JEE Main · 2004 · 24 Apr 2004 · Offline · Vector Algebra

A velocity

\frac{1}{4} m / s

is resolved into two components along OA and OB making angles

3 0^{\circ}

and

4 5^{\circ}

respectively with the given velocity. Then the component along OB is

Answer: (D)

Solution

Resolving a velocity into components along two directions is a triangle-of-velocities problem, handled by the sine rule. The component along

O B

, set at

4 5^{\circ}

, is

\frac{\frac{1}{4} s i n 3 0 ^{\circ}}{s i n ( 3 0 ^{\circ} + 4 5 ^{\circ} )}

. With

sin 3 0^{\circ} = \frac{1}{2}

and

sin 7 5^{\circ} = \frac{6 + 2}{4}

, this becomes

\frac{\frac{1}{4} \cdot \frac{1}{2}}{\frac{6 + 2}{4}} = \frac{1}{2 ( 6 + 2 )} = \frac{6 - 2}{8}

m/s.

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option