The length of a seconds pendulum is $1 m$ . The length of a seconds pendulum at a depth equal to one fourth of the radius of the Earth is

0.5 m

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0.75 m

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0.25 m

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1 m

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Solution

The correct option is C $0.75 m$
$g = g_{o} (1 - \frac{h}{R})$
$T = 2 π \sqrt{\frac{1}{g_{o}}}$
$T = 1$
$T = 2 π \sqrt{\frac{l_{2}}{g_{2}}}$
$= 2 π \sqrt{\frac{l_{2}}{g_{o} (1 - \frac{R / 4}{R})}}$
$∴ \sqrt{\frac{l_{2}}{g_{o} (3 / 4)}} = \sqrt{\frac{1}{g}}$
$∴ l_{2} = 3 / 4 = 0.75 m$

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The length of a seconds pendulum is 1 m. The length of a seconds pendulum at a depth equal to one fourth of the radius of the Earth is

The correct option is C 0.75 mg=go(1−hR)T=2π√1goT=1T=2π√l2g2=2π  ⎷l2go(1−R/4R)∴√l2go(3/4)=√1g∴l2=3/4=0.75m

The length of a seconds pendulum is $1 m$ . The length of a seconds pendulum at a depth equal to one fourth of the radius of the Earth is

The correct option is C $0.75 m$
$g = g_{o} (1 - \frac{h}{R})$
$T = 2 π \sqrt{\frac{1}{g_{o}}}$
$T = 1$
$T = 2 π \sqrt{\frac{l_{2}}{g_{2}}}$
$= 2 π \sqrt{\frac{l_{2}}{g_{o} (1 - \frac{R / 4}{R})}}$
$∴ \sqrt{\frac{l_{2}}{g_{o} (3 / 4)}} = \sqrt{\frac{1}{g}}$
$∴ l_{2} = 3 / 4 = 0.75 m$