Discussion

Explanation:

The radius of the bucket is 3 cm and the radius of the original water level is (3 + 1) = 4 cm

Original volume = 200 cc

Let the original height of water = h cm

Using the formula for volume of frustum of cone:

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

∴ h = 600/37π

Now, the height of the liquid increases by 0.3 times itself i.e. it becomes 1.3h

In a frustum of cone, ratio of radii is the same as ratio of heights (applying the concepts of similarity).

∴ New radius = 4.3 cm

∴ Volume of ball = new volume – original volume

​​​​​​​$=\frac{1.3h}{3}\pi [4.3^2+3^2+(4.3\left)\right(3\left)\right]-200$

$=\frac{1.3h}{3}\times \pi \times \frac{600}{37\pi }\times 40.4-200$

= 283.9 - 200 = 83.9 cc

The closest value in the options is 80 cc.

Hence, option (b).

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