Physics Circle home Student Login

Home›Chapters›Units, Dimensions and Measurement PYQs›Errors in Measurement›Question 95

JEE Main 2019 (10 Jan, Shift 2) · Errors in Measurement

Errors in Measurement question from JEE Main 2019 (10 Jan, Shift 2)

Q95JEE Main 2019 · 10 JAN · SHIFT 2Medium

Errors in Measurement

The diameter and height of a cylinder are measured by a meter scale to be 12.6 ±\pm 0.1 cm and 34.2 ±\pm 0.1 cm, respectively. What will be the value of its volume in appropriate significant figures?

Show answer and solution

Answer: (d)

  1. V=πd2h4=π(12.6)2(34.2)4≈4264 cm3V = \frac{\pi d^2 h}{4} = \frac{\pi (12.6)^2 (34.2)}{4} \approx 4264\text{ cm}^3
  2. ΔVV=2Δdd+Δhh=2(0.112.6)+0.134.2=0.0159+0.0029=0.0188\frac{\Delta V}{V} = 2\frac{\Delta d}{d} + \frac{\Delta h}{h} = 2\left(\frac{0.1}{12.6}\right) + \frac{0.1}{34.2} = 0.0159 + 0.0029 = 0.0188
  3. ΔV=0.0188×4264≈80 cm3\Delta V = 0.0188 \times 4264 \approx 80\text{ cm}^3
  4. Data have 3 significant figures, so V=4260±80 cm3V = 4260 \pm 80\text{ cm}^3.
  5. So the answer is (d) 4260±80 cm34260 \pm 80\text{ cm}^3.
Report a mistake

More questions on Errors in Measurement