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AIIMS 2015
Exam
AIIMS
Year
2015
Subject
Physics
Chapter
Laws of Motion

AIIMS 2015 · Physics · Laws of Motion

A mass M is tied to the top of two identical poles of height H...

A mass MM is tied to the top of two identical poles of height HH using massless strings of equal length. The mass is at height hh above the ground at equilibrium. If the distance between poles is LL the tension in each string will beA mass M is tied to the top of two identical poles of height H using massless strings of equal length. The mass is atA mass M is tied to the top of two identical poles of height H using massless strings of equal length. The mass is at

Options

  1. AMg(L2)2+(H−h)22(H−h)\frac{M g \sqrt{\left(\frac{L}{2}\right)^2+(H-h)^2}}{2(H-h)}
  2. BMgL2+H22(H−h)\frac{M g \sqrt{L^2+H^2}}{2(H-h)}
  3. CMgL2+H22H\frac{M g \sqrt{L^2+H^2}}{2 H}
  4. DMgh2+L22L\frac{M g \sqrt{h^2+L^2}}{2 L}

Correct answer

A. Mg(L2)2+(H−h)22(H−h)\frac{M g \sqrt{\left(\frac{L}{2}\right)^2+(H-h)^2}}{2(H-h)}

Step-by-step solution

At equilibrium, Mg=2Tsin⁡θM g=2 T \sin \theta A mass M is tied to the top of two identical poles of height H using massless strings of equal length. The mass is at
T=Mg2sin⁡θT=\frac{M g}{2 \sin \theta} Here, sin⁡θ=(H−h)(L/2)2+(H−h)2\sin \theta=\frac{(H-h)}{\sqrt{(L / 2)^2+(H-h)^2}} From equation (i) T=Mg(L/2)2+(H−h)22(H−h)\frac{M g \sqrt{(L / 2)^2+(H-h)^2}}{2(H-h)}
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