**A hollow right prism of equilateral triangular base of side 4cm is filled with w...**

### Question

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### Options

A) 1cm

B) \(\sqrt{\frac{3\pi}{24}}\)

C) \(\frac{\pi}{24\sqrt{3}}\)

D) 24\(\sqrt{3}\)

The correct answer is C.

### 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 \(\frac{1}{4\sqrt{3}}\)

= \(\frac{\pi}{24\sqrt{3}}\)

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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 \(\frac{1}{4\sqrt{3}}\)

= \(\frac{\pi}{24\sqrt{3}}\)