The block shown in the diagram has a mass of $2 \mathrm{\mu g}$ and a charge of $2\mathrm{ }\times \mathrm{ }{10}^{–9}$ $\mathrm{C}$. If the block is given an initial velocity of $2 \mathrm{m} {\mathrm{s}}^{-1}$ in the x-direction at $t=\mathrm{ }0$ and an electric field of $2 \mathrm{N} {\mathrm{C}}^{-1}$ is switched on in the x-direction, then the range ( $\text{R}$ ) of the block on the ground will be

• A

  $\mathrm{4 }\mathrm{m}$
• B

  $\mathrm{1 }\mathrm{m}$
• C

  $\mathrm{2 }\mathrm{m}$
• Correct answer
  D

  $\mathrm{3 }\mathrm{m}$

Solution

At first, we will find time taken by block to hit ground

using,$S_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$

$-5=0+\frac{1}{2}\left(-g\right)\times t^2$

$\Rightarrow t=\sqrt{\frac{10}{g}}=\mathrm{1 }\text{s}$

Now, in this time, block will move in x-direction.

$S_x=u_x.t+\frac{1}{2}{a}_x.t^2$

Here,$S_x=R,u_x=2,$

$a_x=\frac{2E}{m}=\frac{2\times {10}^{-9}\times 2}{2\times {10}^{-9}}=2$

$\Rightarrow R=2\times 1+\frac{1}{2}\times 2\times 1^2$

$=2+1=3 \mathrm{m}$

Hence, the correct answer is option: 'D'.