Question

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

• A. $\frac{1}{6}m,\frac{2}{3}m,\frac{1}{3}m$
• B. $\frac{1}{3}m,\frac{1}{3}m,\frac{1}{3}m$
• C. $\frac{2}{3}m,\frac{1}{3}m,\frac{1}{6}m$
• D. $\frac{1}{3}m,\frac{4}{3}m,\frac{2}{3}m$

Answer 1

Answer: (D)

$k=73.5N{m}^{-1}$ Force $=5\times 9.8$
In figure (1)
$5\times 9.8=(2k)x_1$
$\therefore x_1=\frac{5\times 9.8}{2\times 73.5}=\frac{1}{3}$
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)
$5\times 9.8=k{x}_3$
$x_3=\frac{5\times 9.8}{73.5}=\frac{2}{3}$