JEE Physics 2023 Question with Solution

Given: $P = \frac{a^2b^3}{c\sqrt{d}}$ and the percentage errors are $1\%$ in $a$, $2\%$ in $b$, $3\%$ in $c$, and $4\%$ in $d$.

Find: The percentage error in $P$.

For multiplication, division, powers, and roots, percentage errors add with their respective powers. Therefore,

Substitute the given values:

Now add all contributions:

Therefore, the percentage error in $P$ is $13\%$. The correct option is B.

Given: $P$ depends on $a^2$, $b^3$, $c^{-1}$, and $d^{-1/2}$.

Find: Total percentage error in $P$.

The error contribution from each factor is obtained by multiplying the percentage error by the magnitude of its power:

• From $a^2$: $2 \times 1\% = 2\%$
• From $b^3$: $3 \times 2\% = 6\%$
• From $c^{-1}$: $1 \times 3\% = 3\%$
• From $d^{-1/2}$: $\frac{1}{2} \times 4\% = 2\%$

Adding them gives:

Hence, the required percentage error is $13\%$, so the correct option is B.