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Question

A train A runs from east to west and another train B of the same mass and speed runs from west to east along the equator. A presses the track with a force F1 and B with F2. Then:

A
F1=F2
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B
F1<F2
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C
F1>F2
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D
F1F2
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Solution

The correct option is C F1>F2
Figure shows the train moving along earth's equator, comparing it with motion on a circular path with radius of curvature of circular path equal to earth's radius (R).


At every point on the equator, applying equation of dynamics in radial direction,
mgN=mv2nR
[vn=vt+vtp]
Here,vt=velocity of train
vtp= velocity of train's position, due to rotation of earth (Earth rotates in anticlockwise direction i.e from West to East)
N=mgmv2nR ...(i)
[m and R are constant]

Case 1 :
Train running from east to west


vn=vtvtp ...(ii)

Case 2 :
When train runs from west to east


vn=vt+vtp ...(iii)
From Eq.(i), we can see that higher value of vn will yield a lower normal reaction (N) from earth on train.
From Eq.(ii) and (iii),
vn(case-2)>vn(case-1)
N(case-1)>N(case-2)
From Newton's 3rd law, train A presses the track with larger force than that train B.
F1>F2

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