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QUESTION: 1

N molecules each of mass m of a gas A and 2N molecules each of mass 2m of gas B are contained in the same vessel which is maintained at temperature T. The mean square velocity of molecules of B type is v^{2} and the mean square rectangular component of the velocity of A type is denoted by ω^{2}. Then ω^{2}/v^{2} [1991]

Solution:

Mean kinetic energy of the two types of molecules should be equal. The mean square velocity of A type molecules = ω^{2} + ω^{2} + ω^{2} = 3ω^{2}

Therefore,

This gives ω^{2} / v^{2 }= 2/3

QUESTION: 2

10 gm of ice cubes at 0°C are released in a tumbler (water equivalent 55 g) at 40°C.Assuming that negligible heat is taken from the surroundings, the temperature of water in the tumbler becomes nearly (L = 80 cal/g) [1988]

Solution:

Let the final temperature be T Heat gained by ice = mL + m × s × (T – 0) = 10 × 80 + 10 × 1 × T

Heat lost by water = 55 × 1× (40 – T)

By using law of calorimetery, 800 + 10 T = 55 × (40 – T)

⇒ T = 21.54°C = 22°C

QUESTION: 3

A centigrade and a Fahrenheit thermometer are dipped in boiling water. The water temperature is lowered until the Fahrenheit thermometer registers 140°. What is the fall in temperature as registered by the Centigrade thermometer ?

Solution:

Using

Temperature of boiling water = 100°C

We get, fall in temperature = 100 – 60 = 40°C

QUESTION: 4

Thermal capacity of 40 g of aluminium (s = 0.2 cal /g K) is [1990]

Solution:

Thermal capacity = ms = 40 × 0.2 = 8 cal/°C = 4.2 × 8 J = 33.6 joules/°C

QUESTION: 5

Mercury thermometer can be used to measure temperature upto [1992]

Solution:

Mercury thermometer is based on the principle of change of volume with rise of temperature and can measure temperatures ranging from –30°C to 357°C.

QUESTION: 6

If the temperature of the sun is doubled, the rate of energy received on earth will be increased by a factor of [1993]

Solution:

Amount of energy radiated ∝ T^{4}

QUESTION: 7

A body cools from 50.0°C to 48°C in 5s. How long will it take to cool from 40.0°C to 39°C?Assume the temperature of surroundings to be 30.0°C and Newton's law of cooling to be valid.

Solution:

Rate of cooling ∝ temperature difference

between system and surrounding.

As the temperature difference is halved, so the rate of cooling will also be halved.

So time taken will be doubled

t = 2 × 5 sec. = 10 sec.

QUESTION: 8

A beaker full of hot water is kept in a room. If it cools from 80°C to 75°C in t_{1} minutes, from 75° C to 70°C in t_{2} minutes and from 70°C to 65°C in t_{3} minutes, then [1995]

Solution:

Let θ_{0} be the temperature of the surrounding.

Then

… (1)

Similarly, …(2)

and … (3)

From (1), (2) & (3), it is obvious that

t_{1} < t_{2} < t_{3}

QUESTION: 9

A black body is at temperature of 500 K. It emits energy at rate which is proportional to [1997]

Solution:

According to Stefan's Law

QUESTION: 10

The radiant energy from the sun, incident normally at the surface of earth is 20 k cal/m^{2} min. What would have been the radiant energy, incident normally on the earth, if the sun had a temperature, twice of the present one? [1998]

Solution:

According to Stefan’s law

E ∝ T^{4}

⇒ E_{2} = 320 kcal/m^{2}. min.

QUESTION: 11

If 1 g of steam is mixed with 1 g of ice, then the resultant temperature of the mixture is [1999]

Solution:

Heat required by ice a t 0° C to reach a temperature of 100°C = mL + mcΔθ = 1 × 80 + 1 × 1 × (100 – 0) = 180 cal

Heat available with 1 g steam to condense into 1 g of water at 100°C = 536 cal.

Obviously the whole steam will not be condensed and ice will attain a temperature of 100°C; so the temperature of mixture = 100°C.

QUESTION: 12

The coefficient of linear expansions of brass and steel are α_{1} and α_{2} respectively. When we take a brass rod of length ℓ_{1} and a steel rod of length ℓ_{2} at 0ºC, then the difference in their lengths (ℓ_{2} – ℓ_{1}) will remain the same at all temperatures if [1995, 1999]

Solution:

Let ΔT be increase in the temperature of brass wire.Then length of brass wire,

Similarly, length of steel wire when temperature is increased by According to question , l' – l'' = l_{1} – l_{2}

QUESTION: 13

The presence of gravitational field is required for the heat transfer by [2000]

Solution:

In convection, the temperature gradient exists in the vertical direction and not in the horizontal direction. So, up and down movement of particles takes place which depends on the weight and gravity.

QUESTION: 14

A black body has maximum wavelength λ_{m} at temperature 2000 K. Its corresponding wavelength at temperature 3000 K will be

Solution:

According to Wein's displacement law,

λ_{m}T = 2.88 × 10^{–3} When T = 2000 K,

λ_{m} (2000) = 2.88 × 10^{–3} ....(1)

When T = 3000 K, λ'_{m} (3000) = 2.88 × 10^{–3} ....(2)

Dividing (1) by (2),

QUESTION: 15

A cylindrical rod having temperature T_{1} and T_{2} at its end. The rate of flow of heat is Q_{1} cal/sec.If all the linear dimensions are doubled keeping temperature constant, then the rate of flow of heat Q_{2} will be [2001]

Solution:

Rate of heat flow

Dimensions of area A = [L^{2}], dimensions of distance l = [L]

QUESTION: 16

Wien's law is concerned with [2002]

Solution:

According to Wein's displacement law, product of wavelength belonging to maximum intensity and temperature is constant i.e., λ_{m}T = constant.

QUESTION: 17

Radiation from which of the following sources, approximates black body radiation best? [2002]

Solution:

This hole in the cavity is also called Fery's black body, it is made such that when light enters the cavity it suffers multiple reflection inside it and since with every reflection some part is absorbed so almost all the radiant light get absorbed.

QUESTION: 18

Two rods of thermal conductivities K_{1} and K_{2}, cross-sections A_{1} and A_{2 }and specific heats S_{1 }and S_{2} are of equal lengths. The temperatures of two ends of each rod are T_{1} and T_{2}. The rate of flow of heat at the steady state will be equal if [2002]

Solution:

Rate of heat flow for one rod

.Rate of heat flow for other rod

In steady state,

QUESTION: 19

Consider a compound slab consisting of two different materials having equal thicknesses and thermal conductivities K and 2K, respectively.

The equivalent thermal conductivity of the slab is [2003]

Solution:

In series, equivalent thermal conductivity

QUESTION: 20

If λ_{m} denotes the wavelength at which the radiative emission from a black body at a temperature T K is maximum, then [2004]

Solution:

From Wein’s displacement law

QUESTION: 21

Which of the following circular rods (given radius r and length l), each made of the same material and whose ends are maintained at the same temperature will conduct most heat? [2005]

Solution:

We know that

Also, Thermal resistance

Heat flow will be maximum when thermal resistance is minimum.

From given option

It is clear that for option (2) resistance is minimum, hence, heat flow will be maximum. [**Alt **: (i) Rate of heat flow is directly proportional to area

(ii) inversely proportional to length.

∴ Heat flow will be maximum when r is maximum and ℓ is minimum. ]

QUESTION: 22

The temperature of in version of a thermocouple is 620^{0}C and the neutral temperature is 300^{0}C. What is the temperature of cold junction ? [2005]

Solution:

QUESTION: 23

A black body at 1227°C emits radiations with maximum intensity at a wavelength of 5000Å. If the temperature of the body is increased by 1000°C, the maximum intensity will be observed at

Solution:

Applying Wein's displacement law, λ_{m}T = constant

= ( 5000 * 1500 ) / 2500 = 3000 Å

QUESTION: 24

A black body is at 727° C. It emits energy at a rate which is proportional to [2007]

Solution:

According to Stefan's law, E ∝ T^{4}

∝ (t + 273)^{4} K [727°C = (727 + 273)K]

∝ (727 + 273)^{4} K

∝ (1000)^{4} K

QUESTION: 25

If the cold junction of a thermo-couple is kept at 0°C and the hot junction is kept at T°C then the relation between neutral temperature (T_{n}) and temperature of inversion (T_{i}) is [2007]

Solution:

Since = Neutral temperature

[T_{c }= 0°C = temperature of cold junction]

QUESTION: 26

Assuming the sun to have a sph erical outer surface of radius r, radiating like a black body at temperature t°C, the power received by a unit surface, (normal to the incident rays) at a distance R from the centre of the sun is [2007] where σ is the Stefan’s constant.

Solution:

Power radiated by the sun at t°C

= σ(t + 273)^{4}4πr^{2}

Power received by a unit surface

QUESTION: 27

On a new scale of temperature (which is linear) and called the W scale, the freezing and boiling points of water are 39° W and 239° W respectively. What will be the temperature on the new scale, corresponding to a temperature of 39° C on the Celsius scale? [2008]

Solution:

For different temperature scales, we have

= constant

Where L.F.P ⇒ Lower Fixed point

U.H.F. ⇒ Upper fixed point

where x is the measurement at that scale.

Here, if C and W be the measurements on Celsius and W scale then,

= 117° W

QUESTION: 28

An electric kettle takes 4A current at 220 V. How much time will it take to boil 1 kg of water from temperature 20° C? The temperature of boiling water is 100° C. [2008]

Solution:

Heat required to raise the temperature of 1kg water from 20°C to 100°C is given by Q = msΔθ = (1× 4200 × 80) J

Power of kettle (P) = VI = (220 × 4)W

∴ Time taken =

= 381.81 sec = 6.36 min

QUESTION: 29

The two ends of a rod of length L and a uniform cross-sectional area A are kept at two temperatures T_{1} and T_{2} (T_{1} > T_{2}). The rate of heat transfer, through the rod in a steady state is given by: [2009]

Solution:

[(T_{1}–T_{2}) is the temperature difference]

QUESTION: 30

A black body at 227°C radiates heat at the r ate of 7 cals/cm^{2}s. At a temperature of 727°C, the rate of heat radiated in the same units will be:

Solution:

Accordin g to Stefan’s law E = σT^{4},

T_{1} = 500 K

T_{2 }= 1000 K

∴ E_{2} = 16 × 7 = 112 cal / cm^{2}s

QUESTION: 31

A cylindrical metallic rod in therrnal contact with two reservoirs of heat at its two ends conducts an amount of heat Q in time t. The metallic rod is melted and the material is formed into a rod of half the radius of the original rod. What is the amount of heat conducted by the new rod, when placed in thermal contact with the two reservoirs in time t? [2010]

Solution:

The rate of heat flow is given by

Area of Original rod A = πR^{2} ;

Areal of new rod A' =

Volume of original rod will be equal to the volume of new rod.

QUESTION: 32

The total radiant energy per unit area, normal to the direction of incidence, received at a distance R from the centre of a star of radius r, whose outer surface radiates as a black body at a temperature T K is given by: [2010]

Solution:

QUESTION: 33

If the radius of a star is R and it acts as a black body, what would be the temperature of the star, in which the rate of energy production is Q ? [2012] (σ stands for Stefan’S constant)

Solution:

Stefan’s law for black body radiation Q = σe AT^{4}

Here e = 1

A = 4πR^{2}

QUESTION: 34

Liquid oxygen at 50 K is heated to 300 K at constant pressure of 1 atm. The rate of heating is constant. Which one of the following graphs represents the variation of temperature with time? [2012]

Solution:

Initially liquid oxygen will gain the temperature up to its boiling temperature then it change its state to gas. After this again its temperature will increase, so corresponding graph will be

QUESTION: 35

A slab of stone of area 0.36 m^{2} and thickness 0.1 m is exposed on the lower surface to steam at 100°C. A block of ice at 0°C rests on the upper surface of the slab. In one hour 4.8 kg of ice is melted. The thermal conductivity of slab is : (Given latent heat of fusion of ice = 3.36 × 105 Jkg^{–1}.) : [2012M]

Solution:

Rate of heat given by steam = Rate of heat taken by ice where

K = Thermal conductivity of the slab

m = Mass of the ice

L = Latent heat of melting/fusion

A = Area of the slab

K =1.24 J/m/s/°C

QUESTION: 36

A piece of iron is heated in a flame. It first becomes dull red then becomes reddish yellow and finally turns to white hot. The correct explanation for the above observation is possible by using [NEET 2013]

Solution:

Wein’s displacement law According to this law

or, λ_{max} × T = constant

So, as the temperature increases λ decreases.

QUESTION: 37

The molar specific heats of an ideal gas at constant pressure and volume are denoted by C_{p} and C_{v}, respectively. If and R is theuniversal gas constant, then C_{v} is equal to [NEET 2013]

Solution:

QUESTION: 38

The density of water at 20°C is 998 kg/m^{3} and at 40°C 992 kg/m^{3}. The coefficient of volume expansion of water is [NEET Kar. 2013]

Solution:

From question, Δρ = (998 – 992) kg/m^{3} = 6 kg/m^{3}

= 995 kg/m^{3}

∴ Coefficient of volume expansion of water,

QUESTION: 39

Two metal rods 1 and 2 of same lengths have same temperature difference between their ends.Their thermal conductivities are K_{1} and K_{2} and cross sectional areas A_{1} and A_{2}, respectively. If the rate of heat conduction in rod 1 is four times that in rod 2, then [NEET Kar. 2013]

Solution:

Q_{1} = 4Q_{2} (Given)

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