JEE MAIN - Physics (2016 - 9th April Morning Slot - No. 6)
A uniformly tapering conical wire is made from a material of Young’s modulus
Y and has a normal, unextended length L. The radii, at the upper and lower ends of this conical wire, have values R and 3 R, respectively. The upper end of the wire is fixed to a rigid support and a mass M is suspended from its lower end. The equilibrium extended length, of this wire, would equal :
L $$\left( {1 + {2 \over 9}{{Mg} \over {\pi Y{R^2}}}} \right)$$
L $$\left( {1 + {1 \over 3}{{Mg} \over {\pi Y{R^2}}}} \right)$$
L $$\left( {1 + {1 \over 9}{{Mg} \over {\pi Y{R^2}}}} \right)$$
L $$\left( {1 + {2 \over 3}{{Mg} \over {\pi Y{R^2}}}} \right)$$
Explanation
_9th_April_Morning_Slot_en_6_1.png)
Here r = 3R $$-$$ $${{2R} \over L}$$ x
$$ \therefore $$ Extension in the wire of length dx,
dl = $${{Fdx} \over {AY}}$$
= $${{Mg\,dx} \over {\pi {r^2}\,Y}}$$
= $${{Mg\,dx} \over {\pi {{\left( {3R - {{2R} \over L}x} \right)}^2}Y}}$$
$$ \therefore $$ Change in wire length,
$$\Delta $$L = $$\int\limits_0^L {dl} $$
= $$\int\limits_0^L {{{Mg\,dx} \over {\pi {{\left( {3R - {{2R} \over L}x} \right)}^2}Y}}} $$
= $${{Mg} \over {\pi Y}}\int\limits_0^L {{{dx} \over {{{\left( {3R - {{2R} \over L}x} \right)}^2}}}} $$
= $${{Mg} \over {\pi Y}}\left[ { - {1 \over {\left( {3R - {{2R} \over L}x} \right)}} \times \left( { - {L \over {2R}}} \right)} \right]_0^L$$
= $${{Mg} \over {\pi Y}}$$ $$\left[ {\left( {{L \over {2{R^2}}} - {L \over {6{R^2}}}} \right)} \right]$$
= $${{Mg} \over {\pi Y}}\left( {{{2L} \over {6{R^2}}}} \right)$$
= $${{MgL} \over {3\pi {R^2}Y}}$$
$$ \therefore $$ The equilibrium extended length of the wire,
= L + $$\Delta $$L
= L + $${{MgL} \over {3\pi {R^2}Y}}$$
= L (1 + $${1 \over 3}$$ $${{Mg} \over {\pi {R^2}Y}}$$)
Comments (0)
