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Rate of heat flow through a cylindrical ...

Rate of heat flow through a cylindrical rod is `H_(1)`. Temperatures of ends of rod are `T_(1)` and `T_(2)`. If all the dimensions of rod become double and temperature difference remains same and rate of heat flow becomes `H_(2)`. Then `(H_(1))/(H_(2))` is `0.x`. Find value of x.

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A cylindrical rod having temperature T_(1) and T_(2) at its ends. The rate of flow of heat is Q_(1) cal//sec . If all the linear dimensions are doubled keeping temperature constant, then rate of flow of heat Q_(2) will be

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Knowledge Check

  • Heat is flowing through two cylindrical rods of the same material. The diamters of the rods are in the ratio 1: 2 and the length in the ratio 2 : 1 . If the temperature difference between the ends is same then ratio of the rate of flow of heat through them will be

    A
    `1 : 1`
    B
    `1 : 8`
    C
    `2 : 1`
    D
    `8 : 1`

  • The order of heat of fusion of T_(2),D_(2) and H_(2) is

    A
    `T_(2) gt D_(2) gt H_(2)`
    B
    `H_(2) gt T_(2) gt D_(2)`
    C
    `D_(2) gt T_(2) gt H_(2)`
    D
    `D_(2)=T_(2)gtH_(2)`

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