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1.

Three objects coloured black, gray and white can withstand hostile conditions upto `2800^(@)C`. These objects are thrown into a furance where each of them attains a temperature of `2000^(@)C`. Which object will glow brightest?A. The white objectB. The black objectC. All glow with equal brightnessD. Grey object

Answer» Correct Answer - (b)
Discussion
2.

A black body maintained at a certain temperature radiates heat energy at the rate `Q` watt. If its surface is smoothened so as to lower its emissivitiy by 10%, what will be the increase in its rate of radiation at double the initial temperature?A. `(0.9xx102^(4)-1)Q` wattB. `0.9xx2^(4)Q` wattC. `(0.9xx2)^(4)Q` wattD. `(0.9)^(4)xx2Q` watt

Answer» (a) For a black body,
Rate of radiation `Q=sigmaT^(4)`
After smoothing and doubling the temperature `=`Rate `Q`
`=0.9sigma(2T)^(4)=0.9xx10^(4)Q`
Charge `=(0.9xx2^(4)-1)Q` watt
Discussion
3.

A uniform metal ring with centre `C` have two points `A` and `B` such that angle `ACB` is `theta`. `A` and `B` are maintained at two different constant temperature. If the angle between `A` and `B` i.e. `theta=180^(@)` the rate of heat flow from `A` and`B` is `1.2W`, then what will be the rate, when `theta=90^(@)`?A. `0.6W`B. `0.9W`C. `1.6W`D. `1.8W`

Answer» (c) Let `DeltaT=` temperature difference between the rings
`R=` total internal resistance of the ring when `theta=180^(@)`
The resistance will be `R//2, R//2` and both are in parallel so, equivalent resistance `=R//4`
Rate of total heat flow `=(Q_(1))/t=1.2=(DeltaT)/((R//4))implies`
When `theta=90^(@)` there are two sections with resistances `R//4` adn `3R//4` in parallel.
So, equivalent resistance `=(3R)/16`
Rate of heat flow `=(Q_(2))/t=(DeltaT)/(((3R)/16))`
`l_(2)=16/3((DeltaT)/R)`
`l_(2)=1.6W`
`impliesl_(2)=1.6W`
Discussion
4.

What amount of ice at `-14^(@)C` required to cool `200gg` of water from `25^(@)C` to 10CC? (Given `C_("ice")=0.5cal/g.^(@)C,L_(f)` for ice `=80cal//g`)A. `31g`B. `41g`C. `51g`D. `21g`

Answer» (A) Heat given by water in cooling from `25^(@)` to `10^(@)C`
`Q_(1)=(cm DeltaT)_(w)=200xx1xx(25-10)=3000cal`
Heat absorbed by `mg` of ice at `-14^(@)C`
`Q_(2)=(mc DeltaT)_("ice")+mL_(f)+(mcDeltaT)_(w)`
`=m[(0.5)[0-{-14}]+80+1(10-0)]`
`=97mcal`
`Q_(2)=Q_(1)`
`97m=3000=30.93g`
`m=31g`
Discussion
5.

An aluminium rod of length `L` and cross-sectional area `2A` is joined with a copper rod of length `2L` and area of cross-section is `A` as shown in figure. Find the temperature of aluminium-copper junction in the steady state of the system. Given thermal conductivity `K_(Al)=240J//m//s//.^(@)C, K_(Cu)=400J//m//s//.^(@)C`A. `300^(@)C`B. `400^(@)C`C. `288.24^(@)C`D. `275.4^(@)C`

Answer» (c) In steaky state condition rate of heat flow in both rods must be equal Let junction temperature `=T`
`H=Q/T=(K_(Al)A_(Al)(400^(@)-T))/(L_(Al))=(K_(Cu)A_(Cu)(T_20^(@)))/(L_(Cu))`
`=(240(2A)(400^(@)-T))/L=(400xxAxx(T-20^(@)))/(2L)`
`T=288.24^(@)C`
Discussion
6.

Find the time in which a layer of ice thickness `h` will grow on the surface of the pond of surface area `A`, when the surrounding temperature falls to `-T^(@)C`. (Assume `K=` thermal conductivity of ice, `rho-` densityy of water `L=` latent heat of fusion)A. `t=(rhoL)/(2KT)h^(2)`B. `t-(rhoL)/(KT)h^(2)`C. `t=(rhoLh^(2))/(3KT)`D. `t=(rhoLh^(2))/(4KT)`

Answer» Correct Answer - (a)
Discussion
7.

A sphere a cube and a thin circular plate all of same material having same mass are initially heated to `200^(@)C`. Which of these will cool fastest?A. Circular plateB. SphereC. CubeD. All of these

Answer» (a) Radiation is direction proportional to the radiating area coooling fast.
For a given mass, area of circular plate is maximum. So, it cools first.
Discussion
8.

Two spheres of the same material having radii `r` and `4r` and temperature `2T_(0)` and `T_(0)` respectively. The ratio of rate of radiation of energy by the sphere isA. `1:1`B. `1:2`C. `2:1`D. `3:1`

Answer» (a) Energy radiated per second `P=esigmaAT^(4)`
`:.(P_(1))/(P_(2))=(A_(1)T_(1)^(4))/(A_(2)T_(2)^(4))=((4pir^(2))/(4pi(4r)^(2)))(16T_(0)^(4))/(T_(0)^(4))=(r^(2))/(16r^(2))xx(16T_(0)^(4))/(T_(0)^(4))=1:1`
Discussion