Discussion

Explanation:

For frustum of a cone we know, $\frac{\mathrm{Volume} \mathrm{of} \mathrm{smaller} \mathrm{cone}}{\mathrm{Volume} \mathrm{of} \mathrm{bigger} \mathrm{cone}}$ = ${\left(\frac{r}{R}\right)}^3$

Since the original cone is cut at half its height from the base, hence ratio of height of smaller cone and original cone = 1 : 2

Ratio of heights of these cones will be same as ratio of their radii as well.

∴ (Volume of smaller cone)/(Volume of bigger cone) = (1/2)³ = 1/8

⇒ If the volume of bigger cone is 8V, volume of smaller cone will be V.

∴ Volume of the frustum = 8V – V = 7V.

⇒ $\frac{\mathrm{Volume} \mathrm{of} \mathrm{frustum}}{\mathrm{Volume} \mathrm{of} \mathrm{bigger} \mathrm{cone}}$ = $\frac{7V}{8V}$ = $\frac{7}{8}$.

Hence, option (a).

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