CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

A pendulum clock consists of a 1 mm rod connected to a small, heavy bob. If it is designed to keep correct time at 30C, how fast or slow will it go in 12 hours at 50C? Coefficient of linear expansion of iron =1.2×105/C.
  1. 4 sec, fast 
  2. 4 sec,slow
  3. 5.2 sec, fast 
  4. 5.2 sec, slow 


Solution

The correct option is D 5.2 sec, slow 
Time lost/gained due to temperature difference Δθ is given by
ΔT=T0×12αΔθ
So, time period at temperature θ is 
T=T0(1+αθ2)
T30=T0[1+α(30)2]=T0(1+15α)
T50=T0[1+α(50)2]=T0(1+25α)
T50T30=T0(1+25α)T0(1+15α)=(1+25α)(1+15α)1
=(1+25α)(115α)
=(1+25α15α375α2)=(1+10α)
(neglecting square term as α is very small)

T50T30T50=T50T301=(1+10α)1=10α
=10α=10×1.2×105=1.2×104
T50T30T50 shows fractional loss in time and this is (+) ve. Thus the clock goes slow.
Time lost in 12 hours
Δt=12×60×60×1.2×104=5.185.2 sec

flag
 Suggest corrections
thumbs-up
 
0 Upvotes


Similar questions
View More...



footer-image