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

In the given figure, a battery of emf \(E\) is connnected across a conductor \(P Q\) of length \('Y'\) and different area of cross-sections having radii \(r_{1}\) and \(r_{2}\left(r_{2}\,<\,r_{1}\right)\). Choose the correct option as one moves from \(P\) to \(Q\) :

A Electric field decreases.
B Drift velocity of electron increases.
C Electron current decreases.
D All of these

Verified Solution

Answer & Solution

Correct Answer

(B) Drift velocity of electron increases.

Step-by-step Solution

Detailed explanation

Current is constant in conductor \(i=\) constant Resistance of element \(d R=\frac{\rho d x}{\pi r^{2}}\) \(d V=i d R=\frac{i \rho d x}{\pi r^{2}}\) \(E=\frac{d V}{d x}=\frac{i \rho}{\pi r^{2}}\) \(V_{d}=\frac{e E \tau}{m}\) \(\therefore V_{d} \propto E\)…

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

Explore more questions on app

From JEE Mains

Explore more questions on app