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201.

A mercury thermometer is to be made with glass tubing of internal bore 0.5 mm diameter and the distance between the fixed point is to be 20 cm. Estimate the volume of the bulb below the lower fixed point, given that the coefficient of cubical expansion of mercury is `0.00018//K`. and the coefficient of linear expansion of glass is `0.000009//K`.

Answer» Let V be the required volume,
Apparent increase in volume `=vxxgamma_(alpha)xx100cm^3`
or `Vxx(0.00018-0.000027)xx100`
`=pixx625xx10^-6xx20`
or `V=(pixx625xx10^-6xx20)/(1.53xx10^-4)=2.57cm^3`
202.

Statement I: The bulb of one thermometer is spherical while that of the other is cylindrical . Both have equal amounts of mercury. The response of the cylindrical bulb thermometer will be quicker. Statement II: Heat conduction in a body is directly proportional to cross-sectional area.A. Statement I is true, Statement II is true and Statement II is the correct explanation for statement I.B. Statement I is true, statement II is true and statement II NOT the correct explanation for Statement IC. Statement I is true, Statement II is false.D. Statement I is false, statement II is true.

Answer» Correct Answer - A
Both are true and Statement II explains Statement I because for same volume surface area of the cylindrical bulb will be more.
203.

A pendulum clock has an iron pendulum 1 m long `(alpha_("iron")=10^(-5) //^(@)C)` . If the temperature rises by `10^(@)C`, the clockA. will lose 8 seconds per dayB. will lose 4.32 seconds per dayC. will gain 8 seconds per dayD. will gain 4.32 second per day

Answer» Correct Answer - B
`(DeltaT)/(T_(0))=1/2 alpha.Delta theta`
In one day the change in time `=86400xx(DeltaT)/(T_(0))`
`=86,400xx1/2xx10^(-5)xx10=4.32 sec `(lose)
204.

A uniform metal rod is used as a bar pendulum. If the room temperature rises by `10^(@)C`, and the coefficient of linear expansion of the metal of the rod is `2 xx 10^(-6) per^(@)C`, the period of the pendulum will have percentage increase ofA. `-2xx10^-3`B. `-1xx10^-3`C. `2xx10^-3`D. `1xx10^-3`

Answer» Correct Answer - D
We know that `T=2pisqrt((L)/(g))`
`:. DeltaT=(2pi)/(2)((1)/(Lg))^(1//2)DeltaL`
or `(DeltaT)/(T)=(DeltaL)/(2L)=(1)/(2)alphaDeltatheta=(2xx10^-6xx10)/(2)=10^(-5)`
`:. (DeltaT)/(T)xx100=1xx10^-5xx100=10^-3%`
205.

A metallic bar is heated from `0^(@)C` to `100^(@)C`. The coefficient of linear expansion is `10^(-5)K^(-1)`. What will be the percentage increase in lengthA. `0.01%`B. `0.1%`C. `1%`D. `10%`

Answer» Correct Answer - B
`alpha=10^(-5) //K`
`(Deltal)/lxx100=alpha. DeltaTxx100=0.1 %`
206.

If a bimetallic strip is heated it willA. bend towards the metal with lower themal expansion coefficient.B. bend towards the metal with higher thermal expansion coefficient.C. twist itself into helix.D. have no bending.

Answer» Correct Answer - A
In bimetallic strips the two melts have different thermal expansion coefficient. Hence on heating it bents towards the metal with lower thermal expansion coefficient
207.

Can heat be added to a substance without causing the temperature of the body of rise? If so does this contradict the concept of heat as energy in the process of transfer because of temperature differece?

Answer» Yes, heat can be added to a body without causing the temperature of the body to rise as in the case of latent heat being added to cause a change of state of the body or isothermal process. This does not contradict the concept of heat as energy in transit due to temperature difference because there is a big difference of temperature between the substance and the source from which heat is received by the substance and in this case, the heat supplied increases the intermolecular distance.
208.

The graph of elongation of rod of a substance A with temperature rise is shown if Fig. A liquid B contained in a cylindrical cessel made up of substance A, graduated in millilitres at `0^@C` is heated gradually. The readings of the liquid level in the vessel corresponding to different temperatures are shown in the figure. the real volume expansivity of liquid isA. `2.7xx10^(-5)//^@C`B. `15.4xx10^(-5)//^@C`C. `16.2xx10^(-5)//^@C`D. `151.2xx10^(-5)//^@C`

Answer» Correct Answer - C
From Fig. (a) the slope of line is
`(DeltaL)/(DeltaT)=((1002.2-1000.0)mm)/((300-0)^@C)`
`=4xx10^-6(m)//^@C`
But `(DeltaL)/(DeltaT)=Lalpha`
`alpha=(4xx10^-6)/(1m)(m//^@C)`
`4xx10^(-6)//^@C`
209.

The specific heat of a substance varies as `(3 theta^(2) + theta) xx 10^(3) cal g^(-10) C^(-1)`. What is the amount of heat required to rise the temperature of `1 kg` of substance frm `10^@C` to `20^@ C` ?

Answer» For small change in temperature `d theta`, heat required, `dQ = mSd theta`.
`:. Q = underset(theta_(1))overset(theta_(2))int mSd theta`
`:. Q = underset(10) overset(20) int 1000(3 theta^(2) +theta) xx 10^(-3) d theta = |theta^(3)+ theta^(2)/(2)|_(10)^(20)`
=`(20^(3) +(20^(2))/(2)) -(10^(3)+(10^(2))/(2)) = 8200 -1050 = 7150 cal`.
210.

A substance is in the solid form at `0^@`C. The amount of heat added to this substance and its temperature are plotted in the following graph. If the relative specific heat capacity of the solid substance is 0.5, find from the graph (i) the mass of the substance, (ii) the specific latent heat of the melting process and (ii) the specific heat of the substance in the liquid state. Specific heat capacity of water `=1000 cal//kg//K`

Answer» 1000 cal of heat raises the temperature of the substance from `0^@` C to `80^@` C.
`1000=m(1000xx0.5)xx80`
(`because` specific heat`=`relative sp. Heat`xx`of water)
or `m=0.025kg`
Latent heat `=200xx5=1000cal` (`because` 1 div reads 200cal) `=0.025xxL`
`L=40000 cal//kg`
In the liquid state teperature rises from `80^@`C to `120^@` C, that is by `40^@` C after absorbing 600 cal.
`0.025Sxx40=600` or `S=600cal//kg//K`
211.

The amount of heat required to change the state of 1 kg of substance at constant temperature is calledA. kilocalB. calorieC. specific heatD. latent heat

Answer» Correct Answer - D
latent heat
212.

The power of a system which can convert `10 kg` of water at `30^@ C` into ice `0^@ C` in one minute `(L_(ice) = 336000 J//Kg` and `S_(water) = 4200 J//kg//K`) will beA. 77 kWB. 55 kWC. 38.5 kWD. 40 kW

Answer» Correct Answer - A
`P=(Q)/(t) , Q =m xx S_(w) xx 30+ m xx L_(ice)`.
213.

The amount of steam at `100^@ C` that should be passed into `600 g` of water at `10^@ C` to make the final temperature as `40^@ C` will beA. 40 gB. 30 gC. 20 gD. 45 g

Answer» Correct Answer - B
Heat lost by steam = Heat gained by water
`m_(steam) xxL_(v)+m_(steam)S_(w)(100^@-40^@) = m_(water) S_(w)(40^@-10^@)`.
214.

Heat required to convert 1 g of ice at `0^(@)C` into steam at `100 ^(@)C` isA. 100 calB. `0.01 cal//^(@)C`C. 720 calD. 1 kilocal

Answer» Correct Answer - C
heat required `=mxx80+mxx1xx100+mxx540`
`=m(720)=1xx720=720` Calories
215.

One kg of ice at `0^(@)C` is mixed with 1 kg of water at `10^(@)C`. The resulting temperature will beA. between `0^(@)C` and `10^(@)C`B. `0^(@)C`C. less than `0^(@)C`D. greater than `0^(@)C`

Answer» Correct Answer - B
Resultant temperature is `0^(@)C` while ice will not melt.
216.

Water is used as an effective coolant. Give reason.

Answer»

By allowing water to flow in pipes around the heated parts of a machine, heat energy from such part is removed. Water in pipes extracts more heat from surrounding without much rise in its temperature because of its large specific heat capacity. So, Water is used as an effective coolant.

217.

Why does it become very cold when ice starts melting in the cold countries? Explain.

Answer»

Reasons used : 1kg of ice on meting absorbs 336000J of heat energy and 1kg of water to freeze will absorb 336000J of heat energy.

When ice start meting heat is absorbed from the atmosphere (336000J for every 1kg of ice) and temp, falls in the surrounding and it becomes very cold.

218.

Why is melting of ice a better coolant than water at zero degree Celsius? Explain.

Answer»

Reasons used : 1kg of ice on meting absorbs 336000J of heat energy and 1kg of water to freeze will absorb 336000J of heat energy.

 Sp. Latent heat of ice is 336000J for every 1kg ice. Hence to change ice at 0°C to water at 0°C, it will extract 336000J of heat from the hot engine and will cool the engine for longer time. 

219.

Give one example where high specific heat capacity of water is used as a heat reservoir.

Answer»

 In cold countries water is used as heat reservoir for wine and juice bottle to avoid freezing. Due to high specific heat capacity imports a large amount of heat before reaching to the freezing temp. Hence bottles kept in water remains warm and do not freeze. 

220.

Give one example each where high specific heat capacity of water is used (i) as coolant, (ii) as heat reservoir.

Answer»

(1) Radiator in car.

(2) To avoid freezing of wine and juice bottles.

221.

The S.I. unit of specific heat capacity is : (a) JKg-1 (b) JK-1 (c) JKg-1 K-1 (d) kJkg-1 K-1

Answer»

The S.I. unit of specific heat capacity is Jk-1.

222.

The base of cooking pans is made thicker and heavy because: (a) it lowers the heat capacity of pan (b) it increases the heat capacity of pan (c) the food does not get charred and keeps hot for long time (d) both (a) and (c)

Answer» (d) both (a) and (c)
223.

What is a calorimeter? Name the material of which it is made of. Give two reasons for using the material stated by you.

Answer»

A calorimeter is a cylindrical vessel which is used to measure the amount of heat gained or lost by a body when it is mixed with other body.

It is made up of thin copper sheet because:

(i) Copper is a good conductor of heat, so the vessel soon acquires the temperature of its contents.

(ii)Copper has low specific heat capacity so the heat capacity of calorimeter is low and the amount of heat energy taken by the calorimeter from its contents to acquire the temperature of its contents is negligible.

224.

Why is the base of a cooking pan made thick and heavy.

Answer»

By making the base of a cooking pan thick, its thermal capacity becomes large and it imparts sufficient heat energy at a low temperature to the food for its proper cooking. Further it keeps the food warm for a long time, after cooking.

225.

What is the principle of method of mixture (or principle of calorimetry)? Name the law on which this principle is based.

Answer»

The principle of method of mixture:

Heat energy lost by the hot body = Heat energy gained by the cold body.

This principle is based on law of conservation of energy.

226.

Discuss the role of high specific heat capacity of water with reference to climate in coastal areas.

Answer»

The specific heat capacity of water is very high. It is about five times as high as that of sand.

Hence the heat energy required for the same rise in temperature by a certain mass of water will be nearly five times than that required by the same mass of sand. Similarly, a certain mass of water will give out nearly five times more heat energy than that given by sand of the same mass for the same fall in temperature. As such, sand gets heated or cooled more rapidly as compared to water under the similar conditions. Thus a large difference in temperature is developed between the land and the sea due to which land and sea breezes are formed. These breezes make the climate near the sea shore moderate.The specific heat capacity of water is very high. It is about five times as high as that of sand. Hence the heat energy required for the same rise in temperature by a certain mass of water will be nearly five times than that required by the same mass of sand. Similarly, a certain mass of water will give out nearly five times more heat energy than that given by sand of the same mass for the same fall in temperature. As such, sand gets heated or cooled more rapidly as compared to water under the similar conditions. Thus a large difference in temperature is developed between the land and the sea due to which land and sea breezes are formed. These breezes make the climate near the sea shore moderate.

227.

 Name the law on which this principle is based.

Answer»

The principle of calorimetry is based on the law of conservation of energy.

228.

Two bodies at different temperatures are mixed in a calorimeter. Which of the following quantities remains conserved ? (a) sum of the temperatures of the two bodies (b) total heat of the two bodies (c) total internal energy of the two bodies (d) internal energy of each body.

Answer»

(c) total internal energy of the two bodies

229.

What is the other name given to principle of methods of mixtures?

Answer»

The other name given to the principle of mixture is the principle of calorimetry. 

230.

Why do the farmers fill their fields with water on a cold winter night?

Answer»

In the absence of water, if on a cold winter night the atmospheric temperature falls below 0oC, the water in the fine capillaries of plant will freeze, so the veins will burst due to the increase in the volume of water on freezing. As a result, plants will die and the crop will be destroyed. In order to save the crop on such cold nights, farmers fill their fields with water because water has high specific heat capacity, so it does not allow the temperature in the surrounding area of plants to fall up to 0oC.

231.

The mechanical equivalent of heat (a) has the same dimension as heat (b) has the same dimension as work (c) has the same dimension as energy (d) is dimensionless.

Answer»

(d) is dimensionless.

232.

What do you understand by the following statements :(i) The heat capacity of water is 60JK-1. (ii) The specific heat capacity of lead is 130Jkg-1K-1

Answer»

 (i) Heat capacity is the amount of heat required to raise the temperature of a body by 1°C or 1K. Thus, 60JK-1 of energy is required to raise the temperature of the given body by 1K. 

(ii) Specific heat capacity is the amount of heat energy required to raise the temperature of unit mass of a substance through 1°C or IK. Thus, 130JKg-1K-1 of heat energy required to raise the temperature of unit mass of lead through 1K. 

233.

What do you mean by the following statements:(i) the heat capacity of a body is 50JK-1 ?(ii) The specific heat capacity of copper is 0.4Jg-1K-1 

Answer»

(i) The heat capacity of a body is 50JK-1 means to increase the temperature of this body by 1K we have to supply 50 joules of energy. 

(ii) The specific heat capacity of copper is 0.4Jg-1K-1 means to increase the temperature of one gram of copper by 1K we have to supply 0.4 joules of energy.

234.

 What is the principle of methods of mixtures?

Answer»

The principle of method of mixture says that the heat lost by a hot body is equal to the heat gained by a cold body. 

235.

State two factors on which heat absorbed by a body depends. 

Answer»

Heat absorbed by a body is directly proportional to : 1. its mass 

2. Rise in temperature 

3. Specific heat capacity

236.

The heat capacity of a body depends on (a) the heat given (b) the temperature raised (c) the mass of the body (d) the material of the body.

Answer»

The correct answer is (c) (d)

(c) the mass of the body 

(d) the material of the body.

237.

Write the expression for the heat energy Q received by the substance when m kg of substance of specific heat capacity c J kg-1 k-1 is heated through Δt° C.

Answer»

The expression for the heat energy Q

Q= mc Δ t (in joule)

238.

Name three factors on which heat energy absorbed by a body depends and state how does it depend on them.

Answer»

The quantity of heat energy absorbed by a body depends on three factors :

(i)Mass of the body - The amount of heat energy required is directly proportional to the mass of the substance.

(ii)Nature of material of the body - The amount of heat energy required depends on the nature on the substance and it is expressed in terms of its specific heat capacity c.

(iii)Rise in temperature of the body - The amount of heat energy required is directly proportional to the rise in temperature.

239.

Assume that the thermal conductivity of copper is twice that of aluminium and four times that of brass. Three metal rods made of copper, aluminium and brass are each 15 cm long and 2 cm in diameter. These rods are placed end to end, with aluminium between the other two. The free ends of the copper and brass rods are maintained at `100^@C` and `0^@C` respectively. The system is allowed to reach the steady state condition. Assume there is no loss of heat anywhere. Under steady state condition the equilibrium temperature of the copper aluminium junction will beA. `57^@C`B. `35^@C`C. `18.8^@C`D. `28.5^@C`

Answer» Correct Answer - A
See solution above.
240.

A body cools in a surrounding of constant temperature `30^@C` Its heat capacity is `2 J//^@C`. Initial temeprature of cooling is valid. The body cools to `38^@C` in 10 min The temperature of the body In `.^@C` denoted by `theta`. The veriation of `theta` versus time t is best denoted asA. B. C. D.

Answer» Correct Answer - A
Self explanatory.
241.

A body cools in a surrounding of constant temperature `30^@C` Its heat capacity is `2 J//^(@)C`. Initial temeprature of cooling is valid. The body cools to `38^@C` in 10 min In further 10 min it will cools from `38^@C` to _____A. `36^@C`B. `36.4^@C`C. `37^@C`D. `37.5^@C`

Answer» Correct Answer - B
We have `theta-theta_s=(theta_0-theta_s)e^(-kt)`
Where `theta_0=`initial temperature of body `=40^@C`
`theta=`temperature of body after time t.
Since body cools from 40 to 38 in 10 min, we have
`38-30=(40-30)e^(-10k)` .(i)
Let after 10 min, the body temperature be `theta`
`theta-30^@=(38-30)e^(-10k)` .(ii)
`(Eq(i))/(Eq(ii))`gives `(8)/(theta-30)=(10)/(8)`,`theta-30=6.4`
`theta=36.4^@C`
242.

A thin brass rectangular sheet of sides 15.0 and 12.0 cm is heated in a furnace to `600^@` C and taken out. How much electric power is needed to maintain the sheet at this temperature, given that its emissivity is 0.250? Neglect heat loss due to convection (Stefan-Boltzmann constant, `sigma=5.67xx10^-8 W//m^(2)-K^(4)`).

Answer» Area of both sides of the plate
`A=2xx(15.0)xx(12.0)xx10^-4m^2=3.60xx10^-2m^2`
The energy radiated by the plate
`=esigmaAT^4=0.250xx5.67xx10^-8xx3.60xx10^-2xx(600+273)^4`
`=5.10xx10^(-12)xx873^4=296.4W`
243.

`30` gram copper is heated to increase its temperature by `20^@ C` if the same quantity of heat is given to `20` gram of water the rise in its temperature. `(S_(w)=4200J//kg-K & S_(cu) = 420 J//kg-K)`.A. `5^@C`B. `6^@C`C. `3^@C`D. `8^@C`

Answer» Correct Answer - C
`Q_(1)=Q_(2)rArr m_(cu)S_(cu) xx Delta theta_(1) =m_(w)S_(w) xx Delta theta_(2)`.
244.

The quantity of heat which can rise the temperature of `x gm` of a substance through `t_(1)^@ C` can rise the temperature of `y gm` of water through `t_(2)^@C` is same. The ratio of specific heats of the substances isA. `y t_(1)//xt_(2)`B. `xt_(2)//yt_(1)`C. `yt_(2)//xt_(1)`D. `xt_(1)//yt_(2)`

Answer» Correct Answer - C
`Q_(1)=Q_(2) rArr m_(1)S_(1)theta_(1)=m_(2)S_(2)theta_(2)`.
245.

The densities of two substances are in the ratio `5 : 6` and the specific heats are in the ratio `3:5` respectively. The ratio of their thermal capacities per unit volume isA. `2:1`B. `1 : 2`C. `4 : 1`D. `1 : 4`

Answer» Correct Answer - B
`H=mS = rho VS rArr (H_(1))/(H_(2)) = ((rho_(1))/(rho_(2)))((S_(1))/(S_(2)))`.
246.

The ratio of densities of two substances is `2 :3` and that of specific heats is `1 : 2`. The ratio of thermal capacities per unit volume isA. `1 : 2`B. `2 : 1`C. `1 : 3`D. `3 :1`

Answer» Correct Answer - C
`H=mS = rhoVS rArr(H_(1))/(H_(2))=((rho_(1))/(rho_(2))) ((S_(1))/(S_(2)))`.
247.

The ratio of the densities of the two bodies is `3:4` and the ratio of specific heats is `4:3` Find the ratio of their thermal capacities for unit volume?

Answer» `(rho_1)/(rho_2)=(3)/(4)`,`(s_1)/(s_2)=(4)/(3)`
`theta=(mxxs)/((m)/(rho))implies(theta_1)/(theta_2)=(s_1)/(s_2)xx(rho_1)/(rho_2)=1:1`
248.

Two sphere with radii in the ratio `1 : 2` have specific heats in the ratio `x : y` and densities in the ratio `z : x`. The ratio of their thermal capacities isA. `z : 2y`B. `zy : 8`C. `z : 8y`D. `xy : 2z`

Answer» Correct Answer - C
`H=mS = rho (4)/(3) pi r^(3) S rArr (H_(1))/(H_(2)) = (rho_(1))/(rho_(2)) xx (S_(1))/(S_(2)) xx ((r_(1))/r_(2))^(3)`.
249.

Specific heat of aluminium is `0.25 cal//g-.^(0)c`. The water equivalent of an aluminium vessel of mass one kilogram isA. `40 cal//^(0)C`B. `250 g`C. `250 cal//^(0) C`D. `40 g`

Answer» Correct Answer - B
water equivalent = `mS` gram.
250.

Equal masses of three liquids A,B and C are taken. Their initial temperature are `10^@C,25^@C` and `40^@C` respectively. When A and B are mixed the temperature of the mixutre is `19^@C`. When B and C are mixed, the temperature of the mixture is `35^@C`. Find the temperature if all three are mixed.

Answer» Let the specific heat capacities of A,B,C be `S_1`,`S_2`,`S_3`.
`theta_(A+B)=19=(10S_1+25S_2)/(S_1+S_2)` or `S_1=(2)/(3)S_2`
`theta_(B+C)=35=(25S_2+40S_3)/(S_2+S_3)` or `S_3=2S_2`
`theta_(A+B+C)=(10S_1+25S_2+40S_3)/(S_1+S_2+S_3)`
`=(10xx(2)/(3)S_2+25S_2+40xx2S_2)/((2)/(3)S_2+S_2+2S_2)= 30.5^(@)C`.