Pendulum beating seconds at the equator (g = 978 cm / sec2) is taken to Antarctica (g=983 cm/sec2). If initial length is 100cm its length is to be increased by

T=2πℓg; ℓ1=100cm( seconds pendulum) ℓ2ℓ1=g2g1=983978⇒ℓ2−ℓ1ℓ1=983978−1=983−978978ℓ2−ℓ1=100×983−978978=0.5cm

Pendulum beating seconds at the equator (g = 978 cm / sec2) is taken to Antarctica (g=983 cm/sec2). If initial length is 100cm its length is to be increased by

T=2πℓg; ℓ1=100cm( seconds pendulum) ℓ2ℓ1=g2g1=983978⇒ℓ2−ℓ1ℓ1=983978−1=983−978978ℓ2−ℓ1=100×983−978978=0.5cm

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Pendulum beating seconds at the equator (g = 978 cm / sec²) is taken to Antarctica (g=983 cm/sec²). If initial length is 100cm its length is to be increased by

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0.2 cm

0.1 cm

0.4 cm

0.5 cm

answer is D.

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Detailed Solution

$\begin{array}{l} T = 2 π \sqrt{\frac{ℓ}{g}}; ℓ_{1} = 100 cm (seconds pendulum) \\ \frac{ℓ_{2}}{ℓ_{1}} = \frac{g_{2}}{g_{1}} = \frac{983}{978} \Rightarrow \frac{ℓ_{2} - ℓ_{1}}{ℓ_{1}} = \frac{983}{978} - 1 = \frac{983 - 978}{978} \\ ℓ_{2} - ℓ_{1} = 100 \times \frac{983 - 978}{978} = 0 .5 cm \end{array}$