Q32. The resistance of a potentiometer wire of length 4 m is 10Ω. A resistance box and 2 V accumulat

Q: What is the correct answer for CUET UG Physics Q32?

The correct answer is d. Given:Length of potentiometer wire (L) = 4 mResistance of potentiometer wire (R_pot) = 10 ΩEMF of accumulator (E) = 2 VDesired potential gradient (k) = 1 µV/mmFirst, convert the potential gradient to V/m:k = 1 µV/mm = 1 × 10⁻⁶ V / 10⁻³ m = 1 × 10⁻³ V/m.Let R_box be the resistance introduced in the resistance box.The total resistance in the primary circuit (R_total) = R_pot + R_box = 10 Ω + R_box.T

Q: Q32. The resistance of a potentiometer wire of length 4 m is 10Ω. A resistance box and 2 V accumulator are connected in series with it. What resistance should be introduced in the box to have a…

The correct answer is d. Given:Length of potentiometer wire (L) = 4 mResistance of potentiometer wire (R_pot) = 10 ΩEMF of accumulator (E) = 2 VDesired potential gradient (k) = 1 µV/mmFirst, convert the potential gradient to V/m:k = 1 µV/mm = 1 × 10⁻⁶ V / 10⁻³ m = 1 × 10⁻³ V/m.Let R_box be the resistance introduced in the resistance box.The total resistance in the primary circuit (R_total) = R_pot + R_box = 10 Ω + R_box.T

Q: What are the options for this CUET UG Physics question?

A. 799.9Ω · B. 790Ω · C. 9.995Ω · D. 4990Ω

Question

Q32. The resistance of a potentiometer wire of length 4 m is 10Ω. A resistance box and 2 V accumulator are connected in series with it. What resistance should be introduced in the box to have a potential drop of 1µV/mm of the potentiometer wire?
(a) 799.9Ω
(b) 790Ω
(c) 9.995Ω
(d) 4990Ω

FirstInTest · Accepted Answer

Given:Length of potentiometer wire (L) = 4 mResistance of potentiometer wire (R_pot) = 10 ΩEMF of accumulator (E) = 2 VDesired potential gradient (k) = 1 µV/mmFirst, convert the potential gradient to V/m:k = 1 µV/mm = 1 × 10⁻⁶ V / 10⁻³ m = 1 × 10⁻³ V/m.Let R_box be the resistance introduced in the resistance box.The total resistance in the primary circuit (R_total) = R_pot + R_box = 10 Ω + R_box.The current (I) flowing through the potentiometer wire is I = E / R_total = 2 V / (10 + R_box) Ω.The potential drop across the potentiometer wire (V_pot) is V_pot = I × R_pot.V_pot = [2 / (10 + R_box)] × 10 = 20 / (10 + R_box) V.The potential gradient (k) is also given by k = V_pot / L.k = [20 / (10 + R_box)] / 4 m = 5 / (10 + R_box) V/m.Now, equate the desired potential gradient with the calculated one:1 × 10⁻³ V/m = 5 / (10 + R_box) V/m10 + R_box = 5 / (1 × 10⁻³)10 + R_box = 5000R_box = 5000 - 10R_box = 4990 Ω.

Answer

799.9Ω

Answer

790Ω

Answer

9.995Ω

Answer

4990Ω

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