A pendulum clock is 10 seconds fast at a temperature of 15°C and 20 seconds slow at temperature of 30°C. The temperature at which it gives
the correct time is
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1
Answer:
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Answered by
0
Answer:
T=2π
g
l
⇒T∝l
1/2
⇒
T
ΔT
=
2
1
l
Δl
and Δl=l∝Δt, linear expansion formula
⇒
l
Δl
=∝Δt
Δt= change in temperature=Final−initial
5 sec fast⇒1 hour alarm rises 5 sec before i.e., time period reduced
⇒ΔT=−5
Case I:
5 sec fast at 15
o
C
T
−5
=
z
1
∝Δt=
2
1
∝(15−t) ……..(1)
t= temperature at which correct reading.
15
o
<t<30
o
Case II:
10 sec slow at 30
o
C
⇒ΔT=10sec
⇒
T
10
=
2
1
∝(30−t) ………(2)
Heated to 30
o
C
dividing (1) by (2)
We get
−0.5=
30−t
15−t
⇒30−t=2(t−15)
⇒t=20 c
Explanation:
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