A 10 m potentiometer wire is attached to a battery with a steady voltage. A leclanche cell is balanced on 4 m wire. With same length, but doubled cross-section, find null point.

Considering that the length of the potentiometer is same, the null point will be acquired at some point due to the fact that the potential gradient will not change.

Therefore, Potential gradient = V/L

Here,

  • V = Voltage (Potentiometer)
  • L = Length (Potentiometer)

Hence, the null point won’t alter and will be acquired at 4m.


Related Questions

  1. How does a potentiometer measure EMF?
  2. Copper is not used as potentiometer wire because?
  3. Sensitivity of potentiometer can be increased by__________.
  4. What are the types of potentiometers?
  5. What is balance point in potentiometer?
  6. Why Potentiometer Is Preferred Over Voltmeter?
  7. What is a DC Potentiometer?
  8. What are the advantages of a potentiometer?
  9. Which Metal Is Used In Potentiometer?
  10. Why Do We Prefer A Potentiometer With A Longer Bridge Wire?
  11. Why High Resistance Is Used In Potentiometer?
  12. A Potentiometer Contains A Wire Of Length 4m And Resistance 10ohms (Connected To EMF 3V). Find Potential Gradient Of Wire (1) 0.75 V/M (2) 2 V/M (3) 6 V/M (4) 10 V/M
  13. How is potential gradient measured?
  14. State the Working Principle of a Potentiometer.
  15. What is a null point in a potentiometer?
  16. What is the symbol for a potentiometer?
  17. What Is Balancing Length In Potentiometer?

Read More:

CBSE CLASS XII Related Questions

  • 1.
    A current of \(4.0\,\text{A}\) flows through a wire of length \(1\,\text{m}\) and cross-sectional area \(1.0\,\text{mm}^2\), when a potential difference of \(2\,\text{V}\) is applied across its ends. Calculate the resistivity of the material of the wire.


      • 2.
        Draw the plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions that can be drawn from this plot.


          • 3.
            Two point charges -Q and Q are located at points (d, 0) and (0, d) respectively, in x-y plane. The electric field E at the origin will be:

              • \(\frac{1}{4\pi\epsilon_0} \frac{\sqrt{2}Q}{d^2} (\hat{i} - \hat{j})\)
              • \(\frac{1}{4\pi\epsilon_0} \frac{\sqrt{2}Q}{d^2} (-\hat{i} - \hat{j})\)
              • \(\frac{1}{4\pi\epsilon_0} \frac{Q}{d^2} (\hat{i} - \hat{j})\)
              • \(\frac{1}{4\pi\epsilon_0} \frac{Q}{d^2} (-\hat{i} - \hat{j})\)

            • 4.

              A rectangular loop of size 5 cm $\times$ 8 cm is lying in the $x$-$y$ plane in a uniform magnetic field $\vec{B} = (2.0\ \text{T})\hat{k}$. The total magnetic flux linked with the loop is:

                • 80 Wb
                • 16 Wb
                • \(8 \times 10^{-2}\) Wb
                • \(8 \times 10^{-3}\) Wb

              • 5.
                A wire of length \(L\) is bent round into (i) a square coil having \(N\) turns and (ii) a circular coil having \(N\) turns. The coil in both cases is free to turn about a vertical axis coinciding with the plane of the coil, in a uniform, horizontal magnetic field and carry the same currents. Find the ratio of the maximum value of the torque acting on the square coil to that on the circular coil.


                  • 6.
                    In a region, the electric potential varies as \(V = 10 - 50x\), where \(V\) is in volts and \(x\) is in meters. The electric field in the region is

                      • \(10\ \text{N/C}\) along \(+x\)
                      • \(10\ \text{N/C}\) along \(-x\)
                      • \(50\ \text{N/C}\) along \(+x\)
                      • \(50\ \text{N/C}\) along \(-x\)
                    CBSE CLASS XII Previous Year Papers

                    Comments


                    No Comments To Show