A steam boat goes across a lake and comes back (a)  on a quiet day when the water is still and (b) on a rough day when there is uniform current so as to help the journey onward and to impede the journey back. If the speed of launch on both days was same, in which case it will complete the journey in lesser time?

If the breadth of the lake is  l  and velocity of boat is vb , then time in going and coming back on a quite daytQ=lvb+lvb=2lvb …(1)Now, if va  is the velocity of air-current then time taken in going across laket1=lvb+va  (as current helps the motion)And time taken in coming backt2=lvb−va  (As current opposes the motion)So, tR=t1+t2=2lvb[1−(vavb)2] …(2)From equation (1) and (2),tRtQ=1[1−(vavb)2]>1    [as 1−va2vb2<1]i.e., tR>tQi.e., time taken to complete journey on quite day is lesser than on rough day.