Dear Amrit Pal,

Ans:- Hi first of all I must say that you see the answer of the question. because it is a conceptual and if the solution is not correct then I will also have to attempt it in another way.

Let us consider some terms

Mass of the earth=M

Density of the earth=d

Universal gravitational constant=G

radious of the earth=R

mass of the particle is=m

Now you must know that the value of the gravitational acc g′ at a depth h inside the earth is=(1- h/R)g

where g is it's value on the surface of the earth. g=GM/R² and hence  the total force is= 4/3 ∏GmRd(1 - h/R)

F=4/3 ∏Gmd(R-h)..............(1)

Now the value we are getting is due to the superposition of the cut out mass and the remaining mass.

HenceF=Fr+Fc..............(2)

Fr=force due to the remaining and Fc=force due to the cut out

Using eq 1 we can find out the force due to the cut out portion which is Fc=4/3∏ Gmd (R/2 - h)....................(3)

Putting these values of eq 1 and 3 in 2 we get,

Fr=F-Fc

=4/3 ∏Gmd{(R-h)-(R/2 -h)}

=2/3 ∏GmRd

Which is constant i.e h independent

then the acceleration due to this portion is a=Fr/m=2/3 ∏ GRd

So the time taken is t=√{(2R/2)/(2/3 ∏ GRd)}

=√{3/(2GR∏d)}

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Soumyajit Das IIT Kharagpur

Dear Saumyajit,

we know that radius of earth is 6400km or 6400000m.

So, in the mouth of the cavity, the ball is at a potential=mgh.

and, at the centre, it will have the KE=mv²/2.

Equating both.

mgh=mv²/2, where h=6400000m.

by this, we get, v=11200m/s.

now, u may apply the first eq. of motion to find the time taken.