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A students measures the distance traversed in free fall of a body, the 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 $e_1$ and $e_2$ respectively, the percentage error in the estimation of $g$ is
$e_2-e_1$
$e_1+2{e_2}$
$e_1+e_2$
$e_1-2{e_2}$
Solution
From, the, relation
$h = ut = \frac{1}{2}g{t^2}$
$h = \frac{1}{2}g{t^2}$$ \Rightarrow g = \frac{{2h}}{{{t^2}}}$ (body, initially, at, rest)
Taking, natural, log aritham, on, both, sides, we, get
$In\,g = In\,h – 2\,In\,t$
Differentiating, $\frac{{\Delta h}}{g} = \frac{{\Delta h}}{h}\, – 2\,\frac{{\Delta t}}{t}$
For, max imum, permissible, error,
or,${\left( {\frac{{\Delta g}}{g} \times 100} \right)_{\max }} = \left( {\frac{{\Delta h}}{h} \times 100} \right) + 2 \times \left( {\frac{{\Delta t}}{t} \times 100} \right)$
According, to, problem
$\frac{{\Delta h}}{h} \times 100{ = _{{e_1}}}\,and\,\frac{{\Delta t}}{t} \times 100{ = _{{e_2}}}$
Therefore, $( {\frac{{\Delta g}}{g} \times 100} )_{\max } = {e_1} + 2{e_2}$
Similar Questions
A student determined Young's Modulus of elasticity using the formula $Y=\frac{M g L^{3}}{4 b d^{3} \delta} .$ The value of $g$ is taken to be $9.8 \,{m} / {s}^{2}$, without any significant error, his observation are as following.
| Physical Quantity | Least count of the Equipment used for measurement | Observed value |
| Mass $({M})$ | $1\; {g}$ | $2\; {kg}$ |
| Length of bar $(L)$ | $1\; {mm}$ | $1 \;{m}$ |
| Breadth of bar $(b)$ | $0.1\; {mm}$ | $4\; {cm}$ |
| Thickness of bar $(d)$ | $0.01\; {mm}$ | $0.4 \;{cm}$ |
| Depression $(\delta)$ | $0.01\; {mm}$ | $5 \;{mm}$ |
Then the fractional error in the measurement of ${Y}$ is
