A tunnel is dug along a chord of the earth at a perpendicular distance $\frac{R}{2}$ from the earth’s centre. The wall of the tunnel may be assumed to be frictionless. A particle is released from one end of the tunnel. The pressing force by the particle on the wall and the acceleration of the particle varies with x (distance of the particle from the centre of the earth) according 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

answer is B, C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Net force towards centre of earth $= m g' = \frac{m g x}{R}$
Normal force $N = m g' \sin θ$
Thus pressing force $N = \frac{m g x}{R} \frac{R}{2 x}$
$N = \frac{M g}{2}$ constant and independent of x.
Hence (B)
Question Image
tangential force $F = m a = m g' \cos θ$
$Q = g' \cos θ = \frac{g x}{R} \frac{\sqrt{\frac{R}{4} - x^{2}}}{x}$
$a = \frac{g x}{R} \sqrt{R^{2} - 4 x^{2}}$
Curve is parabolic and at $x = \frac{R}{2}, α = 0$
Hence (C)