The relative density of material of a body is found by weighing it first in air and then in water. If the weight in air is (5.00 ± 0.05 ) Newton and weight in water is (4.00±0.05) Newton. Then the relative density along with the maximum permissible percentage error is

The correct option is 5.0 ± 11%.

Step-by-step Explanation:

Relative density (R.D.) of a material is defined as the ratio of its weight in air to the loss of its weight in water.
Let:
Weight in air, $W_1=(5.00\pm 0.05)\text{N}$
Weight in water, $W_2=(4.00\pm 0.05)\text{N}$

The loss of weight in water ($W$) is given by:
$W=W_1-W_2$

Substituting the values:
$W=5.00-4.00=1.00\text{N}$

The absolute error in the loss of weight ($\Delta W$) is the sum of the absolute errors in the individual measurements (since errors add up during subtraction):
$\Delta W=\Delta W_1+\Delta W_2$
$\Delta W=0.05+0.05=0.10\text{N}$

So, the loss of weight in water is:
$W=(1.00\pm 0.10)\text{N}$

Now, the relative density ($\text{R.D.}$) is calculated as:
$\text{R.D.}=\frac{W_1}{W}$
$\text{R.D.}=\frac{5.00}{1.00}=5.0$

The relative error in relative density ($\frac{\Delta \text{R.D.}}{\text{R.D.}}$) is given by:
$\frac{\Delta \text{R.D.}}{\text{R.D.}}=\frac{\Delta W_1}{W_1}+\frac{\Delta W}{W}$

Substituting the values into the error formula:
$\frac{\Delta \text{R.D.}}{\text{R.D.}}=\frac{0.05}{5.00}+\frac{0.10}{1.00}$
$\frac{\Delta \text{R.D.}}{\text{R.D.}}=0.01+0.10=0.11$

To find the maximum permissible percentage error, multiply the relative error by 100:
$\text{Percentage Error}=0.11\times 100\%=11\%$

Therefore, the relative density along with the maximum permissible percentage error is:
5.0 ± 11%