Question

A iron sphere diameter 42cm is dropped into a cylindrical drum partly filled with water .If the radius of the drum is1.4 m, how much will the surface of the water be raised​

Answer 1

$\blue{\bold{\underline{\tt{\underline{Answer :}}}}}$

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$\green{\underline \bold{\tt{Given :-}}}$

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• Diameter of iron sphere = 42cm

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• Radius of drum = 1.4m

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$\red{\underline \bold{\tt{To \: Find :-}}}$

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• How much will the surface of the water be raised when iron sphere is dropped.

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$\large{\orange{\underline{\tt{Solution :-}}}}$

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1 m = 100 cm

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1.4 m = 140 cm

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$\underline{\bold{\texttt{Volume of cylinder :}}}$

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$\bf \dag \ \ \ \pi r ^2 h$

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$\sf r = \dfrac { 140 } { 2 } = 70$

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$\purple\longrightarrow$ $\sf Volume = \pi 70^2 h$

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$\underline{\bold{\texttt{Volume of sphere :}}}$

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$\bf \dag \ \ \ \dfrac { 4 } { 3 } \pi r^3$

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$\sf r = \dfrac { 42 } { 2 } = 21$

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$\purple\longrightarrow$ $\sf Volume = \dfrac { 4 } { 3 } \pi 21^3$

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Volume of water rises = Volume of iron sphere

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$\bf \longmapsto \pi 70^2 h = \dfrac { 4 } { 3 } \pi 21^3$

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$\sf \longmapsto 70 ^2 h = \dfrac { 4 } { 3 } 21^3$

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$\sf \longmapsto 70 \times 70 \times h = \dfrac { 4 } { 3 } \times 21 \times 21 \times 21$

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$\sf \longmapsto 70 \times 70 \times h = 4 \times 21 \times 21 \times 7$

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$\sf \longmapsto 10 \times 700 \times h = 4 \times 21 \times 21$

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$\sf \longmapsto 5 \times 5 \times h = 21 \times 3$

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$\sf \longmapsto h = \dfrac { 21 \times 3} { 5 \times 5 }$

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$\sf \longmapsto h = \dfrac { 63 } { 25 }$

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$\bf \dashrightarrow h = 2.52 cm$

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• Hence when the sphere is dropped the water will rise upto height of 2.52 cm from the original point.

Answer 2

Answer:

diameter of Square =42cm

Radius of sphere =42/2=21cm

volume of sphere =4/3πr3

=4/3 multiplied by 22/7 multiplied by 21cm multiplied by 21cm multiplied by 21cm

=38808cmcube