Verified Solution • 2025–26 Session moderate

QJEEM21Phf83e: JEE Main 2021 Physics - Class 11 JEEM21Phf83e Solution: 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 (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 :

Multiple Choice Options

  • A 2πRg2\pi \sqrt {{R \over g}}
  • B g2πR{g \over {2\pi R}}
  • C 2πRg{{2\pi R} \over g}
  • D 12πgR{1 \over {2\pi }}\sqrt {{g \over R}}

Step-by-Step Verified Solution

JEE Main 2021 (Online) 26th February Morning Shift Physics - Simple Harmonic Motion Question 92 English Explanation

Value of g on the particle of mass m,

g=GMdR3g = {{GMd} \over {{R^3}}}

Force acting on the particle towards the center of the earth,

F = mg

Force along the tunnel = F cos θ\theta = F1

= mg cos θ\theta

= m . GMdR3{{GMd} \over {{R^3}}} (xd)\left( {{x \over d}} \right)

= GMmR3{{GMm} \over {{R^3}}} . x

= gsmR. x{{{g_s}m} \over R}.\,x [as gs = GMR2{{GM} \over {{R^2}}} at earth surface]

∴\therefore acceleration along the tunnel

α\alpha = F1m{{{F_1}} \over m}

= gsR.x{{{g_s}} \over R}.x

Time period = 2πω=2πxα{{2\pi {} } \over \omega } = 2\pi {\sqrt {{x \over \alpha }} } [as ω2=αx{\omega ^2} = {\alpha \over x} ]

=2πRgs= 2\pi {\sqrt {{R \over {{g_s}}}} }

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