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JEE Advanced Model Paper 18 with solutions for online practice
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© examsnet.com
Question : 19
Total: 60
Column-I shows certain situations and Column-II shows information about forces
List-I
List-II
P) Situation
Front view of a car rounding a curve with constantspeed
1)
→
F
1
+
→
F
2
= 0
Q)
Passengers in a rotor not sliding relative to rotor wallcylindrical rotor is rotating with constant angularvelocity about its symmetry axis
2)
→
F
1
+
→
F
2
+
→
F
3
is centripetal force
R)
Particle kept on rough surface of a bowl , no relativemotion of particle in bowl , bowl has constant angularvelocity
3)
→
F
1
is static friction
S)
Car parked on a banker road with no sidewaysskidding
4)
→
F
1
+
→
F
2
+
→
F
3
= 0
P - 2 , 3 ; Q - 1 , 2 , 3 ; R - 2 , 3 ; S - 3 , 4
P - 1 , 2 , 3 ; Q - 2 , 3 ; R - 2 , 3 ; S - 3 , 4
P - 1 , 2 ; Q - 1 , 2 ; R - 1 , 2 , 3 ; S - 4
P - 1 , 2 , 3 ; Q - 1 , 2 ; R - 2 , 3 ; S - 2 , 3
Validate
Solution:
P) speed constant & moving in circle ⇒
→
F
1
+
→
F
2
+
→
F
3
=
m
v
2
R
=
→
f
c
&
→
F
3
+
→
F
2
= 0
(for equilibrium in vertical) &
→
F
1
must be static friction.
Q)
F
1
&
F
2
will cancel each other
F
1
= friction (static) &
→
F
1
+
→
F
2
+
→
F
3
=
→
F
c
(as circular motion)
R)
→
F
1
+
→
F
2
+
→
F
3
=
→
F
c
F
1
= static friction
S) Car at rest ⇒
F
1
+
F
2
+
F
3
= 0
© examsnet.com
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