AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

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

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

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

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

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

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

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

Watch 3-min video & get full concept clarity

courses

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!