Question

The drift of the man along the direction of flow, when he arrives at the opposite bank is:
(A) $\dfrac{1}{{6\sqrt 3 }}km$
(B) $6\sqrt 3$
(C) $3\sqrt 3$
(D) $\dfrac{1}{{3\sqrt 3 }}km$

Answer 1

Hint: We should know that when a point which is in production originates at an early stage in the processes, then it is known as upstream. When the direction of an individual is against the stream then the direction describes the upstream. Based on this concept we can solve this problem.

Complete step by step answer
We know that we have to solve this question. We have to consider the concept of drift velocity, which comes into consideration when the position of the man is calculated.
We know that the value of t is given as:
$t = \dfrac{d}{{v\sin \theta }}$
Now we have to put the values in the expression to get:
$\dfrac{{{\text{0}}{\text{.5 km}}}}{{{\text{3sin120}}^\circ {\text{ km}}{{\text{h}}^{{\text{ - 1}}}}}}{\text{ = }}\dfrac{{\text{1}}}{{{\text{3}}\sqrt {\text{3}} }}{\text{h}}$
Now we have to find the value as:
$x = (u + v\cos \theta )t = (2 + 3\cos 120^\circ )\dfrac{1}{{3\sqrt 3 }}$
So, after evaluation we get the answer as:
= $\dfrac{1}{{6\sqrt 3 }}$
So, the drift of the man along the direction of flow, when he arrives at the opposite bank is $\dfrac{1}{{6\sqrt 3 }}$.

So, the correct answer is option A.

Note: Let us consider that the speed of a boat or any individual is u km/hr and the speed of the stream is v km/hr. In case of the downstream speed, we have to solve a question by (u + v) km/hr. On the other hand, in case of the upstream speed the question is solved as (u - v)km/hr.