| ⇦ |
| ⇨ |
When a current of (2.5±0.5) A flows through a wire, it develops a potential difference of (20±1) V, then the resistance of wire is
Options
(a) (8±2)Ω
(b) (8±1.6)Ω
(c) (8±1.5)Ω
(d) (8±3)Ω
Correct Answer:
(8±2)Ω
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - When ₃Li⁷ nuclei are bombarded by protons, and the resultant nuclei are ₄Be⁸,
- Two concentric coils each of radius equal to 2π cm are placed at right angles
- Current in a coil changes from 4 A to zero in 0.1 second and the emf induced is 100V
- Application of Bernoulli’s theorem can be seen in
- If the velocity of charged particle has both perpendicular and parallel components
Topics: Current Electricity
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- When ₃Li⁷ nuclei are bombarded by protons, and the resultant nuclei are ₄Be⁸,
- Two concentric coils each of radius equal to 2π cm are placed at right angles
- Current in a coil changes from 4 A to zero in 0.1 second and the emf induced is 100V
- Application of Bernoulli’s theorem can be seen in
- If the velocity of charged particle has both perpendicular and parallel components
Topics: Current Electricity (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends

R=V/I
R=20/2.5
R= 8 ohm
Now,
∆ R/R=∆ V/V+∆I/I
=1/20 +0.5/2.5
= 1/4
Therefore,
∆ R÷R=1÷4
∆ R=1÷4×R
∆R=1÷4×8
∆R=2
Therefore,
Resistance with error limits=R+ – ∆R
=(8+ – 2)ohm