A wire has a resistance of R. If its length is increased by a factor of 8 and its crosssectional area is decreased by a factor of 4, what is its new resistance (R')?

Correct option is C

Correct Option:  (c),R' = 32 R

The formula for the resistance of a wire:

$$R = p\frac{L}{A}$$ 

• ​R is the resistance,
  
• ρ is the resistivity of the material (which remains constant), 
  
• A is the cross-sectional area of the wire.
  
• L is the length of the wire.

​The length is increased by a factor of 8, so the new length L is: 

$$L' = 8L$$ 

The cross-sectional area is decreased by a factor of 4, so the new area $$A'$$ is:

$$A' = \frac{A }{4}$$ 

The resistance of the wire with the new length and area $$R'$$ 

$$R' = p \frac{L'}{A'}$$ 

$$R' = p 8L \div \frac{A}{L}$$  $$p \times \frac{8L \times4}{A} = p \frac{32L}{A}$$ 

The original resistance is:

$$R = p \frac{L}{A}$$ 

$$R' = 32 \times p \frac{L}{A} = 32R$$  

$$R' = 32R$$ 

The new resistance R is  32 times. 

L′L'