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TS EAMCET 14-Sep-2020 Shift 2 Solved Paper

Section: Physics

Question : 86 of 160

Marks: +1, -0

\vec{A}_1 + \vec{A}_2 = \vec{5A_3}

\vec{A}_1 - \vec{A}_2 = \vec{3A_3}

\vec{A}_3 = \hat{2i} + \hat{4j}

\frac{|\vec{A}_1|}{|\vec{A}_2|}

Solution:

Given that,

A_1 + A_2 = 5 A_3

A_1 - A_2 = 3 A_3

On adding Eqs. (i) and (ii), we get

2 A_1 = 8 A_3 \Rightarrow A_1 = 4 A_3

A_1 = 4(2\hat{i} + 4\hat{j}) [ \because A_3 = 2\hat{i} + 4\hat{j} (\text{given}) ]

A_1 = 8\hat{i} + 16\hat{j}

On subtracting Eqs. (ii) from (i), we get

2 A_2 = 2 A_3 \Rightarrow A_2 = A_3 = 2\hat{i} + 4\hat{j}

\frac{|A_1|}{|A_2|} = \frac{|8\hat{i} + 16\hat{j}|}{|2\hat{i} + 4\hat{j}|} = \frac{\sqrt{(8)^2 + (16)^2}}{\sqrt{(2)^2 + (4)^2}}

= \frac{\sqrt{64+256}}{\sqrt{4+16}} = \frac{\sqrt{320}}{\sqrt{20}} = \sqrt{16} = 4

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