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A steel ball of diameter Lb = 4.196 cm is placed on a hole of diameter Lh = 4.042 cm on an aluminium plate. The initial temperature is 14 ¢XC. What would be the minimum temperature so that the ball will fit through the hole if both are heated? The linear expansion coefficients for steel and aluminium are £\steel = 1.100 ¡Ñ 10¡V5 K¡V1 and £\alu = 2.200 ¡Ñ 10¡V5 K¡V1, respectively. ans 3610 degrees celcius

  • Physics -

    First let's use symbols people can read.
    a1 = 1.1*10^-5 C^-1 (thermal expansion cefficient of steel)
    a2 = 2.2*10^-5 C^-1 (thermal expnsion cefficient of aluminum)

    The hole and the ball will both expand in proportion to the linear expansion coefficient and the temperature rise, dT. The diameters will be equal when
    4.042(1 + a2*dT) = 4.196(1 + a1*dT)
    0.154 = (8.892 - 4.616)*10^-5*dT
    = 4.276*10^-5*dT

    dT = 3602 C

    Add that to the initial temperature of 14 C and you get 3616 C

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