NEET UG PHYSICS MCQ: MOTION IN PLANE

1 Hard

1 of 1

Application +4 | -1

At what angle must the two forces

(x+y)

and

(x-y)

act, so that the resultant may be

\sqrt{x^2+y^2}

View answer and explanation

Correct option: A

\cos ^{-1}\left[-\frac{x^2+y^2}{2\left(x^2-y^2\right)}\right]

Option A (

\cos ^{-1}\left[-\frac{x^2+y^2}{2\left(x^2-y^2\right)}\right]

) is correct because it is derived from the law of cosines, solving for the angle between the two forces.

More Questions from MOTION IN PLANE

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\mathrm{\vec a}\,and\,\mathrm{\vec b}

\mathrm{\vec a}\,and\,\mathrm{\vec b}

\overrightarrow P \,\,.\,\,\overrightarrow Q = 0\,and\,\overrightarrow P \,\, \times \,\,\overrightarrow Q = 0

\overrightarrow Q \,and\,\overrightarrow R

\vec{A}=3 \hat{i}+4 \hat{j}

\vec{B}=6 \hat{i}+8 \hat{j}

A

B

\overrightarrow{\mathrm{A}}

\overrightarrow{\mathrm{B}}

{\cot ^{ - 1}}\left( 2 \right)