changed, then iron draws a current of 2 A, what is the new voltage applied? Ans. 110V
10. A copper wire has a diameter 0.5 mm and resistivity 1.6 x 10-8
Ω τη.
(a) What will be the length of this wire to make the resistance 129?
(b) How much will be the resistance of another copper wire of same length but half the
diameter?
[Ans.(a) 147.3 m (b) 482)
step by step please solve
Answers
Answer:
8. A copper wire has a diameter of 0.5 mm and resistivity of 1.6 x 10-8 2 m. ?
ANSWERS
Solution
Given that, the diameter of the cooper wire is 0.5 mm and it's resistivity is 1.6 × 10^-8 Ωm.
{ Diameter = 0.5 mm
= 0.5/1000 m
= 5 × 10^-4 m }
Now,
2(Radius) = Diameter
Radius = Diameter/2
= (5 × 10^-4)/2
= 2.5 × 10^-4 m
Given resistivity (p) = 1.6 × 10^-8 Ωm.
R = p l/A
Also, A = πr²
A = 22/7 × (2.5 × 10^-4)^2
A = 22/7 × 6.25 × 10^-8
A = 1.964 × 10^-7 m^2
Using formula:
R = p l/A
→ l = (R × A)/p
{ R = 10 Ω (given) }
→ l = (10 × 1.964 × 10^-7)/(1.6 × 10^-8)
→ l = (1.964 × 10^-7 × 10^8)/(1.6)
→ l = 122.75 m
According to question,
Diameter is doubled.
D = 2D'
{ Diameter = 2(New diameter) }
So, R = 2R'
Also,
New Area = π(2r)²
= 4πr²
= 4 × (Area or Old Area)
Let's denote the new area by A' and old area by A
→ A' = 4A
Similarly,
New Resistance by R' and Old Resistance by R.
→ R' = p l/A'
→ R' = p l/4A
→ R' = 1/4 × p l/A
→ R'= 1/4 × R
If the diameter of the wire is doubled then the resistance becomes the one-fourth or
ANSWER(a)
Let n be the number of turns in the coil, Then total length of wire used,
l=2πr×n=2π×7×10
−2
×n
Total resistance, R=ρ
A
l
or 4=
π(0.7×10
−3
)
2
2×10
−7
×2π×7×10
−2
×n
∴n=70.
Ans (b) picture
Hope this helps
