JEE PYQ: Properties Of Matter Question 1
Question 1 - 2021 (16 Mar 2021 Shift 2)
In order to determine the Young’s Modulus of a wire of radius 0.2 cm (measured using a scale of least count = 0.001 cm) and length 1 m (measured using a scale of least count = 1 mm), a weight of mass 1 kg (measured using a scale of least count = 1 g) was hanged to get the elongation of 0.5 cm (measured using a scale of least count 0.001 cm). What will be the fractional error in the value of Young’s Modulus determined by this experiment?
(1) 0.14%
(2) 0.9%
(3) 9%
(4) 1.4%
Show Answer
Answer: (4)
Solution
$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{FL}{AL} = \frac{mgL}{\pi R^2 \ell}$
$\frac{\Delta Y}{Y} = \frac{\Delta m}{m} + \frac{\Delta L}{L} + 2 \cdot \frac{\Delta R}{R} + \frac{\Delta \ell}{\ell}$
$\frac{\Delta Y}{Y} \times 100 = 100\left[\frac{1}{1000} + \frac{1}{1000} + 2\left(\frac{0.001}{0.2}\right) + \frac{0.001}{0.5}\right] = \frac{1}{10} + \frac{1}{10} + 1 + \frac{1}{5} = \frac{14}{10} = 1.4%$
BETA
