Q133 / 155 · WBJEE 2016 · Physics
Work done for a certain spring when it is streched to 1mm from mean position is 10J amount of work that must be done on the spring to stretch it further by $1 \mathrm{mm}$ is
- Correct answerA$30 \mathrm{J}$
- B$40 \mathrm{J}$
- C$10 \mathrm{J}$
- D$20 \mathrm{J}$
Solution
The work done in stretching the spring, $W=\frac{1}{2} K x^{2}$
$K=$ force constant of spring
$x=$ extension in length of the spring
$$
\begin{array}{l}
W_{1}=\frac{1}{2} K x_{1}^{2}. \\
W_{2}=\frac{1}{2} K x_{2}^{2} \\
x_{1}=1 \\
x_{2}=1+1=2 \\
\frac{W_{1}}{W_{2}}=\frac{\frac{1}{2} K(1)^{2}}{\frac{1}{2} K(2)^{2}}=\frac{1}{4} \\
W_{2}=4 W_{1} \\
W_{1}=10 \mathrm{J} \\
W_{2}=40 \mathrm{J}
\end{array}
$$
$\therefore$ More work required $=40 \mathrm{J}-10 \mathrm{J}=30 \mathrm{J}$
Hence, the correct answer is option: 'A'.