A measuring jar of internal diameter 10 cm is partially filled with water.

Given that,

Diameter of jar = 10 cm

Radius of jar = 5 cm

Let the level of water be raised by h

Diameter of the spherical bowl = 2 cm

Radius of the ball = 1 cm

Volume of jar = 4 (Volume of spherical ball)

$\pi r_{1}^{2} h=4\left(\frac{4}{3} \pi r_{2}^{3}\right)$

$r_{1}^{2} h=4\left(\frac{4}{3} r_{2}^{3}\right)$

$5 \times 5 \times \mathrm{h}=4 \times \frac{4}{3} \mathrm{r}_{2}^{3}$

$5 \times 5 \times \mathrm{h}=4 \times \frac{4}{3} \times 1 \times 1 \times 1$

$h=\frac{4 \times 4 \times 1}{3 \times 5 \times 5}$

Height of water in jar = 16/75 cm