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SSC CGL Model Paper 1

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Question : 67 of 100

Marks: +1, -0

A right triangle with hypotenuse of 20cm and other two sides of variable length is rotated about its longest side thus giving rise to a solid. Find the max. possible volume (in cm

^3

) of such solid?

$\frac{100}{3\pi}$
$\frac{2000}{3\pi}$
$\frac{500}{7\pi}$
$\frac{1500}{7\pi}$

Solution:

For max. volume

AB = BC

Applying Pythagoras

AB^2+BC^2 = (20)^2

\Rightarrow AB = BC = 10\sqrt{2}

Figure is rotated along AC, a double cone will form with radius BD

In

\triangle ABC; AB \times BC = BD \times AC

10\sqrt{2} \times 10\sqrt{2} = BD \times 20

BD = 10

Hence, Volume of cone of radius 10cm and height 10 cm is given by

= \frac{1}{3}\pi r^2 h

= \frac{1}{3}\pi \times 100 \times 10 = \frac{1000\pi}{3}

Volume of required double cone

= 2 \times \frac{1000\pi}{3}

= \frac{2000}{3\pi}

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