A particle of mass 0.1 kg is subjected to a force which varies with distance as shown in figure. It starts its journey from rest at $\mathrm{x}=0$ , then, its velocity at$\mathrm{x}=12\mathrm{m}$  is

Complete Solution:

Work done = Area under the F-x graph

=12×10×4+10×4+12×10×4=80J

Now, work done = increase in kinetic energy. If v is the velocity at x = 12 m, then increase in K.E. =12mv².

1/2mv²=80

v²=80×2m

=80×20.1=1600

v=40ms⁻¹

Hence, the correct choice is (d).