question_answer

A 5 m aluminium wire \[\text{(Y=7 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{10}}}\text{N/}{{\text{m}}^{\text{2}}}\text{)}\] of diameter 3 mm supports a 40 kg mass. In order to have the same elongation in a copper wire \[\text{(Y=12 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{10}}}\text{N/}{{\text{m}}^{\text{2}}}\text{)}\] of the same length under the same weight, the diameter should be in mm

A)  1.75      

B)                         2.0

C)  2.3                        

D)         5.0

Correct Answer: C

Solution :

When strain is small, the ratio of the longitudinal stress to the corresponding longitudinal strain is called Youngs modulus \[(Y)\]of the material of the body.                 \[Y=\frac{stress}{strain}=\frac{F/A}{l/L}=\frac{F\cdot L}{\pi {{r}^{2}}l}\] Given,   \[{{Y}_{1}}=7\times {{10}^{10}}N/{{m}^{2}}\]                 \[{{Y}_{2}}=12\times {{10}^{10}}N/{{m}^{2}}\]                 \[{{r}_{1}}=\frac{{{D}_{1}}}{2}=\frac{3}{2}mm\],                 \[{{r}_{2}}=\frac{{{D}_{2}}}{2}\] \[\therefore \]  \[\frac{{{Y}_{2}}}{{{Y}_{1}}}={{\left( \frac{{{D}_{1}}}{{{D}_{2}}} \right)}^{2}}\]                 \[\frac{12\times {{10}^{10}}}{7\times {{10}^{10}}}={{\left( \frac{3}{{{D}_{2}}} \right)}^{2}}\] \[\Rightarrow \]               \[\frac{3}{{{D}_{2}}}=\sqrt{\frac{12}{7}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,{{D}_{2}}=3\sqrt{\frac{7}{12}}\,\approx \,2.3\,mm\]