Question

A smooth tunnel is dug along the chord of earth at a perpendicular distance R/2 from the centre of earth the pressing force by the particle on the wall and the acceleration of the particle varies with X
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Answer 1

Correct question :

A smooth tunnel is dug along the chord of the earth at a perpendicular distance 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 vary with x.

Solution :

Let, the particle of mass m be at a point P.

Where, The point P lies at distance x from the centre of the earth.

Now,

$\implies \bf x^2 \ = \ b^2 \ + \ \dfrac {R^2}{4}$

The mass of the eart which have the radius x,

$\implies \bf M' \ = \ \rho \bigg( \dfrac {4 \pi}{3} \ \pi x^3 \bigg) \ = \ \dfrac {M}{\dfrac {4 \pi}{3} \times R^3} \times \dfrac {4 \pi}{3} \ x^3$

$\implies \bf M' \ = \ M \ \dfrac {X^3}{R^3}$

Now,

Net force experienced by the particle = ?

$\implies \bf F \ = \ \dfrac {GM'm}{x^2} \ = \ \dfrac {GMm}{R^3} \ x$

Here, The passing force,

$\implies \bf F_{\perp} \ = \ FCos \theta$

$\implies \bf \dfrac {GMmx}{R^3} \times \dfrac {\dfrac {R}{2}}{x}$

$\implies \bf \dfrac {GMm}{2R^2} \ is \ constant \ here$

Now,

Let ‘a’ be acceleration of the particle.

$\implies \bf ma \ = \ F_11 \ = \ FSin \theta$

Therefore,

$\implies \bf ma \ = \ \dfrac {GMm}{R^3} \ x \times \dfrac {b}{x}$

$\implies \bf a \ = \ \dfrac {GMm}{R^3} \ b$

$\implies \bf \dfrac {GM}{R^3} \ {\sqrt x^2 \ - \ \dfrac {R^4}{4}}$