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Choose the correct option · Q3

Q.The length of second's pendulum on the surface of earth is nearly 1 m. Its length on the surface of moon should be [Given: acceleration due to gravity

(g) on moon is 1/6th of that on the earth's surface] (A) 16\frac{1}{6} m
(B) 6 m
(C) 136\frac{1}{36} m
(D) 16\frac{1}{\sqrt6} m
Maharashtra MsbshseTextbookMCQImportance★★★★★
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A second's pendulum has a fixed period T=2T=2 s, so from T=2πL/gT=2\pi\sqrt{L/g}, L=gT24π2L=\frac{gT^2}{4\pi^2}. Since TT (and hence T2/4π2T^2/4\pi^2) is the same on the earth and the moon, L∝gL\propto g at each location. Given gmoon=gearth/6g_{moon}=g_{earth}/6 and $L_{earth}\approx …

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