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A small disc of radius r is cut out from a disc of radius R. The weight of the disc which now has a hole in it, is reduced to $$\frac{{24}}{{25}}$$ of the original weight. If R = xr, what is the value of x ?
Answer & Solution
Correct Answer:
Option
E
Since weight of the disc is proportional to its area, we have :
$$\eqalign{ & \pi \left( {{R^2} - {r^2}} \right) = \frac{{24}}{{25}}\pi {R^2} \cr & \Rightarrow {R^2} - {r^2} = \frac{{24}}{{25}}{R^2} \cr & \Rightarrow {r^2} = \frac{1}{{25}}{R^2} \cr & \Rightarrow {R^2} = 25{r^2} \cr & \Rightarrow R = 5r \cr} $$
$$\eqalign{ & \pi \left( {{R^2} - {r^2}} \right) = \frac{{24}}{{25}}\pi {R^2} \cr & \Rightarrow {R^2} - {r^2} = \frac{{24}}{{25}}{R^2} \cr & \Rightarrow {r^2} = \frac{1}{{25}}{R^2} \cr & \Rightarrow {R^2} = 25{r^2} \cr & \Rightarrow R = 5r \cr} $$
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