A metallic sphere with an internal cavity weighs 40 g weight in air and 20 g weight in water. If the density of the material with cavity be 8 g per $c m^{3}$ then the volume of cavity is:

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Solution

The correct option is D 15 $c m^{3}$
$V_{s} = 5 c c$ As given
When it is placed in water net force $=$ new downward force
$\Rightarrow H o g - (V + V_{s}) g = 20 g \Rightarrow V + V_{s} = 20 ∴ V = 20 - V_{s} c c V = 20 - 5 c m^{2} = 15 c m^{2}$

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A metallic sphere with an internal cavity weighs 40 g weight in air and 20 g weight in water. If the density of the material with cavity be 8 g per cm3 then the volume of cavity is:

The correct option is D 15 cm3Vs=5cc As givenWhen it is placed in water net force= new downward force⇒Hog−(V+Vs)g=20g⇒V+Vs=20∴V=20−VsccV=20−5cm2=15cm2

A metallic sphere with an internal cavity weighs 40 g weight in air and 20 g weight in water. If the density of the material with cavity be 8 g per $c m^{3}$ then the volume of cavity is:

The correct option is D 15 $c m^{3}$
$V_{s} = 5 c c$ As given
When it is placed in water net force $=$ new downward force
$\Rightarrow H o g - (V + V_{s}) g = 20 g \Rightarrow V + V_{s} = 20 ∴ V = 20 - V_{s} c c V = 20 - 5 c m^{2} = 15 c m^{2}$