Question

A test tube loaded with lead shots weighs 50 gf and floats upto the mark 'X' in water. The test tube is then made to

float alcohol. It is found that 10 gf of lead shots have to be removed, so as to float it to level 'X'. Find RD of alcohol.​

Answer 1

Answer:-

Relative Density (R.D.) of alcohol is 0.8 .

Explanation:-

In this problem, we will use 'W' for the term 'weight'.

Also let :-

• W₁ = 'W' of water displaced .

• W₂ = 'W' of alcohol displaced.

When test tube floats on water :-

• 'W' of the test tube = 50 gf

Now according to Archimedes Principle, weight of water displaced by the test tube will be :-

'W' of fluid displaced = 'W' of test tube

Thus, weight of water displaced by the test tube (W₁) is 50 gf.

When test tube floats on alcohol :-

As the test tube floats on alcohol, 10 gf of lead shots have to be removed, so that it can float up to the level 'X'.

∴ 'W' of test tube in alcohol :-

= 50 gf - 10 gf

= 40 gf

Thus, weight of alcohol displaced by the test tube (W₂) is 40 gf.

______________________________

Now as in both cases the test tube is floating up to the mark 'X', thus volume also remains same.

R.D. of alcohol :-

= W₂/W₁

= 40/50

= 0.8