60. A force time graph for a linear motion is shown in figure. The segment shown are circular. The  linear momentum gained between 0 and 8 second .


$$\begin{gathered} \left(A\right) -2\mathrm{\pi } \mathrm{N}-\mathrm{s} \left(\mathrm{B}\right) \mathrm{Zero} \mathrm{N}-\mathrm{s} \\ \left(\mathrm{C}\right) +4\mathrm{\pi N}-\mathrm{s} \left(\mathrm{D}\right) +6\mathrm{\pi } \mathrm{N}-\mathrm{s} \end{gathered}$$

Dear Student,


For a given Force-Time graph the net momentum acquired by the particle can be obtained by estimating the total area under the graph.

Let's divide the graph into three sections as shown




Each of the section is a circular section of radius 2 units. The net area for the graph is

Area under section II - Area under sections I and III and that is obtained as zero.

Thus the net momentum acquired by the particle is zero for the given time interval.

Option B is the right answer





Hope the explanation was clear enough for you.




Regards