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JEE Main 2019 (12 Jan, Shift 2) · Wire Combinations & Special Cases

Wire Combinations & Special Cases question from JEE Main 2019 (12 Jan, Shift 2)

Q25JEE Main 2019 · 12 JAN · SHIFT 2Medium

Wire Combinations & Special Cases

A load of mass M kg is suspended from a steel wire of length 2m and radius 1.0 mm in Searle's apparatus experiment. The increase in length produced in the wire is 4.0 mm. Now the load is fully immersed in a liquid of relative density 2. The relative density of the material of load is 8. The new value of increase in length of the steel wire is:

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Answer: (c)

  1. In liquid the tension becomes Mg−Mg - buoyancy =Mg(1−ρlρ)=Mg(1−28)=34Mg= Mg\left(1 - \frac{\rho_l}{\rho}\right) = Mg\left(1 - \frac{2}{8}\right) = \frac{3}{4}Mg
  2. Extension ∝\propto tension: Δl′=34×4.0=3.0 mm\Delta l' = \frac{3}{4} \times 4.0 = 3.0\text{ mm}
  3. So the answer is (c) 3.0 mm.
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