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Time period of a pendulum clock at 2525°c is 22 seconds. If it looses 88.6464 seconds per day on a day of temperature 4545°c,the coefficient of linear expansion of pendulum isAisA 22×10power510power-5°cBcB 10power510power-5°cCcC 55×10power510power-5°cDcD 1010 power6power-6°c

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Kondaveti prahakar , 7 Years ago
Grade 11
anser 2 Answers
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Subho Pal

Last Activity: 7 Years ago

T2 is directly proportional to L [T: period L: length of the pendulum]
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At 250C T25=2 s
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At 450C T45=24*3600+8.64/24*3600/2 s
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=2 + 8.64/12*3600 s
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=2.0002 s
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L45/L25 =2.0002/22 =1.0002=1+α.t=1+α.45-25=1+20*α
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or, 20*α=0.0002
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or, α=0.0002/20
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therefore, α=10-5
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option B10power-5 0C-1 is correct

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Phanisivani

Last Activity: 4 Years ago

∆T=1/2a∆theta86400. 45-25=1/2a8.64-286400. 20=1/2a464/10086400. a=20/464×432. a=20/232×432. a=5/232×216. a=10^-5/°c approximately

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