The percentage error in the above problem is

The correct option is 8.95%.

Context:
This question refers to finding the percentage error in the velocity of a body when the distance covered is measured as $d=(13.8\pm 0.2)\text{m}$ and the time taken is measured as $t=(4.0\pm 0.3)\text{s}$.

Step-by-step Derivation:

1. Identify the given values and their absolute errors:
Measured distance, $d=13.8\text{m}$
Absolute error in distance, $\Delta d=0.2\text{m}$
Measured time, $t=4.0\text{s}$
Absolute error in time, $\Delta t=0.3\text{s}$

2. Formulate the relationship for error propagation in velocity ($v=\frac{d}{t}$):
When two quantities are divided, their relative errors add up to give the relative error in the resulting quantity. Thus, the relative error in velocity $v$ is given by:

$\frac{\Delta v}{v}=\frac{\Delta d}{d}+\frac{\Delta t}{t}$

3. Calculate the percentage error by multiplying the relative error by 100:

$\text{Percentage Error in}v=\left(\frac{\Delta d}{d}+\frac{\Delta t}{t}\right)\times 100$

4. Compute the individual percentage errors:
Percentage error in distance:

$\frac{\Delta d}{d}\times 100=\frac{0.2}{13.8}\times 100\approx 1.449\%$

Percentage error in time:

$\frac{\Delta t}{t}\times 100=\frac{0.3}{4.0}\times 100=7.5\%$

5. Sum the individual percentage errors to find the total percentage error in velocity:

$\text{Total Percentage Error}=1.449\%+7.5\%=8.949\%\approx 8.95\%$

Therefore, the percentage error in the velocity is approximately 8.95%.