Question

If error in measurement in mass and velocity is 1 % then error in measurement of kinetic energy is?

Answer 1

- Hint: We can solve it by error progression method in which when quantities are in multiplication in any formula we simply add their individual error but if there is an exponent of quantities in a given formula, first we multiply each quantity with its exponent then we add.

Complete step-by-step solution -
Given:
Error in velocity(v) 1%; Error is mass (m) 1%
We know formula of kinetic energy,$$K.E={\displaystyle \frac{1}{2}}\left(m{v}^2\right)$$
Now according to the error progression rule if quantities are in multiplication in any formula with exponent then we first multiply individual error of each quantity with its exponent then we add.
using above rules of error progression,
Percentage error in Kinetic Energy =$$(\mathrm{\%}\text{error}\text{in}m)+2\left(\mathrm{\%}\text{ error in }v\right)$$
Percentage error in Kinetic Energy= 1% + 2%
or error in Kinetic Energy= 3%
Thus, percentage error in kinetic energy is 3%.

Note: When two quantities are multiplied or divided, maximum percentage error is equal to sum of percentage errors
Percentage error due to power of measured quantity is equal to multiple of power value i.e
For quantity $$Z=A^n$$
Percentage error in $$Z=n$$ (percentage error in A)