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Question: Answered & Verified by Expert
A mass $1 \mathrm{~kg}$ is suspended by a thread. It is
(i) lifted up with an acceleration $4.9 \mathrm{~m} / \mathrm{s}^2$
(ii) lowered with an acceleration $4.9 \mathrm{~m} / \mathrm{s}^2$
The ratio of the tensions is
PhysicsLaws of MotionJEE Main
Options:
  • A $3: 1$
  • B $1: 3$
  • C $1: 2$
  • D $2: 1$
Solution:
2063 Upvotes Verified Answer
The correct answer is: $3: 1$
$\begin{aligned} & T_1=m(g+a)=1 \times\left(g+\frac{g}{2}\right)=\frac{3 g}{2} \\ & T_2=m(g-a)=1 \times\left(g-\frac{g}{2}\right)=\frac{g}{2} \quad \therefore \quad \frac{T_1}{T_2}=\frac{3}{1}\end{aligned}$

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