Two simple pendulums of lengths L₁ and L₂ have periodic time T₁ and T₂ respectively (T₁ > T₂). The time period of the pendulum of length (L₁ - L₂) is [(L₁ - L₂) > 60 cm ] (a) √(T₁² + T₂²) (b) √(T₁² - T₂²) (c) T₁ + T₂ (d) T₁ - T₂

Answer: (b). T² = (4π²)/(g)L, so T² ∝ L. For length (L₁ - L₂): T² = T₁² - T₂² So the answer is (b) √(T₁² - T₂²).

Two simple pendulums of lengths L₁ and L₂ have periodic time T₁ and… | MHT CET 2026 (17 Apr, Shift 1)

Q10MHT CET 2026 · 17 APR · SHIFT 1Medium

Simple Pendulum

Two simple pendulums of lengths $L_1$ and $L_2$ have periodic time $T_1$ and $T_2$ respectively $(T_1 > T_2)$ . The time period of the pendulum of length $(L_1 - L_2)$ is

$[(L_1 - L_2) > 60$ cm $]$

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Answer: (b)

$T^2 = \frac{4\pi^2}{g}L$ , so $T^2 \propto L$ .
For length $(L_1 - L_2)$ : $T^2 = T_1^2 - T_2^2$
So the answer is (b) $\sqrt{T_1^2 - T_2^2}$ .

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