Question

A Hall effect flow meter is used to measure the volumetric flow through a blood vessel. The system calculated the flow rate to be 100 cm\(^3\cdot\text{s}^{-1}\). Later, calibration revealed that (i) the voltmeter was reading 40% higher than actual, and (ii) the caliper was measuring the vessel diameter 10% smaller than actual. Assuming a uniform flow profile and ignoring viscosity, the actual blood flow is _________ cm\(^3\cdot\text{s}^{-1}\) (rounded off to two decimal places).

Hint:

In flow meters, diameter errors affect flow measurement linearly—not quadratically—when using Hall voltage–based calculations.

Answer 1

Correct Answer: 78.71

The Hall effect flow meter formula is: \[ Q \propto V \times D \] where \(V\) is measured voltage and \(D\) is vessel diameter.
The system used incorrect readings: - Voltmeter reading was 40% higher → measured value \[ V_{\text{measured}} = 1.4\,V_{\text{actual}} \] - Caliper measured diameter 10% smaller → \[ D_{\text{measured}} = 0.9\,D_{\text{actual}} \] The system-reported flow rate is: \[ Q_{\text{reported}} = k \, V_{\text{measured}} \, D_{\text{measured}} \] Actual flow should be: \[ Q_{\text{actual}} = k \, V_{\text{actual}} \, D_{\text{actual}} \] Taking ratio: \[ \frac{Q_{\text{actual}}}{Q_{\text{reported}}} = \frac{V_{\text{actual}}\,D_{\text{actual}}} {1.4V_{\text{actual}} \times 0.9D_{\text{actual}}} = \frac{1}{1.26} \] Thus, \[ Q_{\text{actual}} = \frac{100}{1.26} \approx 79.37\ \text{cm}^3\text{s}^{-1} \] The value lies in the range: \[ \boxed{78.71\ \text{to}\ 79.91} \] Final Answer: 79.37 cm\(^3\cdot\text{s}^{-1}\)