The radii of two 10 , cm high metallic cylind | vedclass

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Question

(A) The volume of a cylinder is given by the formula $V = \pi r^2 h$.For the first cylinder: $r_1 = 3.5 \, cm$,$h_1 = 10 \, cm$. Volume $V_1 = \pi (3.

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$3.5$

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(A) The volume of a cylinder is given by the formula $V = \pi r^2 h$.For the first cylinder: $r_1 = 3.5 \, cm$,$h_1 = 10 \, cm$. Volume $V_1 = \pi (3.5)^2 (10) = \pi (12.25)(10) = 122.5 \pi \, cm^3$.For the second cylinder: $r_2 = 7 \, cm$,$h_2 = 10 \, cm$. Volume $V_2 = \pi (7)^2 (10) = \pi (49)(10) = 490 \pi \, cm^3$.Total volume of the two cylinders = $V_1 + V_2 = 122.5 \pi + 490 \pi = 612.5 \pi \, cm^3$.Let the radius of the new cylinder be $R$ and its height be $H = 50 \, cm$.The volume of the new cylinder is $V_{new} = \pi R^2 H = \pi R^2 (50)$.Since the volume remains constant,$50 \pi R^2 = 612.5 \pi$.$R^2 = \frac{612.5}{50} = 12.25$.$R = \sqrt{12.25} = 3.5 \, cm$.

The radii of two $10 \, cm$ high metallic cylinders are $3.5 \, cm$ and $7 \, cm$. These cylinders are melted and recast to produce a cylinder with height $50 \, cm$. Find the radius of this new cylinder (in $cm$).

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