| ⇦ | 
| ⇨ |
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: - A particle moves along with x-axis.The position x of particle with respect to time
- Efficiency of a Carnot engine is 50% when temperature of sink is 500K. In order
- A transistor is operated in common-emitter configuration at Vᵥ = 2 V such that a change
- Cohesive force is experienced between
- A paramagnetic sample shows a net magnetisation of 0.8 Am⁻¹, when placed
Topics: Current Electricity
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A particle moves along with x-axis.The position x of particle with respect to time
- Efficiency of a Carnot engine is 50% when temperature of sink is 500K. In order
- A transistor is operated in common-emitter configuration at Vᵥ = 2 V such that a change
- Cohesive force is experienced between
- A paramagnetic sample shows a net magnetisation of 0.8 Am⁻¹, when placed
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