Two masses $90 \mathrm{~kg}$ and $160 \mathrm{~kg}$ are separated by a distance of $5 \mathrm{~m}$. The magnitude of intensity of the gravitational field at a point which is at a distance $3 \mathrm{~m}$ from the $90 \mathrm{~kg}$ mass and $4 \mathrm{~m}$ from the $160 \mathrm{~kg}$ mass is…

Question

Two masses $90 \mathrm{~kg}$ and $160 \mathrm{~kg}$ are separated by a distance of $5 \mathrm{~m}$. The magnitude of intensity of the gravitational field at a point which is at a distance $3 \mathrm{~m}$ from the $90 \mathrm{~kg}$ mass and $4 \mathrm{~m}$ from the $160 \mathrm{~kg}$ mass is (Universal gravitational constant, $\left.G=6.67 	imes 10^{-11} \mathrm{~N}-\mathrm{m}^2 \mathrm{~kg}^{-2}ight)$

Aicharya · Accepted Answer

A system of two masses is shown in the figure,

Here, $m_1=90 \mathrm{~kg}$ and $m_2=160 \mathrm{~kg}$ Gravitational field intensity due to mass $A$, $ \mathbf{E}_A=\frac{G M_A}{\mathbf{r}_{C A}^2}=G \frac{90}{\left(3^2\right)}=10 G \hat{\mathbf{r}}_{C A} $ Similarly, $\quad \mathbf{E}_B=G \times \frac{160}{4^2}=10 G \hat{\mathbf{r}}_{C B}$ In $\triangle A B C$, $ \begin{aligned} (A B)^2 & =(A C)^2+(B C)^2 \\ (5)^2 & =(3)^2+(4)^2 \end{aligned} $ Hence $\triangle A B C$ is a right angle triangle. Hence, the resultant of $\mathbf{E}_A$ and $\mathbf{E}_B$, $ E=\sqrt{E_A^2+E_B^2}=\sqrt{(10 G)^2+(10 G)^2}=\sqrt{2} \times 10 G $ Putting the value of $G$, we get $ \begin{aligned} \Rightarrow \quad E & =\sqrt{2} \times 10 \times 6.67 \times 10^{-11} \\ & =9.43 \times 10^{-10} \mathrm{Nkg}^{-1} \end{aligned} $ Hence, the correct option is (b)

Two masses $90 \mathrm{~kg}$ and $160 \mathrm{~kg}$ are separated by a distance of $5 \mathrm{~m}$. The…

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