QJEEM24Phd991: JEE Main 2024 Physics - Class 11 JEEM24Phd991 Solution: A light string passing over a smooth light pulley connects two…

Given that the acceleration of the system is $\frac{g}{\sqrt{2}}$ , let's analyze the forces acting on both masses to find the ratio $\frac{m_1}{m_2}$ .

The tension force in the string is equal for both masses since the pulley and string are light and smooth. Assuming the pulley only changes the direction of the tension force without affecting its magnitude, we can analyze the forces in the vertical direction for both masses.

The net force acting on $m_1$ (towards the pulley) would be the tension $T$ minus its weight component in the direction of motion, which we will assume is $m_1g$ , and for $m_2$ , it would be its weight component in the direction of motion, which we can assume is $m_2g$ minus the tension $T$ . Given that $m_2$ is moving downwards and $m_1$ is moving upwards, and considering that m_2>m_1 , the acceleration $a$ of both masses will be the same and can be described as:

For $m_1$ :

$T - m_1g = m_1a$

For $m_2$ :

$m_2g - T = m_2a$

Adding these equations to eliminate $T$ gives:

$m_2g - m_1g = m_1a + m_2a$

Given that the acceleration $a$ of the system is $\frac{g}{\sqrt{2}}$ , we can substitute $a$ with $\frac{g}{\sqrt{2}}$ in the equation:

$m_2g - m_1g = m_1\left(\frac{g}{\sqrt{2}}\right) + m_2\left(\frac{g}{\sqrt{2}}\right)$

Simplifying and factoring out $g$ , we get:

$m_2 - m_1 = \frac{m_1}{\sqrt{2}} + \frac{m_2}{\sqrt{2}}$

Multiplying every term by $\sqrt{2}$ to clear the denominator, we get:

$\sqrt{2}(m_2 - m_1) = m_1 + m_2$

Rearranging the terms to isolate the masses on one side, we obtain:

$\sqrt{2}m_2 - m_2 = m_1(\sqrt{2} + 1)$

This simplifies to:

$m_2(\sqrt{2} - 1) = m_1(\sqrt{2} + 1)$

Hence, the ratio $\frac{m_1}{m_2}$ is:

$\frac{m_1}{m_2} = \frac{\sqrt{2} - 1}{\sqrt{2} + 1}$

This corresponds to Option C $\frac{\sqrt{2}-1}{\sqrt{2}+1}$ .