C.

125 rad/s

Given:-   

            $\mathrm{\alpha } = {\mathrm{\alpha }}_0 \left(1 - \frac{\mathrm{t}}{5}\right)$

At t = 0,  $\mathrm{\alpha } = {\mathrm{\alpha }}_0$

∴          ${\mathrm{\alpha }}_0$= 50 rad/s²

       $\frac{\mathrm{d\omega }}{\mathrm{dt}} = \mathrm{\alpha } \left(1 - \frac{\mathrm{t}}{5}\right)$

∴      ${\int }_0^{\mathrm{\omega }}\mathrm{d\omega } = {\mathrm{\alpha }}_0 {\int }_0^5\left(1 - \frac{\mathrm{t}}{5}\right)$

⇒            ω =  ${\mathrm{\alpha }}_0 {\left[\mathrm{t} - \frac{{\mathrm{t}}^2}{10}\right]}_0^5$ 

⇒                 = 50 $\left(5 - \frac{25}{10}\right)$ rad/s

⇒              ω = 125 rad/s  

C.

∼ 0.58 × 10¹⁴ kg m² s^-1

As the satellite is moving in equatorial plane with orbital radius 4 x 10⁷ m.

∴        Satellite is geostationary satellite.

Hence, the time taken by satellite to complete its one revolution

       T = 24 h

            = 24 × 3600 s

       T = 86400 s

Velocity of satellite 

       v = $\frac{2\mathrm{\pi r}}{\mathrm{T}}$

Angular momentum, L = mvr

       L = m $\left(\frac{2 \mathrm{\pi } \mathrm{r}}{\mathrm{T}}\right)$ r

           = $\left(\frac{2\mathrm{\pi } \mathrm{m}}{\mathrm{T}}\right)$r²

∴     L = $\frac{2 \times 3.14 \times 500}{86400}$× ( 4 × 10⁷ )²

        L = 0.58 × 10¹⁴ kg m² s^-1

A.

$\left(\frac{\mathrm{n}}{\mathrm{n} + 1}\right)$ mgR

Gravitational potential energy of mass m at any point at a distance r from the centre of earth is

              U = $- \frac{\mathrm{G} \mathrm{Mm}}{\mathrm{r}}$

At the surface of earth r = R,

∴          U_s = $- \frac{\mathrm{G} \mathrm{Mm}}{\mathrm{R}}$

                 = - mgR                              $\left(\because \mathrm{g} = \frac{\mathrm{GM}}{{\mathrm{R}}^2}\right)$

At the height h = nR from the surface of earth

                 r = R + h                  

                   = R + nR

               r = R ( 1 + n )

∴       U_h = $- \frac{\mathrm{G} \mathrm{Mm}}{\mathrm{R} \left(1 + \mathrm{n}\right)}$

              = $-\frac{\mathrm{mg} \mathrm{R}}{\left(1 + \mathrm{n}\right)}$

Change in gravitational potential energy is

          ΔU = U_h $-$ U_s

                = $- \frac{\mathrm{m} \mathrm{gR}}{1+\mathrm{n}} - \left(- \mathrm{m} \mathrm{gR} \right)$

                 = $- \frac{\mathrm{m} \mathrm{gR}}{1 + \mathrm{n}} \left(1 - \frac{1}{1 + \mathrm{n}}\right)$

            ΔU = m gR $\left(\frac{\mathrm{n}}{1 + \mathrm{n}}\right)$

C.

$\frac{1}{\sqrt{2}}$

Let u₁ and v₁ be the initial and final velocities of ball 1 and u₂ and v₂ be the similar
quantities for ball 2.

Here, u₂ = 0 and v₁ = 0

∴    initial K.E 

          K_i =  $\frac{1}{2} {\mathrm{mu}}_1^2 + \frac{1}{2} {\mathrm{mu}}_2^2$

          K_i = $\frac{1}{2} {\mathrm{mv}}_1^2$

and  final K.E

          K_f = $\frac{1}{2} {\mathrm{mv}}_1^2 + \frac{1}{2} {\mathrm{mv}}_2^2$

          K_f = $\frac{1}{2} {\mathrm{mv}}_2^2$

Loss of KE

         ΔK = K_i $-$ K_f

              = $\frac{1}{2} {\mathrm{m}}_1^2 - \frac{1}{2} {\mathrm{mv}}_2^2$

According to question

        $\frac{1}{2} \left(\frac{1}{2} {\mathrm{mv}}_1^2\right) = \frac{1}{2} {\mathrm{mu}}_1^2 - \frac{1}{2} {\mathrm{mu}}_2^2$

( since half KE is lost by impact )

        ${\mathrm{u}}_1^2 = 2 {\mathrm{v}}_2^2$

 ⇒     ${\mathrm{v}}_2 = \frac{{\mathrm{u}}_1}{\sqrt{2}}$

The coefficient of restitution is defined as the ratio of the final velocity to the initial velocity after their collision.

∴  Coefficient of restitution,

    e = $\left|\frac{{\mathrm{v}}_2 - {\mathrm{v}}_1}{{\mathrm{u}}_1 - {\mathrm{u}}_2}\right|$

        = $\frac{{\mathrm{v}}_2}{{\mathrm{v}}_1}$

   e = $\frac{1}{\sqrt{2}}$

A.

40 A and 16 A

Given:-

   η = 80%

    P_i = 4 kW

⇒  P_i = 4000 W

     V_p = 100 V

     V_s = 200 V

      I_p = $\frac{4000}{100}$

      I_p = 40 A

The efficiency of the transformer is defined as the ratio of useful power output the input power, the two being measured in the same unit

Transformer efficiency given by,

        η = $\frac{{\mathrm{V}}_{\mathrm{s}} {\mathrm{I}}_{\mathrm{s}}}{{\mathrm{V}}_{\mathrm{p}} {\mathrm{I}}_{\mathrm{p}}}$

Where V_s → Secondary voltage

           V_p → Primary voltage

          I_s → secondary current

          I_p → primary current

⇒    $\frac{80}{100} = \frac{200 {\mathrm{I}}_{\mathrm{s}}}{4000}$

               = $\frac{{\mathrm{I}}_{\mathrm{s}}}{20}$

⇒    I_s = $\frac{20 \times 80 }{100}$

⇒   I_s = 16 A