Question

Figure shows three identical springs $A,B,C$. When a $4\text{kg}$ mass is hung on $A$, it descends by $1\text{cm}$. When a $6\text{kg}$ mass is hung on $C$, it will descend by

• A. $1.5\text{cm}$
• B. $3.0\text{cm}$
• C. $4.5\text{cm}$
• D. $6.0\text{cm}$

Answer 1

The correct option is B $3.0\text{cm}$

Let $k$ be the force constant of each spring. The force $F$ needed to extend the spring $A$ by an amount $x$ is
$F=kx$ or $x=\frac{F}{k}$
In Fig (b), the springs are connected in series. Therefore, the effective force constant $k^{{}^{\prime }}$ of the combination is given by
$\frac{1}{k{}^{\prime }}=\frac{1}{k}+\frac{1}{k}$
or $k^{{}^{\prime }}=\frac{k}{2}$
If a force $F^{{}^{\prime }}$ is applied, the combination will extend by
$x^{{}^{\prime }}=\frac{F^{{}^{\prime }}}{k^{{}^{\prime }}}$
Thus, $\frac{x^{{}^{\prime }}}{x}=\frac{F^{{}^{\prime }}}{F}.\frac{k}{k^{{}^{\prime }}}=\frac{6g}{4g}\times \frac{k}{\frac{k}{2}}=3$
or $x^{{}^{\prime }}=3x=3\times 1\text{cm}=3\text{cm}$
Hence, the correct option is (b)