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Grade 12th passGeneral Physics

Question no. 22 solution please . How i get the answer

Question image for Question no. 22 solution please . How i get the a
Profile image of dilip kumar
8 Years agoGrade 12th pass
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3 Answers

Profile image of Eshan
ApprovedApproved Tutor Answer8 Years ago
Dear student,

Since the vectors form a right angled triangle,Bsin\theta=R=\dfrac{B}{\sqrt{2}}

\implies sin\theta=\dfrac{1}{\sqrt{2}}
\implies \theta=45^{\circ}
Profile image of prateek
8 Years ago
We have the vertical component of vector B = B sinθ
The vertica component will be equal to the resultant vector.
 B Sinθ = B/√2
we get   
Sinθ= 1/√2 
Angle b/w vectors is 
 
Profile image of prateek
8 Years ago
 
 
We have the vertical component of vector B = B sinθ
The vertica component will be equal to the resultant vector.
 B Sinθ = B/√2
we get   
Sinθ= 1/√2 
Angle b/w vectors is 45°