Question

3. A wire of length L is drawn such that its diameter is reduced to half of its original

diameter. If the resistance of the wire were 10Ω, its new resistance would be.

​

Answer 1

Answer:

Let the new length be ℓ  

1

​  

, keeping volume constant,

πr  

2

L=π(  

2

r

​  

 

​  

)  

2

ℓ  

1

​  

 

⇒ℓ  

1

​  

=4L

Now, 10=  

πr  

2

 

ρL

​  

;R=  

π(  

2

r

​  

 

​  

)  

2

 

ρℓ  

1

​  

 

​  

 

10

R

​  

=  

πr  

2

 

ρℓ  

1

​  

×4

​  

×  

ρL

πr  

2

 

​  

=  

r  

2

×L

4L×4×r  

2

 

​  

=  

1

16

​  

 

R=160Let the new length be ℓ  

1

​  

, keeping volume constant,

πr  

2

L=π(  

2

r

​  

 

​  

)  

2

ℓ  

1

​  

 

⇒ℓ  

1

​  

=4L

Now, 10=  

πr  

2

 

ρL

​  

;R=  

π(  

2

r

​  

 

​  

)  

2

 

ρℓ  

1

​  

 

​  

 

10

R

​  

=  

πr  

2

 

ρℓ  

1

​  

×4

​  

×  

ρL

πr  

2

 

​  

=  

r  

2

×L

4L×4×r  

2

 

​  

=  

1

16

R=160