Question

Solid spherical balls each of diameter 1.4 cm are dropped into a cylindrical beaker containing water upto a height of 28 cm and are fully submerged. The diameter of the beaker is 5.6 cm. If the water in the beaker rises by 17.5 cm, then find the number of balls dropped in it

Answer 1

Answer:

Volume of water rise =

$\pi \times ( \frac{5.6}{2} ) {}^{2} \times 17.5$

Volume of one spherical ball =

$\frac{4}{3} \times \pi \times ( \frac{1.4}{2} ) {}^{3}$

so number of balls =

$\frac{volume \: of \: water \: rise}{vol. \: of \: one \: ball} = \frac{\pi \times 2.8 {}^{2} \times 17.5}{ \frac{4}{3} \times \pi \times 0.7 {}^{3} } = \frac{3}{4} \times 4 \times 4 \times 25 = 300$

hence required balls = 300. Ans