- JEE Main 2019:
We know that:
$$\frac{\Delta P}{P} = \frac{\Delta A^3B^2}{A^3B^2} + \frac{\Delta B^2}{B^2} + \frac{\Delta C^2}{C^2} + \frac{\Delta D}{D}$$
Substituting the values, we get:
$$\frac{\Delta P}{P} = 3\left(\frac{1}{100}\right) + 2\left(\frac{2}{100}\right) + 3\left(\frac{3}{100}\right) + 4\left(\frac{4}{100}\right) = 18%$$
Therefore, the percentage error in the measurement of P is 18%.
- JEE Main 2018:
We know that:
$$\text{Percentage error} = \frac{\text{Absolute error}}{\text{Actual value}} \times 100$$
In this case, the absolute error in the length of
