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JEE Main
Exam
JEE Main
Subject
Physics
Chapter
Mechanical Properties of Solids

JEE Main · Physics · Mechanical Properties of Solids

A 5 ~m long aluminium wire (Y=7 × 10^ 10 Nm^ -2 . ) of diameter 3...

A 5 m5 \mathrm{~m} long aluminium wire \left(Y=7 \times 10^{10} \mathrm{Nm}^{-2}\right.)ofdiameter) of diameter3 \mathrm{~mm}supportsasupports a40 \mathrm{~kg}mass.Inordertohavethesameelongationinthecopperwire(Y=12×1010Nm−2)mass. In order to have the same elongation in the copper wire \left(Y=12 \times 10^{10} \mathrm{Nm}^{-2}\right) of the same length under the same weight, the diameter should now be (in mm)\mathrm{mm} )

Options

  1. A1.751.75
  2. B1.51.5
  3. C2.52.5
  4. D5.05.0

Correct answer

C. 2.52.5

Step-by-step solution

l=FLπr2Y⇒r2∝1Y(F,Ll=\frac{F L}{\pi r^{2} Y}\Rightarrow r^{2} \propto \frac{1}{Y}(F, L and ll are constant) r2r1=[Y1Y2]1/2=[7×101012×1010]1/2⇒r2=1.5×(712)1/2=1.145 mm∴\frac{r_{2}}{r_{1}}=\left[\frac{Y_{1}}{Y_{2}}\right]^{1 / 2}=\left[\frac{7 \times 10^{10}}{12 \times 10^{10}}\right]^{1 / 2}\Rightarrow \quad r_{2}=1.5 \times\left(\frac{7}{12}\right)^{1 / 2}=1.145 \mathrm{~mm}\therefore diameter =2.29 mm=2.29 \mathrm{~mm}.
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