A force of 200N acts between two objects at a certain distance apart. The value of the force when the distance is halved is
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Correct Answer: Option D
Explanation:
let the masses of the two objects be M1 and M2
and their distance apart = r
Therefore the force acting F1 = \( \frac{GM1M2}{r^2} \)
for a new distance of r/2, we have a force of
F2 = \( \frac{GM2M2}{r/2} \\
\text{Thus } F1 \times r^2 = GM1M1 \text{and} \\
F2 \times (r/2)^2 = GM1M1 \\
\text{Since } GM1M1 = GM1M1 \\
\implies F1 \times r^2 = F2 \times (r/2)^2 \\
\text{Therefore } 200 \times r^2 = F2 \times r^2/4 \\
F2 = \frac{4 \times 200 \times r^2}{r^2} = 800N \)
let the masses of the two objects be M1 and M2
and their distance apart = r
Therefore the force acting F1 = \( \frac{GM1M2}{r^2} \)
for a new distance of r/2, we have a force of
F2 = \( \frac{GM2M2}{r/2} \\
\text{Thus } F1 \times r^2 = GM1M1 \text{and} \\
F2 \times (r/2)^2 = GM1M1 \\
\text{Since } GM1M1 = GM1M1 \\
\implies F1 \times r^2 = F2 \times (r/2)^2 \\
\text{Therefore } 200 \times r^2 = F2 \times r^2/4 \\
F2 = \frac{4 \times 200 \times r^2}{r^2} = 800N \)