# 560 - Physics Questions Answers

which of the following phenomenon cannot be observed with sound waves?

(1) diffraction

(2) polarisation

(3) reflection

(4) interference

Joshi sir comment

polarisation

a source of sound of frequency f emits sound wave with speed v is at rest. if an observer is moving towards source with speed Vo(V not), then wavelength of sound waves observed by the observer will be

(1) v/f

(2) (v - Vo)/f

(3) [v(v - Vo)]/[f(v + Vo)]

(4) [v(v + Vo)]/[f(v - Vo)]

Joshi sir comment

since velocity and frequency are changing proportionally so wavelength will be constant = v/f

a sonometer wire supports a 4kg load and vibrates in fundamental mode with a tuning fork of frequency 416 Hz. the length of the wire between the bridges is now doubled. in order to maintain fundamental mode the load should be changed to

Joshi sir comment

for sonometer wire

in fundamental mode l = λ/2 = v/2f       here l, λ and f are length , wavelength and frequency respectively

now v = √(T/µ)      here T and µ are tension and mass per unit length

so finally we will get l = √(T/µ)/2f

make two equations by using this formula and given conditions

in the two equations f and µ will be same

in a stationery wave, there is no energy current, but there is energy density. (TRUE/FALSE)

Joshi sir comment

TRUE

at STP, the speed of sound in hydrogen is 1324ms‾¹ then the spped of sound in air (at STP) is

Joshi sir comment

formula for speed of sound is v = √(γRT/M)

you have to form two equations for the two mediums, for both the mediums γ will be same because Hydrogen and Air both are diatomic gases, T will also be same for both the cases are taken in STP, only M will be different

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

Joshi sir comment

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

Joshi sir comment

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

Joshi sir comment

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

Joshi sir comment

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 Qis 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