NEET Practice Questions (MCQs) with Answers & Solutions

Time $\propto c^x{G}^y{h}^z\Rightarrow T=k{c}^x{G}^y{h}^z$

Putting the dimensions in the above relation

$\Rightarrow$$\left[M^0{L}^0{T}^1\right]={\left[L{T}^{-1}\right]}^x{\left[M^{-1}{L}^3{T}^{-2}\right]}^y{\left[M{L}^2{T}^{-1}\right]}^z$

$\Rightarrow \left[M^0{L}^0{T}^1\right]=\left[M^{-y+z}{L}^{x+3y+2z}{T}^{-x-2y-z}\right]$

Comparing the powers of M, L and T

$-y+z=0$ …(i)

$x+3y+2z=0$ …(ii)

$-x-2y-z=1$ …(iii)

On solving equations (i) and (ii) and (iii)

$x=\frac{-5}{2},y=z=\frac{1}{2}$

Hence dimension of time are $\left[G^{1/2}{h}^{1/2}{c}^{-5/2}\right]$ 

Let radius of gyration $\left[k\right]\propto {\left[h\right]}^x{\left[c\right]}^y{\left[G\right]}^z$

By substituting the dimension of $\left[k\right]=\left[L\right]$

$\left[h\right]=\left[M{L}^2{T}^{-1}\right],\left[c\right]=\left[L{T}^{-1}\right],\left[G\right]=\left[M^{-1}{L}^3{T}^{-2}\right]$

and by comparing the power of both sides

we can get $x=1/2,y=-3/2,z=1/2$

So dimension of radius of gyration are ${\left[h\right]}^{1/2}{\left[c\right]}^{-3/2}{\left[G\right]}^{1/2}$ 

$$\begin{gathered} \left(4\right) \mathrm{C}=\frac{\mathrm{Q}}{\mathrm{V}} \\ \mathrm{C}=\frac{\mathrm{Q}}{\mathrm{W}/\mathrm{Q}} = \frac{{\mathrm{Q}}^2}{\mathrm{W}} \\ \left[\mathrm{C}\right]={\mathrm{M}}^{-1} {\mathrm{L}}^{-2} {\mathrm{T}}^4 {\mathrm{A}}^2 \\ \mathrm{F}=\mathrm{ilB} \\ \left[\mathrm{B}\right]={\mathrm{MA}}^{-1} {\mathrm{T}}^{-2} \\ \mathrm{Y}=\frac{\mathrm{X}}{3{\mathrm{Z}}^2}=\frac{{\mathrm{M}}^{-1}{\mathrm{L}}^{-2}{\mathrm{T}}^4{\mathrm{A}}^2}{{\left[{\mathrm{MT}}^{-2}{\mathrm{A}}^{-1}\right]}^2}=\left[{\mathrm{M}}^{-3}{\mathrm{L}}^{-2}{\mathrm{T}}^8{\mathrm{A}}^4\right] \end{gathered}$$

In given equation, $\frac{\alpha z}{k\theta }$ should be dimensionless

∴ $\alpha =\frac{k\theta }{z}\Rightarrow \left[\alpha \right]=\frac{[M{L}^2{T}^{-2}{K}^{-1}\times K]}{\left[L\right]}=\left[ML{T}^{-2}\right]$

and $P=\frac{\alpha }{\beta }\Rightarrow \left[\beta \right]=\left[\frac{\alpha }{p}\right]=\frac{\left[ML{T}^{-2}\right]}{\left[M{L}^{-1}{T}^{-2}\right]}=\left[M^0{L}^2{T}^0\right]$. 

$\nu =\frac{P}{2l}{\left[\frac{{\displaystyle F}}{{\displaystyle m}}\right]}^{1/2}$$\Rightarrow {\nu }^2=\frac{P^2}{4l^2}\left[\frac{{\displaystyle F}}{{\displaystyle m}}\right]\therefore m\propto \frac{F}{l^2{\nu }^2}$

$\Rightarrow \left[m\right]=\left[\frac{ML{T}^{-2}}{L^2{T}^{-2}}\right]=\left[M{L}^{-1}{T}^0\right]$

(i) G, (ii) H, (iii) I, (iv) B, (v) C

$\because$ Density, $\rho =\frac{M}{V}=\frac{M}{\pi r^{2}L}$

$\Rightarrow \frac{\Delta \rho }{\rho }=\frac{\Delta M}{M}+2\frac{\Delta r}{r}+\frac{\Delta L}{L}$

$=\frac{0.003}{0.3}+2\times \frac{0.005}{0.5}+\frac{0.06}{6}$

$=0.01+0.02+0.01=0.04$

∴ Percentage error $=\frac{\Delta \rho }{\rho }\times 100=0.04\times 100=4\%$ 

Significant figure=$\frac{97.52}{2.54}$=38.4

Light year and wavelength both represents the distance, so both has dimension of length not of time.

Light year measures distance and year measures time. One light year is the distance traveled by light in one year.