Laws of Motion

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.

C.

0.05 m

As the block A moves with velocity 0.15 ms^-1  it compresses the spring which pushes B towards right. A goes on compressing the spring till the velocity acquired by B becomes equal to the velocity of A, i.e., 0.15 ms^-1.Let this velocity be v.

Now, spring is in a state of maximum compression. Let x be the maximum compression at this stage.

According to the law of conservation of linear momentum, we get

             ${\mathrm{m}}_{\mathrm{A}} \mathrm{u} = \left({\mathrm{m}}_{\mathrm{A}} + {\mathrm{m}}_{\mathbf{B}}\right) \mathrm{v}$

⇒               v = $\frac{{\mathrm{m}}_{\mathrm{A}} \mathrm{u}}{{\mathrm{m}}_{\mathrm{A}} + {\mathrm{m}}_{\mathrm{B}}}$        

⇒                   = $\frac{2 \times 0.15}{2 + 3}$

                  v = 0.06 ms^-1

According to the law of conservation of energy

         $\frac{1}{2} \mathrm{m}{ }_{\mathrm{A}} {\mathrm{u}}^2 = \frac{1}{2} \left({\mathrm{m}}_{\mathrm{A}} + {\mathrm{m}}_{\mathrm{B}}\right) {\mathrm{v}}^{2 }+ \frac{1}{2} \mathrm{k} {\mathrm{x}}^2$

          $\frac{1}{2} {\mathrm{m}}_{\mathrm{A}} {\mathrm{u}}^{2 }- \frac{1}{2} \left({\mathrm{m}}_{\mathrm{A}} + {\mathrm{m}}_{\mathrm{B}}\right) {\mathrm{v}}^2 = \frac{1}{2} \mathrm{k} {\mathrm{x}}^2$   

   $\frac{1}{2}\times 2 \times {\left(0.15\right)}^2 - \frac{1}{2} \left(2 + 3\right) {\left(0.06\right)}^{2 } = \frac{1}{2 } \mathrm{k} {\mathrm{x}}^2$

                         0.0225 - 0.009 = $\frac{1}{2} \mathrm{k} {\mathrm{x}}^2$

                          0.0135 = $\frac{1}{2} \mathrm{k} {\mathrm{x}}^2$

⇒                              x = $\sqrt{\frac{0.027}{\mathrm{k}}}$

⇒                                 = $\sqrt{\frac{0.027}{10.8}}$

                           x = 0.05 m

B.

63.90^o

A body that travels an equal distance in equal amounts of time along a circular path has a constant speed but not constant velocity. This is because velocity is a vector and thus it has magnitude as well as direction.

               v = ( Rg tanθ )^1/2

               $\tan \mathrm{\theta } = \frac{{\mathrm{v}}^2}{\mathrm{r} \mathrm{g}}$

               = $\frac{400}{20 \times 900}$

         tan θ = 63.70^o 

         tanθ ≈ 63.90^o