Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance (R)/(2) from the earth's centre, where R is the radius of the earth. The wall of the tunnel is frictionless. If a particle is released

Correct answer: B. 2π (R)/(g) Force along the tunnel F = - ( GMmr R ^ 3 )cosθ F = - (gm)/(R)x ( GM R ^ 2 = g, rcosθ = x) a = - (g)/(R)x ω ^ 2 = (g)/(R) T = 2π (R)/(g)

JEE Main 202126 FebruaryMorning Shift

Exam: JEE Main
Year: 2021
Date: 26 February
Shift: Morning Shift
Subject: Physics
Chapter: Oscillations

JEE Main 2021 · 26 February · Morning Shift · Physics · Oscillations

Assume that a tunnel is dug along a chord of the earth, at a...

Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance

\dfrac{R}{2}

from the earth's centre, where

R

is the radius of the earth. The wall of the tunnel is frictionless. If a particle is released in this tunnel, it will execute a simple harmonic motion with a time period: Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance (R)/(2) from the earth's centre,

Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance (R)/(2) from the earth's centre,

Options

A $\dfrac{1}{2\pi}\sqrt{\dfrac{g}{R}}$
B $2\pi\sqrt{\dfrac{R}{g}}$
C $\dfrac{g}{2\piR}$
DImage is virtual, opposite side of convex mirror.

Correct answer

B. $2\pi\sqrt{\dfrac{R}{g}}$

Step-by-step solution

Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance (R)/(2) from the earth's centre,

Force along the tunnel

$F = - \left(\dfrac{GMmr}{{R}^{3}}\right)cos\theta$

$F = - \dfrac{gm}{R}x \left(\dfrac{GM}{{R}^{2}} = g, rcos\theta = x\right)$

$a = - \dfrac{g}{R}x$

${\omega}^{2} = \dfrac{g}{R}$

$T = 2\pi\sqrt{\dfrac{R}{g}}$

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