Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Confused! kindly explain, - Properties of Solids and Liquids - JEE Main-2

A massless spring (k=800\:N/m) , attached with a mass (500\:g) is completely immersed in 1\:kg of water .The spring stretched  by  2\:cmand released so that it starts vibrating . What would be the order of magnitude of the change in the temperature of water when the vibrations stop completely ? ( Assume that water container and spring receive negligible heat and specific  heat of mass = 400\:J/kg\:K , specific heat of water 4184\:J/kg\:K )

 

  • Option 1)

    10^{-4}\:K

     

  • Option 2)

    10^{-5}\:K

  • Option 3)

  • Option 4)

Answers (1)

best_answer

Given

 K=800\:\:N/m

m=0.5\:\:kg

\frac{1}{2}KA^{2}=m_{1} \:s_{1}\Delta T+ m_{2} \:s_{2}\Delta T

\Rightarrow \frac{1}{2}\times 800\times \left ( \frac{2}{100} \right )^{2}=\frac{1}{2}\times 400\times\Delta T+4184\Delta T

\Rightarrow \Delta T=\frac{400\times\left ( \frac{2}{100} \right )^{2}}{200+4184}=3.64\times10^{-5}K

 

 


Option 1)

10^{-4}\:K

 

Option 2)

10^{-5}\:K

Option 3)

Option 4)

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