Question

Find the maximum possible percentage error in the measurement of force on an object(on mass m) travelling at velocity v in a circle of radius r, if m=(4.0 plus minus 0.1)kg, v=(10 plus minus 0.1)m/s and r=(8.0 plus minus 0.2)m

• A. ( 50 ± 1) N
• B. ( 50 ± 2.5) N
• C. ( 50 ± 1.5) N
• D. ( 50 ± 3.5) N

Answer 1

The correct option is D ( 50 ± 3.5) N

Force$\left(f\right)=\frac{m{v}^2}{r}=\frac{4\times (10{)}^2}{8}=50\therefore \frac{\Delta F}{F}=\frac{\Delta m}{m}+2\left(\frac{\Delta V}{V}\right)+\left(\frac{\Delta r}{r}\right)=\frac{0.1}{4}+2\frac{\left(0.1\right)}{10}+\frac{0.8}{8}\therefore \frac{\Delta F}{F}=0.07$

Percentage error in $F=25\%+2+2.5\%$

$\therefore$ Percentage error in $F=7\%$

Eror in $\Delta F$ in $F=F\times 0.07=0\times 0.07$

$\therefore$ Error in $\Delta F$ in $F=3.5N$

$\therefore F=(50\pm 3.5)N$