JEE Mains · Physics · STD 12 - 3. current electricity

A potentiometer wire of length \(300\,cm\) is connected in series with a resistance \(780\,\Omega\) and a standard cell of emf \(4\,V\). A constant current flows through potentiometer wire. The length of the null point for cell of emf \(20\,mV\) is found to be \(60\,cm\). The resistance of the potentiometer wire is\(...\Omega\)

A \(78\)
B \(200\)
C \(2\)
D \(20\)

Verified Solution

Answer & Solution

Correct Answer

(D) \(20\)

Step-by-step Solution

Detailed explanation

Let resistance of potentiometers wire is \(R\) \(i=\frac{4}{R+780}\) Potential difference across \(AB\) \(=\frac{4 R}{R+780}\) Potential difference across \(AC\) \(=\frac{4 R \times 60}{(R+780) \times 300}=\frac{4 R}{5(R+780)}\) This should be equal to \(20\,mV\)…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Match the thermodynamic processes taking place in a system with the correct conditions. In the table: \(\Delta Q\) is the heat supplied, \(\Delta W\) is the work done and \(\Delta U\) is change in internal energy of the system

Process	Condition
\((I)\) Adiabatic	\((A)\; \Delta W =0\)
\((II)\) Isothermal	\((B)\; \Delta Q=0\)
\((III)\) Isochoric	\((C)\; \Delta U \neq 0, \Delta W \neq 0 \Delta Q \neq 0\)
\((IV)\) Isobaric	\((D)\; \Delta U =0\)

JEE Mains 2020 Medium

Match List - I with List - II.

List - I	List - II
(A) Permeability of free space	(I) \(\left[\mathrm{M} \mathrm{~L}^2 \mathrm{~T}^{-2}\right]\)
(B) Magnetic field	(II) \(\left[\mathrm{M} \mathrm{~T}^{-2} \mathrm{~A}^{-1}\right]\)
(C) Magnetic moment	(III) \(\left[\mathrm{M} \mathrm{~L} \mathrm{~T}^{-2} \mathrm{~A}^{-2}\right]\)
(D) Torsional constant	(IV) \(\left[\mathrm{L}^2 \mathrm{~A}\right]\)

Choose\(\text{ }\)the\(\text{ }\)correct\(\text{ }\)answer\(\text{ }\)from\(\text{ }\)the\(\text{ }\)options\(\text{ }\)given\(\text{ }\)below\(\text{ }\):

JEE Mains 2025 Easy

The surface charge density of a thin charged disc of radius \(R\) is \(\sigma \). The value of the electric field at the centre of the disc is \(\frac{\sigma }{{2\,{ \in _0}}}\). With respect to the field at the centre, the electric field along the axis at a distance \(R\) from the centre of the disc

JEE Mains 2013 Hard

Explore more questions on app

From JEE Mains

Explore more questions on app