JAMB - Mathematics (1978 - No. 29)
A hollow right prism of equilateral triangular base of side 4cm is filled with water up to a certain height. If a sphere of radius \(\frac{1}{2}\)cm is immersed in the water, then the rise of water is
1cm
\(\sqrt{\frac{3\pi}{24}}\)
\(\frac{\pi}{24\sqrt{3}}\)
24\(\sqrt{3}\)
Explanation
The rise of water is equivalent to the volume of the sphere of radius \(\frac{1}{2}\)cm immersed x \(\frac{1}{\text{No. of sides sq. root 3}}\)
Vol. of sphere of radius = 4\(\pi\) x \(\frac{1}{8}\) x \(\frac{1}{3}\) - (\(\frac{1}{2}\))3
= \(\frac{1}{8}\)
= \(\frac{\pi}{6}\) x \(\frac{1}{4\sqrt{3}}\)
= \(\frac{\pi}{24\sqrt{3}}\)
Vol. of sphere of radius = 4\(\pi\) x \(\frac{1}{8}\) x \(\frac{1}{3}\) - (\(\frac{1}{2}\))3
= \(\frac{1}{8}\)
= \(\frac{\pi}{6}\) x \(\frac{1}{4\sqrt{3}}\)
= \(\frac{\pi}{24\sqrt{3}}\)
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