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NISHITH Multimedia India (Pvt.) Ltd., 5 9 JEE MAINS - CW - VOL - I JEE ADVANCED - VOL - IV THERMAL PHYSICS NISHITH Multimedia India (Pvt.) Ltd., LEVEL-V SINGLE ANSWER QUESTIONS 1. Three identical rods of equal length L are joined to form an equilateral triangle ABC as shown in figure. D is the midpoint of AB. The coefficient of linear expansion is 1 for AB, and 2 for AC and BC. If 1 2    4 , the change in time period of the systam is C A D 1 B 2 3 (A) 1 2 ( )L t    (B) 2 1 2 L t 2    (C) 1 2 ( 2 )L t 2     (D) Zero 2 A heated peanut (at time t  0 and temperature T T  0  is taken out of the oven to cool and placed on a table near an open window. Write an expression for its temperature as a function of time. (A)  0  kt s e T T T    (B)  0  kt s s e T T T T     (C) 0 0   kt s e T T T T     (D) 0 kt e T T   3. In a vertical U-tube containing a liquid, the two arms are maintained at different temperatures, t1 and t2 . The liquid columns in the two arms have heights l1 and l2 respectively. The coefficient of volume expansion of the liquid is equal to L2 t1 t 12 L1 (A) 1 2 2 1 1 2 l l l t l t   (B) 1 2 1 2 2 2 l l l t l t   (C) 1 2 2 1 1 2 l l l t l t   (D) 1 2 1 1 2 2 l l l t l t   4. A calorimeter contains 50 g of water at 0 50 C , The temperature falls to 0 45 C in 10 minutes. When the calorimeter contains 100 g of water at 0 50 C , it takes 18 min for the temperature to fall to 0 45 C . The ‘water equivalent’ of the calorimeter is (A) 2.5gm (B) 5gm (C) 12.5gm (D) 8.5gm 5. A compensated pendulum as shown in fig is in the form of an iscolated triangle of base length 5cmand -6 0 1 =25 10 ( ) C   .The side length is l and its  is -6 0 1 =16 10 ( ) C   .If the pendulum records correct time at all temperatures, then l  A) 25 8 cm B) 14 3 cm C) 20 7 cm D) 12 5 cm 6. A cube of coefficient of linear expension s is floating in a bath containing a liquid of coefficient of volume expertion l  . When the temperature is raised by T , the depth upto which the cube is submerged in the liquid remains the same. Then the relation between c and l  is A) l 3 s    B) l 3 / 2 s    C) l 2 s    D) l / 2 s   7. A heavy brass bar has projections at its ends as shown in the figure. Two fine steel wires, fastened between the projections, are just taut (zero tension) when the whole system is at 0 C  . What is the tensile stress in the steel wires when the temperature of the sytem is raised to 300 C  ? Given that 6 0 1 20 10 brass α C     6 0 1 12 10 steel α C     11 2 2 10 Y N m steel    (A) 7 2 48 10 Nm  (B) 7 2 84 10 Nm  (C) 4 2 32 10 Nm  (D) 4 2 24 10 Nm  THERMAL EXPANSION
THERMAL EXPANSION 6 0 NISHITH Multimedia India (Pvt.) Ltd., JEE ADVANCED - VOL - IV NISHITH Multimedia India (Pvt.) Ltd., 8. The variation of lengths of two metal rods Aand B with change in temperature are shown in Fig. The coefficients of linear expansion αA for the metal A will be nearly: (Given 6 0 9 10 / α C B    ) (A) 6 0 13 10 / C   (B) 6 0 27 10 / C   (C) 6 0 18 10 / C   (D) 6 0 43 10 / C   9. Fig. shows the graphs of elongation versus temperature for two different metals. If these metals be employed to form a straight bimetallic strip of thickness 6 cm and heated it bends in the form of an arc, the radius of curvature changing with temperature approximately as shown in Fig. The linear expansivities of the two metals are: a) 6 0 24 10 / C   and 6 0 12 10 / C   b) 6 0 20 10 / C   and 6 0 10 10 / C   c) 6 0 18 10 / C   and 6 0 9 10 / C   d) 6 0 16 10 / C   and 6 0 8 10 / C   MULTIPLE ANSWER QUE STIONS 10. A metallic circular disc having a circular hole at its centre rotates about an axis passing through its centre and perpendicular to its plane. When the disc is heated: (A) its speed will decrease (B) its diameter will decrease (C) its moment of inertia will increase (D) its speed will increase 11. A bimetallic strip is formed out of two identical strips one of copper and the other of brass. The co-efficients of linear expansion of the two metals are  C and  B . On heating, the temperature of the strip goes up by T and the strip bends to form an arc of radius of curvature R. Then R is [IIT - 1999] (A) proportional to T (B) inversely proportional to T (C) proportional to B C | |   (D) Inversely proportional to B C | |   (a) internal energies at A and B are the same (b) word done by the gas is process AB is 0 0 PV n  4 (c) pressure at C is 0 4 P (d) temperature at C is 0 4 T MATRIX MATCHING TYPE QUESTIONS 12. Whenever a liquid is heated in a container, expansion in liquid as well as container takes place. If  is the volume expansion coefficient of liquid and  is coefficient of linear expansion of the container match the entries of Column I and Column II Column I Column II (i) Liquid level rises with (A)    2 respect to container (ii) Liquid level remains (B) 2 3      same with respect to container (iii) Liquid level drops with (C)    3 respect to container (iv) Liquid level remains (D)    3 same with respect to ground
NISHITH Multimedia India (Pvt.) Ltd., 6 1 JEE MAINS - CW - VOL - I JEE ADVANCED - VOL - IV THERMAL PHYSICS NISHITH Multimedia India (Pvt.) Ltd., 13. Column I gives some devices and Column II gives some processes on which the functioning of these devices depend. Match the devices in Column I with the processes in Column II. (IIT 2007) Column I A. Bimetallic strip B. Steam engine C. Incandescent lamp D. Electric fuse Column II p. Radiation from a hot body q. Energy conversion r. Melting s. Thermal expansion of solids INTEGER TYPE QUESTIONS 14. A cube of coefficient of linear expansion s is floating in a bath containing a liquid of coefficient of volume expansion l  . When the temperature is raised by T,the depth up to which the cube is submerged in the liquid remains the same. The relation between s and 1  is s x l    . Find the value of x. 15. A steel rod of length 5 m is fixed between two support. The coefficient of linear expansion of steel is 12.5 × 10–6/°C. Calculate the stress (in 108 N/m2) in the rod for an increase in temperature of 40°C. Young’s modulus for steel is 11 2 2 10 Nm  16. A cylindrical wire of length l with radius r is suspended vertically from a rigid support and carries a bob of mass M at the other and. If Y is young’s modulus of elasticity of the wire, S is specific heat and  is density of the wire and the bob gets snapped find change in temperature of wire 17. A cubical block of co-efficient of lienar expansion s is submerged partially inside a liquid of co-efficient of volume expansion 1  . On increasing the temperature of the system T , the height of the cube inside the liquid remains unchanged. Find the relation between s and 1  . [IIT-2004] Passage for Q no 18,19 Two rods of different metals having the same area of cross section A are placed between two rigid walls. For the first rod 1 1 1 l Y , ,  , and for the sec- ond rod 2 2 2 l Y , ,  are the physical quantitities with usual meanings. Now the temp of the sys- tem is increased by 0 t C . Given   6 1 1 1 [ 20 , 18 10 0 l cm C       ; 11 2 y n l cm 1 2 10 10 , tan 18       10 11 1 2 2 20 10 01 9 10 ] l  y Nm     18. The with which the rods act on each other is 9F KN then F = 19. The length of the smallar rod at 0 100 C is 3.6 L cm then L = LEVEL - V KEY SINGLE ANSWER QUESTIONS 1) D 2)B 3)A 4)C 5)A 6)C 7) A 8) B 9) D MULTIPE ANSWER QUESTIONS 10) A,C 11) B. D MATRIX MATCHING TYPE 12) (i)(D), (ii)(C), (iii)(B), (iv)  (A) 13) (As, Bq, Cp, q, Dq, r) INTEGER TYPE QUESTIONS 14) 2 15) 1 16) 2 2 2 4 M g 2 r Ys   17)    2  18) 4 19) 6
THERMAL EXPANSION 6 2 NISHITH Multimedia India (Pvt.) Ltd., JEE ADVANCED - VOL - IV NISHITH Multimedia India (Pvt.) Ltd., LEVEL - V - HINTS SINGLE ANSWER QUESTIONS 1. 2 2 2 2 2 2 Dc (1 t) (1 t) 2 1 2                     l l l l 2 neglecting 2 2   2 1 & & solving above we get 1 2   4 2. If T is the instantaneous temerature and Ts be the temperature of the surroundings, then form netwon’s law of cooling,  s  dT K T T dt    On separating the variables and integrating the expression, we have 0   0 T t T s dT K dt T T      ln ln T T T T Kt      s s   0  0 ln s s T T Kt T T     0 s Kt s T T e T T      0  Kt T T T T e s s     3. Let 0 , 1 and 2 be the densities of the liquid at temperatures 0, t­1 and t2 respectively To balance pressure, 1 l 1 g = 2 l 2 g or 0 0 1 2 1 2 l l 1 t 1 t                     1 1 2 2 2 1 l  l t  l  l t             2 1 1 2 1 2 l t l t l l 4. Assume the surrounding temperature to be 0 T C 0 , and that the water equivalent of the calorimeter is ‘w’ g. In the first case, there is 100  w g of water. We have:       10 50 45 50 0 0 k w s T T e      and       18 50 45 50 0 0 k w s T T e           10 18 50 100 w w     100 90 1.8    w w or 10 0.8  w  w g 12.5 5. Distance of the cetre of mass should remain the same i.e., the length and hight = corst - 5 25 2 16 l  = 5 25 2 8  6. 2 mg l d g  1 1 2 mg l d g  2 2    2 1 2 1 1 ; 1 s l l l t t             2 2 2 1 1 1 1 2 1 1 s dg l d g l t r t           2 1 1 l s        t t 2 l s    7. T S y .    α α T b s  Δ 8. Slope of line   0 1006 1000 Δ Δ A mm L A Lα T C T     i.e,   6 0 / 1000 mm C mm αA T  ....(i) Similarly, for line B   2 0 / 1002 mm C mm αB T  ....(ii) Dividing E.q. (i) by Eq. (ii) 1000 3 3 1002 B A A B α α α α    From Eq. (iii), 6 αA 3 9 10    6 0 27 10 / C    9. The slope of lines are: 1 Δ 4 0 0 Δ 4 0 L Lα T      [For the first metal] and 1 Δ 2 0 1 Δ 4 0 2 L Lα T     