Correct Answer - B
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` tan beta = (Q sin theta)/( P + Q cos theta) and `
` R^2= P^2 + Q P Q cos theta` ….(i)
When ` (Q) is doubled, resultant ` vec R_1` is perpendicular to ` vec P` . Fig. 2 (CF). 51 (b). The
` R_1^2 = P^2 + 4 Q^2 + 4 PQ cos thea` ....(ii)
From right angled triangle ` BAD_1`, we have
` R-1^2 = ( 2 Q)^2 - P^2 = 4 Q^2 -P^2`
Putting this value in (ii), we get
` Q^@ + 2 - P^2 =P^2 + 4 Q^2 + 4 PQ cos theta` ltbRgt or ` P^2 + 2 PQ cos theta =0`
Putting this value in (ui), we get
` R^2 = (P^2 +2 PQ cos theta =0`
Putting this value in (i), we get .
` R^1 = (P^2 + 2 PQ cos theta) + Q^2 =0 + Q^2 =Q^2 ` or ` R=Q`.