Question

16 There is water to a height of 14 cm in a cylindrical glass jar of radius 8 cm. Inside the
water there is a sphere of diameter 12 cm completely immersed. By what height will
the water go down when the sphere is removed?
​

Answer 1

$\large{\underline{\rm{\pink{\bf{Question:-}}}}}$

There is water to a height of 14 cm in a cylindrical glass jar of radius 8 cm. Inside the water there is a sphere of diameter 12 cm completely immersed. By what height will the water go down when the sphere is removed?

$\large{\underline{\rm{\pink{\bf{Given:-}}}}}$

Radius of the cylindrical jar = 8 cm

Height of the water level = 14 cm

Diameter of the sphere = 12 cm

$\large{\underline{\rm{\pink{\bf{To \: Find:-}}}}}$

The height that the water goes down when the sphere is removed.

$\large{\underline{\rm{\pink{\bf{Solution:-}}}}}$

We know that,

• R = Radius
• H = Height
• d = Diameter

Given that,

Radius of the cylindrical jar (r) = 8 cm

Height of the water level (h) = 14 cm

Diameter of the sphere (d) = 12 cm

Volume of water = $\sf \pi R^{2}H$

Substituting them, we get

$\sf \longrightarrow \pi \times 8 \times 8 \times 14 \: cm^{3} =896 \pi \: cm^{3}$

Since, the diameter of the sphere is 12 cm

∴ Radius (r) = $\sf \dfrac{Diameter}{2}$

Radius = $\sf \dfrac{12}{2}$

$\sf =6 \:cm$

Volume = $\sf \dfrac{4}{3} \: \pi r^{3}$

Substituting the values,

Volume = $\sf \dfrac{4}{3} \: \pi \times 6 \times 6 \times 6 \: cm^{3}$

$\sf =288 \pi \: cm^{3}$

By immersing the sphere in the cylinder water rose up = $\sf 288 \pi \: cm^{3}$

Let us consider the height of the water rose as 'h' cm

$\sf \pi \times 8 \times 8 \times h=288 \pi$

$\sf h=\dfrac{288 \pi}{\pi \times 8 \times 8}$

$\sf \longrightarrow \dfrac{9}{2} \: cm$

∴ Water rises = $\sf \dfrac{9}{2} \: cm$

$\sf \longrightarrow 4.5 \: cm$

Therefore, water will go down by 4.5 cm when the sphere is removed.