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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
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B
( 50 ± 2.5) N
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C
( 50 ± 1.5) N
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D
( 50 ± 3.5) N
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Solution

The correct option is D ( 50 ± 3.5) N
Force$$(f)=\cfrac{mv^2}{r}=\cfrac{4\times(10)^2}{8}=50\\ \therefore\cfrac{\Delta F}{F}=\cfrac{\Delta m}{m}+2(\cfrac{\Delta V}{V})+(\cfrac{\Delta r}{r})\\ \quad=\cfrac{0.1}{4}+2\cfrac{(0.1)}{10}+\cfrac{0.8}{8}\\ \therefore \cfrac{\Delta F}{F}=0.07$$
Percentage error in $$F=25\%+2+2.5\%$$
$$\therefore$$ Percentage error in $$F=7\%$$
Eror in $$\Delta F$$ in $$F=F\times0.07\\ \quad=0\times0.07$$
$$\therefore$$ Error in $$\Delta F$$ in $$F=3.5N$$
$$\therefore F=(50\pm3.5)N$$


Physics

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