17. 11. A student measures the dist...

17. 11. A student measures the distance traversed in free fall of a body, initially at rest in a given time. He uses this data to estimate g, the acceleration due to gravity. If the maximum percentage errors in measurement of the distance and the time are e, and e, respectively, the percentage error in the estimation of g is [AIPMT (Mains)-2010] (1) Eze, (2) e,+ 2e (3) e, te (4) e-20,

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We know that, ( h=u t+frac{1}{2} g t^{2} ) ( Rightarrow h=frac{1}{2} g t^{2} quad[u=0, text { as the body is initially at rest }] ) ( Rightarrow g=frac{2 h}{t^{2}} ) Taking natural logarithm on both sides, we get ( ln g=ln h-2 ln t ) Different iating, ( frac{Delta g}{g}=frac{Delta h}{h}-2 frac{Delta t}{t} ) For maximum permissible error, ( Rightarrowleft(frac{Delta g}{g} times 100right)_{max }=left(frac{Delta h}{h} times 100right)+2left(frac{Delta t}{t} times 100right) ) According to question, ( left(frac{Delta h}{h} times 100right)=e_{1} ) and ( left(frac{Delta t}{t} times 100right)=e ) or ( ,left(frac{Delta g}{g} times 100right)_{max }=e_{1}+2 e_{2} )
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