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Q.

A simple pendulum has time period T₁. The point of suspension is now moved upward according to equation 2ms^-2. If new time period is T₂, then ratio $\frac{T_{1}^{2}}{T_{2}^{2}}$ will be ( g=10ms^-2)

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a

2/3

b

5/6

c

6/5

d

3/2

answer is C.

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Detailed Solution

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$\begin{array}{l} T_{1} = 2 π \sqrt{\frac{1}{g}} \\ T_{2} = 2 π \sqrt{\frac{1}{g + a}} \\ T_{q} = 2 π \sqrt{\frac{1}{g + 2}} \\ T_{1}^{2} = 4 π^{2} \frac{1}{g} \end{array}$
$\begin{array}{l} T_{2}^{2} = 4 π^{2} \frac{1}{g + 2} \\ \frac{T_{1}^{2}}{T_{2}^{2}} = \frac{1}{g} \times \frac{g + 2}{1} \\ = \frac{12}{10} = \frac{6}{5} \end{array}$

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