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A spring of force constant K is cut into 2 peices such that one part is double of other. Find force constant of longer part

  • Option 1)

    \frac{2}{3} K

  • Option 2)

    \frac{3}{}2 K

  • Option 3)

    3K

  • Option 4)

    6K

 

Answers (1)

best_answer

K_{1} = \frac{(n+1)K}{2} = \frac{3}{2}K

 

Cutting of Spring -

If a spring is cut into 2 peices of length  l_{1} & l_{2} where  l_{1} = nl_{2}

Spring const of

* First part     K_{1} = \frac{K(n+1)}{n}

                     K_{2} = (n+1)K

- wherein

Formula

\frac{K_{1}}{K_{2}} = \frac{1}{n}

 

 

 


Option 1)

\frac{2}{3} K

This is incorrect

Option 2)

\frac{3}{}2 K

This is correct

Option 3)

3K

This is incorrect

Option 4)

6K

This is incorrect

Posted by

Avinash

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