Question

When a toy of irregular shape is immersed in a measuring cylinder containing water of $20ml$, the water rises to $37ml$, then the volume of the toy is:

• A. $20\times$${10}^{-6}$$m^3$
• B. $17\times$${10}^{-6}$$m^3$
• C. $37\times$${10}^{-6}$$m^3$
• D. $27\times$${10}^{-6}$$m^3$

Answer 1

The correct option is B $17\times$${10}^{-6}$$m^3$

Initial volume of water, $V_1=20ml$
Final volume of water, $V_2=37ml$
Hence, volume of toy, $V$
$=V_2-V_1=37ml-20ml=17ml$
Since, $1ml=1c{m}^3$
$\therefore$$V=17ml=17c{m}^3$
Also, $1c{m}^3={10}^{-6}$$m^3$
$\therefore$ Volume of toy $=17\times$${10}^{-6}$$m^3$.