Question

If ${\overrightarrow{v}}_{m\omega }$ = velocity of a man relative to water $\overrightarrow{v_{\omega }}=$ velocity of water, $\overrightarrow{v_m}$ = velocity of man relative to ground, match the following.
$\begin{array}{cc}\text{Column I}& \text{Column II}\\ \text{i. Minimum distance for}{v}_{m\omega }>v_{\omega }& a.\theta =si{n}^{-1}\left(\frac{v_{m\omega }}{v_{\omega }}\right)\\ \text{ii Minimum time for}{v}_{m\omega }\ge v_{\omega }& b.\overrightarrow{v_m}\perp \overrightarrow{v_{\omega }}\\ \text{iii. Minimum distance for}{v}_{m\omega }<v_{\omega }& c.\overrightarrow{v_{m\omega }}\perp \overrightarrow{v_{\omega }}\\ \text{iv. Minimum time for}{v}_{m\omega }<v_{\omega }& d.\theta =si{n}^{-1}\frac{v_{\omega }}{v_{m\omega }}\end{array}$

• A. i- b,d ii- c iii- a iv- c
• B. i- b,d ii- a iii- a iv- c
• C. i- b,d ii- c iii- a iv- a
• D. i- b,c ii- c iii- a iv- c

Answer 1

The correct option is A i- b,d ii- c iii- a iv- c

Distance will be minimum because the man will reach from point A to point B directly.

$\text{sin}\theta =\frac{v_{\omega }}{v_{m\omega }}$ and
$\overrightarrow{v_m}\perp \overrightarrow{v_{\omega }}$
Hence $i.\to b.,d.$
Time taken is minimum if $\overrightarrow{v_{m\omega }}$ is perpendicular to $\overrightarrow{v_{\omega }}$.
Hence $ii.\to c.,iv.\to c.$
(ii) If $\overrightarrow{v_{m\omega }}<\overrightarrow{v_{\omega }}$, then drift or distance is shortest if $\text{sin}\theta =\frac{v_{m\omega }}{v_{\omega }}$. Hence $\left(iii\right)\to a$.