Q204 / 225 · JEE Main 2003 · Physics

A spring of spring constant $5 \times 10^3 \mathrm{~N} / \mathrm{m}$ is stretched initially by $5 \mathrm{~cm}$ from the unstretched position. Then the work required to stretch it further by another $5 \mathrm{~cm}$ is

  • A
    $12.50 \mathrm{~N}-\mathrm{m}$
  • Correct answer
    B
    $18.75 \mathrm{~N}-\mathrm{m}$
  • C
    $25.00 \mathrm{~N}-\mathrm{m}$
  • D
    $6.25 \mathrm{~N}-\mathrm{m}$

Solution

Required work done $=\frac{1}{2} \mathrm{~K}\left(\mathrm{x}_2^2-\mathrm{x}_1^2\right)=\frac{1}{2} \times 5 \times 10^3\left[10^2-5^2\right] \times 10^{-4}$ $=\frac{1}{2} \times 5 \times 75 \times 10^3 \times 10^{-4}=18.75$

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

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