Q92 / 125 · BITSAT 2010 · Physics

The potential energy for a force field $\overrightarrow{\mathrm{F}}$ is given by $U(x, y)=\cos (x+y)$. The force acting on a particle at position given by coordinates $(0, \pi / 4)$ is $-$

  • A
    $-\frac{1}{\sqrt{2}}(\hat{\mathrm{i}}+\hat{\mathrm{j}})$
  • Correct answer
    B
    $\frac{1}{\sqrt{2}}(\hat{i}+\hat{j})$
  • C
    $\left(\frac{1}{2} \hat{\mathrm{i}}+\frac{\sqrt{3}}{2} \hat{\mathrm{j}}\right)$
  • D
    $\left(\frac{1}{2} \hat{\mathrm{i}}-\frac{\sqrt{3}}{2} \hat{\mathrm{j}}\right)$

Solution

$\quad \mathrm{F}_{\mathrm{x}}=-\frac{\partial \mathrm{U}}{\partial \mathrm{x}}=\sin (\mathrm{x}+\mathrm{y})$ $$ \begin{array}{l} \mathrm{F}_{\mathrm{y}}=-\frac{\partial \mathrm{U}}{\partial \mathrm{x}}=\sin (\mathrm{x}+\mathrm{y}) \\ \left.\mathrm{F}_{\mathrm{x}}=\sin (\mathrm{x}+\mathrm{y})\right]_{(0, \pi / 4)}=\frac{1}{\sqrt{2}} \\ \left.\mathrm{~F}_{\mathrm{y}}=\sin (\mathrm{x}+\mathrm{y})\right]_{(0, \pi / 4)}=\frac{1}{\sqrt{2}} \\ \therefore \mathrm{F}=\frac{1}{\sqrt{2}}[\hat{\mathrm{i}}+\hat{\mathrm{j}}] \end{array} $$

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

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