IIT JAM Physics MCQ Test 8

30 Questions MCQ Test Mock Test Series for IIT JAM Physics | IIT JAM Physics MCQ Test 8

This mock test of IIT JAM Physics MCQ Test 8 for Physics helps you for every Physics entrance exam. This contains 30 Multiple Choice Questions for Physics IIT JAM Physics MCQ Test 8 (mcq) to study with solutions a complete question bank. The solved questions answers in this IIT JAM Physics MCQ Test 8 quiz give you a good mix of easy questions and tough questions. Physics students definitely take this IIT JAM Physics MCQ Test 8 exercise for a better result in the exam. You can find other IIT JAM Physics MCQ Test 8 extra questions, long questions & short questions for Physics on EduRev as well by searching above.

The latent heat of water diminishes by 0.695 cal./gm. for each degree centigrade rise in temperature at temperature 100and latent of water vapour at 100°C, is 540 cal.gm. The specific heat for saturated steam is


By second latent heat equation, the specific heat Cof saturated steam is given by

T = 1000C = 373K 
L = 540 cal./gm.
and C= specific heat in liquid state (water) 
= 1.0 cal./gm.

= 1 - .695 - 1.448 
= -1.143 cal/gm.0C.


The portion AB of the indicator diagram representing the state of matter denotes


The volume of matter in portion AB of the curve is almost constant and pressure is decreasing. These are the characteristics of the liquid state.


Mercury melts at-38.87°C at 1 atmosphere pressure and it's density is 13.59 g/ c.c. The density of the solid is 14.19 g/c.c. The heat of fusion is 2.33 cal/g. What would be the melting point of mercury at 1000 atmosphere?


By the problem
dp = ( 1000 - 1) atoms = 999 * 1.03 * 106 dyne / cm2 
L = 2.33 cal = 2.33*4.2*107 erg
vl = (1/13.59 ) c.c. = 0.073 c.c; vs = (1 /14.19) c.c = 0.070 c.c 
T = - 38.87°C = (- 38.87 + 273)K = 234.13K
From Clapeyion s equation . the change in the melting point 
  265 .740C
∴ Required melting point = ( - 38 .87 - 265.74 )°C = - 304.61 °C


Calculate the latent heat of ice given that change of pressure by one atmosphere changes the melting point of ice by 0.0074oC and when one gm. of ice melts its volume changes by 0.0907 c.c.


From Clausius Clapeyron equation

Here, dP = 1 atm osphere = 1.1013 * 106 dynes/cm2 
T = 0°C = 273K, V2 - V1 =0.0907 c.c. 
dT = 0.00740C = 0.0074 K
  3.39 * 109 ergs/gm.  
= 81.1cal/gm.


A system of non-interacting fermi particles with fermi energy εf has the density of states proportional to √ε where ε is the energy of the particle. Then the average energy per particle at temperature T = 0 K is


n(ε) dε =  
At T = 0 K, all non-integrating particles have energy < εr
Then average energy per particle at T = 0 K.


The coordinates of the triple point of water are t = 0.0075oC and p = 0.0060 atmosphere. Calculate the slope of the ice line in atoms /oC.


Let O be the triple point of water. Then the coordinates of point O are 
P = 0.0060 atoms, t = 0.0075°C.
Now we know that the freezing point of water into ice at 1 atmosphere pressure is 0°C. Hence the coordinates of point D on the ice line are 
P' = 1 atoms., t ' = 0o
dP = P'- P = 1 - 0.006 = 0.994 atmos.

and dT = dt = t'— 1 = 0 — 0.0075 = - 0. 0075:C 
Therefore, the slope of ice line is

= - 132.5 atmos/0C.


Fermi function  gives the probability of occupation of electrons per energy state. Then the probability of number of electrons at absolute temperature (T = 0 K) when ε=εf is


 T = 0 K and  



Fermi energy of conduction electron in silver is 5.48 eV. Then the number of such electrons per cm3 given that h = 6.62 x 10-2 erg-sec. and 1 eV = 1.6 * 10-12 erg will be


Fermi energy, 
Number of electrons per cm3 


A mass m of water at temperature T1 K is isobarically and adiabatically mixed with an equal mass of water at T2 K. The entropy change of the inverse is


A point to note is that, since the process is adiabatic there is no change of the surroundings. We know that for isobaric (constant pressure process), the change in entropy is given by

Also, the higher temperature loses heat and the water at the lower temperature gains heat, we can write

Total entropy of entropy change of universe



1 g of perfect gas at volume v A, pressure pA and temperature TA changes from state A to state B when volume is V0, pressure is pB and temperature TB. The change in entropy is


From first law of thermodynamics 
dQ = dU + dW 
= CvdT + pdV
Change in entropy 

pV = RT


A body P(temperature Tp) has twice the mass and twice the specific heat compared to that of the body Q (temperature TQ). If the bodies are supplied equal amount of heat, the relationship between their resulting temperature changes (Δ Tp and Δ TQ) will be



A flask containing 1 mole of oxygen gas is known to contain a mixture of O and O2 molecules. The specific heat of the gas is found to be 17/10 R, where R is the gas constant. The number n1 of O1 atoms and the number nof O2 molecules are in the ratio (n1: n2)


Let N1 and N2 are the number o f moles of O atoms andO2 molecules

Where Cv1 and Cv2 are the specific heats o f O atoms and O2 molecules

Number ratio is same as the mole ratio 
So, n1 : n2 = 4 :1


A heating element weighing 100 g made of a material with specific heat of 1J/ (gC) is exposed to ambient air at 30°C and attains a steady-state temperature of 130°C while absorbing 100 W of electric power. Neglecting radiation, the time taken for the heating element to cool to 129° C after the power is switched off is most nearly equal to


100 W is required to maintain 130°C temperature. It means it lost 100J of heat in 1 second which is compensated by electric power.
Heat lost by heating element to cool to 129°C. = mCΔT = 100 x 1 x = 100 J
So , power is switched off for 1 second.


A block made of a metal with specific heat C resting on a horizontal wooden board is imparted a speed v by an impulse. The metal block comes to rest due to friction with the wooden board (coefficient of kinetic friction μ). If all the heat generated by the friction is absorbed by the metal block, its temperature rises by


Heat generated during work done against friction 
= KE possessed by metal block


Two vessels separately contain two ideal gases A and B at the same temperature, the pressure of A being twice that of B. Under these conditions, the density of A is found to be one and halftimes the density of B. The ratio of molecular weights of A and B is


Ideal gas equation
PV = nRT 


*Multiple options can be correct

An ideal gas completes a cycle consisting of two isochores as shown in fig. Which one of the following are incorrect?


(i) Process 1 - 2 is isothermal and heat is absorbed so its entropy increases
(ii) In process 4-1, heat is evolved so entropy increases

*Multiple options can be correct

Which one among the following are false for Helmholtz free energy when the process is isothermal?


F = U - TS
dF = dU - TdS - SdT
dT = 0 for isothermal process
dF = dU - TdS  
= dU - dQ = - PdV [Since dQ = dU + PdV ]
dF = - dW 
So, dW = - dF

*Multiple options can be correct

At higher temperature, which one of the following statement are false?


 for M.B 
 for B.E 
 for F.B and 
At higher temperature

[For higher T, terms  and higher terms can be neglected.]

 [Since, eα vary at higher temperature ]
 [factor T is neglected Since ]
Thus both Fermi-Dirac and Bose-Einstein statistics approach Maxwell

*Multiple options can be correct

If  are unit vectors, then the value of  is possible


We know that 


= 9

*Multiple options can be correct

If the scalar function is f = x2 y z3 than at the point (2, 1,-1)


at the point (2, 1,-1) 
grad f = 
Hence the direction of the maximum directional derivative at the point (2, 1,-1) 

Again the magnitude of the directional derivative at the point (2, 1,-1)

*Answer can only contain numeric values

1 m3 of an ideal gas with v = 1.4 is at a pressure of 100 kPa and a temperature 300 K. Initially the state of the gas is at the point a of the PV diagram shown. The gas is taken through a reversible cycle a → b → c → a. The pressure at point b is 200 kPa and the line ba, when extended , passes through the origin.

The work done by the gas in each steps are  and  Calculate the value of n if 


At point a Pa = 100 kPa 
Ta = 300 K
At point b Pb = 200 kPa 

Since, ab when produced passes through origin. 
So, For 

Wab = Area of trapezium under ab = 
= 1/2 x 3 x 10-3 x 300 x 103 = 450 R Joule 
wbc = o

*Answer can only contain numeric values

Two thermally isolated systems have heat capacities which vary as Cv = βT2 (where β > 0). Initially one system is at 300 K and the other at 600 K. The systems are then brought into thermal contact and the combined system is allowed to reach thermal equilibrium. Then calculate the final temperature (in Kelvin) of the combined system.


Let the final temperature of combined system is T 
Heat lost = Heat gained

T3 - 6003 = 3003 - T3
2T3 = 3003 + 6003 
2T3 = 243 x 106
T = 495 K  

*Answer can only contain numeric values

The P-V-diagram below represents an ideal monoatomic gas cycle for 1 mole of a gas. Calculate the work done (in Joule) during the cycle.


Heat absorbed Q1 = 23400 J
Heat rejected Q2 = 20400 J 
Work done W = Q1 - Q2 = 3000 J

*Answer can only contain numeric values

1 Kg of a liquid (specific heat = 2000 JK-1kg-1), independent of temperature, is heated from 300 K to 600 K by either of the following process P1 :
P1: bringing it in contact with a reservoir at 600 K.
Calculate the change in the entropy (in J/K) of the universe in process P1.


*Answer can only contain numeric values

A carnot engine operating between two temperatures 727°C and 27°C is supplied heat energy at the rate of 500 joule/cycle. 60% of the work output is used to derive a refrigerator, which rejects heat to the surrounding at 27°C. If the refrigerator removes 1050 Joule of heat per cycle from the low temperature reservoir, determine the temperature (in Kelvin) of reservoir.



Work output 
In refrigeration Q1 + W = Q2
Work input = 60% of work output of heat engine
Work input = 0.6 * 350 = 210 J
Heat removed Q1 = 1050 J
Heat rejected Q2 = Q1 + W = 1050 + 210 = 1260J

*Answer can only contain numeric values

Consider a car not engine operating between temperatures of 800 and 400 k. The engine perform 2000 J of work per cycle. The heat extracted per cycle from the high temperature reservoir is_____


Q = 4000J

*Answer can only contain numeric values

A given quantity of gas is taken from the state A → C reversibly, by two paths A → C directly and A → B → C as shown in the figure

During the process A→ C the work done by the gas in 200J and the heat absorbed is 300 J. If during the porocess A → B → C the work done by the gas is 50J , the heat absorbed is____ .


During path AC
dU = dQ - dW = 300 - 200 = 100 J
Hence internal energy is point function dU will same in all path 
In path ABC dQ = dU + dW 
dQ = 100 + 50 = 150 J

*Answer can only contain numeric values

30 gm of ice at 0°C is added to a beaker containing 60gm of water at 35°C. What is the final temperature of the system when it comes to internal equailibrium? (The specfic heat of water is 1 cal / gm°c and latent heat of melting of ice is 80 cal / gm)


The amount of heat required to melt the ice of mass 10gm at 0°C is 
Q = ML = 30X 80 = 2 4 0 0 cal.
Where L is the latent heat of melting of ice and m is the mass of the ice. The amount of heat availabel in water of mass 60gm at 35°C is Q = MCvdT = 60 x 1 x 35 = 2100 cal. Since the heat available is less than the heat required to melt the ice therefore ice will not melt as a result the temperature of the system will be at 0°C.

*Answer can only contain numeric values

find the rate of change of  in the direction normal to the surface x2y + y1x+yz2 = 3 at the point (1,1,1)


Rate of change of 
Rate of change o f at (1,1,1,) = 
Normal to the surface 

Unit normal = 
Required rate of change of 
Required rate of change of 

*Answer can only contain numeric values

find the directional derivative of  in the direc­tion of  Find the the greatest rate of incrase of 


Here ,  

 unit veetor = 
So, the required directional derivative at (1.-2.1)

Greatest rate of interest of 

Related tests