Physics

This is a Multiple Choice  type Questions as classified in NCERT Exemplar

Answer- c

Explanation- given u= (3? +4? )m/s

And v= - (3? +4? )m/s, and mass of ball is 150g=0.15kg

Change in momentum = final momentum – initial momentum

 = mv-mu

= m (v-u) = 0.15 [- (3? +4? )- (3? +4? )]

= - [0.15 $*6i+0.15*8j$ ]

= - [0.9? +1.2? ]

$?$ P =- [0.9? +1.2? ]

This is a Multiple Choice type Questions as classified in NCERT Exemplar

Answer-b

Explanation- to solve this question we have to apply newton's law of motion, in terms of force and change in momentum.

as we know F= dp/dt

As body moving uniform velocity so dp= 0

So F=0

As all part of scale is moving with uniform velocity and total force is zero . hence torque will also be zero.

This is a Multiple Choice Type Questions as classified in NCERT Exemplar

Answer- c

Explanation- in a uniform translatory motion, all parts of the ball have the same velocity in magnitude and direction and this velocity is constant . in this situation body A will be in translatory motion.

This is a multiple choice answer as classified in NCERT Exemplar

(a, b, d, e) a) clearly every particle at x will have amplitude =asinkx=fixed

b) for mean position =0

coswt=0

wt= (2n-1) $\frac{\pi }{2}$

hence for a fixed of n all particles are having same value of time t= (2n-1) $\frac{\pi }{2w}$

c) amplitude of all the particles are asinkx which is different for different particles at different values of x

d) the energy is a stationary wave is confined between two nodes.

e) particles at different nodes are always at rest.

This is a multiple choice answer as classified in NCERT Exemplar

(a, b) v_o=400Hz, v=340m/s

V_m=10m/s

(a) as both source and observer are stationary, hence frequency observed will be same as natural frequency v_o=400Hz

(b) the speed of sound v=v+v_w= 340+10=350m/s

(c) there will be on effect on frequency because there is no relative motion between source and observer hence c, d are incorrect.

This is a multiple choice answer as classified in NCERT Exemplar

(a, b, d) As we know by y (x, t) = 0.06 sin (2πx/3) cos (120πt)

By comparing the equation with general equation N denotes nodes and A denotes antinodes.

(a) Clearly frequency is common for all the points

(b) Consider all the particles between two nodes they are having same phase at given time

(c) But are having different amplitude of 0.06sin (2 $\frac{\pi }{3}x$ ) and because of different amplitudes they are having different energies.

This is a multiple choice answer as classified in NCERT Exemplar

(b, c, d)

during propagation of a plane progressive mechanical wave

(i) clearly the particles O, A and B are having different phase.

(ii) Particles of the wave are having up and SHM.

(iii) For a progressive wave propagating in a fluid

V= $\sqrt[]{\frac{B}{\rho }}$ , V= $\sqrt[]{\frac{1}{\rho }}$

This is a multiple choice answer as classified in NCERT Exemplar

(c, d) Speed of sound waves in a fluid is given by

V= $\sqrt[]{\frac{B}{\rho }}$

V= $\frac{1}{\sqrt[]{\rho }},$ V $\propto \sqrt[]{B}$ so when we increase velocity bulk modulus also increases.

This is a multiple choice answer as classified in NCERT Exemplar

(b, c) y (x, t) = 0.06 sin (2πx/3) cos (120πt)

(a) y (x, t)=asinkxcoswt

(b) w=120 $\pi$ , f=60hz

(c) k=2 $\frac{\pi }{3}=\frac{2\pi }{\lambda },\lambda =3m,f=60hz$ , v =60 (3)=180m/s

(d) since in stationary wave all particles of the medium executes SHM with varying amplitude nodes.

This is a multiple choice answer as classified in NCERT Exemplar

(a, b, c) y (x, t)=30sin (36t+0.018x+ $\frac{\pi }{4}$ )

Y=asin (wt+kx+ $\varnothing$ )

(a) as the equation involves positive sign with x . hence the wave is travelling from right to left. Hence option a is correct.

(b) W=36

2 $\pi v=36$

$v=\frac{36}{2\pi }$ = $\frac{18}{\pi }$

$k=0.018=\frac{2\pi }{\lambda }$

$\frac{2\pi v}{v\lambda }=0.018$

w/v=0.018

v=2000cm/s=20m/s

(c) 2 $\pi v=36$

$v=\frac{18}{\pi }$ = 5.7Hz

(d) $\frac{2\pi }{\lambda }=0.018$ , $\lambda =2\pi$ /0.0018cm=3.48cm