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QUESTION: 1

The latent heat of water diminishes by 0.695 cal./gm. for each degree centigrade rise in temperature at temperature 100^{0 }and latent of water vapour at 100°C, is 540 cal.gm. The specific heat for saturated steam is

Solution:

By second latent heat equation, the specific heat C_{2 }of saturated steam is given by

Here

T = 100^{0}C = 373K

L = 540 cal./gm.

and C_{1 }= specific heat in liquid state (water)

= 1.0 cal./gm.

= 1 - .695 - 1.448

= -1.143 cal/gm.^{0}C.

QUESTION: 2

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

Solution:

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.

QUESTION: 3

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?

Solution:

By the problem

dp = ( 1000 - 1) atoms = 999 * 1.03 * 10^{6} dyne / cm^{2}

L = 2.33 cal = 2.33*4.2*10^{7} erg

v_{l} = (1/13.59 ) c.c. = 0.073 c.c; v_{s} = (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 .74^{0}C

∴ Required melting point = ( - 38 .87 - 265.74 )°C = - 304.61 °C

QUESTION: 4

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

Solution:

From Clausius Clapeyron equation

or

Here, dP = 1 atm osphere = 1.1013 * 10^{6} dynes/cm^{2}

T = 0°C = 273K, V_{2} - V_{1} =0.0907 c.c.

dT = 0.0074^{0}C = 0.0074 K

3.39 * 10^{9} ergs/gm.

= 81.1cal/gm.

QUESTION: 5

A system of non-interacting fermi particles with fermi energy ε_{f} 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

Solution:

n(ε) dε =

At T = 0 K, all non-integrating particles have energy < ε_{r}

So,

NOW,

So,

Then average energy per particle at T = 0 K.

QUESTION: 6

The coordinates of the triple point of water are t = 0.0075^{o}C and p = 0.0060 atmosphere. Calculate the slope of the ice line in atoms /^{o}C.

Solution:

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 ' = 0^{o}C

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/^{0}C.

QUESTION: 7

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=ε_{f} is

Solution:

T = 0 K and

Probability

QUESTION: 8

Fermi energy of conduction electron in silver is 5.48 eV. Then the number of such electrons per cm^{3 }given that h = 6.62 x 10^{-2} erg-sec. and 1 eV = 1.6 * 10^{-12} erg will be

Solution:

Fermi energy,

So,

Therefore,

Number of electrons per cm^{3}

QUESTION: 9

A mass m of water at temperature T_{1} K is isobarically and adiabatically mixed with an equal mass of water at T_{2} K. The entropy change of the inverse is

Solution:

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

Now,

and

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

⇒

Since,

and

Total entropy of entropy change of universe

QUESTION: 10

1 g of perfect gas at volume v_{ A}, pressure p_{A} and temperature T_{A} changes from state A to state B when volume is V_{0}, pressure is p_{B} and temperature T_{B}. The change in entropy is

Solution:

From first law of thermodynamics

dQ = dU + dW

= C_{v}dT + pdV

Change in entropy

pV = RT

or

∴

QUESTION: 11

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

Solution:

QUESTION: 12

A flask containing 1 mole of oxygen gas is known to contain a mixture of O and O_{2} molecules. The specific heat of the gas is found to be 17/10 R, where R is the gas constant. The number n_{1} of O_{1} atoms and the number n_{2 }of O_{2} molecules are in the ratio (n_{1}: n_{2})

Solution:

Let N_{1} and N_{2} are the number o f moles of O atoms andO_{2} molecules

Where C_{v1} and C_{v2} are the specific heats o f O atoms and O_{2} molecules

Number ratio is same as the mole ratio

So, n_{1} : n_{2 }= 4 :1

QUESTION: 13

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

Solution:

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.

QUESTION: 14

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

Solution:

Heat generated during work done against friction

= KE possessed by metal block

QUESTION: 15

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

Solution:

Ideal gas equation

PV = nRT

So,

=

*Multiple options can be correct

QUESTION: 16

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

Solution:

(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

QUESTION: 17

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

Solution:

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

QUESTION: 18

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

Solution:

∴

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 ]

∴

Hence,

Thus both Fermi-Dirac and Bose-Einstein statistics approach Maxwell

*Multiple options can be correct

QUESTION: 19

If are unit vectors, then the value of is possible

Solution:

We know that

⇒

Now

≤

= 9

i.e.

*Multiple options can be correct

QUESTION: 20

If the scalar function is f = x^{2 }y z^{3} than at the point (2, 1,-1)

Solution:

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

QUESTION: 21

1 m^{3} 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

Solution:

At point a P_{a} = 100 kPa

T_{a} = 300 K

At point b P_{b} = 200 kPa

Since, ab when produced passes through origin.

So, For

W_{ab} = Area of trapezium under ab =

= 1/2 x 3 x 10^{-3} x 300 x 10^{3} = 450 R Joule

w_{bc} = o

*Answer can only contain numeric values

QUESTION: 22

Two thermally isolated systems have heat capacities which vary as C_{v} = βT^{2} (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.

Solution:

Let the final temperature of combined system is T

Heat lost = Heat gained

T^{3} - 600^{3} = 300^{3} - T^{3}

2T^{3} = 300^{3} + 600^{3}

2T^{3} = 243 x 10^{6}

T = 495 K

*Answer can only contain numeric values

QUESTION: 23

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.

Solution:

Heat absorbed Q_{1} = 23400 J

Heat rejected Q_{2} = 20400 J

Work done W = Q_{1} - Q_{2} = 3000 J

*Answer can only contain numeric values

QUESTION: 24

1 Kg of a liquid (specific heat = 2000 JK^{-1}kg^{-1}), independent of temperature, is heated from 300 K to 600 K by either of the following process P_{1 }:

P_{1}: bringing it in contact with a reservoir at 600 K.

Calculate the change in the entropy (in J/K) of the universe in process P_{1.}

Solution:

*Answer can only contain numeric values

QUESTION: 25

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.

Solution:

Efficiency

Work output

In refrigeration Q_{1} + W = Q_{2}

Work input = 60% of work output of heat engine

Work input = 0.6 * 350 = 210 J

Heat removed Q_{1} = 1050 J

Heat rejected Q_{2} = Q_{1} + W = 1050 + 210 = 1260J

Now,

*Answer can only contain numeric values

QUESTION: 26

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_____

Solution:

Q = 4000J

*Answer can only contain numeric values

QUESTION: 27

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____ .

Solution:

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

QUESTION: 28

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)

Solution:

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 = MC_{v}dT = 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

QUESTION: 29

find the rate of change of in the direction normal to the surface x^{2}y + y^{1}x+yz^{2} = 3 at the point (1,1,1)

Solution:

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

QUESTION: 30

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

Solution:

Here ,

unit veetor =

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

Greatest rate of interest of

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