Two blocks of mass M_1=20 ~kg and M_2=12 ~kg are connected by a metal rod of mass 8 ~kg . The system is pulled vertically up by applying a force of 480 ~N as shown. The tension at the mid-point of the rod is:

Correct answer: D. 192 ~N Acceleration produced in upward direction aligned & a= F M_1+M_2+ Mass of metal rod & =(480)/(20+12+8)=12 ms^ -2 aligned Tension at the mid point aligned T& = ( mathrm M _ 2 + Mass of rod 2 )a & =(12+4) ×12=192 N aligned

JEE Main 201322 AprilOnline

Exam: JEE Main
Year: 2013
Date: 22 April
Mode: Online
Subject: Physics
Chapter: Laws of Motion

JEE Main 2013 · 22 April · Online · Physics · Laws of Motion

Two blocks of mass M_1=20 ~kg and M_2=12 ~kg are connected by a...

Two blocks of mass

M_1=20 \mathrm{~kg}

and

M_2=12 \mathrm{~kg}

are connected by a metal rod of mass

8 \mathrm{~kg}

. The system is pulled vertically up by applying a force of

480 \mathrm{~N}

as shown. The tension at the mid-point of the rod is: Two blocks of mass M_1=20 ~kg and M_2=12 ~kg are connected by a metal rod of mass 8 ~kg . The system is pulled

Options

A $144 \mathrm{~N}$
B $96 \mathrm{~N}$
C $240 \mathrm{~N}$
D $192 \mathrm{~N}$

Correct answer

D. $192 \mathrm{~N}$

Step-by-step solution

Acceleration produced in upward direction

\begin{aligned}& \mathrm{a}=\frac{\mathrm{F}}{\mathrm{M}_1+\mathrm{M}_2+\text{ Mass\text{ of }\text{ metal }\text{ rod } }}\\ & =\frac{480}{20+12+8}=12 \mathrm{ ms}^{-2} \end{aligned}

Tension at the mid point

\begin{aligned}\mathrm{T}& = \left(\m mathrm{M}}_{2}+\frac{\text{ Mass of rod } }{2}\right)\mathrm{a}\\ & =(12+4) \times12=192 \mathrm{ N} \end{aligned}

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