JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Laws of Motion

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A hanging mass M is connected to a four times bigger mass by using a string-pulley arrangement, as shown in the figure. The bigger mass is placed on a horizontal ice-slab and being pulled by 2 Mg force. In this situation, tension in the string is

\frac{X}{5}

Mg for

x =

____. Neglect mass of the string and friction of the block (bigger mass) with ice slab. (Given

g =

acceleration due to gravity)

Answer: 6

Solution

For the system, apply Newton's second law to both masses. For the 4M mass:

2 M g - T = 4 M a

. For the hanging mass M:

T - M g = M a

. Adding these equations:

M g = 5 M a

, so

a = \frac{g}{5}

. Substituting back:

T = M g + M a = M g + M \frac{g}{5} = \frac{6 M g}{5}

. Therefore,

x = 6

JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Integer answer

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