A.

96 J

Total mass of the shell = 20 kg

Ratio of the masses of the fragments are 8 kg and 12 kg 

Now according to the conservation of momentum 

             m₁ v₁ = m₂ v₂

∴            8 × 6 = 12  × v

v (velocity of the larger fragment) = 4 m/s

  Kinetic energy = $\frac{1}{2}{\mathrm{mv}}^2$

                        = $\frac{1}{2} \times 12 \times {\left(4\right)}^2$

   Kinetic energy = 96 J

D.

1.1 H

Power P = I × V

             = V R × V

           P = V² R

Resistance of bulb

          R = $\frac{{\mathrm{V}}^2}{\mathrm{P}}$

           R = $\frac{{\left(100\right)}^2}{50}$

               = 200 Ω 

Current through bulb

            (I) = $\frac{\mathrm{V}}{\mathrm{R}}$

                 = $\frac{100}{200}$

              I  =  0.5 A

In a circuit containing inductive reactance (X_L ) and resistance (R ), impedance (Z) of the circuit is

             Z = $\sqrt{{\mathrm{R}}^2 + {\mathrm{\omega }}^2 {\mathrm{L}}^2}$              ...... (i)

             Z = $\frac{200}{0.5}$

             Z = 400 Ω

Now,      ${\mathrm{X}}_{\mathrm{L}}^2 = {\mathrm{Z}}^2 - {\mathrm{R}}^2$

             ${\mathrm{X}}_{\mathrm{L}}^2$  = ( 400 )²  - ( 200 )²

           ${\left(2 \mathrm{\pi f} \mathrm{L}\right)}^2 = 12 \times {10}^4$

             $\mathrm{L} = \frac{2\sqrt{3} \times 100}{2\mathrm{\pi } \times 50}$

                  = $\frac{2\sqrt{3} }{\mathrm{\pi }}$

              L = 1.1 H

C.

cannot keep the body in equilibrium

If we keep 1 N and 2 N forces act in the same direction then these are balanced by 3 N force, but this is against statement of question.

Hence, options (c) is correct.

B.

200 J

Work done W = Area ABCEFDA

                     = Area ABCD + Area CEFD

                    = Area ABCD + Area CEFD

                    =  $\frac{1}{2} \times \left(15 + 10\right) \times 10 + \frac{1}{2} + \left(10 + 20 \right) \times 5$

Work done W = 125 + 75

                     = 200 J

B.

7.2 m

Here, A= 4 i + 10 j

  $\left|\mathrm{A}\right| = \sqrt{{\left(4\right)}^2 + {\left(10\right)}^2}$

       = $\sqrt{16 + 100}$

  $\left|\mathrm{A}\right| = \sqrt{116}$ m

Now, according to question,

   A = 8 i + n j

⇒ $\left|\mathrm{A}\right| = \sqrt{{\left(8\right)}^2 + {\mathrm{n}}^2}$

    $\sqrt{116} = \sqrt{{\left(8\right)}^2 + {\mathrm{n}}^2}$

Squaring on both sides

   116 = 64 + n²

⇒ ${\mathrm{\pi }}^2 = 116 - 64$

     n = 7.2 m