Q87 / 100 · BITSAT 2023 (Memory Based Paper 1) · Physics
Electric field in a certain region is given by $\vec{E}=\left(\frac{A}{x^2} \hat{i}+\frac{B}{y^3} \hat{j}\right)$. The SI unit of $A$ and $B$ are:
- A$\mathrm{Nm}^3 \mathrm{C}^{-1} ; \mathrm{Nm}^2 \mathrm{C}^{-1}$
- Correct answerB$\mathrm{Nm}^2 \mathrm{C}^{-1} ; \mathrm{Nm}^3 \mathrm{C}^{-1}$
- C$\mathrm{Nm}^3 \mathrm{C} ; \mathrm{Nm}^2 \mathrm{C}$
- D$\mathrm{Nm}^2 \mathrm{C} ; \mathrm{Nm}^3 \mathrm{C}$
Solution
Electric field in a certain region is given by,
$\vec{E}=\frac{A}{x^2} \hat{i}+\frac{B}{y^3} \hat{j}$
$\left[\frac{\mathrm{A}}{\mathrm{x}^2}\right]=\mathrm{NC}^{-1} \Rightarrow[\mathrm{A}]=\mathrm{Nm}^2 \mathrm{C}^{-1}$
$\left[\frac{\mathrm{B}}{\mathrm{y}^3}\right]=\mathrm{NC}^{-1} \Rightarrow[\mathrm{B}]=\mathrm{Nm}^3 \mathrm{C}^{-1}$
$\vec{E}=\frac{A}{x^2} \hat{i}+\frac{B}{y^3} \hat{j}$
$\left[\frac{\mathrm{A}}{\mathrm{x}^2}\right]=\mathrm{NC}^{-1} \Rightarrow[\mathrm{A}]=\mathrm{Nm}^2 \mathrm{C}^{-1}$
$\left[\frac{\mathrm{B}}{\mathrm{y}^3}\right]=\mathrm{NC}^{-1} \Rightarrow[\mathrm{B}]=\mathrm{Nm}^3 \mathrm{C}^{-1}$
Hence, the correct answer is option: 'B'.