Question

Blocks $A$ and $B$ of mass $m$ are placed inside an elevator which is going upward with an acceleration of $2{\text{m/s}}^2$ and a convex mirror having focal length $12\text{cm}$ is placed below block $B$. All the surfaces are smooth and the pulley is light. The mass pulley system is released from rest with respect to the elevator at time $t=0\text{s}$, when the distance of the block $B$ from the mirror is $42\text{cm}$. Find the distance between the image of the block $B$ and mirror at $t=0.2\text{s}$.

• A. $5.6\text{cm}$
• B. $6.6\text{cm}$
• C. $7.6\text{cm}$
• D. $8.6\text{cm}$

Answer 1

The correct option is D $8.6\text{cm}$

FBD of blocks $A$ and $B$ in elevator frame.

$F_P\to$ Pseudo Force.
For block $A$,
$T=ma$ ............$\left(1\right)$
For block $B$,
$mg+F_P-T=ma$
$\Rightarrow mg+m{a}_e-T=ma$ ............$\left(2\right)$
$[$$a_e\to$ acceleration of elevator$]$
On adding $\left(1\right)$ and $\left(2\right)$ and solving, we get,
$a=6{\text{m/s}}^2$
Distance travelled by block $B$ in $t=0.2\text{s}$,
$S=\frac{1}{2}a{t}^2=\frac{1}{2}\times 6\times (0.2{)}^2=0.12\text{m}=12\text{cm}$
Distance of block $B$ from mirror $=42-12=30\text{cm}$
Here, $u=-30\text{cm}$
$f=+12\text{cm}$
So, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{1}{v}-\frac{1}{30}=\frac{1}{12}$
$\Rightarrow \frac{1}{v}=\frac{7}{60}$
$\Rightarrow v=\frac{60}{7}\text{cm}=8.6\text{cm}$