A wire’s resistance is 'R' ohm. If the wire is melted and stretched 'n' times the original length, find new resistance

The given data:

  • Wire of Radius = r
  • Length = l1 
  • Resistance = R

This wire is now expanded to a new length l2, that is n times l1.

Because there is zero change in the wire’s radius, then it can be considered that the volume before and after stretching is constant, that is, V1 = V2.

Here,

  • V1 = Volume before stretching
  • V2 = Volume after stretching

It is already known that Volume is the product of area and height.

In this case, height can be considered the length of the wire, A1 × l1 = A2 × l2. We are aware that, l2 = n × l1.

After replacing, we can get,

⇒ A1 × l1 = A2 × n × l1 

⇒ A1/n = A2

It is already known that resistance R of a material is estimated by means of the formula:

R = ρ l/A

Where,

  • ρ = Resistivity of the material
  • A = Area
  • l = length of the wire

Before the extension, the resistivity was expressed as R1 = ρl1/A1

Now, after the extension, resistivity can be expressed as R2 =ρl2 /A2

Upon value substitution for A2 in the equation for R2, we can obtain:

⇒ R2 = ρl2 × n/A1

Now after replacing, we get: l2= n ×l1,

⇒ R2 = ρl1 × n2/A1 

Therefore, the term ρl1A1 = R1 and R1 (value given in the question is R )

⇒ R2 = R × n2

Read Also: Physics Practical


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CBSE CLASS XII Related Questions

  • 1.
    Consider a cylindrical conductor of length \( l \) and area of cross-section \( A \). Current \( I \) is maintained in the conductor and electrons drift with velocity \( \vec{v}_d \, (|\vec{v}_d| = \frac{eE}{m} \tau) \), where symbols have their usual meanings. Show that the conductivity of the material of the conductor is given by \[ \sigma = \frac{n e^2 \tau}{m}. \]


      • 2.
        Read the following paragraph and answer the questions that follow.
        A p-type or n-type semiconductor can be converted into a p-n junction by doping it with suitable impurity. The motion of majority charge carriers causes diffusion current across the junction while the barrier electric field causes motion of minority carriers for drift current. In case of unbiased diode, the diffusion and drift currents are equal. This equilibrium is disturbed by the biasing batteries. Diodes, therefore, allow currents in one direction. This property of diode is used in making rectifiers.


          • 3.
            A student sets up the circuit as shown in the figure to find the value of unknown resistance X and records a set of readings of the voltmeter and the ammeter by using the rheostat.


              • 4.
                Two air-filled capacitors of capacitances $C_1$ and $C_2$ are connected in parallel with a dc battery. After the capacitors are fully charged, a slab of dielectric constant K is inserted between the plates of each capacitor. How will the (i) charge on each capacitor and (ii) energy stored in the capacitor affected after the slab is introduced.


                  • 5.
                    A charged particle $+q$ in an electric field $\vec{E}$ experiences a force in the direction of the electric field. As a result, its kinetic energy changes. Similarly, the charged particle also experiences a force when it moves in a magnetic field $\vec{B}$. But this magnetic force is perpendicular to both velocity $\vec{v}$ of the charged particle and the magnetic field $\vec{B}$, so it cannot change the kinetic energy of the charged particle. Consider two charged particles 1 and 2 of masses $m$ and $\frac{m}{2}$ having charges $-q$ and $+2q$ respectively. They are accelerated from rest through the same potential difference $V$ and acquire kinetic energy $K_1$ and $K_2$. Then they enter in a region of uniform magnetic field $\vec{B}$ perpendicular to their velocities.


                      • 6.
                        With the help of a labelled diagram, explain the principle, construction and working of an a.c. generator.

                          CBSE CLASS XII Previous Year Papers

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