Question

Find the length of a seconds pendulum at a place where g=10ms-2(Take π=3.14)

Answer 1

Solution :

⏭ Given:

✏ Gravitational acceleration = 10m/s$^2$

✏ Value of π = 3.14

⏭ To Find:

✏ Length of a second pendulum

⏭ Concept:

✏ We know that, Time period of second pendulum = 2sec

⏭ Formula:

✏ Formula of time period of simple pendulum in terms of length of pendulum and gravitational acceleration is given by...

$\boxed{\sf{\pink{\large{T = 2\pi\sqrt{\dfrac{l}{g}}}}}}$

• T denotes time period
• l denotes length of pendulum
• g denotes gravitational acceleration

⏭ Calculation:

$\implies\sf\:2 = 2\times 3.14\sqrt{\dfrac{l}{10}}\\ \\ \implies\sf\:l = \dfrac{10}{(3.14)^2}\\ \\ \implies\:\boxed{\sf{\orange{\large{l=1.01\:m}}}}$

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✒ Length of second pendulum = 1.01m

Answer 2

$\mathtt{\huge{ \fbox{Solution :)}}}$

We know that , the time period for simple pendulum is given by

$\large \mathtt{ \fbox{T =2\pi \sqrt{ \frac{l}{g} } }}$

Where ,

T = time period

l = length of pendulum

g = acceleration due to gravity

Since , the time period of second pendulum is 2 sec

Thus ,

$\sf \hookrightarrow 2 = 2 \times 3.14 \sqrt{ \frac{l}{10} } \\ \\\sf \hookrightarrow 1 = 3.14 \sqrt{ \frac{l}{10} } \\ \\\sf Squaring \: on \: both \: sides \: , \: we \: get \\ \\\sf \hookrightarrow {(1)}^{2} = {(3.14)}^{2} \times \frac{l}{10} \\ \\\sf \hookrightarrow l = \frac{10}{9.85} \\ \\\sf \hookrightarrow l =1.01 \: \: cm$

Hence , the length of second pendulum is 1.01 cm

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