Problem 87 - A Complete Resource Book in Physics for JEE Main

Question

\(A B\) is a wire of uniform resistance. The galvanometer \(G\) shows zero current when the length \(A C=20 \mathrm{~cm}\) and \(C B=80 \mathrm{~cm}\). The resistance \(R\) is equal to (A) \(2 \Omega\) (B) \(8 \Omega\) (C) \(20 \Omega\) (D) \(40 \Omega\)

Step-by-Step Solution

Answer

The resistance R of the entire wire AB can be determined using the ratio of lengths and resistances. From the given information, we have the length ratio \(\frac{L_{AC}}{L_{CB}} = \frac{20}{80}\). As the wire has a uniform resistance, their resistances have the same ratio \(\frac{R_{AC}}{R_{CB}} = \frac{20}{80}\). Thus, the total resistance R can be expressed as \(R = R_{AC} + R_{CB}\), where \(R_{AC} = \frac{R}{5}\) and \(R_{CB} = \frac{4R}{5}\). Solving for R, we find that it is an arbitrary multiple of 5, and the answer that matches this criterion is choice (C) 20Ω.

Step 1: Understand the problem and what we are given

We are given a wire of uniform resistance, AB. We are also given the lengths of AC and CB when the galvanometer G shows zero current (AC=20cm and CB=80cm). We need to find the resistance R of the entire wire AB.

Step 2: Write the formula for resistance

The formula for resistance of a uniform wire (R) is as follows: \(R = \rho \frac{L}{A}\), where \(\rho\) is resistivity, L is the length, and A is the cross-sectional area. Since the wire has uniform resistance, the lengths of AC and CB are in proportion to their resistance: \(R_{AC} = k\cdot L_{AC}\) and \(R_{CB} = k\cdot L_{CB}\), where k is a constant of proportionality. The total resistance R can be expressed as: \(R = R_{AC} + R_{CB}\)

Step 3: Use the given lengths to find the resistance ratio

The length of AC is 20cm, and the length of CB is 80cm. Thus, we have the length ratio: \(\frac{L_{AC}}{L_{CB}} = \frac{20}{80}\)

Step 4: Find the resistance ratio

Since the lengths are proportional to their resistance, we can find the resistance ratio: \(\frac{R_{AC}}{R_{CB}} = \frac{L_{AC}}{L_{CB}} = \frac{20}{80}\)

Step 5: Solve for R using the given ratio

Now, we can use the ratio to solve for the total resistance R using the formula we derived in step 2: \(R = R_{AC} + R_{CB}\) We know that \(R_{AC} = \frac{R}{5}\) and \(R_{CB} = \frac{4R}{5}\) from the resistance ratio. Substitute these expressions into our equation: \(R = \frac{R}{5} + \frac{4R}{5}\) Combine the terms: \(R = \frac{5R}{5}\) Divide by 5 on both sides: \(R = R\) Therefore, R is an arbitrary multiple of 5.

Step 6: Match the answer to the given choices

The choices provided are: (A) 2Ω (B) 8Ω (C) 20Ω (D) 40Ω From our solution in Step 5, we know that R is an arbitrary multiple of 5. The only multiple of 5 in the choices is (C) 20Ω. Hence, the correct answer is (C) 20Ω.