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Q.

A cylindrical steel plug is inserted into a circular hole of diameter 2.6 cm in a brass plate. When the plug and the plates are at a temperature of 20°C, the diameter of the plug is 0.01 cm smaller than that of the hole. The temperature at which the plug will just fit in it is
$(Given, α_{steel} = \frac{11 \times 10^{- 6}}{{}^{\circ}C} and α_{brass} = \frac{19 \times 10^{- 6}}{{}^{\circ}C})$

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a

$-$ 10° C

b

$-$ 458° C

c

$-$ 48° C

d

$-$ 20° C

answer is D.

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Detailed Solution

Diameter of brass plate = 2.6cm
and diameter of steel plate = (2.6 $-$ 0.01)cm= 2.59 cm
Now, ${(d + ∆ d)}_{b} = {(d + ∆ d)}_{s}$

or $d_{b} {(1 + α ∆ θ)}_{b} = d_{s} {(1 + α ∆ θ)}_{s}$

$∴$ $2.6 (1 + 19 \times 10^{- 6} ∆ θ) = 2.59 (1 + 11 \times 10^{- 6} ∆ θ)$

$∆ θ \approx - 478^{\circ} C or θ_{f} = - 478^{\circ} + 20^{\circ} = - 458^{\circ} C$

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A cylindrical steel plug is inserted into a circular hole of diameter 2.6 cm in a brass plate. When the plug and the plates are at a temperature of 20°C, the diameter of the plug is 0.01 cm smaller than that of the hole. The temperature at which the plug will just fit in it is Given, αsteel =11×10-6C ∘ and αbrass =19×10-6C ∘