Question

If the error in the measurement of a body be 3% and in the measurement of velocity be 2%. What will be the possible error in calculation of KE​

Answer 1

$\huge \underline \mathtt\red {Answer :-}$

$KE = \frac{1}{2} m {v}^{2}$

Therefore,

$\frac{Δk}{k} = \frac{Δm}{m} + \frac{2Δv}{v}$

$\frac{Δk}{k} \times 100 = \frac{Δm}{m} \times 100 + 2( \frac{Δv}{v} ) \times 100$

Hence Percentage error in KE

$= 3\% + 2 \times 2\%$

$= 7\%$