Question 3.6: A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10 −7 m 2 , and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment?
Class 12 - Physics - Current Electricity Current Electricity Page-127
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Answered by
5
Given, resistance of wire , R = 5Ω
Cross section area of wire , A = 6 × 10⁻⁷ m²
And length of wire, L = 15m
We know, the relation ,
here R is the resistance of wire,
is the resistivity , L the length of wire and A is cross section area of wire.
so, ρ = RA/L
= 5 × 6 × 10⁻⁷/15 Ωm
= 2 × 10⁻⁷ Ωm
Cross section area of wire , A = 6 × 10⁻⁷ m²
And length of wire, L = 15m
We know, the relation ,
here R is the resistance of wire,
so, ρ = RA/L
= 5 × 6 × 10⁻⁷/15 Ωm
= 2 × 10⁻⁷ Ωm
Answered by
2
We now that
R = ρl / A
Where L= length= 15m
A= area= 6 x 10-7 m2
R=resistance= 5ohm
ρ= resistivity
ρ= RA/l
= 5 x 6 x 10-7/15
=2 x 10-7 ohm-m
Now specific conductance or electrical conductivity (σ) is the reciprocal of resistivity
Therefore σ = 1/ρ
σ= 0.5 x 107 S⋅m−1
=5 x 106 S⋅m−1
R = ρl / A
Where L= length= 15m
A= area= 6 x 10-7 m2
R=resistance= 5ohm
ρ= resistivity
ρ= RA/l
= 5 x 6 x 10-7/15
=2 x 10-7 ohm-m
Now specific conductance or electrical conductivity (σ) is the reciprocal of resistivity
Therefore σ = 1/ρ
σ= 0.5 x 107 S⋅m−1
=5 x 106 S⋅m−1
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