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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
Options:
(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
Solution:
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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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