Question

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Q.64. The period of oscillation of' a simple pendulum is given by $T=2\mathrm{\pi }\sqrt{\frac{\mathrm{l}}{\mathrm{g}}}$, where l is about 100 cm and is known to have 1 mm accuracy. The period is about 2 s. The time of 100 oscillations is measured by a stop watch of least count 0.1 s. The percentage error in g is
(1) 0.1 %
(2) 1 %
(3) 0.2 %
(4) 0.8 %

Answer 1

Dear Student,
As we know the formula for time period is given by,
$T=2\mathrm{\pi }\sqrt{\frac{\mathrm{l}}{\mathrm{g}}}$
So, value of $g=\frac{4{\mathrm{\pi }}^2\mathrm{l}}{T^2}$
we get from the above formula, percentage error in g,


$\frac{∆g}{g}\times 100=\frac{{\displaystyle ∆l}}{{\displaystyle l}}\times 100+1\frac{{\displaystyle ∆T}}{{\displaystyle T}}\times 100$ .............(1)
Now, given in the question, $l=100cm\&∆l=1mm=0.1cm.$We know, Time period = total time / no. of oscillations
Here, time period is given = 1 sec. and no of oscillations = 100.
So, using equation (1) we get,
$$\begin{gathered} \Rightarrow \frac{∆g}{g}\times 100=\frac{{\displaystyle 0.1}}{{\displaystyle 100}}\times 100+1\times \frac{{\displaystyle 0.1}}{{\displaystyle (1\times 100)}}\times 100 \\ =\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{\%} \end{gathered}$$ [ given least count of stopwatch=0.1s]


Regards