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The period of oscillation of a simple pendulum of length \(l\) suspended from the roof of a vehicle, which moves without friction down an inclined plane of inclination \(\alpha\), is given by
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Verified Answer
The correct answer is:
\(2 \pi \sqrt{\frac{1}{g \cos \alpha}}\)
Hints: 
\(\begin{aligned}
& \mathrm{g}_{\text {eff }}=\mathrm{g} \cos \alpha \\
& \mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}_{\text {eff }}}}
\end{aligned}\)
\(\begin{aligned}
& \mathrm{g}_{\text {eff }}=\mathrm{g} \cos \alpha \\
& \mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}_{\text {eff }}}}
\end{aligned}\)
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