To solve this problem, we need to understand how the acceleration due to gravity is affected by changes in the density of the Earth while keeping its radius constant.
The acceleration due to gravity \( g \) on the surface of the Earth is given by the formula:
\(g = \dfrac{G \cdot M}{R^2}\)
Where:
The mass \( M \) of the Earth can be expressed in terms of density \(\rho\) and volume \( V \) since mass is the product of density and volume:
\(M = \rho \cdot V\)
The volume \( V \) of a sphere, like Earth, is calculated as:
\(V = \dfrac{4}{3}\pi R^3\)
Substituting the expression for \( M \) and \( V \) into the formula for \( g \), we get:
\(g = \dfrac{G \cdot \rho \cdot \dfrac{4}{3}\pi R^3}{R^2} = \dfrac{4}{3}\pi G \rho R\)
From this equation, we can see that \( g \) is directly proportional to the density \( \rho \) of the Earth when the radius \( R \) is constant.
If the density of the Earth is doubled, the new density \( \rho' = 2\rho \). So the new acceleration due to gravity \( g' \) becomes:
\(g' = \dfrac{4}{3}\pi G (2\rho) R = 2 \left(\dfrac{4}{3}\pi G \rho R\right) = 2g\)
Given that the initial acceleration due to gravity is approximately \( 10 \, \text{m/s}^2 \), the new acceleration due to gravity when the density is doubled will be:
\(g' = 2 \times 10 \, \text{m/s}^2 = 20 \, \text{m/s}^2\)
Therefore, the correct answer is 20 m/s\(^2\).
A particle is moving in a straight line. The variation of position $ x $ as a function of time $ t $ is given as:
$ x = t^3 - 6t^2 + 20t + 15 $.
The velocity of the body when its acceleration becomes zero is: