D.

increase with time

Given:- 

        x = a e^-αt + be^βt

We know that

   velocity v = $\frac{\mathrm{dx}}{\mathrm{dt}}$

              v  =  $\frac{\mathrm{d} }{\mathrm{dt}}\left( {\mathrm{ae}}^{-\mathrm{\alpha t}} + {\mathrm{be}}^{\mathrm{\beta t}} \right)$

                  = -a$\mathrm{\alpha }$ e^-αt + bβ e^βt

              v = $-\mathrm{a} {\mathrm{\alpha e}}^{-\mathrm{\alpha t}} + \mathrm{b\beta } {\mathrm{e}}^{\mathrm{\beta t}}$

               v = A + B

where A = - a$\mathrm{\alpha } {\mathrm{e}}^{-\mathrm{\alpha t}}$ , B = bβ e^-βt

The value ot term A = -a$\mathrm{\alpha } {\mathrm{e}}^{-\mathrm{\alpha t}}$ increases and of term B = bβ e^-βt increases with time. As a result velocity goes on increasing with time. 

C.

$\frac{3 \mathrm{\pi }}{4}$ rad

Given:-

      A + B = C

Since, B = C $-$ A

Also

   $\left|\mathrm{C}\right| \perp \left|\mathrm{A}\right|$

    B² = 2 A²

⇒   B = $\sqrt{2}$ A

Now

    A² + B² + 2A B cosθ = C² = A²

∴          $\cos \mathrm{\theta } = - \frac{1}{\sqrt{2}}$

This gives

       $\mathrm{\theta } = \frac{3\mathrm{\pi }}{4}$ rad

C.

3.25 m/s

Given:- r = 25 cm =  m

  ω = 13 rad/s

   Linear speed  = Radius × angular speed

            V = rω

          = 0.25 × 13

        3.25 m/s

A.

2 : 1

As W = $\frac{1}{2} {\mathrm{kx}}^2$

If both wires are stretched through same distance, then   

     W ∝ k

    $\frac{{\mathrm{W}}_1}{{\mathrm{W}}_2} = \frac{{\mathrm{k}}_1}{{\mathrm{k}}_2}$

            = $\frac{{\mathrm{k}}_1}{{\displaystyle \frac{{\mathrm{k}}_1}{2}}}$

            = $\frac{2}{1}$

     $\frac{{\mathrm{W}}_1}{{\mathrm{W}}_2}$ = 2 : 1

C.

$\frac{4}{3} \sqrt{\frac{\sqrt{3}}{2}}$s

In the frame of elevator, 

a = acceleration of the particle with respect to the elevator

      m sin30 ( g + 2) = ma

                    a  = ( g + 2 ) sin30^o

                        = ( 10 + 2 ) $\frac{1}{2}$

                         = 6 m/s²

The distance travelled by the particle from the top to the bottom

     d = $\frac{4}{\cos {30}^{\mathrm{o}}}$

        = $\frac{4}{{\displaystyle \raisebox{1ex}{$\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$

   d = $\frac{8 \sqrt{3}}{3}$ m

Using kinematic equation

   s = ut + $\frac{1}{2} {\mathrm{at}}^2$

   $\frac{8 \sqrt{3}}{3} = 0 \times \mathrm{t} + \frac{1}{2} {\mathrm{at}}^2$

⇒     t² = $\frac{16 \sqrt{3}}{3 \times 6}$s

⇒     t = $\sqrt{\frac{16}{18}\sqrt{3}}$

⇒    t = $\sqrt{\frac{16}{9 \times 2}\sqrt{3}}$

⇒  t =  $\frac{4}{3} \sqrt{\frac{\sqrt{3}}{2}}$s