Let ' l₁ ' be the length of simple pendulum. Its length changes to ' l₂ ' to increase the periodic time by 20%. The ratio (l₂)/(l₁) = (a) 1.22 (b) 1.33 (c) 1.44 (d) 1.55

Answer: (c). T ∝ √(l) ⇒ (l₂)/(l₁) = ((T₂)/(T₁))² A 20% increase means (T₂)/(T₁) = 1.2. (l₂)/(l₁) = 1.2² = 1.44 So the answer is (c) 1.44.

Let ' l₁ ' be the length of simple pendulum. Its length changes to '… | MHT CET 2024 (10 May, Shift 1)

Q35MHT CET 2024 · 10 MAY · SHIFT 1Medium

Simple Pendulum

Let ' $l_1$ ' be the length of simple pendulum. Its length changes to ' $l_2$ ' to increase the periodic time by $20\%$ . The ratio $\frac{l_2}{l_1} =$

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Answer: (c)

$T \propto \sqrt{l} \Rightarrow \frac{l_2}{l_1} = \left(\frac{T_2}{T_1}\right)^2$
A 20% increase means $\frac{T_2}{T_1} = 1.2$ .
$\frac{l_2}{l_1} = 1.2^2 = 1.44$
So the answer is (c) 1.44.

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