If the accelerating potential is increased to 2V, the electron in the same magnetic field will experience a force F (a) F (b) (c) 2 F (d) 2F 2 76
PHYSICS FOR YOU | OCTOBER ‘16
4. An electric current I enters and leaves a uniform circular wire of radius a through diametrically opposite points. A charged particle q moving along the axis of the circular wire passes through its centre at speed v. The magnetic force acting on the particle when it passes through the centre has a magnitude μ0 I μ0 I (a) qv (b) qv 2a 2 πa μ0 I (c) qv (d) zero a 5. A beam consisting of protons and electrons moving at the same speed goes through a thin region in which there is a magnetic field perpendicular to the beam. The protons and the electrons (a) will go undeviated (b) will be deviated by the same angle and will not separate (c) will be deviated by different angles and hence separate (d) will be deviated by the same angle but will separate. 6. A charged particle of mass 10–3 kg and charge 10–5 C enters a magnetic field of induction 1 T. If g = 10 m s–2, for what value of velocity will it pass straight through the field without deflection? (a) 10–3 m s–1 (b) 103 m s –1 6 –1 (c) 10 m s (d) 1 m s–1 7. Mark out the correct options. (a) Diamagnetism does not occur in all materials. (b) Diamagnetism results from the partial alignment of permanent magnetic moment.
(c) The magnetising field intensity is always zero in free space. (d) The magnetic field of induced magnetic moment is opposite to the applied field. 8. A particle is moving with velocity v = i + 3 j and it produces an electric field at a point given by E = 2k. It will produce magnetic field at that point equal to (all quantities are in SI units) (a) (6i − 2 j ) μ 0 ε0 (b) (6i + 2 j ) μ 0 ε0 (c) zero (d) cannot be determined from the given data 9. A point charge is moving in clockwise direction in a circle with constant speed. Consider the magnetic field produced by the charge at a fixed point P (not at the center of circle) on the axis of the circle. Then, (a) it is constant in magnitude only (b) it is constant in direction only (c) it is constant both in direction and magnitude (d) it is constant neither in magnitude nor in direction 10. A dip circle is so that its needle moves freely in the magnetic meridian. In this position, the angle of dip is 40°. Now the dip circle is rotated so that the plane in which the needle moves makes an angle of 30° with the magnetic meridian. In this position, the needle will dip by an angle (a) 40° (b) 30° (c) more than 40° (d) less than 40° 11. What should be the current in a circular coil of radius 5 cm to annul BH = 5 × 10–5 T? (a) 0.4 A (b) 4 A (c) 40 A (d) 1 A 12. A particle of mass 2 × 10–5 kg moves horizontally between two horizontal plates of a charged parallel plate capacitor between which there is an electric field of 200 N C–1 acing upward. A magnetic induction of 2.0 T is applied at right angles to the electric field in a direction normal to both B and v . If g is 9.8 m s–2 and the charge on the particle is 10–6 C, then find the velocity of charge particle so that it continues to move horizontally. (a) 2 m s–1 (b) 20 m s–1 (c) 0.2 m s–1 (d) 100 m s–1
Assertion & Reason Type
Directions : In the following questions, a statement of assertion is followed by a statement of reason. Mark the correct choice as : (a) If both assertion and reason are true and reason is the correct explanation of assertion. (b) If both assertion and reason are true but reason is not the correct explanation of assertion. (c) If assertion is true but reason is false. (d) If both assertion and reason are false. 13. Assertion : The sensitivity of a moving coil galvanometer is increased by placing a suitable magnetic material as a core inside the coil. Reason : Soft iron has a high magnetic permeability and cannot be easily magnetized or demagnetized. 14. Assertion : The true geographic north direction is found by using a compass needle. Reason : The magnetic meridian of the earth is along the axis of rotation of the earth. 15. Assertion : The net force on a closed circular current carrying loop placed in a uniform magnetic field is zero. Reason : The torque produced in a conducting circular ring is zero when it is placed in a uniform magnetic field such that the magnetic field is perpendicular to the plane of loop. JEE MAIN / JEE ADVANCED / PETs
Only One Option Correct Type
16. A charged particle of specific charge (charge/mass) α is released from origin at time t = 0 with velocity v = v0 (i + j ) in a uniform magnetic field B = B0 i. π Coordinates of the particle at time t = are ( B0α) ⎛ 2v0 −v0 ⎞ ⎞ ⎛ − v0 , 0, 0 ⎟ , (a) ⎜ 0, ⎟ (b) ⎜ ⎠ ⎝ 2 B0α ⎝ αB0 B0 α ⎠ −2v0 ⎞ ⎛ v0 π v π ⎞ ⎛ 2v , 0, (c) ⎜ 0, 0 , 0 ⎟ (d) ⎜ B0α ⎟⎠ ⎝ B0α ⎝ B0α 2 B0α ⎠ 17. A small block of mass m, having charge q, is placed on a frictionless inclined plane making an angle θ with the horizontal as shown in figure. There exists a uniform magnetic field B parallel to the inclined plane but perpendicular to the length of a spring. PHYSICS FOR YOU | OCTOBER ‘16
77
If m is slightly pulled on the incline in downward direction, the time period of oscillation will be (assume that the block does not leave contact with the plane) (a) 2π m / k (b) 2 π 2m / k
(c) 2π qB / K
(d) 2 π qB / 2K
18. A thin, plastic disk of radius R has a charge q uniformly distributed over its surface. If the disk rotates at an angular frequency ω about its axis, then magnetic dipole moment of the disk is ωqR 2 ωqR 2 (b) (c) ωqR2 (d) 2ωqR2 2 4 19. From a cylinder of radius R, a cylinder of radius R/2 is removed, as shown in the figure. Current flowing in the remaining cylinder is I. Then, magnetic field strength is (a)
(a) zero at point A (b) zero at point B μ I (c) 0 at point A 2 πR
(d)
μ0 I at point B 3 πR
More than One Options Correct Type 20. A conductor ABCDEF, shaded as shown, carries a current I. It is placed in the xy plane with the ends A and E on the x-axis. A uniform magnetic field of magnitude B exists in the region. The force acting on it will be
(a) zero, if B is in the x-direction (b) λBI in the z-direction, if B is in the y-direction (c) λBI in the negative y-direction, if B is in the z-direction (d) 2aBI, if B is in the x-direction 21. A microammeter has a resistance of 100 Ω and a full scale range of 50 μA. It can be used as a voltmeter or as a higher range ammeter provided a resistance is added to it. Pick the correct range and resistance combination(s) 78
PHYSICS FOR YOU | OCTOBER ‘16
(a) 50 V range with 10 kΩ resistance in series (b) 10 V range with 200 kΩ resistance in series (c) 5 mA range with 1 Ω resistance in parallel (d) 10 mA range with 1 Ω resistance in parallel 22. A direct current I flows along a long straight wire as shown in the figure. From point O the current spreads radially all over on infinite conducting plane perpendicular to the wire. Then (a) Magnetic field in region 1 is non-uniform (b) Magnetic field in region 2 in non-uniform (c) Magnetic field in region 3 is non-uniform (d) Magnetic field in region 3 is zero. 23. A current I flows in a long round uniform cylindrical wire made of paramagnetic material with susceptibility χ. Which of following statements are correct regarding the surface molecular current (Is) and the volume molecular current (Iv) ? (a) Both the currents Is and Iv have same magnitude. (b) Both the currents Is and Iv have different magnitude. (c) Both the currents Is and Iv flow in the same direction. (d) Both the currents Is and Iv flow in the opposite directions. Integer Answer Type 24. A current I = 10 A flows in a ring of radius r0 = 15 cm made of a very thin wire. The tensile strength of the wire is equal to T = 1.5 N. The ring is placed in a magnetic field, which is perpendicular to the plane of the ring so that the forces tend to break the ring. Find B (in T) at which the ring is broken. 25. An elevator carrying a charge of 0.5 C is moving down with a velocity of 5 × 103 m s–1. The elevator is 4 m from the bottom and 3 m horizontally from P as shown in figure. What magnetic field (in μT) does it produce at point P? 26. An iron of volume 10–4 m3 and relative permeability 1000 is placed inside a long solenoid wound with 5 turns per cm. If a current of 0.1 A is passed through the solenoid, find the magnetic moment of the rod.
Comprehension Type
Column I Column II (A) In the given situation, (P) Resultant force is acting along Pm (B) If loop is rotated (Q) Resultant force is acting opposite such that Pm is along positive z-direction to Pm (C) If loop is rotated (R) Fx = 0, Fy = 0 such that Pm is along negative z-direction (D) If loop is rotated (S) Fx = 0, Fz = 0 such that Pm is along positive y-direction A B C D (a) P P, R S Q (b) S P, R P, R Q, S (c) Q P, R P, R P, S (d) R R, S Q, R P, Q
A thin, uniform rod with negligible mass and length 0.200 m is attached to the floor by a frictionless hinge at point P as shown in the figure. A horizontal spring with force constant k = 4.80 N m–1 connects the other end of the rod to a vertical wall. The rod is in a uniform magnetic field B = 0.340 T directed into the plane of the proper. There is current I = 6.50 A in the rod, in the direction shown. 27. Calculate the torque due to the magnetic force on the rod, for an axis at P. (a) 0.0442 N m–1, clockwise (b) 0.0442 N m–1, anticlockwise (c) 0.022 N m–1, clockwise (d) 0.022 N m–1, anticlockwise 28. When the rod is in equilibrium and makes an angle of 53.0° with the floor, is the spring stretched or compressed? (a) 0.05765 m, stretched (b) 0.05765 m, compressed (c) 0.0242 m, stretched (d) 0.0242 m, compressed Matrix Match Type 29. An elementary current loop is placed in a nonuniform magnetic field as shown in the figure. Where, Pm is magnetic moment of loop.. In column I different orientations of loop are described and in column II, the corresponding forces experienced by the loop. Match the entries of column I with entries of column II.
30. Three wires are carrying same constant current I in different directions. Four loops enclosing the wires in different manners as shown in the figure. The direction of dl is shown in each loop. Match the entries of column I with entries of column II. Column I Column II (A) Along closed loop 1 (P) ∫ B ⋅ dl = μ 0 I
(B) Along closed loop 2 (Q)
∫ B ⋅ dl
(C) Along closed loop 3 (R)
∫ B ⋅ dl = 0
= −μ 0 I
(D Along closed loop 4 (S) Net work done by the magnetic force to move a unit charge along the loop is zero A B C D (a) P, Q Q, R Q, S R, S (b) Q, S P, S R, S R, S (c) P, Q P, R Q, S P, S (d) P, S Q, S R, S R, S Keys are published in this issue. Search now!
Check your score! If your score is > 90%
EXCELLENT WORK !
You are well prepared to take the challenge of final exam.
No. of questions attempted
……
90-75%
GOOD WORK !
You can score good in the final exam.
No. of questions correct
……
74-60%
SATISFACTORY !
You need to score more next time
Marks scored in percentage
……
< 60%
NOT SATISFACTORY! Revise thoroughly and strengthen your concepts.
PHYSICS FOR YOU | OCTOBER ‘16
79
P
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.
SINGLE OPTION CORRECT TYPE
1. A smooth cannon of mass M attached with an ideal spring of stiffness k fires a shell of mass m with muzzle velocity u. The recoil velocity of the cannon is
mu cos θ mu sin θ (b) M +m 2M + m mu cos θ 2mu cos θ (d) (c) M +m M +m 2. A uniform disc of mass m and radius R with a spring connected (spring constant K) to its centre stands in static equilibrium on a very rough inclined plane. Now a constant moment of couple M0, is applied at the centre as shown in the figure such that the disc rolls down the incline. The maximum distance the disc’s centre of mass will go down as measured from initial position is (a)
M0 2mg sin θ (a) + Rk 3k (c)
2 M0 Rk
(b)
2 M0 2mg sin θ + Rk k
(d)
2 M0 mg sin θ + Rk k
3. In a resonance tube experiment, a closed organ pipe of length 120 cm is used and tuned with a tuning fork of frequency 340 Hz. If water is poured into the pipe, then which of the following statements is incorrect? [Velocity of sound in air is 340 m s–1, neglect end correction] (a) Minimum length of water column to have the resonance is 45 cm. (b) The distance between two successive nodes is 50 cm. (c) The maximum length of water column to create the resonance is 95 cm. (d) The distance between two successive nodes is 25 cm. 4. A ball is hung vertically by a thread of length l from a point P of an inclined wall that makes an angle β with the vertical. The thread with ball is deviated through a small angle α(α > β) and set free. Assuming the wall to be perfectly elastic, the period of such pendulum is (a)
l g
⎡π −1 ⎛ β ⎞ ⎤ ⎢ 2 − sin ⎜⎝ α ⎟⎠ ⎥ ⎣ ⎦
(b)
l g
⎡ −1 ⎛ β ⎞ ⎤ ⎢ sin ⎜⎝ α ⎟⎠ ⎥ ⎣ ⎦
(c)
l g
⎡ −1 ⎛ β ⎞ ⎤ ⎢ cos ⎜⎝ α ⎟⎠ ⎥ ⎣ ⎦
(d) 2
l g
⎡ −1 ⎛ β ⎞ ⎤ ⎢ cos ⎜⎝ − α ⎟⎠ ⎥ ⎣ ⎦
By Akhil Tewari, Author Foundation of Physics for JEE Main & Advanced, Professor, IITians PACE, Mumbai.
80
PHYSICS FOR YOU | OCTOBER ‘16
COMPREHENSION TYPE
Direction for question 5 to 8 : The pressure of a monoatomic gas increases linearly from 4 × 105 N m–2 to 8 × 105 N m–2 when its volume increases from 0.2 m3 to 0.5 m3. Calculate 5. Work done by the gas (a) 2.8 × 105 J (b) 1.8 × 106 J 5 (c) 1.8 × 10 J (d) 1.8 × 102 J 6.
Increase in internal energy (a) 4.8 × 105 J (b) 4.8 × 104 J 5 (c) 6.8 × 10 J (d) 4.8 × 106 J
7.
Amount of heat supplied (a) 8.6 × 105 J (b) 12.6 × 105 J 5 (c) 6.6 × 10 J (d) 10.6 × 105 J
8. Molar heat capacity of the gas (R = 8.31 J mol–1 K–1) (a) 20.1 J mol–1 K–1 (b) 17.14 J mol–1 K–1 (c) 18.14 J mol–1 K–1 (d) 20.14 J mol–1 K–1 SUBJECTIVE TYPE
9. A body is projected vertically upwards from the surface of the earth with a velocity sufficient to carry it to infinity. Calculate the time taken by it to reach height h. 10. In Young's double slit experiment if the source consists of two wavelengths λ1 = 4000 Å and λ2 = 4002 Å. Find the distance from the centre where the fringes disappear, if d =1cm ; D =1 m.
PHYSICS FOR YOU | OCTOBER ‘16
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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.
Object is released from rest at y = 0, so u = 0 Equations of motion become at time t1 v1 = 0 – gt1 = – 9.8t m s–1 1 h = 0 − gt 2 = − 4.9 t 2 m 2 v12 = 0 − 2 gy = −19.6h m2s −2 Hence signs of u, v, a and h are negative as they are chosen along negative y-direction.
Q1. Are photons always moving? What about their rest mass? –Basavraj S. Awatiger, Hubballi (Karnataka)
Ans. Yes, photons always move with speed of light (c). In relativistic dynamics, rest mass of a particle, m0 = m 1 − v 2 /c 2 ; m = moving mass of particle,
(a) a – t graph (b) v – t graph (c) h – t graph
For photon, v = c ∴ m0 = m 1 − c 2 /c 2 = 0 Rest mass of photon is zero. Q2. Explain why Gauss’s law is not very useful in calculating the electric field of a charged disc. –Aquil Ahmed, Banda (U.P.)
Ans. Gauss’s law is useful as a calculational tool only in case of high symmetry, where one can produce a Gaussian surface on which the electric field is either constant or has no perpendicular component. It is not possible to do so in any simple way for the case of a charged disc. Gauss’s law still applies but it is just not particularly useful to find electric field. Q3. What are the sign used for initial velocity (u), final velocity (v), acceleration (a), and height (h) during free fall motion of an object and why? –Priyanka Soren
Ans. Assume motion of the object is in y-direction. Choose upward direction as positive so motion of an object under free fall is in negative y-direction. Since acceleration due to gravity is always downward so a = – g = – 9.8 m s–2 . 82
PHYSICS FOR YOU | OCTOBER ‘16
Q4. A magnetised needle in a uniform magnetic field experiences a torque but no net force. An iron nail near a bar magnet, however, experiences a force of attraction in addition to a torque. Why? –Inova Singh, Dispur (Assam)
Ans. There is only a uniform magnetic field so that a magnetised needle experiences no net force. An iron nail near a bar magnet is in the nonuniform magnetic field of the magnet. The nail is magnetised due to magnetic induction. This nail, which behaves like a small magnet, apart from experiencing a torque, does experience a force. I don’t know anything, but I do know that everything is interesting if you go into it deeply enough. – Richard Feynman
SOLUTION SET-38
1 2 2 2 1. (c) : Since, kinetic energy K = mω A cos ωt 2 1 2 2 2 and potential energy, U = mω A sin ωt 2 1 2 2 2 2 or K − U = mω A [cos ωt − sin ωt ] 2
∴ Angular frequency = 2ω, 2π π π × T ⎡ 2π ⎤ = = ∴ω = ⎥ time period, = ⎢ 2ω ω 2π T ⎦ ⎣ π×4 = =2 s 2π 2. (b) : Linear momentum is conserved. ∴ p1 = p2
where K2 is for α-particle and K1 is for nucleus. K2 = 54 K1
...(i)
Given K1 + K2 = 5.5 MeV From (i) and (ii), K1 + 54 K1 = 5.5 MeV 55 K1 = 5.5 MeV K1 =
1 10
or
K1 =
...(ii)
5.5 55
MeV 54
∴
K2 = 54K1 or K 2 =
or ∴
K2 = 5.4 MeV Kinetic energy of α-particle = 5.4 MeV.
10
MeV
⎛ dM ⎞ (L ) 3. (d) : Heat of radiator, P = ⎜ ⎝ dt ⎟⎠ Pt M L= ∴ L M t 4. (c) : Missing ones are dark fringes 1⎞ ⎛ d 2 + b2 − d = ⎜ n − ⎟ λ ∴ ⎝ 2⎠ ⎛ 1 b2 ⎞ ⎛ 1⎞ 1⎞ b2 ⎛ d ⎜1 + = ⎜n − ⎟ λ ⎟ − d = ⎜⎝ n − ⎟⎠ λ or 2 2 2 ⎝ d ⎠ 2d 2⎠ ⎝ or
P=
2 b2 3 = λ ⇒ λ=b 2d 2 3d
5. (c) : As υ =
nv n T = 2l 2l μ
Here, TI > TII, lI = lII , μI = μII and υI = υII ∴ nI < nII
Gm(dM ) x2
⎡ A + Bx 2 ⎤ = Gm ⎢ ⎥ dx ⎣ x2 ⎦
a+L ⎛ 1 ⎞ F = ∫ Gm ⎜ 2 ⎟ dx ( A + Bx 2 ) ⎝ x ⎠ a
and 2(216 m) K1 = 2(4 m) K 2
or
For n = 2,
dF =
But p = 2mK where K = kinetic energy.
216K1 = 4K2 or
b2 λ b2 ⇒ λ= = 2d 2 d
6. (b) : Mass per unit length of the rod = A + Bx2. So the mass of length dx is dM = dx(A + Bx2)
1 = mω 2 A2 cos 2ωt 2
or
For n = 1,
a+L ⎛A ⎞ A ⎡A ⎤ = ∫ Gm ⎜ 2 + B ⎟ dx = Gm ⎢ − + BL ⎥ ⎝ ⎠ ⎣a a+L ⎦ x a
7. Given, the pitch of screw gauge = 1 mm and total number of division on the circular scale = 50 1 mm = 0.02 mm Least Count, L.C = 50 The instruments has a positive zero error e = n × L.C = 6 × 0.02 = 0.12 mm i.e. zero correction = – 0.12 mm Linear scale reading (LSR) = 3 × 1 mm = 3 mm Circular scale reading (CSR) = 31 × (0.02 mm) = 0.62 mm Measured reading = LSR + CSR = 3 + 0.62 = 3.62 mm True reading = 3.62 – 0.12 = 3.50 mm
Solution Senders of Physics Musing SET-38
1. Suman Tiwari, Patna (Bihar) 2. Trideep Jyotida, New Delhi PHYSICS FOR YOU | OCTOBER '16
83
8. Consider dN number of turns of radius r and thickness dr. Let dε be the corresponding induced emf, then ⎛ dφ ⎞ dε = (dN ) ⎜ ⎟ ⎝ dt ⎠
L= =
⎛N ⎞ dε = ⎜ ⎟ dr (πr2ω × B0 cos ωt) ⎝a⎠
L=
⎛ N πωB0 cos ωt ⎞ 2 ε = ∫ dε = ⎜ ⎟⎠ ∫ r dr ⎝ a 0 a
N πω(B0 cos ωt )a N πωB0 cos ωta = 3a 3 3
=
εmax =
2
πNa 2 B0 ω
∴
t0
μ
1/2 ∫ (F0 − kt ) dt
0
t0 2 ⎡ (F0 − kt )3/2 ⎤⎦ 0 ⎣ μ (−3k)
1
2 −3k μ 2 3k μ k= =
3
1
∫ dx =
0
d × (πω2 × B0 sin ωt) dt
dε = dN
L
∴
[(F0 − kt 0 )3/2 − (F0 + k(0))3/2 ]
F03/2
(Using (i))
3 3×3×3 2 F0 2 = 3 L μ 3 × 1 3 × 10 −2
2×3 3 × 0. 1
= 20 N s −1
We get, n = 3 9. When outer surface is grounded charge –Q resides on the inner surface of sphere B. Now sphere A is connected to earth, potential on its surface becomes zero. Let the charge on the surface A becomes q kq kQ a − =0 ⇒ q= Q a b b In this position energy stored
SOLUTION OF SEPTEMBER 2016 CROSSWORD
2
U1 =
Q2 1 ⎡a ⎤ 1 ⎡a ⎤ Q Q (− Q ) + + ⎢ ⎥ 8 π ε0 a ⎣ b ⎦ 8 π ε0 b 4 π ε0 b ⎣⎢ b ⎥⎦
When S3 is closed, total charge will appear on the outer surface of shell B. In this position energy stored 2
U2 =
1 ⎛a ⎞ − 1⎟ Q 2 ⎠ 8 π ε0 b ⎜⎝ b
Heat produced = U1 – U2 =
Q 2 a (b − a) 8 π ε0 b3
= 1. 8 J
10. Time taken by the pulse to reach from P to Q, F ...(i) t0 = 0 k where F0 = 3 N dx F − kt T ⇒ = 0 Now, v = μ dt μ 84
PHYSICS FOR YOU | OCTOBER' 16
Winner (September 2016) s Preeti Dahiya, Sonipat (Haryana)
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ACROSS 1. A diffraction grating made of parallel glass plates, each of which extends slightly beyond the next, used to examine extremely fine structures through interferometry. [7, 7] 2. A propulsion engine in which a fuel burns in air that has been compressed by the forward motion of the engine only. [6] 7. A thermometer designed for the measurement of very low temperatures. [9] 10. The removal of any unwanted a.c. components from a circuit or circuit element. [10] 11. The dissociation of molecules by nuclear radiation. [10] 14. A tradename for an alloy of iron, cobalt and nickel with an expansivity similar to that of glass. [5] 15. A mixture of two substances that solidifies as a whole when cooled, without change in composition. [8] 18. Absence of an electron in a semiconductor. [4] 21. A quantum mechanical description of an elementary vibrational motion in which a lattice of atoms or molecules uniformly oscillates at a single frequency. [6] 23. An instrument for measuring the total solar radiation intensity received on a horizontal surface. [11] 25. A non SI unit of luminance. [3] 27. Process of connecting a charged object to earth to remove object’s unbalanced charge. [9] 28. Point of maximum displacement of two superimposed waves. [8] 29. The study of the behaviour and characterstics of nucleons or atomic nuclei. [10] 30. A now-discarded hypothetical medium once thought to fill all space and to be responsible for carrying light waves and other electromagnetic waves. [6] 31. The process of preventing the plasma from coming into contact with the walls of the reaction vessel in a controlled thermonuclear reaction. [11] DOWN 3. The molecular attraction exerted between the surfaces of bodies in contact. [8] 4. A grid or pattern placed in the eyepiece of an optical instrument, used to establish scale or position. [7]
5. 6. 8. 9. 12. 13. 16. 17. 19. 20. 22. 24. 26.
The SI unit of dose equivalent. [7] Study of motion of particles acted on by forces. [8] An instrument for studying thin films on solid surfaces. [12] An apparatus for generating very high frequency currents at high potential. [5, 4] Roughly spherical ice particles, usually a few millimeters in radius, produced in very turbulent clouds. [4] Number of decays per second of a radioactive substance. [8] Two or more sounds that, when heard together, sound pleasant. [10] The amount of potential energy stored in an elastic substance by means of elastic deformation. [10] A hypothetical quantum of gravitational energy, regarded as a particle. [8] Defect of eye, in which distant objects focus in front of the retina. [6] An exposed ring surroundings a strongly illuminated spot on a photographic emulsion. [8] A portable insulation tester calibrated directly in mega ohms. [6] Point where disturbances caused by two or more waves result in no displacement. [4] PHYSICS FOR YOU | OCTOBER ‘16
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