The volume of a solid cylinder whose base dia | vedclass

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(B) The volume of a cylinder is given by $V = \pi r^2 h$.Let the initial radius be $r_1$ and height be $h_1$. Then $V_1 = \pi r_1^2 h_1 = 3850\, mm^3$

Vedclass · Accepted Answer

$1925$

Vedclass · Answer

(B) The volume of a cylinder is given by $V = \pi r^2 h$.Let the initial radius be $r_1$ and height be $h_1$. Then $V_1 = \pi r_1^2 h_1 = 3850\, mm^3$.For the new cylinder,the length is doubled $(h_2 = 2h_1)$ and the diameter is halved,which means the radius is also halved $(r_2 = r_1 / 2)$.The new volume $V_2$ is given by $V_2 = \pi r_2^2 h_2$.Substituting the new values: $V_2 = \pi (r_1 / 2)^2 (2h_1) = \pi (r_1^2 / 4) (2h_1) = \frac{1}{2} \pi r_1^2 h_1$.Since $V_1 = \pi r_1^2 h_1 = 3850$,we have $V_2 = \frac{1}{2} 	imes 3850 = 1925\, mm^3$.

The volume of a solid cylinder whose base diameter is $14\, mm$ and length is $25\, mm$ is $3850\, mm^3$. If the length of the cylinder is doubled but the diameter is halved,what will be the volume of the resulting cylinder? (in $mm^3$)

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