MHT CET · Physics · Work Power Energy
The string of pendulum of length ' \(L\) ' is displaced through \(90^{\circ}\) from the vertical and released. Then the maximum strength of the string in order to withstand the tension, as the pendulum passes through the mean position is ( \(\mathrm{m}=\) mass of pendulum, \(\mathrm{g}=\) acceleration due to gravity)
- A mg
- B 3 mg
- C 5 mg
- D 6 mg
Answer & Solution
Correct Answer
(B) 3 mg
Step-by-step Solution
Detailed explanation
By conservation of energy
\(\begin{aligned}
& \text { K.E. }=\frac{1}{2} \mathrm{mv}^2=\mathrm{mg} l \\
& \mathrm{v}=\sqrt{2 \mathrm{~g} l}
\end{aligned}\)
The forces acting on the pendulum at mean position are tension in string, centripetal force and weight of pendulum.
\(\begin{aligned}
& \mathrm{T}-\frac{\mathrm{mv}^2}{l}=\mathrm{mg} \\
& \mathrm{~T}-\frac{\mathrm{m}(\sqrt{2 \mathrm{~g} l})^2}{l}=\mathrm{mg} \\
& \therefore \quad \mathrm{~T}=3 \mathrm{mg}
\end{aligned}\)
...[From (i)]
\(\begin{aligned}
& \text { K.E. }=\frac{1}{2} \mathrm{mv}^2=\mathrm{mg} l \\
& \mathrm{v}=\sqrt{2 \mathrm{~g} l}
\end{aligned}\)
The forces acting on the pendulum at mean position are tension in string, centripetal force and weight of pendulum.
\(\begin{aligned}
& \mathrm{T}-\frac{\mathrm{mv}^2}{l}=\mathrm{mg} \\
& \mathrm{~T}-\frac{\mathrm{m}(\sqrt{2 \mathrm{~g} l})^2}{l}=\mathrm{mg} \\
& \therefore \quad \mathrm{~T}=3 \mathrm{mg}
\end{aligned}\)
...[From (i)]
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