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

Hint: 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.