MHT CET · Physics · Work Power Energy
A body of mass ' \(\mathrm{m}\) ' attached at the end of a string is just completing the loop in a vertical circle. The apparent weight of the body at the lowest point in its path is ( \(\mathrm{g}=\) gravitational acceleration)
- A zero
- B \(\mathrm{mg}\)
- C \(3 \mathrm{mg}\)
- D \(6 \mathrm{mg}\)
Answer & Solution
Correct Answer
(D) \(6 \mathrm{mg}\)
Step-by-step Solution
Detailed explanation
The tension at the lowest point is \(\mathrm{T}=\frac{\mathrm{mv}^2}{\mathrm{r}}+\mathrm{mg}\) To complete the vertical circle, the minimum velocity should be \(\mathrm{v}=\sqrt{5 \mathrm{gr}}\)
\( \begin{aligned} \therefore \mathrm{T} & =\frac{\mathrm{m}(\sqrt{5 \mathrm{gr}})^2}{\mathrm{r}}+\mathrm{mg} \\ \mathrm{R} & =5 \mathrm{mg}+\mathrm{mg} \\ \therefore \mathrm{T} & =6 \mathrm{mg} \end{aligned} \)
\( \begin{aligned} \therefore \mathrm{T} & =\frac{\mathrm{m}(\sqrt{5 \mathrm{gr}})^2}{\mathrm{r}}+\mathrm{mg} \\ \mathrm{R} & =5 \mathrm{mg}+\mathrm{mg} \\ \therefore \mathrm{T} & =6 \mathrm{mg} \end{aligned} \)
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