A fighter plane of length $20 m,$ wing span (distance from tip of one wing to the tip of the other wing) of $15 m$ and height $5 m$ is flying towards east over Delhi. Its speed is $240 m s^{- 1} .$ The earth's magnetic field over Delhi is $5 x 10^{- 5} T$ with the declination angle $~ 0^{°}$ and dip of $θ$ such that $\sin θ = \frac{2}{3} .$ If the voltage developed is $V_{B}$ between the lower and upper side of the plane and $V_{W}$ between the tips of the wings then $V_{B}$ and $V_{W}$ are close to

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$V_{B} = 40 m V; V_{W} = 135 m V$ with left side of pilot at higher voltage

$V_{B} = 45 m V; V_{W} = 120 m V$ with right side of pilot at higher voltage

$V_{B} = 40 m V; V_{W} = 135 m V$ with right side of pilot at higher voltage

$V_{B} = 45 m V; V_{W} = 120 m V$ with left side of pitot at higher voltage.

answer is D.

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Detailed Solution

Length of the plane, $l = 20 m$

Wing span, $l^{'} = 15 m$

Height of plane, $h = 5 m$

Velocity of plane $= 240 m s^{- 1}$ towards east

$\sin θ = \frac{2}{3}, B = 5 \times 10^{- 5} T, V_{B} = ?, V_{W} = ?$

$V_{B} =$ Voltage developed between the lower and upper side of the plane

$= ν h B \cos θ = 240 \times 5 \times 5 \times 10^{- 5} \times \frac{\sqrt{5}}{3} = 44.72 \times 10^{- 3} V = 45 m V$

$B_{ν} = B \sin θ$

$= 5 \times 10^{- 5} \times \frac{2}{3} = \frac{1}{3} \times 10^{- 4} T$

Therefore, $V_{W} =$ Voltage developed between tips of the wings

$= B_{ν} l^{'} ν = \frac{1}{3} \times 10^{- 4} \times 15 \times 240 = 1200 \times 10^{- 4} = 120 m V$

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