MHT CET · Physics · Laws of Motion
A bob of a simple pendulum of mass ' \(\mathrm{m}\) ' is displaced through \(90^{\circ}\) from mean position and released. When the bob is at lowest position, the tension in the string is
- A 4mg
- B 2mg
- C mg
- D 3mg
Answer & Solution
Correct Answer
(D) 3mg
Step-by-step Solution
Detailed explanation
When it is displaced through \(90^{\circ}\) from mean position, it is at a height ' \(r\) ' and has potential energy mgr. At the lowest position this potential energy is converted into kinetic energy.
\(\begin{aligned}& \therefore \frac{1}{2} \mathrm{mv}^2=\mathrm{mgr} \\& \therefore \mathrm{mv}^2=\mathrm{mgr}
\end{aligned}\)
At the lowest position the tension in the string
\(\mathrm{T}=\mathrm{mg}+\frac{\mathrm{mv}^2}{\mathrm{r}}=\mathrm{mg}+2 \mathrm{mg}=3 \mathrm{mg}\)
\(\begin{aligned}& \therefore \frac{1}{2} \mathrm{mv}^2=\mathrm{mgr} \\& \therefore \mathrm{mv}^2=\mathrm{mgr}
\end{aligned}\)
At the lowest position the tension in the string
\(\mathrm{T}=\mathrm{mg}+\frac{\mathrm{mv}^2}{\mathrm{r}}=\mathrm{mg}+2 \mathrm{mg}=3 \mathrm{mg}\)
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