Question

A pendulum of length 36 cm has a time period of 1.2 s.Find the time period of another pendulum,whose length is 81 cm. *
1 point
1.6 s
2.2 s
1.8 s
2.0 s

Answer 1

Answer :

➥ The time period of another pendulum = 1.8 s

Given :

➤ Pendulum of length (l₁) = 36 cm

➤ Another pendulum of length (l₂) = 81 cm

➤ Time period of pendulum (T₁) = 1.2 s

To Find :

➤ Time period of another pendulum (T₂) = ?

Solution :

As we know that

$\tt{ :\implies \dfrac{T_1}{T_2} = \sqrt{ \dfrac{l_1}{l_2} } }$

$\tt{: \implies \dfrac{1.2}{T_2} = \sqrt{ \dfrac{36}{81} } }$

$\tt{: \implies \dfrac{1.2}{T_2} = \dfrac{6}{9}}$

$\tt{: \implies 6T_2 = 9 \times 1.2}$

$\tt{: \implies 6T_2 = 10.8}$

$\tt{: \implies T_2 = \cancel{\dfrac{10.8}{6} }}$

$: \implies \green{\underline{ \overline{ \boxed{ \purple{ \bf{ \: \: T_2 = 1.8 \: sec \: \: }}}}}}$

Hence, the time period of another pendulum is 1.8 sec.

Answer 2

Answer:

Answer :

➥ The time period of another pendulum = 1.8 s

Given :

➤ Pendulum of length (l₁) = 36 cm

➤ Another pendulum of length (l₂) = 81 cm

➤ Time period of pendulum (T₁) = 1.2 s

To Find :

➤ Time period of another pendulum (T₂) = ?

Solution :

As we know that

\tt{ :\implies \dfrac{T_1}{T_2} = \sqrt{ \dfrac{l_1}{l_2} } }:⟹

T

2

T

1

=

l

2

l

1

\tt{: \implies \dfrac{1.2}{T_2} = \sqrt{ \dfrac{36}{81} } }:⟹

T

2

1.2

=

81

36

\tt{: \implies \dfrac{1.2}{T_2} = \dfrac{6}{9}}:⟹

T

2

1.2

=

9

6

\tt{: \implies 6T_2 = 9 \times 1.2}:⟹6T

2

=9×1.2

\tt{: \implies 6T_2 = 10.8}:⟹6T

2

=10.8

\tt{: \implies T_2 = \cancel{\dfrac{10.8}{6} }}:⟹T

2

=

6

10.8

: \implies \green{\underline{ \overline{ \boxed{ \purple{ \bf{ \: \: T_2 = 1.8 \: sec \: \: }}}}}}:⟹

T

2

=1.8sec

Hence, the time period of another pendulum is 1.8 sec.

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