Question

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 $\left(g=9.8 \,ms ^{-2}\right.$ )

• 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)

$k=73.5 \,Nm ^{-1} \,\,$ 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)
$5 \times 9.8=k x_{3}$
$x_{3}=\frac{5 \times 9.8}{73.5}=\frac{2}{3}$