JEE PYQ: Mathematics In Physics Question 4
Question 4 - 2021 (24 Feb Shift 2)
The period of oscillation of a simple pendulum is $T = 2\pi\sqrt{\frac{L}{g}}$. Measured value of ‘$L$’ is 1.0 m from meter scale having a minimum division of 1 mm and time of one complete oscillation is 1.95 s measured from stopwatch of 0.01 s resolution. The percentage error in the determination of ‘$g$’ will be:
(1) 1.33%
(2) 1.30%
(3) 1.13%
(4) 1.03%
Show Answer
Answer: (3)
Solution
$T = 2\pi\sqrt{\frac{L}{g}}$ $T^2 = 4\pi^2 \frac{L}{g}$ $g = 4\pi^2 \frac{L}{T^2}$ $\frac{\Delta g}{g} = \frac{\Delta L}{L} + \frac{2\Delta T}{T}$ $= \frac{1 \text{ mm}}{1 \text{ m}} + \frac{2(10 \times 10^{-3})}{1.95} \times 100$ $= 1.13%$
BETA
