| ⇦ | 
| ⇨ |
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: - The minimum velocity (in ms⁻¹) with which a car driver must traverse a flat curve
- A simple pendulum has a time period T₁ when on the earth’s surface and T₂ When taken
- A satellite S is moving in an elliptical orbit around the earth. The mass
- In N-type semi-conductor current is due to
- If ∆U and ∆W represent the increase in internal energy and work done by the system
Topics: Current Electricity
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The minimum velocity (in ms⁻¹) with which a car driver must traverse a flat curve
- A simple pendulum has a time period T₁ when on the earth’s surface and T₂ When taken
- A satellite S is moving in an elliptical orbit around the earth. The mass
- In N-type semi-conductor current is due to
- If ∆U and ∆W represent the increase in internal energy and work done by the system
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