Thermal Properties Of Matter

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Question
CBSEENPH11020257

‘n’ moles of an ideal gas undergoes a process A→B as shown in the figure. The maximum temperature of the gas during the process will be:
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  • fraction numerator 9 straight p subscript straight o straight V subscript straight o over denominator 4 space nR end fraction
  • fraction numerator 3 straight p subscript straight o straight V subscript straight o over denominator 2 nR end fraction
  • 9 over 2 fraction numerator straight p subscript straight o straight V subscript straight o over denominator nR end fraction
  • fraction numerator 9 straight p subscript straight o straight V subscript straight o over denominator nR end fraction

Solution

A.

fraction numerator 9 straight p subscript straight o straight V subscript straight o over denominator 4 space nR end fraction

As, T will be maximum temperature where product of pV is maximum
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Equation of line AB, we have
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Question
CBSEENPH11020317

100g of water is heated from 30°C to 50°C ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is 4184 J/Kg/K)

  • 8.4 kJ

  • 84 kJ

  • 2.1 kJ

  • 4.2 kJ

Solution

A.

8.4 kJ

ΔQ = M,S,ΔT
= 100 × 10-3 × 4.184 × 20 = 8.4 × 103
ΔQ = 84 kJ, ΔW = 0
ΔQ = ΔV + ΔW
ΔV = 8.4 kJ

Question
CBSEENPH11020344

A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75°C. T is given by:(Given : room temperature = 30° C, specific heat of copper = 0.1 cal/gm°C

  • 1250°C

  • 825°C

  • 800°C

  • 885° C

Solution

D.

885° C

Heat given = Heat taken
(100) (0.1)(T – 75) = (100)(0.1)(45) + (170)(1)(45)
10(T – 75) = 450 + 7650 = 8100
T – 75 = 810
T = 885 °C

Question
CBSEENPH11020456

A heater coil is cut into two equal parts and only one part is now used in the heater. The heat generated will now be

  • doubled

  • four times

  • one fourth

  • halved

Solution

A.

doubled

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Question
CBSEENPH11020496

A light ray is incident perpendicular to one face of a 90° prism and is totally internally reflected at the glass-air interface. If the angle of reflection is 45°, we conclude that the refractive index n

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  • n<2

  • straight n greater than square root of 2
  • straight n space greater than space fraction numerator 1 over denominator square root of 2 end fraction
  • n<1/2

Solution

B.

straight n greater than square root of 2

Angle of incidence i > C for total internal reflection.
Here i = 45° inside the medium. ∴ 45° > sin−1 (1/n)
⇒ n > √2.