Mathematics in Physics

JEE Main level, single correct, medium

87 of 121

Question 87 of 121

If the length of the pendulum in pendulum clock increases by 0.10.1%, then the error in time per day is:

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Answer: (A) 43.2s43.2s

T=2πlgT=2\pi \sqrt{\dfrac{l}{g}}
T∝l12T\propto {l}^{\dfrac{1}{2}}
ΔTT=12Δll\dfrac{\Delta T}{T}=\dfrac{1}{2}\dfrac{\Delta l}{l}
ΔT24×3600=12×0.1100\dfrac{\Delta T}{24\times 3600}=\dfrac{1}{2}\times \dfrac{0.1}{100}
ΔT=12×0.1100×24×3600\Delta T=\dfrac{1}{2}\times \dfrac{0.1}{100}\times 24\times 3600
ΔT=43.2s\Delta T=43.2\text{s}

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