| ⇦ | 
| ⇨ |
A body takes 5 minute for cooling from 50⁰C to 40⁰C. Its temperature comes down to 33.33⁰C in next 5 minute. Temperature of surroundings is
Options
(a) 15⁰C
(b) 20⁰C
(c) 25⁰C
(d) 10⁰C
Correct Answer:
20⁰C
Explanation:
Apply Newton’s law of cooling
In first case, (50 – 40)/5 = K [(50+40)/2 – θ]
⇒ 2 = K [45 – θ] ——(i)
In second case, take 33.33 = 33 x (1/3) = 100/3
[40 – (100/3)]/5 = K {[40 + (100/3)/2] – θ}
⇒ 20/(5×3) = K [(110/3) – θ] ——(ii)
(i)/(ii) = (2 x 5 x 3)/20 = (45 – θ) / [(110/3) – θ]
⇒ 3/2 = (45 – θ) / [(110/3) – θ]
= 110 – 30 = 90 – 20 ⇒ θ = 20⁰C.
Related Questions:
- The e.m.f. of a battery is 2 V and its internal resistance is 0.5 Ω. The maximum
- A 50 ohm galvanometer gets full scale deflection when a current of 0.01 A passes
- If two charges +4e and +e are at a distance x apart, then at what distance charge q
- The tension in a piano wire is 10 N. What should be the tension in the wire to produce
- The time of reverberation of a room A is one second. What will be the time (in seconds)
Topics: Properties of Bulk Matter
(130)
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
Leave a Reply