JEE Main 2017 (Offline) | Heat and Thermodynamics Question 323 | Physics

An external pressure P is applied on a cube at 0^oC so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and $$\alpha$$ is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by:

$${P \over {3\alpha K}}$$

$${P \over {\alpha K}}$$

$${3 \alpha \over {P K}}$$

3PK$$\alpha$$

JEE Main 2017 (Offline)

MCQ (Single Correct Answer)

-1

The temperature of an open room of volume 30 m³ increases from 17^oC to 27^oC due to the sunshine. The atmospheric pressure in the room remains 1 $$ \times $$ 10⁵ Pa. If N_i and N_f are the number of molecules in the room before and after heating, then N_f – N_i will be :

- 1.61 $$ \times $$ 10²³

1.38 $$ \times $$ 10²³

2.5 $$ \times $$ 10²⁵

- 2.5 $$ \times $$ 10²⁵

JEE Main 2017 (Offline)

MCQ (Single Correct Answer)

-1

C_P and C_v are specific heats at constant pressure and constant volume respectively. It is observed that
C_P – C_v = a for hydrogen gas
C_P – C_v = b for nitrogen gas
The correct relation between a and b is

a = 28 b

a = 1/14 b

a = b

a = 14 b

JEE Main 2016 (Online) 10th April Morning Slot

MCQ (Single Correct Answer)

-1

Out of Syllabus

A Carnot freezer takes heat from water at 0^oC inside it and rejects it to the room at a temperature of 27^oC. The latent heat of ice is 336×10³ J kg⁻¹. If 5 kg of water at 0^oC is converted into ice at 0^oC by the freezer, then the energy consumed by the freezer is close to :

1.67 $$ \times $$ 10⁵ J

1.68 $$ \times $$ 10⁶ J

1.51 $$ \times $$ 10⁵ J

1.71 $$ \times $$ 10⁷ J

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