Units & Dimensions

JEE Main level, single correct, medium

9 of 149

Question 9 of 149

Electric field in a certain region is given by E⃗=(Ax2i^+By3j^)\vec{E}=\left(\dfrac{A}{{x}^{2}}\hat{i}+\dfrac{B}{{y}^{3}}\hat{j}\right). The SI unit of AA and BB are :

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Answer: (B) Nm2C−1;Nm3C−1N{m}^{2}{C}^{-1};N{m}^{3}{C}^{-1}

E⃗=Ax2i^+By3j^\vec{E}=\dfrac{A}{{x}^{2}}\hat{i}+\dfrac{B}{{y}^{3}}\hat{j}
Here, the units of Ax2\dfrac{A}{{x}^{2}} and By3\dfrac{B}{{y}^{3}} is equal to unit of electric field.
The unit of electric field is NC−1N{C}^{-1}.
So, the SI unit of AA is [Ax2]=NC−1⇒[A]=Nm2C−1\left[\dfrac{A}{{x}^{2}}\right]=N{C}^{-1}\Rightarrow \left[A\right]=N{m}^{2}{C}^{-1}
And the SI unit of BB is [By3]=NC−1⇒[B]=Nm3C−1\left[\dfrac{B}{{y}^{3}}\right]=N{C}^{-1}\Rightarrow \left[B\right]=N{m}^{3}{C}^{-1}

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