MCQ
Two springs $A$ and $B(k_A = 2k_B)$ are stretched by applying forces of equal magnitudes at the four ends. If the energy stored in $A$ is $E,$ that in $B$ is:
- A$\frac{\text{E}}{2}$
- ✓$2\text{E}$
- C$\text{E}$
- D$\frac{\text{E}}{4}$
✓
Answer
Correct option: B.
$2\text{E}$
Let $x_A$ and $x_B$ be the extensions produced in springs $A$ and $B,$ respectively.
Restoring force on spring $A, F = k_Ax_A ...(1)$
Restoring force on spring $B, F = k_Bx_B ...(2)$
From $(1)$ and $(2),$ we get:
$k_Ax_A = k_Bx_B$
It is given that $k_A = 2k_B$
$\therefore\ \text{x}_\text{B}=2\text{x}_\text{A}$
Energy stored in spring $A:$
$\text{E}=\frac{1}{2}\text{k}_\text{A}\text{x}_\text{A}^2\ \dots(3)$
Energy stored in spring $B:$
$\text{E}'=\frac{1}{2}\text{k}_\text{B}\text{x}_\text{B}^2=\frac{1}{2}\Big(\frac{\text{k}_\text{A}}{2}\Big)(2\text{x}_\text{A})^2$
$\therefore\ \text{E}'=2\times\Big(\frac{1}{2}\text{k}_\text{A}\text{x}_\text{A}^2\Big)=2\text{E} [$From $(3)]$
Restoring force on spring $A, F = k_Ax_A ...(1)$
Restoring force on spring $B, F = k_Bx_B ...(2)$
From $(1)$ and $(2),$ we get:
$k_Ax_A = k_Bx_B$
It is given that $k_A = 2k_B$
$\therefore\ \text{x}_\text{B}=2\text{x}_\text{A}$
Energy stored in spring $A:$
$\text{E}=\frac{1}{2}\text{k}_\text{A}\text{x}_\text{A}^2\ \dots(3)$
Energy stored in spring $B:$
$\text{E}'=\frac{1}{2}\text{k}_\text{B}\text{x}_\text{B}^2=\frac{1}{2}\Big(\frac{\text{k}_\text{A}}{2}\Big)(2\text{x}_\text{A})^2$
$\therefore\ \text{E}'=2\times\Big(\frac{1}{2}\text{k}_\text{A}\text{x}_\text{A}^2\Big)=2\text{E} [$From $(3)]$
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