Question

A load suspended by a massless spring produces an extension of $x\text{cm}$ in equilibrium. When it is cut into two unequal parts, the same load produces an extension of $7.5\text{cm}$ when suspended by the larger part of length $60\text{cm}$. When it is suspended by the smaller part, the extension is $5.0\text{cm}$. Then

• A. $x=12.5\text{cm}$
• B. $x=3.0\text{cm}$
• C. The original length of the spring is $90\text{cm}$.
• D. The original length of the spring is $80\text{cm}$

Answer 1

The correct option is A $x=12.5\text{cm}$

Assume original length of spring $=l$
At equilibrium,
Weight suspended = Spring force
$mg=kx$ ... (1)
When spring is cut into two unequal parts, let their stiffness be $k_1$ and $k_2$.
$\therefore k_1\left(60\right)=k_2(l-60)=kl$ ... (2)
Now according to question
$mg=k_1\left(7.5\right)$ ... (3)
and $mg=k_2\left(5.0\right)$ ... (4)
$\therefore k_1=\frac{kl}{60},k_2=\frac{kl}{(l-60)}$
Thus, $\frac{k_1}{k_2}=\frac{5.0}{7.5}=\frac{(l-60)}{60}$ or $\frac{2}{3}=\frac{(l-60)}{60}$
$\therefore l=100\text{cm}$ ... (5)
From eqs. (1) and (3),
$kx=k_1\times 7.5$ ... (6)
From eqs. (2) and (5)
$k_1\left(60\right)=k_2(100-60)=k\times 100$
$\therefore k_1=\frac{5}{3}k$ ... (7)
From Eqs. (6) and (7)
$kx=\frac{5k}{3}\times 7.5$ or $x=12.5\text{cm}$
Hence, the correct answer is option (a).