JEE MAIN - Physics (2017 - 9th April Morning Slot - No. 19)
A steel rail of length 5 m and area of cross section 40cm2
is prevented from expanding along its length while the temperature rises
by 10oC. If coefficient of linear expansion and Young’s modulus of steel are 1.2×10−5 K−1 and 2×1011 Nm−2 respectively, the force developed in the rail is approximately :
2 $$ \times $$ 107 N
1 $$ \times $$ 105 N
2 $$ \times $$ 109 N
3 $$ \times $$ 10$$-$$5 N
Forklaring
Young's modulus (Y) = $${{{F \over A}} \over {{{\Delta L} \over L}}}$$
as $${{{\Delta L} \over L}}$$ = $$\alpha $$ $$\Delta $$$$\theta $$
$$\therefore\,\,\,$$ Y = $${{F \over {A\alpha \Delta \theta }}}$$
$$ \Rightarrow $$$$\,\,\,$$ F = YA$$\alpha $$$$\Delta $$$$\theta $$
= 2 $$ \times $$ 1011 $$ \times $$ 40$$ \times $$10$$-$$4 $$ \times $$ 1.2 $$ \times $$ 10$$-$$5 $$ \times $$ 10
= 9.6 $$ \times $$ 104 N
$$ \simeq $$ 1 $$ \times $$ 105 N
as $${{{\Delta L} \over L}}$$ = $$\alpha $$ $$\Delta $$$$\theta $$
$$\therefore\,\,\,$$ Y = $${{F \over {A\alpha \Delta \theta }}}$$
$$ \Rightarrow $$$$\,\,\,$$ F = YA$$\alpha $$$$\Delta $$$$\theta $$
= 2 $$ \times $$ 1011 $$ \times $$ 40$$ \times $$10$$-$$4 $$ \times $$ 1.2 $$ \times $$ 10$$-$$5 $$ \times $$ 10
= 9.6 $$ \times $$ 104 N
$$ \simeq $$ 1 $$ \times $$ 105 N