Question

A metallic sphere of radius 21 CM is dropped into a cylinder vessel, which is partly, filled with water. the diameter of vessel is 1.68 m. if the sphere is completely submerged, find by how much the surface of water will rise.

Answer 1

Here is your answer......

Answer 2

Answer:

The surface of water is rises 1.74 cm .

Step-by-step explanation:

Explanation:

Given , radius of metallic sphere = 21cm

and the diameter of the vessel = 1.68m = 1.68 ×100 cm = 168cm

So , the radius of the vessel = $\frac{168}{2}$ = 84

Volume of sphere = $\frac{4}{3} \pi r^{3}$

and volume of cylinder =  $\pi r^{2} h$

Let height of the rise  water be h cm.

Step 1:

According to the question , metallic sphere is completely submerged .

Therefore , the volume of  metallic sphere and volume  of water  risen.

⇒Volume of water displaced  = volume of  the sphere

⇒$\pi r^{2} h$ = $\frac{4}{3} \pi r^{3}$

⇒$\pi$×$84^{2}$×h = $\frac{4}{3}$×$\pi$ ×$21^{3}$

⇒h = $\frac{ 4}{3}$ ×$\frac{1}{84 .(84) }$ ×$21^{3}$

⇒h= 1.33 ×21 ×21×21×$\frac{1}{84 .(84) }$ = 1.74cm

Final answer :

Hence , the surface of water is rises 1.74 cm .

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