Question

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

• A. be $4$ times its present value
• B. be doubled
• C. remain same
• D. be halved

Answer 1

The correct option is B

be doubled

Step 1: Given data

The density of the earth increased $4$ times$\Rightarrow \rho \text{'}=4\rho$

Radius became half$\Rightarrow R\text{'}=\frac{R}{2}$

We have to find the change in our weight

Step 2: Formula used

We know,

Weight,

$$\begin{gathered} W=mg \\ W=m\left(\frac{GM}{R^2}\right)\left(\because g=\frac{GM}{R^2}\right) \\ W=m\left(\frac{GV\rho }{R^2}\right)\left(\because M=V\rho \right) \\ W=m\left(\frac{G\left({\displaystyle \frac{4}{3}}{\mathrm{\pi R}}^3\right)\rho }{R^2}\right)\left(\because V=\frac{4}{3}{\mathrm{\pi R}}^3\right) \\ W=mG\frac{4}{3}\mathrm{\pi R}\rho \end{gathered}$$

Where

$m\to$our mass

$M\to$mass of earth

$g\to$acceleration due to gravity

$G\to$gravitational constant

$R\to$radius of earth

$V\to$volume of earth

$\rho \to$density of earth

Step 3: Calculating weight

Initial weight, $W=\frac{4}{3}\mathrm{\pi }mGR\rho$

Final weight, $W\text{'}=\frac{4}{3}\mathrm{\pi }mGR\text{'}\rho \text{'}$

$$\begin{gathered} W\text{'}=\frac{4}{3}\mathrm{\pi }mGR\text{'}\rho \text{'} \\ W\text{'}=\frac{4}{3}\mathrm{\pi }mG\frac{R}{2}\left(4\rho \right) \\ W\text{'}=\frac{8}{3}\mathrm{\pi }mGR\rho \\ W\text{'}=2W \end{gathered}$$

Therefore, the weight will be doubled

Hence, option B is correct.