Current Electricity and Electrical InstrumentHard
Question
A silver wire of length 10 metre and cross-sectional area 10-8 m2 is suspended vertically and a weight of 10 N is attached to it. Young′s modulus of silver and its resistivity are 7 × 1010 N/m2 and 1.59 × 10-8 Ω-m respectively. The increase in its resistance is equal to (keeping volume constant)
Options
A.0.0455 Ω
B.0.455 Ω
C.0.91 Ω
D.0.091 Ω
Solution
Y = 
7 × 1010 =
⇒ x =
= 
m
By volume conservation A1 l1 = A2 l2
10 × 10-8 =
A
ᐃR = R − R0 =
ᐃR =
− 
ᐃR =
= 15.9 ×
= 0.454 Ω
7 × 1010 =
⇒ x =
= mBy volume conservation A1 l1 = A2 l2
10 × 10-8 =
AᐃR = R − R0 =
ᐃR =
− ᐃR =
= 15.9 ×
= 0.454 ΩCreate a free account to view solution
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