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Question -

The diameters of internal and external surfaces of a hollow spherical shell are 10cm and 6 cm respectively. If it is melted and recast into a solid cylinder of length of 8/3, find the diameter of the cylinder?



Answer -

Given,

Internal diameter ofthe hollow sphere = 6 cm

So, the internalradius of the hollow sphere = 6/2 cm = 3 cm = r

External diameter ofthe hollow sphere = 10 cm

So, the externalradius of the hollow sphere = 10/2 cm = 5 cm = R

We know that,

Volume of the hollowspherical shell = 4/3 ╧А ├Ч (R3┬атАУ r3)

= 4/3 ╧А ├Ч (53┬атАУ33)┬а┬а┬а┬а┬а┬а┬а┬а тАж..┬а (i)

And given, the lengthof the solid cylinder = 8/3 cm

Let the radius of thesolid cylinder be r cm

We know that,

Volume of the cylinder= ╧А ├Ч r2┬а├Ч h

= ╧А ├Ч r2┬а├Ч8/3┬а ┬а┬а┬а┬а┬атАж.. (ii)

Now equating both (i)and (ii), we have

4/3 ╧А ├Ч 53┬атАУ33┬а= ╧А ├Ч r2┬а├Ч 8/3

4/3 x (125 тАУ 27) = r2┬а├Ч8/3

98/2 = r2

r2┬а=49

r = 7

So, d = 7 x 2 = 14 cm

Therefore, thediameter of the cylinder is 14 cm

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