Rankers Physics

Relation between Current and Drift Velocity: Practice Problem & Solution

A wire is stretched so that its length increases by 10%. The resistance of the wire increases by :
11%
15%
21%
28%

Solution Explained:

To solve this problem, we apply the core principles of Relation between Current and Drift Velocity. Understanding the underlying formula is key to arriving at the correct answer below:

When a wire is stretched, its length increases, and its cross-sectional area decreases. The resistance of a wire is given by the formula:

 

R=ρLAR = \rho \frac{L}{A}

 

Where:


  • RR
     

    is the resistance,


  • ρ\rho
     

    is the resistivity of the material (constant),


  • LL
     

    is the length of the wire,


  • AA
     

    is the cross-sectional area of the wire.

When the wire is stretched:

  1. Length increases by 10%: The new length
    L′L'
     

    is given by: 

    L′=L+0.1L=1.1LL' = L + 0.1L = 1.1L 

  2. Volume remains constant: The volume of the wire before and after stretching remains the same. Volume is the product of length and area: 

    Volume before=L×A\text{Volume before} = L \times A 

    Volume after=L′×A′=1.1L×A′\text{Volume after} = L' \times A' = 1.1L \times A'Since the volume remains constant:

     

    L×A=1.1L×A′L \times A = 1.1L \times A'Solving for

    A′A', the new cross-sectional area:

     

    A′=A1.1A' = \frac{A}{1.1} 

Step 2: New Resistance

The new resistance

R′R'

of the stretched wire is given by:

 

R′=ρL′A′=ρ1.1LA1.1=ρ1.1L×1.1A=ρ1.21LAR' = \rho \frac{L'}{A'} = \rho \frac{1.1L}{\frac{A}{1.1}} = \rho \frac{1.1L \times 1.1}{A} = \rho \frac{1.21L}{A}

 

So, the new resistance

R′R'

is 1.21 times the original resistance

RR

.

Step 3: Conclusion

The resistance increases by 21% when the wire is stretched by 10%.

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