Question

Solve me 6 and 7 with workout please.........​

Answer 1

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Answer 2

$\large\bf\underline \red{Question:-}$

A copper wire has diameter 0.5mm and resistivity of 1.6×10^-8 ohm m .what will be the length of this wire to make it's resistance 10ohm?

how much does the resistance change if the diameter is doubled.

$\large\bf\underline \orange{Given:-}$

• Diameter of copper wire = 0.5mm

• Resistivity of copper wire = $\rm 1 {0}^{ - 8} \Omega\:m$

• Resistance = $\rm 10 \Omega\:$

$\large\bf\underline \orange{To \: find:-}$

• length of wire

• how much resistance change if the diameter is doubled.

$\huge\bf\underline \green{Solution:-}$

Diameter = 0.5mm

⠀⠀⠀⠀⠀⠀=0.5/1000 m

⠀⠀⠀⠀⠀⠀=5×10 -⁴m

then,

Radius = diameter/2

Radius = 5×10-⁴/2

Radius = 2.5×10 -⁴ m

Area of cross section of wire = πr²

⠀⠀⠀⠀⠀⠀= 22/7 × (2.5×10-⁴)²

⠀⠀⠀⠀⠀⠀= 22/7 ×(2.5)²×(10-⁴)²

⠀⠀⠀⠀⠀⠀= 22/7 × 6.25 × 10^-8

⠀⠀⠀⠀⠀⠀= 0.1964×10^-6 m²

we know that,

$\large \bf \blue{ R = \rho\frac{l}{A}}$

• R = resistance
• p = resistivity ( rho )
• l = length of wire
• A = cross section of wire.

$\mapsto\rm\:l = \frac{10 \times 0.1964 \times {10}^{ - 6} }{1.6 \times 10 {}^{ - 8} } \\ \\\mapsto\rm\:l = \frac{1.964 \times {10}^{ - 6} }{1.6 \times {10}^{ - 8} } \\ \\ \mapsto\rm\:l = \frac{1.964 \times 10 {}^{ - 6 - ( - 8)} }{1.6} \\ \\\mapsto\rm\:l = \frac{1.964 \times 10 {}^{ - 6 + 8} }{1.6} \\ \\\mapsto\rm\:l = \frac{1.964 \times 10 {}^{2} }{1.6} \\ \\ \mapsto\rm\:l = \frac{196.4}{1.6} \\ \\\mapsto\bf\blue{l = 122.75m}$

So,

Length of wire = 122.75m.

When diameter of wire is doubled :-

Then,

Area of cross section when diameter is doubled

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀= π(2r)²

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀= 4πr²

Radius = 2×old radius

• R = pl/A

⠀⠀⠀⠀⠀⠀⠀➝ R = Pl/4πr²

⠀⠀⠀⠀⠀⠀⠀➝ R = pl/4A

⠀⠀⠀⠀⠀⠀⠀➝ R = 1/4 ×pl/A

⠀⠀⠀⠀⠀⠀⠀➝ R = 1/4 × R1

So,

If diameter of wire is doubled then the resistance of wire becomes 1/4 times to it's old resistance