General Physics Ques 36
Question 15
- A student uses a simple pendulum of exactly $1 \mathrm{~m}$ length to determine $g$, the acceleration due to gravity. He uses a stop watch with the least count of $1 \mathrm{~s}$ for this and records $40 \mathrm{~s}$ for 20 oscillations. For this observation, which of the following statement(s) is/are true?
(2010)
(a) Error $\Delta T$ in measuring $T$, the time period, is $0.05 \mathrm{~s}$
(b) Error $\Delta T$ in measuring $T$, the time period, is $1 \mathrm{~s}$
(c) Percentage error in the determination of $g$ is $5 %$
(d) Percentage error in the determination of $g$ is $2.5 %$
Numerical Value Based Questions
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Answer:
Correct Answer: 36.(a, c)
Solution:
Formula:
- $T=\frac{40 \mathrm{~s}}{20}=2 \mathrm{~s}$
Further, $t=n T=20 T$ or $\Delta t=20 \Delta T$
$\therefore \quad \frac{\Delta t}{t}=\frac{\Delta T}{T}$ or $\Delta T=\frac{T}{t} \cdot \Delta t=\frac{2}{40} \quad(1)=0.05 \mathrm{~s}$
Further, $\quad T=2 \pi \sqrt{\frac{l}{g}}$ or $\quad T \propto g^{-1 / 2}$
$\therefore \quad \frac{\Delta T}{T} \times 100=-\frac{1}{2} \times \frac{\Delta g}{g} \times 100$
or $%$ error in determination of $g$ is
$$ \begin{aligned} \frac{\Delta g}{g} \times 100 & =-200 \times \frac{\Delta T}{T} \ & =-\frac{200 \times 0.05}{2} \ & =-5 % \end{aligned} $$
$\therefore$ Correct options are (a) and (c).
BETA
