Q135 / 160 · TS EAMCET 2016 · Physics

An elastic spring of wnstretched length $L$ and force constant $k$ is stretched by a small length $x$. It is further stretched by another small length $y$. Work done dwring the second stretching is

  • A
    $\frac{k y}{2}(x+2 y)$
  • B
    $\frac{k}{2}(2 x+y)$
  • C
    ky $(x+2 y)$
  • Correct answer
    D
    $\frac{k y}{2}(2 x+y)$

Solution

According to the question, In the first condition, $$ W_1=\frac{1}{2} k x^2 $$ In the second condition, $$ W_2=\frac{1}{2} k(x+y)^2 $$ Work for $x$ to $y$ expansion $$ \begin{aligned} & =\frac{1}{2} k(x+y)^2-\frac{1}{2} k x^2 \\ & =\frac{1}{2} k\left[x^2+y^2+2 x y-x^2\right] \\ & =\frac{1}{2} k\left[y^2+2 x y\right] \\ & =\frac{1}{2} k y[y+2 x] \end{aligned} $$

Hence, the correct answer is option: 'D'.

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