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JEE Mains Model Paper 24
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© examsnet.com
Question : 3 of 90
Marks:
+1
,
-0
Two springs of force constant 100 N/m and 150 N/m are in series as shown. The blockis pulled by a distance of 2.5 cm to the right from equilibrium position. What is theratio of work done by the spring at left to the work done by the spring at right?
3
2
2
3
0.2
0.5
Validate
Solution:
k
e
q
=
100
×
150
250
= 60 N/m
F =
k
e
q
x = 60 ×
2.5
100
=
3
2
N
For left spring
x
1
=
3
2
(
100
)
For right spring
x
2
=
3
2
(
150
)
So
1
2
(
100
)
(
3
2
)
2
(
1
100
)
2
1
2
(
150
)
(
3
2
)
2
(
1
150
)
2
=
150
100
= 3/2
© examsnet.com
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