Search any question & find its solution
Question:
Answered & Verified by Expert
A mass of $1 \mathrm{kg}$ is suspended by means of a thread. The system is
(i) lifted up with an acceleration of $4.9 \mathrm{ms}^{-2}$
(ii) lowered with an acceleration of $4.9 \mathrm{ms}^{-2}$. The ratio of tension in the first and second case is
Options:
(i) lifted up with an acceleration of $4.9 \mathrm{ms}^{-2}$
(ii) lowered with an acceleration of $4.9 \mathrm{ms}^{-2}$. The ratio of tension in the first and second case is
Solution:
1090 Upvotes
Verified Answer
The correct answer is:
3: 1
When the body of mass m lifted up the forces acting on the body
$$
\begin{aligned}
T_{1}-m g &=\frac{m g}{2} \\
T_{1} &=m g+\frac{m g}{2}=\frac{3 m g}{2}
\end{aligned}
$$
(ii) When the body lowered the acting forces
$$
\left(m g-T_{2}\right)=\frac{m g}{2}
$$
$$
T_{2}=m g-\frac{m g}{2}=\frac{m q}{2}
$$
$\therefore$
$$
\begin{aligned}
\frac{T_{1}}{T_{2}} &=\frac{3 m g / 2}{m g / 2}=\frac{3}{1} \\
&=3: 1
\end{aligned}
$$
$$
\begin{aligned}
T_{1}-m g &=\frac{m g}{2} \\
T_{1} &=m g+\frac{m g}{2}=\frac{3 m g}{2}
\end{aligned}
$$
(ii) When the body lowered the acting forces
$$
\left(m g-T_{2}\right)=\frac{m g}{2}
$$
$$
T_{2}=m g-\frac{m g}{2}=\frac{m q}{2}
$$
$\therefore$
$$
\begin{aligned}
\frac{T_{1}}{T_{2}} &=\frac{3 m g / 2}{m g / 2}=\frac{3}{1} \\
&=3: 1
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.