Question

For a simple pendulum values are given as length of pendulum $l=\left( 50\pm 0.1 \right)\text{cm}$, Period of oscillation $T=\left( 2\pm 0.1 \right)\text{s}$. The maximum permissible error in the measurement of acceleration due to gravity g will be nearly.
(a) $\pm 1\ m/{{s}^{2}}$
(b) $\pm 2\ m/{{s}^{2}}$
(c) $\pm 3.5\ m/{{s}^{2}}$
(d) $\pm 0.4\ m/{{s}^{2}}$

Answer 1

- Hint: We know that acceleration of simple pendulum can be given by the formula, $T=2\pi \sqrt{\dfrac{l}{g}}$ then we have to convert in the form of acceleration. Then by using the formula of error i.e. $X=a{{b}^{2}}\Rightarrow \dfrac{\Delta X}{X}=\dfrac{\Delta a}{a}+\dfrac{2\Delta b}{b}$, we can find the value of g and find the maximum permissible error.

Formula used: $T=2\pi \sqrt{\dfrac{l}{g}}$, $X=a{{b}^{2}}\Rightarrow \dfrac{\Delta X}{X}=\dfrac{\Delta a}{a}+\dfrac{2\Delta b}{b}$

Complete step-by-step solution:
In question it is given that for a simple pendulum values are given as length of pendulum $l=\left( 50\pm 0.1 \right)\text{cm}$, Period of oscillation $T=\left( 2\pm 0.1 \right)\text{s}$. Now, the formula of time period can be given as,
$T=2\pi \sqrt{\dfrac{l}{g}}$ …………………..(i)
Where, $T$ is the time period of the pendulum, $l$ is length of pendulum and g is gravitational acceleration of the pendulum.
Now, on converting the equations in the terms of g we will get,
$g=4{{\pi }^{2}}\dfrac{l}{{{T}^{2}}}$ ………………….(ii)
Now, the error in any equation can be found by using the formula,
$X=a{{b}^{2}}\Rightarrow \dfrac{\Delta X}{X}=\dfrac{\Delta a}{a}+\dfrac{2\Delta b}{b}$
In the same way, we can also write equation (ii) in terms of error as,
$\Rightarrow \dfrac{\Delta g}{g}=\pm \left( \dfrac{\Delta l}{l}+\dfrac{2\Delta T}{T} \right)$
Where, $l=50$, $\Delta l=0.1$, $T=2$ and $\Delta T=0.1$, is given in the question, so, on substituting these values in the equation we will get,
$\Rightarrow \dfrac{\Delta g}{g}=\pm \left( \dfrac{0.1}{50}+\dfrac{2\left( 0.1 \right)}{2} \right)$
$\Rightarrow \dfrac{\Delta g}{g}=\pm \left( 0.002+0.1 \right)$
$\Rightarrow \Delta g=\pm \left( 0.102 \right)\times 9.8$
$\Rightarrow \Delta g=\pm 0.9996\cong \pm 1\ m/{{s}^{2}}$
Thus, the maximum permissible error in gravitational acceleration is $\pm 1\ m/{{s}^{2}}$.
Hence, option (a) is correct.

Note: In using the formula of error student might take negative sign in front of $\Delta T$ i.e. $\Rightarrow \dfrac{\Delta g}{g}=\pm \left( \dfrac{\Delta l}{l}-\dfrac{2\Delta T}{T} \right)$ as T is in division in the formula but, it is wrong here the division or multiplication does not matter for considering the positive or negative sign. So, students must remember this while solving the sums of errors.