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13.Oscillations
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In an experiment for determining the gravitational acceleration $g$ of a place with the help of a simple pendulum, the measured time period square is plotted against the string length of the pendulum in the figure. What is the value of $g$ at the place? ...... $m/s^2$
A
$9.81$
B
$9.87$
C
$9.91$
D
$10$
(JEE MAIN-2014)
Solution
From graph it is clear that wher
$\mathrm{L}=1 \mathrm{m}, \mathrm{T}^{2}=4 \mathrm{s}^{2}$
As we know,
$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$
$\Rightarrow \mathrm{g}=\frac{4 \pi^{2} \mathrm{L}}{\mathrm{T}^{2}}$
$=4 \times\left(\frac{22}{7}\right)^{2} \times \frac{1}{4}=\left(\frac{22}{7}\right)^{2}$
$\therefore \mathrm{g}=\frac{484}{49}=9.87 \mathrm{m} / \mathrm{s}^{2}$
Standard 11
Physics
