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Sanskrit ka question ha please answer the question . |
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Answer» The \: radius \: of \: upper \: cone (R _{1}) = \frac{4}{2 } = 2Theradiusofuppercone(R 1 )= 24 =2Bottom radius 4cmVolume \: of \: upper \: cone (V _{1}) = \frac{1}{3)} \pi r^{2} hVolumeofuppercone(V 1 )= 3)1 πr 2 h= \frac{1}{3} \pi \times ({2}^{2}) \times ( \frac{h}{2} )= 31 π×(2 2 )×( 2h )Vol \: of \: BOTTOM \: FRUSTUM (V_{2}) = \frac{\pi h}{3} (R \frac{2}{1} R \frac{2}{2} + R _{1}R _{2})Volofbottomfrustum(V 2 )= 3πh (R 12 R 22 +R 1 R 2 )= \frac{\pi( \frac{h}{2}) }{3} (28)= 3π( 2h ) (28)Vol \: of \: bottom \: frustum (V _{2}) = \frac{\pi h}{6} (28)Volofbottomfrustum(V 2 )= 6πh (28)\frac{V _{1} }{ V_{2} } = \frac{2\pi h}{3} \times \frac{6}{\pi h \times 28} V 2 V 1 = 32πh × πh×286 \frac{V _{1}}{V _{2}} = \frac{2\pi h}{3} \times \frac{6}{\pi h \times 28} V 2 V 1 = 32πh × πh×286 \frac{V_{1} }{ V_{2} } = \frac{1}{7} V 2 V 1 = 71 ❥ the above is the answer of your question❥ hope that's helps you |
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