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QJEEM24Phcbb6: JEE Main 2024 Physics - Class 11 JEEM24Phcbb6 Solution: A light unstretchable string passing over a smooth light pulley …

A light unstretchable string passing over a smooth light pulley connects two blocks of masses m1m_1 and m2m_2 . If the acceleration of the system is g8\frac{g}{8} , then the ratio of the masses m2m1\frac{m_2}{m_1} is :

Multiple Choice Options

  • A 5:35: 3
  • B 8:18: 1
  • C 9:79: 7
  • D 4:34: 3

Step-by-Step Verified Solution

To find the ratio of the masses m2m1\frac{m_2}{m_1} given that the acceleration of the system is g8\frac{g}{8} , where gg is the acceleration due to gravity, we need to analyze the forces acting on each mass and apply Newton's second law of motion.

For mass m1m_1 , the force acting downward (towards the centre of the Earth) is the gravitational force m1gm_1g , and the tension TT acts upward. Its equation of motion, considering downward as positive, can be given by:

m1g−T=m1am_1g - T = m_1a

For mass m2m_2 , the force acting downward is m2gm_2g , but since it is on the opposite side of the pulley, the tension TT is upwards (considering upward movement as positive direction), so we have:

T−m2g=m2aT - m_2g = m_2a

Since the pulley is light and smooth, the tension TT is the same on both sides of the pulley. Also, the system accelerates together, so a=g8a = \frac{g}{8} .

Adding the two equations to eliminate TT gives us:

m1g−m2g=(m1−m2)am_1g - m_2g = (m_1 - m_2)a

g(m1−m2)=(m1−m2)g8g(m_1 - m_2) = (m_1 - m_2)\frac{g}{8}

Canceling out gg and dividing both sides by m1−m2m_1 - m_2 (assuming m1≠m2m_1 \neq m_2 ), we get:

1=181 = \frac{1}{8}

This simplification doesn't align with finding the ratio directly, indicating a mistake in handling the simultaneous equations relation to aa and TT . Let's correct our approach to finding the ratio of the masses based on the acceleration and tension.

Since the acceleration a=g8a = \frac{g}{8} , and considering the system as a whole, the net force causing the acceleration is the difference in gravitational forces on the two masses. This net force provides the system's entire acceleration. Thus, correctly setting up the equations for the system should give:

m2g−m1g=(m1+m2)g8m_2g - m_1g = (m_1 + m_2)\frac{g}{8}

g(m2−m1)=g8(m1+m2)g(m_2-m_1) = \frac{g}{8}(m_1 + m_2)

8(m2−m1)=m1+m28(m_2 - m_1) = m_1 + m_2

8m2−8m1=m1+m28m_2 - 8m_1 = m_1 + m_2

7m2=9m17m_2 = 9m_1

m2m1=97\frac{m_2}{m_1} = \frac{9}{7}

Therefore, the correct ratio of the masses m2m1\frac{m_2}{m_1} is 9:79:7 , which corresponds to Option C.

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