Cups A and B are cone shaped and have heights of $28 c m$ $28 cm$ and $23 c m$ $23 cm$ and openings with radii of $11 c m$ $11 cm$ and $9 c m$ $9 cm$ , respectively. If cup B is full and its contents are poured into cup A, will cup A overflow? If not how high will cup A be filled?

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Answer 1 · 2017-12-26T13:39:03

Cup A will be filled upto 15.3967 cm

Volume of cone B $V_{b} = (\frac{1}{3}) π r_{2}^{2} h_{2}$ $V_b= (1/3) pi r_2^2 h_2$

Given $r_{2} = 9 c m, h_{2} = 23 c m$ $r_2 = 9 cm, h_2 = 23 cm$

$V_{b} = (\frac{1}{3}) \cdot π \cdot 9^{2} \cdot 23 = 1950.929 π$ $V_b = (1/3) * pi * 9^2 * 23 = 1950.929 pi$ $c m^{3}$ $cm^3$

Similarly Volume of cone A $V_{a} = (\frac{1}{3}) π r_{1}^{2} h_{1}$ $V_a = (1/3) pi r_1^2 h_1$

Given $r - 1 = 11 c m, h_{1} = 28 c m$ $r-1 = 11 cm, h_1 = 28 cm$

$V_{a} = (\frac{1}{3}) \cdot π \cdot 11^{2} \cdot 37 = 3547.9053 π$ $V_a = (1/3) * pi * 11^2 * 37 = 3547.9053 pi$ $c m^{3}$ $cm^3$

As volume of cone A is greater than the volume of cone B, cup A will not overflow.

Volume of partly filled cone A $V_{p} = (\frac{1}{3}) π r_{1}^{2} h_{3}$ $V_p = (1/3) pi r_1^2 h_3$

$V_{p} = = V_{b}$ $V_p = = V_b$

$(\frac{1}{3}) π \cdot 9^{2} \cdot 23 = (\frac{1}{3}) \cdot π \cdot 11^{2} \cdot h_{3}$ $(1/3) pi * 9^2 * 23 = (1/3) * pi * 11^2 * h_3$

$h_3 = (cancel((1/3) * pi) * 9^2 * 23) / (cancel((1/3) * pi) * 11^2)$

$h_3 =( 1863) / 121 = 15.3967$ $cm$

Cups A and B are cone shaped and have heights of 28 cm28cm28 cm and 23 cm23cm23 cm and openings with radii of 11 cm11cm11 cm and 9 cm9cm9 cm, respectively. If cup B is full and its contents are poured into cup A, will cup A overflow? If not how high will cup A be filled?