A light string passing over a smooth light pulley connects two block of masses m₁ and m

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4-1.Newton's Laws of Motion

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A light string passing over a smooth light pulley connects two block of masses $m_1$ and $m_2$ (vertically). If the acceleration of the system is $\left( {\frac{g}{8}} \right)$, then the ratio of masses is

A

$8 : 1$

B

$9 : 7$

C

$4 : 3$

D

$5 : 3$

Solution

$\mathrm{a}=\left(\frac{\mathrm{m}_{2}-\mathrm{m}_{1}}{\mathrm{m}_{1}+\mathrm{m}_{2}}\right) \mathrm{g}$

$\frac{g}{8}=\left(\frac{m_{2}-m_{1}}{m_{1}+m_{2}}\right) g$

$m_{1}+m_{2}=8 m_{2}-8 m_{1}$

$9 m_{1}=7 m_{2}$

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

Standard 11

Physics

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