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Units and Measurements question

2019 · 10 Jan · Shift 2 · Q69
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Units and Measurements question

2019 · 10 Jan · Shift 2 · Q69

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1
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 ?
  1. A
    4264.4 ±\pm± 81.0 cm3
  2. B
    4264 ±\pm± 81 cm3
  3. C
    4300 ±\pm± 80 cm3
  4. D
    4260 ±\pm± 80 cm3
View written solutionFree

Correct answer: D

  1. Given data
  • Diameter of cylinder: d=12.6±0.1 cmd = 12.6 \pm 0.1\,\text{cm}d=12.6±0.1cm
  • Height of cylinder: h=34.2±0.1 cmh = 34.2 \pm 0.1\,\text{cm}h=34.2±0.1cm

Radius is

r=d2=12.62=6.3 cmr = \frac{d}{2} = \frac{12.6}{2} = 6.3\,\text{cm}r=2d​=212.6​=6.3cm

with error

Δr=0.12=0.05 cm\Delta r = \frac{0.1}{2} = 0.05\,\text{cm}Δr=20.1​=0.05cm
  1. Formula for volume

Volume of a cylinder:

V=πr2h=πd2h4V = \pi r^2 h = \frac{\pi d^2 h}{4}V=πr2h=4πd2h​

Using d=12.6d=12.6d=12.6 cm and h=34.2h=34.2h=34.2 cm,

V=π4(12.6)2(34.2)V = \frac{\pi}{4}(12.6)^2(34.2)V=4π​(12.6)2(34.2)

Now,

(12.6)2=158.76(12.6)^2 = 158.76(12.6)2=158.76 158.76×34.2=5429.592158.76 \times 34.2 = 5429.592158.76×34.2=5429.592 V=π4×5429.592≈4264.39 cm3V = \frac{\pi}{4} \times 5429.592 \approx 4264.39\,\text{cm}^3V=4π​×5429.592≈4264.39cm3

So the calculated volume is

V≈4264.4 cm3V \approx 4264.4\,\text{cm}^3V≈4264.4cm3
  1. Error calculation

Since

V∝d2hV \propto d^2 hV∝d2h

maximum fractional error is

ΔVV=2Δdd+Δhh\frac{\Delta V}{V} = 2\frac{\Delta d}{d} + \frac{\Delta h}{h}VΔV​=2dΔd​+hΔh​

Substitute values:

ΔVV=2(0.112.6)+0.134.2\frac{\Delta V}{V} = 2\left(\frac{0.1}{12.6}\right) + \frac{0.1}{34.2}VΔV​=2(12.60.1​)+34.20.1​ =2(0.00794)+0.00292= 2(0.00794) + 0.00292=2(0.00794)+0.00292 =0.01588+0.00292=0.01880= 0.01588 + 0.00292 = 0.01880=0.01588+0.00292=0.01880

Hence absolute error:

ΔV=V×0.01880\Delta V = V \times 0.01880ΔV=V×0.01880 ΔV≈4264.4×0.01880≈80.16 cm3\Delta V \approx 4264.4 \times 0.01880 \approx 80.16\,\text{cm}^3ΔV≈4264.4×0.01880≈80.16cm3

So,

ΔV≈80 cm3\Delta V \approx 80\,\text{cm}^3ΔV≈80cm3
  1. Reporting in appropriate significant figures

When reporting a measured quantity with error:

  • The error is usually given to one significant figure (or at most two if the first digit is 1 or 2).
  • The value is rounded to the same place as the error.

Here,

V=4264.4 cm3,ΔV≈80 cm3V = 4264.4\,\text{cm}^3, \qquad \Delta V \approx 80\,\text{cm}^3V=4264.4cm3,ΔV≈80cm3

Since the error is in the tens place, the volume should also be rounded to the tens place:

V≈4260 cm3V \approx 4260\,\text{cm}^3V≈4260cm3

Therefore,

V=4260±80 cm3V = 4260 \pm 80\,\text{cm}^3V=4260±80cm3
  1. Option check
  • A: 4264.4±81.04264.4 \pm 81.04264.4±81.0 — too many significant figures in both value and error.
  • B: 4264±814264 \pm 814264±81 — value not rounded consistently with error.
  • C: 4300±804300 \pm 804300±80 — over-rounded central value.
  • D: 4260±804260 \pm 804260±80 — correct.

Hence the correct option is D.

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