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Question: Two identical springs have the same force constant \( 73.5N{m^{ - 1}} \) . The elongation produced i...

Two identical springs have the same force constant 73.5Nm173.5N{m^{ - 1}} . The elongation produced in each spring in three cases shown in figure (a), figure (b), and figure (c). ( g=9.8ms2g = 9.8m{s^{ - 2}} ):

(A) 16m,23m,13m\dfrac{1}{6}m,\dfrac{2}{3}m,\dfrac{1}{3}m
(B) 13m,13m,13m\dfrac{1}{3}m,\dfrac{1}{3}m,\dfrac{1}{3}m
(C) 23m,13m,16m\dfrac{2}{3}m,\dfrac{1}{3}m,\dfrac{1}{6}m
(D) 13m,23m,23m\dfrac{1}{3}m,\dfrac{2}{3}m,\dfrac{2}{3}m

Explanation

Solution

Hint : Here, the spring constant is given and we have to find the elongation produced in each spring. To find elongation we have to find the resultant of each spring constant in each figure with weight hanging on the spring due to which elongation is possible.

Complete Step By Step Answer:
Here, the spring constant is 73.5Nm173.5N{m^{ - 1}} and let us denote it as kk .
Now, let us first consider the figure (a), the effective spring constant is given by:
Let effective spring constant be k1{k_1}
k1=k+k{k_1} = k + k
k1=2k=2(73.5)=147Nm1\Rightarrow {k_1} = 2k = 2(73.5) = 147N{m^{ - 1}}
In figure (a), the elongation in the spring, is given by
Let us consider elongation in figure (a) as y1{y_1}
y1=mgk=5×9.8147=13{y_1} = \dfrac{{mg}}{k} = \dfrac{{5 \times 9.8}}{{147}} = \dfrac{1}{3}
y1=13m\therefore {y_1} = \dfrac{1}{3}m
Now, in figure (b), let us find the effective spring constant of the springs given in the above figure.
Let effective spring constant be k2{k_2} , it is given by:
k2=k×kk+k=k22k=k2=73.52Nm1{k_2} = \dfrac{{k \times k}}{{k + k}} = \dfrac{{{k^2}}}{{2k}} = \dfrac{k}{2} = \dfrac{{73.5}}{2}N{m^{ - 1}}
The elongation of the spring be y2{y_2} in figure (b) as
y2=5×9.8×273.5=23{y_2} = \dfrac{{5 \times 9.8 \times 2}}{{73.5}} = \dfrac{2}{3}
y2=23m\therefore {y_2} = \dfrac{2}{3}m
Now, in figure (c), the effective spring constant is as follows:
Let k3{k_3} be the effective spring constant in figure (c),
k3=k=73.5Nm1{k_3} = k = 73.5N{m^{ - 1}}
The elongation of the spring be y3{y_3} in figure (c)
y3=5×9.873.5=23m{y_3} = \dfrac{{5 \times 9.8}}{{73.5}} = \dfrac{2}{3}m
y3=23m\therefore {y_3} = \dfrac{2}{3}m
Thus, we have concluded that the elongations in the figures with springs and given spring constant is as follows 13m,23m,23m\dfrac{1}{3}m,\dfrac{2}{3}m,\dfrac{2}{3}m respectively.
The correct answer is option D.

Note :
Here, we have been given the spring constant of springs in the above figures using that we calculated the effective spring constant and found their elongations in each figure. The elongation is calculated by using formula yE = l0 + mgk{y_E}{\text{ }} = {\text{ }}{l_0}{\text{ }} + {\text{ }}\dfrac{{mg}}{k}