MHT CET · Physics · Mathematics in Physics
Let the two forces have equal magnitude ' \(A^{\prime}\). If the magnitude of the resultant is \(\frac{2 \mathrm{~A}}{3}\) then the angle between those two forces is
- A \(\cos ^{-1}\left(+\frac{7}{9}\right)\)
- B \(\cos ^{-1}\left(-\frac{7}{9}\right)\)
- C \(\cos ^{-1}\left(-\frac{5}{9}\right)\)
- D \(\cos ^{-1}\left(+\frac{5}{9}\right)\)
Answer & Solution
Correct Answer
(B) \(\cos ^{-1}\left(-\frac{7}{9}\right)\)
Step-by-step Solution
Detailed explanation
rexultant \(=2 A / 3\)
\(2 A / 3=\sqrt{A^{2}+A^{2}+2 A A \cos \phi}\)
\(\frac{4 A^{2}}{9}=2 A^{2}+2 A^{2} \cos \phi\)
\(\frac{4 A^{2}-18 A^{2}}{9}=2 A^{2} \cos \phi\)
\(\cos \phi=\frac{-14}{9 \times 2} \Rightarrow \phi=\) \(\cos ^{-1}\)(-7 / 9)
\(2 A / 3=\sqrt{A^{2}+A^{2}+2 A A \cos \phi}\)
\(\frac{4 A^{2}}{9}=2 A^{2}+2 A^{2} \cos \phi\)
\(\frac{4 A^{2}-18 A^{2}}{9}=2 A^{2} \cos \phi\)
\(\cos \phi=\frac{-14}{9 \times 2} \Rightarrow \phi=\) \(\cos ^{-1}\)(-7 / 9)
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