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Byju's Answer
Standard XII
Physics
Young's Modulus of Elasticity
A force F dou...
Question
A force F doubles the length of wire of cross-section a. The Young modulus of wire is
A
F
a
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B
F
3
a
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C
F
2
a
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D
F
4
a
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Solution
The correct option is
D
F
2
a
As youngsters modulus is given as
,
Y
=
f
/
a
.
L
/
Δ
L
From above we can conclude that
Y
′
=
2
Y
That is youngs modulus becomes
2
times to its original value
.
Hence,
option
(
C
)
is correct answer.
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Similar questions
Q.
To determine the Youngs modulus of a wire,the formula is
Y
=
F
A
×
Δ
L
L
; where L = length, A = area of cross-section of the wire,
Δ
L = change in length of the wire when stretched with a force F. The conversion factor to
change it from CGS to MKS system is :
Q.
A metal wire of length L, area of cross section A and Young's modulus Y is stretched by a variable force F such that F is always slightly greater than the elastic forces of resistance in the wire. Then the elongation of the wire is
l
.
Q.
A metal wife of length L area of cross section A and Young modulus Y is streched by a variable force F such that F is always slightly greater than the elastic forces of resistance in the wife when the elongation of the wire is
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then correct option(s)
Q.
A metal wire length
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, cross sectional area
A
and Young's modulus
Y
is stretched by a variable force
F
.
F
is varying in such a way that
F
is always slightly greater than the elastic forces or resistance in the wire. When the elongation in the wire is
l
, up to this instant
Q.
Two wires are made of the same material and have the same volume. The first wire has cross-sectional area
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