| ⇦ | 
| ⇨ |
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: - Thin uniform rod of length L, cross-sectional area A and density ρ is rotated
- A monoatomic gas at a pressure P, having a volume V expands isothermally
- In an ammeter, 0.2% of main current passes through the galvanometer
- Figure below shows two paths that may be taken by a gas to go from a state A
- What is the nature of Gaussian surface involved in Gauss’s law of electrostatics?
Topics: Current Electricity
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Thin uniform rod of length L, cross-sectional area A and density ρ is rotated
- A monoatomic gas at a pressure P, having a volume V expands isothermally
- In an ammeter, 0.2% of main current passes through the galvanometer
- Figure below shows two paths that may be taken by a gas to go from a state A
- What is the nature of Gaussian surface involved in Gauss’s law of electrostatics?
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