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Question: Block is at rest in equilibrium. Now suddenly support is removed so that upper end of spring is only...

Block is at rest in equilibrium. Now suddenly support is removed so that upper end of spring is only connected with one end of spring having length 88cm88\,cm , If acceleration of block and tension ii string just after removing the support is aa and TT ,
then-

(A) a=Zero,T=14a = Zero,\,T = 14
(B) a=3m/s2,T=14a = 3\,m/{s^2},\,T = 14
(C) a=3m/s2,T=0a = 3\,m/{s^2},\,T = 0
(D) T=20N,a=ZeroT = 20\,N,\,a = Zero

Explanation

Solution

The spring is the elastic object that does not have deformation under a certain limit. It can withstand certain force and after releasing return back to its original position. Use the formula of the restoring force and calculate it. Apply this value in the formula of the tension, to find the value of tension of spring.
Useful formula:
The formula of the restoring force of the spring is given by
F=kxF = kx
Where FF is the restoring force of the spring, kk is the spring tension and the xx is the displacement of the spring.
(2) The tension of the string is given by
T=WFT = W - F
Where TT is the tension of the string and WW is the weight of the block suspended.

Complete step by step answer
It is given that the
Sum of the length of the spring and the string, L=88cmL = 88\,cm
Length of the string, l=85cml = 85\,cm
Tension when the support is connected, K=200Nm1K = 200\,N{m^{ - 1}}
Mass of the block, m=2Kgm = 2\,Kg
The length of the spring is obtained by subtracting the length of the spring and the string and the length of the string.
l=8885=3cml = 88 - 85 = 3\,cm
Since the spring does not undergo deformation, there will be no acceleration of the block. Hence the acceleration of the block is zero.
Using the formula of the restoring force,
F=kxF = kx
Substituting the known values in it,
F=200×0.03=6NF = 200 \times 0.03 = 6\,N
Using the formula of the tension,
T=WFT = W - F
Substituting the weight as the product of the mass and the acceleration due to gravity and the calculated restoring force in it.
T=mg6T = mg - 6
T=2×106=14NT = 2 \times 10 - 6 = 14\,N
Hence the tension of the string is calculated as 14N14\,N .

Thus the option (A) is correct.

Note: The spring connected at the end of the string balances the support connected in between. So when the support is suddenly removed, this change in the tension and the force is balanced by the spring itself, since it is an elastic body. So no change in velocity will take place by the block. Hence its acceleration is zero and remains equilibrium.