IIT JAM Physics MCQ Test 7

30 Questions MCQ Test Mock Test Series for IIT JAM Physics | IIT JAM Physics MCQ Test 7

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Vessel A is filled with hydrogen while vessel B, whose volume is twice that of A, is filled with the same mass of oxygen at the same temperature. The ratio of the mean kinetic energies of hydrogen and oxygen is


Mean molecular energy 


At what temperature will the r.m.s. velocity of gas be half its value at 0°C ?


Since the r..m.s. velocity  we have  
Here T1 =273K and  So. we have, 


A gas has molar heat capacity C = 37.35J mol-1 k-1 in the process PT = constant. Find the number of degrees of freedom of molecules in the gas.


For one mole of an ideal gas,
PV = RT and pt = k (constant)
∴ Work done, when temperature changes from T to (T + dT),

NOW, dQ-dU + dW
CcTT - CvdT +2R dT
Hence, degree of freedom 


Calculate the difference of heat capacities of He 4 if the specific volume of it at 6 K and at a pressure of 19.7atm be 2.64* 10-2 m3 k mol-1. Given KT = 9.42* 10-8 m2N-1 and α = 5.35*10-2 K-1.


We have    4813 J.k mol-1 K-1


If 250 g of Ni at 120° C is dropped into 200 g of water at 100C contained by a calorimeter of 20 cal/0C heat capacity, what will be the final temperature of the mixture? (Given CNi = .106k cal/0C)


Heat lost by Ni = heat gained by water + Calorimeter 
We know heat capacity of Ni. CNi = 0.106 kcal/0

So. 0.250 (0.106) (1200 - t ) = 
3.18 - 0 .071 = 0.2201 - 2.20  
0.247t= 5.38


In the following indicator diagram, the net amount of work done will be


The cyclic process 1 is clockwise and the process 2 is anti clockwise. Therefore W1, will be positive and W2 will be negative area 2 > area 1, Hence the net work will be negative.
Hence the correct answer will be (b)


One mole of an ideal gas is takes from an initial state (P. V. T) to a final state (2P. 2V, 4T) by two different paths as shown in the Fig.A and B given above. If the changes in internal energy between the final and the initial states of the gas along the paths I and II are denoted by  respectively, then :


The change in internal energy dU is independent of the path. i.e.. if initial and final states of change are same then dU will be same.

Here, in both cases initial states and final states are respectively same. Hence. dU will be same in both ⇒ 


A carnot engine accepts 1000 calorie heat at 500K and give back 400 calorie heat to sink. What temperature is of sink?


Q1 =1000 calories
Q2 =400 calories
T= 500 k


If (1) represents isothermal and (2) represents adiabatic, which of the graphs given above in respect of an ideal gas are correct?


Slope of adiabatic curve = γ x slope of isothermal curve.

This is shown by every cuive given in question . Hence . all curves represent the ideal gas.


Consider a reversible isothermal expansion of a photon gas. Determine the entropy S for this gas at temperature T and volume V


The energy density of a photon gas is 
U = σT4
Thus the energy is

Now use the first law of thermodynamics
dE = dQ - dW
dQ = dE + dW = dE + pdV


Find Cp - Cv =


But pressure P is a function of T and V and dP is perfect differential, hence

If the change takes place at constant pressure. dP = 0. Then

Substituting this value in equation (1) above

Putting these values in equation (2), we get 


The Joule-Thomson coefficient and change in temperature during Joule- Thomson effect is given by 

which one is correct about αT to produce cooling and heating effect respectively?


dT = 
dP: represents the fall in temperature which is always negative.
Now dT is negative if αT > 1
So, cooling is produced
dT is positive if  or heating is produced.


A system of N localized, non-interacting spin 1/2 ions of magnetic moment μ each is kept in an external magnetic field H. If the system is in equilibrium at temperature T, then Helmholtz free energy of the system is


Magnetic moment for spin 1/2 system
∴ Partition function.     ...(ii)
from (1) and (2). we get

For N spins 
Helmholtz free energy, 


For a system comprising a very large number of particles N of similar subsystem . and each having partition function Z. if the system is to be considered non-interacting, then the partition function for the composite system is given by


Partition function

for composite system,
  Given Z1 = Z2
So, Z = Z2N


The change in vapour pressure of water as the boiling temperature changes from 100oC to 103oC. Given latent heat of steam = 540 cal./gm. and specific volume of steam = 1670 c.c./gm.


The Clapeyronss equation describing the rate of change of vapour pressure with temperature is

Here L = 540 cal. = 540 - 4.2 - 107 ergs., T = 100 + 273 = 373 K
dT = ( 103 - 100 ) = 3°, V2 = 1670 c.c./gm., V1= 1 c.c./gm. 
∴ Change in vapour pressure

= 0.17091 atmospheres.

*Multiple options can be correct

The diameter of 4He atom is 1 . One mole of the gas occupies 20 litres at 20 K. Given R = 8.4 J mol-1 K-1 and NA = 6 x 1023 mol-1 Then


From eq. we know that


= 7.5 x 10-7 m
The number of collisions per unit distance is the inverse of the mean free path.

The average speed of a molecule obeying maxwellian distribution is given by

Collision frequency is

The mean free time, or the average time between collisions is the inverse of the collision frequency and in this case will be equal to

*Multiple options can be correct

There are two vessels; each of them contains one mole of a monatomic ideal gas. Initial volume of the gas in each vessel is 8.3 * 10-3 m3 at 27oC. Equal amount of heat is supplied to each vessel. In one of the vessels the volume of the gas is doubled without change in internal energy, whereas the volume of the gas is held constant in the second vessel. The vessels are now connected to allow free mixing of the gas. (R = 8.3 J/mol K.) Then


According to 1st law of thermodynamics,
So for the vessel for which internal energy (and hence, temperature) remains constant

and for the vessel for which volume is kept constt., 
According to given problem 

Now when the free mixing of gases is allowed
u1+u2 = u
Here  and 
so 1 x 300 + 1 x 438.6 = 2T, i.e., T = 369.3 K
Further for the mixture from PV = μRT with V = V + 2V = 3V and μ = μ1 + μ2 = 2.
we have 

*Multiple options can be correct

One gram of water (1 cm3) becomes 1671 cm3 of steam when boiled at a constant pressure of 1 atm (1.013 * 105 Pa). The heat of vaporization at this pressure is Ly= 2.256 x 106 J/kg. Then


(a) The work done by the vaporizing water is 
W = p(V2 - V1)
= (1.013 x 105 Pa) (1671 x 10-6 m3 -1 x 10-6m3)
= 169 J
(b) The heat added to the water to vaporize it is 
Q = mLv = (10-3 kg) (2.256 x 106 J/kg) = 2256 J
(c) From the first law of thermodynamics, the change in internal energy is 
ΔU= Q — W = 2256 J - 169 J = 2087 J

*Multiple options can be correct

Which of the following relations are true?


(P,V,T,S) are four thermodynamic variables but only two of them are idependent variables while the other two may in be expressed as their function.
(i) Representing the pressure P as a function of twp omde[emdemt varoan;es S amd T i.e.
P = P (S,T)
If the process is isobaric i.e., change takes place at constant pressure, dP = 0. 

(ii) Representing P as a function of T and V i.e.
P = P (T,V)
But for isobaric process dP = 0. Hence


*Multiple options can be correct

Which of the following statements are correct?


The atoms or molecules having an even number of fundamental particles have integral spin value and are called Bosons. Similarly, the atoms or molecules having odd number of fundamental particles have half integral spin value and are called Fermions. Accordingly, the classification of particles is given below:
Proton, Neutron, electron and positron are fundamental particles having half spin value. Hence they all are Fermions.
Photon has spin 1 and hence it is Boson

Hydrogen-molecule = 
Lithium ion 

*Answer can only contain numeric values

At what temperature (in Kelvin) will the average speed of hydrogen molecules be the same as that of nitrogen molecules kept at 35oC. Molecular weights of nitrogen and hydrogen are 28 and 2 atomic mass units respectively. Given Boltzmann’s constraint = 1.38 * 10-23 joule per degree.


If  are the average speeds of hydrogen and nitrogen molecules at temperatures  T1K and T2K respectively, we have
where m1 and m2 are the masses of hydrogen and nitrogen molecules respectively.
Dividing eq. (1) and (2), we get
Given when  
Substituting these values in (3), we get

The most probable speed of nitrogen at temperature T is given by

Here k = 1.38 * 10-23 joule per degree, T = 22 K 
and m = 28 a.m.u. =
= 1.14 * 102m/s.

*Answer can only contain numeric values

At what temperature (in Kelvin) is the root mean square velocity equal to the escape velocity from the surface of the earth for hydrogen and for oxygen?


K.E. of a molecule of mass m and mean square speed 

Escape velocity.  where R is the radius of the earth. 
∴ K.E. of a molecule =  
Temperature for hydrogen

Temperature for oxvaen.

= 16 x 104K.

*Answer can only contain numeric values

Two glass bulbs of equal volumes are connected by a narrow tube and are filled with a gas at OoC and a pressure of 76 cm of Hg. One of the bulbs is then placed in a water bath maintained at 62oC . What is the new value of pressure (in cm) inside the bulbs in terms of Hg? The volume of connecting tube is negligible.


We have PV = nkT
As total number of air molecules in the bulbs would remain constant, we have
n1 + n2 = n1 + n2
As temperature of second bulb is unchanged, we have

This gives new pressure

Here P0 = 76 cm o f Hg,

 = 83.75 cm of Hg

*Answer can only contain numeric values

A vessel of volume V = 30 litre contains an ideal gas at temperature. T = 0°C. Keeping temperature constant, a part of the gas is allowed to escape from the vessel causing the pressure to fall down by ΔP = 0.78 atm. Find the mass of the gas released. Its density under normal conditions is p= 1.3 g/it


Let m and P be the initial mass and pressure of the gas inside the vessel. Therefore,
PV = (m/M) RT ...(1)
where M is the molecular weight of the gas in the vessel.
After a part of the gas is released , we have
where , m is the mass of the remaining gas in the vessel.
Hence, mass of the gas released is equal to (subtracting (2) from (1)

Now, under normal conditions (P0 = 1 atm. T = 273K), density of the gas is given  to be p . Therefore, we find.

or, M / RT = P/ PO
Now here  V = 30 * 10-3 m3

and, Po =1atm.

*Answer can only contain numeric values

A vertical hollow cylinder contains an ideal gas. The gas is enclosed by a 5 kg movable piston with area of cross section 5x10-3m2. Now the gas is heated from 300 K to 350K and the piston rises by 0.1 m. The piston is now clamped at this position and the gas is cooled back to 300K. Find the difference between the heat energy added during the heating process and the energy lost during the cooling process. (1 atm = 105 N/m2)


Initially when the gas is heated, pressure remains constt.
SO  ...(i)
Now when the piston is clamped, volume becomes constt. 
So heat withdrawn ..(ii)
So the difference between heat added and removed,

But here 
i.e., Ans.

*Answer can only contain numeric values

During an integral number of complete cycles, a reversible engine (shown by a circle) absorbs 1200 joules from reservoir at 400 K and performs 200 joules of mechanical work, find the sum of quantities of heat (in Joules) exchanged with the other two reservoirs.


By conservation of energy Q1 = W + Q2 + Q3
i.e., Q2 + Q3 = Q1 - W = 1200 - 200 = 1000 (1)
And as change in entropy in a reversible-process is zero
i.e., 2Q2 + 3Q3 = 1800 (2)
Solving equation (1) and (2) for Q2 and Q3, we get 
Q2 = 1200 J and Q3 = -200 J
i.e.. the reservoir at temperature T2 absorbs 1200 J of heat while the reservoir at temperature T3 lose 200 J of heat. Hence their sum will be 1400 J

*Answer can only contain numeric values

Suppose 1.00 kg of water at 1000C is placed in thermal contact with 1.00 kg of water at O0C. What is the total change in entropy (in J/K)? Assume that the specific heat of water is constant at 4190 J/kgk over this temperature range.


This process involves irreversible heat flow because of the temperature differences.
Let final temp is 

So. the final temperature is 500C = 323 K. The entropy change of the hot water is

The entropy change of the cold water is

The total entropy change of the system is

An irreversible heat flow in an isolated system is accompanied by an increase in entropy. We could have reached the same end state by simply mixing the two quantities of water. This. too. is an irreversible process; because the entropy depends only on the state of the system . the total entropy change would be the same, 102J/K.

*Answer can only contain numeric values

Calculate the critical temperature (in °C) of a gas for which the Vander Waal’s constants are a = 0.00374. b = 0.0023 and the gas constant is given by 273 R = 1.00646.


The critical temperature is given by

given a = 0.00874, b = 0.0023 ..(1)
Substituting these values in equation (1). we get

= 305.50C.

*Answer can only contain numeric values

Calculate the pressure (in atmospheres) required to make water freeze at - 10C. Change of specific volume when 1 gm of water freezes into ice = .091 cc.,J =4.2 x 107 ergs/cal..1 atmosphere 106 dynes / cm2 and the latent heat of ice = 80 cal./ gm.


Given dT = 10C . V2 - V1 = -0.091 c.c.
L = 80 cal. = 80 * 4.2 * 107 ergs, T = 0 + 273 = 273 K 
Applying Clapeyrons latent heat equation

or  dynes/cm2.

= 135.2 Atmos.
Hence to lower the melting of ice by 1°C, the pressure must be raised by 135.2 atmospheres.
Now melting point of ice at atmospheric pressure is 0°C. Hence pressure required to make ice freeze at - 1 °C 
= 135.2 + 1 = 136.2 atmospheres

*Answer can only contain numeric values

A weightless piston divides a thermally insulated cylinder into two parts of volumes V and 3V. 2 moles of an ideal gas at pressure P = 2 atmosphere are confined to the part with volume V = 1 litre. The remainder of the cylinder is evacuated. Initially the gas is at room temperature. The piston is now released and the gas expands to fill the entire space of the cylinder. The piston is then pressed back to the initial position. Find the increase of internal energy (in Joule) in the process. The ratio of the specific heat of the gas γ = 1.5.


As the expansion is against zero pressure, so no work is done. Now, for given mass of gas,
PV = P1V1
= P1 x 4V
∴ P1 = P/4 is the pressure of gas after expansion.
In the second process, compression is adiabatic, so 

Change in internal energy = work done in adiabatic compression.

= 8P1V = 2PV
= 2 * 2 * 105 * 10-3 = 400J
= 12.05
∴ Final temperature = T1 + (T2 - T2)
= 300 + 12.05 = 312.05 K

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