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कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान इयत्ता ११

Suppose the Bent Part of the Frame of the Previous Problem Has a Thermal Conductivity of 780 J S−1 M−1 °C−1 Whereas It is 390 J S−1 M−1°C−1 for the Straight Part. Calculate the Ratio of the Rate of

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प्रश्न

Suppose the bent part of the frame of the previous problem has a thermal conductivity of 780 J s−1 m−1 °C−1 whereas it is 390 J s−1 m−1°C−1 for the straight part. Calculate the ratio of the rate of heat flow through the bent part to the rate of heat flow through the straight part.

बेरीज
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उत्तर


Resistance of any branch, `R = l/{KA}`

Here, K is the thermal conductivity, A is the area of cross section and l is the length of the conductor.

`R_{BC} = 1/{780.A} = {5xx10^-2}/{780.A}`

`R_{CD} = {60xx10^-2}/"780.A"`

`R_{DE} ={5xx10^-2}/{780.A}`

`R_{AB} = {20xx10^-2}/{390.A}`

`R_{EF} = {20xx10^-2}/{390.A}`

`R_{BE} = {60xx10^-2}/{390.A}`

`R_(BE) = R_2 = (60xx10^-2)/(390xxA)`

since R1 and R2 are in parallel, the amount of heat flowing through them will be same.

`{q_1}/{q_2} = {R_2}/{R_1}`

`= {60xx10^-2xx780xxA}/{390xxAxx70xx10}`

`= 12/7`

`⇒ {q_1}/{q_2 }= 12/7`

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Thermal Expansion of Solids
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पाठ 28: Heat Transfer - Exercises [पृष्ठ १००]

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एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
पाठ 28 Heat Transfer
Exercises | Q 27 | पृष्ठ १००

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