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Two identical springs have the same force constant $73.5 \,Nm ^{-1}$. The elongation produced in each spring in three cases shown in Figure-$1$, Figure-$2$ and Figure-$3$ are $\left(g=9.8 \,ms ^{-2}\right)$
$\frac{1}{6} \,m, \frac{2}{3} \,m, \frac{1}{3} \,m$
$\frac{1}{3} \,m, \frac{1}{3} \,m, \frac{1}{3} \,m$
$\frac{2}{3} \,m, \frac{1}{3} \,m, \frac{1}{6} \,m$
$\frac{1}{3} \,m, \frac{4}{3} \,m, \frac{2}{3} \,m$
Solution
(d)
$k=73.5 \,Nm ^{-1} \quad \text { Force }=5 \times 9.8$
In figure $(1)$
$5 \times 9.8=(2 k) x_1$
$\therefore x_1=\frac{5 \times 9.8}{2 \times 73.5}=\frac{1}{3}$
In figure $(2)$
$5 \times 9.8=\frac{k \times k}{k+k} \times x_2$
or $5 \times 9.8=\frac{k}{2} \times x_2$
$x_2=\frac{98}{73.5}=\frac{4}{3}$
In figure $(3)$
$5 \times 9.8=k x_3$
$x_3=\frac{5 \times 9.8}{73.5}=\frac{2}{3}$
