Four uniform wires having equal length 1 m each and resistance in the ratio 1:2:3:4 are connected together to form a square CDEF in that order i.e. CD, DE, EF, FC and resistance of wire CD is ‘r’. This loop surrounds a cylindrical region of circular cross section in which uniform magnetic field whose net flux depends on time according to relation $ϕ = 40 t$ weber, where ‘t’ is in second. The magnetic field is perpendicular to the plane of square loop. Centre of circular cross section and square loop coincides.( $V_{1} & V_{2}$ are ideal voltmeters)

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Reading of voltmeter $V_{1}$ is 8 Volt

The potential difference between point C and F is 16 Volt

Reading of voltmeters $V_{1}$ and $V_{2}$ are equal

Reading of voltmeter $V_{2}$ is 16 Volt

answer is B, D.

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Detailed Solution

$\begin{array}{l} ε = \frac{- d ϕ}{d t} = 40 V \\ i = \frac{ε}{10 r} = \frac{40}{10 r} = \frac{4}{r} \\ V_{1} = 2 r \times i = 8 V \\ V_{2} = 8 r \times i = 32 V \\ V_{C F} = 4 r \times i = 16 V \\ V_{1} and V_{2} are different because electric field in the loop is non-conservative. \end{array}$