JEE MAIN - Physics (2020 - 7th January Morning Slot - No. 13)
Speed of a transverse wave on a straight wire (mass 6.0 g, length 60 cm and area of cross-section 1.0 mm2) is 90 ms-1. If the Young's modulus of wire is 16 $$ \times $$ 1011 Nm-2, the extension of wire over its natural length is :
0.03 mm
0.04 mm
0.02 mm
0.01 mm
Explanation
Velocity of the wave, v = $$\sqrt {{T \over \mu }} $$
$$ \Rightarrow $$ T = v2$$\mu $$
We know, Youngs modulus,
Y = $${{{F \over A}} \over {{{\Delta l} \over l}}}$$ = $${{{T \over A}} \over {{{\Delta l} \over l}}}$$
[As here F = T]
$$ \Rightarrow $$ $${Y{{\Delta l} \over l}}$$ = $${{T \over A}}$$ = $${{{{v^2}\mu } \over A}}$$
$$ \Rightarrow $$ $$\Delta $$l = $${{{{v^2}\mu l} \over {AY}}}$$
= $${{90 \times 90 \times {{60 \times {{10}^{ - 3}}} \over {60 \times {{10}^{ - 2}}}} \times 60 \times {{10}^{ - 2}}} \over {1 \times {{10}^{ - 6}} \times 16 \times {{10}^{11}}}}$$
= 3 $$ \times $$ 10-5 m
= 0.03 mm
$$ \Rightarrow $$ T = v2$$\mu $$
We know, Youngs modulus,
Y = $${{{F \over A}} \over {{{\Delta l} \over l}}}$$ = $${{{T \over A}} \over {{{\Delta l} \over l}}}$$
[As here F = T]
$$ \Rightarrow $$ $${Y{{\Delta l} \over l}}$$ = $${{T \over A}}$$ = $${{{{v^2}\mu } \over A}}$$
$$ \Rightarrow $$ $$\Delta $$l = $${{{{v^2}\mu l} \over {AY}}}$$
= $${{90 \times 90 \times {{60 \times {{10}^{ - 3}}} \over {60 \times {{10}^{ - 2}}}} \times 60 \times {{10}^{ - 2}}} \over {1 \times {{10}^{ - 6}} \times 16 \times {{10}^{11}}}}$$
= 3 $$ \times $$ 10-5 m
= 0.03 mm
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