575 - Physics Questions Answers
a person is standing between two vertical walls. he heard the echo of his sound after t = 1s and t = 1.5s. distance between walls is (speed of sound in air = 340ms‾¹
time taken by the sound to reach upto the first wall = 0.5 sec
similarly time taken by the sound to reach upto the second wall = 0.75
so total time = 1.25 sec. so distance between the walls = 340 * 1.25 = 425
which of the following is minimum audible wavelength at room temperature?
(1) 20mm
(2) 20cm
(3)20m
(4) 20km
maximum audible frequency = 20000 Hz
Velocity of sound = 340 m/s
so minimum audible wavelength = 340/20000 m = 34000/20000 cm = 340/20 mm = 17 mm approx 20 mm
two identical conducting rods are first connected independently to two vessels, one containing water at 100°C and the other containing ice at 0°C. In the second case, the rods are joined end to end connected to the same vessels. Let M1 and M2 g/s be the rate of melting of ice in the two cases respectively, the ratio M1/M2 is
(1) 1:2
(2) 2:1
(3) 4:1
(4) 1:4
in first case rods are taken in parellel so net length and area will be l and 2A with conductivity coefficient K
similarly in second case rods are taken in series so net length and area will be 2l and A with conductivity coefficient K
now use H = KA(θ1- θ2)/L and H = dQ/dt = mL/t L is latent heat which is fixed in both the cases
Three discs A,B and C having radii 2m, 4m and 6m respectively are coated with carbon black on their outer surfaces. the wavelenghts corresponding to maximum intensity are 300 nm, 400nm and 500 nm respectively. if the power radiated by them are Qa, Qb and Qc respectively then
(1) Qa is maximum
(2) Qb is maximum
(3) Qc is maximum
(4) Qa= Qb = Qc
according to Wein's displacement law λT = constant
so ratio of temperatures of the three discs will be 1/3:1/4:1/5 = 20:15:12
and ratio of area of the discs = 1:4:9
now according to Stephan's law Q α A * T4
so ratio Qa:Qb:Qc = 1*(20)4 : 4*(15)4 : 9*(12)4
on solving we will get that Qb is maximum
A bullet of mass 1 gm moving with a speed of 20m/s hits an ice block of mass 990 gm kept on a frictionless floor and gets tuck in it. the amount of ice that melts, if 50% of the lost kinetic energy goes to ice will be
(1) 0.030 g
(2) 0.30 g
(3) 0.0003 g
(4) 3.0 g
amount of energy given by bullet = 0.5 (1/2) m v2 / 4.2 cal.
and amount of energy received by the ice block = ML here L is latent heat and M is mass of ice melted
on comparing we will get m v2/(4*4.2) = M*80 or M = 1*400/16.8*80 = 5/16.8 = 0.30 gm
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