JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Laws of Motion

Three masses

M = 100 kg, m_{1} = 10 kg

and

m_{2} = 20 kg

are arranged in a system as shown in figure. All the surfaces are frictionless and strings are inextensible and weightless. The pulleys are also weightless and frictionless. A force F is applied on the system so that the mass

m_{2}

moves upward with an acceleration of

2 ms^{- 2}

. The value of F is :
(Take

g = 10 ms^{- 2}

)

(A)3360 N✓ Correct
(B)3380 N
(C)3120 N
(D)3240 N

Answer: (A)

Solution

For the 20 kg block moving upward with acceleration

a_{2} = 2 m/s^{2}

, the tension is

T - 20 g = 20 (2)

, giving

T = 240 N

.

For the 10 kg block relative to the 100 kg block:

10 a_{1} - 240 = 10 (2)

, so

a_{1} = 26 m/s^{2}

.

For the 100 kg block:

F - T - N_{1} = 100 a_{1}

, where

N_{1} = 20 a_{1} = 520 N

.

Therefore:

F - 240 - 520 = 2600

, giving

F = 3360 N

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Single correct option

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JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Laws of Motion

Solution

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