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Time period of a pendulum clock at 25°c is 2 seconds. If it looses 8.64 seconds per day on a day of temperature 45°c, the coefficient of linear expansion of pendulum is
A) 2×10power-5°c
B) 10power-5°c
C) 5×10power-5°c
D) 10 power-6°c

Kondaveti prahakar , 8 Years ago
Grade 11
anser 2 Answers
Subho Pal
 Tis directly proportional to L  [T: period L: length of the pendulum]
At 250C T25=2 s
At 450C T45=(24*3600+8.64)/(24*3600/2) s 
=2 + 8.64/(12*3600) s
=2.0002 s
L45/L25 =(2.0002/2)2 =1.0002=1+α.t=1+α.(45-25)=1+20*α
or,   20*α=0.0002
or, α=0.0002/20
therefore,  α=10-5
option (B)10power-5  0C-1 is correct
Last Activity: 8 Years ago
Phanisivani
∆T=1/2a∆theta(86400).                                                  (45-25)=1/2a(8.64-2)(86400).                                       (20)=1/2a(464/100)(86400).                                       a=20/464×432.                                                              a=20/232×432.                                                                a=5/232×216.                                                                a=10^-5/°c approximately
Last Activity: 5 Years ago
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