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Two masses $\mathrm{M_{1}}$ and $\mathrm{M_{2}}$ are tied together at the two ends of a light inextensible string that passes over a frictionless pulley. When the mass $\mathrm{ M_{2}}$ is twice that of $\mathrm{ M_{1}, }$ the acceleration of the system is $\mathrm{a_{1}. }$ When the mass $\mathrm{M_{2}}$ is thrice that of $\mathrm{M_{1},}$ the acceleration of the system is $\mathrm{ a_{2}. }$ The ratio $\mathrm{\frac{a_{1}}{a_{2}} }$ will be :

Option: 1
$\frac{1}{3}$

Option: 2
$\frac{2}{3}$

Option: 3
$\frac{3}{2}$

Option: 4
$\frac{1}{2}$

Answers (1)

We know that,

$\mathrm{ a=\frac{\left(M_2-M_1\right) }{\left(M_1+M_2\right)}g }$

For case 1:

$\mathrm{ a_{1}=\left ( \frac{\left(2M_2)-(M_1\right) }{\left(2M_1)+M_2\right)} \right )g }$

for case 2:
$\mathrm{ a_2=\left(\frac{3 M_1-M_1}{3 M_1+M_1}\right) g }$

$\mathrm{a_{2}=\frac{9}{2}}\rightarrow (2)$

From the eq (1) & (2)

$\mathrm{\frac{a_{1}}{a_{2}}=\frac{2}{3}}$

Hence (2) is correct option.

Posted by

Ramraj Saini

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