While measuring the acceleration due to gravity by a simple pendu… - Vidyadip

MCQ

While measuring the acceleration due to gravity by a simple pendulum, a student makes a positive error of $1\%$ in the length of the pendulum and a negative error of $3\%$ in the value of time period. His percentage error in the measurement of $g$ by the relation $g = 4{\pi ^2}\left( {l/{T^2}} \right)$ will be ........ $\%$

A
$2$
B
$4$
✓
$7$
D
$10$

✓

Answer

Correct option: C.

$7$

c
Error always gets added so,

$\mathrm{g}=4 \pi^2\left(\frac{\mathrm{l}}{\mathrm{T}^2}\right) $

$\frac{\Delta \mathrm{g}}{\mathrm{g}}=\frac{\Delta 1}{\mathrm{~L}}+\frac{2 \Delta \mathrm{T}}{\mathrm{T}} $

$\frac{\Delta \mathrm{g}}{\mathrm{g}} \times 100=1+2(3) $

$\frac{\Delta \mathrm{g}}{\mathrm{g}} \times 100=7 $

Percentage error $=7 \%$

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