A simple pendulum of length $l_{1}$ has time period $T_{1}$ and another pendulum of length $l_{2}$ has time period $T_{2}$ then time period of pendulum of length $(l_{1} + l_{2})$ is

T_{1} + T_{2}

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\sqrt{T_{1} + T_{2}}

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\sqrt{T_{1} T_{2}}

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\sqrt{T_{1}^{2} + T_{2}^{2}}

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Solution

The correct option is C $\sqrt{T_{1}^{2} + T_{2}^{2}}$
$\begin{matrix} Simple pendulum of length ℓ_{1} & Time period T_{1} Simple pendulum of length ℓ_{2} & Time period T_{2} \end{matrix}$
$T = 2 π \sqrt{\frac{l}{g}}$
$\begin{matrix} \Rightarrow l = T^{2} \times \frac{g}{4 π^{2}} \Rightarrow l_{1} = T_{1}^{2} \cdot \frac{g}{4 π^{2}} and l_{2} = T_{2}^{2} \frac{g}{4 π^{2}} \Rightarrow l_{1} + l_{2} = (T_{1}^{2} + T_{2}^{2}) \times \frac{g}{4 π^{2}} \end{matrix}$
$\begin{matrix} \sqrt{l_{1} + l_{2}} = \sqrt{T_{1}^{2} + T_{2}^{2}} \cdot \frac{\sqrt{g}}{2 π} \Rightarrow \sqrt{T_{1}^{2} + T_{2}^{2}} & = 2 π \sqrt{\frac{l_{1} + l_{2}}{g}} \end{matrix}$
$\begin{matrix} Now comparing with time period of a pendulum of length (ℓ_{1} + l_{2}) \Rightarrow T^{'} = 2 π \sqrt{\frac{l_{1} + l_{2}}{g}} = \sqrt{T_{1}^{2} + T_{2}^{2}} \Rightarrow Time period = \sqrt{T_{1}^{2} + T_{2}^{2}} \end{matrix}$

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