Question

Diameter and mass of a planet is double that of the earth. Then time period of a pendulum at surface of planet is how much times of time period at earth surface:
\[\left( {\text{A}} \right){\text{ }}\dfrac{1}{{\sqrt 2 }}\] $Times$
\[\left( {\text{B}} \right){\text{ }}\sqrt 2 \] $Times$
\[\left( {\text{C}} \right)\] $Equal$
\[\left( {\text{D}} \right)\] $\text{None of these}$

Answer 1

Hint: A direct line passing via the middle of a circle or sphere and meeting the circumference or floor at every cease. A straight line which passes from side to side of any parent or frame, via its center. The diameter is the duration of the line thru the center that touches two points on the edge of the circle.
The orbital duration of a satellite relies upon the mass of the planet being orbited and the space of the satellite tv for pc from the center of the planet.
Because the Sun keeps 'burn' hydrogen into helium in its middle, the middle slowly collapses and heats up, causing the outer layers of the Sun to develop larger.
It is a very sluggish technique, and inside the remaining \[4\] billion years, the Sun has barely grown using possibly \[20\] percentage at maximum.

Formula used:
\[T = 2\pi \sqrt {\dfrac{l}{g}} \]
Here, \[T\] is the time period of pendulum
\[l\] Is the length of a string
\[g\] Is the acceleration due to gravity

Complete step by step answer:
Let, us consider the time period of simple pendulum, \[\tau = 2\pi \sqrt {\dfrac{l}{g}} \]
For earth: \[g = \dfrac{{G{M_E}}}{R}\]
For planet: \[{M_p} = 2{{\rm M}_E}\]and\[{R_p} = 2R\]
Thus, \[{g_p} = \dfrac{{G(2{M_E})}}{{{{(2R)}^2}}}\]
\[ \Rightarrow {g_p} = \dfrac{g}{2}\]
Now,\[\dfrac{{{t_p}}}{{{t_E}}} = \sqrt {\dfrac{g}{{g / 2}}} \]
Cancel out \[g\],\[ \Rightarrow {t_p} = \sqrt 2 {t_E}\]

Hence the correct option is B.

Note: The longer the period of string, the farther the pendulum falls; and consequently, the longer the length, or to and fro swing of the pendulum.
The length of a pendulum does not now depend on the mass of the ball, but simplest on the length of the string.
With the assumption of small angles, the frequency and length of the pendulum are unbiased of the initial angular displacement amplitude.
A pendulum can have an equal period regardless of its preliminary perspective.