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

JEE Main · Physics · Mechanical Properties of Solids

String AB of unstretched length L is stretched by applying a force...

String ABAB of unstretched length LL is stretched by applying a force FF at the mid-point CC~such that the segments ACAC and BCBC make an angle~θ\theta~with ABAB as shown in the figure. The string may be considered as an elastic element with a force to elongation ratio KK. The force FF is given by

String AB of unstretched length L is stretched by applying a force F at the mid-point C ~such that the segments AC and

Options

  1. AKL(1−tanθ)sinθKL\left(1 - tan\theta\right)sin\theta
  2. B2KL(1−cosθ)tanθ2KL\left(1 - cos\theta\right)tan\theta
  3. CKL(1−cosθ)tanθKL\left(1 - cos\theta\right)tan\theta
  4. D2KL(1−sinθ)tanθ2KL\left(1 - sin \theta\right)tan\theta

Correct answer

C. KL(1−cosθ)tanθKL\left(1 - cos\theta\right)tan\theta

Step-by-step solution

Refer the diagram applying Lami's thorem,

Fsin⁡(180−2θ)=\dfrac{F}{\sin \left(180 - 2\theta\right)} =\dfrac{T}{\sin \left(90+90 +\theta\right)}

∴Fsin⁡(2θ)=Tcos⁡(θ)⇒∴ \dfrac{F}{\sin \left(2\theta\right)}=\dfrac{T}{\cos \left(\theta\right)} ⇒\dfrac{F}{2\sin \left(\theta\rright)\cos \left(\theta\right)}}=\dfrac{T}{\cos \left(\theta\right)}</p><p></p><p>∴ F = 2\sin \left(\theta\right) \timesT . . . \left(1\right)</p><p></p><p>butT = K \times∆ L = K \times \dfrac{L}{2} \times \left(\dfrac{1 - \cos \left(\theta\right)}{\cos \left(\theta\right)}\right)

put in equation~ \left(1\right),

F=2sin⁡(θ)\timesK×F = 2\sin \left(\theta\right) \timesK \times\dfrac{L}{2} \times\left(\dfrac{1 - \cos \left(\theta\right)}{\cos \left(\theta\right)}\right)

∴ F = KL \left(1 - \cos \left(\theta\right)\right)\tan \left(\theta\right)

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