Question
एक नाव की शांत जल में चाल 5 km /h है ,तथा 1 km चोडी नदी को न्यूनतम पथ के अनुदिश 15 मिनट में पार करती है तो नदी के जल प्रवाह का वेग km/h होगा..?
let boat is moving in a direction making an angle θ with the perpendicular to the flow, then for minimum drift
5sinθ = r
and 5cosθ = 1/.25 = 4
so cosθ = 4/5
and sinθ = 3/5
so r = 3 km/hr
You can see video related to this question in you tube (iit brain) as the name irodov problems
Read 3 Solution.
its answer should be 0 because it is given in the ques that water is still, means water is not moving.i.e 0 speed.
But ans is 3 km/h ....
condition for which the boat will reach the straight opposite bank is D=0
,Also in the ques it is given that drifted distance is min, so D=0,
formula for time to cross the river in minimum drifted distance is,
t=d/√v square - u square, v =velocity of boat, u= velocity of river,
so, In ques it is given - t=15min = 15/60 hr =3/12hr =1/4 hr.
v=5km/h, d=width of river= 1km, put the above values in formula then uyou will get the correct ans i.e 3km/h