Physics MCQs for NEET — Practice Questions with Answers

$\frac{1}{\sqrt{{\epsilon }_0{\mu }_0}}=C=$ velocity of light

According to problem muscle × speed = power

$\therefore$ muscle = $\frac{\text{power }}{\text{speed}}=\frac{M{L}^2{T}^{-3}}{L{T}^{-1}}$ = $ML{T}^{-2}$

Wave number = $\frac{1}{\lambda }$

$\therefore$dimension is $\left[M^0{L}^{-1}{T}^0\right]$

[Pressure] =[stress] = $\left[M{L}^{-1}{T}^{-2}\right]$

$M{L}^{-1}{T}^{-2}$

$F=\frac{{\mu }_0}{4\pi }\frac{2I_1{I}_{2}l}{r}$ $\Rightarrow {\mu }_0=\left[F\right]{\left[A\right]}^{-2}=\left[ML{T}^{-2}{A}^{-2}\right]$

$\phi =BA=\frac{F}{I\times L}A=\frac{\left[ML{T}^{-2}\right]\left[L^2\right]}{\left[A\right]\left[L\right]}=\left[M{L}^2{T}^{-2}{A}^{-1}\right]$

By substituting the dimension of given quantities ${\left[M{L}^{-1}{T}^{-2}\right]}^x{\left[M{T}^{-3}\right]}^y{\left[L{T}^{-1}\right]}^z={\left[MLT\right]}^0$

$$\begin{gathered} \mathrm{x}+\mathrm{y}=0 - \left(\mathrm{i}\right) \\ -\mathrm{x}+\mathrm{z}=0 -\left(\mathrm{ii}\right) \\ -2\mathrm{x}-3\mathrm{y}-\mathrm{z}=0 -\left(\mathrm{iii}\right) \\ \mathrm{We} \mathrm{will} \mathrm{try} \mathrm{to} \mathrm{express} \mathrm{y} \mathrm{and} \mathrm{z} \mathrm{in} \mathrm{terms} \mathrm{of} \mathrm{x} \& \mathrm{substitute} \mathrm{in} \left(\mathrm{iii}\right) \\ \mathrm{From} \left(\mathrm{i}\right) \\ \mathrm{y} = -\mathrm{x} \\ \mathrm{From} \left(\mathrm{ii}\right) \\ \mathrm{z} = \mathrm{x} \\ \mathrm{Also} \mathrm{comparing} \mathrm{these} 2 \mathrm{euqation} \\ \mathrm{z}=-\mathrm{y} \\ \mathrm{Putting} \mathrm{in} \left(\mathrm{iii}\right) \\ -2\mathrm{x}-3\left(-\mathrm{x}\right)-\mathrm{x}=0 \\ 3\mathrm{x}-3\mathrm{x}=0 \\ 0=0 \end{gathered}$$

As you can see, in equation (iii) everything is cancelling. We need one more equation to find values. But if one variable is given, others can be found.

$$\begin{gathered} \mathrm{Ex}: \mathrm{If} \mathrm{x}=1 \\ \mathrm{Then} \mathrm{z}=1 \mathrm{and} \mathrm{y}=-1 \\ \mathrm{So} \mathrm{in} \mathrm{Question}, \mathrm{if} \mathrm{options} \mathrm{are}, \mathrm{for} \mathrm{values} \mathrm{of} \mathrm{x}, \mathrm{y} \mathrm{and} \mathrm{z} \\ \left(1\right) 1, 1, -1 \\ \left(2\right) 1, -1, 1 \\ \left(3\right) -1, 1, 1 \\ \left(4\right) 1, 1, 1 \\ \mathrm{Then} \mathrm{option} \left(2\right) \mathrm{is} \mathrm{correct}. \end{gathered}$$

$E=\frac{1}{2}L{i}^2$ hence $L=\left[M{L}^2{T}^{-2}{A}^{-2}\right]$

Strain is dimensionless.