Q61JEE Main 2023MCQ

The number of air molecules per ${\text{cm}}^3$ increased from $3 \times 10^{19}$ to $12 \times 10^{19}$ . The ratio of collision frequency of air molecules before and after the increase in the number respectively is: [6-Apr-2023 shift 1]

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📖 Explanation

Collision frequency is given by $Z=n \pi d^2 V_{\;\text{avg }\;}$ , where ${\text{n}}$ is number of molecules per unit volume. $\;\frac{z_1}{z_2}=\;\frac{n_1}{n_2}=\;\frac{3}{12}=\;\frac{1}{4}=0.25$

Q62JEE Main 2023NAT

A steel rod of length $1 {\text{m}}$ and cross sectional area $10^{-4} {\text{m}}^2$ is heated from $0^{\circ} {\text{C}}$ to $200^{\circ} {\text{C}}$ without being allowed to extend or bend. The compressive tension produced in the rod is ________ of steel $=2 \times 10^{11} {\text{Nm}}^{-2}$ , coefficient of linear expansion $=10^{-5} {\text{K}}^{-1}$ ) $\times 10^4 {\text{N}}$ . (Given Young's modulus [8-Apr-2023 shift 2]

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📖 Explanation

Thermal stress $= {\text{Y}} \alpha \Delta {\text{T}}$ $\; {\text{F}}= {\text{YA}} \alpha \Delta {\text{T}}$ $\;=2 \times 10^{11} \times 10^{-4} \times 10^{-5} \times 200$ $\;=4 \times 10^4$ $\; {\text{x}}=4$

Q63JEE Main 2023NAT

Two plates A and B have thermal conductivities $84 {\text{Wm}}^{-1} {\text{K}}^{-1}$ and $126 {\text{Wm}}^{-1} {\text{K}}^{-1}$ respectively. They have same surface area and same thickness. They are placed in contact along their surfaces. If the temperatures of the outer surfaces of ${\text{A}}$ and ${\text{B}}$ are kept at $100^{\circ} {\text{C}}$ and $0^{\circ} {\text{C}}$ respectively, then the temperature of the surface of contact in steady state is ______ ${ }^{\circ} {\text{C}}$ . [13-Apr-2023 shift 2]

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📖 Explanation

Let the temperature of contact surface is ${\text{T}}$ then, $\;\;{ {\text{K}}_A {\text{A}} ( {\text{T}}_{ {\text{A}}}-T) }/{ {\text{L}}}=\;{ {\text{K}}_{ {\text{B}}} {\text{A}} ( {\text{T}}- {\text{T}}_{ {\text{B}}}) }/{ {\text{L}}}$ $\;84(100-T)=126( {\text{T}}-0)$ $\;T=40^{\circ} {\text{C}}$

Q64JEE Main 2023MCQ

Work done by a Carnot engine operating between temperatures $127^{\circ} {\text{C}}$ and $27^{\circ} {\text{C}}$ is $2 {\text{kJ}}$ . The amount of heat transferred to the engine by the reservoir is: [8-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{n}}=1-\;\frac{300}{400}=\;\frac{1}{4}$ $\; {\text{n}}=\;{ {\text{w}}}/{ {\text{Q}}_1}=\;\frac{1}{4} \Rightarrow {\text{Q}}_1=8 {\text{kJ}}$

Q65JEE Main 2023MCQ

Heat is given to an ideal gas in an isothermal process.A. Internal energy of the gas will decrease.B. Internal energy of the gas will increase.C. Internal energy of the gas will not change.D. The gas will do positive work.E. The gas will do negative work.Choose the correct answer from the options given below: [30-Jan-2023 Shift 1]

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📖 Explanation

$\;\text{ Sol. }\;\; {\text{dQ}}= {\text{dU}}+ {\text{dW}} \Rightarrow {\text{dU}}= {\text{nC}}_{ {\text{V}}} {\text{dT}}$ $\; {\text{dU}}=0 \;\; \;\text{ (for isothermal) }\;$ $\therefore \;\; {\text{U}}=\;\text{ constant }\;$ $\;\;\text{ Also }\; {\text{dQ}} > 0 \;\; \;\text{ (supplied) }\;$ $\;\;\text{ Hence }\; {\text{dW}} > 0$

Q66JEE Main 2023MCQ

Given below are two statements. One is labelled as Assertion $A$ and the other is labelled as Reason R. Assertion ${\text{A}}$ : If ${\text{dQ}}$ and ${\text{dW}}$ 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 option given below : [29-Jan-2023 Shift 1]

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📖 Explanation

First law of thermodynamics is based on law of conservation of energy and it can be written as ${\text{dQ}}= {\text{dU}}- {\text{dW}}$ .where ${\text{dW}}$ is work done on the system

Q67JEE Main 2023MCQ

The temperature of an ideal gas is increased from $200 {\text{K}}$ to $800 {\text{K}}$ . If r.m.s. speed of gas at $200 {\text{K}}$ is ${\text{v}}_0$ . Then, r.m.s. speed of the gas at $800 {\text{K}}$ will be: [6-Apr-2023 shift 2]

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📖 Explanation

$\;\;\text{ using }\; v_{ {\text{rms}}}=\sqrt{\;{3 R T}/{ {\text{m}}}}$ $\; {\text{v}}_0=\sqrt{\;{3 R \times 200}/{ {\text{m}}}}$ . . . (1) $\; ( {\text{v}}^{'}) =\sqrt{\;{3 R \times 800}/{ {\text{m}}}}$ . . . (2)dividing (2) by (1) $\;\;{ {\text{v}}^{'}}/{ {\text{v}}_0}=\sqrt{\;\frac{800}{200}}={\sqrt{4}}=2$ $\;\;\text{ or }\; {\text{v}}^{'}=2 {\text{v}}_0$

Q68JEE Main 2023MCQ

A sample of gas at temperature $T$ is adiabatically expanded to double its volume. The work done by the gas in the process is $(.$ given, $\gamma =\;\frac{3}{2})$ : [1-Feb-2023 Shift 1]

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📖 Explanation

$\;T_1 V_1^{\gamma -1}=T_2 V_2^{\gamma -1}$ $\;T V^{1 / 2}=T_2(2 V)^{1 / 2}$ $\;T_2=\;{T}/{{\sqrt{2}}}$ $\;W=\;\frac{R (T_1-T_2) }{\gamma -1}=\;{R (T-{T}/{{\sqrt{2}}}) }/{\frac{1}{2}}=R T(2-{\sqrt{2}})$

Q69JEE Main 2023MCQ

The pressure of a gas changes linearly with volume from A to B as shown in figure. If no heat is supplied to or extracted from the gas then change in the internal energy of the gas will be [31-Jan-2023 Shift 1]

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📖 Explanation

As $\Delta {\text{q}}=0$ $\Delta {\text{u}}\;=- {\text{W}}$ ${\text{W}}\;=∫ {\text{PdV}}$ $\Delta {\text{u}}\;=- {\text{W}}=30 \times 10^3 \times 150 \times 10^{-6}$ $\;=4500 \times 10^{-3}$ $\;=4.5 {\text{J}}$

Q70JEE Main 2023MCQ

Given below are two statements:Statement I : If heat is added to a system, its temperature must increase.Statement II : If positive work is done by a system in a thermodynamic process, its volume must increase. In the light of the above statements, choose the correct answer from the options given below [8-Apr-2023 shift 1]

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📖 Explanation

St I FalseEx. in isothermal process temp. is constant but heat can be added. ST II True ${\text{w}}=∫ {\text{PdV}}$ If volume increases the ${\text{w}}=+$ ve

Q71JEE Main 2023NAT

A hole is drilled in a metal sheet. At $27^{\circ} {\text{C}}$ , the diameter of hole is $5 {\text{cm}}$ . When the sheet is heated to $177^{\circ} {\text{C}}$ , the change in the diameter of hole is ${\text{d}} \times$ $10^{-3} {\text{cm}}$ . The value of ${\text{d}}$ will be __________if coefficient of linear expansion of the metal is $1.6 \times$ $10^{-5} /{ }^{\circ} {\text{C}}$ . [24-Jan-2023 Shift 1]

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📖 Explanation

$\; {\text{d}}_0 \;\text{ at }\; 27^{\circ} {\text{C}} & {\text{d}}_1 \;\text{ at }\; 177^{\circ} {\text{C}}$ $\; {\text{d}}_1= {\text{d}}_0(1+\alpha \Delta {\text{T}})$ $\; {\text{d}}_1- {\text{d}}_0=5 \times 1.6 \times 10^{-5} \times 150 {\text{cm}}$ $\;=12 \times 10^{-3} {\text{cm}}$

Q72JEE Main 2023MCQ

Match List I with List II : List I List II (A) 3 Translational degrees of freedom (I) Monoatomic gases (B) 3 Translational, 2 rotational degrees of freedoms (II) Polyatomic gases (C) 3 Translational, 2 rotational and 1 vibrational degrees of freedom (III) Rigid diatomic gases (D) 3 Translational, 3 rotational and more than one vibrational degrees of freedom (IV) Nonrigid diatomic gasesChoose the correct answer from the options given below : [10-Apr-2023 shift 1]

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📖 Explanation

Fact Based Type of gas No of degree of freedom 1 Monoatomic 3 (Translational) 2. Diatomic + rigid 3 (Translational +2 Rotational $=5)$ 3. Diatomic + non - rigid 3 (Trans) +2 (Rotational) +1 (vibrational) 4. Polyatomic 3 (Trans) +2 (Rotational) + more than 1 (vibrational)

Q73JEE Main 2023MCQ

The temperature at which the kinetic energy of oxygen molecules becomes double than its value at $27^{\circ} {\text{C}}$ is [8-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{KE}} \;\text{ of }\; {\text{O}}_2 \;\text{ molecules }\;=5 \times (\;\frac{1}{2} {\text{KT}})$ $\;( {\text{KE}})_{27^{\circ} {\text{C}}}=5 \times \;\frac{1}{2} {\text{k}}(27+273)=\;\frac{5}{2} {\text{k}} \times 300$ $\;( {\text{KE}})_{ {\text{T}}}=2 (\;\frac{5}{2} {\text{k}}) \times 300=\;\frac{5}{2} {\text{k}}(600)$ $\;\;\text{ i.e. }\; {\text{T}}=600 {\text{K}}$ $\;=600-273$ $\; {\text{T}}=327^{\circ} {\text{C}}$

Q74JEE Main 2023MCQ

If the r. ${\text{m}}$ . s speed of chlorine molecule is $490 {\text{m}} / {\ s}$ at $27^{\circ} {\text{C}}$ , the ${\text{r}} . {\text{m}} . {\ s}$ speed of argon molecules at the same temperature will be (Atomic mass of argon $=39.9 {\text{u}}$ , molecular mass of chlorine $=70.9 {\text{u}}$ ) [12-Apr-2023 shift 1]

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📖 Explanation

$\;\;\text{ Molar mass of chlorine }\;=70.9 {\text{u}}$ $\;\;\text{ Atomic mass of argon }\;=39.9 {\text{u}}$ $\; {\text{V}}_{\;\text{rms }\;}=\sqrt{\;{3 {\text{RT}}}/{ {\text{m}}}}$ $\; {\text{V}}_{\;\text{rms }\;} \propto \sqrt{\;{1}/{ {\text{m}}}}$ $\;\;{ {\text{V}}_{\;\text{rmschlorine }\;}}/{ {\text{V}}_{\;\text{rmsarg on }\;}}=\sqrt{\;{ {\text{m}}_{\;\text{arg on }\;}}/{ {\text{m}}_{\;\text{chlorine }\;}}}$ $\;\;{490}/{ {\text{V}}_{\;\text{rmsargon }\;}}=\sqrt{\;\frac{39.9}{70.9}}$ $\; {\text{V}}_{\;\text{rmsargon }\;}=490 \times 1.33$ $\; {\text{V}}_{\;\text{msargon }\;}=651.7 {\text{m}} / {\ sec}$

Q75JEE Main 2023MCQ

In an Isothermal change, the change in pressure and volume of a gas can be represented for three different temperature; $T_3>T_2>T_1$ as : [24-Jan-2023 Shift 2]

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📖 Explanation

For isothermal process ${\text{P}}- {\text{V}}$ graph is rectangular hyperbolaAs dotted line is isobaric line which implies $T_3 > T_2 > T_1$ as volume is increasing.

Q76JEE Main 2023MCQ

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : Efficiency of a reversible heat engine will be highest at $-273^{\circ} {\text{C}}$ temperature of cold reservoir.Reason R : The efficiency of Carnot's engine depends not only on temperature of cold reservoir but it depends on the temperature of hot reservoir too and is given as $η= (1-\;{ {\text{T}}_2}/{ {\text{T}}_1})$ .In the light of the above statements, choose the correct answer from the options given below : [30-Jan-2023 Shift 2]

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📖 Explanation

Both ${\text{A}}$ and ${\text{R}}$ are true and ${\text{R}}$ is the correct explanation of A

Q77JEE Main 2023MCQ

A gas is compressed adiabatically, which one of the following statement is NOT true. [10-Apr-2023 shift 2]

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📖 Explanation

In Adiabatic process, $\Delta {\text{Q}}=0$ If gas is compressed, then w (by gas) $\ne 0$ Therefore by $1^{\;\text{st }\;}$ law $\;\Delta {\text{Q}}=\Delta {\text{u}}+ {\text{w}}$ $\;0=\Delta {\text{u}}+ {\text{w}}$ $\;\Delta {\text{u}}=- {\text{w}} \ne 0$ It implies in adiabatic compression, internal energy of gas changes.

Q78JEE Main 2023MCQ

Let $\gamma _1$ be the ratio of molar specific heat at constant pressure and molar specific heat at constant volume of a monoatomic gas and $\gamma _2$ be the similar ratio of diatomic gas. Considering the diatomic gas molecule as a rigid rotator, the ratio, $\;\frac{\gamma _1}{\gamma _2}$ is [24-Jan-2023 Shift 2]

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📖 Explanation

For monoatomic gas $\gamma _1=\;\frac{5}{3}$ For diatomic gas at low temperatures $\gamma _2=\;\frac{7}{5}$ $\therefore \;\frac{\gamma _1}{\gamma _2}=\;\frac{\;\frac{5}{3}}{\frac{7}{5}}=\;\frac{25}{21}$

Q79JEE Main 2023MCQ

A source supplies heat to a system at the rate of $1000 {\text{W}}$ . If the system performs work at a rate of $200 {\text{W}}$ . The rate at which internal energy of the system increase is [6-Apr-2023 shift 1]

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📖 Explanation

From Ist law of thermodynamics, $d Q=d U+d W$ Also, we can write this as, $\;\frac{d Q}{d t}=\;\frac{d U}{d t}+\;\frac{d W}{d t}$ $\;\Rightarrow 1000 W=\;\frac{d U}{d t}+200 {\text{W}}$ $\;\Rightarrow \;\frac{d U}{d t}=800 W$

Q80JEE Main 2023MCQ

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 :- [25-Jan-2023 Shift 2]

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📖 Explanation

Diatomic gas molecules have three translational degree of freedom, two rotational degree of freedom & it is given that it has one vibrational mode so there are two additional degree of freedom corresponding to one vibrational mode, so total degree of freedom $=7$ ${\text{C}}_{ {\text{v}}}=\;{ {\text{fR}}}/{2}=\;{7 {\text{R}}}/{2}$