Measurement MCQs for NEET — Physics Questions with Answers

Using dimensional analysis

By substituting the dimensions of mass [M], length [L] and coefficient of rigidity $\eta$ as  $\left[M{L}^{-1}{T}^{-2}\right]$ we get $T=2\pi \sqrt{\frac{M}{\eta L}}$ is the right formula for time period of oscillations 

Pairs (1, 2, 3) have same dimensions

Reynolds number and coefficient of friction are dimensionless.

Latent heat and gravitational potential both have dimension $\left[L^2{T}^{-2}\right]$.

Curie and frequency of a light wave both have dimension $\left[T^{-1}\right]$. But dimensions of Planck's constant h is $\left[M{L}^2{T}^{-1}\right]$ and torque is $\left[M{L}^2{T}^{-2}\right]$

Energy = h X frequency

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