physics ncert solutions class 11th

This is a short answer type question as classified in NCERT Exemplar

No, work done is not always zero in circular path it is zero only when all the forces are zero.

This is a short answer type question as classified in NCERT Exemplar

No total mechanical energy of the body falling freely under gravity is not conserved because a small part of its energy is utilised against resistive force of air, which is non conservative force. Gain in KE< loss in PE.

This is a short answer type question as classified in NCERT Exemplar

Force of gravity acts on the car vertically downward while car is moving along horizontal direction. So angle between them is 90. So work done is zero.

W= fscos90=0

This is a short answer type question as classified in NCERT Exemplar

(a) Force is applied on the body to lift it in the upward direction and displacement of the body is also in upward direction. So angle is zero.

Work done = fscos $\theta$ = fs

W= positive

(b) But the gravitational force in downward direction and displacement in upward direction so angle is 180. So work done will be negative. W= fscos 180= -fs

This is a short answer type question as classified in NCERT Exemplar

When the elevator is descending then electric power is required to prevent it from falling freely under gravity.

Also as the weight inside the elevator increases, its speed of descending increases, therefore there should be a limit on the number of passengers in the elevator to prevent the elevator from descending with a large velocity.

So when we increase the weight it fall with more speed . that is why there is limit in elevator.

This is a short answer type question as classified in NCERT Exemplar

As the block M is at rest and frictional force =Mgsin $\theta$

The force of friction acting between the blocks and incline opposes the tendency of sliding of the block . since block not in motion therefore no work is done. Hence no dissipation of energy.

This is a long answer type question as classified in NCERT Exemplar

V= volume of ballon

${\rho }_{air}=$  density of air

${\rho }_{He}=$ density of helium

V( ${\rho }_{air}-{\rho }_{He}$ )g= ma= mdv/dt= upthrust

Integrating with respect to t

V( ${\rho }_{air}-{\rho }_{He}$ )gt=mv

½ mv²= ½ m v²/m² ( ${\rho }_{air}-{\rho }_{He}$ )²g²t²

= ½v²/m ( ${\rho }_{air}-{\rho }_{He}$ )²g²t²

= if the ballon rises to height h then s= ut +1/2at²

h=1/2at²= ½ $\frac{\mathrm{}\mathrm{V}\mathrm{}({\rho }_{air}-{\rho }_{He})\mathrm{g}{t}^2}{m}$

so from above equation

1/2mv²= [V( ${\rho }_{air}-{\rho }_{He}$ )g][ $\frac{1}{2m}V\mathrm{}({\rho }_{air}-{\rho }_{He})\mathrm{g}{t}^2$ ]

= V( ${\rho }_{air}-{\rho }_{He}$ )gh

So ½ mv²+V ${\rho }_{He}$ gh= ${\rho }_{air}$ hg

KE_ballon+PE_ballon= change in PE of air

This is a long answer type question as classified in NCERT Exemplar

m =50g = 50 $*{10}^{3}kg$

Side = 1cm = 0.01m

Speed v = 0.1m/s

Young's modulus= 2 $*{10}^{11}N/m$ ²

According to the formula

F/A= Y $\frac{?L}{L}$

And F= K $?L$  where K is the compression in the spring.

K= YA/L = YL

Initial KE= 2 (1/2mv²)= 5 $*{10}^{-4}J$

Final PE= 2 (1/2)K ( $?L$ )²

$?L=\sqrt[]{\frac{KE}{E}}=\sqrt[]{\frac{KE}{YL}}$  = $\sqrt[]{\frac{5*{10}^{-4}}{2*{10}^{11}*0.1}}=1.58*{10}^{-7}m$

This is a long answer type question as classified in NCERT Exemplar

Let M be the mass of the rocket at any time t and v₁ the velocity of the rocket at the same time t

Let? m = mass of gas ejected in? t time

Relative speed of the gas ejected =u

KE +? t = KE of rocket +KE of gas

= ½ (M-? m) (v+? v)² + ½? m (v-u)²

KE_t= KE of rocket at time t= ½ Mv²

So? K = KE_t+? t -KE_t

= (M? v=? mu)v+1/2? mu²

Since action and reaction forces are equal

M? v/? t=? m/? t|u|

M? v=? m u

So? K= ½? mu²

? K=? W

? W=1/2? mu²

This is a long answer type question as classified in NCERT Exemplar

(a) As ball 1 is rolling down without slipping there is no dissipation of energy hence, total mechanical energy is conserved. Ball 3 is having negligible friction hence, there is no loss of energy.

(b) Ball 1 acquires rotational energy, ball 2 loses energy by friction. They cannot cross at C. Ball 3 can cross over.

(c) Ball 1, 2 turn back before reaching C. Because of loss of energy, ball 2 cannot reach back to A. Ball 1 has a rotational motion in “wrong” sense when it reaches B. It cannot roll back to A, because of kinetic friction.