Physics

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

m =100kg

Height h= 10m

Power= rate of work done =change of energy /time= mgh/t

= $\frac{100*9.8*10}{20}=5*98=490W$

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

Total linear momentum of the system of two balls is always conserved. While balls are in contact, there may be deformation which means elastic PE which came from part of KE .But they are in contact while collision so according to above case kinetic energy will not conserved.

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

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