A velocity V = 1/4 m/s is resolved into two components along OA and OB making angles 30^ and 45^ respectively with the given velocity. Then the compon - Mathematics MCQ AIEEE 2004 2026 | ExamTest.live

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A velocity V = \frac{1}{4} m/s is resolved into two components along O A and OB making angles 3 0^{\circ} and 4 5^{\circ} respectively with the given velocity. Then the component along OB is

Solution & Step-by-step Explanation

Let the components be V_{O A} and V_{OB} . By the law of sines in the velocity triangle: \frac{V _{OB}}{s i n 3 0 ^{\circ}} = \frac{V}{s i n ( 3 0 ^{\circ} + 4 5 ^{\circ} )} . V_{OB} = \frac{V s i n 3 0 ^{\circ}}{s i n 7 5 ^{\circ}} = \frac{( 1/4 ) ( 1/2 )}{s i n ( 4 5 ^{\circ} + 3 0 ^{\circ} )} . sin 7 5^{\circ} = sin 4 5^{\circ} cos 3 0^{\circ} + cos 4 5^{\circ} sin 3 0^{\circ} = \frac{1}{2} \frac{3}{2} + \frac{1}{2} \frac{1}{2} = \frac{3 + 1}{2 2} . V_{OB} = \frac{1/8}{( 3 + 1 ) /2 2} = \frac{2 2}{8 ( 3 + 1 )} = \frac{2}{4 ( 3 + 1 )} . Rationalizing the denominator: V_{OB} = \frac{2 ( 3 - 1 )}{4 ( 3 - 1 )} = \frac{6 - 2}{8} = \frac{1}{8} (6 - 2) m/s .

Try it yourself before checking the explanation above.

A velocity V = \frac{1}{4} m/s is resolved into two components along O A and OB making angles 3 0^{\circ} and 4 5^{\circ} respectively with the given velocity. Then the component along OB is

\frac{1}{8} m/s

\frac{1}{4} (3 - 1) m/s

\frac{1}{4} m/s

\frac{1}{8} (6 - 2) m/s

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