Question

6.
A second's pendulum is taken to a planet where acceleration due to gravity is 9 times that on the earth. What is the time period of the
pendulum on that planet?
2 Marks
O 6.6. s
O 0.66 s
O 0.33 s
015​

Answer 1

Given :

➳ Acceleration due to gravity is 9 times that on the earth.

To Find :

⟶ Time period of second pendulum on that planet.

SoluTion :

➠ Second pendulum is the simple pendulum, having a time period of 2 second. Its effective length is 99.992 cm or approximate one metre on earth.

➠ Formula of time period in terms of length of simple pendulum and acceleration due to gravity is given by

$\bigstar\:\boxed{\bf{\gray{T=2\pi\sqrt{\dfrac{L}{g}}}}}$

➠ ATQ, length of simple pendulum remains same on both planets, so we can say that time period of simple pendulum is inversely proportional to the square root of acceleration due to gravity.

$\bigstar\:\boxed{\bf{\gray{T\propto\:\dfrac{1}{\sqrt{g}}}}}$

Let,

▪ Time period on earth = T

▪ Acc. due to gravity of earth = g

▪ Time period on planet = T'

▪ Acc. due to gravity of planet = g'

⇒ T/T' = √g'/√g

⇒ 2/T' = √9g/√g

⇒ T' = 2/√9

⇒ T' = 2/3

⇒ T' = 0.66 s

Answer 2

Required answer:

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As per the given data using the simple pendulum formula, we can solve this question.

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Simple pendulum formula:

$\longrightarrow\rm{T = 2 \: \pi \sqrt{\dfrac{L}{g}}}$

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Now, about the forumula:

• T1 denotes the time period on Earth.
• T2 denotes the time on the unknown planet.
• g1 denotes the gravity on Earth.
• g2 denotes the gravity on the unknown planet.

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Using the above forumula:

$\longrightarrow\rm{\dfrac{T1}{T2} = \dfrac{\sqrt{g2}}{\sqrt{g1}}}$

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$\longrightarrow\rm{\dfrac{2}{T2} = \dfrac{\sqrt{9g}}{\sqrt{g1}}}$

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$\longrightarrow\rm{T2 = \dfrac{2}{\sqrt{9}}}$

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$\longrightarrow\rm{T2 = \dfrac{2}{3}}$

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$\longrightarrow\rm{T2 = 0.66 \: seconds}$

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Therefore, option (B) is the answer.