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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Mechanical Properties of Solids

NTA Abhyas 2020 · Physics · Mechanical Properties of Solids

A rigid bar of mass 15 kg is supported symmetrically by three...

A rigid bar of mass 1515 kg is supported symmetrically by three wires, each of length 2m2m. The wires at the endpoints are made of copper and the middle& one is made of steel. If the tension in each wire is the same, then the ratio of the diameter of copper wire to the diameter of steel wire is [Given, Ycopper=1.1×1011Nm−2{Y}_{copper} = 1. 1 \times {10}^{11}Nm^{- 2} and Ysteel=1.9×1011Nm−2{Y}_{steel} = 1. 9 \times {10}^{11}Nm^{- 2}]

Options

  1. A1911\sqrt{\dfrac{19}{11}}
  2. B1119\sqrt{\dfrac{11}{19}}
  3. C1119\dfrac{11}{19}
  4. D1911\dfrac{19}{11}

Correct answer

A. 1911\sqrt{\dfrac{19}{11}}

Step-by-step solution

Young's modulus of copper (Y1)=1.1×1011Nm−2( {Y}_{1} ) = 1 . 1 \times {10}^{11}Nm^{- 2}

Young's modulus of steel (Y2)=1.9×1011Nm−2( {Y}_{2} ) = 1 . 9 \times {10}^{11}Nm^{- 2}

Let d1{d}_{1} and d2{d}_{2} be the diameters of copper and steel wires respectively.

Since the tension is the same in each wire, therefore each wire has the same extension. Also, as each wire is of the same length, the strain in each of them is the same.

Y=FA∆llY = \dfrac{\dfrac{F}{A}}{\dfrac{∆ l}{l}}

Y1Y2=(d2d1)2\dfrac{Y1}{Y2} = \left(\dfrac{d2}{d1}\right)2

d1d2=Y2Y1=1911\dfrac{d1}{d2} = \sqrt{\dfrac{Y2}{Y1}} = \sqrt{\dfrac{19}{11}}

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