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JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Airforce X Y / Indian Navy SSR MCQ


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30 Questions MCQ Test Physics for Airmen Group X - JEE Advanced (Single Correct MCQs): Heat & Thermodynamics

JEE Advanced (Single Correct MCQs): Heat & Thermodynamics for Airforce X Y / Indian Navy SSR 2024 is part of Physics for Airmen Group X preparation. The JEE Advanced (Single Correct MCQs): Heat & Thermodynamics questions and answers have been prepared according to the Airforce X Y / Indian Navy SSR exam syllabus.The JEE Advanced (Single Correct MCQs): Heat & Thermodynamics MCQs are made for Airforce X Y / Indian Navy SSR 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Advanced (Single Correct MCQs): Heat & Thermodynamics below.
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JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 1

A constant volume gas thermometer works on (1980)

Detailed Solution for JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 1

Note :  At constant volume, Charle's law is used.

JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 2

A metal ball immersed in alcohol weighs W1 at 0°C and W2 at 50°C. The coefficient of expansion of cubical the metal is less than that of the alcohal. Assuming that the density of the metal is large compared to that of alcohol, it can be shown that

Detailed Solution for JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 2

W1 = mg – Vdag

 

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JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 3

A wall has two layers A and B, each made of different material. Both the layers have the same thickness. The thermal conductivity of the meterial of A is twice that of B.Under thermal equilibrium, the temperature difference across the wall is 36°C. The temperature  difference across the layer A is

Detailed Solution for JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 3

JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 4

An ideal monatomic gas is taken round the cycle ABCDA as shown in the P – V diagram (see Fig.). The  work done during the cycle is

Detailed Solution for JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 4

The work done during the cycle = area enclosed in the curve

JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 5

If one mole of a monatomic gas   is mixed with one mole of a diatomic gas  , the value of γ for mixture is

Detailed Solution for JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 5

JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 6

From the following statements concerning ideal gas at any given temperature T, select the correct one(s)

Detailed Solution for JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 6

For an ideal gas PV = nRT

⇒ Coefficient of volume expansion

Note :  Average translation K.E. for O2 is 3/2 KT
(Three degrees of freedom for translational motion).
Now decrease in pressure increases the volume.

⇒ It increases mean free path of the molecules. Also average K.E. does not depend on the gas, so molecules of each component of mixture of gases have same average translational energy.

JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 7

Three rods of identical cross-sectional area and made from the same metal from the sides of an isosceles traingle ABC, right-angled at B. The points A and B are maintained at temperatures T and (√2) T respectively. In the steady state, the temperature of the point C is Tc. Assuming that only heat conduction takes place, Tc / T is

Detailed Solution for JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 7

Heat flow from B to A, A to C and C to B (for steady state condition, ΔQ/Δt is same)

 

JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 8

Two metallic spheres S1 and S2 are made of the same material and have got identical surface finish. The mass of S1 is thrice that of S2. Both the spheres are heated to the same high temperature and placed in the same room having lower temperature but are thermally insulated from each other. The ratio of the initial rate of cooling of S1 to that of S2 is 

Detailed Solution for JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 8

According to Stefan's law

JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 9

The average translational kinetic energy of O2 (relative molar mass 32) molecules at a particular temperature is 0.048 eV.
The translational kinetic energy of N2 (relative molar mass 28) molecules in eV at the same temperature is

Detailed Solution for JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 9

Average translational kinetic energy of an ideal gas molecule is 3/2 KT which depends on temperature only. Therefore, if temperature is same, translational kinetic energy of O2 and N2 both will be equal.

JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 10

A vessel contains 1 mole of O2 gas (relative molar mass 32) at a temperature T. The pressure of the gas is P. An identical vessel containing one mole of He gas (relative molar mass 4) at a temperature 2T has a pressure of

Detailed Solution for JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 10

Therefore, if T is doubled, pressure also becomes two times, i.e.., 2P.

JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 11

A spherical black body with a radius of 12 cm radiates 450 W power at 500 K. if the radius were halved and the temperature doubled, the power radiated in watt would be

Detailed Solution for JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 11

The energy radiated per second by a black body is given by Stefan's Law

Dividing (ii) and (i), we get

 

JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 12

A closed compartment containing gas is moving with some acceleration in horizontal direction. Neglect effect of gravity.Then the pressure in the compartment is 

Detailed Solution for JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 12

When a enclosed gas is accelerated in the positive x-direction then the pressure of the gas decreases along the positive x-axis and follows the equation ΔP = – r a dx where r is the density and a the acceleration of the container.
The result will be more pressure on the rear side and less pressure on the front side.

JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 13

 A gas mixture consists of 2 moles of oxygen and 4 moles of argon at temperature T. Neglecting all vibrational modes, the total internal energy of the system is

Detailed Solution for JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 13

The internal energy of n moles of a gas is

where F = number of degrees of freedom.
Internal energy of 2 moles of oxygen at temperature T is

Internal energy of 4 moles of argon at temperature T is

Total internal energy = 11 RT

JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 14

The ratio of the speed of sound in nitrogen gas to that in helium gas, at 300 K is

Detailed Solution for JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 14

JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 15

 A monatomic ideal gas, initially at temperature T1, is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature T2 by releasing the piston suddenly. If L1 and L2 are the length of the gas column before and after expansion respectively, 

then is given by

Detailed Solution for JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 15

JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 16

A block of ice at –10°C is slowly heated and converted to steam at 100°C. Which of the following curves represents the phenomenon qualitatively ?

Detailed Solution for JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 16

1. The temp. of ice changes from –10°C to 0°C.

2. Ice at 0°C melts into water at 0°C.

3. Water at 0°C changes into water at 100°C.

4. Water at 100°C changes into steam at 100°C.

JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 17

An ideal gas is initially at temperature T and volume V. Its volume is increased by ΔV due to an increase in temperature ΔT, pressure remaining constant. The quantity
varies with temperature as

Detailed Solution for JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 17

We know that V/T = constant

JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 18

Starting with the same initial conditions, an ideal gas expands from volume V1 to V2 in three different ways. The work done by the gas is W1 if the process is purely isothermal, W2 if purely isobaric and W3 if purely adiabatic. Then 

Detailed Solution for JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 18

Work done is equal to area under the curve on PV diagram.

JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 19

The plots of intensity versus wavelength for three black bodies at temperature T1, T2 and T3  respectively are as shown. Their temperatures are such that

Detailed Solution for JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 19

According to Wien's law,λT = constant From graph λ1 < λ3 < λ2

∴T1 > T3 > T2

JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 20

Three rods made of same material and having the same cross-section have been joined as shown in the figure. Each rod is of the same length. The left and right ends are kept at 0oC and 90oC respectively. The temperature of the junction of the three rods will be

Detailed Solution for JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 20

Let θ°C be the temperature at B. Let Q is the heat flowing per second from A to B on account of temperature difference.

By symmetry, the same will be the case for heat flow from C to B.

∴ The heat flowing per second from B to D will be

Dividing eq. (ii) by eq. (i)

JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 21

In a given process on an ideal gas, dW = 0 and dQ < 0. Then for the gas 

Detailed Solution for JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 21

From the first law of thermodynamics

dQ = dU + dW

Here dW = 0  (given)

∴ dQ = dU

Now since dQ < 0  (given)

∴ dQ is negative

⇒ dU = – ve   ⇒ dU decreases.

⇒ Temperature decreases.

JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 22

P-V plots for two gases dur ing adiabatic pr ocesses are shown in the figure. Plots 1 and 2 should correspond respectively to

Detailed Solution for JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 22

For adiabatic process 

Also for monoatomic gas 

for diatomic gas

⇒ Graph 1 is for diatomic and graph 2 is for mono atomic.

 

JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 23

When a block of iron floats in mercury at 0oC, fraction k1 of its volume is submerged, while at the temperature 60oC, a fraction k2 is seen to be submerged. If the coefficient of volume expansion of iron is γFe and  that of mercury is γHg, then the ratio k1/k2 can be expressed as

Detailed Solution for JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 23

For equilibrium in case 1 at 0° C Upthrust = Wt. of body

For equilibrium in case 2 at 60° C Note : When the temperature is increased the density will decrease.

JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 24

An ideal gas is taken through the cycle  as shown in the figure. If the net heat supplied to the gas in the cycle is 5J, the work done by the gas in the process C⇒A is

Detailed Solution for JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 24

For cyclic process;

JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 25

Which of the followin g graphs cor rectly represents the variation of with P for an ideal gas at constant temperature ?

Detailed Solution for JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 25

PV = constant. Differentiating,

∴ Graph between b and P will be a rectangular hyperbola.

JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 26

An ideal Black-body at room temperature is thrown into a furnace. It is observed that

Detailed Solution for JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 26

Note :  According to Kirchoff's law, good absorbers are good emitters as well.
At high temperature (in the furnace), since it absorbs more energy, it emits more radiations as well and hence is the brightest.

JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 27

The graph, shown in the adjacent diagram, represents the variation of temperature (T) of two bodies, x and y having same surface area, with time (t) due to the emission of radiation. Find the correct relation between the emissivity and absorptivity power of the two bodies

Detailed Solution for JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 27

The graph sh ows that for the same temperature difference (T2 – T1), less time is taken for x. This means the emissivity is more for x. According to Kirchoff's law, a good emitter is a good absorber as well.

JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 28

Two rods, one of aluminum and the other made of steel, having initial length l1 and l2 are connected together to form a single rod of length l1 + l2. The coefficients of linear expansion for aluminum and steel are αa and αs and respectively. If the length of each rod increases by the same amount when their temperature are raised by t0C, then find the ratiol1/(l1 + l2)

Detailed Solution for JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 28

The lengths of each rod increases by the same amount

JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 29

The PT diagram for an ideal gas is shown in the figure, where AC is an adiabatic process, find the corresponding PV diagram.

Detailed Solution for JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 29

If we study th e P – T gr aph we fin d AB to be a isothermal process, AC is adiabatic process given. Also for an expansion process, the slope of adiabatic curve is more (or we can say that the area under the P – V graph for isothermal process is more than adiabatic process for same increase in volume).
Only graph (b) fits the above criteria.

JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 30

2 kg of ice at –200C is mixed with 5kg of water at 200C in an insulating vessel having a negligible heat capacity. Calculate the final mass of water remaining in the container. It is given that the specific heats of water & ice are 1kcal/kg/0C & 0.5 kcal/kg/0C while the latent heat of fusion of ice is 80 kcal/kg

Detailed Solution for JEE Advanced (Single Correct MCQs): Heat & Thermodynamics - Question 30

Heat required to convert 5 kg of water at 20°C to 5 kg of water at 0°C

Heat released by 2 kg. Ice at – 20°C to convert into 2 kg of ice at 0°C

How much ice at 0°C will convert into water at 0°C for giving another 80 kcal of heat

Therefore the amount of water at 0°C

Thus, at equilibrium, we have, [6 kg water at 0°C + 1kg ice at 0°C].

 

 

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