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JEE Advanced 2020 Paper 2
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© examsnet.com
Question : 8
Total: 54
A student skates up a ramp that makes an angle
30
°
with the horizontal. He/she starts (as shown in the figure) at the bottom of the ramp with speed
v
0
and wants to turn around over a semicircular path xyz of radius R during which he/she reaches a maximum height h (at point y) from the ground as shown in the figure. Assume that the energy loss is negligible and the force required for this turn at the highest point is provided by his/her weight only. Then (g is the acceleration due to gravity)
[JEE Adv 2020 P2]
v
0
2
−
2
g
h
=
1
2
g
R
v
0
2
−
2
g
h
=
√
3
2
g
R
the centripetal force required at points x and z is zero
the centripetal force required is maximum at points x and z
Validate
Solution:
At point y
m
g
‌
sing
‌
30
°
=
m
v
2
R
⇒
v
2
=
g
R
2
Using conservation of Mechanical Energy
1
2
m
v
0
2
=
m
g
h
+
1
2
m
v
2
v
0
2
−
2
g
h
=
g
R
2
© examsnet.com
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