NEET-XII-Physics
23: Heat and Temperature
- #20An aluminium plate fixed in a horizontal position has a hole of diameter 2.000 cm. A steel sphere of diameter 2.005 cm rests on this hole. All the lengths refer to a temperature of 10 °C. The temperature of the entire system is slowly increased. At what temperature will the ball fall down? Coefficient of linear expansion of aluminium is 23 × 10-6 °C-1 and that of steel is 11 × 10-6 °C-1.Ans : Given:
Diameter of the steel sphere at temperature (T1 = 10 °C) , dst = 2.005 cm
Diameter of the aluminium sphere, dAl = 2.000 cm
Coefficient of linear expansion of steel, αst = 11 × 10`` -6`` °C`` -1``
Coefficient of linear expansion of aluminium, αAl = 23 × 10`` -``6 °C`` -``1
Let the temperature at which the ball will fall be T2 , so that change in temperature be ΔT.
d'st = 2.005(1 + αst ΔT)
`` \Rightarrow d{\text{'}}_{\,\mathrm{\,st\,}}=2.005+2.005\times 11\times {10}^{-6}\times ∆T``
`` d{\text{'}}_{\,\mathrm{\,Al\,}}=2\left(1+{\alpha }_{\,\mathrm{\,Al\,}}\times ∆T\right)``
`` \Rightarrow d{\text{'}}_{\,\mathrm{\,Al\,}}=2+2\times 23\times {10}^{-6}\times ∆T``
The steel ball will fall when both the diameters become equal.
So, d'st = d'Al
`` \Rightarrow 2.005+2.005\times 11\times {10}^{-6}∆T=2+2\times 23\times {10}^{-6}∆T``
`` \Rightarrow \left(46-22.055\right)\times {10}^{-6}∆T=0.005``
`` \Rightarrow ∆T=\frac{0.005\times {10}^{6}}{23.945}=208.81``
`` \,\mathrm{\,Now\,},∆T={T}_{2}-{T}_{1}={T}_{2}-10°\,\mathrm{\,C\,}``
`` \Rightarrow {T}_{2}=∆T+{T}_{1}=208.81+10``
`` \Rightarrow {T}_{2}=218.8\cong 219°\,\mathrm{\,C\,}``
Therefore, the temperature at which the ball will fall is 219 °C.
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