JEE Advance - Physics (2011 - Paper 1 Offline - No. 7)
Steel wire of lenght ‘L’ at 40oC is suspended from the ceiling and then a mass ‘m’ is hung from its free
end. The wire is cooled down from 40oC to 30oC to regain its original length ‘L’. The coefficient of linear
thermal expansion of the steel is 10−5 /oC, Young’s modulus of steel is 1011 N/m2 and radius of the wire is 1 mm. Assume that L >> diameter of the wire. Then the value of ‘m’ in kg is nearly
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3
Объяснение
Change in length $$\Delta L = L\,\alpha \,\Delta T$$ ..... (i)
Also $$Y = {{mgL} \over {A\Delta L}} \Rightarrow \Delta L = {{mgL} \over {YA}}$$ ..... (ii)
Equation (i) and (ii) we get ($$\because$$ $$A = \pi {r^2}$$)
$$m = {{\alpha \Delta TY \times \pi {r^2}} \over g}$$
$$ = {{({{10}^{ - 5}}) \times (10) \times ({{10}^{11}}) \times 3.14 \times {{(1 \times {{10}^{ - 3}})}^2}} \over {9.8}}$$
= 3.2 kg $$ \simeq $$ 3 kg