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(D) Given: Force constant $k = 73.5 \,Nm^{-1}$,Mass $m = 5 \,kg$,Acceleration due to gravity $g = 9.8 \,ms^{-2}$.The weight acting on the springs is $

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$\frac{1}{3} \,m, \frac{4}{3} \,m, \frac{2}{3} \,m$

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(D) Given: Force constant $k = 73.5 \,Nm^{-1}$,Mass $m = 5 \,kg$,Acceleration due to gravity $g = 9.8 \,ms^{-2}$.The weight acting on the springs is $F = mg = 5 	imes 9.8 = 49 \,N$.In Figure-$1$,the two springs are in parallel. The equivalent spring constant is $k_{eq} = k + k = 2k$.The elongation $x_1$ is given by $F = k_{eq} x_1 \implies 49 = (2 	imes 73.5) x_1 \implies x_1 = \frac{49}{147} = \frac{1}{3} \,m$.In Figure-$2$,the two springs are in series. The equivalent spring constant is $k_{eq} = \frac{k 	imes k}{k + k} = \frac{k}{2}$.The elongation $x_2$ is given by $F = k_{eq} x_2 \implies 49 = (\frac{73.5}{2}) x_2 \implies x_2 = \frac{49 	imes 2}{73.5} = \frac{98}{73.5} = \frac{4}{3} \,m$.In Figure-$3$,there is only one spring. The elongation $x_3$ is given by $F = k x_3 \implies 49 = 73.5 x_3 \implies x_3 = \frac{49}{73.5} = \frac{2}{3} \,m$.Thus,the elongations are $\frac{1}{3} \,m, \frac{4}{3} \,m, \frac{2}{3} \,m$.

Two identical springs have the same force constant $73.5 \,Nm^{-1}$. The elongation produced in each spring in the three cases shown in Figure-$1$,Figure-$2$,and Figure-$3$ are (given $g=9.8 \,ms^{-2}$):

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