Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Please,please help me - Newtons Laws of Motion - JEE Main

A uniform cylinder of length L and mass M having cross - sectional area A is suspended, with its length vertical,from a fixed point by a massless spring,such that it is half submerged in a liquid of density \sigma at equilibrium position. The extension x_{0} of the spring when it is in equilibrium is :

 

 

 

  • Option 1)

    \frac{Mg}{k}\left ( 1+\frac{LA\sigma }{M} \right )

    (Here k is spring constant)

  • Option 2)

    \frac{Mg}{k}

    (Here k is spring constant)

  • Option 3)

    \frac{Mg}{k}\left ( 1-\frac{LA\sigma }{M} \right )

    (Here k is spring constant)

  • Option 4)

    \frac{Mg}{k}\left ( 1-\frac{LA\sigma }{2M} \right )

    (Here k is spring constant)

 

Answers (1)

best_answer

As we have learned @2013

At equilibrium F_{net } = 0

\Rightarrow Mg = k x_0 + \sigma \left ( \frac{AL}{2} \right )g

 

\Rightarrow kx_0 = Mg - \frac{\sigma AL}{2M}


Option 1)

\frac{Mg}{k}\left ( 1+\frac{LA\sigma }{M} \right )

(Here k is spring constant)

Option 2)

\frac{Mg}{k}

(Here k is spring constant)

Option 3)

\frac{Mg}{k}\left ( 1-\frac{LA\sigma }{M} \right )

(Here k is spring constant)

Option 4)

\frac{Mg}{k}\left ( 1-\frac{LA\sigma }{2M} \right )

(Here k is spring constant)

Posted by

SudhirSol

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 Main 2024 session 2 final answer key soon at jeemain.nta.ac.in; result on April 25
anam-khan

JEE Main 2024 Session 2: Last day to challenge answer key today
anam-khan

BTech admission 2024 without JEE; top government engineering colleges, upcoming entrance exams

Latest Question