 # NCERT Solutions Class 10 Mathematics Solutions for Surface Areas and Volumes - Exercise 13.4 in Chapter 13 - Surface Areas and Volumes

A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained is drawn into a wire of diameter 1/16 cm, find the length of the wire.

The diagram will be as follows Consider AEG

\frac{E G}{A G}=\tan 30^{\circ}\\ \mathrm{EG}=\frac{10}{\sqrt{3}} \mathrm{~cm}=\frac{10 \sqrt{3}}{3}\\ \text { In } \triangle \mathrm{ABD} \text { . }\\ \frac{\mathrm{BD}}{\mathrm{AD}}=\tan 30^{\circ}\\ \mathrm{BD}=\frac{20}{\sqrt{3}}=\frac{20 \sqrt{3}}{3} \mathrm{~cm}

Radius (r1) of upper end of frustum = (10√3)/3 cm

Radius (r2) of lower end of container = (20√3)/3 cm

Height (r3) of container = 10 cm

Now,

Volume of the frustum = (⅓)×π×h(r12+r22+r1r2)

=\frac{1}{3}\times\pi\times10\left[\left(\frac{10\sqrt{3}}{3}\right)^2+\left(\frac{20\sqrt{3}}{3}\right)^2+\frac{(10\sqrt{3})(20\sqrt{3})}{3\times3}\right]

Solving this we get,

Volume of the frustum = 22000/9 cm3

The radius (r) of wire = (1/16)×(½) = 1/32 cm

Now,

Let the length of wire be “l”.

Volume of wire = Area of cross-section x Length

= (πr2)xl

= π(1/32)2x l

Now, Volume of frustum = Volume of wire

22000/9 = (22/7)x(1/32)2x l

Solving this we get,

l = 7964.44 m

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