Question

If density of earth increased $4$ times and its radius becomes half of what it is then our weight will be,

• A. Four times it present value
• B. doubled
• C. Remain same
• D. halved

Answer 1

The correct option is B

doubled

Step 1: Given Data

${\rho }_2=4{\rho }_1$

$R_2=\frac{R_1}{2}$

Step 2: Formula used,

Since we know that our weight is a pull force acting on us by the earth is given by
$W=mg$,
Where $m$ is the mass
And $g$ is the acceleration due to gravity.

Here, we have used the terms,
$G$- Gravitational constant,
$R$- Radius of the earth,
$\rho$ - is the density,
$V$- is the volume of the earth,
$g_1$ - acceleration due to gravity before variation,
$g_2$ - acceleration due to gravity after changing the densities and radius,
Acceleration due to gravity is given by
$$g={\displaystyle \frac{GM}{R^2}}$$
The mass and density relation,
Mass = density x volume
$m$ = $\rho \times V$
Therefore,
$$\begin{gathered} g=\frac{G\rho V}{R^2}=\frac{G\rho }{R^2}\times \frac{4\pi R^3}{3} \\ g=\frac{4}{3}\pi R\rho G \end{gathered}$$

Step 3: Calculate the weight if ${\rho }_2=4{\rho }_1\mathrm{\&}{R}_2={\displaystyle \frac{R_1}{2}}$,
From the above equation, it can be seen that $g\propto \rho R$
$$\begin{gathered} {\displaystyle \frac{g_2}{g_1}}={\displaystyle \frac{{\rho }_2{R}_2}{{\rho }_1{R}_1}} \\ {g}_2=g_1\left[{\displaystyle \frac{{\rho }_2{R}_2}{{\rho }_1{R}_1}}\right] \end{gathered}$$

Substitute the value of ${\rho }_2=4{\rho }_1\mathrm{\&}{R}_2={\displaystyle \frac{R_1}{2}}$ in the above equation,
$$\begin{gathered} g_2=g_1\left[4\times {\displaystyle \frac{1}{2}}\right] \\ {g}_2=2g_1 \end{gathered}$$
Multiply both sides by m, and we get,
$$\begin{gathered} m{g}_2=2m{g}_1 \\ {w}_2=2w_1 \end{gathered}$$
Therefore the weight is doubled.
Hence, the correct option is B.