Question

If the length of a simple pendulum is recorded as $\left( {90.00 \pm 0.02} \right)cm$ and period as $\left( {1.90 \pm 0.02} \right)s$, the percentage error in the measurement of acceleration due to gravity is:
(A) $4.2$
(B) $2.1$
(C) $1.5$
(D) $2.8$

Answer 1

Hint
To solve this question, we need to find the expression of the acceleration due to gravity in terms of the pendulum’s length and its time period of oscillation. Then, applying the concept of relative errors on that expression will give the answer.
Formula Used: The formula used to solve this question is
$\Rightarrow T = 2\pi \sqrt {\dfrac{l}{g}}$, where $T$ is the time period of a simple pendulum of length $\operatorname{l}$ and $g$ is the acceleration due to gravity.

Complete step by step answer
The length of the simple pendulum is recorded as $\left( {90.00 \pm 0.02} \right)cm$
This means that the true value of the length is $l = 90.00cm$ and the error in its measurement is $\Delta l = 0.02cm$.
Also, the time period is recorded as $\left( {1.90 \pm 0.02} \right)s$
This means that the true value of the length is $T = 1.90s$ and the error in its measurement is $\Delta T = 0.02s$.
We know that the time period of oscillation of a simple pendulum is given by
$\Rightarrow T = 2\pi \sqrt {\dfrac{l}{g}}$
Taking square on both the sides, we get
$\Rightarrow {T^2} = 4{\pi ^2}\dfrac{l}{g}$
Multiplying with $\dfrac{g}{{{T^2}}}$ on both the sides, we have
$\Rightarrow g = 4{\pi ^2}\dfrac{l}{{{T^2}}}$
Taking relative errors on both the sides of this equation
$\Rightarrow\dfrac{{\Delta g}}{g} = \dfrac{{\Delta l}}{l} + 2\dfrac{{\Delta T}}{T}$
Substituting the values on the RHS from above
$\Rightarrow \dfrac{{\Delta g}}{g} = \dfrac{{0.02}}{{90.00}} + 2\dfrac{{0.02}}{{1.90}}$
On solving, we get
$\Rightarrow \dfrac{{\Delta g}}{g} = 0.021$
Now, the percentage error will be equal to
$\Rightarrow \dfrac{{\Delta g}}{g} \times 100$
$\Rightarrow 0.021 \times 100$
$\Rightarrow 2.1\%$
Thus, the percentage error in the measurement of the acceleration due to gravity is equal to $2.1\%$.
Hence, the correct answer is option (A).

Note
Do not understand the relative error concept as simply the multiplication of the exponents of the quantities with their corresponding relative errors. Instead, the modulus of the exponents is multiplied. This is done to ensure that the relative errors of all the quantities are being added. Always remember, the errors are always added, never subtracted.