NCERT Question 6 - Chapter 12 Class 10 - Electricity

Last updated at Aug. 6, 2019 by Teachoo

A copper wire has a diameter 0.5 mm and resistivity of 1.6 × 10–8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?

NCERT Question 6 A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10–8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?
Diameter = 0.5 mm
= 0.5 × 10−3 m
= 5 × 10−4 m
Radius = r = 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟/2
= (5 × 〖10〗^(−4))/2 m
= 2.5 × 10−4 m
Resistivity = 𝜌 = 1.6 × 10−8 Ω m
Resistance = 10 Ω
We need to find length of the wire (l)
We know that
R = 𝜌 𝑙/𝐴
∴ l = R (𝐴/𝜌)
To find l, first we need to find A
Finding Area of cross - section (A)
A = 𝜋r2
A = 22/7 × (2.5 × 10−4)2
A = 22/7 × (2.5)2 × (10−4)2
A = 22/7 × 6.25 × 10−8
A = 22/7 × 625/100 × 10−8
A = 13750/7 × 10−10
A = 1964.28 × 10−10 m2
A = 1.964 × 10−7 m2
Finding length
l = R (𝐴/𝜌)
l = (10 × 1.964 × 〖10〗^(−7))/(1.6 × 〖10〗^(−8) ).
l = (10 × 1964 × 〖10〗^(−7))/(16 × 〖10〗^(−8) ) × 10/1000
l = (1964 × 〖10〗^(−7))/(16 × 〖10〗^(−8) ) × 100/1000
l = 491/4 × 10−7 − (−8) × 1/10
l = (122.7 × 〖10〗^(−7 + 8))/10
l = (122.7 × 10)/10
l = 122.7 m
When diameter is doubled
Since
New Diameter = 2 × Old Diameter
New Radius = 2 × Old Radius
And,
New Area = 𝜋 (New Radius)2
= 𝜋 × (2r)2
= 4𝜋r2
= 4 × Old Area
= 4 × A
Thus,
New Resistance = (𝜌 𝑙)/(𝑁𝑒𝑤 𝐴𝑟𝑒𝑎)
= (𝜌 𝑙)/4𝐴
= 1/4 × (𝜌 𝑙)/𝐴
= 1/4 × Old resistance
If diameter doubles, resistance becomes 𝟏/𝟒 times

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.