| ⇦ | 
| ⇨ |
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: - In the Bohr’s model of hydrogen atom, the ratio of the kinetic energy to the total
- A block is moving up an inclined plane of inclination 60° with velocity of 20 ms⁻¹
- The S.I. unit of electric flux is
- The component of vector A=2i+3j along the vector i+j is
- An insulated container of gas has two champers seperated by an insulating partition.
Topics: Current Electricity
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In the Bohr’s model of hydrogen atom, the ratio of the kinetic energy to the total
- A block is moving up an inclined plane of inclination 60° with velocity of 20 ms⁻¹
- The S.I. unit of electric flux is
- The component of vector A=2i+3j along the vector i+j is
- An insulated container of gas has two champers seperated by an insulating partition.
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