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Test Index
JEE Main 29-Jan-2024 Shift 2 Solved Paper
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Section:
Physics
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© examsnet.com
Question : 51
Total: 90
Two metallic wires
P
and
Q
have same volume and are made up of same material. If their area of cross sections are in the ratio
4
:
1
and force
F
1
is applied to
P
, an extension of
∆
l
is produced. The force which is required to produce same extension in
Q
is
F
2
.
The value of
F
1
F
2
is____
[29-Jan-2024 Shift 2]
Your Answer:
Validate
Solution:
👈: Video Solution
Y
=
Stress
Strain
=
F
∕
A
∆
ℓ
∕
ℓ
=
F
ℓ
A
∆
ℓ
∆
ℓ
=
F
ℓ
AY
V
=
A
ℓ
⇒
ℓ
=
V
A
∆
ℓ
=
FV
A
2
Y
Y
&
V
is same for both the wires
∆
ℓ
∝
F
A
2
∆
ℓ
1
∆
ℓ
2
=
F
1
A
1
2
×
A
2
2
F
2
∆
ℓ
1
=
∆
ℓ
2
F
1
A
2
2
=
F
2
A
1
2
F
1
F
2
=
A
1
2
A
2
2
=
(
4
1
)
2
=
16
© examsnet.com
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