| ⇦ | 
| ⇨ |
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: - Carbon,silicon and germanium have four valence electrons each. At room temperature
- The term liquid crystal refers to a state that is intermediate between
- A light emitting diode(LED) has a voltage drop of 2 volt across it
- The e.m.f. of a battery is 2 V and its internal resistance is 0.5 Ω. The maximum
- An electron is moving in a circular path under the influence of a transverse magnetic
Topics: Current Electricity
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Carbon,silicon and germanium have four valence electrons each. At room temperature
- The term liquid crystal refers to a state that is intermediate between
- A light emitting diode(LED) has a voltage drop of 2 volt across it
- The e.m.f. of a battery is 2 V and its internal resistance is 0.5 Ω. The maximum
- An electron is moving in a circular path under the influence of a transverse magnetic
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