Two masses M1 and .M2 are tied together at the two ends of a light inextensible string that passes over a frictionless pulley. When the mass M2 is twice that of M1. the acceleration of the system is a1. When the mass M2 is thrice that of M1. The acceleration of The system is a2. The ratio a1a2 will be:

a=m2g−m1gm1+m2Case 1 M2=2m1 a1=2m1g−m1g3m1 a1=g/3Case - 2 M2=3m1a2=3m1g−m1g4m1a2=g2a1a2=g3g2=23

Two masses M1 and .M2 are tied together at the two ends of a light inextensible string that passes over a frictionless pulley. When the mass M2 is twice that of M1. the acceleration of the system is a1. When the mass M2 is thrice that of M1. The acceleration of The system is a2. The ratio a1a2 will be:

a=m2g−m1gm1+m2Case 1 M2=2m1 a1=2m1g−m1g3m1 a1=g/3Case - 2 M2=3m1a2=3m1g−m1g4m1a2=g2a1a2=g3g2=23

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Two masses M₁ and .M₂ are tied together at the two ends of a light inextensible string that passes over a frictionless pulley. When the mass M₂ is twice that of M₁. the acceleration of the system is a₁. When the mass M₂ is thrice that of M₁. The acceleration of The system is a₂. The ratio $\frac{a_{1}}{a_{2}}$ will be:

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$\frac{2}{3}$

$\frac{1}{2}$

$\frac{1}{3}$

$\frac{3}{2}$

answer is B.

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Detailed Solution

$\begin{array}{l} a = \frac{m_{2} g - m_{1} g}{m_{1} + m_{2}} \\ Case 1 M_{2} = 2 m_{1} \\ a_{1} = \frac{2 m_{1} g - m_{1} g}{3 m_{1}} \\ a_{1} = g / 3 \end{array}$

Case - 2

$\begin{array}{l} M_{2} = 3 m_{1} \\ a_{2} = \frac{3 m_{1} g - m_{1} g}{4 m_{1}} \\ a_{2} = \frac{g}{2} \\ \frac{a_{1}}{a_{2}} = \frac{\frac{g}{3}}{\frac{g}{2}} = \frac{2}{3} \end{array}$