Question

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

• A. $94.3 \times 10^{-10} \; N \; kg^{-1}$
• B. $9.43 \times 10^{-10} \; N \; kg^{-1}$
• C. $9.43 \times 10^{-12} \; N \; kg^{-1}$
• D. $94.3 \times 10^{-12} \; N \; kg^{-1}$

Answer 1

Answer: (B)

A system of two masses is shown in the figure,

Here, $m_{1}=90\, kg$ and
$m_{2}=160\, kg$
Gravitational field intensity due to mass $A$,
$E _{A} =\frac{G M_{A}}{ r _{C A}^{2}}$
$=G \frac{90}{\left(3^{2}\right)}=10 G \hat{ r }_{C A} $
Similarly, $ E _{B}=G \times \frac{160}{4^{2}}=10 G \hat{ r }_{C B}$
In $\triangle A B C$,
$(A B)^{2} =(A C)^{2}+(B C)^{2}$
$(5)^{2} =(3)^{2}+(4)^{2}$
Hence $\Delta A B C$ is a right angle triangle.
Hence, the resultant of $E _{A}$ and $E _{B}$,
$E=\sqrt{E_{A}^{2}+E_{B}^{2}}$
$=\sqrt{(10 G)^{2}+(10 G)^{2}}=\sqrt{2} \times 10\, G$
Putting the value of $G$, we get
$\Rightarrow E =\sqrt{2} \times 10 \times 6.67 \times 10^{-11}$
$=9.43 \times 10^{-10} \,Nkg ^{-1}$