Verified Solution • 2025–26 Session moderate

QNEET-2022-Ph-527: NEET 2022 Physics - Class 12 NEET-2022-Ph-527 Solution: The sliding contact C is at one fourth of the length of the…

The sliding contact C is at one fourth of the length of the potentiometer wire (AB) from A as shown in the circuit diagram. If the resistance of the wire AB is R0R_0 then the potential drop (V) across the resistor R is 2V0R2R0+3R\frac{2V_0R}{2R_0+3R} 4V0R3R0+16R\frac{4V_0R}{3R_0+16R} 4V0R3R0+R\frac{4V_0R}{3R_0+R} 2V0R4R0+R\frac{2V_0R}{4R_0+R}

Multiple Choice Options

  • A 2V0R2R0+3R{{2{V_0}R} \over {2{R_0} + 3R}} 2 R 0 _ + 3 R 2 V 0 _ R _
  • B 4V0R3R0+16R{{4{V_0}R} \over {3{R_0} + 16R}} 3 R 0 _ + 16 R 4 V 0 _ R _
  • C 4V0R3R0+R{{4{V_0}R} \over {3{R_0} + R}} 3 R 0 _ + R 4 V 0 _ R _
  • D 2V0R4R0+R{{2{V_0}R} \over {4{R_0} + R}} 4 R 0 _ + R 2 V 0 _ R _

Step-by-Step Verified Solution

Equivalent resistance across point AC R A C = R 0 4 _ R R 0 4 + R = R R 0 R 0 + 4 R From voltage divider rule V A C = R A C R A C + R C B V 0 = R R 0 R 0 + 4 R V 0 R R 0 R 0 + 4 R + 3 R 0 4 = 4 R R 0 V 0 4 R R 0 + 3 R 0 2 + 12 R R 0 = 4 R V 0 3 R 0 + 16 R

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