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Velocity of COM

A light strin...

Question

A light string passing over a smooth light pulley connects two blocks of masses $m_{1}$ and $m_{2}$ (vertically). If the acceleration of the system is $\frac{g}{8}$ , then the ratio of the masses is

8 : 1

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9 : 7

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4 : 3

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5 : 3

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Solution

The correct option is B $9 : 7$
Considering $m_{1} > m_{2}$ , drawing the FBD of the situation,

For mass $m_{1} :$
$m_{1} g - T = m_{1} a . . . . (i)$
For mass $m_{2} :$
$T - m_{2} g = m_{2} a . . . . (i i)$
Adding the equations we get

$a = \frac{(m_{1} - m_{2}) g}{m_{1} + m_{2}}$

Here, $a = \frac{g}{8}$

$∴ \frac{1}{8} = \frac{\frac{m_{1}}{m_{2}} - 1}{\frac{m_{1}}{m_{2}} + 1}$

$\Rightarrow \frac{m_{1}}{m_{2}} + 1 = 8 \frac{m_{1}}{m_{2}} - 8$

$\Rightarrow \frac{m_{1}}{m_{2}} = \frac{9}{7}$

Hence, $(B)$ is the correct answer.

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A light string passing over a smooth light pulley connects two blocks of masses m1 and m2 (vertically). If the acceleration of the system is g8, then the ratio of the masses is

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A light string passing over a smooth light pulley connects two blocks of masses $m_{1}$ and $m_{2}$ (vertically). If the acceleration of the system is $\frac{g}{8}$ , then the ratio of the masses is