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Question

Three identical cars A,B and C are moving at the same speed on three bridges. The car A goes on a plane bridge, B on a bridge convex upward and C goes on a bridge concave upward. Let FA, FB and FC be the normal force exerted by the cars on the bridges when they are at the middle of bridges then :

A
FA is maximum of the three forces
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B
FB is maximum of the three forces
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C
FC is maximum of the three forces
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D
FA=FB=FC
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Solution

The correct option is C FC is maximum of the three forces
FA, FB and FC are the magnitude
of the normal contact forces with which car and bridge exert forces on
each other.
for car 'A'
FA=mg ......... (I)
for car 'B'
mv2r=mgFB
FB=mgmv2r ......... (II)
for car 'C'
mv2r=Fcmg
Fc=mg+mv2r ......... (III)
from equations (III), (II) and (I), we see that
Fc>FA>Fb
156921_136708_ans_4c73ae17f1a84be896824afa4485a1b7.png

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