Question

☑Answer the 5th question ❤

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Answer 1

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•Refer to the Questionar attachment....

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$<font color=red><marquee behavior=alternate>Solution</marquee></font>$

From the above figure....

AP=(20/2)cm=10 cm

now from the ∆APD..we get....

tan60°=PD/AP

=>PD=(tan60°)AP

=>PD=10√3 cm

the base radius of the frustum is=10√3 cm

volume of the frustum is

$= (\frac{1}{3} \times \pi \times (10 \sqrt{3} ) {}^{2} \times 10)cm {}^{3}$

now if it is drawn into a wire of diameter (1/16)cm and of length (l)cm then it's volume(as cylinder)..

$= (\pi \: \times ( \frac{ \frac{1}{16} }{2} ) {}^{2} \times l)cm {}^{3}$

now...so according to the problem...

$\frac{1}{3} \times \pi \: \times (10 \sqrt{3} ) {}^{2} \times 10 = \pi \: \times ( \frac{1}{32} ) {}^{2} \times l \\ = > 1000 \times 32 {}^{2} = l \\ = > l = 1024000 \: cm$

therefore the length of the wire is

therefore the length of the wire is =1024000cm=10240m=10.240km

$<font color=green><marquee direction="Right">hope this help you$

Answer 2

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HoPE It Helps..

* not sure abt answer❗️

THNKS!!♥️