Two unequal masses are connected on two sides of light string pas… - Vidyadip

MCQ

Two unequal masses are connected on two sides of light string passing over a light and smooth pulley as shown in the figure. The system is released from rest. The larger mass is stopped $1\, sec$ after the system is set into motion and then released. The time elapsed before the string is tight again $(g = 10\, m/sec^2)$

A
$\frac{1}{4}sec$
B
$\frac{1}{2}sec$
C
$\frac{2}{3}sec$
✓
$\frac{1}{3}sec$

✓

Answer

Correct option: D.

$\frac{1}{3}sec$

d
Net pulling force $=2 \mathrm{g}-1 \mathrm{g}=10 \mathrm{N}$

Mass being pulled $=2+1=3 \mathrm{kg}$

Acceleration of the system is,

$a=\frac{10}{3} \mathrm{m} / \mathrm{s}^{2}$

Velocity of both the blocks at $t=1$ s will be.

$v_{0}=a t=\frac{10}{3} \times 1=\frac{10}{3} \mathrm{m} / \mathrm{s}$

Now at this moment, velocity of $2\, kg$ block becomes zero, while that of $1 \mathrm{kg}$ block is

$\frac{10}{3} \mathrm{m} / \mathrm{s}$ upwards. Hence, string becomes tight

again when

displacement of $1 \mathrm{kg}$ block $=$ displacement of $2 \mathrm{kg}$ block

or $\quad v_{0} t-\frac{1}{2} g t^{2}=\frac{1}{2} g t^{2}$

$\mathrm{t}=\frac{\mathrm{v}_{0}}{\mathrm{g}}=\frac{(10 / 3)}{10}=\frac{1}{3} \mathrm{sec}$

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