A solid right circular cone of radius 4, cm a | vedclass

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(C) Volume of the cylindrical vessel $V_{cyl} = \pi r_1^2 h_1 = \pi 	imes 5^2 	imes 8 = 200\pi\, cm^3$.Volume of the solid cone $V_{cone} = \frac{1}

Vedclass · Accepted Answer

$511.24$

Vedclass · Answer

(C) Volume of the cylindrical vessel $V_{cyl} = \pi r_1^2 h_1 = \pi 	imes 5^2 	imes 8 = 200\pi\, cm^3$.Volume of the solid cone $V_{cone} = \frac{1}{3} \pi r_2^2 h_2 = \frac{1}{3} 	imes \pi 	imes 4^2 	imes 7 = \frac{112}{3}\pi\, cm^3$.Volume of water required = $V_{cyl} - V_{cone} = 200\pi - \frac{112}{3}\pi = \frac{600\pi - 112\pi}{3} = \frac{488}{3}\pi\, cm^3$.Using $\pi \approx 3.14159$,Volume $\approx \frac{488}{3} 	imes 3.14159 \approx 162.666 	imes 3.14159 \approx 511.03\, cm^3$.Given the options provided,the closest value is $511.24\, cm^3$.

$A$ solid right circular cone of radius $4\, cm$ and height $7\, cm$ is placed inside a cylindrical vessel of radius $5\, cm$ and height $8\, cm$. How much water (in $cm^3$) will be required to fill the cylindrical vessel completely?

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