Question

By melting the solid cylindrical metal, a few conical materials are to be made. If three times the radius of the cone is equal to twice the radius of the cylinder and the ratio of the height of cylinder and the height of the cone is 4:3 , then find numbers of cones which can be made.​

Answer 1

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Let R be the radius and H be the height of the cylinder and let rand h be the radius and height of the cone respectively. Then,

$3r = 2R$

$and, H:h = 4:3 \: .....(1)$

$\frac{H}{h} = \frac{4}{3}$

$3H = 4h \: .....(2)$

Let n be the required number of cones which can be made from the materials of the cylinder. Then, the volume of the cylinder will be equal to the sum of the volumes of n cones. Hence, we have

$\pi \: {R}^{2} H = \frac{n}{3} \pi \: {r}^{2} h$

$3 {R}^{2} H = n {r}^{2} h$

$n = \frac{3R²H}{r²H}$

$→ 3× \frac{9r²}{4} \times \frac{4h}{3} \div r²h$

From (1) and (2)

$→ R = \frac{3r}{2} and \: H = \frac{4h}{3}$

$→n = \frac{3 \times 9 \times 4}{3 \times 4}$

$→n = 9$

Hence, the required number of cones is 9.

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Answer 2

Answer:

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