Q281JEE Main 2019MCQ

A 15 g mass of nitrogen gas is enclosed in a vessel at a temperature 27°C. Amount of heat transferred to the gas, so that rms velocity of molecules is doubled, is about : [Take R = 8.3 J/ K mole]

🚀 Solve in Practice Mode

📖 Explanation

Q = $nC_v\Delta T$ as gas in closed vessel Q = $\frac{15}{28}$ × ${5\times R}/2$ × (4T - T) Q = 10000 J = 10 kJ

Q282JEE Main 2019MCQ

A thermometer graduated according to a linear scale reads a value x0 when in contact with boiling water, and $x_0/3$ when in contact with ice. What is the temperature of an object in 0 °C, if this thermometer in the contact with the object reads $x_0/2$ ?

🚀 Solve in Practice Mode

📖 Explanation

⇒ T°C = $x_0/6$ & $(x_0-x_0/3)$ = (100 - 0°C) $x_0$ = ${300}/2$ \Rightarrow T°C = ${150}/6$ = 25°C

Q283JEE Main 2019MCQ

When M1 gram of ice at $-10° C$ (specific heat = 0.5 cal $\ {g}^{-1 } \ {°C}^{-1}$ is added to $\ {M}_{2}$ is added to of water at 50°C, finally no ice is left and the water is at 0°C. The value of latent heat of ice, in ${calg^{-1}}$ is:

🚀 Solve in Practice Mode

📖 Explanation

$M_{1} \times 10 \\times 0.5+M_{1} L=M_{2} \\times 50$ $L =\ \frac{50 M_{2}-5 M_{1}}{M_{1}}$ $L =\ \frac{50 M_{2}}{M_{1}}-5 \$

Q284JEE Main 2019MCQ

Half mole of an ideal monoatomic gas is heated at constant pressure of 1atm from 20 °C to 90°C. Work done by gas is close to : ( Gas constant R = 8.31 J /mol.K)

🚀 Solve in Practice Mode

📖 Explanation

WD = PΔV = nRΔT = $1/2$ × 8.31 × 70

Q285JEE Main 2019MCQ

One mole of an ideal gas passes through a process where pressure and volume obey the relation $p=P_{0}\ \$ [1-\ \frac{1}{2}\ ( \frac{v_{0}}{v}\ )^{2}\ ]$I_1 $/$ I_2 $.here$ P_0 $and$ V_0 $are constants, calculate the change in the temperature of the gas if its volume changes from$ V_0 $to$ 2V_0$

🚀 Solve in Practice Mode

📖 Explanation

Given $P=P_{0}\ [1-1 / 2\ (\ \frac{V_{0}}{V} )^{2}\ ]$ As PV $\;\;$ = $\;\;$ nRT $\;\;$ ( n = 1 mole ) $P= \frac{R T}{V}$ $\ \frac{R T}{V}=p_{0}-\ \frac{p_{0}}{2} \ \frac{V_{0}^{2}}{V^{2}}$ $\ T =\ \frac{1}{R}\ [P_{0} V-\ \frac{P_{0} V_{0}^{2}}{2 V}\ ]$ $V =V_{0}$ $T_{1} = \frac{1}{R}\ [\ \frac{P_{0} V_{0}}{2}\ ]$ $V =2 V_{0}$ $T_{2}=\ \frac{1}{R} [2 P_{0} V_{0}- \frac{P_{0} V_{0}}{4} ]$ $T_{2}= \frac{7 P_{0} V_{0}}{4 R}$ $\Delta T=T_{1} \sim T_{2}= \frac{5 P_{0} V_{0}}{4 R}$

Q286JEE Main 2019MCQ

One kg of water at $20 °C$ , is heated in an electric kettle whose heating element has a mean (temperature averaged) resistance of $20 \Omega$ . The rms voltage in the mains is $200 \ V$ . Ignoring heat loss from the kettle, time taken for water to evaporate fully, is close to:

🚀 Solve in Practice Mode

📖 Explanation

$Q=P t \ \ msat +m L$ $= \ \frac{V_{0}^{2}}{R} t(1 \\times 4200 \times 80)+ (1 \\times 2260 \\times 10^{3}\ )$ $= \frac{(\sqrt{2} \\times 200)^{2}}{20} t$ $t=649 \ {sec}$ $t=10.8 \ {min}$ $t \approx 10 \ {min}$

Q287JEE Main 2019MCQ

In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation VT = K, where K is a constant. In this process the temperature of the gas is incresed by ΔT. The amount of heat absorbed by gas is (R is gasconstant) :

🚀 Solve in Practice Mode

📖 Explanation

VT = K ⇒ V $(\frac{PV}{nR})$ = k \Rightarrow $PV^2$ = K Since C = $R/{1-x}$ + $C_v$ (For polytropic process) C = $R/{1-2}+{3R}/2$ = $R/2$ ∴ ΔQ = nCΔT = $R/2$ ΔT

Q288JEE Main 2019MCQ

For a given gas at 1 atm pressure, rms speed of the molecules is 200 m/s at 127ºC. At 2 atm pressure and at 227ºC, the rms speed of the molecules will be:

🚀 Solve in Practice Mode

📖 Explanation

$V_{r m s} \propto \sqrt{T}$ $\frac{V_{2}}{200}= \frac{500}{400}$ $V_{2}=100 \sqrt{5}$

Q289JEE Main 2019MCQ

Following figure shows two process A and B for a gas. If $\Delta Q_A$ and $\Delta Q_B$ are the amount of heat absorbed by the system in two cases, and $\Delta U _A$ and $\Delta U_B$ are changes in internal energies, respectively, then:

🚀 Solve in Practice Mode

📖 Explanation

$\Delta U_{A}=\Delta U_{B}$ is $U$ is state function $\Delta Q_{A} =\Delta U_{A}+W_{A}$ $\Delta Q_{B} =\DeltaU_{B}+w_{B} \$ $\DeltaQ_{A}-\Delta Q_{B}=w_{A}-w_{B}$ $\Rightarrow 0$ $\Delta Q_{A}>\Delta Q_{B}$

Q290JEE Main 2019MCQ

A thermally insulated vessel contains 150 g of water at 0°C. Then the air from the vessel is pumped out adiabatically. A fraction of water turns into ice and the rest evaporates at 0°C itself. The mass of evaporated water will be closest to: (Latent heat of vaporization of water = $=2.10 \times 10^{6} {J} {kg}^{-1}$ and Latent heat of Fusion of water $=3.36\times 10^{5} {J} {kg}^{-1}$ )

🚀 Solve in Practice Mode

📖 Explanation

$m L_{v}=(150-m) L_{f}$ $m \times 2.1 \times 10^{6}=(150-m) (3.36 \times 10^{5} )$ $m \approx 20 g$

Q291JEE Main 2019MCQ

Two kg of a monoatomic gas is at a pressure of 4 × $10^4$ N/ $m^2$ . The density of the gas is 8 kg / $m^3$ . What is the order of energy of the gas due to its thermal motion ?

🚀 Solve in Practice Mode

📖 Explanation

Thermal energy of N molecule = N $(3/2 kT)$ = $N/N_A 3/2 RT$ = $3/2$ (nRT) = $3/2$ PV = $3/2 P(m/8)$ = $3/2$ × 4 × $10^4\times 2/8$ = 1.5 × $10^4$ order will $10^4$

Q292JEE Main 2019MCQ

For the given cyclic process CAB as shown fora gas, the work is :

🚀 Solve in Practice Mode

📖 Explanation

Since P-V indicator diagram is given, so workdone by gas is area under the cyclic diagram. ∴ ΔW = Work done by gas = $1/2$ × 4 × 5 J = 10 J

Q293JEE Main 2019MCQ

An ideal gas is enclosed in a cylinder at pressure of 2 atm and temperature, 300 K. The mean time between two successive collisions is 6 × $10^{-8}$ s. If the pressure is doubled and temperature is increased to 500 K, the mean time between two successive collisionswill be close to:

🚀 Solve in Practice Mode

📖 Explanation

t α $\frac{\text{Volume}}{\text{velocity}}$ volume α $T/P$ ∴ t α $\text{sqrtT}/P$ $t_1/{6\times 10^{-8}}$ = $\sqrt\frac{500}{2P}$ × $P/\sqrt{300}$ $t_1$ = 3.8 × $10^{8}$ ~ 4 × $10^{-8}$

Q294JEE Main 2019MCQ

Temperature difference of 120°C is maintained between two ends of a uniform rod AB of length 2L. Another bent rod PQ, of same crosssection as AB and length ${3L}/2$ , is connected across AB (See figure). In steady state, temperature difference betweenP and Q will be close to : [9-Jan-2019 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

${\Delta T}/R_{eq}$ = I = $\frac{(120)5}{8R}$ = $\frac{120\times 5}{8R}$ $\Delta T_{PQ}$ = $\frac{120\times 5}{8R}$ × $3/5$ R = ${360}/8$ = 45°C

Q295JEE Main 2019MCQ

A $25 \times 10^{-3} m^{3}$ volume cylinder is field with 1 mol $O_2$ gas room temperature (300k) The molicular diameter of $O_2$ ,and its root mean square speed , are found to be 0.3 nm and 200m/s, respectively.what is the average collision rate (per second)for an $O_2$ Molecule?

🚀 Solve in Practice Mode

Q296JEE Main 2019MCQ

The temperature, at which the root mean square velocity of hydrogen molecules equals their escape velocity from the earth, is closest to: $[Boltzmann Constant $k_B =1.38 × 10^{-23}$ J/K Avogadro Number $NA = 6.02 × 10^{26}$ /kg Radius of Earth : $6.4 × 10^6$ m Gravitational acceleration on Earth = 10 $ms^{-2}$ ]$

🚀 Solve in Practice Mode

📖 Explanation

(But according to given data in the question no answer, avogadro number is given wrong ) $\frac{3 R T}{M}=2 g R$ $\frac{3 \times \frac{25}{3} \times T}{2\times 10^{-3}}=2 \times 10 \times 6.4 \times 10^{6}$ $\Rightarrow T \approx 10^{4} K$

Q297JEE Main 2019MCQ

A rod, of length L at room temperature and uniform area of cross section A, is made of a metal having coefficient of linear expansion α/ °C. It is observed that an external compressive force F, is applied on each of its ends, prevents any change in thelength of the rod, when its temperature rises by ΔT K. Young's modulus, Y, for this metal is : [9-Jan-2019 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

Young's modulus y = $\frac{\text{Stress}}{\text{Strain}}$ = $\frac{F/A}{(\Delta l/l)}$ = $F/{A(\alpha \Delta T)}$

Q298JEE Main 2019MCQ

n moles of an ideal gas with constant volume heat capacity Cv undergo an isobraric expansion by certain volume . The ratio of the work done in the process, to the heat supplied is :

🚀 Solve in Practice Mode

📖 Explanation

n = moles cv = neat capacity $\ {\ {P}\ (\ {v}_{2}-\ {v}_{1}\ )}/{ {C}_{\ {\text{v}}}\ \Delta\ {t}+ {P} ( {\text{v}}_{2}- {\text{v}}_{1} )}=\ {\ {nR}}/{\ {C}_{\ {\text{v}}}-\ {nR}}$

Q299JEE Main 2019MCQ

A heat source at T= $10^3$ K is connected to another heat reservoir at T = $10^2$ K by a copper slab which is 1 m thick. Given that the thermal conductivity of copper is 0.1 $WK^{-1} m^{-1}$ , the energy flux through it in the steady state is :

🚀 Solve in Practice Mode

📖 Explanation

$(\frac{dQ}{dt})$ = ${kA\Delta T}/l$ \Rightarrow $1/A$ $(\frac{dQ}{dt})$ = ${(0.1)(900)}/1$ = 90 W/ $m^2$

Q300JEE Main 2019MCQ

At $40° C$ ,a brass wire of 1 mm radius is hung from the ceiling. A small mass, M is hung from the free end of the wire. When the wire is cooled down from $40° C$ to $20° C$ it regains its original length of 0.2 m. The value of M is close to. (Coefficientof linear expansion and Young's modulus of brass are $10^{-5}$ \ $^{0}{C}$ and $10^{11}$ $N$ / $m_2$ respectively ; g = 10 ms $^{-2}$ )

🚀 Solve in Practice Mode

📖 Explanation

$Y A \alpha \Delta t=M g$ $M=\ \frac{Y \pi r^{2} \\times \Delta t}{g}$ $M=\ \frac{10^{11} \times \pi \times 10^{-6} \times 10^{-5} \\times 20}{10}$ $M=6.28 \ {ks}$