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Mechanical Properties Of Fluids

Question
CBSEENPH11019415

A clock with a metallic pendulum gains 5s in a day at 15°C and loses 10s in a day at 30°C. Find the temperature at which the clock will read correct time.

Solution
Let L be the length and straight alpha the temperature coefficient of linear expansion of pendulum clock. 
If the temperature of the clock is increased or decreased by θ, the length of the pendulum will increase or decrease by Lstraight alpha . And the clock runs slow or fast.
Let the clock keep correct time when the length is L and temperature is t°C.
We know time period of the pendulum is given by, 
                        straight T equals 2 straight pi square root of straight L over straight g end root 
The time period of oscillation of pendulum when the temperature is changed by θ is, 

straight T apostrophe space equals space 2 straight pi square root of fraction numerator straight L left parenthesis 1 plus αθ right parenthesis over denominator straight g end fraction end root space equals space straight T left parenthesis 1 plus 1 half αθ right parenthesis 
The change in time period of oscillation of pendulum when the temperature is changed by θ is, 
straight T apostrophe minus straight T equals dT equals 1 half αθT 
The change in the time in a day, 
                increment straight T equals 1 half αθT subscript straight o 
where, straight T subscript straight o is the duration of the day. 
At space space 15 degree straight C comma the gain is 5s.
Therefore,
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#6 {main}</pre>               ...(1) 
At 30 degree straight C comma the loss is 10s.
Therefore,
           space minus 10 equals 1 half straight alpha left parenthesis straight t minus 30 right parenthesis straight T subscript straight o              ...(2) 
From (1) and (2), we get 
 space straight t equals 20 degree straight C, is the temperature at which the clock will read correct time.