A copper wire has a diameter of 0.6 mm and resistivity of 1.7 × 10-8 ohm metre (1)What will be the length of this wire to make its resistance 17 ohm ? (2)How much does resistance change if the diameter is doubled ? (3)How much does the resistance change if the diameter is Halved?
Please give step wise solution...
Answers
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★1) length = 282.8 m approx ★
★ 2) increased 1/4 times ★
★ 3) increased 4 times ★
EXPLAINATION IS REFFERED TO THE ATTACHMENT
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Answer:
Explanation:
1) R = p L / A
17 = 1.7 × 10-8 x L / pi (0.3)^2
L = 282.8 metres
2) Let us first do the calculation in general
A conducting wire of resistance R ohm is stretched in such a way that its area becomes 1/nth part of it's original area, now what will be the resistance of wire
Answer : New resistance is N^2 times original resistance;
Explanation :
Resistance of a wire is given as;
R = pL/A
Where p is the resistivity of wire. It will be same for same material
A is the cross sectional area
L is the length.
Now if the wire is stretched, whereas the area of wire becomes 1/n;
A’ = A/n
V = A x L
Since stretching of wire wont change the volume of wire;
V’=V
A’ x L’ = A x L
A/n x L’ = A x L
L’ = nL
Calculating new resistance of wire after it is stretched;
R’ = pL"/A"
R’ = p nL⁄(A/n)
R’= n^2 p L/A
R’= n^2 R
Hence we Diameter is doubled,
R' = 2^2 R
R' = 4 R
3) When diameter is halved
R' = 0.5^2 R
R' = 0.25R