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There are two solid cylinders, one with diameter $60 cm$ and height $30 cm$ and the other with radius $30 cm$ and height $60 cm$ . are melted and recast into a third solid cylinder of height $10 cm$ . Then the diameter of the cylinder formed is?

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45 cm

200 cm

180 cm

90 cm

answer is C.

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Detailed Solution

We have, Diameter of the first cylinder

= 60 cm

Radius of first cylinder

= \frac{d}{2} = \frac{60}{2} = 30 cm

Height of first cylinder

= 30 cm

Radius of second cylinder

= 30 cm

Height of second cylinder

= 60 cm

Volume of first cylinder

(V_{1})

V_{1} = π r^{2} h

V_{1} = π (30)^{2} (30) (∵ R = 30 cm, h = 30 cm)

V_{1} = 27000 π

—(|)
Volume of second cylinder

(V_{2})

V_{2} = π r^{2} h

V_{2} = π (30)^{2} (60)

V_{2} = 27000 π

—(||)
As we know that two cylinders are melted to made a one new cylinder. So, we add both the volumes from (|) and (||)

V_{1} + V_{2} = 27000 π + 54000 π

V_{1} + V_{2} = 81000 π

—(|||)
Now, Volume of third cylinder

(V_{3})

V_{3} = π r^{2} h

From eq (|||) and (IV)

π r^{2} h = 81000 π

π r^{2} (10) = 81000 π

r^{2} = \frac{81000 π}{10 π} r^{2}

r^{2} = 8100 cm

r = 90 cm

As we know,

D = 2 r

= (2 \times 90) cm

= 180 cm

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