Q126 / 160 · AP EAMCET 2011 · Physics

A certain vector in the $x y$-plane has an $x$-component of $4 \mathrm{~m}$ and a $y$-component of $10 \mathrm{~m}$. It is then rotated in the $x y$-plane so that its $x$-component is doubled. Then its new $y$-component is (approximately)

  • A
    $20 \mathrm{~m}$
  • Correct answer
    B
    $7.2 \mathrm{~m}$
  • C
    $5.0 \mathrm{~m}$
  • D
    $4.5 \mathrm{~m}$

Solution

Here, $\mathbf{A}=4 \mathbf{i}+10 \mathbf{j}$ $\begin{aligned} \therefore \quad|\mathrm{A}|=\sqrt{(4)^2+(10)^2} & =\sqrt{16+100} \\ & =\sqrt{116} \mathrm{~m} \end{aligned}$ Now, according to question, So, $\begin{aligned} \mathbf{A} & =8 \mathbf{i}+n \mathbf{j} \\ |\mathbf{A}| & =\sqrt{(8)^2+n^2} \\ \sqrt{116} & =\sqrt{(8)^2+n^2} \end{aligned}$ squaring both sides $\begin{aligned} 116 & =64+n^2 \\ or \quad n^2 & =116-64=52 \\ or \quad n & =7.2 \mathrm{~m} \end{aligned}$

Hence, the correct answer is option: 'B'.

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