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Show that the percentage error in the me...

Show that the percentage error in the measurement of resistance by a metre bridge is minimum when the null point is near about the centre of the wire.

A

20 cm

B

50 cm

C

80 cm

D

100 cm

Text Solution

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The correct Answer is:
To show that the percentage error in the measurement of resistance by a metre bridge is minimum when the null point is near about the center of the wire, we can follow these steps: ### Step 1: Define the variables Let: - \( R \) = the resistance to be measured - \( L \) = the length of the wire from one end to the null point - The total length of the wire = 100 cm ### Step 2: Establish the relationship using the metre bridge principle According to the principle of the metre bridge, we have: \[ \frac{x}{R} = \frac{L}{100 - L} \] Where \( x \) is the resistance in the segment of the wire corresponding to the length \( L \). ### Step 3: Rearranging the equation From the above equation, we can express \( x \) as: \[ x = R \cdot \frac{L}{100 - L} \] ### Step 4: Calculate the differential error in \( x \) To find the error in the measurement of resistance \( x \), we differentiate \( x \): \[ \Delta x = R \cdot \left( \frac{\Delta L (100 - L) - L \Delta L}{(100 - L)^2} \right) \] This simplifies to: \[ \Delta x = R \cdot \frac{\Delta L \cdot (100 - 2L)}{(100 - L)^2} \] ### Step 5: Calculate the relative error in \( x \) The relative error in \( x \) is given by: \[ \frac{\Delta x}{x} = \frac{\Delta L \cdot (100 - 2L)}{L(100 - L)} \] ### Step 6: Express the percentage error To express this as a percentage error, we multiply by 100: \[ \text{Percentage Error} = \frac{\Delta L \cdot (100 - 2L)}{L(100 - L)} \cdot 100 \] ### Step 7: Minimize the percentage error To minimize the percentage error, we need to maximize the term \( L(100 - L) \). This is a quadratic function that achieves its maximum value when: \[ L = 100 - L \implies 2L = 100 \implies L = 50 \] ### Conclusion Thus, the percentage error in the measurement of resistance by a metre bridge is minimum when the null point is at the center of the wire, i.e., \( L = 50 \) cm.
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