Question on: JAMB Physics - 1998
The force of attraction between two point masses is 10-4N when the distance between them is 0.18m. If the distance is reduced to 0.06m, calculate the force.
A
1.1 x 10-5N
B
3.3 x 10-5N
C
3.0 x 10-4N
D
9.0 x 10-4N
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Correct Option: D
F \(\alpha \frac{1}{r^2}\)
therefore, \(\frac{F_1}{F_2}\) = \(\frac{r_1}{(r_2)^2}\)
\(\frac{10^{-4}}{F_2}\) = \(\frac{(0.06)^2}{(0.18)^2}\)
F = 9 x 10-4N
therefore, \(\frac{F_1}{F_2}\) = \(\frac{r_1}{(r_2)^2}\)
\(\frac{10^{-4}}{F_2}\) = \(\frac{(0.06)^2}{(0.18)^2}\)
F = 9 x 10-4N
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