Question

Assume that a mango and its seed, both are spherical, now if the radius of seed is 2/5 of the thickness of the pulp. The seed lies exactly at the centre of the fruit. What percent of the total volume of the mango is its pulp. (1)318/5% (2) 97.67 % (3)206/3 (4) None of these

Answer 1

The percent of volume of pulp to the total volume of mango is given by,

$\small\longrightarrow\delta V_p=\dfrac{V_p}{V}\times100$

where $\smallV_p$ is volume of pulp and $\smallV$ is total volume. Since both are supposed to be spherical,

$\small\text{ \longrightarrow\delta V_p=\dfrac{\dfrac{4}{3}\pi(r_p)^3}{\dfrac{4}{3}\pi r^3}\times100 }$

$\small\longrightarrow\delta V_p=\left(\dfrac{r_p}{r}\right)^3\times100\quad\dots(1)$

where $\smallr_p$ is thickness of pulp and $\smallr$ is total radius of mango.

Given that radius of seed is 2/5 times the thickness of the pulp.

• $\smallr_s=\dfrac{2}{5}r_p$

And seed lies exactly at the centre of the fruit so thickness of pulp is uniform.

Then radius of mango is given by,

$\small\longrightarrow r=r_s+r_p$

$\small\longrightarrow r=\dfrac{2}{5}r_p+r_p$

$\small\longrightarrow r=\dfrac{7}{5}r_p$

$\small\longrightarrow\dfrac{r_p}{r}=\dfrac{5}{7}$

Then (1) becomes,

$\small\longrightarrow\delta V_p=\left(\dfrac{5}{7}\right)^3\times100$

$\small\longrightarrow\delta V_p=\dfrac{12500}{343}$

$\small\text{ \longrightarrow\underline{\underline{\delta V_p=36.44\%}} }$