Question

If the weight of a spherical shell is $\dfrac{7}{8}$ th of what it would be if it were a solid shell. The ratio of inner to outer radii of the shell is?
A) $1:2$
B) $1:3$
C) $2:3$
D) $3:4$

Answer 1

Hint: In this problem, we have to find the ratio of inner to outer radii of the given shell. First of all, must understand the given data in the problem. And then using the formulae volume of the spherical shell and volume of a solid shell we can solve this problem.

Complete step-by-step solution:
Here in this figure,
$R$$ = $ the outer radii of a given spherical.
$r$ $ = $ the inner radii of the given spherical.
In the given problem, we have some data to solve the problem.
That is, the weight of the spherical shell is $\dfrac{7}{8}$ part of the given solid shell.
Now applying this we get,
$\dfrac{4}{3}\pi \left( {{R^3} - {r^3}} \right) = \dfrac{7}{8} \times \dfrac{4}{3}\pi {R^3}$
Now we have to simplify the equation.
In this equation, the term $\dfrac{4}{3}\pi $ is the same on the right and left-hand sides. So, cancel the common terms.
Therefore, we have
$\left( {{R^3} - {r^3}} \right) = \dfrac{7}{8}{R^3}$
Now simplify this, we get
$ - {r^3} = \dfrac{7}{8}{R^3} - {R^3}$
On the right-hand side take the least common multiple to solve the equation.
$ - {r^3} = \dfrac{{7{R^3} - 8{R^3}}}{8}$
$\Rightarrow - {r^3} = - \dfrac{{{R^3}}}{8}$
Simplify this, we get
${r^3} = \dfrac{{{R^3}}}{8}$
Now use cross multiplication,
\[\dfrac{{{r^3}}}{{{R^3}}} = \dfrac{1}{8}\]
Now we will take cube root on both sides, we get
\[\sqrt[3]{{\dfrac{{{r^3}}}{{{R^3}}}}} = \sqrt[3]{{\dfrac{1}{8}}}\]
Now we get,
\[\dfrac{r}{R} = \dfrac{1}{2}\]
Now the ratio is
\[r:R = 1:2\]
Hence the ratio of inner to outer radii of the shell is \[1:2\]
Hence the answer is option A.
Note: In the above problem, we proved this using the given data and formulae. This is the proper method to solve this problem. The amount of space that contains something is called volume. A sphere is a three-dimensional shape.