Q101JEE Main 2022NAT

At a certain temperature, the degrees of freedom per molecule for gas is 8 . The gas performs $150 {\text{J}}$ of work when it expands under constant pressure. The amount of heat absorbed by the gas will be ________ J. [28-Jul-2022-Shift-2]

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📖 Explanation

${\text{W}}= {\text{nR}} \Delta {\text{T}}=150 {\text{J}}$ $Q=(\;{ {\text{f}}}/{2}+1) {\text{nR}} \Delta {\text{T}}=(\;\frac{8}{2}+1) 150=750 {\text{J}}$

Q102JEE Main 2022MCQ

A $100 {\text{g}}$ of iron nail is hit by a $1.5 {\text{kg}}$ hammer striking at a velocity of $60 {\text{ms}}^{-1}$ . What will be the rise in the temperature of the nail if one fourth of energy of the hammer goes into heating the nail?$[Specific heat capacity of iron $=0.42 {\text{Jg}}^{-1}{ }^{\circ} {\text{C}}^{-1}$ ]$ [24-Jun-2022-Shift-2]

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📖 Explanation

$\;\frac{1}{2} \times 1.5 \times 60^2 \times \;\frac{1}{4}=100 \times 0.42 \times \Delta T$ $\Delta T=\;\frac{1.5 \times 60^2}{8 \times 100 \times 0.42}=16.07^{\circ} {\text{C}}$

Q103JEE Main 2022MCQ

A Carnot engine takes $5000 {\text{kcal}}$ of heat from a reservoir at $727^{\circ} {\text{C}}$ and gives heat to a sink at $127^{\circ} {\text{C}}$ . The work done by the engine is [24-Jun-2022-Shift-2]

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📖 Explanation

Efficiency $η=1-\;\frac{T_L}{T_H}$ $=1-\;\frac{400}{1000}=0.6$ $\Rightarrow 0.6=\;\frac{W}{Q}$ $\Rightarrow W=0.6 Q=3000 {\text{kcal}}=12.6 \times 10^6 {\text{J}}$

Q104JEE Main 2022MCQ

A sample of an ideal gas is taken through the cyclic process ${\text{ABCA}}$ as shown in figure. It absorbs, $40 {\text{J}}$ of heat during the part $A B$ , no heat during $B C$ and rejects $60 {\text{J}}$ of heat during $C A$ . A work of $50 {\text{J}}$ is done on the gas during the part $B C$ . The internal energy of the gas at $A$ is $1560 {\text{J}}$ . The workdone by the gas during the part ${\text{CA}}$ is : [28-Jun-2022-Shift-2]

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📖 Explanation

$\Delta {\text{U}}_{ {\text{AB}}}=40 {\text{J}}$ as process is isochoric. $\Delta {\text{U}}_{ {\text{BC}}}=+50 ( {\text{W}}_{ {\text{BC}}}=-50 {\text{J}})$ ${\text{U}}_{ {\text{C}}}= {\text{U}}_{ {\text{A}}}+\Delta {\text{U}}_{ {\text{AB}}}+\Delta {\text{U}}_{ {\text{BC}}}=1650$ For CA process, $Q_{C A}=-60 {\text{J}}$ $\Delta U_{C A}+W_{C A}=-60$ $-90+W_{C A}=-60$ $\Rightarrow W_{C A}=+30 {\text{J}}$ The graph given is inconsistent with the statement ${\text{BC}}$ may be adiabatic and ${\text{CA}}$ cannot be like isobaric as shown, as increasing volume while rejecting heat at same time.

Q105JEE Main 2022MCQ

A gas has $n$ degrees of freedom. The ratio of specific heat of gas at constant volume to the specific heat of gas at constant pressure will be : [26-Jul-2022-Shift-2]

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📖 Explanation

${\text{C}}_{ {\text{v}}}=\;{ {\text{nR}}}/{2} \;\; {\text{C}}_{ {\text{p}}}=\;{( {\text{n}}+2) {\text{R}}}/{2}$ $\;{ {\text{C}}_{ {\text{v}}}}/{ {\text{C}}_{ {\text{p}}}}=\;{ {\text{n}}}/{ {\text{n}}+2}$

Q106JEE Main 2022MCQ

The root mean square speed of smoke particles of mass $5 \times 10^{-17} {\text{kg}}$ in their Brownian motion in air at NTP is approximately. $[Given ${\text{k}}=1.38 × 10^{-23} {\text{JK}}^{-1}$ ]$ [29-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

At NTP, $T=298 {\text{K}}$ $v_{ {\text{rms}}} \;\; =\sqrt{\;\frac{3 R T}{M}}$ $\;\; =\sqrt{\;\frac{3 k N_A \times 298}{5 \times 10^{-17} \times N_A}}$ $≃ 15 {\text{mm}} / {\ s}$

Q107JEE Main 2022NAT

A steam engine intakes $50 {\text{g}}$ of steam at $100^{\circ} {\text{C}}$ per minute and cools it down to $20^{\circ} {\text{C}}$ . If latent heat of vaporization of steam is $540 {\text{cal}} {\text{g}}^{-1}$ , then the heat rejected by the steam engine per minute is _____ $\times$ $10^3 {\text{cal}}$ (Given : specific heat capacity of water : $1 {\text{cal}} {\text{g}}^{-1}{ }^{\circ} {\text{C}}^{-1}$ ) [25-Jun-2022-Shift-1]

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📖 Explanation

$\Delta {\text{Q}}_{ {\text{rej}}}=50 \times 540+50 \times 1 \times (100-20)$ $=50 \times [540+80]$ $=50 \times 620$ $=31000 {\text{cal}}$ $=31 \times 10^3 {\text{cal}}$

Q108JEE Main 2022MCQ

A monoatomic gas at pressure ${\text{P}}$ and volume ${\text{V}}$ is suddenly compressed to one eighth of its original volume. The final pressure at constant entropy will be : [26-Jul-2022-Shift-1]

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📖 Explanation

$P V^\gamma =$ constant $\Rightarrow P V^\gamma =(P^{'})(\;\frac{v}{8})^\gamma \;\text{ where }\; \gamma =5 / 3$ $\Rightarrow P^{'}=32 P$

Q109JEE Main 2022MCQ

Let $η_{1}$ is the efficiency of an engine at $T_{1}=447^{\circ} {C}$ and ${T}_{2}=147^{\circ} {C}$ while $η_{2}$ is the efficiency at ${T}_{1}=947^{\circ} {C}$ and ${T}_{2}=47^{\circ} {C}$ The ratio $\;\frac{η_{1}}{η_{2}}$ will be: [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$η_{1}=1-\;\frac{420}{720}=\;\frac{300}{720}$ And $η_{2}=1-\;\frac{320}{1220}=\;\frac{900}{1220}$ $\Rightarrow \;\frac{η_{1}}{η_{2}}=\;\frac{300}{720} \times \;\frac{1220}{900}$ $≃ 0.56$

Q110JEE Main 2022MCQ

An ice cube of dimensions $60 {\text{cm}} \times 50 {\text{cm}} \times 20 {\text{cm}}$ is placed in an insulation box of wall thickness $1 {\text{cm}}$ . The box keeping the ice cube at $0^{\circ} {\text{C}}$ of temperature is brought to a room of temperature $40^{\circ} {\text{C}}$ The rate of melting of ice is approximately : (Latent heat of fusion of ice is $3.4 \times 10^5 {\text{J}} {\text{kg}}^{-1}$ and thermal conducting of insulation wall is $.0.05 {\text{Wm}}^{-1 \circ } {\text{C}}^{-1})$ [26-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\;{ {\text{dQ}}}/{ {\text{dt}}}=\;{ {\text{KA}} \Delta {\text{T}}}/{ℓ}$ $A=2(0.6 \times 0.5+0.5 \times 0.2+0.2 \times 0.6)$ $=2(0.3+0.1+0.12)$ $=2(0.4+0.12)$ $=2(0.52)$ $=1.04 {\text{m}}^2$ ${\text{R}}_{ {\text{th}}}=\;{ℓ}/{ {\text{KA}}} \Rightarrow \;\frac{1 \times 10^{-2}}{0.05 \times 1.04} \Rightarrow \;\frac{10^{-2}}{0.052}$ $\;{ {\text{dQ}}}/{ {\text{dt}}}=\;{\Delta {\text{T}}}/{ {\text{R}}_{ {\text{th}}}} \Rightarrow \;\frac{40 \times 0.052}{10^{-2}} \Rightarrow 2.08 \times 10^2 {\text{J}} / {\text{s}}$ $2.08 \times 10^2= {\text{m}} \times 3.4 \times 10^5$ ${\text{m}}=\;\frac{2.08}{3.4 \times 10^3} \Rightarrow 0.61 \times 10^{-3} {\text{kg}} / {\text{s}}$ $=61 \times 10^{-5} {\text{Kg}} / {\text{s}}$

Q111JEE Main 2022MCQ

Following statements are given :(A) The average kinetic energy of a gas molecule decreases when the temperature is reduced.(B) The average kinetic energy of a gas molecule increases with increase in pressure at constant temperature.(C) The average kinetic energy of a gas molecule decreases with increase in volume.(D) Pressure of a gas increases with increase in temperature at constant pressure.(E) The volume of gas decreases with increase in temperature.Choose the correct answer from the options given below : [25-Jul-2022-Shift-1]

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📖 Explanation

Because ${KE} \propto {T}$ so ${A}$ is correct, ${B}$ is incorrect, statement ${C}$ cannot be said, statement ${D}$ is contradicting itself, statement ${E}$ is incorrect (Isothermal process)So no answer correct (Bonus)If the statement of $D$ would have been."Pressure of gas increases with increase in temperature at constant volume, "then statement ${D}$ would have been correct, so in that case answer would have been 'A'.

Q112JEE Main 2022MCQ

Starting with the same initial conditions, an ideal gas expands from volume $V_1$ to $V_2$ in three different ways. The work done by the gas is ${\text{W}}_1$ if the process is purely isothermal, ${\text{W}}_2$ , if the process is purely adiabatic and ${\text{W}}_3$ if the process is purely isobaric. Then, choose the correct option [29-Jun-2022-Shift-2]

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📖 Explanation

Comparing the area under the PV graph $A_3>A_1>A_2$ $\Rightarrow W_3>W_1>W_2$

Q113JEE Main 2022MCQ

A sample of monoatomic gas is taken at initial pressure of $75 {\text{kPa}}$ . The volume of the gas is then compressed from 1200 ${\text{cm}}^3$ to $150 {\text{cm}}^3$ adiabatically. In this process, the value of workdone on the gas will be : [30-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

For monoatomic gas degree of freedom ${\text{f}}=3$ and $\gamma =\;\frac{5}{3}$ Here for gas,Initial pressure $( {\text{P}}_1) =75 {\text{kPa}}$ Initial volume $( {\text{V}}_1) =1200 {\text{cm}}^3$ Final volume $(V_2) =150 {\text{cm}}^3$ Final pressure $(P_2) =$ ?For adiabatic process, $P_1 V_1^\gamma =P_2 V_2^\gamma$ $\Rightarrow 75 \times (1200)^\gamma =P_2(150)^\gamma$ $\Rightarrow P_2=75 \times (\;\frac{1200}{150}) ^{\;\frac{5}{3}}$ $=75 \times (8)^{\;\frac{5}{3}}$ $=75 \times 32 {\text{kPa}}$ $=2400 {\text{kPa}}$ Work done in adiabatic process, $W=\;\frac{P_2 V_2-P_1 V_1}{1-\gamma }$ $=\;\frac{2400 \times 10^3 \times 150 \times 10^{-6}-75 \times 10^3 \times 1200 \times 10^{-6}}{1-\;\frac{5}{3}}$ $=\;\frac{(2400 \times 150-75 \times 1200) \times 10^{-3}}{-\;\frac{2}{3}}$ $=\;\frac{-810000 \times 10^{-3}}{2}$ $=-405 {\text{kJ}}$

Q114JEE Main 2022MCQ

A thermally insulated vessel contains an ideal gas of molecular mass ${\text{M}}$ and ratio of specific heats $1.4$ . Vessel is moving with speed $v$ and is suddenly brought to rest. Assuming no heat is lost to the surrounding and vessel temperature of the gas increases by : $( {\text{R}}=$ universal gas constant) [26-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Let there be ${\text{n}}$ moles of gas ${\text{E}}_{\;\text{loss }\;}= {\text{E}}_{ {\text{gain}}}$ $\;\frac{1}{2}(n M) v^2=n C_v \Delta T$ $\;\frac{1}{2} M v^2=C_v \Delta T$ here, $\gamma =1.4=\;\frac{7}{5}$ i.e. diatomic gas $\therefore C_v=\;\frac{5 R}{2}$ Now, $\;\frac{1}{2} M v^2=\;\frac{5 R}{2} \Delta T$ $\Delta T=\;\frac{M v^2}{5 R}$

Q115JEE Main 2022MCQ

Read the following statements : A. When small temperature difference between a liquid and its surrounding is doubled, the rate of loss of heat of the liquid becomes twice. B. Two bodies $P$ and $Q$ having equal surface areas are maintained at temperature $10^{\circ} {\text{C}}$ and $20^{\circ} {\text{C}}$ . The thermal radiation emitted in a given time by ${\text{P}}$ and ${\text{Q}}$ are in the ratio $1: 1.15$ . C. A Carnot Engine working between $100 {\text{K}}$ and $400 {\text{K}}$ has an efficiency of $75 \%$ . D. When small temperature difference between a liquid and its surrounding is quadrupled, the rate of loss of heat of the liquid becomes twice. Choose the correct answer from the options given below : [27-Jul-2022-Shift-1]

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📖 Explanation

Heat TransferA. by Newton's low of colling $\;{ {\text{d}} \theta }/{ {\text{dt}}}=\propto \Delta {\text{T}}$ B. ${\text{H}}=\;{ {\text{d}} \theta }/{ {\text{dt}}}=\sigma {\text{eAT}} {\text{T}}^4 \Rightarrow \;{ {\text{H}}_{ {\text{P}}}}/{ {\text{H}}_{ {\text{Q}}}}=(\;{ {\text{T}}_{ {\text{P}}}}/{ {\text{T}}_{ {\text{Q}}}})^4=(\;\frac{283}{293})^4$ ${\text{H}}_{ {\text{P}}}: {\text{H}}_{ {\text{Q}}}=1(1.03)^4=1:(1.03)^4=1: 1.15$ $\Rightarrow {\text{B}}$ is correctC. $η=1-\;\frac{100}{400}=\;\frac{3}{4}=75 \%$ D. is wrong as $\;{ {\text{d}} \theta }/{ {\text{dt}}} \propto \Delta {\text{T}}$

Q116JEE Main 2022MCQ

If $K_1$ and $K_2$ are the thermal conductivities, $L_1$ and $L_2$ are the lengths and $A_1$ and $A_2$ are the cross sectional areas of steel and copper rods respectively such that $\;\frac{K_2}{K_1}=9, \;\frac{A_1}{A_2}=2, \;\frac{L_1}{L_2}=2$ . Then, for the arrangement as shown in the figure, the value of temperature ${\text{T}}$ of the steel - copper junction in the steady state will be: [27-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

$\;{ {\text{d}} \theta }/{ {\text{dt}}}=\;{ {\text{K}}_1 {\text{A}}_1}/{ {\text{l}}_1}( {\text{T}}_1- {\text{T}})=\;{ {\text{K}}_2 {\text{A}}_2}/{ {\text{l}}_2}( {\text{T}}- {\text{T}}_2)$ $\Rightarrow \;{450- {\text{T}}}/{ {\text{T}}-0}=\;{ {\text{K}}_2 {\text{A}}_2 {\text{l}}_1}/{ {\text{K}}_1 {\text{A}}_1 {\text{l}}_2}=9 \times \;\frac{1}{2} \times 2$ $\Rightarrow 450- {\text{T}}=9 {\text{T}} \Rightarrow {\text{T}}=45^{\circ} {\text{C}}$

Q117JEE Main 2022NAT

A monoatomic gas performs a work of $\;\frac{Q}{4}$ where $Q$ is the heat supplied to it. The molar heat capacity of the gas will be____ ${\text{R}}$ during this transformation. Where ${\text{R}}$ is the gas constant. [24-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

By $1^{\;\text{st }\;}$ law, $\Delta U=\Delta Q-\;\frac{\Delta Q}{4}=\;\frac{3}{4} \Delta Q$ $\Rightarrow n C_v \Delta T=\;\frac{3}{4} n C \Delta T$ $\Rightarrow C=\;\frac{4 C_v}{3}=2 R$

Q118JEE Main 2022NAT

As per the given figure, two plates ${\text{A}}$ and ${\text{B}}$ of thermal conductivity ${\text{K}}$ and $2 {\text{K}}$ are joined together to form a compound plate.The thickness of plates are $4.0 {\text{cm}}$ and $2.5 {\text{cm}}$ respectively and the area of cross-section is $120 {\text{cm}}^2$ for each plate. The equivalent thermal conductivity of the compound plate is $(1+\;\frac{5}{\alpha }) {\text{K}}$ , then the value of $\alpha$ will be______ [29-Jun-2022-Shift-1]

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📖 Explanation

$\;\frac{L_1}{K_1 A_1}+\;\frac{L_2}{K_2 A_2}=\;{L_1+L_2}/{K_{\;\text{eff }\;} A_{\;\text{eff }\;}}$ $\Rightarrow \;\frac{4}{K}+\;\frac{2.5}{2 K}=\;{6.5}/{K_{\;\text{eff }\;}}$ $\Rightarrow \;\frac{10.5}{2 K}=\;{6.5}/{K_{\;\text{eff }\;}}$ $\Rightarrow K_{e f f}=\;\frac{13 K}{10.5}= (1+\;\frac{5}{21}) K$ $\Rightarrow \alpha =21$

Q119JEE Main 2022NAT

A block of ice of mass $120 {g}$ at temperature $0^{\circ} {C}$ is put in $300 {g}$ of water at $25^{\circ} {C}$ . The $\times {g}$ of ice melts as the temperature of the water reaches $0^{\circ} {C}$ . The value of $x$ is____$[Use specific heat capacity of water $=4200 {Jkg}^{-1} {K}^{-1}$ , Latent heat of ice $=3.5 × 10^{5} {Jkg}^{-1}$ ]$ [25-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Heat lost by water $=$ Heat gained by ice $0.3 \times 4200 \times 25=x \times 3.5 \times 10^{5}$ $x=\;\frac{0.3 \times 4200 \times 25}{3.5 \times 10^{5}}$ $=90 \times 100 \times 10^{5} \times 10^{3} {gram}=90 {gm}$

Q120JEE Main 2022MCQ

A vessel contains $14 {\text{g}}$ of nitrogen gas at a temperature of $27^{\circ} {\text{C}}$ . The amount of heat to be transferred to the gas to double the r.m.s speed of its molecules will be :Take ${\text{R}}=8.32 {\text{J}} {\text{mol}}^{-1} {\text{k}}^{-1}$ [28-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$n=0.5$ $T=300$ For ${\text{v}}_{ {\text{rms}}}$ to be doubled ${\text{T}}^{'}=4 \times 300=1200$ $\Rightarrow$ Heat transferred $=(0.5)(\;\frac{5}{2})(8.32)(900)$ $=9360 {\text{J}}$