A) 0.5 cm
B) 2cm
C) 4cm
D) 1.5cm
Correct Answer: C
Solution :
[c] Young's modulus of elasticity is \[Y=\frac{F/A}{\Delta L/L}\] \[\therefore \Delta L=\frac{FL}{AY}\] So, \[\Delta L\propto \frac{L}{A}\] \[\therefore \,\,\frac{\Delta {{L}_{2}}}{\Delta {{L}_{1}}}=\frac{{{L}_{2}}}{{{L}_{1}}}\times \frac{{{A}_{1}}}{{{A}_{2}}}=\frac{2}{1}\times \frac{2}{1}=4\] \[\Delta {{L}_{2}}=4\times \Delta {{L}_{1}}=4\times 1=4cm\]
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