Question

Two point masses of 4 kg and 2 kg kept at a distance of 1 m in air attract each other by a force F. These two masses are now dipped in kerosene keeping the distance same. What is the force of attraction between the masses in kerosene?

Answer 1

Step 1: Given data,

$$\begin{gathered} \mathrm{Mass}\mathrm{of}\mathrm{first}\mathrm{object}\mathrm{is}{\mathrm{m}}_1=2\mathrm{kg} \\ \mathrm{Mass}\mathrm{of}\mathrm{second}\mathrm{object}\mathrm{is}{\mathrm{m}}_2=4\mathrm{kg} \\ \mathrm{Distance}\mathrm{between}\mathrm{the}\mathrm{two}\mathrm{point}\mathrm{masses}\mathrm{is}\mathrm{r}=1\mathrm{m} \\ \mathrm{Gravitational}\mathrm{force}\mathrm{constant}\mathrm{is}\mathrm{G}=6.67\times {10}^{-11}{\mathrm{Nm}}^2{\mathrm{kg}}^{-2} \end{gathered}$$

Step 2: Formula

Gravitational force: The force of attraction between two objects depends upon their masses and is directly proportional to their masses. Also, it depends upon the distance between them and is inversely proportional to the square of the distance between them.

The formula is $\mathrm{F}=\mathrm{G}\frac{{\mathrm{m}}_1{\mathrm{m}}_2}{{\mathrm{R}}^2}$ where ${\mathrm{m}}_1,{\mathrm{m}}_2$are the masses of the body, $\mathrm{r}$ is the distance between them and $\mathrm{G}$ is the gravitational constant.

Step 3: Calculating Force

$\mathrm{F}=\mathrm{G}\frac{{\mathrm{m}}_1{\mathrm{m}}_2}{{\mathrm{R}}^2}$

Putting the values ${\mathrm{m}}_1=2\mathrm{kg},{\mathrm{m}}_2=4\mathrm{kg},\mathrm{r}=1\mathrm{m},\mathrm{G}=6.67\times {10}^{-11}{\mathrm{Nm}}^2{\mathrm{kg}}^{-2}$ in the above equation,

$$\begin{gathered} \mathrm{F}=6.67\times {10}^{-11}\frac{2\times 4}{{\left(1\right)}^2} \\ =6.67\times 8\times {10}^{-11}=53.36\times {10}^{-11} \\ \mathrm{F}=5.3\times {10}^{-10}\mathrm{N} \end{gathered}$$

Step 4: Force between the masses in kerosene

1. The Force of attraction between any two objects in the whole universe depends upon their masses and the distance between them.
2. This force is independent of the medium in which objects are placed.
3. If the force between two masses is “$\mathrm{F}$” in the air then the force will remain the same if the masses are placed in kerosene i.e. “$\mathrm{F}$”.
4. Thus, the force of attraction between two masses in kerosene $=\mathrm{F}=5.3\times {10}^{-10}\mathrm{N}$

Hence, the force of attraction between two masses in kerosene is $5.3\times {10}^{-10}\mathrm{N}$.