Question

A hanging wire is $185\,{\text{cm}}$ long having a bob of $1.25\,{\text{kg}}$. It shows a time period of $1.42\,{\text{s}}$ on planet Newtonia. If the circumference of Newtonia is $51400\,{\text{km}}$, find the mass of the planet.
A. $3.5 \times {10^{25}}\,{\text{kg}}$.
B. $9.08 \times {10^{24}}\,{\text{kg}}$.
C. $2.6 \times {10^{25}}\,{\text{kg}}$.
D.$3.14 \times {10^{24}}\,{\text{kg}}$.

Answer 1

To solve this question we have to use the formula of time period of simple pendulum and formula of gravitational.
We know that the time period of a simple pendulum is given by
$T = 2\pi \sqrt {\dfrac{l}{g}}$ …… (1)
Here, $T$ is the time period, $l$ is the length of the wire, $g$ is the acceleration due to gravity its value is $9.8\,{\text{m/}}{{\text{s}}^2}$.

Complete step by step answer:
Given,
Length of the wire is $l = 1.85\,{\text{cm}}$
Time period $T = 1.42\,{\text{s}}$
Circumference of the planet ${\text{51400}}\,{\text{km}}$

Now from the equation$T = 2\pi \sqrt {\dfrac{l}{g}}$, $g$ can be written as

{T^2} = {\left( {2\pi \sqrt {\dfrac{l}{g}} } \right)^2} \\\ {T^2} = 4{\pi ^2}\left( {\dfrac{l}{g}} \right) \\\ g = \dfrac{{4{\pi ^2}l}}{{{T^2}}} \\\ $$ …… (2) Again $$g$$ is given as $$g = \dfrac{{GM}}{{{R^2}}}$$ Here, $$G$$ is gravitational constant and its value is given by $$6.67 \times {10^{ - 11}}$$, $$M$$ is mass of the planet and $$R$$ is the radius of the planet. Now equation (2) can be rewritten as

\dfrac{{GM}}{{{R^2}}} = \dfrac{{4{\pi ^2}l}}{{{T^2}}} \\
\Rightarrow M = \dfrac{{4{\pi ^2}I{R^2}}}{{G{T^2}}} \\
\Rightarrow M = \dfrac{{{{\left( {2\pi R} \right)}^2}I}}{{G{T^2}}} \\
\Rightarrow M = \dfrac{{{{\left( {5.14 \times {{10}^7}} \right)}^2} \times 1.85}}{{6.67 \times {{10}^{ - 11}} \times {{1.42}^2}}} \\
= 3.5 \times {10^{25}},{\text{kg}} \\

**Hence the correct option is (A) $$3.5 \times {10^{25}}{\text{kg}}$$.** **Note:** A pendulum is a body suspended from a fixed support such that, under the force of gravity, it swings freely back and forth. Owing to gravity, when a pendulum is displaced sideways from its resting, equilibrium state, it is subject to a restoring force that will speed it back to the position of equilibrium.