Some lead spheres, each of diameter 6 cm, are dropped into a beaker containing some water and are fully submerged. The diameter of the beaker is 18 cm. Calculate, the number of lead spheres dropped into it, if the water level rises by 40 cm.

Given,

Diameter of lead spheres = 6 cm

Radius, r = $\dfrac{\text{diameter}}{2} = \dfrac{6}{2}$ = 3 cm

Diameter of beaker = 18 cm

Radius of beaker, R = $\dfrac{\text{diameter}}{2} = \dfrac{18}{2}$ = 9 cm

Increase in water level, h = 40 cm

Let n spheres are dropped.

∴ Volume of water increased in beaker = n × Volume of one sphere

$\Rightarrow π\text{R}^2 \text{h} = \text{n} \times \dfrac{4}{3} π\text{r}^3 \\[1em] \text{Divide by π on both sides, we get:} \Rightarrow \text{R}^2 \text{h} = \text{n} \times \dfrac{4}{3} \text{r}^3 \\[1em] \Rightarrow 9^2 \times 40 = \text{n} \times \dfrac{4}{3} \times 3^3 \\[1em] \Rightarrow 81 \times 40 = \text{n} \times \dfrac{4}{3} \times 27 \\[1em] \Rightarrow 3240 = \text{n} \times 36 \\[1em] \Rightarrow \text{n} = \dfrac{3240}{36} \\[1em] \Rightarrow \text{n} = 90.$

Hence, the number of lead spheres dropped into the beaker are 90.