A conical iron piece having diameter 28 , cm | vedclass

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(A) The volume of the conical iron piece is given by $V = \frac{1}{3} \pi r^2 h$. Here,the radius $r = \frac{28}{2} = 14 \, cm$ and height $h = 30 \,

Vedclass · Accepted Answer

$35$

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(A) The volume of the conical iron piece is given by $V = \frac{1}{3} \pi r^2 h$. Here,the radius $r = \frac{28}{2} = 14 \, cm$ and height $h = 30 \, cm$.Volume of cone $= \frac{1}{3} \pi (14)^2 (30) = 10 \pi (196) = 1960 \pi \, cm^3$.When this piece is immersed in a cylindrical vessel,the volume of water displaced is equal to the volume of the cone. The displaced water takes the shape of a cylinder with height $H = 6.4 \, cm$ and radius $R$.Volume of displaced water $= \pi R^2 H = \pi R^2 (6.4)$.Equating the volumes: $1960 \pi = \pi R^2 (6.4)$.$R^2 = \frac{1960}{6.4} = \frac{19600}{64} = 306.25$.$R = \sqrt{306.25} = 17.5 \, cm$.The diameter of the vessel is $2R = 2 	imes 17.5 = 35 \, cm$.

$A$ conical iron piece having diameter $28 \, cm$ and height $30 \, cm$ is totally immersed into the water of a cylindrical vessel,resulting in the rise of water level by $6.4 \, cm$. The diameter,in $cm$,of the vessel is

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