JEE PYQ: Experimental Physics Question 15
Question 15 - 2020 (08 Jan 2020 Shift 2)
A simple pendulum is being used to determine the value of gravitational acceleration $g$ at a certain place. The length of the pendulum is 25.0 cm and a stop watch with 1 s resolution measures the time taken for 40 oscillations to be 50 s. The accuracy in g is:
(1) 5.40%
(2) 3.40%
(3) 4.40%
(4) 2.40%
Show Answer
Answer: (3)
Solution
$\frac{\Delta g}{g} = \frac{\Delta l}{l} + \frac{2\Delta T}{T} = \frac{0.1}{25.0} + 2\left(\frac{1}{50}\right) = 0.004 + 0.04 = 0.044$. Percentage error $= 4.4%$.
BETA
