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How to solve this problem- - Properties of Solids and Liquids - JEE Main-2

At 40^{\circ}C, a brass wire of 1 mn radius is hung from the ceiling. A small mass, M is hung from the free end of the wire. When the wire is cooled down from 40^{\circ}C to 20^{\circ}C it regains its original length of 0.2 m. The value of M is close to :

( Coefficient of linear expansion and Young's modulus of brass are 10^{-5}/^{\circ}C and 10^{11}N/m^{2}, respectively ; g=10ms^{-2})

  • Option 1)

    9kg

  • Option 2)

    0.9kg

  • Option 3)

    1.5kg

  • Option 4)

    0.5kg

 
Answers (1)
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Mg=\left ( \frac{Ay}{l} \right )\triangle l

also we know \Rightarrow \frac{\triangle l}{l} = \alpha \triangle T

\Rightarrow Mg=Ay \alpha \triangle T

\Rightarrow M= \frac{Ay\alpha\triangle T}{g}=6.28

As 6.28 is closet to 9

hence option (1) is correct


Option 1)

9kg

Option 2)

0.9kg

Option 3)

1.5kg

Option 4)

0.5kg

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