Verified Solution • 2025–26 Session moderate

QNEET-2024-Ph-55: NEET 2024 Physics - Class 12 NEET-2024-Ph-55 Solution: Given below are two statements: one is labelled as Assertion A…

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: The potential (V) at any axial point at 2 m distance (r) from the centre of the dipole of dipole moment vector P of magnitude 4×10−6 Cm4 \times 10^{-6}\ \text{Cm} is ±9×103 V\pm 9 \times 10^3\ \text{V}. (Take 14πϵ0=9×109 SI units\frac{1}{4\pi\epsilon_0} = 9 \times 10^9\ \text{SI units}) Reason R: V=±2P4πϵ0r2V = \pm \frac{2P}{4\pi\epsilon_0 r^2} where r is the distance of any axial point situated at 2 m from the centre of the dipole. In the light of the above statements choose the correct answer from the options given below

Multiple Choice Options

  • A Both A and R are true and R is the correct explanation of A.
  • B Both A and R are true and R is NOT the correct explanation of A.
  • C A is true but R is false.
  • D A is false but R is true.

Step-by-Step Verified Solution

The potential V at any point, at distance r from centre of dipole = K P cos _ _ r 2 At axial point where _ = 0 _ , V = K P r 2 = 9 _ 10 9 _ 4 _ 10 _ 6 2 2 = 9 _ 10 3   V At axial point where _ = 180 _ , V = _ K P r 2 = _ 9 _ 10 3   V

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