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JEE Main 202226 JulyMorning Shift
Exam
JEE Main
Year
2022
Date
26 July
Shift
Morning Shift
Subject
Physics
Chapter
Laws of Motion

JEE Main 2022 · 26 July · Morning Shift · Physics · Laws of Motion

Three masses M = 100kg, m _ 1 = 10kg and m _ 2 = 20kg are arranged...

Three masses M=100kg,m1=10kgM = 100kg, {m}_{1} = 10kg and m2=20kg{m}_{2} = 20kg 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 FF is applied on the system so that the mass m2{m}_{2} moves upward with an acceleration of 2ms−22{ms}^{- 2}. The value of FF is

(Take g=10ms−2g = 10{ms}^{- 2})

Three masses M = 100kg, m _ 1 = 10kg and m _ 2 = 20kg are arranged in a system as shown in figure. All the surfaces are

Options

  1. A3360N3360N
  2. B3380N3380N
  3. C3120N3120N
  4. D3240N3240N

Correct answer

A. 3360N3360N

Step-by-step solution

Let acceleration of 100 kg100 \mathrm{~kg} block =a1=a_1
FBD of 100 kg100 \mathrm{~kg} block w.r.t ground
Three masses M = 100kg, m _ 1 = 10kg and m _ 2 = 20kg are arranged in a system as shown in figure. All the surfaces are
F−T−N1=100a1\mathrm{F}-\mathrm{T}-\mathrm{N}_1=100 \mathrm{a}_1 ... (i)
FBD of 20 block wrt 100 kg100 \mathrm{~kg}
Three masses M = 100kg, m _ 1 = 10kg and m _ 2 = 20kg are arranged in a system as shown in figure. All the surfaces are
T−20 g=20(2)\mathrm{T}-20 \mathrm{~g}=20(2)
T=240\mathrm{T}=240 ... (ii)
 N1=20a1\mathrm{~N}_1=20 \mathrm{a}_1 ... (iii)
FBD of 10 kg10 \mathrm{~kg} block wrt 100 kg100 \mathrm{~kg}
Three masses M = 100kg, m _ 1 = 10kg and m _ 2 = 20kg are arranged in a system as shown in figure. All the surfaces are
10a1−240=10(2)10 \mathrm{a}_1-240=10(2)
a1=26 m/s2\mathrm{a}_1=26 \mathrm{~m} / \mathrm{s}^2
 F−240−20(26)=100×26\mathrm{~F}-240-20(26)=100 \times 26
⇒F=3360 N\Rightarrow \mathrm{F}=3360 \mathrm{~N}
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