When a golf ball is lowered into a measuring cylinder containing water,the water level rises by 30 $c m^{3}$ when the ball is completely submerged.If the mass of ball in air is 33 g,find its density.

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Solution

Given, change in volume of water when the ball is submerged = $30 c m^{3}$
Now, the volume of water displaced = volume of ball immersed fully
Hence volume of ball = $30 c m^{3}$
Mass = 33 g
So $D e n s i t y = \frac{m a s s}{v o l u m e} = \frac{33}{30} g c m^{- 3}$

$= 1.1 g c m^{- 3}$

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When a golf ball is lowered into a measuring cylinder containing water,the water level rises by 30 cm3 when the ball is completely submerged.If the mass of ball in air is 33 g,find its density.

Given, change in volume of water when the ball is submerged = 30 cm3 Now, the volume of water displaced = volume of ball immersed fully Hence volume of ball = 30 cm3 Mass = 33 g So Density=massvolume=3330 g cm−3 =1.1 g cm−3

When a golf ball is lowered into a measuring cylinder containing water,the water level rises by 30 $c m^{3}$ when the ball is completely submerged.If the mass of ball in air is 33 g,find its density.

Given, change in volume of water when the ball is submerged = $30 c m^{3}$
Now, the volume of water displaced = volume of ball immersed fully
Hence volume of ball = $30 c m^{3}$
Mass = 33 g
So $D e n s i t y = \frac{m a s s}{v o l u m e} = \frac{33}{30} g c m^{- 3}$

$= 1.1 g c m^{- 3}$