A bucket contains 200cc of liquid. A solid ball is dropped in the bucket resulting in the rise of liquid level to 1.3 times of its original level. If the radius of the base of the bucket is 3 cm and the radius of the surface of the liquid level is 1 cm more than the radius of the base of the bucket before the ball is dropped. Find the volume of the solid metal ball.

Let height of liquid level before the ball was dropped be h

The radius of the surface of the liquid level then is 4cm

We know that the volume of the bucket or frustum is V = $$\frac{1}{3}\pi h(R^2+r^2+Rr)$$

$$\frac{h}{3}\pi\ \left(4^2+3^2+4\cdot3\right)\ =200cc$$

=> h = $$\frac{600}{37\pi\ }$$

The liquid level raises by 0.3h as a result of the solid metal ball being dropped.

By similarity, the radius of the surface of the liquid level after metal ball is dropped is 4.3cm

Volume of he solid metal ball = $$\frac{1.3h}{3}\left(4^2+3^2+4.3\left(3\right)\right)\ -\ 200$$ = 83.82cc