C.

2500 km

Velocity of the S wave is v₁ = 4.5 km/s

The velocity of the P waves is v₂ = 8.0 km/s

Let the time taken by the S and P waves to reach the seismograph be t₁ and  t₂

It is given that

    t₁ $-$ t₂ = 4 min

               = 4 × 60 sec

    t₁ $-$ t₂ = 240 sec              .....(i)

Let the distance of the epicentre (km) be S. Then 

    S = v₁t₁ = v₂t₂

⇒  4.5 × t₁ $-$ 8t₂ = 0

⇒              t₁ = $\frac{4.5}{8}$t₁            ....(ii)

Using (i) and (ii)

      t₁ $-$ t₂ = 240

⇒   t₁ $\left(1 - \frac{4.5}{8}\right)$ = 240

⇒     t₁ = $\frac{240 \times 8}{3.5}$

⇒          = 548.5 s

∴      S = v₁t₁ 

           = 4.5 × 548.5

       S = 2468.6 

       S ≈ 2500 km

We can represent in the form of a figure as

Now if we use the conservation of momentum we get

o = (2m) v cos 45° + (2m) v cos 45° + mv₁

So, v₁ = − 4 cos 45° v = − $2\sqrt{2}$ , thus the velocity of the third part has a magnitude equal to $2\sqrt{2} \mathrm{v}$, this makes an angle of 135° with respect to either of the three masses.

D.

During the upward motion, the speed of the body decreases and will be zero at the highest point (since the gravitational force acting downward), afterward, the body starts downward motion and its speed increases.

B.

2

Height of the cylinder from the level PQ is

   $\mathrm{H} = \mathrm{h} + \frac{\mathrm{h}}{2} = \frac{3\mathrm{h}}{2}$

To cover maximum distance along the plane PQ,

Height of the hole is $= \frac{3 \mathrm{h}}{2} \times \frac{1}{2} = \frac{3 \mathrm{h}}{4}$

This height corresponds to hole (2).