C.

electrical conductivity

Resistivity $\mathrm{\rho } =\frac{\mathrm{m}}{{\mathrm{ne}}^2\mathrm{\tau }}$

$\because \left|\mathrm{\rho }\right| = \frac{\left[\mathrm{M}\right]}{\left[{\mathrm{L}}^{-3}\right] {\left[\mathrm{AT}\right]}^2 \left[\mathrm{T}\right]}$

          = [ M L³ A^-2 T^-3 ]

So, electrical conductivity

       σ = $\frac{1}{\mathrm{\rho }}$

       [ σ ] = $\frac{1}{\left[\mathrm{\sigma }\right]}$

               = [M^-1 L^-3 A² T³ ]

D.

straight line

When the ball goes upwards the direction of velocity of the ball and the direction of acceleration of gravity are in the opposite directions of acceleration of gravity are in the opposite directions so the velocity of the ball decrease until it reaches the top and stops momentarily.

 v = u - gt          ... (i)

Then from the peak position, the ball will start to descend downwards due to the acceleration due to gravity that is g.

 v = -gt               ....(ii)

( negative sign is due to motion in downward direction)   

From both equations, the velocity and time are linearly proportional to each other which is the same as the equation of a straight line. So this graph will be a straight line starting from initial velocity (u) on the y-axis and slope g with time as the x-axis.

C.

equal to or less than one

Distance is scalar, it is never negative. Therefore distance can be equal or greater than displacement. That implies the ratio of displacemrnt to distance is always equal to or less than one.

B.

west-north direction

Let O be the origin, then

passenger in the train at P observes the car at Q along direction PQ, where direction PQ is west north direction.

B.

by rolling them simultaneously on an inclined plane

The acceleration of a body rolling down the plane $= \mathrm{\alpha }$

∴                $\mathrm{\alpha } = \frac{\mathrm{g} \mathrm{sin\theta }}{1 + {\displaystyle \frac{{\mathrm{K}}^2}{{\mathrm{R}}^2}}}$

where K is radius of gyration and R the radius of sphere.

For solid sphere,

                      $\frac{{\mathrm{K}}^2}{{\mathrm{R}}^2} = \frac{2}{5}$

                       $\mathrm{\alpha } = \frac{\mathrm{g} \mathrm{sin\theta }}{1 + {\displaystyle \frac{2}{5}}}$

                       $\mathrm{\alpha } = \frac{\mathrm{g} \sin \mathrm{\theta }}{{\displaystyle \frac{7}{5}}}$                         

 ∴                       $\mathrm{\alpha } = \frac{5}{7}\mathrm{g} \mathrm{sin\theta }$

For hollow sphere,

                        $\frac{{\mathrm{K}}^2}{{\mathrm{R}}^2} = \frac{2}{3}$

                         $\mathrm{\alpha } = \frac{\mathrm{g} \sin \mathrm{\theta }}{1 + {\displaystyle \frac{2}{3}}}$

                          $\mathrm{\alpha } = \frac{3}{5} \mathrm{g} \sin \mathrm{\theta }$

                           $\mathrm{\alpha } = 0.6 \left(\mathrm{g} \mathrm{sin\theta } \right)$

Since, acceleration of solid sphere is more than of hollow sphere, it rolls faster, and reaches the
bottom of the inclined plane earlier. Hence, solid sphere and hollow sphere can be distinguished by rolling them simultaneously on an inclined plane.