Volume 24 Managing Editor Mahabir Singh Editor Anil Ahlawat (BE, MBA)
No. 12
December 2016
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CONTENTS
Class 11 NEET | JEE Essentials
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Ace Your Way CBSE
22
JEE Workouts
31
MPP-6
35
Brain Map
46
Class 12 Brain Map
47
NEET | JEE Essentials
39
Ace Your Way CBSE
54
JEE Workouts
63
MPP-6
66
Competition Edge Physics Musing Problem Set 41
70
Key Concept
72
Olympiad Problems
78
Physics Musing Solution Set 40
80
You ask, We Answer
83
Crossword
85
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Physics For you | december ‘16
7
6
MECHANICAL PROPERTIES OF SOLIDS AND FLUIDS developed inside the body. The restoring force per unit area of the body is called stress.
) some important Terms Deforming force : External force which tries to change in the length, volume or shape of the body is called deforming force. Elasticity : It is that property of the material of a body by virtue of which the body opposes any change in its shape or size when deforming forces are applied to it, and recovers its original state as soon as the deforming forces are removed. Perfectly elastic body : The body which perfectly regains its original form on removing the external deforming force, is defined as a perfectly elastic body, e.g., quartz is very nearly to a perfectly elastic body. Plastic body : The bodies which remain in deformed state even after removal of the deforming force are defined as plastic bodies. Internal restoring force : When an external force acts at any substance then due to the intermolecular force there is an internal resistance produced into the substance called internal restoring force. ¾ At equilibrium the numerical value of internal restoring force is equal to the external force.
•
•
•
•
•
) stress When deforming force is applied on the body then the equal restoring force in opposite direction is
•
8
Physics For you | DECEmbEr ‘16
Stress =
Restoring force
F F = internal = external Area of the body A A
•
The effect of stress is to produce distortion or a change in size, volume and shape (i.e., configuration of the body). There are three types of stress
•
Longitudinal or normal stress : When object is one dimensional then force acting per unit area is called longitudinal stress. It is of two types : (a) Compressive stress (b) Tensile stress Compressive stress = F/A
Tensile stress = F/A F ¾
F
F
Consider a block of solid as shown in figure. Let a force F be applied to the face which has area A. Resolve F into two components :
F Fn
F Ft
(a)
Fn = F sinq called normal force and Ft = F cosq called tangential force. F F sin q \ Normal (tensile) stress = n = A A
•
•
Tangential or shear stress : It is defined as the restoring force acting per unit area tangential to the surface of the body. Refer to figure (a). F F cos q Tangential (shear) stress = t = A A Bulk stress or volume stress : F = PA When the force is acting all along the surface normal to the area, then force acting per unit area is known as volume stress. The effect of pressure is to produce change in volume. The shape of the body may or may not change depending upon the homogeneity of body. ¾ Difference between pressure and stress :
¾
Longitudinal strain = ¾
¾
Stress Stress can be either Pressure is always normal or tangential to normal to the area. the area. Pressure on a body is Stress can be compressive always compressive. or tensile or shear. Breaking stress : The stress required to cause actual fracture of a material is called the breaking stress or ultimate strength. F Breaking stress = A ¾ Dependence of breaking stress : Nature of material, temperature, impurities. ¾ Independence of breaking stress : Cross sectional area or thickness, applied force. ¾ Maximum load (force) applied on the wire depends on cross sectional area or thickness, nature of material, temperature, impurities. The ratio of the change in configuration (i.e., shape, length or volume) to the original configuration of the body is called strain. Strain =
Change in configuration Original configuration
There are three types of strain 10
Physics For you | DECEmbEr ‘16
=
Dl l
V V
Change in volume
x L
Original volume
=
DV V
x
F
L
x L
F
) stress-strain Graph •
Proportional limit : The limit in which Hook’s law is valid and stress is directly proportional to strain is called proportional limit. B A
C Yield point
Permanent set
) strain •
F
Shear strain : It is defined as the angle q (in radian), through which a face originally perpendicular to the fixed face gets turned on applying tangential deforming force.
Stress
•
l
Original length
Shear strain = q = tan q =
Stress is a tensor.
l
Change in length
Volumetric strain : It is defined as the change in volume per unit original volume, when the body is deformed by external forces. Volumetric strain =
Pressure
Pressure is a scalar.
Longitudinal strain : It is defined asthe increase in length per unit original length, when the body is deformed by external forces.
•
•
D
Ultimate E tensile Fracture strength point Proportional limit
Strain
Elastic limit : The maximum stress which on removing the deforming force makes the body to recover completely its original state. Yield point : The point beyond elastic limit, at which the length of wire starts increasing without increasing stress, is defined as the yield point.
•
•
•
Breaking point : The position when the strain becomes so large that the wire breaks down at last, is called breaking point. At this position the stress acting in that wire is called breaking stress and strain is called breaking strain. Elastic after effect or Elastic relaxation : The property, by virtue of which a body does not regain its original form immediately after removing the deforming force but gains it after some time, is defined as elastic relaxation. Elastic fatigue : It is defined as the loss in the strength (elasticity) of a material caused due to repeated alternating strains to which the material is subjected. Elastic hysteresis : The strain persists even when the stress is removed. This lagging behind of strain is called elastic hysteresis. This is the reason why the values of strain for same Extension or strain stress are different while increasing the load and while decreasing the load. ¾ Metals with small plastic deformation are called brittle while metals with large plastic deformation are called ductile. ¾ Elasticity restoring forces are strictly conservative only when the elastic hysteresis is zero, i.e., the loading and unloading stress strain curves are identical. ¾ The material which have low elastic hysteresis have also low elastic relaxation time. Load or stress Lo ad inc rea sin Loa g d de crea sing
•
• •
¾
Longitudinal stress Longitudinal strain
=
F /A F ×l = Dl A × Dl l
Physics For you | DECEmbEr ‘16
Bulk modulus (B) : The moldulus of elasticity of volume Normal stress − F / A − PV B= = = Volumetric strain DV / V DV Negative sign indicates a decrease in volume with an increase in pressure. For ideal gases bulk modulus is of two types : — Isothermal bulk modulus, Biso = P — Adibatic bulk modulus, Badi = gP The reciprocal of bulk modulus of elasticity is defined as compressibility (K). 1 K= B Modulus of rigidity (G) : Modulus of elasticity of shape G=
Tangential stress F / A F = = Shearing strain q Aq
) Poisson’s ratio •
Within the elastic limit, the ratio of lateral strain to the longitudinal strain is called Poisson’s ratio. Lateral strain − DD / D s= = Longitudinal strain Dl / l
•
•
D
l l
l DD D – D ⋅ D Dl The negative sign indicates that longitudinal and lateral strains are in opposite sense. Relation between Y, B, G and s 9 3 1 Y = 3B(1 – 2s), Y = 2G(1 + s), = + Y G B Elongation of a wire by its own weight equals to (Mgl/2AY). Work done in stretching a wire, 1 F W = × load × extension= × Dl 2 2 This work done is stored in the wire as elastic potential energy. So, elastic potential energy density, or
Within elastic limit, modulus of elasticity is defined as the ratio of the stress to the strain. It depends on the nature of the material of the body and is independent of its dimensions. There are three types of moduli of elasticity ¾ Young’s modulus (Y) : The modulus of elasticity of length Y=
12
¾
•
) Modulus of Elasticity •
¾
s=−
u=
W 1 = × stress × strain V 2
) Applications of Elasticity •
The thickness of metallic ropes used in cranes to lift heavy loads is derived by the knowledge of the
•
elastic limit of the material and for a safety factor of 10. In case of twisting of a cylinder (or wire) of length L and radius r, elastic restoring couple per unit twist is given by C=
•
pGr 4 2L
WL3 48YIg
where L is the length of a beam, Y is the Young’s modulus for the material of the beam, and Ig is the geometrical moment of inertia. ¾ For a beam of circular cross section of radius r, pr 4 Ig = 4 ¾ For a beam of rectangular cross section of breadth b and thickness d, Ig =
•
•
¾
) Buoyancy and Archimede’s Principle •
bd 3 12
) Pressure •
¾
Depression of a beam loaded at the middle by a load W and supported at the ends is δ=
If a uniform force is exerted normal to an area (A), then pressure (P) is defined as the normal force (F) F per unit area, i.e., P = . A Practical units : atmospheric pressure (atm), bar and torr 1 atm = 1.01325 × 105 Pa = 1.01325 bar = 760 torr = 10.33 m of water 1 bar = 105 Pa 1 torr = pressure exerted by 1 mm of mercury column = 133 Pa Pressure is of three types ¾ Atmospheric pressure : Upto top of Force exerted by air atmosphere Air column on unit cross column Sea Area = 1 m2 section area of sea level level called atmospheric pressure (Po). F Po = = 101.3 kN m −2 A \ Po = 1.013 × 105 N m–2
Barometer is used to measure atmospheric pressure. It was discovered by Torricelli. — Atmospheric pressure varies from place to place and at a particular place from time to time. Gauge pressure : Excess pressure (P – Patm) measured with the help of pressure measuring instrument called Gauge pressure. Pgauge = hrg or Pgauge ∝h — Gauge pressure is always measured with help of manometer. Absolute pressure : Sum of atmospheric and Gauge pressure is called absolute pressure. Pabs = Patm + Pgauge ⇒ Pabs = Po + hrg —
•
•
•
•
Buoyancy : If a body is partially or wholly immersed in a fluid, it experiences an upward force due to the fluid surrounding it. This phenomenon of force exerted by fluid on the body is called buoyancy and force is called buoyant force or upthrust. Buoyant force, FB = rVg r = density of fluid, V = volume of displaced fluid Archimede’s principle : It states that the buoyant force on a body that is partially or totally immersed in a liquid is equal to the weight of the fluid displaced by it. Buoyant force acts vertically upward through the centre of gravity of the displaced fluid. This point is called centre of buoyancy. It depends upon the effective acceleration. ¾ If a lift is accelerated downwards with acceleration a(a < g) then FB = rV(g – a) ¾ If a lift is accelerated downwards with a = g then FB = rV(g – a) = 0 ¾ If a lift is accelerated upward with acceleration a then FB = rV(g + a) Due to upthrust, the weight of the body decreases. Wapp = W – FB (W is the true weight of the body) Decrease in weight = W – Wapp = FB = Weight of the fluid displaced Using Archimede’s principle, we can determine relative density (R.D.) of a body as Density of body R.D. = Density of pure water at 4 °C Physics For you | DECEmbEr ‘16
13
•
•
If a body is weighed in air (WA), in water (WW) and in a liquid (WL), then Loss of weight in liquid Specific gravity of liquid = Loss of weight in water W − WL = A WA − WW Weight of liquid displaced = upthrust = weight of body This is known as law of floatation. ¾ The floating body will be in stable equilibrium when the metacentre lies above centre of gravity of body. ¾ The floating body will be in unstable equilibrium when the metacentre lies below centre of gravity of body. ¾ The floating body will be in neutral equilibrium when the metacentre coincides with centre of gravity of body. ¾ If a person floats on his back on the surface of water, the apparent weight of the person is zero.
•
) Equation of continuity •
• •
•
•
14
Steady flow is defined as that type of flow in which the fluid characteristics like velocity, pressure and density at a point do not change with time. ¾ In steady flow, all the particles passing through a given point follow the same path and hence a unique line of flow. This line or path is called a streamline. Streamlines do not intersect each other. In an unsteady flow, the velocity, pressure and density at a point in the flow varies with time. Laminar flow is the flow in which the fluid particles move along well-defined streamlines which are straight and parallel. In this flow, the velocities at different points in the fluid may have different magnitudes, but their directions are parallel. Turbulent flow is an irregular flow in which the particles can move in zig-zag way due to which eddies formation take place which are responsible for high energy losses. In compressible flow, the density of fluid varies from point to point, whereas in incompressible flow, the density of the fluid remains constant throughout. Liquids are generally incompressible while gases are compressible. Physics For you | DECEmbEr ‘16
The continuity equation is the mathematical expression of the law of conservation of mass in fluid dynamics.
A1
v1
v2
A2 v2t
v1t
•
) Different Types of Fluid Flow •
Rotational flow is the flow in which the fluid particles while flowing along path-lines also rotate about their own axis. In irrotational flow, particles do not rotate about their axis.
In the steady flow, the mass of fluid entering into a tube of flow in a particular time interval is equal to the mass of fluid leaving the tube. m1 m2 = or r1A1v1 = r2 A2v2 Dt Dt Here r = r1 = r2 = density of fluid, v = velocity of fluid, A = Area of cross section of tube or A1v1 = A2v2 or Av = constant
) Bernoulli’s Theorem •
Bernoulli’s equation is mathematical expression of the law of mechanical energy conservation of fluid dynamics.
•
This theorem is applied to the ideal fluids. Characteristics of an ideal fluid are : ¾ The fluid is incompressible. ¾ The fluid is non-viscous. ¾ The fluid flow is steady. ¾ The fluid flow is irrotational. The sum of pressure energy, kinetic energy and potential energy per unit volume remains constant along a streamline in an ideal fluid flow i.e.,
•
1 P + rv 2 + rgh = constant 2 (Energy per unit volume) or
P v2 + + gh = constant r 2
(Energy per unit mass) P v2 or + + h = constant rg 2 g (Energy per unit weight)
In this equation,
P is called pressure head, rg
v2 is called velocity head and h is called 2g gravitational/potential head.
) Applications of Bernoulli’s Theorem •
Venturimeter : It is a device to measure the flow of speed of incompressible fluid. ¾ Volume of the fluid flowing out per second 2hrm g
Q = a1v1 = a1a2 ¾
v1 P2
P1
Speed of fluid at wide neck v1 =
•
r(a12 − a22 )
Area = a1
h
Area = a2 v2 Liquid of density m
The viscous force is directly It is independent of the proportional to the surface area of solid surfaces in area of contact of liquid contact. layers. The viscous force is directly proportional to the relative velocity between two layers of a liquid.
It is independent of the relative velocity of one body with respect to another body in contact.
It does not depend upon the normal reaction between the two layers of the liquid.
It is directly proportional to the normal reaction between the surfaces in contact.
•
2hrm g a22 × r a12 − a22
A2
P2 = P h H
(H – h)
¾
• A1
1 Pa
•
Horizontal range, R = v1 × t 2(H − h) = 2 gh × = 2 h(H − h) g H R will be maximum if h = and Rmax = H. 2 In general as shown in figure, speed of outflow, v1 = 2 gh +
2(P − Pa ) r
) Viscosity
•
Viscosity is the property of the fluid by virtue of which it opposes the relative motion between its adjacent layers. It is the fluid friction or internal friction. Difference between viscosity and solid friction Viscosity
Solid friction
It is internal friction as the It is external friction as layer exerting the friction the friction force is due force is internal to the to an external body. liquid.
velocity
•
dv dy dv is velocity gradient, h is coefficient of viscosity dy of fluid and A is contact area of the layers. Stoke’s law : When a sphere of radius r moves uniformly through a viscous liquid then retarding force experienced by the sphere is, Fv = 6phrv. Terminal velocity : When a solid sphere 4 FB = 3 r3g falls in a liquid then its accelerating velocity is Fv = 6rvT controlled by the viscous force of liquid and hence it attains a constant 4 velocity which is known W= 3 r3g as terminal velocity (vT). At equilibrium, FB + Fv = W 4 3 4 or pr sg + 6 phrvT = pr 3rg 3 3 2 2 r (r − s) g or vT = 9 h vT ¾ The variation of velocity with time (or distance) is shown in the adjacent graph. Time or distance F = −hA
Torricelli’s law : ¾ If the container is open at the top to the atmosphere then speed of efflux v1 = 2 gh 2
Newton’s law of viscosity : Viscous force acting between two layers of a liquid flowing in streamlined motion is given by
) Poiseuille’s Formula •
Volume of liquid flowing per second V through a horizontal capillary tube of length l, radius r, across a pressure difference P, under streamline motion, is given by Physics For you | DECEmbEr ‘16
15
V= • •
•
pPr 4 P = 8 hl R
Liquid resistance R =
•
8 hl
4
pr Two capillary tubes are joined in series. P = P1 + P2 and V is same through the two tubes. ¾ Equivalent liquid resistance, Rs = R1 + R2 Two capillary tubes are joined in parallel. Equivalent RR liquid resistance, RP = 1 2 . R1 + R2 ¾
•
•
In parallel, V = V1 + V2 but pressure difference P is same across both tubes.
) reynold’s Number •
•
•
•
The type of flow pattern (laminar or turbulent) is determined by a non-dimensional number called Reynold’s number (Re). Which is defined as rvd Re = h where r is the density of the fluid having viscosity h and flowing with mean speed v. d denotes the diameter of obstacle or boundary of fluid flow. Although there is not a perfect demarkation for value of Re for laminar and turbulent flow but some authentic references take the value as Re
2000
between to 2000
•
•
16
¾
On gradually increasing the speed of flow at certain speed transition from laminar flow to turbulent flow takes place. This speed is called critical speed. For lower density and higher viscosity fluid, laminar flow is more probable. It is the property of liquid by virtue of which its free surface possess a tendency to contract so as to acquire a minimum possible surface area and behave like a stretched membrane. Quantitatively, surface tension of liquid is F W S= = l A where F is the force acting on imaginary length l drawn tangentially to the liquid surface at rest. Physics For you | DECEmbEr ‘16
Formation of a bigger drop by a number of smaller drops ¾ When n number of smaller drops of a liquid, each of radius r, surface tension S are combined to form a bigger drop of radius R then Volume of bigger drop = volume of n smaller drops
1000
¾
4 3 4 pR = n × pr 3 or R = n1/3r 3 3 The surface area of bigger drop = 4pR2 = 4pn2/3r2, which is less than the area of n smaller drops. In this process energy is released, given by W = S × (4pr2n – 4pR2) = 4pSr2n2/3(n1/3 – 1) 1 1 = 4 pSR2 (n1/3 − 1) = 4 pSR3 − r R
) surface Tension •
W is the work done in increasing the area A of free surface of liquid at rest without any change in temperature. Molecules on the free surface of a liquid at rest experience maximum downward pulling force which gives rise to surface tension. Surface tension depends only on the nature of liquid and is independent of the surface area or length of the imaginary line drawn on the free surface of liquid at rest. Work done in forming a liquid drop/bubble ¾ Work done in forming a liquid drop of radius R, surface tension S is W = 4pR2S. ¾ Work done in forming a soap bubble of radius R, surface tension S is W = 2 × 4 pR2 S = 8pR2S. ¾ Work done in increasing the radius of a liquid drop from R1 to R2 is W = 4 pS(R22 − R12 ). ¾ Work done in increasing the radius of a soap bubble from R1 to R2 is W = 8 pS(R22 − R12 ).
¾
The increase in temperature of bigger drop Dq =
3S 1 1 − Jcr r R
where r = density and c = specific heat of liquid.
) Excess Pressure • • •
Excess pressure inside a liquid drop, P = 2 S/R Excess pressure inside a soap bubble, P = 4 S/R Pressure difference (P) across curved surfaces of radii R1 and R2
¾
If the curvatures are in mutually opposite direction, then 1 1 R1 R2 P =S − R R 1
¾
If the curvatures are in the same direction, then 1 1 P =S + R2 R R R 1
¾
2
2
For a cylindrical surface, P = because R1 = R and R2 = ∞
¾
•
•
S R
•
2S For a spherical surface, P = R because R1 = R2 = R
If there is an air bubble of radius R in a liquid of density r at a depth h below the surface of liquid, then total pressure inside that bubble is P = P0 + hrg + 2S/R where P0 is the atmospheric pressure and hrg is the hydrostatic pressure. If P1 and P2 are pressures inside the two soap bubbles and P0 is pressure outside each bubble, then ratio of 3
1
0
) Angle of contact and capillarity Angle of contact is defined as the angle between the
1. One end of a nylon rope of length 4.5 m and diameter 6 mm is fixed to a tree-limb. A monkey weighing 100 N jumps to catch the free end and stays there. The elongation of the rope and the corresponding change in the diameter respectively are (Young’s modulus of nylon = 4.8 × 1011 N m–2 and Poisson ratio of nylon = 0.2) (a) 3.32 × 10–5 m (b) 4.41 × 10–8 m –9 (c) 8.85 × 10 m (d) 1.94 × 10–9 m. 2. Water flows in a horizontal A B tube as shown in the figure. The pressure of water changes by 600 N m–2 between A and B where the areas of cross section are 30 cm2 and 15 cm2 respectively.
2S cos q 2S r = cos q = rrg Rrg R where S is the surface tension of the liquid, q is the angle of contact, r is the density of liquid, r is the radius of capillary tube, R is the radius of the meniscus and g is the acceleration due to gravity. ¾ If q > 90°, i.e., meniscus is convex, h will be negative, i.e., the liquid will fall in a capillary tube. ¾ If q = 90°, i.e., meniscus is plane, h = 0, so no phenomenon of capillarity. ¾ If q < 90°, i.e., meniscus is concave, h will be positive, i.e., the liquid will rise in the capillary. If a capillary tube is of insufficient length as compared to height to which liquid can rise in the capillary tube, then the liquid rises upto the full length of capillary tube but there is no overflowing of the liquid in the form of fountain. It is so because the liquid meniscus adjusts its radius of curvature so that hR = constant, i.e., hR = h′R′. h=
1
P −P their volumes is 2 0 . P −P •
•
tangents to solid and liquid surfaces at a point of contact inside the liquid. It depends on the nature of solid and liquid both and for concave meniscus, it is acute while for convex it is obtuse. The phenomenon of rise or fall of liquid in a capillary tube is known as capillarity. The rise or fall in a capillary tube is given by
•
The rate of flow of water (in cm3 s–1) through the tube will be (a) 6300 (b) 1890 (c) 2315 (d) 1680. 3. Water and mercury are filled in two cylindrical vessels up to same height. Both vessels have a hole in the wall near the bottom. If the velocity of water and mercury coming out of the holes are v1 and v2 respectively, then (a) v1 = v2 (b) v1 = 13.6 v2 (c) v1 = v2/13.6
(d) v1 = 13.6 v2 .
4. A ring is cut from a platinum tube has 8.5 cm internal and 8.7 cm external radius. It is supported horizontally from a pan of a balance so that it comes Physics For you | DECEmbEr ‘16
17
in contact with the water in a glass vessel. What is the surface tension of water if an extra weight = 3.97 g wt is required to pull it away from water ? (Take g = 980 cm s–2) (a) 72.04 dyne cm–1 (b) 35.98 dyne cm–1 (c) 54.03 dyne cm–1 (d) 18.02 dyne cm–1 5. A 45 kg boy whose leg bones are 5 cm2 in area and 50 cm long falls through a height of 2 m without breaking his leg bones. If the bones can stand a stress of 0.9 × 108 N m–2, then the Young’s modulus for the material of the bone is (Take g = 10 m s–2) (a) 2.25 × 109 N m–2 (b) 2.76 × 108 N m–2 (c) 4.42 × 107 N m–2 (d) 3.17 × 109 N m–2. 6. Two rods A and B of the same material and same length have radii r1 and r2 respectively. When they are rigidly fixed at one end and twisted by the same couple applied at the other end, then the ratio of the angles of twist at the ends of A and B is (a) r2 : r1 (b) r14 : r24 (c) r24 : r14 (d) r13 : r23. 7. A large tank filled with water to a height h is said to be emptied through a small hole at the bottom. Find the ratio of time taken for the level of water to fall down from h to h/2 and h/2 to zero. (a) 1 : 2 (b) 2 + 1 (c) 2 (d) 2 − 1 8. The lower end of a capillary tube is dipped in water. Water rises to a height of 8 cm. The tube is then broken at a height of 6 cm. The height of water column and angle of contact will be 3 3 (a) 6 cm, sin −1 (b) 6 cm, cos −1 4 4 1 1 (c) 4 cm, sin −1 (d) 4 cm, cos −1 . 2 2 9. The rate of steady volume of water through a capillary tube of length l and radius r under a pressure difference of P is V. This tube is connected with another tube of same length but half the radius in series. Then the rate of increase of steady volume through them is (the pressure difference across the combination is P) 16 V 17 V V V (a) (b) (c) (d) . 17 16 16 17 10. A cylindrical vessel is filled with water up to height H. A hole is bored in the wall at a depth h from the free surface of water. For maximum range, h is equal to H 3H H (a) (b) (c) (d) H. 2 4 4 18
Physics For you | DECEmbEr ‘16
11. Two blocks of masses 1 kg and 2 kg are connected by a metal wire going over a smooth pulley as shown in the figure. The breaking stress of the metal is 1 kg 2 × 109 N m–2. What should be the minimum radius of the wire used if it is not to break? (Take g = 10 m s–2) (a) 4.6 × 10–5 m (b) 9.2 × 10–5 m –4 (c) 2.3 × 10 m (d) 2.3 × 10–5 m
2 kg
12. If two soap bubbles of different radii are connected by a tube, (a) air flows from bigger bubble to the smaller bubble till the sizes become equal (b) air flows from bigger bubble to the smaller bubble till the sizes are interchanged (c) air flows from the smaller bubble to the bigger bubble (d) there is no flow of air. 13. A rectangular film of liquid is extended from (4 cm × 2 cm) to (5 cm × 4 cm). If the work done is 3 × 10–4 J, the value of the surface tension of the liquid is (a) 0.250 N m–1 (b) 0.125 N m–1 –1 (c) 0.2 N m (d) 8.0 N m–1. [NEET Phase II 2016] 14. Three liquids of densities r1, r2 and r3 (with r1 > r2 > r3), having the same value of surface tension T, rise to the same height in three identical capillaries. The angles of contact q1, q2 and q3 obey p > q1 > q2 > q3 ≥ 0 2 p (b) 0 ≤ q1 < q2 < q3 < 2 p (c) < q1 < q2 < q3 < p 2 p (d) p > q1 > q2 > q3 > . [NEET Phase II 2016] 2 (a)
15. Two non-mixing liquids of densities r and nr (n > 1) are put in a container. The height of each liquid is h. A solid cylinder of length L and density d is put in this container. The cylinder floats with its axis vertical and length pL (p < 1) in the denser liquid. The density d is equal to (a) {2 + (n – 1)p}r (b) {1 + (n – 1)p}r (c) {1 + (n + 1)p}r (d) {2 + (n + 1)p}r. [NEET Phase I 2016]
16. The Young’s modulus of steel is twice that of brass. Two wires of same length and of same area of cross section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weights added to the steel and brass wires must be in the ratio of (a) 4 : 1 (b) 1 : 1 (c) 1 : 2 (d) 2 : 1. [AIPMT 2015] 17. Consider a water jar of radius R R that has water filled up to height H H and is kept on a stand of height h. 2r Through a hole of radius r (r r3, so for positive value of cos q cos q1 > cos q2 > cos q3 p For 0 ≤ q < , q1 < q2 < q3 2 p Hence, option is (b) i.e., 0 ≤ q1 < q2 < q3 < 2 15. (b) : h h
d = density of cylinder A = area of cross section of cylinder Using law of floatation, Weight of cylinder = Upthrust by two liquids L × A × d × g = nr × (pL × A)g + r(L – pL)Ag d = npr + r(1 – p) = (np + 1 – p)r d = {1 + (n – 1)p} r 16. (d) : Let L and A be length and area of cross section of Steel Brass each wire. In order to have L, A L, A the lower ends of the wires to be at the same level (i.e., Ws Wb same elongation is produced in both wires), let weights Ws and Wb are added to steel and brass wires respectively. Then by definition of Young’s modulus, the elongation produced in the steel wire is Ws L Ys A and that in the brass wire is WL DLb = b Yb A DLs =
W/ A as Y = DL / L
But DLs = DLb (given) Ws L Wb L = Ys A Yb A Y As s = 2 Yb
\
or
17. (a) : Let v1 and v2 be the velocities of water when it leaks out through the hole and when it hits the ground respectively. Then, as per Bernoulli’s theorem, v12 + 2gh = v22 Now, according to Torricelli’s law, v1 = 2 gH ...(i) \ 2gH + 2gh = v22 ...(ii) According to continuity equation, a1v1 = a2v2 or pr 2 ⋅ 2 gH = px 2 ⋅ 2 g (H + h) [Using (i) and (ii)]
x2 = r 2
(L – pL) pL
d
Ws 2 = Wb 1
Ws Ys = Wb Yb (given)
1/ 4 H H or x = r H + h H +h
18. (c) : Equation of motion for the point mass ...(i) ma = mg – kv dv dt dv mg − kv ⇒ = or = mg − kv m dt m Integrating, v
t
0
0
dv 1 ∫ mg − kv = m ∫ dt
(
− kt t 1 v mg − [ln(mg − kv )]0 = m ; v = 1 − e k m k Putting (ii) in (i), we get
)
...(ii)
− kt
mg ( −kt /m ) or a = ge m k 1− e Hence option (c) represents the correct variation. ma = mg − k ×
19. (d) : Bulk modulus, B =
Normal stress Volumetric strain
N N = A (2pa)b 2 pa Da × b 2Da Volumetric strain = = a pa2 × b N a × \ B= 2 pab 2Da N = 4pb Da × B \ Required force = Frictional force = mN = (4pmBb)Da P=
20. (a) Physics For you | DECEmbEr ‘16
21
CLASS XI Series 6
CBSE
Thermodynamics Kinetic Theory
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 calculator is not allowed.
section-A
1. Can a system be heated and its temperature remains constant? 2. Air pressure in a car tyre increases during driving. Explain. 3. The volume of a given mass of a gas at 27 °C and 1 atm is 100 cc. What will be its volume at 327 °C? 4. Calculate the number of atoms in 39.4 g of gold. Molar mass of gold is 197 g mol–1. 5. During adiabatic changes, the volume of a gas is found to depend inversely on the square of its absolute temperature. Find how its pressure will depend on the absolute temperature. section-B
6. What happens to the change in internal energy of a gas during (a) isothermal expansion and (b) adiabatic expansion? 7. A monatomic ideal gas, initially at temperature T1 is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature T2 by releasing the piston suddenly. If L1 and L2 are the lengths of the gas column before 22
Physics For you | DECEmbEr ‘16
and after expansion respectively then, what is the ratio T1/T2? 8. Calculate the number of degrees of freedom of molecules of hydrogen in 1 cc of hydrogen gas at NTP. 9. Write any four fundamental postulates of the kinetic theory of an ideal gas. OR At what temperature does all molecular motion cease ? Explain. 10. Discuss whether the following phenomena are reversible : (a) Water fall (b) Rusting of iron section-c 11. A vessel A contains hydrogen and another vessel B whose volume is twice of A contains same mass of oxygen at the same temperature. Compare (a) average kinetic energies of hydrogen and oxygen molecules, (b) root mean square speeds of the molecules and (c) pressures of gases in A and B. Molecular weights of hydrogen and oxygen are 2 and 32 respectively.
12. Explain why (a) There is no atmosphere on the moon. (b) There is fall in temperature with altitude. 13. Estimate the average thermal energy of a helium atom at (a) room temperature 27 °C, (b) the temperature on the surface of the sun 6000 K, (c) the temperature of 10 million kelvin (the typical core temperature in case of a star). 14. Show that the slope of an adiabatic curve at any point is g times the slope of an isothermal curve at the corresponding point. 15. Explain why (a) Two bodies at different temperatures T1 and T2 if brought in thermal contact do not necessarily settle to mean temperature (T1 + T2)/2. (b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat. (c) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude.
19. 1 g of water at 100 °C is converted into steam at the same temperature. If the volume of steam is 1671 cm3, find the change in the internal energy of the system. Given latent heat of steam = 2256 J g–1 and 1 atmospheric pressure = 1.013 × 105 N m–2. 20. Ten small planes are flying at a speed of 150 km h–1 in total darkness in an air space that is 20 × 20 × 1.5 km3 in volume. You are in one of the planes, flying at random within this space with no way of knowing where the other planes are. On the average about how long a time will elapse between near collision with your plane. Assume for this rough computation that a safety region around the plane can be approximated by a sphere of radius 10 m.
16. Two cylinders A and B of equal capacity are connected to each other via a stopcock. The cylinder A contains a gas at standard temperature and pressure, while the cylinder B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following : (a) What is the final pressure of the gas in A and B? (b) What is the change in internal energy of the gas? (c) What is the change in temperature of the gas? (d) Do the intermediate states of the system (before settling to final equilibrium state) lie on its P-V-T surface?
OR (a) The difference between two specific heats of a gas is 5000 J kg–1 K–1 and the ratio of specific heats is 1.6. Find the two specific heats. (b) A cylinder containing one gram molecule of the gas was compressed adiabatically until its temperature rose from 27 °C to 97 °C. Calculate the work done on the gas. Given g = 1.5. 21. (a) When a molecule (or an elastic ball) hits a (massive) wall of container, it rebounds with the same speed. When a ball hits a massive bat held firmly, the same thing happens. However, when the bat is moving towards the ball, the ball rebounds with a different speed. Does the ball move faster or slower? (b) When gas in a cylinder is compressed by pushing in a piston, its temperature rises. Guess an explanation of this in terms of kinetic theory. (c) What happens when a compressed gas pushes a piston out and expands ? What would you observe?
17. Two perfect gases at absolute temperatures T1 and T2 are mixed. There is no loss of energy. Find the temperature of the mixture if the masses of the molecules are m1 and m2 and the number of molecules in the gases are n1 and n2 respectively.
22. Two thermally insulated vessels 1 and 2 are filled with air at temperature (T1, T2), volumes (V1, V2) and pressures (P1, P2) respectively. If the valve joining the two vessels is opened, what will be the temperature inside the vessel at equilibrium ?
18. A thermos flask contains coffee. It is vigorously shaken. Consider the coffee as the system. (a) Has any heat been added to it? (b) Has any work been done on it? (c) Has its internal energy changed? (d) Does its temperature rise?
section-D
23. One day Ramesh was pumping air into his cycle tyre. He noticed that both the volume and pressure of the air in the tyre were increasing simultaneously. He was a bit confused as he had Physics For you | DECEmbEr ‘16
23
learnt in his Physics class that pressure varies inversely with volume as per Boyle's law. Next day he talked to his Physics teacher about this phenomenon. The teacher thought it to be an important phenomenon and therefore explained to the whole class. Answer the following questions : (a) What are the values displayed by Ramesh and the teacher? (b) What is your opinion to correct the explanation for the given observation? section-e
24. State first law of thermodynamics and use it to find the expression for work done during adiabatic expansion. Write two limitations of first law of thermodynamics. OR What is an isothermal process? State two essential conditions for such a process to take place. Show analytically that work done by one mole of an ideal gas during isothermal expansion from volume V1 to volume V2 is given by W = RT loge
V2 V1
What is the change in internal energy of a gas, which is compressed isothermally? 25. What is Carnot engine ? Derive an expression for the efficiency of a Carnot engine. On what factors does it depend? OR Consider a P-V diagram in which the path followed by one mole of perfect gas in a cylindrical container is shown in figure.
(c) Given the internal energy for one mole of gas at temperature T is 3/2 RT, find the heat supplied to the gas when it is taken from state 1 to 2, with V2 = 2V1. 26. Show that the pressure exerted by an ideal gas is 1 P = r v2, where r is the density and v is the root 3 mean square velocity. OR Using the law of equipartition of energy, determine the values of CP, CV and g for (a) monatomic, (b) diatomic and (c) triatomic gases. solutions
1. From first law of thermodynamics, DQ = DU + DW ⇒ DQ = nCV DT + DW (Q DU = nCVDT) If system has constant temperature inspite of heat supplied then DT = 0 and DQ = DW. It means heat supplied (DQ) to the system is used in doing work (DW) against the surrounding. 2. A car tyre has a fixed volume (V). When car is driven, the temperature (T) of air in its tyre increases. So, according to Charles’ law, P ∝ T at constant V, pressure of the air in the car tyre increases. 3. Here, T1 = 27 °C = 300 K, V1 = 100 cm3 T2 = 327 °C = 600 K, V2 = ? At constant pressure, V ∝ T or
V1 T1 = V2 T2
⇒
V2 =
V1T2 100 × 600 = = 200 cm3 T1 300
4. Given, mass of gold, m = 39.4 g Molar mass of the gold, M = 197 g mol–1 Number of gold atoms in 39.4 g of gold NA 6.023 × 1023 mol −1 ×m= × 39.4 g M 197 g mol −1 = 1.205 × 1023 1 constant 5. Given, V ∝ 2 \ V = T T2 PV But = constant T =
(a) Find the work done when the gas is taken from state 1 to state 2. (b) What is the ratio of temperature T1/T2, if V2 = 2V1? 24
Physics For you | DECEmbEr ‘16
\
P constant = constant or P ∝ T3 ⋅ 2 T T
6. (a) Isothermal expansion : Temperature remains constant during an isothermal change. As internal energy is a function of temperature only, so it will remain constant during an isothermal change. As DT = 0, so DU = CV DT = 0 (b) Adiabatic expansion : For an adiabatic change, DQ = 0, so from first law of thermodynamics, DQ = DU + DW = 0 or DW = – DU During adiabatic expansion, work is done by a gas i.e., DW is positive, so DU must be negative. Hence internal energy of a gas decreases during an adiabatic expansion. 7. For an adiabatic process, TV g – 1 = constant For a monatomic gas, g = 5/3 \ TV2/3 = constant Let A = area of cross-section of the cylinder. Then, T1(L1A)2/3 = T2(L2A)2/3 \
T1 L2 = T2 L1
2 /3
8. At NTP, volume occupied by 1 g mole of gas = 22400 cc Number of molecules in 1 cc of H2
6.023 × 1023 = 2.688 × 1019 22400 Since each diatomic molecule has 5 degrees of freedom. \ Total number of degrees of freedom = 5 × 2.688 × 1019 = 1.344 × 1020 =
9. (i) All gases consist of molecules. The molecules are rigid elastic spheres and identical in all respects for a given gas and different for different gases. (ii) The size of a molecule is negligible as compared to the average distance between molecules. (iii) The molecules are in a state of continuous random motion, moving in all directions with all possible velocities. (iv) The molecules exert no force on each other or on the walls of the container except during collision. OR All molecular motion ceases at absolute zero or at 0 K. According to the kinetic interpretation of temperature internal energy of an ideal gas is purely kinetic, 3 2 E E = kBT or T = 2 3 kB
absolute temperature ∝ average kinetic energy of molecules \ For temperature = 0 K, average kinetic energy = 0 Thus at 0 K, the velocity of molecules becomes zero.
or
10. (a) Water fall : It is not a reversible process. During fall of the water, the major part of its potential energy is converted into kinetic energy of the water. However, on striking the ground, a part of it is converted into heat and sound. It is not possible to convert the heat and the sound produced along with the kinetic energy of water into potential energy so that the water can not rise back to its initial height. Therefore, water fall is not a reversible process. (b) Rusting of iron : During rusting, iron gets oxidised by the oxygen of the air. Since it is a chemical change so it is not a reversible process. 11. (a) For all gases at the same temperature, average kinetic energy per molecule is same and is 3 E = kBT 2 As the gases in both vessels are at the same temperature, so the ratio of their average kinetic energy per molecule = 1 : 1. (b) As vrms = \
vH = vO
3RT M
MO 32 = = 4 :1 MH 2
(c) According to kinetic theory, P =
1m 2 v 3 V rms
where m is the mass and V is the volume of the gas. Masses of both gases are equal. So the ratio of their pressure is 2
V PH v H 16 2 = × O = × = 32 : 1 PO vO VH 1 1
12. (a) Gravitational force of the moon is roughly 1/6th of the gravitational force of the earth, so the escape velocity of air molecules on the moon is smaller than that on the earth. As the moon is in the proximity of the Earth as seen from the sun so solar irradiance on the moon surface is same as that of the earth. Escape velocity of the particle on the moon is 2.38 km s–1. The rms speed of the constituents of air (O2, N2, CO2 and water vapour) lies between 0.4 km s–1 to 0.8 km s–1, which is smaller than 2.38 km s–1. Inspite of this a significant number of molecules have speeds greater than 2.38 km s–1 Physics For you | DECEmbEr ‘16
25
and hence escape easily. Now rest of molecules arrange the speed distribution for the equilibrium temperature. Again a significant number of molecules get their speeds greater than 2.38 km s–1 and escape. Due to this continuous process, the moon has lost its atmosphere over a long period of time. (b) We know, E = U + K = constant As the molecules move higher their potential energy (U) increase but and kinetic energy (K) decrease and hence fall in temperature. At greater height more volume is available for gas to expand and hence some cooling takes place i.e., a fall in temperature. 13. The average kinetic energy of the gas at a 3 temperature T, E = kT 2 where k = 1.38 × 10–23 J K–1 T = Temperature in kelvin (a) T = 27 °C = 27 + 273 = 300 K 3 –21 E = × 1.38 × 10−23 × 300 = 6.21 × 10 J 2 (b) T = 6000 K 3 –19 E = × 1.38 × 10−23 × 6000 = 1.242 × 10 J 2 (c) T = 10 × 106 K = 107 K 3 –16 E = × 1.38 × 10−23 × 107 = 2.07 × 10 J 2 14. For an isothermal change, PV = constant Differentiating both sides, we get P ⋅ dV + V ⋅ dP = 0 or V ⋅ dP = – PdV \ Slope of an isothermal curve, P dP =− V dV isothermal For an adiabatic change, PV g = constant Differentiating both sides, we get P . gVg–1 . dV + V g . dP = 0 \ Slope of an adiabatic curve, gP dP =− V dV adiabatic
Clearly, slope of an adiabatic curve = g × slope of an isothermal curve. As g > 1, so an adiabatic P-V curve is steeper than the corresponding isothermal P-V curve. 15. (a) The two bodies may have different masses and different materials i.e., they may have different 26
Physics For you | DECEmbEr ‘16
thermal capacities. In case the two bodies have equal thermal capacities, they would settle at the mean temperature (T1 + T2)/2. (b) The purpose of a coolant is to absorb maximum heat with least rise in its own temperature. This is possible only if specific heat is high because Q = mc DT. For a given value of m and Q the rise in temperature DT will be small if c is large. This will prevent different parts of the nuclear reactor from getting too hot. (c) The relative humidity of a harbour town is more than that of a desert town. Due to high specific heat of water, the variations in the temperature of humid air are less. Hence the climate of a harbour town is without the extreme of hot or cold. 16. (a) When the stopcock is suddenly opened, the volume available to the gas at 1 atm becomes twice the original volume and hence pressure becomes half the original volume (Boyle's law). Hence the pressure of the gas in each of the cylinders A and B is 0.5 atm. (b) As the system is thermally insulated, so DQ = 0. Also, the gas expands against zero pressure, so DW = 0. Hence by first law of thermodynamics, DU = 0 i.e., there is no change in the internal energy of the gas. (c) As there is no change in the internal energy of the gas, so the temperature of the gas remains unchanged. (d) No. The free expansion of the gas is very rapid and hence cannot be controlled. The intermediate states are non-equilibrium states and do not satisfy the gas equation. In due course, the gas returns to equilibrium state which lies on P-V-T surface. 17. According to the kinetic theory, the average kinetic 3 energy of a gas molecule = kBT . Before mixing 2 the two gases, the average kinetic energy of all the molecules of the gas 3 3 = kB n1T1 + kB n2T2 2 2 After mixing, the mean kinetic energy of both the 3 gases = kB(n1 + n2)T 2 where T is the temperature of the mixture. If there is no energy loss, then
3 3 3 kB(n1 + n2)T = kB n1T1 + kB n2T2 2 2 2 n T +n T T= 1 1 2 2 n1 + n2
18. (a) No. As the thermos flask is insulated, heat has not been added to the coffee (DQ = 0). (b) Yes. Some work is done by the man in shaking the coffee against the forces of viscosity i.e., DW is negative. (c) By first law of thermodynamics, DQ = DU + DW. As DQ = 0 and DW is negative, so DU is positive i.e., internal energy of the coffee increases. (d) Because of the increase in internal energy of the coffee, the temperature of the coffee will also increase. 19. Mass of water, m = 1 g = 10– 3 kg Latent heat of steam L = 2256 J g– 1 = 2256 × 103 J kg–1 Atmospheric pressure, P = 1.013 × 105 N m– 2 Volume of steam, Vs = 1671 cm3 = 1671 × 10–6 m3 Mass 10−3 Volume of water, Vw = = 10–6 m3 = Density 103 According to first law of thermodynamics, dQ = dU + PdV or mL = dU + P(Vs – Vw) \ Change in internal energy is dU = mL – P(Vs – Vw) = 10–3 × 2256 × 103 – 1.013 × 105 (1671 × 10–6 – 10–6) 5 = 2256 – 1.013 × 10 × 10–6 × 1670 = 2256 – 0.1013 × 1670 = 2256 – 169.171 = 2086.829 J 20. Here, v = 150 km h–1, N = 10 V = 20 × 20 × 1.5 km3. Diameter of plane, d = 2 R = 2 × 10 = 20 m = 20 × 10–3 km N 10 n= = = 0.0167 km–3 V 20 × 20 × 1.5 Mean free path of a plane 1 λ= 2 πd 2n Time elapse before collision of two planes randomly, λ 1 t= = v 2 πd 2nv 1 = 2 1.414 × 3.14 × (20) × 10−6 × (0.0167) × (150) ≈ 225 h
OR Here, CP – CV = 5000 J kg–1 K–1 C and P = 1.6 or CP = 1.6 CV CV \
1.6 CV – CV = 5000 or 0.6 CV = 5000
or
CV =
5000 = 8333.33 J kg–1 K–1 0. 6
CP = CV + 5000 J kg–1 K–1 = 13333.33 J kg–1 K–1 Here, T1 = 27 + 273 = 300 K, T2 = 97 + 273 = 370 K Work done in adiabatic compression of the gas is given by W=
R 8.3 × (370 − 300) (T2 − T1 ) = = −1162 J 1− g 1 − 1.5
21. (a) Let the speed of the ball be u relative to the wicket behind the bat. If the bat is moving towards the ball with a speed v relative to the wicket, then the relative speed of the ball to bat is v + u towards the bat. When the ball rebounds (after hitting the massive bat) its speed, relative to bat, v + (v + u) = 2v + u, moving away from the wicket. So the ball speeds up after the collision with the bat. For a molecule, this would imply an increase in temperature. (b) When a gas in cylinder is compressed by pushing in a piston, the speed of the molecules or their kinetic energy increases. This increases the temperature of the gas. (c) When a compressed gas pushes a piston out, the speed of the molecules or their kinetic energy decreases. This decreases the temperature of the gas. 22. As PV = nRT \
n=
PV RT
For first vessel, n1 =
P1V1 RT1
For second vessel, n2 =
P2V2 RT2
For the combined vessel, n = But n = n1 + n2 \ P (V1 + V2 ) = P1V1 + P2V2 RT RT1 RT2
P (V1 + V2 ) RT
Physics For you | DECEmbEr ‘16
27
or
T=
T1T2 P (V1 + V2 ) P1V1T2 + P2V2T1
As V = constant so DW = 0 Also, DQ = 0 so DU = 0 or DT = 0 i.e., T = constant Using Boyle's law, P(V1 + V2) = P1V1 + P2V2 T T (P V + P V ) Hence, T = 1 2 1 1 2 2 P1V1T2 + P2V2T1
23. (a) The values displayed by Ramesh are good observant, highly interested in learning physical phenomena observed in daily life. The values displayed by the teacher are providing good education and undertaking the doubts of student. (b) Boyle's law is valid for a fixed mass of a gas. When we pump air into a cycle tyre, air molecules are pushed into the tyre and so the mass of the air in the tyre increases. Hence Boyle's law is not applicable under the given situation. 24. If some heat is supplied to a system which is capable of doing work, then the quantity of heat absorbed by the system will be equal to the sum of the increase in its internal energy and the external work done by the system on the surroundings. Let DQ = Heat supplied to the system by the surroundings DW = Work done by the system on the surroundings DU = Change in internal energy of the system Then according to the first law of thermodynamics, P DQ = DU + DW Area = A Suppose the system is a gas contained in a cylinder dx provided with a movable piston. Then the gas does work in moving the piston. The work Gas done by the system against a constant pressure P is DW = Force × Distance Q = Pressure × Area × Distance = PA dx or DW = P DV where DV = Adx = the change in the volume of the gas. So, the first law of thermodynamics takes the form, DQ = DU + P DV When the piston moves up through a small distance dx, the work done by the gas will be dW = PA dx = P dV 28
Physics For you | DECEmbEr ‘16
where A is the cross-sectional area of the piston and dV = A dx is the increase in the volume of the gas. Suppose the gas expands adiabatically and changes from the initial state (P1, V1, T1) to the final state (P2, V2, T2). The total work done by the gas will be Wadiabatic =
dx
Gas Insulating wall
V2
∫ P dV
V1
For an adiabatic change PV g = K or P = KV –g \
Area = A
Wadiabatic =
V2
∫ KV
−g
(K = constant)
dV
V1 V2
=K ∫V V1
−g
V
V 1− g 2 dV = K 1 − g V1
(
)
K g g V21− − V11− 1− g 1 g g = KV11− − KV21− g −1 But K = P1V1g = P2V2g 1 g g \ Wadiabatic = P1V1gV11− − P2V2gV21− g −1 1 (P V − P V ) Wadiabatic = g −1 1 1 2 2 Limitations of the first law of thermodynamics are as follows : (i) It does not indicate the direction of transfer of heat. (ii) It does not tell anything about the conditions under which heat can be converted into mechanical work. (iii) It does not indicate the extent to which heat energy can be converted into mechanical work continuously. OR An isothermal process is one in which the pressure and volume of the system change but temperature remains constant. Essential conditions for an isothermal process to take place are (i) The walls of the container must be perfectly conducting to allow free exchange of heat between the system and the surroundings. =
(
)
(
)
Wisothermal =
V2
∫ P dV
V1
For n moles of a gas, PV = nRT or P = \
Wisothermal =
V2
∫
V1
nRT dV = nRT V
V2
nRT V
1
∫ V dV
V1
V
= nRT[lnV ]V2 = nRT(ln V2 – ln V1) 1
= nRT ln or
V2 V1
Wisothermal = 2.303 nRT log
V2 V1
= 2.303 nRT log
P1 P2
The change in internal energy of a gas, which is compressed isothermally is zero because the internal energy of an ideal gas depends only on its temperature. As temperature remains constant, there is no change in internal energy. 25. Carnot engine is an ideal reversible heat engine that operates between two temperatures T1 (source) and T2 (sink). The working substance is carried through a reversible cycle of the following four steps : Step 1 : Isothermal expansion (AB). Place the cylinder on the source so that the gas acquires the temperature T1 of the source. As the gas absorbs the required amount of heat from the source, it expands isothermally. If Q1 heat is absorbed from the source and W1 work is done by the gas in isothermal
expansion which takes its states from (P1, V1, T1) to (P2, V2, T1), then V W1 = Q1 = nRT1 ln 2 = area ABMKA V1 P
A(P1, V1, T1)
batic Adiaression p com
(ii) The process of compression or expansion should be very slow, so as to provide sufficient time for the exchange of heat. Work done by the gas when the piston moves up through a small distance dx is given by dW = PA dx = PdV Area = A where A is the cross-sectional area of the piston and dV = Adx, is the small dx increase in the volume of the gas. Suppose the gas expands isothermally from initial Gas state (P1, V1) to the final state Conducting wall (P2, V2). The total amount of work done will be
Isothermal expansion + Q1
B(P2, V2, T1)
Adiabatic expansion
D(P4, V4, T2) – Q2 O
C(P3, V3, T2)
Isothermal compression K
L
M
N
V
Step 2 : Adiabatic expansion (BC). The gas is now placed on the insulating stand and allowed to expand slowly till its temperature falls to T2. If W2 work is done by the gas in the adiabatic expansion which takes its state from (P2, V2, T1) to (P3, V3, T2), then nR(T1 − T2 ) W2 = = area BCNMB g −1 Step 3 : Isothermal compression (CD). The gas is now placed in thermal contact with the sink at temperature T2. The gas is slowly compressed so that as heat is produced, it easily flows to the sink. The temperature of the gas remains constant at T2. If Q2 heat is released by the gas to the sink and W3 work is done on the gas by the surroundings in the isothermal compression which takes its state from (P3, V3, T2) to (P4, V4, T2), then V W3 = Q2 = nRT2 ln 3 = area CNLDC V4 Step 4 : Adiabatic compression (DA). The cylinder is again placed on the insulating stand. The gas is further compressed slowly till it returns to its initial state (P1, V1, T1). If W4 is the work done in the adiabatic compression from (P4, V4, T2) to (P1, V1, T1), then nR(T1 − T2 ) W4 = = area DAKLD g −1
Total work done by the gas = W1 + W2 (in steps 1 and 2) Total work done on the gas = W3 + W4 (in steps 3 and 4) \ Net work done by the gas in one complete cycle, Physics For you | DECEmbEr ‘16
29
W = W1 + W2 – (W3 + W4) But W2 = W4 \ W = W1 – W3 = Q1 – Q2 Also, W = area ABMKA + area BCNMB – area CNLDC – area DAKLD or W = area ABCDA Hence in a Carnot engine, the mechanical work done by the gas per cycle is numerically equal to the area of the Carnot cycle. Efficiency of Carnot engine is defined as the ratio of the net work done per cycle by the engine to the amount of heat absorbed per cycle by the working substance from the source. Q W Q1 − Q2 = =1− 2 \ h= Q1 Q1 Q1 or
h=1–
nRT2 ln(V3 / V4 ) nRT1 ln(V2 / V1 )
T2 T1
OR Given, PV1/2 = constant = C
(a) Work done by the gas DW =
V2
V2
V1
V1
∫ PdV = ∫
C V
dV
V
2 1 / 2 V =C = 2C( V2 − V1 ) 1 2 V1 \ DW = 2P1 V1 ( V2 − V1 ) 30
Physics For you | DECEmbEr ‘16
⇒ T ∝ \
...(ii)
V
T1 V1 V1 1 = = = 2V1 T2 V2 2
3 (c) Given, U = RT 2 3 3 ⇒ DU = RDT = R(T2 − T1 ) 2 2 3 ⇒ DU = RT1( 2 − 1) 2 From eqn. (i)
...(iii) (as V2 = 2V1)
(using eqn. (iii))
DW = 2P1 V1 ( V2 − V1 ) = 2P1V1 ( 2 − 1)
Now step 2 is an adiabatic expansion, therefore T1V2g–1 = T2V3g–1 ...(i) Similarly, step 4 is an adiabatic compression, therefore T1V1g–1 = T2V4g–1 ...(ii) On dividing (i) by (ii), we get g −1 g −1 V V V3 V2 or 2 = 3 = V1 V4 V1 V4 Hence, h = 1 –
(b) Q PV = nRT C C ⇒ V = nRT ⇒ T = V nR V
...(i)
\
= 2RT1 ( 2 − 1) From first law of thermodynamics, 7 DQ = DU + DW = RT1 ( 2 − 1) 2
26. Consider an ideal gas enclosed in a cubical y vessel of edge L. Also Area = A (vx , vy , vz) there are n molecules (–vx , vy , vz) per unit volume. A vx t z x molecule moving with velocity (vx , vy , vz) hits the planar wall (perpendicular to x-axis) of area A. As the collision is elastic, the molecule rebounds with the same velocity. The y and z components of velocity do not change while the x-component reverses sign. So the velocity after the collision is (–vx , vy , vz). The change in momentum of the molecule = – mvx – mvx = – 2mvx By the conservation of momentum, the momentum imparted to the wall in each collision = 2 mvx \ Number of molecules hitting wall of area A in time Dt 1 = Avx Dt × number of molecules per unit volume 2 1 = Avx Dt n 2 Total momentum transferred to the wall in time Dt is 1 Dp = 2 mvx × Avx Dt n = nmvx2 A Dt 2 Contd. on Page No. 82
integer Answer tyPe
1. A stone is thrown vertically upward. When the stone is at point A, its distance from a certain point O is 6 5 m at t = 0 and the component of velocity along OA is non-zero. When it is at point B(OB = 10 m), the component of velocity along OB is zero. When it is at point C(OC = 6 m), the component of velocity of the particle along OC is zero. If the velocity of projection of the stone is v0 = 5n m s–1, then find the value of n. 2. A cord of length 64 m is used to connect a 100 kg astronaut to a spaceship whose mass is much larger than that of the astronaut. The value of the tension in the cord is x × 10–2 N. Assume that the spaceship is orbiting near the earth’s surface. Also assume that the spaceship and the astronaut fall on a straight line from the earth’s centre. The radius of the earth is 6400 km. Find the value of x. 3. A vessel contains two immiscible 1 liquids of density r1 = 1000 kg m–3 –3 and r2 = 1500 kg m . A solid block 2 of volume V = 10–3 m3 and density d = 800 kg m–3 is tied to one end of a string and the other is tied to the bottom of the vessel as shown in figure. The block is immersed with 2/5th of its volume in the liquid of higher density and 3/5th in the liquid of lower density. The entire system is kept in an elevator which is moving upwards with an acceleration of a = g/2. Find the tension (in N) the string. (Take g = 10 m s–2) 4. A projectile is projected from ground with least velocity to cross a wall 3.6 m high and 4.8 m away from the point of projection. The range of projectile
class-Xi
n is 8.4 m. If angle of projection is a and tan a = . 4 Find the value of n. 5. A circular tube of mass M is placed vertically on a horizontal surface as shown in figure. Two small spheres each of m m mass m, just fit in the tube one released from the top. If M the tube looses contact with the ground at q = 60° then find the value of m/M. 6. A man weighing 60 kg is standing on a trolly weighing 240 kg. The trolly is resting on frictionless horizontal rails. If the man starts walking on the trolly along the rails at a speed 1 m s–1, then after 5 s, his displacement in metre relative to the ground will be 7. A homogeneous disc with a radius 0.2 m and mass 5 kg rotates around an axis passing through its centre. The angular velocity of rotation of the disc as a function of time is given by the formula w = 2 + 6t. The tangential force applied to the rim of the disc (in N) is 8. One end of a steel wire is fixed to ceiling of an elevator moving up with an acceleration 2 m s–2 and a load of 10 kg hangs from other end. Area of cross section of the wire is 2 cm2. The longitudinal strain in the wire is n × 10–6. What is the value of n? (Take g = 10 m s–2 and Y = 2 × 1011 N m–2) 9. There are two pendulums of lengths l1 (= 81 cm) and l2 (= 64 cm) which start oscillating. At some Physics For you | DECEmbEr ‘16
31
instant of time, both the pendulums are passing from mean position in the same phase. After how many oscillations of shorter pendulum, both the pendulums will be in the same phase in mean position? 10. A bus is moving with a velocity of 5 m s–1 towards a huge wall. The driver sounds a horn of frequency 165 Hz. If the speed of sound in air in 335 m s–1, the number of beats per second heard by the passengers in the bus would be 11. A vessel has 6 g of hydrogen at pressure P and temperature 500 K. A small hole is made in it so that hydrogen leaks out. How much hydrogen in gram leaks out if the final pressure is P/2 and temperature falls to 300 K? 12. A body hanging from a massless spring stretches it by 2 cm at earth’s surface. How much will the same body stretch the spring (in cm) at height 2624 km from the surface of earth? (Take radius of earth = 6400 km) 13. The rate of change of position with respect to time gives the velocity of the particle. Figure shows position-time graph for two particles P and Q. The ratio of velocities of both particles P and Q is n. Find the value of n. 14. A transverse wave of amplitude 5 mm is generated at one end (x = 0) of a long string by a vibrating source of frequency 500 Hz. At a certain instant of time, the displacement of a particle A at x = 1 m is –5 mm and that of particle B at x = 2 is +5 mm. The wavelength of the wave is k m. Find the value of k. 15. When the system shown in the diagram is in 450 N m–1 equilibrium, the right m m spring is stretched by 1 cm.
150 N m–1
The coefficient of static friction between the blocks is 0.3. There is no friction between the bottom block and the supporting surface. The force constants of the springs are 150 N m–1 and 450 N m–1 (refer figure). The blocks have equal mass of 2 kg each. Find the maximum amplitude (in cm) of 32
Physics For you | DECEmbEr ‘16
the oscillations of the system shown in the figure that does not allow the top block to slide on the bottom. solutions
1. (4) : From figure, AC = (6 5 )2 − 62 = 6 5 − 1 = 12 m And CB = 102 − 62 = 8 m \ The maximum height attained by the ball is H = AB = 12 m + 8 m = 20 m \ 2 v = u2 – 2gH or 02 = v02 – 2gH \ v0 = 2 gH = 2 × 10 × 20 = 20 m s −1 = 5n m s −1 \ n=4 2. (3) : As according to given problem the mass of satellite M is much greater than that of astronaut m, as the centre of mass of the system will be close to satellite and as the satellite is orbiting close to the surface of earth, the equation of motion of the system (S + A) will be GMe ( M + m) = ( M + m)Rw2 2 h R R GMe r Rw 2 = =g ...(i) Me S R2 M TA and the equation of motion m of the astronaut will be GMem + T = mr w2 2 r GMe T = m r w2 − r2 r R 2 T = mg − R r
(Using eqn. (i))
( R + h) R2 = mg − R (R + h)2 −2 h h T = mg 1 + − 1 + R R =
3 mgh R
(as r = R + h)
−2 h 2h as 1 + ≈ 1 − R R
So substituting the given data,
T=
3 × 100 × 10 × 64 6400 × 103
= 3 × 10
−2
N
\ x=3 3. (6) : We will analyse this problem from the reference frame of elevator. Total buoyant force on the block, 2 3 3 2 Fb = V r2 + V r1 ( g + a) 5 1 5 1 5 5 From the condition of equilibrium, Fb = T + Vd(g + a) T = Fb – Vd(g + a)
v 2 sin 2a v 2 sin 2a R= 0 or 8.4 = 0 g 10 42 sin a cos a
v0
3.6 m
v02 =
y
4.8 m 8.4 m
N
mg
Mg
For any sphere, N + mg cos 60° = mg mg = mg ⇒ N = 2 2 From eqns (i) and (ii), mg m = Mg ⇒ =2 2 M
mv 2 R
N+
...(ii)
6. (4) : Here, m1 = 60 kg, m2 = 240 kg, u1 = 1 m s–1, u2 = 0 If v is combined velocity of trolly and man, then applying the principle of conservation of linear momentum, (m1 + m2) v = m1u1 + m2u2 m u + m2 u2 60 × 1 + 0 = \ v= 1 1 = 0.2 m s–1 m1 + m2 60 + 240
3 2 = ( g + a)V r2 + r1 − d 5 5 As a = g/2 \ (g + a) = 10 + 5 = 15 m s–2 Putting the values, 3 2 = 15 × 10−3 × 1500 + × 1000 − 800 = 6 N 5 5 4. (7) : Here, R = 8.4 m, x = 4.8 m, y = 3.6 m
v02 sin 2a = 84
mv2 R
N
N
x
The equation of trajectory is gx 2 y = x tan a − 2v02 cos2 a
10(4.8)2 42 2× cos2 a sin a cos a 5 3.6 = 4.8 tan a − (4.8)2 tan a 42 7 n \ tan a = 1.75 = = or n = 7 4 4 3.6 = 4.8 tan a −
5. (2) : Conserving energy for any sphere 1 0 = − mg (R − R cos 60°) + mv 2 2 v 2 gR = ⇒ v = gR 2 2 As shown in figure, for tube 2N cos 60° = Mg (when it just lift off) N = Mg ...(i)
Velocity of the man with respect to ground = 1 – 0.2 = 0.8 m s–1 Displacement of the man is 5 s = 0.8 × 5 = 4 m 7. (3) : Here, r = 0.2 m, M = 5 kg, F = ? As w = 2 + 6t dw \ = 6 = a, angular acceleration dt Now, τ = I a = rF sin 90° Ia 1 a = Mr 2 × r 2 r 1 6 = × 5 (0.2)2 × =3N 2 0. 2
\ F=
8. (3) : Tension in wire, F = m(g + a) = 10(10 + 2) = 120 N F 120 Stress, S = = = 60 × 104 N m−2 A 2 × 10−4 Strain =
S 60 × 104 = = 3 × 10−6 Y 2 × 1011
As per question, n × 10–6 = 3 × 10–6 or n = 3 Physics For you | DECEmbEr ‘16
33
9. (9) : As T = 2π \
T1 = T2
l or T ∝ l g
l1 = l2
81 9 = 64 8
...(i)
Let shorter pendulum makes n oscillations, then longer pendulum will make one less oscillation than n to come in phase again, while passing through mean position, i.e., ...(ii) nT2 = (n – 1)T1 or T1 = n T2 n − 1 From (i) and (ii), n 9 = or 8n = 9n – 9 or n = 9 n −1 8
Case II : For reflected sound from the wall, bus acts as listener uL = 5 m s–1 (u + uL ) u′ (335 + 5) × 335 u′′ = = = 170 Hz u 335 × 2 \ Number of beats per second = u′′ – u = 170 – 165 = 5 11. (1) : Here, m = 6 g, T = 500 K m 6 From PV = ...(i) RT , PV = R × 500 M M If x g of hydrogen leaks out, then, P Pressure = , T = 300 K 2 P (6 − x ) ...(ii) \ 2 V = M R × 300 From (i) and (ii), 2 (6 − x ) 6R × 500 = R × 300 M M or 30 = (12 – 2x) 3 ⇒ x = 1 g 12. (1) : Let g′ be the acceleration due to gravity at height h.
or 34
gR 2
6400 = g× 2 + 6400 2624 ( R + h)
g′ ≈ g ×
1 g = 2 2
Physics For you | DECEmbEr ‘16
g′ x g′ 1 = or x = x0 = × 2 = 1 cm g x0 g 2 (\ x0 = 2 m) 13. (1) : The slope of x-t graph gives velocity. 20 − 15 \ vP = = 0.5 m s−1 10 − 0 15 − 10 and vQ = = 0.5 m s −1 10 − 0 v 0.5 \ n= P = =1 vQ 0.5
10. (5) : Case I : The bus acts as the source us = 5 m s–1, u = 335 m s–1, u = 165 Hz u × u 335 × 165 335 u′ = = = Hz u − us 335 − 5 2
Then, g ′ =
If m is the mass of the particle and k is the spring constant of the spring, then At the surface of earth, mg = kx0 ...(ii) At height h, mg′ = kx ...(iii) From (ii) and (iii)
2
...(i)
14. (2) : y = a cos(kx – wt) and a = 5 mm At t = 0, y = a cos kx It is given that at x = 1 m, y = –5 mm = –a ⇒ cos(k) = –1 and at x = 2 m, y = +5 mm = +a ⇒ cos(2k) = 1 2π \ k = π or =π⇒λ=2m λ 15. (3) : Suppose origin is at the equilibrium position and the direction of increasing x is towards the right. If the blocks are at the origin, the net force on them is zero. If the blocks are displaced by a small distance x to the right of the origin, value of the net force on them is –4kx. Applying Newton’s second law to the two-block system gives \ –4kx = 2ma ( k = 150 N m–1) Applying Newton’s second law to the lower block gives k(x1 – x) – f = ma where x1 = initial stretch and f is the magnitude of the frictional force. f = k(x1 + x) The maximum value of x is the amplitude A and the maximum value for f is msmg. Thus, msmg = k(x1 + A) m mg or A = s − x1 k On putting the values, we get A=
0.3 × 2 × 10 1 − = 0.3 m = 3 cm 150 100
MPP-6
Class XI
T
his specially designed column enables students to self analyse their extent of understanding of specified chapter. 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.
Thermal Properties of Matter Total Marks : 120
Time Taken : 60 min
NEET / AiiMs / PMTs Only One Option Correct Type
1. On a temperature scale Y, water freezes at –160 °Y and boils at –50 °Y. On this Y scale, a temperature of 340 K is (a) –106.3 °Y (b) – 96.3 °Y (c) –86.3 °Y (d) –76.3 °Y 2. Six identical conducting rods are joined as shown. The ends A and D are maintained at 200 °C and 20 °C respectively. No heat is lost to surroundings. The temperature of the junction C will be A 200 °C
(a) 60 °C
B
(b) 80 °C
C
D 20 °C
(c) 100 °C (d) 120 °C
3. Two thin rods of lengths L1 and L2 at a certain temperature are joined to each other end to end. The composite rod is then heated through a temperature T. The coefficients of linear expansion of the two rods are a1 and a2 respectively. Then, the effective coefficient of linear expansion of the composite rod is α + α2 (a) 1 (b) α1 + α2 2 L α + L2 α1 L α +L α (c) 1 2 (d) 1 1 2 2 L1 + L2 L1 + L2 4. Temperature of 100 g of water in a thermoflask remains fixed for a pretty long time at 50 °C. An
equal mass of sand at 20 °C is poured in the flask and shaken for some time so that the temperature of the mixture is 40 °C. Now the experiment is repeated with 100 g of a liquid at 50 °C and an equal mass of sand at 20 °C when the temperature of the mixture is found to be 30 °C. The specific heat of the liquid (in kJ kg–1 K–1) is (Specific heat of water = 4200 J kg–1 K–1) (a) 1.05 (b) 2.01 (c) 1.55 (d) 1.95 5. Two slabs A and B of different materials but of the same thickness are joined end to end to form a composite slab. The thermal conductivities of A and B are K1 and K2 respectively. A steady temperature difference of 12 °C is maintained across the K composite slab. If K1 = 2 , the temperature 2 difference across slab A is (a) 4 °C (b) 6 °C (c) 8 °C (d) 10 °C 6. An ideal gas is expanding such that PT2 = constant. The coefficient of volume expansion of the gas is 3 1 2 4 (b) (c) (d) T T T T 7. 22320 cal of heat is supplied to 100 g of ice at 0 °C. If the latent heat of fusion of ice is 80 cal g–1 and latent heat of vaporization of water is 540 cal g–1, the final amount of water thus obtained and its temperature respectively are (a) 8 g, 100 °C (b) 100 g, 90 °C (c) 92 g, 100 °C (d) 82 g, 100 °C (a)
Physics For you | december ‘16
35
8. When a liquid is heated in a glass vessel, its coefficient of apparent expansion is 1.03 × 10–3 °C–1. When the same liquid is heated in a copper vessel, its coefficient of apparent expansion is 1.006 × 10–3 °C–1. If the coefficient of linear expansion of copper is 17 × 10–6 °C–1, then the coefficient of linear expansion of glass is (a) 8.5 × 10–4 °C–1 (b) 9 × 10–6 °C–1 (c) 27 × 10–6 °C–1 (d) 10 × 10–4 °C–1
13. Assertion : For higher temperatures, the peak emission wavelength of a black body shifts to lower wavelengths. Reason : Peak emission wavelength of a black body is proportional to the fourth-power of temperature.
9. A lead bullet of unknown mass is fired with a speed of 180 m s–1 into a tree in which it stops. Assuming that in this process two third of heat produced goes into the bullet and one third into wood. The temperature of the bullet raises by (Specific heat of lead = 0.12 J g–1 °C–1) (a) 140 °C (b) 106 °C (c) 90 °C (d) 100 °C
15. Assertion : When hot water is poured in a beaker of thick glass, the beaker cracks. Reason : Outer surface of the beaker expands suddenly.
10. What fraction of the volume of a glass flask must be filled with mercury so that the volume of the empty space may be the same at all temperatures? (aglass = 9 × 10–6 °C–1, gHg = 18.9 × 10–5 °C–1) 1 1 1 1 (b) (c) (d) 2 7 5 4 11. The temperature of a room heated by a heater is 20 °C, when outside temperature is –20 °C, and it is 10 °C, when the outside temperature is –40 °C. The temperature of the heater is (a) 60 °C (b) 40 °C (c) 80 °C (d) 100 °C (a)
12. A piece of iron is heated in a flame. It first becomes dull red, then becomes reddish yellow and finally turns to white hot. The correct explanation for the above observation is possible by using (a) Newton’s law of cooling (b) Stefan’s law (c) Wien’s displacement law (d) Kirchhoff ’s law 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 the assertion. (b) If both assertion and reason are true but reason is not the correct explanation of the assertion. (c) If assertion is true, but reason is false. (d) If both assertion and reason are false. 36
Physics For you | december ‘16
14. Assertion : While measuring the thermal conductivity of a liquid experimentally, the upper layer is kept hot and the lower layer is kept cold. Reason : This avoids heating of liquid by convection.
JEE MAiN / JEE AdvANcEd / PETs Only One Option Correct Type
16. Two identical thin metal strips, one of aluminium and the other of iron are riveted together to form a bimetallic strip. The temperature is raised by 50 °C. If the central planes of the two strips are separated by 2 mm and the coefficients of thermal expansion for aluminium and iron are respectively 30 × 10–6 °C–1 and 10 × 10–6 °C–1, the average radius of curvature of the bimetallic strip is about (a) 50 cm (b) 100 cm (c) 150 cm (d) 200 cm 17. Parallel rays of light of intensity I = 912 W m–2 are incident on a spherical black body kept in surroundings of temperature 300 K. Take StefanBoltzmann constant s = 5.7 × 10–8 W m–2 K–4 and assume that the energy exchange with the surroundings is only through radiation. The final steady state temperature of the black body is close to (a) 330 K (b) 660 K (c) 990 K (d) 1550 K 18. A 10 W electrical heater is used to heat a container filled with 0.5 kg of water. It is found that the temperature of the water and the container rises by 3 K in 15 minutes. The container is then emptied, dried, and filled with 2 kg of an oil. It is now observed that the same heater raises the temperature of the container-oil system by 2 K in 20 minutes. Assuming no other heat losses in any of the processes, the specific heat capacity of the oil is (Specific heat of water = 4200 J kg–1 K–1) (a) 2550 J kg–1 K–1 (b) 5100 J kg–1 K–1 (c) 3000 J kg–1 K–1 (d) 1500 J kg–1 K–1
heat
1L
0 1L A
B
5L 6L E 3K
2K
C
4K
D
5K
3L 4L
6K
heat flow through A and E slabs are same. heat flow through slab E is maximum. temperature difference across slab E is smallest. heat flow through C = heat flow through B + heat flow through D. 21. The temperature drop through a two layer furnace wall is 900 °C. Each layer is of equal area of cross section. Which of the following actions will result in lowering the temperature T of the interface? (a) (b) (c) (d)
lB corresponding to maximum spectral radiancy in the radiation from B is shifted from the wavelength lA corresponding to maximum spectral radiancy in the radiation from A, by 1 µm. If the temperature of A is 5802 K, then (a) the temperature of B is 1934 K (b) lB = 1.5 µm (c) the temperature of B is 1160 K (d) the temperature of B is 2901 K 23. A solid material is supplied with heat at a constant rate. E The temperature of the C material varies with heat D A CD = 2AB input as shown in figure. B Which of the following O Heat input interpretations from the graph is/are correct? (a) AB represents the change of state from solid to liquid. (b) CD represents change of state from liquid to vapour. (c) Latent heat of fusion is twice the latent heat of vaporization. (d) Latent heat of vaporization is twice the latent heat of fusion. Temperature
19. Ice starts forming in a lake with water at 0 °C, when the atmospheric temperature is –10 °C. If time taken for 1 cm of ice to be formed is 7 hours, the time taken for the thickness of ice to increase from 1 cm to 2 cm is (a) 7 hours (b) less than 7 hours (c) more than 7 hours but less than 14 hours (d) more than 14 hours More than One Options Correct Type 20. A composite block is made of slabs A, B, C, D and E of different thermal conductivities (given in terms of a constant K) and sizes (given in terms of length, L) as shown in the figure. All slabs are of same width. Heat Q flows only from left to right through the blocks. Then in steady state
Inner Outer layer layer 1000 °C
100 °C T
(a) By increasing the thermal conductivity of outer layer. (b) By increasing the thermal conductivity of inner layer. (c) By increasing the thickness of outer layer. (d) By increasing the thickness of inner layer. 22. Two bodies A and B have thermal emissivities of 0.01 and 0.81 respectively. The outer surface areas of the two bodies are equal. The two bodies emit total radiant power at the same rate. The wavelength Physics For you | december ‘16
37
(a) 1000 : 39 (c) 243 : 130
Integer Answer Type –1
–1
24. A piece of ice (specific heat capacity = 2100 J kg °C and latent heat = 3.36 × 105 J kg–1) of mass m grams is at –5 °C at atmospheric pressure. It is given 420 J of heat so that the ice starts melting. Finally, when the ice-water mixture is in equilibrium, it is found that 1 g of ice has melted. Assuming there is no other heat exchange in the process, the value of m is 25. A metal rod AB of length 10x has its one end A in ice at 0 °C and the other end B in water at 100 °C. If a point P on the rod is maintained at 400 °C, then it is found that equal amounts of water and ice evoparate and melt per unit time. The latent heat of evaporation of water is 540 cal g–1 and latent heat of melting of ice is 80 cal g–1. If the point is at a distance of lx from the ice end A, the value of l is 26. The density of a substance at 0 °C is 10 g cm–3 and at 100 °C its density is 9.7 g cm–3. The coefficient of linear expansion of the substance is 10–x °C–1. The value of x is Comprehension Type
A solid aluminium sphere and a solid lead sphere of same radius are heated to the same temperature and allowed to cool under identical surrounding temperatures. The specific heat capacity of aluminium is 900 J kg–1 °C–1 and that of lead is 130 J kg–1 °C–1. The density of lead is 104 kg m–3 and that of aluminium is 2.7 × 103 kg m–3. Assume that the emissitivity of both the spheres is the same. 27. The ratio of rate of heat loss from the aluminium sphere to the rate of heat loss from the lead sphere is (a) 1 : 1 (b) 90 : 13 (c) 100 : 27 (d) 1 : 4 28. The ratio of rate of fall of temperature of the aluminium sphere to the rate of fall of temperature of the lead sphere is
(b) 39 : 1000 (d) 130 : 243 Matrix Match Type
29. Match the entries in column I with the entries of column II. Column I Column II (A) Bimetallic strip (P) Conduction (B) Chimney (Q) Radiation (C) Heat transfer (R) Convection through a metallic wire (D) Sun’s rays reach the (S) Thermal expansion of earth solids A B C D (a) P S Q R (b) Q P R S (c) R Q S P (d) S R P Q 30. Match the entries in column I with the entries of column II. Column I Column II (A) Temperature of the (P) Newton’s law of cooling stars (B) Special case of (Q) Kirchhoff ’s law Stefan’s law (C) Good absorbers are (R) Planck’s radiation law good emitters (D) The energy (S) Wien’s displacement law distribution in black body spectrum A B C D (a) P Q R S (b) Q R S P (c) S P Q R (d) R S P Q Keys are published in this issue. Search now! J
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Physics For you | december ‘16
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Unit
6
DUAL NATURE OF RADIATION AND MATTER ATOMS AND NUCLEI
F Quantum Theory of Light •
•
•
hc 1240 = eV l l(in nm)
E hu h = = c c l If source is 100% efficient, then the number of photons emitted per second by the source can be given by
f
\
Intensity of light is the energy crossing per unit area per unit time perpendicular to the direction of propagation. I= f
E P = At A For a point source, intensity at a distance r from P the source, I = 4 pr 2
Reflected photon p2 =–h
Force on surface F = F=
2Nh l
2Nh Pl 2h = n but n = l tl hc
2h P l 2P × = l hc c
F 2P 2 I = = A cA c Force exerted on perfectly absorbing surface Pressure =
Power of source
•
Incident photon p1 = h
but change in momentum of surface = − Dp =
Momentum of photon, p =
P P Pl n= = = = Energy of photon E hu hc
Force exerted on perfectly reflecting surface f Let N photons are there in time t, change in momentum of photons −2Nh = Dp = l
Energy radiated from a source is in the form of small packets which are known as photon. Photon is not a material particle. According to Planck the energy of a photon is directly proportional to the frequency of the radiation, i.e., E ∝ u; E = hu =
•
•
•
Nh −0 Incident photon Dp Nh F= = l = p1 = h t t tl (There are N photons in time t.) h P ⇒F=N = l c F P I Pressure = = = A Ac c
No reflected photon p2 = 0
Physics For you | december ‘16
39
When a beam of light is incident at an angle q on perfectly reflector surface then force exerted on the surface 2P 2h cos q = n cos q l c 2IA cos q = c F 2 I cos q Pressure = = A c
n to ho tp en cid In
•
n
F=
•
Re
f
te lec
•
to ho p d
f
f
f f
•
•
h h = p mv where p = momentum of the particle, m = mass of the particle, and v = velocity of the particle. p2 h f Also, kinetic energy, K = ⇒l= 2m 2mK
f
l=
f
40
0.286
Å
V For deuterons (md = 2 × 1.67 × 10–27 kg) : 0.202 l= Å V Physics For you | december ‘16
30.835 T
h mvrms For gas molecules at T K For gas molecules : l =
The wavelength of a matter wave given by l = h p has physical significance; its phase velocity has no physical significance. However, the group velocity of the matter wave is physically meaningful and equals the velocity of the particle. Electron diffraction experiments by Davisson and Germer, and by G.P. Thomson, as well as many later experiments, have verified and confirmed the wave-nature of electrons. The de Broglie hypothesis of matter waves supports the Bohr’s concept of stationary orbits.
n=2
λ n=3
For standing wave,
12.27
= Å 2mqV V For protons (mp = 1.67 × 10–27 kg) :
2mkT
=
λ
For electrons (me = 9.1 × 10–31 kg) : l=
h
λ
de Broglie wavelength associated with charged particles h
2mK
6.62 × 10−34
=
3 h E = kT ⇒ l = 2 3mkT
l=
f
h
2 × 1.67 × 10−27 K For thermal neutrons at ordinary temperatures : K = kT l=
•
•
For neutrons (mn = 1.67 × 10–27 kg) : l=
When some part of incident light on the plate is absorbed and the remaining part is reflected IA then net force on the plate F = (1 + r ) and c F I pressure P = = (1 + r ), where r is the reflection A c coefficient of the plane and 0 < r < 1.
A perfectly reflecting solid sphere of radius r is kept in the path of a parallel beam of light of large aperture. If the beam carries an intensity I, the force pr 2 I exerted by the beam on the sphere is . c F Matter Waves (de Broglie Waves) • The waves associated with moving particle are called matter waves or de Broglie waves. The wavelength associated with a moving particle is known as de Broglie wavelength and it is given by
For a-particles (ma = 4 × 1.67 × 10–27 kg) : 0.101 Å l= V de Broglie wavelength associated with uncharged particles
f
2 pr = nl =
nh nh or mvr = mv 2p
F Photoelectric Effect •
n=4
When light of sufficiently small wavelength is incident on a metal surface, electrons are ejected from the metal. This phenomenon is called the photoelectric effect. The electrons ejected from the metal are called photoelectrons.
•
•
Photoelectric effect is an instantaneous process, as soon as light is incident on the metal, photoelectrons are emitted. Experimental set-up to study photoelectric effect Light
Cathode
Anode
e
•
•
•
Einstein’s photoelectric equation, maximum kinetic energy of photoelectrons, K max = hu − f = hc − f. l Here, f is the work function of the metal. The work function represents the energy needed to remove the least tightly bound electrons from the surface. It depends only on nature of the metal and independent of any other factors. If the photon energy (say for l = l0) is just sufficient to librate the electron only, then kinetic energy hc of the photoelectron will be zero. Then =f l hc or l0 = = threshold wavelength f Corresponding threshold frequency u0 =
•
f
G
V
f
•
•
c f = l0 h
Variation of photoelectric current with
•
Postulates: f An electron revolves in a certain stable orbit without the emission of radiant energy. f Angular momentum of an electron in the orbit is integral multiple of h/2p. h mvr = n where, n = 1, 2, 3, ..., and is called 2p principal quantum number. f When an electron makes a transition from a higher orbit to a lower stable orbit, the difference in the energy of the electron is radiated as a photon of energy hu.
Photocurrent
Intensity (u and V are constant) and potential difference (I and u are constant) are
Photocurrent
O
Stopping potential does not depend on the distance between cathode and anode. Quantum efficiency number of electron emitted per second = nt per second total number of photon inciden ne = nph
F Bohr’s Model of hydrogen Like Atoms
Another form of photoelectric equation
f
For u ≤ u0, stopping potential is zero.
f
1 1 K max = h(u − u0 ) = hc − l l0 •
Stopping potential varies linearly with frequency. Slope of the graph (h/e) is same for all metals.
Intensity of light
Potential difference
Stopping potential : If the fastest electron just fails to reach the anode, then anode potential is known as stopping potential (V0). hc eV0 = K max = − f = hu − f l
•
hu = Ei – Ef In the nth orbit of hydrogen like atom, mv 2 1 (e)(Ze) = ⋅ rn 4 pε0 rn2 Ln = mvrn =
nh 2p Physics For you | december ‘16
41
f
Radius of orbit, 2
n n2 h 2 ε 0 ⇒ rn = (0.53) Å rn = 2 Z pme Z f
f
f f
Velocity of electron, e2 Z Z vn = ⇒ vn ∝ n 2h ε0 n 1 cZ Also, vn = 137 n Ze 2 Kinetic energy of electron, K = 8pε0 rn 2 Potential energy of electron, U = − Ze 4pε0 rn Total energy of electron, E = K + U = –K me 4 Z 2 Z2 En = − 2 2 2 = − (13.6) 2 eV n 8h ε0 n
f
•
•
•
F hydrogen spectrum
• •
42
In an experimental set up (Coolidge tube) highly energetic electrons are made to strike a metal surface, X-rays come out. Intensity variation of X-rays with wavelength is shown in the figure. Characteristic X-rays are Ka and Kb which have very large intensity. At other wavelength, intensity varies gradually and these X-rays are called continuous X-rays. Physics For you | december ‘16
Total number of emission from nth state to ground n(n −1) . state are 2 Ionisation energy: The minimum energy needed to ionise an atom. Ionisation energy of a hydrogen atom in ground state is 13.6 eV and ionisation potential is 13.6 V. Excitation energy is the energy needed to take the atom from its ground state to an excited state.
Intensity
continuous and characteristic X-rays F
Time period of the electron 2 prn 4ε20n3h3 T1n3 Tn = = 2 4 = 2 vn Z me Z
Kβ
Kα
λmin
30 40 50 60 70 80 90 Wavelength (pm)
•
Cut off wavelength of X-ray spectrum is given by, hc l min = , where V is accelerating potential. eV
l min (in nm) =
1242 V (in volt)
lmin depends only on accelerating voltage (V) not on the material of the target. Characteristic X-rays:
into to fill this vacancy and atoms emits a characteristic X-ray photon. Energy levels of the atom when one electron is knocked out is shown in the figure.
•
f
•
Incident electron
K-electron
e–
hυ Kα X-ray (a) f f
f
•
(b)
f
(c)
Incident electron strikes a target atom. Incident electron knocks out one of its deep seated electrons (n = 1) and there remains a vacancy. One electron from higher shell (n = 2) moves
f
f
Wavelengths and frequencies of the some of the X-rays Transition K→L
Series Line Wavelength Frequency Energy difference K uKa EK – EL = huKa a lKa
K→M
K
b
lKb
uKb
EK – EM = huKb
L→M
L
a
lLa
uLa
EL – EM = huLa
Hard X-rays: short lmin (more energy) • Soft X-rays: longer lmin (less energy) F Moseley’s Law • According to Moseley’s observations, frequency of X-rays spectrum is given by
•
1 0
10
20 30
40 50 60 Position number
hc ( EK − EL )
lKb =
hc ( EK − E M )
l La =
hc ( EL − E M )
3Rc 3Rc (Z − b) so a = 4 4 Thus proportionality constant a does not depend on the nature of target but depends on transition.
•
Bohr’s model
Moseley’s correction
1 1 2 1 2 1 DE 13.6Z 2 − 2 eV 13.6(Z − 1) 2 − 2 eV n1 n2 n1 n2
3 2
lK a =
u=
u = a ( Z − b) ; where a and b are constant.
Bohr’s model will give reasonable results if Z is replaced by (Z – b) with b ≈ 1. Energy released during the transition from n = 2 to n = 1 is given by
Wavelength
1 1 DE = hu = Rhc(Z − b)2 2 − 2 1 2
•
•
Choose E = 0 (reference) when atom is in ground state. EK = energy of the atom when an electron from the K-shell is knocked out EL, EM, EN etc. have same interpretation. X-rays emitted due to electronic transition from a higher energy state to a vacancy created in the K shell are called K X-rays.
u
1 1 RcZ 2 − 2 2 n1 n2
1 1 Rc(Z − 1)2 − 2 2 n1 n2
1 l
1 1 RZ 2 − 2 2 n1 n2
1 1 R(Z − 1)2 − 2 2 n1 n2
Physics For you | december ‘16
43
Bragg’s Law F •
It predicts the conditions under which diffracted X-ray beams from a crystal are possible.
2d sin q = nl; n = 1, 2, 3, … n = order of spectrum, l = wavelength of X-ray • By using a monochromatic X-ray beam and noting the angles of strong reflection, the interplanar spacing d and several information about the structure of the solid can be obtained. Effect of Mass of Nucleus on Bohr Model F • As no external force is acting on a nucleuselectron system, hence the center of mass of the nucleus-electron system, must remain at rest. r1
+Ze Nucleus
+Ze
r2
CM
r
r1
r2 +F e
CM
r
+Ze
e
For the electron in nth orbit f Radius, 2 2
f
Speed, vn =
mN me mN me
•
• •
Ze 2 2nhε0
Physics For you | december ‘16
Z 2e 4 mN me
Atomic number = Number of protons (Z) f Mass number (A) = Number of protons (Z) + number of neutrons (N). f Nuclear mass is the total mass of the protons and neutrons. A nuclide is a specific nucleus of an atom characterised by its atomic number Z and mass number A. A Symbolically, nuclide ≡ Z X ; X = chemical symbol of the element Properties of nucleons Name Symbol Charge
2 2
n h ε0 n h ε0 (mN + me ) ⇒ rn′ = 2 pZe µe pZe2 me mN m rn′ = rn × e µe rn′ =
44
Fcf
mN me Reduced mass of the system, µe = mN + me Relative picture of the atom Rest
•
e–
Energy, En′ = −
8ε20n2h2 (mN + me ) µ Z 2 µe En′ = En × e ⇒ En′ = −(13.6 eV) me n2 me Nuclear Force F • It is the strongest force in the universe and acts only between the nucleons. • Very short range : Only upto size of nucleus (3 or 4 fermi). More than this distance, nuclear force is almost zero. • Very much depends upon distance : Small variation in distance may cause of large change in nuclear force while electrostatic force remains almost unaffected. • Independent of charge : Interacts between n-n as well as between p-p and also between n-p. • Spin dependent : It is stronger between nucleons having same sense of spin than between nucleons having opposite sense of spin. • Nature f Attractive : If distance is about 1 fm or above. f Repulsive. If distance is less than 0.5 fm. composition of Nucleus F • Nucleons are protons and neutrons which are present in the nuclei of atoms.
me r mN r and r2 = mN + me mN + me
r1 =
•
e–
f
Mass
Rest energy
Proton
p
+e
1.007276 u 938.28 MeV
Neutron
n
0
1.008665 u 939.57 MeV
•
Isotopes, Isobars, Isotones and Isomers f Isotopes : The atoms of an element which have same atomic number but different mass number.
•
•
Isobars : The atoms having same mass number but different atomic number. 37 37 Example : 31H and 23He, 17 Cl and 16 S f Isotones : The nuclides having the same number of neutrons. 37 39 Example : 17 Cl and 19 K f Isomers : Nuclei with same atomic number and mass number but existing in different energy state. Nuclear size : The number of protons and neutrons per unit volume is approximately constant over the entire range of nuclei. A i.e., constant 4 3 pR 3 ⇒ A ∝ R3 ⇒ R ∝ A1/3 R = R0A1/3; R0 = 1.2 fermi = 1.2 × 10–15 m f
•
Nuclear density, r =
f
=
mA
=
Mass of nucleus Volume of nucleus
3m ⇒ r ∝ A0 4 pR03
4 × pR03 A 3 And r = 2.23 × 1017 kg m–3 Binding Energy F • The loss in energy which is responsible for binding the nucleons together in a nucleus is called the binding energy. 2
Eb = DMc where, DM = mass defect DM = [ZmP + (A – Z)mN] – M M → mass of nucleus \ Eb = [(ZmP + (A – Z) mN)–M] × c2
Eb A = Average energy needed to separate a nucleus into its individual nucleons. Shape of BE/A is determined primarily by three factors f a constant term, which originates because nucleons interact only with their nearest neighbours. f a sharp decrease for light nuclei. f a gradual decrease for heavy nuclei due to coulomb repulsion of the nuclear protons. Binding energy per nucleon, DEb /n =
Observation from the BE/A graph, energy can be liberated in two different ways Nuclear fission : A heavy nucleus breaks in f two light nuclei of comparable masses. BE per nucleon is greater for two lighter fragments than it is for the original nucleus. f Nuclear fusion : Two light nuclei combine to form a heavier nucleus. BE per nucleon is greater in the final nucleus than it is in the two original nuclei. Law F of radioactive Decay • Rate of decay of nuclei is given by dN dN − ∝ N (t ) ⇒ = − lN (t ) dt dt l → disintegration (decay) constant • Number of undecayed radioactive nuclei at any time t •
N(t) = N0 e–lt
•
N0 → number of nuclei at time t = 0 Number of decayed nuclei, Nd = N0 (1 – e–lt)
Nd
N
N0
N0 t
Physics For you | december ‘16
t
45
• •
•
•
•
Mean life = Average life = τ av =
1 l
A ZX
ln 2 0.693 = = 0.693 τ av Half life = T1/2 = l l 1 f Also, τ av = = 1.44 T1/2 l Number of nuclei present after n half life (t = nT1/2) N = 0n (2) N N N N t /2 t /2 t /2 t /2 N 0 1 → 0 1 → 20 1 → 30 .... 1 → n0 2 2 2 2 If a nuclide can decay simultaneously by two different processes with different decay constants l1 and l2, then effective decay constant of the nuclide is, l = l1 + l2 ; and N = N0 e–(l1 + l2)t Activity of sample: dN = lN = lN 0 e − lt dt = R0 e − lt
R=−
f
f
a−decay
→ ZA−−42Y + 42 He+ Q
Q value of a-decay : Q = (mX – mY – mHe)c2 This energy is shared by the daughter nucleus and the alpha particle. The energy released during the alpha decay of 238 92 U is 4.25 MeV.
The energy supplied to emit proton from 238 92 U is 7.68 MeV. Hence a proton cannot be released by 238 92 U without supply of external energy. Beta decay : It is spontaneous decay of nucleus by emitting an electron or a positron.
f
•
−
f
b − decay : n → p + e − + u ; ZA X b → Z +A1Y + −01 e + u
f
b + decay : p → n + e + + u ; ZA X b → Z −A1Y + +01 e + u
+
The mass number A of a nuclide undergoing beta decay does not change. Gamma decay : It is the phenomenon of emission of one or more g-ray photon by a nucleus in its excited state so as to acquire its ground state.
f
•
Here, R0 = lN0 is decay rate at t = 0. Units of activity f
Co
1 curie = 1 Ci = 3.7 × 1010 dps
1 Rutherford = 1 rd = 106 dps 1 Becquerel = 1 Bq = 1 dps •
Alpha decay : It is spontaneous decay of a heavy nucleus by emitting an alpha particle.
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SYSTEM OF PARTICLES AND ROTATIONAL MOTION
CLASS XI
Centre of Mass and Centre of Gravity l
l
Rotational Motion
The centre of gravity of a body coincides with its centre of mass only if the gravitational field does not vary from one point of the body to other.
l
Mathematically,
l
Perpendicular distance of each particle remains constant from a fixed line or point and particle do not move parallel to the line. Angular displacement, q =
R
l
For continuous body,
Centre of mass of symmetric body w
l
l
q = w0t + at 2
w
w 2 = w02 + 2aq
R
Torque : Turning effect of the force about the axis of rotation.
Semi-circular disc,
l
l
w w
l
Position,
l
Velocity,
w
l
Acceleration, If
= 0, then
l
If the net external torque acting on a system is zero, the angular momentum of the system remains constant, no matter what changes take place within the system.
Perpendicular axes theorem : Iz = Ix + Iy (Object is in x-y plane )
y
R2
= constant; I1w1 = I2w2 (for isolated system)
l
l
Axis 2R
2R
Spherical shell
Solid sphere
CM
A rigid body is said to be in mechanical equilibrium, if both of its linear momentum and angular momentum are not changing with time, i.e., total force and total torque are zero. Linear momentum does not change implies the condition for the translational equilibrium of the body and angular momentum does not change implies the condition for the rotational equilibrium of the body.
l
So,
Current decay in L–R circuit,
Magnetic energy density,
The direction of the induced current is such that it opposes the change that has produced it. If a current is induced by an increasing(decreasing) flux, it will weaken (strengthen) the original flux. It is a consequence of the law of conservation of energy.
Axis
l
Emf induced in the coil/conductor,
l
Coefficient of self induction,
l
Self inductance of a long solenoid,
l
Mutual inductance,
l
Mutual inductance of two long coaxial solenoids,
l
Induced Electric Field l
It is produced by change in magnetic field in a region. This is non-conservative in nature.
l
This is also known as integral form of Faraday’s law.
Coefficient of coupling, For perfect coupling, k = 1 so,
l l
R a
d
b
I = MR2/2
Rolling Motion l
Current growth in L–R circuit l
Magnetic Flux and Faraday's Law Axis
Equilibrium of a Rigid Body
l
Energy stored in the solenoid,
x
B
l
If coils are far away, then M = 0.
Lenz's Law
l
z
= constant.
Conservation of Angular Momentum
l
L–R Circuit
Inductance
For a rigid body,
Parallel axes theorem : IAB = ICM + Md 2
Inductors in parallel,
L
Angular momentum, Work done by torque, W = tdq Power, P = tw
A
w
l
R1
Moment of Inertia
For a system of particles
Energy stored in an inductor,
Axis
Axis
Motion of Centre of Mass
Inductors in series,
Here,
R
L
Axis
Semi-circular ring, l
l
l
Equations of rotational motion w w = w0 + at w
Magnetic Energy
Axis
L l
w
l
I = MR2
Axis
CLASS XII
Combination of Inductors
Angular acceleration, a =
For discrete body, l
w
Axis
R
Angular velocity, w = w
Axis
ELECTROMAGNETIC INDUCTION
Energy Consideration in Motional emf l
Emf in the wire, e = Bvl
l
Induced current,
l
Force exerted on the wire,
K = Ktranslational + Krotational =
w
Induced emf,
w
Induced current,
w
Induced charge flow ,
Electric Generator l
I
For a body rolling without slipping, velocity of centre of mass vCM = Rw Kinetic energy,
Magnetic flux Faraday’s law : Whenever magnetic flux linked with a coil changes, an emf is induced in the coil.
l
B ×
v
F
R l
Motional emf
Power required to move the l
wire, It is dissipated as Joule’s heat.
l
On a straight conducting wire, e = Bvl On a rotating conducting wire about one end, Here,
are perpendicular to each other.
l
l
Mechanical energ y is converted into electrical energy by virtue of electromagnetic induction. Induced emf, e = NABw sinwt = e0sinwt Induced current,
1. Ultraviolet light of wavelength 300 nm and intensity 1.0 W m–2 falls on the surface of a photosensitive material. If two percent of the incident photons produce photoelectrons, then the number of photoelectrons emitted from an area of 1.0 cm2 of the surface is nearly (a) 2 × 1013 s–1 (b) 3 × 1012 s–1 13 –1 (c) 4 × 10 s (d) 4 × 1012 s–1. 2. X-rays are produced in an X-ray tube by electrons accelerated through an electric potential difference of 50.0 kV. An electron makes three collisions in the target before coming to rest and loses half of its remaining kinetic energy in each of the first two collisions. Neglect the recoil of the heavy target atoms, the wavelength of the resulting photons in first two collisions is (Take hc = 1243 eV nm) (a) 49.72 pm ; 99.44 pm (b) 47.38 pm ; 24.86 pm (c) 52.24 pm ; 49.72 pm (d) 49.72 pm ; 24.86 pm. 3. A normal human eye can see an object if it receives a minimum light intensity of 2 × 10–10 W m–2, reflected from an object. What is the minimum number of photons of wavelength 500 nm that must enter the pupil of the eye per second in order to see the object if the appearance of the pupil is about 1 cm2 ? (Given h = 6.63 × 10–34 J s ; c = 3 × 108 m s–1) (a) 2.5 × 104 (b) 2.5 × 105 4 (c) 5.0 × 10 (d) 5.0 × 105. 4. The stopping potential for the photoelectrons emitted from a metal surface of work function 1.7 eV is 10.4 V. Identify the energy levels in hydrogen atom which will emit the wavelength of radiation used (a) n1 = 1, n2 = 2 (b) n1 = 1, n2 = 3 (c) n1 = 2, n2 = 4 (d) n1 = 2, n2 = 5. 5. A radioactive sample decays by two different processes. Half life for the first process is t1 and for the second process is t2. The effective half life is (a) t1 + t2 (b) t1 – t2 t1 t2 (t1 + t2 ) (c) (d) . 2 t1 + t 2
6. Let u1 be the frequency of the series limit of Lyman series, u2 be the frequency of the first line of Lyman series, and u3 be the frequency of the series limit of Balmer series. Then (a) u1 + u2 = u3 (b) u2 – u1 = u3 1 (c) u1 – u2 = u3 (d) υ3 = (υ1 + υ2 ) . 2 7. There is a stream of neutrons with a kinetic energy of 0.0327 eV. If the half life period of neutrons is 700 s, what fraction of neutrons will decay before they travel a distance of 10 m? (a) 3.96 × 10–4 (b) 3.96 × 10–5 –6 (c) 3.96 × 10 (d) 3.96 × 10–7. 8. It is proposed to use nuclear fission reaction: 2 2 1H + 1 H
→ 24 He in a nuclear reactor of 200 MW rating. If the energy from the above reaction is used with 25% efficiency in the reactor, how many grams of deuterium will be needed per day? The masses of 12H and 24He are 2.0141 u and 4.0026 u respectively. (a) 115.3 g (b) 120.3 g (c) 125.3 g (d) 130.3 g. 9. In a hydrogen atom, an electron jumps from the state n to n – 1 where n > > 1. The frequency of the emitted radiation is proportional to (a) n0 (b) n–1 (c) n–2 (d) n–3. 10. When stopping potential is applied to the anode of photocell, no current is observed. This means (a) the emission of photoelectrons stops (b) the photoelectrons are collected near the collector plate (c) the photoelectrons are emitted but are reabsorbed by the photocathode itself (d) the photoelectrons are dispersed from the side of the apparatus. 11. The count rate of a Geiger-Muller counter for the radiation of a radioactive material of half life of 30 min decreases to 5 s–1 after 2 h. The initial count rate was (a) 80 s–1 (b) 625 s–1 (c) 20 s–1 (d) 25 s–1. Physics For you | December ‘16
49
12. A nucleus ruptures into two nuclear parts, which have their velocity ratio equal to 2 : 1. What will be the ratio of their nuclear size (nuclear radius)? (a) 31/2 : 1 (b) 1 : 31/2 1/3 (c) 2 : 1 (d) 1 : 21/3. 13. Electrons of mass m with de-Broglie wavelength l fall on the target in an X-ray tube. The cutoff wavelength (l0) of the emitted X-ray is (a)
λ0 =
(c)
λ0 =
2mcλ2 h
2m2c 2 λ3 h2
2h (b) λ0 = mc (d) l0 = l. [NEET Phase II 2016]
14. The half life of a radioactive substance is 30 min. The time (in min) taken between 40% decay and 85% decay of the same radioactive substance is (a) 15 (b) 30 (c) 45 (d) 60. [NEET Phase II 2016] 15. If an electron in a hydrogen atom jumps from the 3rd orbit to the 2nd orbit, it emits a photon of wavelength l. When it jumps from the 4th orbit to the 3rd orbit, the corresponding wavelength of the photon will be 16 9 (a) (b) λ λ 25 16 20 20 (c) (d) λ λ. 7 13 [NEET Phase II 2016] 16. Given the value of Rydberg constant is 107 m–1, the wave number of the last line of the Balmer series in hydrogen spectrum will be (a) 0.25 × 107 m –1 (b) 2.5 × 107 m –1 (c) 0.025 × 104 m –1 (d) 0.5 × 107 m –1. [NEET Phase I 2016] 17. Radiation of wavelength l, is incident on a photocell. The fastest emitted electron has speed v. 3λ If the wavelength is changed to , the speed of 4 the fastest emitted electron will be (a) (c)
50
1/ 2
1/ 2
4 > v 3
4 (b) < v 3
4 = v 3
3 (d) = v . 4 [JEE Main Offline 2016]
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18. Half lives of two radioactive elements A and B are 20 min and 40 min, respectively. Initially, the samples have equal number of nuclei. After 80 min, the ratio of decayed numbers of A and B nuclei will be (a) 1 : 16 (b) 4 : 1 (c) 1 : 4 (d) 5 : 4. [JEE Main Offline 2016] 19. A hydrogen atom makes a transition from n = 2 to n = 1 and emits a photon. This photon strikes a doubly ionized lithium atom (Z = 3) in excited state and completely removes the orbiting electron. The least quantum number for the excited state of the ion for the process is (a) 2 (b) 4 (c) 5 (d) 3. [JEE Main Online 2016] 20. A neutron moving with a speed v makes a head on collision with a stationary hydrogen atom in ground state. The minimum kinetic energy of the neutron for which inelastic collision will take place is (a) 20.4 eV (b) 10.2 eV (c) 12.1 eV (d) 16.8 eV. [JEE Main Online 2016] SolutionS 1. (b) : Energy incident over 1.0 cm2 = 1.0 × 10–4 J Energy required to produce photoelectrons, E = (1.0 × 10–4) × (2/100) = 2.0 × 10–6 J Energy of each photon,
hc (6.6 × 10−34 ) × 3 × 108 = 6.6 × 10–19 J = λ 300 × 10−9 Number of photoelectrons ejected E0 =
E 2.0 × 10−6 = = 3 × 1012 s–1 E0 6.6 × 10−19 2. (a) : Energy of the incident electron = 50 keV The energy of X-ray photon produced in first collision 50 E1 = 50 − = 25 keV 2 =
Wavelength of photon produced, hc 1243 eV nm = λ1 = E1 25 × 103 eV = 49.72 × 10–3 nm = 49.72 pm In the second collision the electron loses energy, 25 = = 12.5 keV 2
\ \
Energy of X-ray photon produced, E2 = 12.5 keV Wavelength of photon produced, 1243 eV nm = 99.44 × 10−3 nm λ2 = 3 12.5 × 10 eV = 99.44 pm
3. (c) : Power of the light entering the eye, P = IA = 2 × 10–10 × 10–4 = 2 × 10–14 W If n number of photons of given wavelength enter per second the pupil of the eye to see the object, (2 × 10−14 )(500 × 10−9 ) P Pλ = = hc / λ hc (6.63 × 10−34 ) × (3 × 108 ) = 5.0 × 104
then n =
4. (b) : Since stopping potential, Vs = 10.4 V. The maximum kinetic energy of photoelectrons is Kmax = 10.4 eV. The energy of the incident photon is given by E = Kmax + f0 = 10.4 + 1.7 = 12.1 eV We know the energy of electron in energy level of −13.6 hydrogen atom is, En = eV n2 −13.6 When n = 1, E1 = eV = −13.6 eV 12 −13.6 When n = 2, E2 = 2 = −3.4 eV 2 −13.6 When n = 3, E3 = = −1.51 eV 32 Here, E3 – E1 = – 1.51 – (–13.6) = 12.1 eV. Hence the transition corresponds to n1 = 1 and n2 = 3. 5. (d) : As l = l1 + l2 t t 0.693 0.693 0.693 ⇒ = + or t = 1 2 t t1 t2 t1 + t2 6. (c) : We know that series limit means the shortest possible wavelength (maximum photon energy) and first line means longest possible wavelength (i.e., minimum photon energy). 1 1 As υ = Rc − , where R is Rydberg constant. 2 2 n1 n2 For series limit of Lyman series, n1 = 1, n2 = ∞ \ u1 = Rc For first line of Lyman series, n1 = 1, n2 = 2 1 3 1 ∴ υ2 = Rc − = Rc 2 1 22 4
For series limit of Balmer series, n1 = 2, n2 = ∞ Rc 1 ∴ υ3 = Rc − 0 = 2 4 2 Now, υ1 − υ2 = Rc − \
3Rc Rc = = υ3 4 4
u 1 – u 2 = u3
1 2 2E mv , υ = ≈ 2500 m s −1 m 2 Time taken to cover a distance of 10 m i.e., 10 m dt = = 4 × 10−3 s −1 2500 m s
7. (c) : As E =
As − −
dN = λN , dt
dN 0.693 = λdt = (4 × 10−3 s) N T1/ 2 =
0.693 (4 × 10−3 s) 700 s
= 3.96 × 10−6
8. (b) : Energy output of the reactor per day = 200 × 106 (J s–1) × (24 × 60 × 60 s) = 1728 × 1010 J Energy input =
1728 × 1010 = 6912 × 1010 J (25 / 100)
Energy released by the fusion of two 21H nuclei = [(2 × 2.0141) – 4.0026] × 931.5 MeV = 23.85 MeV = 38.15 × 10–13 J Number of 21H nuclei required =
6912 × 1010 J
(38.15 × 10
−13
J) / 2
Mass of fuel required = (362.3 × 1023 ) ×
= 362.3 × 1023 2g
6.023 × 1023
= 120.3 g
9. (d) : According to Bohr’s theory of hydrogen atom 1 1 1 = RZ 2 − n2 n2 λ 1
2
c 1 1 υ = = RcZ 2 − 2 2 λ n1 n2 ∴ υ∝ υ∝
1 n12
−
n2 − n12 = 2 n22 n12 n22 1
n2 − (n − 1)2 2
2
n (n − 1)
=
2n − 1 2
n (n − 1)2
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When n > > 1, 2n 1 υ∝ , i.e., υ ∝ or υ ∝ n−3 4 3 n n 10. (c) : When stopping potential is applied to the anode of photocell, the emitted photoelectrons experience force of repulsion due to electric field set up and they are reabsorbed by the photocathode itself. 11. (a) : Half life time = 30 min; Rate of decrease (N) = 5 s–1 ; Total time = 2 h = 120 min. Relation for initial and N 1 = N0 2
time half life
final
120 30
1 = 2
count 4
rate
1 1 = = 2 16
Therefore N0 = 16 × N = 16 × 5 s–1 = 80 s–1. 12. (d) : Velocity ratio, v1 : v2 = 2 : 1
Mass (m) ∝ Volume ∝ r3. According to law of conservation of momentum, m1v1 = m2v2 \
v1 m2 r23 = = v2 m1 r13
or
r1 v2 = r2 v1
13
or
13
1 = 2
=
1/3
1
13
2
r1 : r2 = 1 : 2
13. (a) : Kinetic energy of electrons,
p2 (h / λ)2 h2 = = 2m 2m 2mλ2 So, maximum energy of emitted X-ray photon = K K=
hc h2 = λ0 2mλ2
∴ λ0 =
2mcλ2 h
14. (d) : N0 = Number of nuclei at time t = 0 N1 = Remaining nuclei after 40% decay = (1 – 0.4) N0 = 0.6 N0 N2 = Remaining nuclei after 85% decay = (1 – 0.85) N0 = 0.15 N0 \
N 2 0.15N 0 1 1 = = = N1 0.6N 0 4 2
2
Hence, two half lives is required between 40% decay and 85% decay of a radioactive substance. \ Time taken = 2T1/2 = 2 × 30 min = 60 min 52
Physics For you | December ‘16
15. (c) : When electron jumps from higher orbit to lower orbit then, wavelength of emitted photon is given by, 1 1 1 =R − 2 2 λ n f ni Transition : 3 → 2, wavelength = l Transition : 4 → 3, wavelength = l′ = ? so,
1 1 5R 1 = R − = 2 2 λ 32 36
and
1 1 7R 1 = R − = 2 3 λ′ 42 144
144 5λ 20 λ × = 7 36 7 16. (a) : Here, R = 107 m–1 The wave number of the last line of the Balmer series in hydrogen spectrum is given by 7 1 1 1 = R 2 − 2 = R = 10 = 0.25 × 107 m −1 2 λ ∞ 4 4 \
λ′ =
17. (a) : According to Einstein’s photoelectric equation maximum kinetic energy of a photoelectron, 1 hc K = mv 2 = − φ λ 2 According to the question, for incident radiation of wavelength l, maximum speed of photoelectron is v. 1 2 hc \ mv = − φ ...(i) λ 2 Assume speed of fastest photoelectron is v′ when 3λ incident photon has wavelength . 4 hc 1 4 \ mv ′2 = −φ 2 3λ or or \
1 4 1 mv ′2 = mv 2 + φ − φ 2 3 2
(from Eqn (i))
4 2 2φ φ 1 2 v + mv ′2 = mv 2 + or v ′ = 3 3m 2 3 3 4 v′ > v 3
18. (d) : Half life of A, T1/2(A) = 20 min Half life of B, T1/2(B) = 40 min Initially, number of nuclei in each sample = N Now, 80 min = 4T1/2(A) = 2T1/2(B) Number of active nuclei after four half lives of A,
N 16 2 15 \ Number of decayed nuclei = N – NA = N 16 Number of active nuclei after two half lives of B, N N NB = = 22 4 3 \ Number of decayed nuclei = N – NB = N 4 15 N 5 \ Required ratio = 16 = 3 4 N 4 19. (b) : Energy of emitted photon, 1 3 1 E = − × 13.6 eV = × 13.6 eV 2 2 4 1 2 NA =
N
4
Energy required to completely remove the electron from nth excited state of doubly ionized lithium, E′ =
13.6 Z 2
n As E ≥ E′
2
3 13.6 × 9 × 13.6 ≥ 4 n2
=
eV =
13.6 × 9 n2
eV
⇒ n2 ≥ 3 × 4 or n ≥ 12 = 3.5 \ Least quantum number for the excited state = 4. 20. (a) : Using conservation of linear momentum, Total momentum before collision = Total momentum after collision v mv = (m + m) v′ ⇒ v ′ = 2 Loss in kinetic energy during the process, 2
1 1 1 v ∆K = mv 2 − (2m) = mv 2 2 2 2 4 For minimum kinetic energy of neutron, lost kinetic energy should be used by the electron to jump from first orbit to second orbit. 1 2 ⇒ mv = (13.6 − 3.4) eV = 10.2 eV 4 1 2 mv = 20.4 eV = Kinetic energy of the 2 neutron for inelastic collision. ∴
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CLASS XII Series 7
CBSE
Semiconductor Electronics : Materials, Devices and Simple Circuits Communication Systems
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 calculator is not allowed.
section-A
1. Show variation of resistivity of Si with temperature in a graph.
Previous Years Analysis 2016 2015 2014 Delhi AI Delhi AI Delhi AI VSA SA-I SA-II
– 1
1 1
1 1
2
VBQ LA
1 –
3 _
3 _
–
–
1 1
– 2
3
2 1 _
– –
1 –
– –
3
7. The block diagram of a receiver is shown in the following figure. Identify X and Y. Write their functions.
2. Why should a photodiode be operated at reverse bias? 3. Why do we need carrier waves of very high frequency in the modulation of signals? 4. Define repeater. What is its function? 5. Which of the following would produce analog signals and which would produce digital signals? (i) A vibrating tuning fork. (ii) Musical sound due to a vibrating sitar string. (iii) Light pulse. (iv) Output of NAND gate. section-B
6. In a transistor, doping level in base is increased slightly. How will it affect (i) collector current and (ii) base current? 54
Physics For you | DECEmbEr ‘16
8. Answer the following questions, giving reasons : (i) Why is the current under reverse bias almost independent of the applied potential upto a critical voltage? (ii) Why does the reverse current show a sudden increase at the critical voltage? 9. Inputs A and B are applied to the logic gate set up as shown in the figure. Complete the given truth table and name the equivalent gate formed by this set-up.
A
B
0 0 1 1
0 1 0 1
A′
B′
Y
OR For the circuit shown in figure find the current flowing through the 1 W resistor. Assume that the two diodes, D1 and D2, are ideal diodes.
16. Mention three different modes of propagation used in communication system. Explain with the help of a diagram how long distance communication can be achieved by ionospheric reflection of radio waves. 17. Distinguish between frequency modulation and amplitude modulation. Why is an FM signal less susceptible to noise than an AM signal? 18. Write the frequency at which T.V. signals are transmitted. Derive an expression for the range upto which signals transmitted by a T.V. tower can be received. 19. Explain, with the help of a circuit diagram, the working of a photodiode. Write briefly how it is used to detect the optical signals.
10. The maximum amplitude of AM wave is found to be 15 V while its minimum amplitude is found to be 3 V. What is the modulation index? section-c
11. Draw a circuit diagram of full wave rectifier. Explain its working principle. Draw the input and output waveforms indicating clearly the functions of the two diodes used. 12. What do we understand by the cut off, active and saturation states of the transistor? 13. In the given circuit diagram, a voltmeter V is connected across a lamp L. How would (i) the brightness of the lamp, and (ii) voltmeter reading V be affected, if the value of resistance R is decreased? Justify your answer ? 14. Express by a truth table the output Y for all possible inputs A and B in the circuit shown in figure
20. A semiconductor has equal electron and hole concentrations of 2.0 × 108 cm–3. On doping with a certain impurity, the hole concentration increases to 4.0 × 1010 cm–3. (i) What type of semiconductor is obtained on doping ? (ii) Calculate the new electron concentration of the semiconductor. (iii) How does the energy gap vary with doping? 21. You are given the two circuits as shown in figure. Show that circuit (i) acts as OR gate while the circuit (ii) acts as AND gate.
(i)
(ii) OR If each diode in figure has a forward bias resistance of 25 W and infinite resistance in reverse bias, what will be the values of the current I1, I2, I3 and I4?
15. Explain briefly the following terms used in communication system : (i) Transducer (ii) Receiver (iii) Transmitter (iv) Bandpass filter (v) Amplification (vi) Demodulator Physics For you | DECEmbEr ‘16
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section-D
23. Arnab was talking on his mobile to his friend for a long time. After his conversation was over, his sister Anita advised him that if his conversation was of such a long duration, it would be better to talk through a landline. (i) Why is it considered harmful to use a mobile phone for a long duration? (ii) Which values are reflected in the advice of his sister Anita? (iii) A message signal of frequency 10 kHz is superposed to modulated a carrier wave of frequency 1 MHz. Determine the sidebands produced. section-e
24. (i) Describe briefly, with the help of a diagram, the role of the two important processes involved in the formation of a p-n junction. (ii) Name the device which is used as a voltage regulator. Draw the necessary circuit diagram and explain its working. OR Draw the circuit diagrams of a p-n junction diode in (i) forward bias, (ii) reverse bias. How are these circuits used to study the V-I characteristics of a silicon diode ? Draw the typical V-I characteristics. 25. With the help of a labelled circuit diagram, explain how an n-p-n transistor can be used as an amplifier in common-emitter configuration. Write an expression for its voltage gain. Explain how the input and output voltages are out of phase by 180° for a common-emitter transistor amplifier. OR (i) Explain the formation of energy bands in solids. On the basis of energy band diagrams, distinguish between metal, insulator and semiconductor. (ii) What is a light emitting diode (LED)? Mention two important advantages of LEDs over conventional lamps. 56
Physics For you | DECEmbEr ‘16
26. (i) Write three important factors which justify the need of modulating a message signal. (ii) What is meant by detection of a modulated signal? Draw block diagram of a detector for AM waves and state briefly, showing the waveforms, how the original message signal is obtained. OR (i) What is space wave propagation? (ii) What is internet? Write the four applications of the internet. soLutions
1. The variation of resistivity of silicon (Si) with temperature is shown as Resistivity ()
22. Figure shows a communication system. What is the output power when input signal is of 1.01 mW ? (gain in dB = 10 log10 (Po/Pi ).
Temperature (T)
2. Because in reverse bias, the fractional change in minority carriers is much larger than the fractional change in majority carriers in forward bias. So, effect of intensity of light on the minority carriers dominated reverse bias current is more easily measurable than that in forward bias current. 3. High frequency carrier waves are used to increase operating range, to reduce antenna length and convert the wide band signal into narrow band signal. Then the signal can be easily recovered and distinguished from other signals at the receiving station. 4. A repeater is a combination of a transmitter, an amplifier and a receiver which picks up a signal from the transmitter, amplifies and retransmits it to the receiver sometimes with a change of carrier frequency. Multiple repeaters help in extending the range of the communication system. 5. Analog signal would be produced by vibrating tuning fork and musical sound due to a vibrating sitar string. Digital signal would be produced by light pulse and NAND gate.
6. If the doping level in base region of a transistor is increased slightly, then greater fraction of charge carriers (electrons or holes) entering into base from the emitter will be neutralised due to recombination of opposite charge carriers in base region. As a result, (i) the collector current IC will decrease, but (ii) base current IB will increase. 7. The component X in the figure is intermediate frequency (IF) stage, which facilitates further processing of the received signal by lowering the carrier frequency. The component Y is an amplifier, which amplifies the detected signal so as to have a strong output.
8. (i) When the diode is reverse biased, a very small current of few microamperes flows due to the drift of minority charge carriers whose number density remains constant, so the current under reverse bias is almost independent of the applied potential upto a critical voltage. (ii) When the reverse voltage across the p-n junction reaches a critical voltage, the reverse current suddenly increases to a large value. It is due to the increase in the number of minority charge carriers because of the breakdown of the diode. The avalanche breakdown occurs in lightly doped diodes due to ionisation by collision. Zener breakdown occurs at low voltages in heavily doped diodes by field emission. 9. Truth table for the given circuit is as follows: A
B
0 0 1 1
0 1 0 1
A′ = A B ′ = B Y ′ = A + B Y = Y ′ 1 1 0 0
1 0 1 0
1 1 1 0
0 0 0 1
Clearly, Y = A⋅B. Hence the given circuit is equivalent to an AND gate. OR Diode D1 is forward biased and offers zero resistance. Diode D2 is reversed biased and offers infinite resistance. The given circuit reduces to the equivalent circuit as shown in figure.
By Ohm's law, the current through 1 W resistor is I=
6 =2A 2 +1
A 10. We know that, µ = m Ac It is given that Am + Ac = 15 V and Ac – Am = 3 V \ 2Ac = 15 + 3 = 18 V ; Ac = 9 V and 2Am = 15 – 3 = 12 V ; Am = 6 V \ µ=6 =2 9 3 11. A full wave rectifier consists of a transformer, two junction diodes D1 and D2 and a load resistance RL. The input a.c. signal is fed to the primary coil P of the transformer. The two ends A and B of the secondary S are connected to the p-ends of diodes D1 and D2. The secondary is tapped at its central point T which is connected to the n-ends of the two diodes through the load resistance RL, as shown in figure.
Working : At any instant, the voltages at the end A (input of D1) and end B( input of D2) of the secondary with respect to the centre tap T will be out of phase with each other. Supposed during the positive half cycle of a.c. input, the end A is positive and the end B is negative with respect to the centre tap T. Then the diode D1 gets forward biased and conducts current along the path AD1XYTA, as indicated by the solid arrows. The diode D2 is reverse biased and does not conduct. During the negative half cycle, the end A becomes negative and the end B becomes positive with respect to the centre tap T. The diode D1 gets reverse biased and does not conduct. The diode D2 conducts current along the path BD2XYTB, as indicated by broken arrows. As during both half cycles of input a.c. the current through load RL flows in the same direction (X → Y), so we get a pulsating d.c. voltage across RL, as shown in figure. Waveforms of input a.c. and output voltage obtained from a full wave rectifier. Physics For you | DECEmbEr ‘16
57
12. Figure shows the circuit diagram of a base-biased n-p-n transistor in CE configuration. Here RB is a resistor in the input circuit and RC in the output circuit.
V0 = VCC – ICRC
...(i)
cutoff region. From equation (i), the output voltage V0 = VCC Active region : When Vi increases slightly above 0.7 V, a current IC flows in the output circuit and the transistor is said to be in the active state. From equation (i), as the term IC RC increases, the output voltage V0 decreases. Now as Vi increases, IC increases almost linearly and so V 0 decreases linearly till its value becomes less then 1.0 V. Saturation region : When Vi is high i.e., the emitterbase junction is heavily forward biased, a large collector current IC flows which produces such a large potential drop across load resistance RC that the emitter-collector junction also gets forward biased. The output voltage V0 decreases to almost zero. The transistor is said to be in the saturation sate because it cannot pass any more collector current IC. 13. In the following figure, the transistor is a n-p-n transistor with baseemitter junction forward biased and the collector reverse biased. Hence, a base current IB and consequently, collector current IC flow in the circuit. If value of resistor R is reduced, then IB and correspondingly IC will increase. Due to increase in IC, the lamp will go more brilliantly. Due to increase in current IC flowing through lamp resistance the potential difference across the lamp increases and consequently, voltmeter reading will increase. 14. The given circuit includes an AND and an OR gate, as shown in figure
Cut off region : When Vi increases from zero to a low value (less then 0.7 V in case of a Si transistor), the forward bias of the emitter-base junction is insufficient to start a forward current. That is, IB = 0 and hence IC = 0. The transistor is said to be in the 58
Physics For you | DECEmbEr ‘16
Inputs A and B are fed to AND gate so that its output is Y′= A ⋅ B Then inputs A and Y′(= A ⋅ B) are fed to OR gate so that the output from it is Y=A+A⋅B
The logic table for the given circuit is as follows : AND gate Inputs Output A B Y′= A ⋅ B 0 0 0 0 1 0 1 0 0 1 1 1 Hence the truth table is Inputs
OR gate Inputs Output A Y′ Y = A + Y′ 0 0 1 1
0 0 0 1
0 0 1 1
Output A B Y 0 0 0 0 1 0 1 0 0 1 1 1 15. (i) Transducer : It is a device which converts energy from one form to another from. (ii) Receiver : It is a device which recovers the original message signal from the signal received at the output of the communication channel. (iii) Transmitter : It is a device which processes the incoming message signal into a form suitable for transmission through a channel and for its subsequent transmission. (iv) Bandpass filter : A bandpass filter blocks lower and higher frequencies and allows only band of frequency (centred around carrier frequency) to pass through it. (v) Amplification : It is the process of increasing the amplitude and hence the strength of an electrical signal by using and electric circuit (consisting of atleast one transistor) called the amplifier. (vi) Demodulator : Demodulator is a device which recovers the original information signal from the modulated wave at the receiver end. 16. (i) Ground wave or surface wave propagation (ii) Sky wave propagation or ionospheric propagation (iii) Space wave propagation/Line of sight propagation. In sky wave propagation, radio waves transmitted by transmitting antenna are directed towards the ionosphere. The radiowaves having frequency range 2 MHz to 30 MHz are reflected back by the ionosphere.
In sky wave propagation, radio signals can be transmitted to the stations which otherwise become inaccessible to the ground due to curvature of earth. Thus due to reflection by ionosphere, radio wave signals can be transmitted virtually from any one place to the other on surface of earth. So it is useful for very long distance radio communication. Thus for long distance radio broadcasts through sky wave propagation, we use short wave bands. 17. In amplitude modulation, the amplitude of the carrier is varied by the modulating signal and the change in amplitude from the unmodulated value is directly proportional to the instantaneous value of the modulating signal but is independent of its frequency. This is represented in figure.
Carrier wave (a)
Modulating sine-wave signal
Amplitude-modulated wave In frequency modulation, the instantaneous frequency of the carrier is varied by the modulating signal. The instantaneous deviation of frequency from the unmodulated values is directly proportional to the instantaneous value of the modulating signal but is independent of its frequency. This is represented in figure.
Frequency-modulation wave Physics For you | DECEmbEr ‘16
59
AM has : (i) low efficiency as the useful power is in the side bands, (ii) noisy reception, (iii) poor quality of reception and (iv) small operating range. FM has : (i) very high efficiency, (ii) noiseless reception, (iii) high fidelity and (iv) large operating range. Practically all natural and man-made noises result in electrical amplitude disturbances (variation). Since in FM modulation, the carrier amplitude is kept constant, all the amplitude-sensitive noises are eliminated because variations in amplitude due to noise are not reproduce. To limit the FM signal to constant amplitude, a limiter circuit is used in an FM receiver. 18. Television frequencies lie in the range 100 – 200 MHz. Let T.V. signals be transmitted from an antenna of height PQ = hT. Due to the curvature of the earth, no direct signals are possible beyond the points S or T, as shown in figure. Let PS = PT = d. From right-angled DOTQ, we get OQ2 = OT2 + QT2 Here, OQ = R + hT , QT = PT = d OT = R = Radius of the earth \ (R + hT)2 = R2 + d2 or R2 + hT2 + 2hTR = R2 + d2 hT or d2 = hT2 + 2hTR = 2hTR 1 + 2R But hT a. Assume relative permeability of the medium surrounding the conductor to be unity.
(a)
3 m0ka 4 m kpa 4 (b) m0kr (c) 0 4r 2r 2
(d)
m0kr 3 4
solutions 1. (b) : The first particle will have a helical path and the second particle will move rectilinearly along the field. For the two particles to meet again and again, v || T = v′T where v′ is the speed of the second particle. y-axis
p
p
p
v1=v sin (q, m) v =v cos
B (uniform)
x-axis axis of helix
\ v′ = v|| = v cosq 1 2 2qV 2qV mv = qV ⇒ v = cosq \ v′ = m 2 m 2. (a) : Both the particles will meet after the time 2pm period of helical motion. T = qB
3. (c) : Distance travelled = pitch of helical path 2qV 2pm cosq × = m qB
= v′T =
2Vm 2p cos q q B
dN dN = q0t − lN ; + lN = q0t dt dt dN 5. (a) : + lN = q0t dt
4. (a) :
On solving the differential equation, we get qt q q N = 0 − 02 + 02 e −lt l l l q q Pinst = lNE0 = q0t − 0 + 0 e −lt E0 l l t q q −lt ∫ q0t − l0 + l0 e E0dt 6. (a) : Pav = 0 t
∫ dt 0
q q q t q = 0 − 0 + 0 − 0 e − lt E0 2 2 2 l l t l t 7. (a) : Electric field on the axis of ring at a distance x is, l Q x E= − + 2pε0x 4pε0 (R2 + x 2)3/2 (Considering right direction as positive) =
x 4 2R l 1 − + 2pε0 x 2(R2 + x 2 )3/2
=
l 2pε0
(As, Q = 4 2lR)
1 2 2 xR − x + 2 (R + x 2 )3/2
Initially x = 3R l 1 2 2 3 \ E= + − 2pε0R 8 3 l −2 2 + 3 l 3−2 2 = = 2pε0R 3 (2 2 ) 2pε0R 2 6 (−e)(E) el 3 − 2 2 Acceleration a = =− m pε0mR 4 6 8. (c) : Force on electron is zero at point where 1 2 2 xR E = 0 or = ; (R2 + x 2 )3/2 = 2 2 x 2R 2 x (R + x 2 )3/2 On solving, x = R 9. (d) : Potential difference between two points DV = – E dx Potential difference due to line charge between x = R and x = 3R VAB = − ∫
R
3R
−
ldx l l ln 3 = ln 3 = 2pε0x 2pε0 pε0 4
Potential difference due to the ring between x = R and x = 3 R −1 4 2lR 4 2lR −l 1 VAB′ = − − 1 = 4pε0 2R 2R pε0 2 Net potential difference VA −VB =
l pε0
1 ln3 − 1 − 4 2
10. (c) : Energy released by H atom in transition from n = 2 to n = 1 is 3 DEH = × 13.6 eV 4 Let He+ ions go to nth state. So energy required 3 1 1 DEHe = 13.6 × 4 − 2 eV = × 13.6 eV 4 n 4 So, n = 4 11. (c) : Visible light lies in the range, l1 = 4000 Å to l2 = 7000 Å. Energy of photons corresponding to these wavelengths (in eV) would be 12375 12375 = 1.77 eV = 3.09 eV and E2 = E1 = 7000 4000 –1.9 eV n=4 –3.4 eV n = 3 n=3 –6.04 eV n = 2 –3.4 eV DE = 10.2 eV DE = 10.2 eV n=2 –13.6 eV n = 1 H atom –13.6 eV n=1 –54.4 eV He+ Z=2 From energy level diagram of He atom we can see that transition from n = 4 to n = 3, energy of photon released will lie between E1 and E2. DE43 = –3.4 –(– 6.04) = 2.64 eV Wavelength of photon corresponding to this energy, 12420 Å = 4.7 × 10–7 m l= 2.64 12. (a) 13. (a) : Magnetic field at any point on Ampere’s loop can be due to all currents passing through inside or outside the loop. But net contribution in the left hand side will come from inside current only. 14. (d) : For r < a, current passing through within the cylinder of radius r is given by r r r kpr 4 2 I = ∫ JdA = ∫ kr 2prdr = 2pk∫ r 3dr = 2 0 0 0 Now using Ampere’s law, m k pr 4 m kr 3 B × 2pr = m0I = 0 ⇒B= 0 4 2
15. (a)
Physics For you | december ‘16
65
6
T
Class XII
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.
Optics
Total Marks : 120
Time Taken : 60 min NEET / AIIMS / PMTs
Only One Option Correct Type
1. A plane mirror coincides with a plane having equation x = 3. A particle is moving along a line with direction ratios 3, 4, 5. If speed of the particle is 2 , the velocity of its image is 3 4 3 4 1 (a) i + j + k (b) − i − j − k 5 5 5 5 5 (c)
3 4 (d) − i + j + k 5 5
3 4 1 i+ j− k 5 5 5
2. What is the optical path of light ray traversing the path POQ across two media of refractive indices 3 4 m1 = and m2 = (PR = a, RS = b, RO = x , SQ = y )? 2 3 P µ1
O
R
µ2 Q
S
(a) (b) (c) (d) 66
8 a 2 + x 2 + 9 b2 + y 2 6 3 a2 + x 2 + 8(b2 + ( y − x )2 ) 6 3 a 2 + b2 + x 2 + y 2 − 4 b2 + y 2 3 9 a2 + x 2 + 8 b2 + ( y − x )2 6 Physics For you | december ‘16
Liquid 3. The effective focal length of the lens combination shown in the figure is – 60 cm. The radii of curvature of the curved surfaces of the planoconvex lenses are 12 cm each and refractive index of the material of 3 the lens is . The refractive index 3 2 of the liquid is 2 3 5 8 4 (a) (b) (c) (d) 2 3 5 3 4. A light ray is travelling between two media as given below. The angle of incidence on the boundary in all the cases is 30°. Identify the correct sequence of increasing order of angles of refraction. (1) Air to water (2) Water to glass (3) Glass to water (Refractive indices of glass and water are respectively 3 4 and ) 2 3 (a) 1, 2, 3 (b) 2, 3, 1 (c) 3, 1, 2 (d) 1, 3, 2 5. Two coherent monochromatic light sources are located at two vertices of an equilateral triangle. If the intensity due to each of the sources independently is 1 W m–2, then the resultant intensity (in W m–2) due to both the sources at third vertex will be (a) zero (b) 2 (c) 2 (d) 4 6. Image of an object at infinity is formed by a convex lens of focal length 30 cm such that the size of the image is 2 cm. If a concave lens of focal length 20 cm is placed in between the convex lens and the image, at a distance 26 cm from the convex lens, the size of the new image is
(a) 2.5 cm (b) 2.0 cm (c) 1.2 cm (d) 1.5 cm 7. A rainbow is formed when a ray of sunlight passes through a spherical raindrop. Then the total angle through which the ray deviates is (i and r denote the angles of incidence and of refraction respectively) (a) 2i – 4r (b) p + 2i – 4r (c) 2(i – r) (d) 2(p + i – 2r) 8. The velocities of light in two different media are 2 × 108 m s–1 and 2.5 × 108 m s–1 respectively. The critical angle for these media is −1 1 −1 4 (a) sin (b) sin 5 5 1 (c) sin −1 2
1 (d) sin −1 4
9. The distance between an object and a divergent lens is n times the focal length of the lens. The linear magnification produced by the lens will be 1 1 (a) n (b) (c) n + 1 (d) n (n + 1) 10. An unpolarized light is travelling along z-axis through three polarizing sheets. The polarizing directions of the first and the third sheet are respectively parallel to x-axis and y-axis whereas that of the second one is at 60° to the y-axis. Then, the fraction of the initial light intensity that emerges from the system is about (a) zero (b) 0.093 (c) 0.031 (d) 0.28 11. An image is formed at a distance of 100 cm from the glass surface when light from point source in air falls on a spherical glass surface with refractive index 1.5. The distance of the light source from the glass surface is 100 cm. The radius of curvature is (a) 20 cm (b) 40 cm (c) 30 cm (d) 50 cm 12. A microscope consists of an objective of focal length 2 cm and eyepiece of focal length 5 cm. The two lenses are kept at a distance of 10.5 cm. If the image is to be formed at the least distance of distinct vision, the distance at which the object is to be placed before the objective is (Least distance of distinct vision is 25 cm) (a) 1.9 cm (b) 2.9 cm (b) 3.9 cm (d) 4.9 cm 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 : If objective and eye lenses of a microscope are interchanged, then it can work as telescope. Reason : The objective lens of telescope has small focal length. 14. Assertion : Blue colour of the sky is due to scattering of blue light. Reason : Blue colour has the shortest wavelength in the visible spectrum. 15. Assertion : Coloured spectrum is seen when we look through a muslin cloth. Reason : It is due to diffraction of white light on passing through fine slits. JEE MAIN / JEE AdvANcEd / PETs
Only One Option Correct Type
16. A convex lens forms the image of an axial point on a screen. A second lens with focal length f cm is placed between the screen and the first lens at a distance of 10 cm from the screen. To view the image, the screen has to be shifted away from the lens by 5 cm. A third lens having focal length of the same magnitude f cm is used to replace the second lens at the same position. But this time to view the image, the screen has to be shifted towards the lens by d cm. The value of f and d respectively, are (a) 30 cm, 2.5 cm (b) 30 cm, 5 cm (c) 7.5 cm, 2.5 cm (d) 7.5 cm, 5 cm 17. The angle of refraction of a very thin prism is 1°. A light ray is incident normally on one of the refracting surfaces. The ray that ultimately emerges from the first surface, after suffering reflection from the second surface makes an angle of 3.32° with the normal. The deviation of the ray emerging from the second surface and the refractive index of the material of the prism respectively are (a) 0.66°, 1.66 (b) 1.66°, 1.5 (c) 1.5°, 1.66 (d) 0.66°, 15 18. P S
i
4 3
Q
3 2
4 3
R
A long thin rectangular slab PQRS having refractive 3 is immersed in a liquid having refractive index 2 Physics For you | december ‘16
67
4 . A ray of light is incident at the edge PQ of 3 the slab as shown in figure. What is the sine of angle of incidence i such that the ray comes out from the slab as shown in figure? 1 4 17 8 (a) (b) (c) (d) 2 8 9 17 19. In Young’s double slit experiment, one of the slits is wider than the other, so that the amplitude of the light from one slit is double that from the other slit. If Im be the maximum intensity, the resultant intensity when they interfere at phase difference f is given by I I f f (a) m 1 + 2 cos2 (b) m 1 + 2 cos2 9 3 2 2 index
(c)
Im 5
2 f 1 + 4 cos 2
(d)
Im 9
2 f 1 + 8 cos 2
More than One Options Correct Type
20. An object and a screen are separated by a distance D. A convex lens of focal length f such that 4f < D, is moved between the object and the screen to get two sharp images. If the two positions of the lens are separated by a distance L, then (a) L is equal to D(D − 4 f ). (b) object distance in one position is numerically equal to image distance in the other position. (D − L ) (c) the ratio of sizes of the two images is . (D + L ) (d) the ratio of sizes of the two image is
(D − L)2 (D + L)2
.
21. A light source, which emits two wavelengths l1 = 400 nm and l2 = 600 nm, is used in a Young’s double slit experiment. If recorded fringe widths for l1 and l2 are b1 and b2 and the number of fringes for them within a distance y on one side of the central maximum are n1 and n2, respectively, then (a) b2 > b1 (b) n1 > n2 (c) from the central maximum, 3rd maximum of l2 overlaps with 5th minimum of l1. (d) the angular separation of fringes for l1 is greater than l2. 22. A transparent thin film of uniform thickness and refractive index m1 = 1.4 is coated on the convex spherical surface of radius R at one end of a long solid glass cylinder of refractive index m2 = 1.5, as shown in the figure. Rays of light parallel to the axis 68
Physics For you | december ‘16
of the cylinder traversing through the film from air to glass get focused at distance f1 from the film, while rays of light traversing from glass to air get focused at distance f2 from the film. Then 1 (a) |f1| = 3R (b) |f1| = 2.8R 2 Air (c) |f2| = 2R (d) |f2| = 1.4R 23. Consider three µ1 converging lenses L1, L2 and L3 O O1 O2 µ2 having identical geometrical L3 construction. The refractive indices of L1 and L2 are m1 and m2 respectively. The upper half of the lens L3 has a refractive index m1 and the lower half has m2. A point object O is imaged at O1 by the lens L1 and at O2 by the lens L2 placed in same position. If L3 is placed at the same place, (a) there will be an image at O1. (b) there will be an image at O2. (c) the only image will form somewhere between O1 and O2. (d) the only image will form away from O2. Integer Answer Type 24. The focal length of a thin biconvex lens is 20 cm. When an object is moved from a distance of 25 cm in front of it to 50 cm, the magnification of its image changes from m25 to m50. The ratio m25/m50 is 25. In Young’s double silt experiment, two slits act as coherent sources of equal amplitude a and same wavelength l. In another experiment with the same set up, the same two slits are incoherent. The ratio of intensity of light at the middle point of the screen in the first case to that in second case is 26. Image of an object approaching a convex mirror of radius of curvature 20 m along its optical axis 25 50 is observed to move from m to m in 30 s. 3 7 The speed of the object in km h–1 is Comprehension Type A prism of refractive index m1 and another prism of refractive index m2 are stuck together without a gap as shown in the figure. The angles of the prisms are as shown. Refractive indices m1 and m2 depends on wavelength l as follows : 10.8 × 104 1.80 × 104 m1 = 1.20 + and m = 1 . 45 + 2 l2 l2
where l is in nm.
D C
A
60°
70°
µ2
µ1 40°
B
27. The wavelength l0 for which rays incident at an angle on the interface BC pass through without bending at the interface is (a) 500 nm (b) 600 nm (c) 650 nm (d) 700 nm 28. For light of wavelength l0, the angle of incidence on the face AC such that the deviation produced by the combination of the prisms is minimum will be (a) 30° (b) 45° 3 − 1 (c) sin (d) sin −1 4 4 3 Matrix Match Type 29. Four combinations of two thin lenses are given in column I. The radius of curvature of all curved surfaces is R and the refractive index of all the lenses is 1.5. Match lens combinations in column I with their focal length in column II. Column I Column II (P) 2R (A) R 2
(B)
(Q)
(C)
(R) –R
(D)
(S) R
(a) (b) (c) (d)
A P Q S Q
B Q S P P
C R R Q R
y 30. A monochromatic S2 parallel beam of light of wavelength l is x O incident normally on S1 Screen the plane containing D slits S1 and S2. The slits are of unequal widths such that intensity only due to one slit on screen is four times that only due to the other slit. The screen is placed along y-axis as shown in figure. The distance between slits is d and that between the screen and slits is D. Match the statement in column I with results in column II. Column I Column II
D S P R S
(A) The distance between two (P) points on the screen having equal intensities, such that intensity at those points is 1 th of maximum intensity. 9 (B) The distance between two (Q) points on the screen having equal intensities, such that intensity at those points is 3 th of maximum intensity. 9 (C) The distance between two (R) points on the screen having equal intensities, such that intensity at those points is 5 th of maximum intensity. 9 (D) The distance between two (S) points on the screen having equal intensities, such that intensity at those point is 7 th of maximum intensity. 9 A B C (a) P, Q, R, S Q, R, S P, Q, R, S (b) P, R, S P, Q, R, S Q, R, S (c) Q, R, S P, Q, R, S Q, R, S (d) P, Q, R P, Q, R, S Q, R, S,
Dl 3d
Dl d
2Dl d
3Dl d
D P, R, S P, Q, R, S P, Q, R, S P, Q, R
Keys are published in this issue. Search now! J
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 !
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Marks scored in percentage
……
< 60%
not satisFactory! Revise thoroughly and strengthen your concepts.
Physics For you | december ‘16
69
P
PHYSICS
MUSING
hysics Musing was started in August 2013 issue of Physics For You with the suggestion of Shri Mahabir Singh. The aim of Physics Musing is to augment the chances of bright students preparing for JEE (Main and Advanced) / AIIMS / NEET / 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.
one or more than one oPtion correct tyPe
1. A long straight wire carries a charge with linear density λ. A particle of mass m and charge q is released at a distance r from the wire. The speed of the particle as it crosses a point at distance 2r from the wire is (a)
qλ ln r π mε0
(b)
qλ ln 2 π mε0
(c)
qλ ln 2 2π mε0
(d)
2qλ ln r . π mε0
2. There are two concentric and coplanar rings of radius a and 3a with uniformly distributed charge –Q and 8Q respectively, fixed in y–z plane with center at origin. A particle of small positive charge q and mass m is at origin as shown in the figure then which of the following statements is/are correct ? (Neglect gravity) (a) If the particle is given a small push along x-axis then it performs oscillatory motion of time period, T = 12
3π3a3mε0 . 19qQ
(b) If the particle is given a small push along x-axis then it performs oscillatory motion of time period, T = 12
7π3a3mε0 . 19qQ
(c) If the particle is given a velocity along x-axis and it performs oscillatory motion then amplitude 7 of oscillation must be less than a. 3 (d) If the particle is given a velocity along x-axis and it performs oscillatory motion then amplitude 5 of oscillation must be less than a. 3 3. To measure velocity of sound wave in air, a student performs air resonance column experiment with a tuning fork of frequency 500 Hz. He records length of air column for first three resonances. First and third resonance lengths are 15.6 cm and 82.8 cm but second resonance length is not clearly readable because of his casual writing. Then, (a) velocity of sound wave is 336 m s–1 (b) velocity of sound wave is 342 m s–1 (c) second resonance length is 49.2 cm (d) second resonance length is 46.8 cm. 4. A solid sphere of mass M and radius R is lying on a rough horizontal plane. A constant force F = 4Mg acts vertically downwards at point P such that OP makes 60° with horizontal. Find the minimum value of coefficient of friction m so that sphere does pure rolling. 3 (a) F = 4Mg 7 P 4 (b) 60° 7 O 2 (c) 7 2 (d) 5
By Akhil Tewari, Author Foundation of Physics for Jee main & Advanced, Professor, IITians PAce, mumbai.
70
Physics For you | december ‘16
5. A uniform rod of mass M and length L is hinged at point A about which it can rotate freely. The rod is kept horizontally by means of a vertical string as shown. A small block is placed at other end B of the rod. Now string is cut, then
(a) just after cutting relative acceleration between block M and end B of the rod is g/2. (b) just after cutting relative acceleration between block m and end B of the rod is zero. (c) tension in the rod at its mid point when the rod 9Mg becomes vertical is . 8 (d) tension in the rod at its mid point when the rod 3 Mg becomes vertical is . 4 6. A solid conducting sphere of radius R is initially uncharged. Charge Q is brought from infinity to point A which is at distance 2R from centre of the sphere. B and D are points on sphere as shown in figure and switch S is initially opened. Which of the following statements is/are correct?
(d) If switch is closed then charge Q flows from sphere to ground. 7. A spherical ball of mass m moving with initial velocity v collides elastically with another ball of mass m, which is fixed at one end of L shaped rigid massless frame as shown in figure. The L shaped frame contains another mass m connected at the other end. The speed of the striking mass after collision is v (a) backwards 2 (b) v in same direction 5 v (c) in same direction 3 v (d) backwards. 7 integer ansWer tyPe 8. Two factories are sounding their sirens at 400 Hz each. A man goes from one factory to the other at a speed of 2 m s–1. The velocity of sound is 320 m s–1. Therefore what will be the number of beats heard by the person in one second ? 9. Where should a convex lens of focal length 9 cm be placed (in cm) from nearest source between two point sources S1 and S2 which are 24 cm apart, so that the images of both sources are formed at the same place? 10. A system consist of a uniformly charged sphere and surrounding medium is filled by charge with a volume charge density ρ = ; where a is a positive r constant and r is distance from centre of the sphere.
VD 5 ; where VB and VD are potential = VB 2 − 5 at B and D respectively due to induced charge on the sphere. Q 1 1 − (b) VB − VD = ; where V B and 4πε0R 2 5 VD are potential at B and D respectively due to induced charge on the sphere. Q (c) If switch is closed then charge flows from 2 sphere to ground.
It is found that electric field outside sphere is 3a then m =? constant and given by E = mε0
(a)
MPP-6 CLASS XI 1. 6. 11. 16. 21. 26.
(c) (c) (a) (d) (a,d) (4)
2. 7. 12. 17. 22. 27.
(b) (c) (c) (a) (a,b) (a)
3. 8. 13. 18. 23. 28.
ANSWER
KEY
4. 9. 14. 19. 24. 29.
5. 10. 15. 20. 25. 30.
(d) (b) (c) (a) (a,b,d) (d)
(a) (c) (a) (d) (8) (d)
Physics For you | december ‘16
(c) (b) (c) (a,c,d) (9) (c) 71
ON
R-C Circuit Er. Sandip Prasad
Initial state
When key is closed, i.e., at t = 0, R-C circuit is in its initial state. In this state, current in the circuit i is maximum because there is no potential difference across capacitor to oppose the applied voltage, i.e., it behaves like being short circuited. If you wish to find initial current, short the terminals of the capacitor. Hence at t = 0 the circuit can be redraw as shown in the figure. C
VR =
VC = 0
R
im
R
im =
R
Transient state
As time passes, i decreases R +q C –q gradually so does instantaneous VR VC voltage across resistance, i VR also decreases gradually, but instantaneous voltage across capacitor VC increases gradually till it reaches at its maximum value or steady state value ε. Although VR and VC are variable, their algebraic summation at any instant always equal to ε, i.e., at any time t, ε = VR + VC . Final state (Steady state)
As charging continues, charging current i decreases gradually and it becomes zero at final state or steady state, hence at steady state condition, the circuit appears as an open circuit (i = 0), which implies VR = 0 and VC = ε. At this state capacitor gets completely charged and the value of charge becomes maximum and its maximum value is q0 = Cε. +q0 C –q0
R
VC =
VR = 0
R
charging of capacitor e C Consider a circuit in which a R + resistor of resistance R is A B connected in series with a i capacitor of capacitance C, the combination is connected K across a battery of emf ε. When key is closed, electrons from plate A swings to plate B as shown in the figure. The voltage across capacitor rise to ε exponentially and not linearly. Charging current i is maximum at the start, i.e., when capacitor is uncharged, then it decreases exponentially and finally ceases when potential difference across capacitor plate becomes equal and opposite to the battery voltage. Hence as charge q on capacitor increases, current i in the circuit decreases. In other words, as the time passes, potential difference across resistor (V R) decreases and difference across the capacitor (VC) increases. Whenever a R-C circuit connected with a DC source, goes from initial state to final state (steady state condition), it passes through a transient state which is of short duration. In fact transient state lies in between initial state and final state (steady state condition).
i=0
Calculation of current in the R-C circuit in transient state R
+q
i K
For the closed loop; As, ∑∆V = 0 dq q q \ + ε − iR − = 0; ε = R + dt C C 1 dq q dq ε − = R; dt = q R C dt ε− C
Sandip Physics Classes, Girish Park-1/1 Shiv Krishna Daw Lane, Kolkata-700007 72
Physics For you | december ‘16
�q
dq i = dt
Integrating both sides, q t dq 1 = ∫ q ∫ R dt 0ε− 0 C q 1 1 q − ε ln = t −1 / C C 0 R q − RC ln ε − − ln(ε − 0) = t C q q ε− ε− t C = e −t /RC C= ln ; ε −RC ε q q ε − = εe −t /RC ; ε − εe −t /RC = C C q −t / RC −t / RC ) ε(1− e ) = ; q = Cε (1 − e C q = q0 (1 – e–t/τ) Here, q0 = Cε is the maximum charge on the capacitor and τ = RC is time constant of the circuit. At t = τ, q = q0(1 – e–1) = 0.63q0 Hence, time constant is defined as the time in which charge in the capacitor grows to 63% of its maximum value. dq d ε Now, i = = [Cε(1 − e −t /RC )] = e −t /RC dt dt R i = i0 e–t/τ ε Here, i0 = = maximum current in the circuit. R Important points Charge on the capacitor at time t is, q = q0 (1 – e–t/RC) q0 is the maximum charge on the capacitor, i.e., charge on the capacitor at steady state. \ q0 = Cε, where ε is the emf of the cell applied to the circuit. Here, τ = RC is the time constant of the R-C circuit. Its unit is second. Hence, charge at time t becomes, q = q0(1 – e–t/τ) We know that charge in the circuit at time t is, q = q0(1 – e–t/τ) If t = τ, i.e., after one time constant, charge in the circuit is, \ q = q0(1 − e −τ /τ )
1 = q0(1 − e −1) = q0 1 − e \ q = 0.632 q0
Hence, the time constant of an R-C circuit is defined as the time during which charge on the capacitor actually rises from zero to 0.632 of its final steady value (maximum value) of the charge. It means after one time constant, 63% of total charge is accumulated on the capacitor. As, q = q0(1 – e–t/τ) At t = 0, q = q0(1 – e0) = q0(1 – 1) \ q=0 Hence at t = 0, capacitor can be treated as short circuit, i.e., capacitor provides zero resistance. At t = ∞, 1 1 q = q0(1 – e–∞) = q0 1 − = q0 1 − ∞ e ∞ \ q = q0 Hence t = ∞, i = 0, i.e., capacitor acts as an open circuit. Just at the start of the charging, charge on the capacitor is zero. As, charging continues, charge on the capacitor changes according to the equation q = q0(1 – e–t/τ). It becomes maximum (q0) when t = ∞, though it is almost charged to this value in about five time constants (5τ). As, q = q0(1 – e–t/τ) Graph showing the variation of charge q with time t As, i = i0e–t/τ ε Here, i0 = = imax R If t = τ, i = i0e −τ /τ = i0e −1
q
C 0.632C O t=
i i0
i = 0.37i0 Graph showing the 0.37 i 0 variation of current i O with time t Voltage across resistance at time t, VR = iR = i0 Re–t/τ ε VR VR = εe −t /τ i0 = R Graph showing the variation of voltage 0.37 across resistance VR O with time t Voltage across capacitor at time t, VC =
t
q q0(1 − e −t /τ ) Cε = (1 − e −t /τ ) = C C C Physics For you | december ‘16
73
Important points
VC
\ VC = ε(1 − e −t /τ ) Graph showing the variation of voltage across capacitor VC with time t
0.632 O
t=
t
Heat dissipated: By energy conservation, Heat dissipated = work done by cell – ∆Ucapacitor 1 1 = Cε(ε) − Cε2 − 0 = Cε2 2 2 Alternatively: ∞
Heat = H = ∫ i 2Rdt 0
=
∞ 2
∫
0
ε
R2
−2t e RC Rdt ∞
=
ε2 ∞ −2t /RC dt ∫e R 0
−2t ε2RC RC ε2C e 0 = = 2R 2 Discharging of capacitor In the previous case, suppose battery is shorted after the capacitor is fully charged and the capacitor is allowed to discharge through a resistor or
+C K R
a capacitor carrying a charge q0 is allowed to discharge through a resistor. Suppose the key is closed at t = 0, let q be the charge on the capacitor and i be the current in the circuit at any instant of time. By Kirchhoff ’s voltage law, q − iR = 0 C dq But i = − dt Negative sign indicates that charge on the capacitor is decreasing. dq −q q dq = \ + R = 0; dt RC C dt q
dq 1 dq 1 t =− dt ; ∫ =− ∫ dt q RC q RC q0 0 q 1 ln = − t ; q = q0e −t /RC q0 RC dq ε Again, i = − = e −t /τ = i0e −t /τ dt R 74
Physics For you | december ‘16
Charge on the capacitor at time t is, q = q0 e–t/RC q0 is the initial charge on the capacitor, i.e., charge on the capacitor at t = 0. Here, τ = RC is the time constant of the R-C circuit. Its unit is second. Hence, charge at time t becomes, As, q = q0 e–t/RC We know that charge in the circuit at time t, q = q0 e–t/τ If t = τ, i.e., after one time constant, charge in the circuit is, \ q = q0 e–τ/τ; q = q0 e–1 = 0.37 q0 Hence, the time constant of an R-C circuit is defined as the time during which charge on the capacitor falls to 0.37 of its initial value (maximum value) of the charge. Time constant can also be defined as the time in which charge on the capacitor falls to 37% of its maximum value during discharging process. As, q = q0 e–t/τ At t = 0, q = q0 e0 = q0 \ q = q0 At t = ∞, 1 1 q = q0e −∞ = q0 ∞ = q0 ∞ e \ q=0 Just at the start of the charging, charge on the capacitor is maximum and equal to q0. As, discharging continues, charge on the capacitor changes according to the equation q = q0 e–t/τ. It becomes zero when t = ∞, though it is almost discharged to zero in about five time constants (5τ). Graph showing the variation of charge q with time t q
q0 0.37 q0 O
As, q = q0(e–t/τ) And i = −
dq i = − q0 (e −t /τ ) ; (−τ) dt
q \ i = 0 (e −t /τ ) τ or i = i0e −t /τ
Graph showing the variation of current i with time t
q ε − i1R + R=0 2R C 2q 2q −2i1R − + ε − i1R = 0 ; 3i1R = ε − C C dq Again, we know that i1 = dt dq 1 dq 2q ⇒ = dt \ 3 R=ε− 2q 3R dt C ε− C Integrating both sides, \ − i1R −
i i0 0.37 i0
Voltage across resistance at time t, q VR = iR = 0 Re −t /τ VR τ VR = εe −t /τ ( q0 = Cε)
q
Graph showing the 0.37 variation of voltage across resistance VR with time t Voltage across capacitor at time t,
q e −t /τ Cε −t /τ q e = 0 = C C C VC \ VC = εe–t/τ Graph showing the variation of voltage across capacitor VC 0.37 with time t As, VC =
Example-1: Initially the capacitor is uncharged, find the charge on the capacitor as a function of time, if switch is closed at t = 0. Soln.: Method-I: By applying KVL in loop (1) As, S ∆V = 0 \ +ε – iR – i2R = 0 ... (i) By applying KVL in loop (2) As, S ∆V = 0
R
C
R R
K C
R i R
t 1 dq = ∫ dt 2q 3R 0 ε− 0 C q 1 2q 1 ln ε − = t −2 / C C 0 3R 2q ε− C = 2t ln ε −3RC 2q ε− C = e −2t /3RC ; ε − 2q = εe −2t /3RC ε C 2q −2t /3RC 2q ε − εe = ; ε(1 − e −2t /3RC ) = C C Cε −2t /3RC q = (1 − e ) 2 Method-II: Another method can be used to find the charge on the capacitor of R-C circuit as a function of time. Before going to the method-II, we need to learn a very important concept related to electrical circuit or network. This concept is not only used to solve the problems of R-C circuit but you can apply it any type of electrical circuit. r1 Consider the circuit as shown in figure. r2 The circuit can be converted A B to a single source of battery r3 by using the following formula : Equivalent emf, ε1 ε2 ε3 r + r + r 2 3 εeq = VA − VB = 1 1 1 1 r + r + r 1 2 3 Where, ε1, ε2 and ε3 are the net emf of the branches 1, 2 and 3 respectively and r1, r2 and r3 are the net resistances respectively. Internal resistance of the equivalent battery can be given by following formula, 1 1 1 1 = + + req r1 r2 r3
∫
R
2 1
i1 i2 i K
q ... (ii) \ − i1R − + i2R = 0 C Hence from both the equations, we can write an expression in terms of i1. From equation (i) ε – (i1 + i2)R – i2R = 0 ε – i1R – 2i2R = 0 ε − i1R ... (iii) \ i2 = 2R From equations (ii) and (iii)
Physics For you | december ‘16
75
Suppose we have to convert the given circuit into an equivalent battery, where ε1 = 3 V, ε2 = –2 V, ε3 = 1 V and r1 = r2 = r3 = 1 W. Equivalent emf of the battery can be written as, +3 −2 +1 + + 1 1 1 =2 V εeq = VA − VB = 1 1 1 3 + + 1 1 1 Internal resistance of the equivalent battery can be given by following formula, 1 1 1 1 1 1 1 1 = + + = = + + \ req = 3 W req r1 r2 r3 req 1 1 1 Hence the given network becomes as a single 2 source of voltage V and having an internal 3 resistance 3 W.
A
r1
3V
r2
2V
r3
B = A
req
A
B
eq =
1V
2 V 3
R
For problems related with R-C circuit with DC source, it is easier to convert the R-C circuit having a number of parallel branches into a R-C circuit of single voltage source with a series resistance. It makes the solution of complicated networks quite quick and easy. Steps for converting an electrical circuit between two terminals into a single battery:
Step-I: First of all, temporarily remove the branch containing capacitor and redraw the remaining circuit. The remaining circuit has a number of parallel branches containing voltage source (battery) and resistance. R
R
C
C
A
R
A
R
K
B B
R
R
K
Step-II: To convert the remaining network into a single equivalent battery, apply the following formula : Equivalent emf, ε1 ε2 ε3 r + r + r 2 3 εeq = VA − VB = 1 1 1 1 r + r + r 1 2 3 76
Physics For you | december ‘16
Internal resistance of the equivalent battery can be given by following formula, 1 1 1 1 A B = + + 1 R req r1 r2 r3 Here there are two parallel branches 1 and 2 in between R K 2 the terminals A and B. So, ε 0 +ε + ε εeq = VA − VB = R R = 1 1 2 + R R Internal resistance of the equivalent battery can be given by following formula, 1 1 1 1 1 R = + \ req = . = + req r1 r2 R R 2 Here req is the resistance between the terminal A and B, i.e., it is considered as the internal resistance of the equivalent battery. B
1
R
A
req = R 2 eq = 2
K
2
B
Step-III: Now, the temporarily removed branch is connected back across from where it was temporarily removed earlier. From this arrangement we get an R-C circuit containing a single source. R
C
R
C
A
R R
K ε
A
B req = R 2
K
εeq = ε 2
B
Hence, from the above circuit it is clear that charging of capacitor takes place, in this case the value of charge at any time t is given by q = q0(1 – e–t/τ) Here, q0 = the maximum charge on the capacitor, i.e., ε charge on the capacitor at steady or final state = C 2
R
and τ = time constant of the circuit = ReqC = R + C 2 3R \ τ= C 2 Hence the equation of charge is given by, Cε q = (1 − e −2t /3RC ) 2 To be continued in next issue…
OLYMPIAD PROBLEMS 1. A heavy particle is tied to the end A of a string of length 1.6 m. Its other end O is fixed. It revolves as a conical pendulum with the string making 60° with the vertical. Which of the following statements is incorrect ? 4π (a) Its period of revolution is s. 7 (b) The tension in the string is double the weight of the particle. (c) The velocity of the particle is 2.8 m s–1. (d) The centripetal acceleration of the particle is 10 m s–2. 2. The gravitational potential changes uniformly from –20 J kg–1 to – 40 J kg–1 as one moves along x-axis from x = – 1 m to x = +1 m. Mark the incorrect statement about gravitational field intensity at the origin. (a) The gravitational field intensity at x = 0 must be equal to 10 N kg–1. (b) The gravitational field intensity at x = 0 may be equal to 10 N kg–1. (c) The gravitational field intensity at x = 0 may be greater than 10 N kg–1. (d) The gravitational field intensity at x = 0 must not be less than 10 N kg–1. 3. A uniform magnetic field of induction B fills a cylindrical volume of radius R. A rod AB of length 2l is placed as shown in figure. If B is changing at the rate of dB/dt, the emf that is produced by the changing magnetic field between the ends of the rod is dB dB l R2 − l 2 (a) (b) l R2 + l 2 dt dt 1 dB 1 dB 2 2 l R −l l R2 + l 2 (c) (d) 2 dt 2 dt 4. A monochromatic beam of light of 6000 Å is used in Young’s double slit experiment set-up. The two slits are covered with two thin films of equal thickness 78
Physics For you | DECEmbEr ‘16
t but of different refractive indices as shown in figure. Considering the intensity of the incident beam on the slits to be I0, find the distance on the screen from central maxima at which intensity is I0 (Assume that there is no change in intensity of the light after passing through the films.)
S1
1 =
3 2
S2
2 =
4 3
d
D
Screen
Consider t = 6 mm, d = 1 mm, and D = 1 m, where d and D have their usual meaning. (a) 20 mm (b) 30 mm (c) 25 mm (d) 15 mm 5. A man of mass m on an initially stationary boat gets off the boat by leaping to the left in an exactly horizontal direction. Immediately after the leap, the boat of mass M, is observed to be moving to the right at speed v. Then, 1 (a) work done by the man on boat is mv 2 2 (b) increase in the mechanical energy of the system 1 M2 + M v2 of man and boat is 2 m (c) velocity of centre of mass of system is v (d) increase in kinetic energy of man is
M2 2 v . m
L 6. A rod of mass m, uniform x cross sectional area A and F length L is accelerated by Smooth applying force F as shown in figure on a smooth surface. Young’s modulus of elasticity of the material of rod is Y. Which of the
following statements is correct? (Consider x is measured from the right end.) (a) Tension in rod as a function of distance x is Fx . 2L F (b) Strain in rod is . AY (c) Elastic potential energy stored in the rod is F 2L . 6AY (d) There is no stress in rod.
7. Figure shows the variation U (J) of the internal energy U A C with density r of one mole of an ideal monatomic B ρ (kg m–3) gas for a thermodynamic cycle ABCA. Here process AB is a part of rectangular hyperbola. Then, (a) process AB is isothermal and net work in the cycle is done by gas (b) process AB is isobaric and net work in the cycle is done by gas (c) process AB is isobaric and net work in the cycle is done on the gas (d) process AB is adiabatic and net work in the cycle is done by gas. 8. A homogeneous rod AB of length L = 1.8 m and mass M is pivoted at the centre O in such a way that it can rotate freely in the vertical plane as shown in figure. The rod is initially in the horizontal position. An insect S of the same mass M falls vertically with speed v on the point C, midway between the points O and B. Immediately after falling, the insect moves towards the end B such that the rod rotates with a constant angular velocity. If the insect reaches the end B when the rod has turned through an angle of 90°, then the value of v will be S
v
A
(a) 3.5 m s–1 (c) 10 m s–1
L 2
O
L 4
C
L 4
B
(b) 7 m s–1 (d) 1.5 m s–1
9. A 100 eV electron collides with a stationary helium ion (He+) in its ground state and excites to a higher level. After the collision, He+ ion emits two photons
in succession with wavelengths 1085 Å and 304 Å. The principal quantum number of the excited state is (a) 2 (b) 3 (c) 5 (d) 7 10. A right angled ⊗ ⊗ ⊗ ⊗ triangular loop as a v v a ⊗ ⊗ ⊗ ⊗ shown in figure enters a a uniform magnetic ⊗ ⊗ ⊗ ⊗ x field (at right angle to a 0 2a 3a the boundary of the field) directed into the paper at constant velocity. Draw the graph between induced emf e and the distance along the perpendicular to the boundary of the field, (say x) along which loop moves.
(a)
a 2a 3a
x
(b)
a 2a 3a
x
(c)
a 2a 3a
x
a
(d)
2a 3a
x
solution oF noVEMBEr 2016 crossword 1
2
R A A P S E R A T E A O B O S 7 P I T O T T U B M U 11 P O L A R I 12 S F C E 15 14 R L A M P L U R U C A U L T P E N N S E T 23 22 S I R D R H E Y F A P I F I C A G E N E C 27 G R A V I T
D I O L Y S I S M I C R O W A V 6 B 8 U B E F A R F R S C O P E R E T B D A P O I N T E 17 D U L U M R 19 I P C A 24 A S T E R S C A O H T G M Y R O M 26 E M A F V E A T I E T O N R E I T 28 29 Y N I S O B A R F L A S H O V E R
H 3 L E 5 P H A T 9 J D 10 Z E T T A N A T S H K 13 Y H N I 16 G F O 18 A G I C S S 21 B S O N E U O S S T O 25 N L I N K C S
4
E
20
I N T E R F A C E
Winner (November 2016) • Jyoti rathour, Haldwani
Solution Senders (October 2016) • Santoshi Rawat, Delhi
• Bhavesh Bisht, Lucknow
Physics For you | DECEmbEr ‘16
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3. (b) : Suppose q is the required angle. At extreme position the velocity of the ball is zero, thus normal l cos ve2 acceleration an = = 0, and l h tangential acceleration at = g sinq. g sin \ Total acceleration at extreme position ae = g sinq At mean position, the velocity of the ball
solution set-40
1. (a) : Mass per unit area of the disc, Mass M M s= = = Area p (4 R)2 − (3R)2 7 pR2
(
)
Consider a ring of radius P x and thickness dx as 4R shown in the figure. Mass of the ring, 3R dx dM = s2pxdx 4R x 2pMxdx = 7 pR2 Potential at point P due to annular disc is VP =
4R
∫
−
GdM
=−
GM 2p 4 R 2
∫
(
18 × 1.8 × 10 −5 × 2 × 10 −3 4 × 900 × 9.8 3 r = × 10 −5 m 7 4 pr 3ρg From equation (i), we get q = 3E Substituting the given values, we get
and displacement of ball, h = (l – l cosq)
xdx
)
r2 =
3
6rvT
Physics For you | December ‘16
v2 The normal acceleration at this position, an = m l 2 g (l − l cos q) = = 2 g (1 − cos q) l and tangential acceleration at = g sin q = g sin0° = 0 Thus total acceleration at mean position am = 2g (1 – cos q) According to given condition, we have ae = am ⇒ g sin q = 2g (1 – cos q) q 1 1 ⇒ tan = ⇒ q = 2 tan −1 2 2 2 4. (c) : Let n number of men are required for the block to just start moving up the plane nF = mg sin q + f nF = mg [sin q + ms cos q] 1 = 200 × 10 sin 45° + cos 45° 2 or n =
mg
200 × 10 × 3 2 2 × 500
≈5
A 5. (b) : Q = Tα A − 4 ⇒ Tα = Q A − 4 = 5.5 216 = 5.4 MeV A 220 6. (c) : Impulse = change in momentum Dp = 50 N s and Dp = p – 0 \ p = 50 N s Velocity along vertical direction p sin60° = 25 3 m s −1 m Velocity along horizontal direction, p cos60° vH = = 25 m s −1 m Since, the impulse is applied tangentially, the ball will take parabolic path. vV =
3 4 × p × × 10 −5 × 900 × 9.8 7 q= = 8 × 10–19 C 81p 5 3× × 10 7 80
g g cos
vm = 2 gh = 2 g (l − l cosq )
7 pR 3R 16R2 + x 2 (4 R)2 + (x )2 Solving, we get, VP = − 2GM 4 2 − 5 7R Work done in moving a unit mass from P to ∞ 2GM (4 2 − 5) = V∞ – VP = 7R 2. (d) : In presence of electric field, in equilibrium, Force on the drop due to electric = Weight of the drop field(E), 4 ...(i) qE = mg = pr 3ρg 3 In the absence of electric field, drop acquires equilibrium. So, Viscous force on the drop = Weight of the drop 4 3 ...(ii) 6phrvT = mg = pr ρg 3 2 18 × h × vT or r = 4×ρ× g Substituting the given values, we get 3R
l
Time of flight, T =
Volume of the gas is constant as it is kept in a closed vessel.
2vV = 5 3s g
L r×p 50 × 0.2 × 5 = Also, w = = = 625 rad s–1 I 2 2 2 × 1 × (0.2)2 mr 5 w 625 = As 2pu = w \ u = 2p 2p Now, total number of rotation during the time of flight 3125 3 = u ×T = 2p V Q 7. (i) VA = VB = V, E = and V = d C 2Q e = 2V = ...(i) C Q′ Q′ Q′ (ii) VA′ = = ; VB′ = C′ K C C Q′ 1 ...(ii) Hence e = 1+ C K From (i) and (ii), we get 2K Q′ = Q , since K > 1, Q′ > Q 1+ K 2K 2 VA′ = V and VB′ = V 1+ K 1+ K V′ 2V Electric field E A′ = A = d (1 + K )d VB′ 2KV EB′ = = d (1 + K )d Change in electric fields 2V V V 1 − K DE A = E A′ − E = − = (1 + K )d d d 1 + K 2KV V V K − 1 DEB = EB′ − E = − = (1 + K )d d d K + 1 The amount of charge that flows into the circuit 2K 1 K − 1 DQ = Q ′ − Q = − 1 Q = Ce 2 K + 1 1+ K 8. According to gas equation, PV = nRT \ n=
PV
=
6 (1.6 × 10 ) × 0.0083
=
16
8.3 × 300 3 3R 5R CP − CV = R \ − CV = R or CV = 2 2 Amount of heat supplied = nCV dT \
RT
dT =
2.49 × 10 4 × 3 × 2 16 × 3 × 8.3
= 375 K
\ Final temperature, T2 = 300 + 375 = 675 K
\
P1V T1
=
P2V
or
T2
P2 = P1
or P2 = 3.6 × 106 N m–2
T2 T1
=
(1.6
× 106 ) × 675 300
9. The motion of the sphere is similar to projectile motion. The components of its acceleration are; 0. 9 ax = = 1.80 m s −2 and a y = 0 0. 5 when the sphere crosses the y-axis, its displacement component along x-axis is zero. ay = 0
y
ax = 1.8 m s–2 30°
x
1 1 0 = ux t + ax t 2 = 3 sin 30° t − (1.8)t 2 or t = 1.66 sec 2 2 10. Let Mmix be the molecular weight of the mixture. m m m1 + m2 = 1 + 2 M mix M1 M2 As v1 =
...(i)
γRT γRT or M1 = 2 M1 v1
γRT γRT or M2 = 2 M2 v2 γRT Similarly, Mmix = v2 m + m2 From (i), Mmix = 1 m1 m2 + M1 M2 m1 + m2 m1v12 + m2v22 γRT or = ⇒ v= m1 + m2 v2 m1v12 m2v22 + γRT γRT and v2 =
Solution Senders of Physics Musing Set-40 1. Nikita Pandey, Dehradun (Uttrakhand) 2. Jisha Nair, Kota (rajasthan) 3. Sabhya Sanchi, Asansol (West bengal) Physics For you | December ‘16
81
CLASS XI Series 6
Specific heat ratio, g = CBSE
Contd. from Page No. 30
Force exerted on the wall of area A Dp = = nmv 2x A Dt 2 Force nmv x A Pressure on the wall = = Area A or P = nmvx2 As the molecules move with different velocities, so we replace vx2 by its average value v x2 in the above equation. \ P = nmv x2 Again, the gas is isotropic. So the molecular velocities are almost equally distributed in different directions. By symmetry, v x2 = v y2 = vz2 =
(
)
1 2 1 2 v + v y2 + vz2 = v 3 x 3
where v 2 is the mean square velocity of the molecules. 1 ...(i) Hence, P = nmv 2 3 Mass nm Density of gas, r = = nm = 1 Volume 1 \ P = rv 2 ...(ii) 3 OR (a) In case of a monatomic gases, like He, Ar, etc., a molecule has three translational degrees of freedom. According to the law of equipartition of energy, average energy associated with each degree 1 of freedom per molecule = kBT 2 Let R = gas constant per mole of a gas NA = Avogadro's number i.e., the number of atoms in one mole of the gas. Then the total internal energy of one mole of a monatomic gas, 3 3 U = kBT × NA = RT (Q kB NA = R) 2 2 The molar specific heat at constant volume will be dU d 3 3 = CV (monatomic) = RT = R dT dT 2 2 The molar specific heat at constant pressure, 3 5 CP (monatomic) = CV + R = R + R = R 2 2 82
Physics For you | DECEmbEr ‘16
CP (5 / 2)R 5 = = = 1.67 CV (3 / 2)R 3
(b) (i) Diatomic molecules such as N2, O2, etc., behave as rigid rotator at moderate temperatures. Such molecules have 5 degrees of freedom : 3 translational and 2 rotational. According to the law of equipartition of energy, the total energy of a mole of such a gas is 5 5 U = kBT × N A = RT 2 2 dU 5 = R \ CV (rigid diatomic) = dT 2 7 CP (rigid diatomic) = CV + R = R 2 (7 / 2)R 7 g(rigid diatomic) = = = 1.4 (5 / 2)R 5 (ii) If the diatomic molecule is not rigid but has also a vibrational mode, then each molecule has 1 an additional energy equal to 2 × kBT = kBT, 2 because a vibrational frequency has both kinetic and potential energy modes. 7 7 5 \ U = kBT + kBT N A = kB N AT = RT 2 2 2 dU 7 = R CV (diatomic with vibrational mode) = dT 2 CP (diatomic with vibrational mode) 9 = CV + R = R 2 g (diatomic with vibrational mode) (9 / 2)R 9 = = = 1.28 (7 / 2)R 7 (c) (i) A non-linear triatomic gas molecule has six degrees of freedom. 6 \ U = kBT × N A = 3RT 2 dU CV = = 3R; CP = CV + R = 4R dT C 4 g = P = = 1.33 CV 3 (ii) A linear triatomic molecule has seven degrees of freedom. 7 7 \ U = kBT × N A = RT 2 2 dU 7 9 CV = = R ; CP = CV + R = R dT 2 2 C (9 / 2)R 9 g= P = = = 1.28 CV (7 / 2)R 7
Y U ASK
WE ANSWER Do you have a question that you just can’t get answered? Use the vast expertise of our mtg team to get to the bottom of the question. From the serious to the silly, the controversial to the trivial, the team will tackle the questions, easy and tough. The best questions and their solutions will be printed in this column each month.
Q1. What are fermions and bosons?
– Aditya Yaligar (Dharwad, Karnataka)
Ans. Fermions are the half spin particles. These particles cannot exist at the same place having exactly the same properties. Electron, proton, neutron etc. are fermions. Bosons have whole number spins. These particles can exist at same place and in same physical state. Photons, mesons etc. are bosons. Q2. We can tune radio only on a AC source, then
how are we able to hear FM radio in our mobile phones? – Seetharaman Ans. Tuner receive a mixture of signals of different frequencies and able us to choose a particular frequency whereas detector is responsible for separating audio information from the carrier wave. For AM signals, this can be done with a diode but for FM signals, the detector circuits are specially designed. Detectors are basically demodulators, FM demodulators convert frequency variations of the input signal into amplitude variations at the output. Q3. Why does pizza develop a nicely melted cheese surface, with lightly browned spots, if topped with real cheese but not with fat-free cheese? – Harlin Engtipi, Guwahati
Ans. A pizza is cooked on the hot pan by conduction of infrared radiation from the oven walls surrounding it, and convection of hot air across its top (especially if the air is being forced to move by a fan). As thermal energy is gradually transferred to the interior, largely to cook the dough, the cheese is supposed to melt uniformly over the top and then lightly brown. The browning occurs where bubbles form in the cheese (where water vaporizes to form bubbles of steam inside
the cheese). As the tops of these bubbles are thin during bubble growth, the tops can absorb enough thermal energy to turn brown. If the pizza is topped with fat-free cheese, the water evaporates very quickly from the cheese, and the dried-out individual strands of cheese never melt and fuse, but instead just burn. To remedy this, fat-free or low-fat cheese is sprayed with an oil film when the pizza is prepared. Then the oil film slows the evaporation of water from the cheese, so that melting, fusing, bubbling, and browning can all occur. Q4. When the sky is overcast, why is snow at the horizon brighter than the adjacent sky? – Nustrat Khan, Hyderabad
Ans. During an overcast day, snow at the horizon is brighter than the adjacent sky for three reasons (1) Drops in the cloud scatter sunlight primarily in a forward direction, and so you intercept more light from the overhead portion of the overcast sky than a portion near the horizon. Thus, the sky near the horizon is comparatively dark. (2) Snow scatters light strongly in all directions, and so you intercept significant light from snow near the horizon. Thus that snow is bright. (3) When you view a border separating regions differing in brightness, your visual system will enhance that difference in order to make the border more distinct. Q5. Why are the hairs on polar bears hollow? – Deepika Tyagi, Surat
Ans. The white hairs on a polar bear trap the visible and infrared portions of sunlight, because those portions are reflected and transmitted down into the pelt to reach the skin. There it is absorbed which increases the thermal energy of the skin. (The ultraviolet portion of sunlight is also absorbed by the hairs, but ultraviolet light contributes little to the warming of a bear). The thermal energy of the skin is maintained partially because the hairs are hollow and conduct thermal energy poorly. (The notion that the hollow hairs somehow function as optical fibres is just a myth.) MPP-6 CLASS XII ANSWER KEY 1. 6. 11. 16. 21. 26.
(d) (a) (a) (a) (a,b,c) (3)
2. 7. 12. 17. 22. 27.
(d) (b) (b) (a) (a,c) (b)
3. 8. 13. 18. 23. 28.
(c) (b) (d) (b) (a,b) (c)
4. 9. 14. 19. 24. 29.
(a) (d) (c) (d) (6) (b)
Physics For you | december ‘16
5. 10. 15. 20. 25. 30.
(d) (b) (a) (a,b,d) (2) (c) 83
Readers can send their responses at
[email protected] or post us with complete address by 25th of every month to win exciting prizes. Winners' name with their valuable feedback will be published in next issue. CUT HERE ACROSS 2 1 2. An instrument for measuring changes in 5 magnetic flux. [9] 3. A semiconductor device that acts as a 10 storage location in processing unit of a 11 computer. [8] 8. A unit of frequency equal to 1012 hertz. [7] 17 11. A mechanical device that prevents any 18 sudden or oscillatory motion of a moving part of any piece of apparatus. [4, 3] 12. A device used for separating two isotopes 21 by thermal diffusion. [7, 6] 15. An inductor that presents a relatively high 22 impedance to alternating current. [5] 17. An electronic instrument for measuring very short time intervals. [11] 20. A particle made from glass is being developed that can absorb pollutants from contaminated water. [7, 8] 27 22. The path that a moving object follows through space as a function of time. [10] 23. An equation that predicts the degree of thermal ionization in a gas. [4, 8] 24. The variable voltage dividers with a shaft or slide control for setting the division ratio. [13] 25. A machine invented by E.O. Lawrence in 1934. [9] 26. The process of boiling or bubbling up of a liquid. [10] 27. The forces which act on a solid object in the direction of the relative fluid flow velocity. [4] 28. The reciprocal of the force constant. [10]
DOWN 1. Father of electrodynamics. [5, 5, 6] 4. A spinning wheel in which the axis of rotation is free to assume any orientation by itself. [9] 5. An electrical device permitting only one way current flow. [5] 6. A mode of computer processing and output in which a large proportion of the output is in pictorial form. [8] 7. A part of the electromagnetic spectrum comprising low energy X-rays. [5, 4]
3
4 7
6 9
8
13
12
14
16
15
19
20
23
24
25 26
28
9. 10. 13. 14. 15. 16. 18. 19. 21. 25.
A highly magnetized, rotating neutron star that emits a beam of electromagnetic radiation. [6] An electronic device in which a single bit of data is stored temporarily. [5] A rule for determining the direction of lines of magnetic force around a wire carrying a current. [9, 4] An instrument for measuring the inclination of a surface to a horizontal plane. [12] The special arrangement of molecules in a liquid crystal. [9] An optical system that produces a beam of parallel light. [10] The reciprocal of capacitance. [9] An instrument for measuring optical transmission or reflection of a material. [12] An instrument that measures the rate of flow of fluids. [9] The streams of gas and dust surrounding the nucleus of a comet. [4] Physics For you | DECEmbEr ‘16
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Physics For you | DECEmbEr ‘16
