# 560 - Physics Questions Answers

a metallic sphere having inner radius a and outer radius b has thermal conductivity k = ko(k not)/r² (a ≤ r ≤ b). the thermal resistance between inner and outer surface for radiant heat flow is

Joshi sir comment

consider a sphere of radius r and thickness dr

formula used is H = KAdT/L so dT/H = L/KA or R = L/KA,  dT/H is thermal resistance

thermal resistance of layer of thickness dr is dr/K4πr2, on putting the value of K we get

dR = r2dr/4πr2k0

or dR = dr/4πk0

now integrate within the proper limits

Assuming the sun to be aspherical body of radius R at a temperature of T K, evaluate the total radiant power incident on earth, at a distance r from the sun. Take radius of earth as Ro(R not)

Joshi sir comment

Total radiant power given by the sun = σAT4 = 4πR2σT4

amount received on the surface of earth = (πR02/4πr2)*4πR2σT4

A body emits radiant energy 1600Js‾¹ when it is at temperature 273 °C. if its temperature decreases to 273 K them it emits radiant energy at the rate of

(1) 0

(2) 800 Js‾¹

(3) 400 Js‾¹

(4) 100 Js‾¹

Joshi sir comment

Temprature of second case is half of the temprature of first case so according to stefan's law radiant power will become 1/16 part = 100

A copper cube of each side 40 cm floats on mercury. the density of copper cube is 3.2g/cc and of mercury is 13.6g/cc at 27° C. the coefficient of volume expansion of Hg and linear expansion of copper are 1.8 * 10^(-4) /°C and 3 * 10^(-6) /°C respectively. Calculate the increase in height i.e. how much block sink further when the temperature rises from 27°C to 100°C?

Joshi sir comment

dipped length of the cube in first case = m/Ad  here m, A, d are mass of cube, area of the cube and density of mercury respectively

after increasing temperature dipped length in the second case = m/A(1+2αt)[d/(1+γt)]

values are m = VD = 403 3.2, A = 402, α = 3 * 10-6, d = 13.6, γ = 1.8 * 10-4, t = 73

all values are in CGS

now solve then substract 1st length from 2nd length

A thermally insulated piece of metal is heated by supplying a constant power P. Due to this, the temperature of the metal starts varying with time as T= at^1/4 + To(T not)

The heat capacity of the metal as a function of temperature is

(1) 4PT³/a^4

(2) 4P(T-To)³/a^4

(3) 4PT²/a³

(4) 4P(T - To)²/a³

Joshi sir comment

P = dQ/dt = mcdT/dt

or mcdT = Pdt

or mc = Pdt/dT

or mc = Pd/dT[(T-T0)4/a4]

here mc is heat capacity

now solve

Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between Ta and Tb?

(1) Ta = Tb/2

(2) Ta = 7Tb/4

(3) Ta = 2Tb

(4) Ta = 4Tb/7

Joshi sir comment

273.15 K = 200 A

and 273.15 K = 350 B

so A = 273.15 K/200 and B = 273.15 K/350

so Ta A = TbB

or Ta/T= B/A = 200/350 = 4/7

so 4th option

The absolute zero temperature is

(1) -273° C

(2) -273.15 K

(3) -273.15° F

(4) -273.15° C

Joshi sir comment

it is -273.15 0

2 moles of a diatomic gas are enclosed in a vessel. when a certain amount of heat is supplied, 50% of the gas molecules get dissociated, bit there is no rise in temperature. What is the heat supplied if the temperature is T?

(1) RT

(2) RT/2

(3) 11RT/2

(4) 5RT

Joshi sir comment

internal energy of 2 mole diatomic gas = 2*5/2 RT = 5RT

now after 50% disociation in monoatomic form total internal energy

of 1 mole diatomic gas = 1*5/2 RT = 2.5 RT

and that of 2 mole monoatomic = 2*3/2 RT = 3RT  ( 2 mole monoatomic gas will be formed by the dissociation of 1 mole diatomic gas)

so internal energy change = 1/2 RT

This internal energy increment will be equal to the heat provided

A gaseous mixture consists of 16g of helium  and 16g of oxygen. the ratio Cp/Cv of the mixture is

(1) 1.59

(2) 1.62

(3) 1.4

(4) 1.54

Joshi sir comment

16 gms He = 4 mole,   16 gms O2= 1/2 mole

so total freedom n = [4(3) + 1/2(5)] / [4 + 1/2]  = 3.22222   here 3 and 5 are freedoms for mono and di atomic gases

so γ = 1 +  2/3.2222222 = 1.621

A vessel of volume 0.3 m³ contains helium at 20.0°C. the average kinetic energy per molecule for the gas is

(1) 6.07 * 10^(-21) J

(2) 7.3 * 10³ J

(3) 14.6 * 10³ J

(4) 12.14 * 10^(-21) J