A conducting loop is shown figure, it is made of seven identical wires each of length a ,Resistivity ‘ $ρ$ ’ , area of cross-section $S_{0}$ . The frame lies in a vertical plane. XX ' is an imaginary boundary separating space in two parts. Right to XX ' , a uniform gravitational field g exists (figure) whereas no gravitational field is present left of XX ' . Instantaneously frame is accelerating in leftwards with acceleration $g m / s^{2}$ . e and m is electronic charge and mass of electron respectively. Ignore inductive and capacitive properties of loop. Instantaneously

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Current in branch RS is $\frac{m g S_{0}}{15 ρ e}$

Current in branch RS is $\frac{4 m g S_{0}}{15 ρ e}$

The magnitude of potential difference between points P and Q is $\frac{7 m g a}{15 e}$

The magnitude of potential difference between points P and Q is $\frac{11}{15} \frac{m g a}{e}$

answer is B, C.

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Detailed Solution

$E e = m a \Rightarrow E = \frac{m a}{e} = \frac{m g}{e} i n (S P) E e = m g \Rightarrow E = \frac{m g}{e} i n (P A) E_{N e t} = \frac{m g a}{e}$

$R_{N e t} = \frac{3 R}{4} + 3 R = \frac{15 R}{4} = \frac{15 P a}{4 S} i = \frac{m g a}{e (\frac{15 R}{4})} = \frac{4 m g a S}{15 e P a} = \frac{4 m g S}{15 e P} ∴ V_{P Q} = \frac{m g a}{e} - (2 R) \frac{4 m g a}{15 Re} = \frac{7 m g a}{15 e} i_{R S} = \frac{i}{4} = \frac{m g S}{15 e P}$