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AIIMS 201826 May
Exam
AIIMS
Year
2018
Date
26 May
Subject
Physics
Chapter
Laws of Motion

AIIMS 2018 · 26 May · Physics · Laws of Motion

In the diagram 100 ~kg block is hanging from a pulley and force F...

In the diagram 100 kg100 \mathrm{~kg} block is hanging from a pulley and force FF is applied on the string to keep the system in equilibrium. If the whole system is moving up with a constant velocity vv, then tension at point PP in the string will be In the diagram 100 ~kg block is hanging from a pulley and force F is applied on the string to keep the system in

Options

  1. A735 N735 \mathrm{~N}
  2. B1050 N1050 \mathrm{~N}
  3. C1470 N1470 \mathrm{~N}
  4. D1200 N1200 \mathrm{~N}

Correct answer

C. 1470 N1470 \mathrm{~N}

Step-by-step solution

Tension in each part of the string will be F. Assume tension at point PP is TT. In the diagram 100 ~kg block is hanging from a pulley and force F is applied on the string to keep the system in
T=3F2F=mg2F=100×9.8 NF=490 N∴T=3×490=1470 N\begin{aligned} & T=3 F \\ & 2 F=m g \\ & 2 F=100 \times 9.8 \mathrm{~N} \\ & F=490 \mathrm{~N} \\ & \therefore \quad T=3 \times 490=1470 \mathrm{~N}\end{aligned}
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