Question

a mass of .2 kg is attached to the lower end of the massless spring of force constt 200 N/m the upper end is fixed to a rigid support. cal T and length by which spribg is attached in quilibrium

Answer 1

Dear student
in equilibrium downwards force on block is balanced by force exerted on itby spring in upward direction so that net force on it is zero.
T=mg=k x (where t is tension in spring ,m is mass of block ,k is spring constant and x is elongation in equilibrium .
$$\begin{gathered} T=mg=kx \\ T=2\times 10=200x \\ T=20newton \\ x=\frac{20}{200}=\frac{1}{10}metre=10cm \end{gathered}$$

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