JEE PYQ: Experimental Physics Question 18
Question 18 - 2019 (08 Apr 2019 Shift 2)
In a simple pendulum experiment for determination of acceleration due to gravity (g), time taken for 20 oscillations is measured by using a watch of 1 second least count. The mean value of time taken comes out to be 30 s. The length of pendulum is measured by using a meter scale of least count 1 mm and the value obtained is 55.0 cm. The percentage error in the determination of g is close to:
(1) 0.7%
(2) 0.2%
(3) 3.5%
(4) 6.8%
Show Answer
Answer: (4)
Solution
$\frac{\Delta g}{g} \times 100 = \frac{\Delta l}{l} \times 100 + 2\frac{\Delta T}{T} \times 100 = \frac{0.1}{55} \times 100 + 2\left(\frac{1}{30}\right) \times 100 = 0.18 + 6.67 = 6.8%$.
BETA
