Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • 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)

best_answer

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

Posted by

solutionqc

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Advanced 2024: Chemical engineering cut-offs at IITs; know opening, closing ranks
anam-khan

IIT JEE Mains Result 2024 Session 2: How is overall merit list prepared?
anam-khan

JEE Mains Session 2 result 2024 soon; List of top NITs in India

Latest Question