Q321JEE Main 2018MCQ

The value closest to the thermal velocity of a Helium atom at room temperature ( 300 K ) in $ms ^{-1}$ is : $[ k _{ B }=1.4 \times 10^{-23} J / K ; m _{ He }=7 \times 10^{-27} kg]$

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Q322JEE Main 2018MCQ

A Carnot's engine works as a refrigerator between 250 K and 300 K. It receives 500 cal heat from the reservoir at the lower temperature. The amount of work done in each cycle to operate the refrigerator is :[Main 15 April 2018 S1]

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Q324JEE Main 2017MCQ

$C_p$ 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 gasThe correct relation between a and b is :[Main 2017]

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📖 Explanation

$C_p-C_v=R$ where $C_p$ and $C_v$ are molar specific heat capacities As per the question $a=R/2\;\;\;b=R/28$ $a=14b$

Q325JEE Main 2017MCQ

An external pressure P is applied on a cube at 0°C so that it is equally compressed from all sides.K is the bulk modulus of the material of the cube and α 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 :[Main 2017]

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📖 Explanation

Due to thermal expansion ${\Delta v}/v=3\alpha \Delta T$ Due to External pressure ${\Delta v}/v=P/K$ Equating both $w_1$ get $3\alpha \Delta T=P/K$ $\Delta T=P/{3\alpha K}$

Q326JEE Main 2017MCQ

An ideal gas has molecules with 5 degrees of freedom. The ratio of specific heats at constant pressure $(C_p)$ and at constant volume $(C_v)$ is :[Main 8 April 2017]

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📖 Explanation

$f=C_p/C_a=1+2/1$ $=7/5$

Q327JEE Main 2017MCQ

The temperature of an open room of volume $30m^3$ increase from 17°C to 27°C due to sunshine.The atmospheric pressure in the room remains $1\times 10^5$ 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 :-[Main 2017]

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📖 Explanation

Using the ideal gas equationPV = NRT(N is number of moles) $P_0V_0=N_iR\times 290$ .....(1) $[T_i=273+17=290K]$ After heating $P_0V_0=N_fR\times 300$ .....(2) $[T_f=273+27=300K]$ from equation (1) and (2) $N_f-N_i=\frac{P_0V_0}{R\times 300}-\frac{P_0V_0}{R\times 290}$ difference in number of moles $-{P_0V_0}/R[10/{290\times 300}]$ Hence $n_f-n_i$ is $={P_0V_0}/R\times [10/{290\times 300}]\times 6.023\times 10^23$ putting $P_0=10^5P_A$ and $V_0-30m^3$ Number of molecules $n_f-n_i=-2.5\times 10^25$

Q328JEE Main 2017MCQ

A compressive force, F is applied at the two ends of a long thin steel rod. It is heated, simultaneously, such that its temperature increases by ∆T. The net change in its length is zero. Let l be the length of the rod, A its area of cross-section, Y its Young's modulus, and α its coefficient of linear expansion. Then, F is equal to :[Main 8 April 2017]

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📖 Explanation

Net change in length = 0Thermal Exp. $=l\propto \Delta t$ $y=\frac{F/A}{{\Delta l}/l}$ ${\Delta l}/l=F/{Ay}$ $\Delta l=\frac{Fl}{Ay}$ $\frac{Fl}{Ay}=l\propto \Delta t$ $F=Ay\propto \Delta t$

Q329JEE Main 2017MCQ

A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75°C. T is given by : (Given : room temperature = 30° C, specific heat of copper = 0.1 cal/gm°C[Main 2017]

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📖 Explanation

Heat given = Heat taken $(100)(0.1)(T-75)=(100)(0.1)(T-75)$ $= (100)(0.1)(45)+(170)(1)(45)10(T-75)$ $=450+7650=8100$ $T-75=810$ T = 885°C

Q330JEE Main 2017MCQ

A steel rail of length 5 m and area of cross section $40cm^2$ is prevented from expanding along its length while the temperature rises by 10 ° C. If coefficient of linear expansion and Young's modulus of steel are $1.2\times 10^{-5}k^{-1}$ and $2\times 10^11Nm^{−2}$ respectively, the force developed in the rail is approximately :[Main 9 April 2017]

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Q331JEE Main 2017MCQ

An engine operates by taking n moles of an ideal gas through the cycle ABCDA shown in figure. The thermal efficiency of the engine is : (Take $C_v=1.5$ R, where R is gas constant)[Main 8 April 2017]

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📖 Explanation

$w=P_0V_0$ Heat given $=Q_{AB}=Q_{BC}$ $=nC_vdT_{AB}+nC_pdT_{BC}$ $=3/2(nRT_B-nRT_A)+5/2(nRT_c-nRT_B)$ $=3/2(2P_0V_0-P_0V_0)+5/2(4P_0V_0-2P_0V)$ $=13/2P_0V_0$ $n=w/\text{Qgiven}=2/13=0.15$

Q332JEE Main 2017MCQ

In an experiment a sphere of aluminium of mass 0.20 kg is heated upto 150°C. Immediately, it is put into water of volume 150 cc at 27°C kept in a calorimeter of water equivalent to 0.025 kg. Final temperature of the system is 40°C. The specific heat of aluminium is : (take 4.2 Joule=1 calorie)[Main 8 April 2017]

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📖 Explanation

Q given=Q used $0.2\times S\times (150-40)=150\times 1\times (40-27)$ $+25\times (40-27)$ $0.2\times S\times 110=150\times 13+25\times 13$ $S=\frac{13\times 25\times 7}{0.2\times 110}$ $S=434$

Q333JEE Main 2017MCQ

N moles of a diatomic gas in a cylinder are at a temperature T. Heat is supplied to the cylinder such that the temperature remains constant but n moles of the diatomic gas get converted into monoatomic gas. What is the change in the total kinetic energy of the gas ?[Main 9 April 2017]

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Q334JEE Main 2017MCQ

For the P-V diagram given for an ideal gas out of the following which one correctly represents the T-P diagram ?[Main 9 April 2017]

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Q335JEE Main 2016MCQ

200 g water is heated from 40°C to 60°C.Ignoring the slight expansion of water, the change in its internal energy is close to(Given specific heat of water=4184 J/kg/K) :[Main 9 Apr 2016]

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📖 Explanation

$\Delta U=mS\Delta T$ $=(2/10)(4184)(20)=16736J$ $=16.7KJ$

Q336JEE Main 2016MCQ

A Carnot freezer takes heat from water at 0°C inside it and rejects it to the room at a temperature of 27°C. The latent heat of iceis $336\times 10^3J kg^{−1}$ . If 5 kg of water at 0°C is converted into ice at 0°C by the freezer, then the energy consumed by the freezer is close to :[Main 10 Apr 2016]

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📖 Explanation

heat required to freeze 5 kg water $5\times 336\times 10^3$ $=1680\times 10^3$ Joule $\Rightarrow Q_1=1680$ KJfor carnot's cycle $Q_2/Q_1=T_2/T_1$ $Q_2/1680==300/273$ $Q_2=1680\times 300/273$ KJ $W=Q_2-Q_1$ $=1680(300/273-1)$ $={1680\times 27}/273\times 10^3J$ $=166.15\times 10^3J$ $=1.66\times 10^5KJ$

Q337JEE Main 2016MCQ

A simple pendulum made of a bob of mass m and a metallic wire of negligible mass has time period 2s at $T=0°C$ . If the temperature of the wire is increased and the corresponding change in its time period is plotted against its temperature, the resulting graph is a line of slope S. If the coefficient of linear expansion of metal is αthen the value of S is :[Main 9 Apr 2016]

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📖 Explanation

$t_0=2=2\pi \sqrt{l_0/g}$ $t={2\pi }\sqrt{{l_0(1+\alpha T)}/g}=2(1+\alpha T)^{1/2}=2+\alpha T$ $t-t_0=\alpha T\Rightarrow \Delta t=\alpha T$

Q338JEE Main 2016MCQ

A pendulum clock loses 12 s a day if the temperature is 40°C and gains 4 s a day if the temperature is 20°C The temperature at which the clock will show correct time, and the co-efficient of linear expansion (α) of the metal of the pendulum shaft are respectively :[Main 2016]

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📖 Explanation

First Method:Time period T $=2\pi \sqrt{l/g}=2\pi \sqrt{l_0/g(1+\alpha \Delta \theta )}$ $T=T_0[1+1/2\alpha \Delta \theta ]$ $N={2\times 86400}/T({2\times 86400}/T_0)(1+1/2\alpha \Delta \theta )=N(1+{\alpha \Delta \theta }/2)$ $\Delta N=N-N_0=1/2\alpha \Delta \theta N_0\Rightarrow \Delta N\propto \Delta \theta$ $\Rightarrow \theta _0=25°$ CPutting $\theta _0$ ,we get $\alpha =18.5\times 10^{-5}/$ °CSecond MethodAccording to given conditions $86412=2\pi \sqrt{l_40/g$ .....(i) $86396=2\pi \sqrt{l_20/g$ .....(ii) $86400=2\pi \sqrt{l/g$ .....(iii)From equation (i) and (iii) $12={2\pi }/\text{sqrtg}[\text{sqrtl}_40-\text{sqrtl}]$ ...(iv)and $4={2\pi }/\text{sqrtg}[\text{sqrtl}-\text{sqrtl}_20]$ .....(v)on dividing (iv) and (v) $3=\frac{\sqrt{1+\alpha (40-\theta )}-1}{1-\sqrt{1+\alpha (20-\theta )}}$ $\Rightarrow 3=\frac{40-\theta }{\theta -20}$ (by Binomial theorem) $\Rightarrow \theta =25°$ on using in θ (i) and (iii) $86412/86400=\sqrt{1+\alpha 15}\Rightarrow \alpha =1.85\times 10^{-5}/$ °C

Q339JEE Main 2016MCQ

The ratio of work done by an ideal monoatomic gas to the heat supplied to itin an isobaric process is :[Main 9 Apr 2016]

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📖 Explanation

$\Delta w=P\Delta V=nR\Delta T$ $\Delta Q=nc_p\Delta T=5/2nR\Delta T$ $\frac{\Delta w}{\Delta Q}=2/5$

Q340JEE Main 2016MCQ

Which of the following shows the correct relationship between the pressure 'P' and density $\rho$ of an ideal gas at constant temperature ?[Main 10 Apr 2016]

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📖 Explanation

$\rho =\frac{PM}{RT}$ $P={\rho RT}/M$ $P\propto \rho$