Question

When an iron piece of irregular shape is immersed in a container containing water of $12ml$, the water rises to $30ml$, then the volume of the iron is .

• A. $18m^3$
• B. $18\times {10}^{-6}{m}^3$
• C. $18\times {10}^{-6}c{m}^3$
• D. $18\times {10}^{-4}{m}^3$

Answer 1

The correct option is B $18\times {10}^{-6}{m}^3$

Initial level of water, $V_1=12ml$
Final level of water, $V_2=30ml$
Volume of iron piece, $V=V_2$- $V_1=30-12=18ml$
Since, $1ml=1c{m}^3$
$\therefore$$V=18ml=18c{m}^3$
Also, $1c{m}^3={10}^{-6}$$m^3$
$\therefore$ Volume of iron piece $=18\times$${10}^{-6}$$m^3$.