Q103 / 150 · MHT CET 2026 (20 April Shift 1) · Physics
A boat crosses a river from one bank A to another bank B which is opposite. The distance between them is D. The speed of water is $V_W$ and that of boat relative to water is $V_B$. If $V_B = 2V_W$, the time taken by the boat to cross the river directly along AB is $\left(\sin 30^\circ = \dfrac{1}{2}\right)$, $\left(\cos 30^\circ = \dfrac{\sqrt{3}}{2}\right)$
- A$\dfrac{D}{V_B\sqrt{2}}$
- B$\dfrac{D\sqrt{2}}{V_B}$
- Correct answerC$\dfrac{2D}{V_B\sqrt{3}}$
- D$\dfrac{\sqrt{3}D}{2V_B}$
Solution
Let the boat be steered at an angle $\theta$ upstream with respect to the line AB to cross the river directly.
For the boat to move exactly along AB, the component of its velocity parallel to the river flow must cancel the river's velocity $V_W$.
$V_B \sin \theta = V_W$
Given $V_B = 2V_W$, substituting this yields:
$2V_W \sin \theta = V_W \Rightarrow \sin \theta = \dfrac{1}{2} \Rightarrow \theta = 30^\circ$
The net velocity of the boat along the line AB is the component of $V_B$ perpendicular to the river flow:
$V = V_B \cos \theta = V_B \cos 30^\circ = V_B \left(\dfrac{\sqrt{3}}{2}\right)$
The time taken to cross the river of width $D$ is:
$t = \dfrac{D}{V} = \dfrac{D}{V_B \dfrac{\sqrt{3}}{2}} = \dfrac{2D}{V_B \sqrt{3}}$
Answer: $\dfrac{2D}{V_B\sqrt{3}}$
For the boat to move exactly along AB, the component of its velocity parallel to the river flow must cancel the river's velocity $V_W$.
$V_B \sin \theta = V_W$
Given $V_B = 2V_W$, substituting this yields:
$2V_W \sin \theta = V_W \Rightarrow \sin \theta = \dfrac{1}{2} \Rightarrow \theta = 30^\circ$
The net velocity of the boat along the line AB is the component of $V_B$ perpendicular to the river flow:
$V = V_B \cos \theta = V_B \cos 30^\circ = V_B \left(\dfrac{\sqrt{3}}{2}\right)$
The time taken to cross the river of width $D$ is:
$t = \dfrac{D}{V} = \dfrac{D}{V_B \dfrac{\sqrt{3}}{2}} = \dfrac{2D}{V_B \sqrt{3}}$
Answer: $\dfrac{2D}{V_B\sqrt{3}}$
Hence, the correct answer is option: 'C'.