JEE Main 2023 (Online) 29th January Morning Shift | Heat and Thermodynamics Question 87 | Physics

JEE Main 2023 (Online) 29th January Morning Shift

MCQ (Single Correct Answer)

-1

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: If $$d Q$$ and $$d W$$ represent the heat supplied to the system and the work done on the system respectively. Then according to the first law of thermodynamics $$d Q=d U-d W$$.

Reason R: First law of thermodynamics is based on law of conservation of energy.

In the light of the above statements, choose the correct answer from the options given below:

Both A and R are correct but R is not the correct explanation of A

Both A and R are correct and R is the correct explanation of A

A is correct but R is not correct

A is not correct but R is correct

JEE Main 2023 (Online) 25th January Evening Shift

MCQ (Single Correct Answer)

-1

Match List I with List II

	List I		List II
A.	Isothermal Process	I.	Work done by the gas decreases internal energy
B.	Adiabatic Process	II.	No change in internal energy
C.	Isochoric Process	III.	The heat absorbed goes partly to increase internal energy and partly to do work
D.	Isobaric Process	IV.	No work is done on or by the gas

Choose the correct answer from the options given below :

A-I, B-II, C-IV, D-III

A-II, B-I, C-III, D-IV

A-II, B-I, C-IV, D-III

A-I, B-II, C-III, D-IV

JEE Main 2023 (Online) 25th January Evening Shift

MCQ (Single Correct Answer)

-1

The graph between two temperature scales P and Q is shown in the figure. Between upper fixed point and lower fixed point there are 150 equal divisions of scale P and 100 divisions on scale Q. The relationship for conversion between the two scales is given by :-

JEE Main 2023 (Online) 25th January Evening Shift Physics - Heat and Thermodynamics Question 78 English

$${{{t_P}} \over {100}} = {{{t_Q} - 180} \over {150}}$$

$${{{t_P}} \over {180}} - {{{t_Q} - 40} \over {100}}$$

$${{{t_Q}} \over {150}} = {{{t_P} - 180} \over {100}}$$

$${{{t_Q}} \over {100}} = {{{t_P} - 30} \over {150}}$$

JEE Main 2023 (Online) 25th January Evening Shift

MCQ (Single Correct Answer)

-1

According to law of equipartition of energy the molar specific heat of a diatomic gas at constant volume where the molecule has one additional vibrational mode is :-

$$\frac{9}{2}R$$

$$\frac{5}{2}R$$

$$\frac{3}{2}R$$

$$\frac{7}{2}R$$

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