Question

A carom board $(4ft\times 4ftsquare)$ has the queen at the centre. The queen, hit by the striker moves to the front edge, rebounds and goes in the hole behind the striking line. Find the magnitude of displacement of the queen
$\left(i\right)$ from the centre to the front edge
$\left(ii\right)$ from the front edge to the hole and
$\left(iii\right)$ from the centre of the hole.

• A. $\left(i\right)$ $\frac{2}{3}\sqrt{10}ft$ $\left(ii\right)$ $\frac{4}{3}\sqrt{10}ft$ $\left(iii\right)$ $2\sqrt{2}ft$
• B. $\left(i\right)$ $\frac{4}{3}\sqrt{10}ft$ $\left(ii\right)$ $\frac{4}{3}\sqrt{10}ft$ $\left(iii\right)$ $2\sqrt{2}ft$
• C. $\left(i\right)$ $\frac{4}{3}\sqrt{10}ft$ $\left(ii\right)$ $\frac{2}{3}\sqrt{10}ft$ $\left(iii\right)$ $2\sqrt{2}ft$
• D. $\left(i\right)$ $\frac{2}{3}\sqrt{10}ft$ $\left(ii\right)$ $\frac{2}{3}\sqrt{10}ft$ $\left(iii\right)$ $2\sqrt{2}ft$

Answer 1

The correct option is A $\left(i\right)$ $\frac{2}{3}\sqrt{10}ft$ $\left(ii\right)$ $\frac{4}{3}\sqrt{10}ft$ $\left(iii\right)$ $2\sqrt{2}ft$

By law of reflection $\angle i=\angle r$

or $\tan i=\tan r$ (from triangle OAB and BDC)

$\frac{2-x}{x}=\frac{4}{2}$

Or $4-2x=x$

$\Rightarrow 3x=2\Rightarrow x=\frac{2}{3}$

We get $BD=2-\frac{2}{3}=\frac{4}{3}$

(i) : Thus displacement of queen from front edge to hole $\overrightarrow{BC}=\frac{2\hat{i}}{3}+2\hat{j}$

Magnitude of displacement $|\overrightarrow{OB}|=\sqrt{(\frac{2}{3}{)}^2+2^2}=\frac{2}{3}\sqrt{10}ft$

(ii) : Thus displacement of queen from center to front edge $\overrightarrow{BC}=\frac{4\hat{i}}{3}-4\hat{j}$

Magnitude of displacement $|\overrightarrow{BC}|=\sqrt{(\frac{4}{3}{)}^2+4^2}=\frac{4}{3}\sqrt{10}ft$

(iii) : Thus displacement of queen from center to hole $\overrightarrow{OC}=2\hat{i}-2\hat{j}$

Magnitude of displacement $|\overrightarrow{OC}|=\sqrt{(2{)}^2+2^2}=2\sqrt{2}ft$