A potentiometer wire AB of length L and resistance 12r is connected to a cell D with emf e and internal resistance 3r. Find length AJ wherein the galvanometer as shown in image shows no deflection.

Potential gradient can be expressed as the potential difference per unit length of wire.

Therefore, the potential gradient = (ε/12r + r) × (12r/L)]

= (ε/13r) × (12r/L)

= 12ε/13L

⇒ (12ε/13L )x = ε/2

⇒ x = \(\frac{13L}{24}\)


Related Questions

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

Read More:

CBSE CLASS XII Related Questions

  • 1.
    Consider the nuclear reaction \( X \to Y + Z \). Let \( M_x \), \( M_y \), and \( M_z \) be the masses of the three nuclei X, Y, and Z respectively. Then which of the following relations hold true?

      • \( (M_x - M_z)<M_y \)
      • \( (M_x - M_y)<M_z \)
      • \( M_x>(M_y + M_z) \)
      • \( M_x<(M_y + M_z) \)

    • 2.
      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.


        • 3.
          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.


            • 4.
              The resistance of a metal wire at \( 20^\circ \text{C} \) is \( 1.05 \, \Omega \) and at \( 100^\circ \text{C} \) is \( 1.38 \, \Omega \). Determine the temperature coefficient of resistivity of this metal.


                • 5.
                  Two metal spheres of radii $r_1$ and $r_2$ ($> r_1$) having charges $q_1$ and $q_2$ respectively kept in air, are brought in contact. Which of the following statements is not correct ?

                    • The total charge of the two spheres is conserved.
                    • Both spheres attain the same potential.
                    • The final potential of the system equals $\frac{1}{4\pi\epsilon_0} \frac{(q_1 + q_2)}{(r_1 + r_2)}$
                    • The final potential of the system equals $\frac{1}{4\pi\epsilon_0} \frac{(q_1 + q_2) (r_1 + r_2)}{r_1 r_2}$

                  • 6.
                    Capacitors are manufactured with certain standard capacitances and working voltages. However, these standard values may not be the ones that are actually needed in a particular application. Two or more capacitors can be grouped in series or in parallel to achieve desired capacitance and voltage. When connected in series, the total capacitance decreases while the voltage rating increases, whereas in parallel connections, the total capacitance increases and maintains the same voltage rating. A capacitor stores energy in the electric field between its plates and stored energy is proportional to the square of the voltage and capacitance $U = \frac{1}{2}CV^2$, where symbols have their usual meanings.
                    Two capacitors, one of $3 \ \mu$F and the other of $6 \ \mu$F, are connected in series in the circuit as shown in the figure, for a long time. }

                      CBSE CLASS XII Previous Year Papers

                      Comments


                      No Comments To Show