Question

A measuring jar has a water level of up to$10mL$. When ten identical spheres are immersed in it, the level rises to $50mL.$ Then the radius of the sphere is _________.

• A. ${\left(\frac{21}{22}\right)}^{\frac{1}{2}}cm$
• B. ${\left(\frac{21}{22}\right)}^{\frac{1}{3}}cm$
• C. ${\left(\frac{22}{21}\right)}^{\frac{1}{3}}cm$
• D. ${\left(\frac{21}{23}\right)}^{\frac{1}{2}}cm$

Answer 1

The correct option is B

${\left(\frac{21}{22}\right)}^{\frac{1}{3}}cm$

Step 1:Given data and formula used

Initial water level$=10mL$, Final water level $=50mL$

Let $V$ be the volume of each sphere.

Then rise in water level = volume of ten identical spheres
Step 2: Calculation of radius

Then volume of ten identical sphere is $=50mL-10mL=40mL$

Volume of each sphere $V=\frac{40}{10}=4c{m}^3$

Formula for volume of sphere $=\frac{4}{3}\pi r^3$

$$\begin{gathered} \frac{4}{3}{\mathrm{\pi r}}^3=4c{m}^3 \\ {r}^3=\frac{4\times 3\times 7}{4\times 22}=\left(\frac{21}{22}\right)c{m}^3\left(\because \pi =\frac{22}{7}\right) \\ r={\left(\frac{21}{22}\right)}^{\frac{1}{3}}\mathrm{cm} \end{gathered}$$

Hence, the correct option is (b).