A) Length =100 cm. Diameter = 1 mm
B) Length = 50 cm, Diameter = 0.5 mm
C) Length = 200 cm, Diameter = 2 mm
D) Length = 300 cm. Diameter = 3 mm
Correct Answer: B
Solution :
Young's modulus \[(Y)=\frac{F.l}{A\Delta l}\] \[\therefore \] \[\Delta l=\frac{F.l}{YA}=\frac{Fl}{Y(\pi {{D}^{2}}/4)}=K\frac{l}{{{D}^{2}}}\] \[\Delta l\propto \frac{l}{{{D}^{2}}}\] For first wire \[\left( \frac{l}{{{D}^{2}}} \right)=\frac{100}{1\times {{10}^{-2}}}=1\times {{10}^{4}}\] For second wire \[\left( \frac{l}{{{D}^{2}}} \right)=\frac{50}{25\times {{10}^{-4}}}=2\times {{10}^{4}}\] For third wire \[\left( \frac{l}{{{D}^{2}}} \right)=\frac{200}{4\times {{10}^{-2}}}=5\times {{10}^{3}}\] For fourth wire \[\left( \frac{l}{{{D}^{2}}} \right)=\frac{300}{9\times {{10}^{-2}}}=\frac{1}{3}\times {{10}^{4}}\] \[=3.33\times {{10}^{3}}\] As \[\left( \frac{l}{{{D}^{2}}} \right)\]is maximum for second wire, therefore increase in its length will be maximum.
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