Question

If the percentage errors in the measurrment of mass,length, and time, are 1%,2%,3% respectively ,then the maximum permissible error in the measurement of the accelaration.of a particle is
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Answer 1

Explanation:

Given percentage errors in mass and velocity = 2% and 3% respectively

The kinetic energy of the particle is given by

KE =

2

mv

2

The percentage error in kinetic energy will be

m

Δm

×100+2

v

Δv

×100

= 2+6

= 8%

Answer 2

Answer: $\frac{\Delta a}{a}= \frac{\Delta L}{L}+ 2.\frac{\Delta T}{T}$

$\frac{\Delta a}{a}= 2\%+ 2.3\%\\=2\%+6\%\\=8\%$

The maximum permissible error in the measurement of the acceleration of a particle is 8%

Explanation:

Given :

the percentage error in the measurement of

mass, $\frac{\Delta M}{M} = 1 \%$

length, $\frac{\Delta L}{L} = 2 \%$

time, $\frac{\Delta T}{T} = 3 \%$

Find,  the maximum permissible error in the measurement of the acceleration of a particle is$acceleration, \frac{\Delta a}{a}=?$

$a = \frac{d^2x}{dt^2}$

unit of acceleration a is $m/s^2$ . Its dimensional formula is $\frac{L}{T^{-2}}$

formula for error in acceleration is

$\frac{\Delta a}{a}\times100= \frac{\Delta L}{L}\times100+ 2.\frac{\Delta T}{T}\times100$

$\frac{\Delta a}{a}= \frac{\Delta L}{L}+ 2.\frac{\Delta T}{T}$