Question

Two blocks each of masses $m$ lie on a smooth table. They are attached to two other masses as shown in figure. The pulleys and strings are light. An object $\text{O}$ is kept at rest on the table and the sides $\text{AB}$ and $\text{CD}$ of the two blocks are made reflecting. The acceleration of the images formed in these two reflecting surfaces with respect to each other is:

• A. $\frac{5g}{6}$
• B. $\frac{g}{3}$
• C. $\frac{17g}{12}$
• D. $\frac{17g}{6}$

Answer 1

The correct option is D $\frac{17g}{6}$

Let $T_1$ be the tension in the right string and $T_2$ in the left string.

$\text{FBD}$ of the blocks is shown below.


Applying second law of motion for the system on right,
$T_1=m{a}_1.......\left(1\right)$
and
$2mg-T_1=2m{a}_1........\left(2\right)$

On solving (1) and (2), we get
$a_1=\frac{2g}{3}$

Thus, acceleration reflecting surface $\text{CD}$,
$\overrightarrow{a_1}=\frac{2g}{3}\hat{i}..........\left(3\right)$

Similarly for system on left,
$T_2=m{a}_2\dots \left(4\right)$
and
$3mg-T_2=3m{a}_2\dots \left(5\right)$
On solving (5) and (6), $a_2=\frac{3g}{4}$

Acceleration of reflecting surface $\text{AB}$,
$\overrightarrow{a_2}=-\frac{3g}{4}\hat{i}......\left(5\right)$

For plane mirror, along perpendicular to the mirror,
$\overrightarrow{a_{i,m}}=-\overrightarrow{a_{o,m}}$
$\Rightarrow \overrightarrow{a_i}-\overrightarrow{a_m}=\overrightarrow{a_m}-\overrightarrow{a_o}$

For reflecting surface $\text{CD}$,
$\overrightarrow{a_i}-\frac{2g}{3}\hat{i}=\frac{2g}{3}\hat{i}-0$ [from Eq.(3)]
$\Rightarrow {\overrightarrow{\left(a_i\right)}}_{CD}=\frac{4g}{3}\hat{i}........\left(5\right)$

For reflecting surface $\text{AB}$,
$\overrightarrow{a_i}-\left(\frac{-3g}{4}\hat{i}\right)=-\frac{3g}{4}\hat{i}-0$
$\Rightarrow {\overrightarrow{\left(a_i\right)}}_{AB}=-\frac{3g}{2}\hat{i}.......\left(6\right)$

Now, acceleration of the two images with respect to each other will be
${\overrightarrow{\left(a_i\right)}}_{CD}-{\overrightarrow{\left(a_i\right)}}_{AB}=\frac{4g}{3}\hat{i}-(-\frac{3g}{2})\hat{i}=\frac{17g}{6}\hat{i}{\text{m/s}}^2$

$\therefore$ Magnitude of acceleration w.r.t each other is $\frac{17g}{6}{\text{m/s}}^2$.
Hence, option $\left(d\right)$ is the correct answer.