The diameter of one of the bases of a truncat | vedclass

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(B) Let the radius of the first base be $R_1 = 50 \, mm$ and the radius of the second base be $R_2 = r \, mm$. The height is $h$. The volume of a trun

Vedclass · Accepted Answer

$55$

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(B) Let the radius of the first base be $R_1 = 50 \, mm$ and the radius of the second base be $R_2 = r \, mm$. The height is $h$. The volume of a truncated cone is given by $V = \frac{\pi h}{3} (R_1^2 + R_1 R_2 + R_2^2)$.Given $R_1 = 50 \, mm$. The new radius $R_1' = R_1 + 0.21 R_1 = 1.21 R_1 = 1.21 	imes 50 = 60.5 \, mm$.The new volume $V' = 1.21 V$. Substituting the values:$1.21 	imes \frac{\pi h}{3} (50^2 + 50r + r^2) = \frac{\pi h}{3} (60.5^2 + 60.5r + r^2)$.Dividing both sides by $\frac{\pi h}{3}$:$1.21 (2500 + 50r + r^2) = 3660.25 + 60.5r + r^2$.$3025 + 60.5r + 1.21r^2 = 3660.25 + 60.5r + r^2$.Subtracting $60.5r$ from both sides:$3025 + 1.21r^2 = 3660.25 + r^2$.$0.21r^2 = 635.25$.$r^2 = \frac{635.25}{0.21} = 3025$.$r = \sqrt{3025} = 55 \, mm$.

The diameter of one of the bases of a truncated cone is $100 \, mm$. If the diameter of this base is increased by $21 \%$ such that it still remains a truncated cone with the height and the other base unchanged,the volume also increases by $21 \%$. The radius of the other base (in $mm$) is

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