The time period of simple pendulum is ‘T’. When the length increases by 10 cm, its period is T₁. When the length is decreased by 10 cm, its period is T₂. Then the relation between T, T₁ and T₂ is

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$2 T^{2} = T_{1}^{2} - T_{2}^{2}$

$\frac{2}{T^{2}} = \frac{1}{T_{1}^{2}} + \frac{1}{T_{2}^{2}}$

$2 T^{2} = {T_{1}}^{2} + T_{2}^{2}$

$\frac{2}{T^{2}} = \frac{1}{T_{1}^{2}} - \frac{1}{T_{2}^{2}}$

answer is C.

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Detailed Solution

$\begin{array}{l} T \propto \sqrt{ℓ} \Rightarrow ℓ = {kT}^{2} \dots \dots \dots (1) \\ ℓ + 10 = {kT}_{1}^{2} \dots \dots . . (2) \\ ℓ - 10 = {kT}_{2}^{2} \dots \dots \dots (3) \\ (2) + (3) \Rightarrow 2 ℓ = k (T_{1}^{2} + T_{2}^{2}) \\ 2 {kT}^{2} = k (T_{1}^{2} + T_{2}^{2}) 2 T^{2} = T_{1}^{2} + T_{2}^{2} \end{array}$