If displacement of a body, s = (200 ± 5) meters and time taken by it t = (20 ± 0.2)seconds, then find the percentage error in the calculation of velocity.

The correct option is 3.5%.

Step-by-Step Explanation:

Velocity ($v$) of a body is defined as the displacement ($s$) divided by the time taken ($t$):
$v=\frac{s}{t}$

When two quantities are divided, the relative error in the result is the sum of the relative errors of the individual quantities. Therefore, the relative error in velocity is given by:
$\frac{\Delta v}{v}=\frac{\Delta s}{s}+\frac{\Delta t}{t}$

To find the percentage error, we multiply both sides of the equation by 100:
$\text{Percentage error in velocity}=\left(\frac{\Delta v}{v}\times 100\right)\%=\left(\frac{\Delta s}{s}\times 100+\frac{\Delta t}{t}\times 100\right)\%$

From the given data:
Displacement, $s=200\pm 5\text{meters}$ (where $s=200$ and absolute error $\Delta s=5$)
Time, $t=20\pm 0.2\text{seconds}$ (where $t=20$ and absolute error $\Delta t=0.2$)

Now, let's calculate the individual percentage errors:
1. Percentage error in displacement:
$\frac{\Delta s}{s}\times 100=\frac{5}{200}\times 100=2.5\%$

2. Percentage error in time:
$\frac{\Delta t}{t}\times 100=\frac{0.2}{20}\times 100=1\%$

Adding the percentage errors together gives the total percentage error in velocity:
$\text{Percentage error in velocity}=2.5\%+1\%=3.5\%$