question_answer

A fighter plane of length 20m, wing span (distance from tip of one wing to the tip of the other wing) of 15 m 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 ONLINE 10-04-2016]

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 :

[c]

\[{{V}_{B}}={{B}_{4}}\left( 5 \right)\left( 240 \right)\]

\[{{B}_{H}}=B\cos \theta \]

\[{{B}_{H}}\frac{5\sqrt{5}\times {{10}^{-5}}}{3}\]

\[{{B}_{v}}=\frac{10}{3}\times {{10}^{-5}}T\]

\[{{V}_{ & B}}=\frac{5\sqrt{5}}{3}\times {{10}^{-5}}\times 5\times 240\]

\[{{V}_{B}}=44.6mV=45mV\]

\[{{V}_{w}}=Bv\ell V\]

\[={{10}^{-4}}\times 1200\]

\[{{V}_{\omega }}=120mV\]

(left side at fighter voltage)