Laws of Motion

JEE Main level, single correct, medium

53 of 210

Question 53 of 210

Two blocks of mass M1=20 kgM_1=20 \mathrm{~kg} and M2=12 kgM_2=12 \mathrm{~kg} are connected by a metal rod of mass 8 kg8 \mathrm{~kg}. The system is pulled vertically up by applying a force of 480 N480 \mathrm{~N} as shown. The tension at the mid-point of the rod is:

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Answer: (D) 192 N192 \mathrm{~N}

Acceleration produced in upward direction
a=FM1+M2+ Mass of metal rod =48020+12+8=12 ms−2\begin{aligned} & \mathrm{a}=\frac{\mathrm{F}}{\mathrm{M}_1+\mathrm{M}_2+\text { Mass of metal rod }} \\ & =\frac{480}{20+12+8}=12 \mathrm{~ms}^{-2} \end{aligned}
Tension at the mid point
T=(M2+ Mass of rod 2)a=(12+4)×12=192 N\begin{aligned} \mathrm{T} & =\left(\mathrm{M}_2+\frac{\text { Mass of rod }}{2}\right) \mathrm{a} \\ & =(12+4) \times 12=192 \mathrm{~N} \end{aligned}

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