UPSEE Solved Model Paper 2

The radius of the circle along which the trolley moves is
r = 40 cm = 0.4 m
When the table is rotated, the tension in the spring is equa to the centripetal force, i.e.
F =

mv2

r

= mrω2 = 2×0.4×(10)2 = 80 N
The extension in the spring is x = 40 - 35 = 5 cm = 0.05 m
∴ Force constant k =

F

x

=

80

0.05

= 1.6×103Nm−1
Hence the correc choice is (2).