A force \(\mathrm{F}=\mathrm{kx}\) (where \(\mathrm{k}\) is positive constant) is acting on a particle. Match column-I and column-II, regarding work done in displacing the particle. $$ \begin{array}{|l|l|} \hline \text { Column - i } & \text { Column - ii } \\ \hline \text { (a) From } \mathrm{x}=-4 \text { to } \mathrm{x}=-2 & \text { (P) Positive } \\ \hline \text { (b) From } \mathrm{x}=-2 \text { to } \mathrm{x}=-4 & \text { (Q) zero } \\ \hline \text { (c) From } \mathrm{x}=-2 \text { to } \mathrm{x}=+2 & \text { (R) negative } \\ \hline \end{array} $$ (A) \(\mathrm{a}-\mathrm{R}, \mathrm{b}-\mathrm{P}, \mathrm{c}-\mathrm{Q}\) (B) \(a-P, b-Q, c-R\) (C) \(a-R, b-Q, c-P\) (D) \(\mathrm{a}-\mathrm{Q}, \mathrm{b}-\mathrm{P}, \mathrm{c}-\mathrm{R}\)

Step by step solution

01

Analyze Case (a)

In this case, the particle is moved from \(x=-4\) to \(x=-2\). As the force and displacement are in opposite directions (the force is positive when \(x<0\) according to \(F=kx\), but the particle moves in the negative direction), the work done should be negative.

02

Analyze Case (b)

The particle moves from \(x=-2\) to \(x=-4\). As the force is positive and the displacement is also in the negative direction, the work done will be positive.

03

Analyze Case (c)

In this situation, the force and the displacement both change their signs as their starting point \(x=-2\) crosses the x=0 point to \(x=2\). This means that for half of the journey, the displacement and the force are in opposite directions, resulting in negative work, and for the other half of the journey, they are in the same direction, resulting in positive work. It is essential to understand that these two effects cancel each other out, resulting in zero work being done overall.

04

Match the Columns

According to the analysis done above, we can now match the columns: (a) from \(x=-4\) to \(x=-2\), the work done is negative, so it matches with R. (b) from \(x=-2\) to \(x=-4\), the work done is positive, so it matches with P. (c) from \(x=-2\) to \(x=2\), the work done is zero, so it matches with Q. So, the correct answer is (A): \(a-R, b-P, c-Q\)

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