question_answer

A fighter plane of length 20m, wing span (distance from tip of one wing to the tip of the other wing) of 15m and height 5m is flying towards east over Delhi. Its speed is \[240\,m{{s}^{-1}}\]The earth's magnetic field over Delhi is \[5\times {{10}^{-5}}\]T with the declination angle \[\sim {{0}^{o}}\] and dip of \[\theta \]such that sin\[\theta =\frac{2}{3}\].If the voltage developed is \[{{V}_{B}}\] between the lower and upper side of the plane and \[{{V}_{w}}\]. between the tips the wings then\[{{V}_{B}}\] and VW are closed to:   JEE Main Online Paper (Held On 10 April 2016)

A) \[{{V}_{B}}=40mV,{{V}_{w}}=135mV\] with left side of pilot at higher voltage.

B) \[{{V}_{B}}=40mV,{{V}_{w}}=135mV\] with right side of pilot at high voltage.

C) \[{{V}_{B}}=45mV,{{V}_{w}}=120mV\] with left side of pilot at higher voltage.

D) \[{{V}_{B}}=45mV,{{V}_{w}}=120mV\]with right side of  pilot at higher voltage.

Correct Answer: C

Solution :

                                                        \[{{V}_{B}}={{B}_{4}}\left( 5 \right)\left( 240 \right)\]              \[{{B}_{H}}=B\cos \theta \]              \[{{B}_{H}}\frac{5\sqrt{5}\times {{10}^{-5}}}{3}\]              \[{{B}_{v}}=\frac{10}{3}\times {{10}^{-5}}T\]                       \[{{V}_{ & B}}=\frac{5\sqrt{5}}{3}\times {{10}^{-5}}\times 5\times 240\]                \[{{V}_{B}}=44.6mV=45mV\]              \[{{V}_{w}}=Bv\ell V\]              \[={{10}^{-4}}\times 1200\]              \[{{V}_{\omega }}=120mV\]                        (left side at fighter voltage)