## THERMAL PROPERTIES OF MATTER QUIZ-9

JEE Advanced Physics Syllabus can be referred by the IIT aspirants to get a detailed list of all topics that are important in cracking the entrance examination. JEE Advanced syllabus for Physics has been designed in such a way that it offers very practical and application-based learning to further make it easier for students to understand every concept or topic by correlating it with day-to-day experiences. In comparison to the other two subjects, the syllabus of JEE Advanced for physics is developed in such a way so as to test the deep understanding and application of concepts.

Q81. One end of a copper rod of uniform cross section and of length 1.5 m is kept in contact with ice and the other end with water at 100°C. At what point along its length should a temperature of 200°C be maintained so that in steady state, the mass of ice melting be equal to that of the steam produced in same interval of time? Assume that the whole system is insulated from surroundings. Latent heat of fusion of ice and vapourization of water are 80 cal/g and 540 cal/g, respectively
•  8.59 cm from ice end
•  10.34 cm from water end
•  10.34 cm from ice end
•  8.76 cm from water end
Solution

82. A bullet of mass 5 g moving at a speed of 200 m/s strikes a rigidly fixed wooden plank of thickness 0.2 m normally and passes through it losing half of its kinetic energy. If it again strikes an identical rigidly fixed wooden plank and passes through it, assuming the same resistance in the two planks, the ratio of the thermal energies produced in the two planks is
•  1:1
•  1:2
•  2:1
•  4:1
Solution

Q83. An iron ball (coefficient of linear expansion =1.2×10(-5)/°C) has a diameter of 6 cm and is 0.010 mm too large to pass through a hole in a brass plate (coefficient of linear expansion =1.9×10^(-5)/°C) when the ball and the plate are both at a temperature of 30°C. At what common temperature of the ball and the plate will the ball just pass through the hole in the plate?
•   23.8°C
•  53.8°C
•  42.5°C
•  63.5°C
Solution

Q84. A black body emits radiation at the rate P when its temperature is T. At this temperature the wavelength at which the radiation has maximum intensity is Î»_0. If at another temperature T' the power radiated is P' and wavelength at maximum intensity is Î»_0/2 then
•  P' T'=32 PT
•  P' T'=16 PT
•  P' T'=8 PT
•  P' T'=4 PT
Solution

Q85. The wavelength of maximum energy released during an atomic explosion was 2.93×10(-10)m. Given that Wien’s constant is 2.93×10^(-3)m-K, the maximum temperature attained must be of the order of
•  10(-7) K
•  107 K
•  10(-13)K
•  (5.86×10)7 K
Solution

Q86. A sphere, a cube and a thin circular plate, all made of the same material and having the same mass are initially heated to a temperature of 1000℃. Which one of these will cool first
•  Plate
•  Sphere
• Cube
•  None of these
Solution

Q87. A thread of liquid is in a uniform capillary tube of length L, as measured by a ruler. The temperature of the tube and thread of liquid is raised by ∆T. If Î³ be the coefficient of volume expansion of the liquid and Î± be the coefficient of linear expansion of the material of the tube, then the increase ∆L in the length of the thread, again measured by the ruler will be
•  ∆L=L(Î³-Î±)∆T
•  ∆L=L(Î³-2Î±)∆T
•  ∆L=L(Î³-3Î±)∆T
•  ∆L=LÎ³ ∆T
Solution

Q88. The only possibility of heat flow in a thermos flask is through its cork which is 75 cm^2 in area and 5 cm thick. Its thermal conductivity is 0.0075 cal/cm-s-°C. The outside temperature is 40°C and latent heat of ice is 80 cal/g. Time taken by 500 g of ice at 0°C in the flask to melt into water at 0°C is
•  2.47 h
•  4.27 h
•  7.42 h
•  4.72 h
Solution

Q89. A heat flux of 4000 J/s is to be passed through a copper rod of length 10 cm and area of cross section 100 cm2. The thermal conductivity of copper is 400 W/m/°C. The two ends of this rod must be kept at a temperature difference of
•  1°C
•  10°C
•  100°C
•  >1000°C
Solution

Q90. A and B are two isolated spheres kept in close proximity so that they can exchange energy by radiation. The two spheres have identical physical dimensions but the surface of A behaves like a perfectly black body while the surface of B reflects 20% of all the radiations it receives. They are isolated from all other sources of radiation
•  If they are in thermal equilibrium and exchange equal amounts of radiation per second, then they will be at same absolute temperature, T_A=T_B
•  If they are in thermal equilibrium and exchange equal amounts of radiation per second, then T_A=(0.8)(1/4) T_B
•  If they are not in thermal equilibrium and are each at t=0 at the same temperature T_A=T_B=T, then the sphere A will lose thermal energy and B will gain thermal energy
• If they are not in thermal equilibrium and are each at t=0 at the same temperature T_A=T_B=T, then the sphere A will gain thermal energy and B will lose thermal energy
Solution
90 (a) Since they are in thermal equilibrium, the temperatures must necessarily be equal. Temperature equality is a necessary and sufficient condition for thermal equilibrium (zeroth law of thermodynamics)

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