B.

cos^-1 $\left(\frac{-1}{3}\right)$

The angle between two vectors is given by

       cosθ = $\frac{\mathrm{a}.\mathrm{b}}{\mathrm{ab}}$

               = $\frac{1 - 1 - 1}{\sqrt{3} \times \sqrt{3}}$

⇒      cosθ = $-\frac{1}{3}$

⇒     $\mathrm{\theta } = {\cos }^{-1} \left(-\frac{1}{3}\right)$

A.

10.7

Given that distance

         $\mathrm{x} = {\mathrm{t}}^2 \left(2 - \frac{\mathrm{t}}{3}\right)$

On differentiating

        $\frac{\mathrm{dx}}{\mathrm{dt}} = 4\mathrm{t} - {\mathrm{t}}^2$

For x to be maximum,

         $\frac{\mathrm{dx}}{\mathrm{dt}} = 0$

⇒     4t - t² = 0

⇒            t = 4

∴ Maximum distance

      x_max = $4^2 \left(2 - \frac{4}{3}\right)$

              = $16 \times \left[\frac{6 - 4}{3}\right]$

              = 16 $\times \frac{2}{3}$

    x_max = 10.7

A.

$\frac{10 \mathrm{F}}{7 \mathrm{m}}$

Let a be the acceleration of centre of sphere. The angular acceleration about the centre of the sphere is $\mathrm{\alpha } = \frac{\mathrm{\theta }}{\mathrm{r}}$, as there is no slipping.

For the linear motion of the centre

   f + F = ma                  .....(i)

and for the rotational motion about the centre

  Fr - fr = I α

            = $\left(\frac{2}{5} {\mathrm{mr}}^2\right) \left(\frac{\mathrm{a}}{\mathrm{r}}\right)$

 ⇒  F - f = $\frac{2}{5} \mathrm{ma}$           ....(ii)

From Eqs. (i) and (ii)

   2F = $\frac{7}{5} \mathrm{ma}$

 ⇒  a  = $\frac{10 \mathrm{F}}{7 \mathrm{m}}$

A.

$\frac{3 \mathrm{cl} + 2 \mathrm{D} {\mathrm{l}}^2}{3 \left( 2\mathrm{c} + \mathrm{D} \mathrm{l} \right)}$

Let the cross-sectional area is $\mathrm{\alpha }$. The mass of an element of length dx located at a distance x away from the left end is (C + Dx)$\mathrm{\alpha }$ dx.  The x-coordinate of the centre of mass is given by

         ${\mathrm{X}}_{\mathrm{cm}} = \frac{\int \mathrm{x} \mathrm{cm}}{\int \mathrm{dm}}$

               = $\frac{{\int }_0^{\mathrm{l}}\mathrm{x}\left(\mathrm{C} + \mathrm{Dx} \right) \mathrm{\alpha } \mathrm{dx}}{{\int }_0^{\mathrm{l}}\left( \mathrm{C} + \mathrm{Dx} \right) \mathrm{\alpha } \mathrm{dx}}$

Where limit is from 0 to l.

              = $\frac{\mathrm{C} {\displaystyle \frac{{\mathrm{l}}^2}{2}}+ {\displaystyle \frac{\mathrm{D} \mathrm{l}}{3}}}{\mathrm{Cl} + {\displaystyle \frac{\mathrm{D} {\mathrm{l}}^2}{2}}}$

             X_cm = $\frac{3 \mathrm{CI} + 2 {\mathrm{DI}}^2}{3 \left(2\mathrm{C} + \mathrm{DI}\right)}$

C.

1 cm

For the circular motion acceleration towards the centre is $\frac{{\mathrm{v}}^2}{\mathrm{r}}$.

The horizontal force on the particle is I due to the spring and kl,

 where, l is the elongation 

           k is spring constant of spring.

     kl = $\frac{{\mathrm{mv}}^2}{\mathrm{r}}$

         = mω² r

     kl = mω² ( l_o + l )

⇒  kl - mω² ( l_o + l ) = 0

⇒ ( k - mω²) l = mω² l_o

⇒             l = $\frac{{\mathrm{m\omega }}^2 {\mathrm{l}}_{\mathrm{o}}}{\mathrm{k} - {\mathrm{m\omega }}^2}$ 

Putting tle values l = $\frac{0.5 \times 4 \times 0.5}{100 - 0.5 \times 4}$

                           = $\frac{1}{98}$m

                            = 0.010 m

                         l = 1 cm

Hence elongation(l) in spring is 1 cm.