| ⇦ | 
| ⇨ |
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 threshold frequency for a metallic surface corresponds to an energy of 6.2 eV,
- An iron rod of volume 10⁻⁴ m³ and relative permeability 1000 is placed inside a long
- The depletion layer in the p-n junction region is caused by
- The angle of a prism is A. One of its refracting surfaces is silvered
- Two lines of force due to a bar magnet
Topics: Current Electricity
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The threshold frequency for a metallic surface corresponds to an energy of 6.2 eV,
- An iron rod of volume 10⁻⁴ m³ and relative permeability 1000 is placed inside a long
- The depletion layer in the p-n junction region is caused by
- The angle of a prism is A. One of its refracting surfaces is silvered
- Two lines of force due to a bar magnet
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