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
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)
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
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?
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
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
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/Tb = 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
it is -273.15 0C
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
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
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
Use the formula 3/2 kT
k is boltzmann constant = 1.38* 10-23