Question

The resultant of two forces 3P nd 2P is R.If the first force is doubled then the resultant is also doubled.The angle b/w two forces is

1)60degree
2)120 degree
3)70degree
4)180degree

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Answer 1

$\blue{Answer}$

$vectors \: be \: 3p \: and \: 2p \: \\ \\ resultant(r) = \sqrt{ {(3p)}^{2} + {(2p)}^{2} + 2 \times 3 \times 2 {(p)}^{2} \cos( \alpha ) } \\ {r}^{2} = 13 { |p| }^{2} + 12 { |p| }^{2} \cos( \alpha ) \\ now \: next \: statement \: \\ 2r = \sqrt{ {(6p)}^{2} + {(2p)}^{2} + 2 \times 6p \times 2p \cos( \alpha ) } - - (1)\\ sq. \: both \: sides \\ 4 {r}^{2} = 36 { |p| }^{2} + 4 { |p| }^{2} + 24 { |p| }^{2} \cos( \alpha ) put \: value \: of \: {r}^{2} - - - (2) \: in \: eq.(2) \\ 40 { |p| }^{2} + 24 { |p| }^{2} \cos( \alpha ) = 4(13 { |p| }^{2} + 12 { |p| }^{2} \cos( \alpha ) ) \\ 40 { |p| }^{2} + 24 { |p| }^{2} \cos( \alpha ) = 52 { |p| }^{2} + 48 { |p| }^{2} \cos( \alpha ) \\ - 12 { |p| }^{2} = 24 { |p| }^{2} \cos( \alpha ) \\ \cos( \alpha ) = \frac{ - 1}{2} \\ so \: \\ \alpha = \frac{2\pi}{3}$
hope it will be helpful for you.