Question

A test tube weighing $17{\text{ gf,}}$ floats in alcohol to the level P. When the test tube is made to float in water to the level P, $3{\text{ gf}}$ of the lead shots are added in it. Find the R.D. of alcohol.

Answer 1

Here, we will use the principle of the floatation and by using the given data will find the weight of the alcohol and the water and take its ratio for the R.D. of the alcohol and simplify the resultant fraction.

Complete step by step solution:
Given that –
When the tube floats in the alcohol-
Weight of the test tube $= 17{\text{ gf}}$
By law of the floatation which states that – the floating object displaces the weight of the fluid which is equal to its own weight.
Weight of the alcohol displaced by the test tube is equal to the weight of the test tube $= 17\,{\text{gf}}$
When the tube starts floating in the water –
When the test tube is made to float in water at level P, as in alcohol then $3\;{\text{gf}}$ lead stones are added in it.
Therefore, the net weight of test tube is $= 17\,{\text{gf + 3 gf = 20 gf}}$
Hence, the weight of the water displaced by the test tube $= 20{\text{ gf}}$
By using the concept of the floatation-
Volume of the alcohol displaced is equal to the volume of the water displaced.
Hence, R.D. of Alcohol can be expressed by the ratio of weight of alcohol displaced to the weight of equal volume of water.
R.D. of alcohol $= \dfrac{{17}}{{20}}$
Multiply and divide the fraction with and simplify the above fraction –
R.D. of alcohol $= \dfrac{{17}}{{20}} \times \dfrac{5}{5}$
$\Rightarrow$ R.D. of alcohol $= \dfrac{{85}}{{100}}$
Since the denominator is in hundreds, shift two points from right to the left of the digit.

Hence, R.D. of alcohol $= 0.85$

Note: Remember the ratio of two similar physical quantities gives the resultant value in unit-less terms. Always remember the properties of the fractions and apply its laws accordingly for the simplification. Know how to convert the fractions into decimals.