Q95 / 160 · JIPMER 2012 · Physics

A certain vector in the $x y$-plane has an $x$-component of 4 m and a $y$-component of 10 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 m
  • Correct answer
    B
    7.2 m
  • C
    5.0 m
  • D
    4.5 m

Solution

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

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

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