Question

A DC potentiometer has a potential gradient of 20 mV/cm. The balancing lengths for a standard cell and an unknown voltage are 75 cm and 120 cm respectively. If the standard cell voltage is 1.50 V, then the unknown voltage is

• A. 1.80 V
• B. 2.00 V
• C. 2.40 V
• D. 3.20 V

Hint:

Notice that the given potential gradient parameter (20 mV/cm) is extra information not required to solve the problem! You can find the answer directly using the ratio of the balancing lengths: \(1.5\text{ V} \times \frac{120\text{ cm}}{75\text{ cm}} = 2.4\text{ V}\).

Answer 1

The Correct Option is C

Concept: A potentiometer operates on the principle that the voltage drop across a uniform wire segment is directly proportional to its physical balancing length, provided the current flowing through it remains constant. The potential gradient (\(x\)) is defined as the voltage drop per unit length: \[ V = x \cdot l \] where \(V\) is the balanced EMF and \(l\) is the corresponding balancing length. When comparing two different voltages using the same potentiometer wire configuration, their values are directly proportional to their respective balancing lengths: \[ \frac{V_{\text{unknown}}}{V_{\text{standard}}} = \frac{l_{\text{unknown}}}{l_{\text{standard}}} \]

Step 1: Extract the given parameters.
• Balancing length for the standard cell (\(l_1\)) = 75 cm
• Balancing length for the unknown cell (\(l_2\)) = 120 cm
• Voltage of the standard cell (\(V_1\)) = 1.50 V

Step 2: Calculate the unknown voltage using the direct proportionality ratio. Using the ratio relationship: \[ V_2 = V_1 \cdot \left(\frac{l_2}{l_1}\right) \] Substitute the given numerical values: \[ V_2 = 1.50 \cdot \left(\frac{120}{75}\right) \] Let's simplify the fraction step-by-step by dividing both numbers by 15: \[ \frac{120}{15} = 8, \quad \frac{75}{15} = 5 \quad \Rightarrow \quad \frac{120}{75} = \frac{8}{5} = 1.6 \] Now evaluate the multiplication: \[ V_2 = 1.50 \times 1.6 = 2.40\text{ V} \] Hence, the value of the unknown voltage source is 2.40 V, matching option (C).