Question

a metal sphere 14 cm in diameter is dropped into a rectangular cistern to , whose base measures 49cm by 42/3cm.If the sphere is totally submerged, by how much will the surface of the water in Custer
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Answer 1

By 2.095 cm the surface of the water will rise in Custer.

We know, the surface area of a sphere of radius 'r' is given as

(4/3)Пr³

Given, diameter = 14cm

So, radius = diameter/2 =14cm/2 = 7cm

Thus, surface area = (4/3)×(22/7)×7³

This, surface area is equal to the base measurements multiplied with height.

Base measurements multiplied by height(h), gives,

49×(42/3)×h = (4/3)×(22/7)×7³

⇒ h = [(4/3)×(22/7)×7³]/[49×(42/3)]

       = 2.095m

This is the rise in water surface.

Answer 2

Answer:

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