Question:

If in a capillary tube the rate of flow of a liquid is measured by collecting 100 ml of liquid in 100 sec in a measuring flask. If the least count of the volume of the measuring flask be 1 ml and the least count of the stop watch used for measuring the time be 0.01 sec, then the uncertainty in the rate of flow of the liquid will be:

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Use relative error formulas for calculations involving division: \[ \frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta t}{t} \]
Updated On: Mar 26, 2025
  • \( 1.01 \times 10^{-2} \)
  • \( 1.01 \)
  • \( 10^{-6} \)
  • \( 10^{-4} \)
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The Correct Option is A

Solution and Explanation

The rate of flow is:
\[ R = \frac{V}{t} \] Given:
\( V = 100 \pm 1 \) ml \( t = 100 \pm 0.01 \) sec Using error propagation:
\[ \frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta t}{t} \] \[ \frac{\Delta R}{R} = \frac{1}{100} + \frac{0.01}{100} \] \[ \frac{\Delta R}{R} = 0.01 + 0.0001 = 0.0101 \] \[ \Delta R = 0.0101 \times 1 = 1.01 \times 10^{-2} \]
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