Question

The difference in the readings shown by a spring balance for a body of mass 40 kg on the surface of the Earth and a planet of mass half of that of the Earth and radius twice that of the Earth is __________ N (Take g = 10 m s^-2).

• A. 320
• B. 3200
• C. 350
• D. 500

Answer 1

The correct option is C

350

Step 1, Given data

Reading of the spring balance = 40 kg = 400 N

Mass of the earth = m

Mass of the planet = $\frac{m}{2}$

The radius of the earth = R

The radius of the planet = 2R

Step 2, Finding the reading of the spring balance on the other planet

We know,

Acceleration due to gravity $g=\frac{GM}{R^2}$

So we can write the equation for finding the acceleration due to gravity on that planet

$\frac{g_p}{g_e}=\frac{M_p}{M_e}\times {\left(\frac{R_e}{R_p}\right)}^2$

Putting all the values

$\frac{g_p}{g_e}=\frac{{\displaystyle \frac{m}{2}}}{m}\times {\left(\frac{R}{2R}\right)}^2$

After solving

$g_p=\frac{1}{8}{g}_e$

Now we know,

Weight of a body $w=mg$

So,

$$\begin{gathered} \frac{w_p}{w_e}=\frac{g_p}{g_e} \\ Or,w_p=\frac{g_p}{g_e}{w}_e \end{gathered}$$

Now putting all the values

$w_p=\frac{g_e}{8\times g_e}\times 400$

After solving

$w_p=50N$

So the difference in the reading $=400-50=350N$

Hence the correct option is (C),