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Class 11 NEET Physics Oscillations Questions
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© examsnet.com
Question : 17
Total: 71
A spring of force constant k is cut into lengths of ratio 1 : 2 : 3. They are connected in series and the new force constant is k'. Then they are connected in parallel and force constant is k" . Then k':k" is : -
[NEET 2017]
1 : 9
1 : 11
1 : 14
1 : 16
Validate
Solution:
👈: Video Solution
Length of the spring segments
=
l
6
,
l
3
,
l
2
As we know
K
∝
1
l
so spring constants for spring segments will be
K
1
=
6
K
,
K
2
=
3
K
,
K
3
=
2
K
so in parallel combination
K
′
′
=
K
1
+
K
2
+
K
3
=
11
K
in series combination
K
′
=
K
(As it will become original spring)
so
K
′
:
K
"
=
1
:
11
© examsnet.com
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