Q341JEE Main 2016MCQ

'n' moles of an ideal gas undergoes a process A → B a shown in the figure. The maximum temperature of the gas during the process will be : [Main 2016]

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📖 Explanation

Equation of line is $PV_0+P_0V=3P_0V_0$ ......(i)Also PV = nRT.......(ii)for $T_{\max},\frac{dT}{dV}=0$ $\Rightarrow V={3V_0}/2,P={3P_0}/2$ $\Rightarrow T_{\max}=\frac{9P_0V_0}{4nR}$

Q342JEE Main 2016MCQ

An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity C remains constant. If during this process the relation of pressure P and volume V is given by $PV^n$ = constant, then n is given by (Here $C_P$ and $C_V$ are molar specific heat at constant pressure and constant volume,respectively):[Main 2016]

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📖 Explanation

$C-C_v-R/{n-1}$ $R/{n-1}=C_v-C$ $n=\frac{C-C_p}{C-C_v}$

Q343JEE Main 2015MCQ

Consider a spherical shell of radius R at temperature T.The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit volume $u=U/V\propto T^4$ and pressure $p=1/3(U/V)$ . If the shell now undergoes an adiabatic expansion the relation between T and R is: [Main 2015]

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📖 Explanation

dQ = dU + dWdU = -pdV $\frac{dU}{dV}=-p=-1/3U/V$ ${dU}/U=-1/3{dV}/V$ $lnU=-1/3lnV+lnC$ $U.V^{1/3}=C$ $VT^4,V^{1/3}=C'$ $T\propto 1/R$

Q344JEE Main 2015MCQ

Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion the average time of collision between molecules increases as $V^q$ , where V is the volume of the gas. The value of q is : $(\gamma =C_p/C_v)$ [Main 2015]

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📖 Explanation

Average time between collision $={\text{Mean free path}}/\text{V}_{\text{rms}}$ $t=1/{{\pi d^2\text{N/V}}/\sqrt{{3RT}/M};t={CV}/\text{sqrtT}$ (where $C=\text\frac{sqrtM}{\pi d^2N\sqrt3R}$ =constant) $\Rightarrow T\propto V^2/t^2$ For adiabatic $TV^{\gamma -1}=k$ $V^2/t^2V^{\gamma -1}=k$ $V^{\gamma -1}/t^2=k$ $t\propto V^{{\gamma +1}/2}$ so, $q={\gamma +1}/2$

Q345JEE Main 2015MCQ

A solid body of constant heat capacity 1 J/°C is being heated by keeping it in contact with reservoirs in two ways:(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same amount of heat.(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat.In both the cases body is brought from initial temperature 100°C to final temperature 200°C. Entropy change of the body in the two cases respectively is:[Main 2015]

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📖 Explanation

Case (i): $∫dS=C\$ [∫^{423}{373}{dT}/T+ ∫^{473}{423}{dT}/T]$=ln(473/373) $Case (ii):$ ∫dS=C$[∫^{385.5}{373}{dT}/T+∫^{398}{385.5}{dT}/T+∫^{410.5}{398}{dT}/T+∫^{423}{410.5}{dT}/T+∫^{435.5}{423}{dT}/T+∫^{448}{435.5}{dT}/T+∫^{460.5}{448}{dT}/T+∫^{473}{460.5}{dT}/T]$=ln(473/373)$Note: If given temperatures are in Kelvin then answer will be option (1).

Q346JEE Main 2014MCQ

one mole of diatomic ideal gas undergoes a cyclic process ABC as shown in figure is adiabatic.The temperature at A,B and C are 400 K,800 K and 600 K respectively.choose the correct statement:[Main 2014]

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📖 Explanation

$\Delta U_{AB}=nC_v(T_B-T_A)=1\times {5R}/2(800-400)=1000R$ $\Delta U_{AB}=nC_v(T_C-T_B)=1\times {5R}/2(600-800)=-500R$ $\Delta U_{total}=0$ $\Delta U_{AB}=nC_v(T_A-T_C)=1\times {5R}/2(400-600)=-500R$

Q347JEE Main 2014MCQ

Three rods of copper,Brass and Steel are welded together to from a Y -shaped structure.Area of cross-section of each rod $=4 cm^2$ ,Each of copper is maintained at 100°C where as ends of brass and steel are kept at 10°C.length of the copper,brass and steel rods are 46,13 and 12 cms respectively.The rods are thermally insulated from surrounding except a ends.Thermal conductive of copper,brass and steel are 0.92,0.26 and 0.12 CGS units respectively. Rate of heat flow through copper rod is:[Main 2014]

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📖 Explanation

${dQ}_1/{dt}={dQ}_2/{dt}+\frac{dQ_3}{dt}$ $\Rightarrow {0.92(100-T)}/46={0.26(T-0)}/13+{0.12(T-0)}/12$ ⇒T=40°C $\frac{dQ_1}{dt}={0.92\times 4(100-40)}/40=4.8$ cal/s

Q348JEE Main 2013MCQ

Assume that a drop of liquid evaporates by decrease in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible? The surface tension is T, density of liquid is $\rho$ and L is its latent heat of vaporization.[Main 2013]

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📖 Explanation

surface Tension $=T \;\;$ Radius $=R$ density $=\rho$ Latent heat $=L$ Let for $R,$ small decrease on $R$ is $\triangle R$ $\therefore$ change in surface energy $\Delta E= \;\; \$ [4 π R^{2}-4 π(R-∆ R)^{2}]$ \times T;; =4 \pi (R-R+\Delta R)(R+R-4 R) \times T;; =4 \pi \cdot \Delta R \cdot 2 R \cdot T ;; \therefore ;; \Delta R \approx 0;; =8 \pi R \cdot \Delta R \cdot T $change in mass$ \Delta m ;; =9 / 3 \pi (R^{3}-(R-\Delta R)^{3}) \times \rho ;; =;\frac{4}{3} \pi (R-R+\Delta R)(R^{2}+(R-\Delta R)^{2}+R(R-\Delta R)) \rho ;; =;\frac{4}{3} \pi \cdot \Delta R(R^{2}+R^{2}+R^{2}-3 R \Delta R+(\Delta R)^{2}) \rho ;; =4 \cdot \pi \cdot \Delta R \cdot R^{2} \cdot \rho $Energy for vaporization$ \Delta V=4 \pi \Delta R \cdot R^{2} \cdot L \rho 8 \pi \cdot R \cdot \Delta R \cdot T=4 \cdot \pi \cdot \Delta R \cdot R^{2} \cdot L \cdot \rho $or$ ;; 2 T=R L \rho $or$ ;; R=;\frac{2 T}{\rho L}$

Q349JEE Main 2013MCQ

The above p-v diagram represents the thermodynamic cycle of an engine, operating with an ideal monoatomic gas. The amount of heat extracted from the source in a single cycle is[Main 2013]

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📖 Explanation

Heat is extracted from the source in path DA and AB is $\Delta Q=3/2R({P_0V_0}/R)+5/2R({2P_0V_0}/R)=13/2P_0V_0$

Q350JEE Main 2012MCQ

A wooden wheel of radius $R$ is made of two semicircular part (see figure). The two parts are held together by a ring made of a metal strip of cross sectional area $S$ and length $L . L$ is slightly less than $2 \pi R$ . To fit the ring on the wheel, it is heated so that its temperature rises by $\Delta T$ and it just steps over the wheel. As it cools down to surrounding temperature, it process the semicircular parts together. If the coefficient of linear expansion of the metal is $\alpha$ , and its Young's modulus is $Y$ , the force that one part of the wheel applies on the other part is:

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📖 Explanation

$\;\gamma =\;\frac{F / S}{\Delta L / L} \Rightarrow \Delta L=\;\frac{F L}{S Y}$ $\; \therefore L \alpha \Delta T=\;\frac{F L}{S Y}$ $\;{[\;\text{ as }\; \Delta L=L \alpha \Delta T]}$ $\; \therefore F=S Y \alpha \Delta T$ $\therefore$ The ring is pressing the wheel from both sides, $\therefore F_{n e t}=2 F=2 Y S \alpha \Delta T$

Q351JEE Main 2012MCQ

A Carnot engine, whose efficiency is $40 \%$ , takes in heat from a source maintained at a temperature of $500 {\text{K}}$ . It is desired to have an engine of efficiency $60 \%$ . Then, the intake temperature for the same exhaust (sink) temperature must be:

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📖 Explanation

$0.4=1-\;\frac{T_2}{500} \;\;$ and $\;\; 0.6=1-\;\frac{T_2}{T_1}$ on solving we get $T_2=750 {\text{K}}$

Q352JEE Main 2012MCQ

Helium gas goes through a cycle $A B C D$ (consisting of two isochoric and isobaric lines) as shown in figure efficiency of this cycle is nearly : (Assume the gas to be close to ideal gas)

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📖 Explanation

Heat given to system $= (n C_v \Delta T) _{A \to B}+ (n C_p \Delta T) _{B \to C}$ $\;= [\;\frac{3}{2}(n R \Delta T)] _{A \to B}+ [\;\frac{5}{2}(n R \Delta T)] _{B \to B} \to$ $\;= [\;\frac{3}{2} \times V_0 \Delta P] _{A \to B}+ [\;\frac{5}{2} \times 2 P_0 \times V_0] _{B \to C}$ $\;=\;\frac{13}{2} P_0 V_0$ $\;\;\text{ and }\; W_0=P_0 V_0$ $\;η=\;\frac{\;\text{ Work }\;}{\;\text{ heat given }\;}$ $\;=\;\frac{P_0 V_0}{\frac{18}{2} P_0 V_0} \times 100$ $\;=15.4 \%$

Q353JEE Main 2011MCQ

Three perfect gases at absolute temperatures $T_1, T_2$ and $T_3$ are mixed. The masses of molecules are $m_1, m_2$ and $m_3$ and the number of molecules are $n_1, n_2$ and $n_3$ respectively. Assuming no loss of energy, the final temperature of the mixture is:

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📖 Explanation

Number of moles of first gas $=\;\frac{n_1}{N_A}$ Number of moles of second gas $=\;\frac{n_2}{N_A}$ Number of moles of third gas $=\;\frac{n_3}{N_A}$ If there is no loss of energy then $P_1 V_1+P_2 V_2+P_3 V_3=P V$ $\;\frac{n_1}{N_A} R T_1+\;\frac{n_2}{N_A} R T_2+\;\frac{n_3}{N_A} R T_3$ $=\;\frac{n_1+n_2+n_3}{N_A} R T_{m i x}$ $\Rightarrow T_{\;\text{mix }\;}=\;\frac{n_1 T_1+n_2 T_2+n_3 T_3}{n_1+n_2+n_3}$

Q354JEE Main 2011MCQ

A thermally insulated vessel contains an ideal gas of molecular mass $M$ and ratio of specific heats $\gamma$ . It is moving with speed $v$ and it's suddenly brought to rest. Assuming no heat is lost to the surroundings, Its temperature increases by:

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📖 Explanation

Here, work done is zero.So, loss in kinetic energy $=$ change in internal energy of gas $\;\;\frac{1}{2} m v^2=n C_v \Delta T=n \;\frac{R}{\gamma -1} \Delta T$ $\;\;\frac{1}{2} m v^2=\;\frac{m}{M} \;\frac{R}{\gamma -1} \Delta T$ $\; \therefore \Delta T=\;\frac{M v^2(\gamma -1)}{2 R} K$

Q355JEE Main 2011MCQ

A Carnot engine operating between temperatures $T_1$ and $T_2$ has efficiency $\;\frac{1}{6}$ . When $T_2$ is lowered by $62 K$ its efficiency increases to $\;\frac{1}{3}$ . Then $T_1$ and $T_2$ are, respectively:

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📖 Explanation

Efficiency of engine $\;\;\frac{1}{6}=1-\;\frac{T_2}{T_1} \;\text{ and }\; η_2=1-\;\frac{T_2-62}{T_1}=\;\frac{1}{3}$ $\; \therefore T_1=372 K \;\text{ and }\; T_2=\;\frac{5}{6} \times 372=310 K$

Q356JEE Main 2011MCQ

$100 {\text{g}}$ of water is heated from $30^{\circ} {\text{C}}$ to $50^{\circ} {\text{C}}$ . Ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is $4184 {\text{J}} / {\text{kg}} / {\text{K}}$ ):

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📖 Explanation

$\Delta U=\Delta Q=m c \Delta T$ $=100 \times 10^{-3} \times 4184(50-30) \approx 8.4 {\text{kJ}}$

Q357JEE Main 2010MCQ

A diatomic ideal gas is used in a Carnot engine as the working substance. If during the adiabatic expansion part of the cycle the volume of the gas increases from $V$ to $32 {\text{V}}$ , the efficiency of the engine is

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📖 Explanation

$\;T_1 V^{\gamma -1}=T_2(32 V)^{\gamma -1}$ $\;\Rightarrow T_1=(32)^{\gamma -1} \cdot T_2$ For diatomic gas, $\gamma =\;\frac{7}{5}$ $\; \therefore \gamma -1=\;\frac{2}{5}$ $\; \therefore T_1=(32)^{\;\frac{2}{5}} \cdot T_2 \Rightarrow T_1=4 T_2$ Now, efficiency $=1-\;\frac{T_2}{T_1}$ $\;=1-\;\frac{T_2}{4 T_2}$ $\;=1-\;\frac{1}{4}$ $\;=\;\frac{3}{4}$ $\;=0.75$

Q358JEE Main 2009MCQ

A long metallic bar is carrying heat from one of its ends to the other end under steadystate. The variation of temperature $\theta$ along the length $x$ of the bar from its hot end is best described by which of the following figures?

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📖 Explanation

The heat flow rate is given by $\;\;\frac{d Q}{d t}=\;\frac{k A (\theta _1-\theta ) }{x}$ $\;\Rightarrow \theta _1-\theta =\;\frac{x}{k A} \;\frac{d Q}{d t}$ $\;\Rightarrow \theta =\theta _1-\;\frac{x}{k A} \;\frac{d Q}{d t}$ where $\theta _1$ is the temperature of hot end and $\theta$ is temperature at a distance $x$ from hot end. The above equation can be graphically represented by option $(a)$ .

Q359JEE Main 2009MCQ

Statement - 1: The temperature dependence of resistance is usually given as $R=R_0(1+\alpha \Delta t)$ . The resistance of wire changes from $100 Ω$ to $150 Ω$ when its temperature is increased from $27^{\circ} {\text{C}}$ to $227^{\circ} {\text{C}}$ . This implies that $\alpha =2.5 \times 10^{-3} / C$ .Statement - 2: $R=R_0(1+\alpha \Delta t)$ is valid only when the change in the temperature $\Delta T$ is small and $\Delta T= (R-R_0) < < R_0$ .

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📖 Explanation

The relation $R=R_0(1+\alpha \Delta t)$ is valid for small values of $\Delta t$ and $R_0$ is resistance at $0^{\circ} {\text{C}}$ and also $(R-R_0)$ should be much smaller than $R_0$ . So, statement (1) is wrong but statement $(2)$ is correct.

Q360JEE Main 2009MCQ

Two moles of helium gas are taken over the cycle $A B C D$ , as shown in the $P$ - $T$ diagram.The net work done on the gas in the cycle $A B C D A$ is:

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📖 Explanation

The net work in the cycle $A B C D$ is $\;W=W_{A B}+W_{B C}+W_{C D}+W_{D A}$ $\;=400 R+2.303 n R T \log \;\frac{P_B}{P_C}+(-400 R)-414 R$ $\;=2.303 \times 2 R \times 500 \log \;\frac{2 \times 10^5}{1 \times 10^5}-414 R$ $\;=693.2 R-414 R=279.2 R$