Physics, asked by chaitanya22787, 1 year ago

A solid shell loses half of its weight in water. If relative density of shell is 5 ,what fraction of its volume is hollow

Answers

Answered by aniket1454
29
Let y fraction of its volume be hollow.

Density of the shell = 5x103 kg/m3

Then

weight of the shell,

V-yV) x (5 x 10 to the power 3) x 9.8

Weight of the water displaced = Vx103x9.8

Loss in the weight of the shell = Weight of the water displaced

1/2 x (V-yV) x 5 x 10 to the power 3 x 9.8 = V x 10 to the power 3 x 9.8

On solving we get,

y = 3/5

Hope this will help you....

chaitanya22787: two substances of densities q1 and q2 are mixed in equal volume and the relative density of mixture is 4.when they are mixed in equal masses,the relative density of the mixture is 3. the values of q1 and q2 are
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