Physics

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

Answer- a

Explanation- x (t)= pt+qt²+rt³

V= dx/dt=p+2qt+3rt²

So a= o+2q+6rt

At t=2s a= 2q+6 $*$ 2 (r)

= 2q+12r

= 2 (4)+12 (5)=68m/s

So F =ma= 2 (68)= 136N

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

Answer- c

Explanation –consider the adjacent diagram

 OA= p₁= initial momentum of player northward

AB= p₂= final momentum of player towards west

OB= OA+AB

Change in momentum = P₂-P₁= AB-OA so we can say that AR will be along south west.

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

Answer- d

Explanation-we know that for a system F_ext=dp/dt

If F_ext=0, dp =0,

P = constant

Hence, momentum of a system will remain conserve if external force on the system is zero.

In case of collision between particles, equal and opposite forces will act on individual particles by newton's third law.

according to newton's 2^nd law F=dp/dt

If F =0 then p = constant so momentum remains constant if external force is zero.

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

Answer- c

Explanation- change in momentum $?p$ =- [0.9? +1.2? ]

Magnitude = $\sqrt[]{{0.9}^2+{1.2}^2}$

$=\sqrt[]{0.81+1.44}=$ 1.5 kg m s^-1       

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.