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Standard X

Physics

Principle of Calorimetry

If 5 litres o...

Question

If 5 litres of kerosene has a mass of 5 kg, then what is the density of kerosene?

500 kg/

m^{3}

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1000 kg/

m^{3}

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100 kg/

m^{3}

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50 kg/

m^{3}

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Solution

The correct option is B 1000 kg/ $m^{3}$
Density of kerosene $= \frac{m a s s}{v o l u m e}$
$= \frac{5 k g}{5 l i t r e s} = \frac{5 k g}{5 \times 10^{- 3} m^{3}} = 1000 k g m^{- 3}$

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Complete the following.

A solid of density $900 k g / m^{3}$ floats in oil as shown in the given diagram. The oil floats on water of density $1000 k g / m^{3}$ as shown. The density of oil in $k g / m^{3}$ could be :

Q. Ratio of the weights of a

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Given: Density of iron

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, density of wood

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and density of air

= 1.293 {kg/m}^{3}

An iceberg of density $900 k g / m 3$ is floating in water of density $1000 k g / m 3$ . the percentage of volume of ice-cube outside the water is

Q. An ice berg 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

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If 5 litres of kerosene has a mass of 5 kg, then what is the density of kerosene?

The correct option is B 1000 kg/m3Density of kerosene =massvolume=5kg5litres=5kg5×10−3m3=1000kgm−3

An iceberg of density 900kg/m3 is floating in water of density 1000kg/m3. the percentage of volume of ice-cube outside the water is

The correct option is B 1000 kg/ $m^{3}$
Density of kerosene $= \frac{m a s s}{v o l u m e}$
$= \frac{5 k g}{5 l i t r e s} = \frac{5 k g}{5 \times 10^{- 3} m^{3}} = 1000 k g m^{- 3}$

An iceberg of density $900 k g / m 3$ is floating in water of density $1000 k g / m 3$ . the percentage of volume of ice-cube outside the water is