An ice-cube of density 900 ~kg / m^3 is floating in water of density 1000 ~kg / m^3 . The percentage of volume of ice-cube outside the water is

Correct answer: (C) 10 \%. The percentage of volume of ice cube outside the water is =fracρ_water -ρ_ice ρ_water × 100 =(1000-900)/1000 × 100=10 \%

Q97 / 160 · JIPMER 2005 · Physics

An ice-cube of density $900 \mathrm{~kg} / \mathrm{m}^3$ is floating in water of density $1000 \mathrm{~kg} / \mathrm{m}^3$. The percentage of volume of ice-cube outside the water is

A
$20 \%$
B
$35 \%$
Correct answer
C
$10 \%$
D
$25 \%$

Solution

The percentage of volume of ice cube outside the water is $\begin{aligned} & =\frac{\rho_{\text {water }}-\rho_{\text {ice }}}{\rho_{\text {water }}} \times 100 \\ & =\frac{1000-900}{1000} \times 100=10 \%\end{aligned}$

Hence, the correct answer is option: 'C'.

An ice-cube of density $900 \mathrm{~kg} / \mathrm{m}^3$ is floating in water of density $1000 \mathrm{~kg} / \mathrm{m}^3$. The percentage of volume of ice-cube outside the water is

Solution

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