Physics M.C.Q (1 Marks)

${[F]=\left[\alpha t^2\right]=[\beta t]}$

${[\alpha]=\frac{[F]}{\left[t^2\right]} \text { and }[\beta]=\frac{[F]}{[t]}}$

$\therefore \quad[\alpha][t]=[\beta]$

$\therefore \quad \frac{\alpha t}{\beta}=\text { dimensionless }$

Strain $=\frac{\Delta l}{l}$ has dimensions $\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0\right]$

Angle measured in radians is also dimensionless $\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0\right]$

$\theta=\frac{l}{r}$

$\text { given }:(N+1) V S D=N M S D$          $........(1)$

$\text { VSD }=\left(\frac{N}{N+1}\right) M S D$

given : $(N+1) \mathrm{VSD}=N \mathrm{MSD}$      $........(2)$

From ($1$) and ($2$)

$V \cdot C=(M S D)-\frac{N}{N+1}(M S D)$

$=M S D\left(1-\frac{N}{N+1}\right)=\frac{M S D}{N+1}$

$=\frac{0.01}{N+1}=\frac{1}{100(N+1)}$

$\rho =\frac{ M }{\pi r ^2 \ell}$

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

$\frac{\Delta \rho}{\rho} \%$ $=\left[\frac{0.002}{0.4}+\frac{2(0.001)}{(0.3)}+\frac{0.02}{5}\right] \times 100 \%$

$=\frac{1}{2} \%+\frac{2}{3} \%+\frac{2}{5} \%$

$=1.6 \%$

$\therefore[ Y ]=[\text { stress }]=[\text { Pressure }]$

$\,\,\,\,\,\,\,\,\,\,\,\,\,=55.3 \times 25$

$\,\,\,\,\,\,\,\,\,\,\,\,\,=1382.5$

$\,\,\,\,\,\,\,\,\,\,\,\,\,=14 \times 10^{2}$

$G=$ Gravitational constant $=\left[\mathrm{M}^{-1} L^{3} \mathrm{~T}^{-2}\right]$

So, $\frac{E}{G}=\frac{[E]}{[G]}=\frac{M L^{2} T^{-2}}{M^{-1} L^{3} T^{-2}}=\left[M^{2} L^{-1} T^{0}\right]$

$\left[\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{2}\right] \propto\left[\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-2}\right]^{\mathrm{a}}\left[\mathrm{LT}^{-2}\right]^{\mathrm{b}}[\mathrm{T}]^{\mathrm{c}}$

$\mathrm{a}=1$

$\mathrm{a}+\mathrm{b}=2 \Rightarrow \mathrm{b}=1$

$-2 \mathrm{a}-2 \mathrm{~b}+\mathrm{c}=-2$

$\Rightarrow \mathrm{c}=2$

$\mathrm{a}=1 \mathrm{~b}=1 \mathrm{c}=2$

$\mathrm{E} \propto[\mathrm{F}][\mathrm{A}]\left[\mathrm{T}^{2}\right]$

$=\frac{1 m m}{100}=0.01 \,\mathrm{~m}=0.001\, \mathrm{~cm}$

Radius $=\mathrm{M.S} .+\mathrm{n}(\mathrm{L}-\mathrm{I})$

$=0+52(0.001)$

$=0.052 \,\mathrm{~cm}$

$t _{\text {mean }}=\frac{t_{1}+ t _{2}+ t _{3}+ t _{4}+ t _{5}}{5}$

$t_{\text {mean }}=\frac{1.25+1.24+1.27+1.21+1.28}{5}$

$t_{\text {mean }}=1.25 s$

$\Delta t_{\text {mean }}=\frac{\left|\Delta t_{1}\right|+\left|\Delta t_{2}\right|+\left|\Delta t_{3}\right|+\left|\Delta t_{4}\right|+\left|\Delta t_{5}\right|}{5}$

$\Delta t_{\text {mean }}=\frac{0+0.01+0.02+0.04+0.03}{5}=0.02 s$

Percentage error $=\frac{\Delta t_{\operatorname{mean}}}{t_{\text {mean }}} \times 100=\frac{0.02}{1.25} \times 100$

$=1.6 \%$

$=\frac{M^{1} L^{1} T^{-2}}{L^{2}}$

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

$=2.91 \times 10^{-4}$ radian

$9.99$

$-0.0099$

___________

$9.98 \quad \rightarrow 3$ siginificant flgures

$\Rightarrow 0.01 mm =\frac{\text { Pitch }}{50}$

$\Rightarrow$ Pitch $=0.5 mm$

$\frac{\Delta x}{x}=\frac{2 \Delta A}{A}+\frac{1}{2} \frac{\Delta B}{B}+\frac{1}{3} \frac{\Delta C}{C}+3 \frac{\Delta D}{D}$

$\frac{\Delta x}{x} \times 100=2(1 \%)+\frac{1}{2}(2 \%)+\frac{1}{3}(3 \%)+3(4 \%)=16 \%$

$\frac{\mathrm{J}}{\mathrm{s}}=(\mathrm{K}) \mathrm{m}^{2} \frac{\mathrm{kelvin}}{\mathrm{m}}$

$(\mathrm{K})= watt \;\mathrm{m}^{-1} \mathrm{K}^{-1}$

Least count $=1 \mathrm{MSD}-1 \mathrm{VSD}=\left[1-\frac{(\mathrm{n}-1)}{\mathrm{n}}\right] \mathrm{MSD}=\frac{1}{\mathrm{n}} \mathrm{MSD}$$=\frac{1}{\mathrm{n}}\left(\frac{1}{\mathrm{n}}\right) \mathrm{cm}=\frac{1}{\mathrm{n}^{2}} \mathrm{cm}$

$b=$ vernier scale division

$L . C .=\left(\frac{ b - a }{ b }\right) M$

$=\frac{50-49}{50} M$    $M=$ Main scale reading

$L C =\frac{ M }{50}$   $M =7.05-7.00$

$L C \cdot=\frac{0.05}{50} \quad M =0.05$

L.C. $=0.001$

Reading $=7+0.001 \times 23=7.023 cm$

On comparing both sides and solving we get

$P = \frac{1}{2},\,q = \frac{1}{2}\,and\,r =  - 2$
$\therefore \,\,l = \frac{1}{{_c2}}{\left[ {\frac{{G{e^2}}}{{4\pi {\varepsilon _0}}}} \right]^{1/2}}$
 

$P - r = 0$         .........($ii$)

${2_p} + q + 3r = 1$        ............($iii$)

$ - P - q - 2r = 0$         ................($iv$)
Solving eqns. ($ii$), ($iii$), and ($iv$), we get
$P = r = \frac{1}{2},q =  - \frac{3}{2}$
From eqn.$\left( i \right)\,l = \frac{{\sqrt {hG} }}{{_c3/2}}$

$⇒$ $dL = \tau \times dt$$ = r \times F \times dt$

i.e. the unit of angular momentum is joule-second.

If physical quantity remains constant then $n \propto 1/u$

$\therefore n_1u_1 = n_2u_2$ .

$ \Rightarrow \sigma = J{m^{ - 2}}{s^{ - 1}}{K^{ - 4}}$

It is also expressed as $JC ^{-1} m ^{-1}$.

Now by putting Sl units of all measurable quantities in above formula

$\Rightarrow \frac{ W b }{ m ^{2}}=$ $\frac{ \mu_{0}Am }{ m ^{2}}$

$\Rightarrow \mu_0 =\frac{ W b }{ A \cdot m }$

   ${N_1}{U_1} = {N_2}{U_2}$

${N_1}\left[ {{M_1}L_1^{ - 3}} \right] = {N_2}\left[ {{M_2}L_2^{ - 3}} \right]$

$\therefore $ ${N_2} = {N_1}\left[ {\frac{{{M_1}}}{{{M_2}}}} \right] \times {\left[ {\frac{{{L_1}}}{{{L_2}}}} \right]^{ - 3}}$ $ = 0.625\left[ {\frac{{1g}}{{1kg}}} \right] \times {\left[ {\frac{{1cm}}{{1m}}} \right]^{ - 3}}$

$ = 0.625 \times {10^{ - 3}} \times {10^6} = 625$

$1 kg =10^{3}$ and $1 m =10^{2} cm$

$1 N =\frac{\left(10^{3} g \right)\left(10^{2} cm \right)}{ s ^{2}}$

$1 N =\frac{100 \times(10 g ) \times 10(10 cm )}{100 \times(0.1 s )^{2}}$

$=10 \times \frac{(10 g )(10 cm )}{(0.1 s )^{2}}$

$1 N =10 \times$ New unit of force Thus, New unit $=1 / 10=0.1 N$

It can be shown that

1 a.m.u $=1.66 \times 10^{27}\,kg$.

According to Einstein, mass energy equivalence

$E = mc ^2$

Where $m =1.66 \times 10^{-27}\, kg$

$C =3 \times 10^8\, m / sec$, we get

$E =1.49 \times 10^{-10}\, J \left(1 Mev =1.6 \times 10^{-13}\, J \right)$

$E =\frac{1.49 \times 10^{-10}\, J }{1.6 \times 10^{-13}}\,Mev$

$E =931.25\,Mev$

Hence a change in mass of $1\,a. m.u$ (called mass defect) releases an energy equal to $931 \,Mev$.

$1\,amu =931\,Mev$ is used as a standard conversion.

Thus, a gradient will be unitless if its numerator has same dimensions as denominator, i.e. $x$ (which has the dimension $L$ ).

Thus, out of the options, Displacement has the dimension $L$ and hence its gradient will be dimensionless.

$(i) \ c t = \lambda (ii)\ x =  \lambda$  $(iii) \,\left[ {\frac{{2\pi c}}{\lambda }} \right] = \left[ {\frac{{2\pi x}}{{\lambda \,t}}} \right]$

${n_2} = {n_1}{\left[ {\frac{{{M_1}}}{{{M_2}}}} \right]^x}\,{\left[ {\frac{{{L_1}}}{{{L_2}}}} \right]^y}\,{\left[ {\frac{{{T_1}}}{{{T_2}}}} \right]^z}$$ = 1 \times {10^6}\,{\left[ {\frac{{1\,kg}}{{10\,kg}}} \right]^1}\,{\left[ {\frac{{1m}}{{1\,dm}}} \right]^2}\,{\left[ {\frac{{1s}}{{1\min }}} \right]^{ - 3}}$ [$\,1MW = {10^6}W$]

$ = {10^6}\left[ {\frac{{1kg}}{{10kg}}} \right]\,{\left[ {\frac{{10\,dm}}{{1\,dm}}} \right]^2}\,{\left[ {\frac{{1sec}}{{60sec}}} \right]^{ - 3}}$$ = 2.16 \times {10^{12}}$unit

$2 \mathrm{cal} / \mathrm{cm}^{2}-\mathrm{min}=\mathrm{X} \mathrm{J} / \mathrm{m}^{2}-\mathrm{sec}$

$X=2$

$\left(\frac{c a l}{J}\right)\left(\frac{\sec }{\min }\right)\left(\frac{m^{2}}{c m^{2}}\right)$

$=2(4.18)\left(\frac{1}{60}\right)(10000)=1393.33$                                                        

$\mathrm{n}_{2}=4\left[\frac{\mathrm{g}}{\mathrm{M}_{2}}\right]\left[\frac{\mathrm{cm}}{\mathrm{L}_{2}}\right]^{-3}$

$=4\left[\frac{\mathrm{g}}{100 \mathrm{g}}\right]\left[\frac{\mathrm{cm}}{10 \mathrm{cm}}\right]^{-3}$

$=40$

$\mathrm{n}_{2}=\mathrm{n}_{1}\left[\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}\right]\left[\frac{\mathrm{L}_{1}}{\mathrm{L}_{2}}\right]^{2}\left[\frac{\mathrm{T}_{1}}{\mathrm{T}_{2}}\right]^{-2}$

$\quad=1\left[\frac{\mathrm{kg}}{10 \mathrm{kg}}\right]\left[\frac{\mathrm{m}}{10 \mathrm{m}}\right]^{2}\left[\frac{10}{60}\right]^{-2}$

$\quad=\left(\frac{1}{10}\right)^{1} \times\left(\frac{1}{10}\right)^{2} \times(60)^{2}=36 \times 10^{-1}$

$=\text { acceration }$

$\mathrm{v} \rightarrow \sqrt{\mathrm{ab}} \Rightarrow \mathrm{v}^{2}=\mathrm{ab} \Rightarrow \mathrm{a}=\frac{\mathrm{v}^{2}}{\mathrm{b}}=\frac{\mathrm{v}^{2}}{\mathrm{acc}}=\frac{\frac{\mathrm{s}^{2}}{\mathrm{t}^{2}}}{\frac{\mathrm{s}}{\mathrm{t}^{2}}}=\mathrm{s}$

$=\mathrm{distance}$

$\mathrm{d}+\mathrm{t} \rightarrow \mathrm{t} \Rightarrow \mathrm{d} \rightarrow \,time$

$\mathrm{v} \rightarrow \frac{\mathrm{c}}{\mathrm{b}+\mathrm{t}}=\frac{\mathrm{c}}{\mathrm{t}} \Rightarrow \mathrm{c}=\mathrm{vt}=\frac{\mathrm{s}}{\mathrm{t}} \times \mathrm{t}=\mathrm{s} \Rightarrow \mathrm{c} \rightarrow \text { distance } $

$p=F \times t$

$p^{\prime}=2 F \times t$

$p^{\prime}=2 p$

$\frac{\text { Impulse }}{\text { Area }}=\frac{M L T^{-1}}{L^2} \Rightarrow\left[ ML ^{-1} T ^{-1}\right]$

Coefficient of viscosity $\Rightarrow \eta=\left[ ML ^{-1} T^{-1}\right]$

So, $\frac{\text { Impulse }}{\text { Area }}=$ coefficient of viscosity

$\text { Work } \rightarrow\left[ ML ^2 T ^{-2}\right]$

$n_1 V_1=n_2 V_2$

$\frac{(8) M _1 L _1^2 T _1^{-2}}{ M _2 L _2^2 T _2^{-2}}=n_2$

$\Rightarrow 8\left[\frac{ M _1}{ M _2}\right]\left[\frac{ L _1}{ L _2}\right]^2\left[\frac{ T _1}{ T _2}\right]^{-2}=n_2$

$\Rightarrow 8\left[\frac{ M _1}{2 M _1}\right]\left[\frac{ L _1}{ L _1}\right]^2\left[\frac{2 T _1}{ T _1}\right]^{-2}=n_2$

$\Rightarrow 8 \times \frac{1}{2} \times \frac{1}{4}=n_2$

$\Rightarrow n_2=1$

So, unit of 8 joule $=1 \times$ new units

$y=x^2 \cos ^2 2 \pi\left(\frac{\beta \gamma}{\alpha}\right)$

The argument of a trigonometric ratio is always dimensionless.

$\frac{\beta \gamma}{\alpha}=\left[ M ^0 L ^0 T ^0\right] \text { or } \beta \gamma=\alpha \Rightarrow \gamma=\frac{ T }{ L ^2}$

$\text { and } y=x^2 \Rightarrow\left[ L ^2\right]$

$\alpha= s ^{-1} \Rightarrow\left[ T ^{-1}\right], \beta=\left[ LT ^{-1}\right]^{-1} \Rightarrow\left[ L ^{-1} T \right]$

$y=m^2$

$\gamma= ms ^{-2}$

$⇒$ ${\mu _0}{\varepsilon _0} = \left( {\frac{1}{{{C^2}}}} \right)$ (where $C =$ velocity of light)

$[{\mu _0}{\varepsilon _0}] = {L^{ - 2}}{T^2}$

Pressure $=f / A$

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

Hiso, $Y=\frac{\text { stress }}{\text { strain }}$ (strain is dimensionliss)

$\therefore Y=f / A=\left[M L^{-1} T^{-2}\right]$

$\therefore$ All the there has seme dimenion equals $\left[ HL ^{-1} T^{-2}\right]$

${\rm{where }} [R] = [M{L^2}{T^{ - 1}}{Q^{ - 2}}]$

$\therefore $ $[\rho ] = [M{L^3}{T^{ - 1}}{Q^{ - 2}}]$

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

$\therefore$ Dimensional formula of Pascal-second is $\left[M L^{-1} T^{-1}\right]$

By the formula of coefficient of viscosity, we have

$\eta=\frac{F}{A(\Delta v / \Delta z)}$

where $F$ is force, $A$ is area and $\frac{\Delta v}{\Delta z}$ is velocity gradient.

$\therefore$ Dimensions of $\eta=\frac{\left[M L T^{-2}\right]}{\left[L^{2}\right]\left[L T^{-1} / L\right]}$

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

Hence, Pascal-second has dimensions of coefficient of viscosity.

On checking the dimensionality the correct relation is

$V=\frac{\pi P r^4}{8 \eta l}$

From the dimensional homogenity

$[x] = [Ay] = [B] \Rightarrow \left[ {\frac{x}{A}} \right] = [y] = \left[ {\frac{B}{A}} \right]$

$[Cz] = [{M^0}{L^0}{T^0}] = $Dimension less

$x$ and $B$; $C$ and ${Z^{ - 1}};y$ and $\frac{B}{A}$ have the same dimension but $x$ and $A$ have the different dimensions.

$M^{0} L^{1} T^{-1}=\left(M^{1}\right)^{\alpha}\left(M^{1} L^{-1} T^{-1}\right)^{\beta}\left(L^{1}\right)^{\gamma}\left(L T^{-2}\right)^{k}$

$0=\alpha+\beta \Rightarrow \alpha=1=k$

$1=-\beta+\gamma+k \beta=\gamma=-1$

$-1=-\beta-2 k$

$\Rightarrow V_{T} \propto \frac{m g}{\eta r}$

$=(\text { Area } \times \text { length }) \times$ density $\times \frac{\mathrm{L}}{\mathrm{T}^{2}}$

$\left[\text { velocity } \mathrm{v}^{2}=\frac{\mathrm{L}^{2}}{\mathrm{T}^{2}}\right]$

$=\left[\mathrm{Av}^{2} \mathrm{D}\right]$

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

Pressure $=\frac{\text { F orce }}{\text { Area }}$

But,Force $=$ Mass $\times$ Accelaration $=[ M ]\left[ LT ^{-2}\right]=\left[ MLT ^{-2}\right]$

Then,

$[\text { Pressure }]=\frac{\left[ MLT ^{-2}\right]}{\left[ L ^2\right]}=\left[ ML ^{-1} T ^{-2}\right]$

[Density $]=\frac{[ Mass ]}{[\text { V olume }]}=\frac{[ M ]}{\left[ L ^3\right]}= ML ^{-3}$

Now,

$[\text { Presure head }]=\frac{[\text { Pressure }]}{[\text { Density }] \times[ g ]}=\frac{\left[ ML ^{-1} T ^{-2}\right]}{\left[ ML ^{-3}\right]\left[ LT ^{-2}\right]}=\left[ M ^0 L ^1 T ^0\right]$

$k =\frac{\left(\frac{ d Q}{ dt }\right)}{ A \left(\frac{ d T }{ d x }\right)}$

$[ k ]=\frac{\left[ ML ^{2} T ^{-3}\right]}{\left[ L ^{2}\right]\left[ KL ^{-1}\right]}=\left[ MLT ^{-3} K ^{-1}\right]$

Strain is dimensionless, so

= $\frac{{{\rm{Force}}}}{{{\rm{Area}}}} = \frac{{ML{T^{ - 2}}}}{{{L^2}}} = [M{L^{ - 1}}{T^{ - 2}}]$

$=\frac{M L^{2} T^{-3} A^{-2} \times L^{2}}{L}=\left[M L^{3} T^{-3} A^{-2}\right]$

luminous flux $=$ power $=\left[M L^{2} T^{-3}\right]$

$\Rightarrow\left[\mathrm{M}^{1} \mathrm{L}^{1} \mathrm{T}^{-2}\right]=\frac{[\alpha]}{\left[\mathrm{M}^{1 / 2} \mathrm{L}^{-3 / 2}\right]}$

$[\alpha]=\left[M^{3 / 2} L^{-1 / 2} T^{-2}\right]$

we get $\sqrt {\frac{{{L^3}}}{{{M^{ - 1}}{L^3}{T^{ - 2}} \times M}}} = T$

$\Rightarrow$ dim $(a^m)$ $= L$ $\Rightarrow m = -1$, dim $[C] = L$

i.e. $\frac{{{v^2}}}{{rg}}$is dimensionless.

Gravitational force $F_1=\frac{G M_P^2}{r^2}$

Electrostatic force $F_2=\frac{1}{4 \pi \varepsilon_0} \frac{e^2}{r^2}$

$\frac{F_2}{F_1}=\frac{e^2}{4 \pi \varepsilon_0 G M_\rho^2}$

$\therefore$ Dimension less $\left[ M ^{\circ} L ^{\circ} T ^{\circ} A ^{\circ}\right]$

For $( 2):$ As $A$ and $B$ have different dimen-

sion so $\exp \left(-\frac{\mathrm{A}}{\mathrm{B}}\right)$ is meaningless.

For $(3)$ $: \mathrm{AB}^{2}$ is meaningful.

For $4)$ $: \mathrm{AB}^{-4}$ is meaningful.

$f = \frac{1}{{2\pi }}\frac{1}{{\sqrt {LC} }}$

$LC = [{T^2}]\, \frac{L}{R} = [T]$. $\left[ {{{(LC)}^2}\left( {\frac{R}{L}} \right)} \right]\, $ $= \,{[{T^2}]^2}[{T^{ - 1}}] = [{T^3}]$ .

dimension of $L=\frac{M L T^{-2}}{Q^{2}} \cdot T^{2}=M L T^{0} Q^{-2}$

$E = i ^{2} R T \Rightarrow R =\frac{ E }{ i ^{2} t }$

dimension of $R =\frac{ M L T ^{-2}}{ Q ^{2}} \cdot T ^{-1}= M L T ^{-1} Q ^{-2}$

$\left(\frac{L}{R}\right)=\frac{M L T^{0} Q^{-2}}{M L T^{-1} Q^{-2}}=[T]$

$\Rightarrow \quad\left[\frac{\mathrm{x}^{3} \mathrm{w}}{\mathrm{y}^{3} \mathrm{z}}\right]=\frac{\mathrm{ML}^{2} \mathrm{T}^{-2}}{\mathrm{AT}}=\frac{[\mathrm{Work}]}{[\mathrm{Charge}]}$

terms in a equation or terms having arithematic operations of addition and subtraction should have same dimensions.

$\Rightarrow \ln a^{2}-x^{n}$

$\operatorname{dim}\left(a^{2}\right)=\operatorname{dim}\left(x^{n}\right)$

$L^{2}=L^{n} \Rightarrow[n=2]$

$\frac{d^2 y}{d x^2}$ will have dimensions of $\frac{y}{x^2}$

$y \rightarrow$ pressure, $x \rightarrow$ velocity gradient

$x \rightarrow \frac{V}{L} \Rightarrow \frac{ LT ^{-1}}{ L } \Rightarrow T ^{-1}$

$\frac{y}{x^2}=\frac{ ML ^{-1} T ^{-2}}{ T ^{-2}} \Rightarrow\left[ ML ^{-1}\right]$

By substituting the following dimensions :

$[E] = [M{L^2}{T^{ - 2}}],\,[v] = [L{T^{ - 1}}],\,[F] = [ML{T^{ - 2}}]$

and by equating the both sides

$x = 1,\,y = - 2,\,z = 0$. So $[m] = [E{v^{ - 2}}]$

$\left[ U \right] = \frac{{[A]\,[{x^{1/2}}]}}{{[{x^2}]\, + \,[B]}}$

$⇒ [M{L^2}{T^{ - 2}}] = \frac{{[A]\,[{L^{1/2}}]}}{{[{L^2}]}}$

$\therefore\,[A] = [M{L^{7/2}}{T^{ - 2}}]$

$[AB]\, = \,[M{L^{7/2}}{T^{ - 2}}] \times [{L^2}]$ $ = [M{L^{11/2}}{T^{ - 2}}]$

Substituting the dimension of

$[F] = [ML{T^{ - 2}}],$ $[C] = [L]\;and\;[T] = [T]$

and comparing both sides, we get

$m = F{L^{ - 1}}{T^{ - 2}}$

$=\left[\frac{\text { Force }}{\text { Area }}\right]=\left[\frac{\mathrm{F}}{(\mathrm{VT})^{2}}\right]=\left[\mathrm{F}^{1} \mathrm{V}^{-2} \mathrm{T}^{-2}\right]$

$\mathrm{ML}^{2} \mathrm{T}^{-2}=\left[\mathrm{MLT}^{-1}\right]^{\mathrm{a}}\left[\mathrm{L}^{2}\right]^{\mathrm{b}}[\mathrm{T}]^{\mathrm{c}}$

$\mathrm{ML}^{2} \mathrm{T}^{-2}=\left[\mathrm{M}^{3} \mathrm{L}^{\mathrm{a}+2 \mathrm{b}} \mathrm{T}^{-\mathrm{a}+\mathrm{c}}\right]$

$a=1 \quad a+2 b=2 \quad-2=-a+c$

                 $\mathrm{b}=1 / 2 \quad \mathrm{c}=-1$

$\mathrm{E}=\left[\mathrm{PA}^{1 / 2} \mathrm{T}^{-1}\right]$

$\left[ {{M^1}{L^1}{T^{ - 2}}} \right]\,\, = \,\,{\left[ {{M^1}{L^{ - 1}}{T^2}} \right]^\alpha }\,{\left[ {L{T^{ - 1}}} \right]^\beta }{\left[ T \right]^\gamma }$

$\,\,\left[ {{M^1}{L^1}{R^{ - 2}}} \right]\,\, = \,\,\left[ {{M^\alpha }{L^{ - \alpha \, + \;\beta }}{T^{ - 2\alpha  - \beta  + \gamma }}} \right]$

$\alpha \,\, = \,\,1\,;\,\, - \alpha \,\, + \;\,\beta \,\, = \,\,1\,$

$\beta \,\, = \,\,\,2,\,\,2 - 2\alpha \,\, - \,\,\beta \,\, + \;\,\gamma \,\, =  - 2$

$\therefore \,\, - 2\,\, - \,\,2\,\, + \;\,\gamma \,\, =  - 2\,\,$

$\,\gamma \,\, = \,\,2$

$ \Rightarrow \,\,F\,\, = \,\,P{v^2}{T^2}$

$F \propto V^a \rho^b g^c$

$F=\left[L^3\right]^a\left[ ML ^{-3}\right]^b\left[ LT ^{-2}\right]^c$

$\left[ ML ^{-2}\right]=F=\left[ M ^{ b } L ^{3 a -3 b +c} T ^{-2 c}\right]$

On comparing.

$b=1,$

$-2 c=-2$

$\Rightarrow c=1$

$3 a-3 b+c=1$

$\Rightarrow 3 a-3+1=1$

$\Rightarrow 3 a-2=1$

$\Rightarrow 3 a=3 \Rightarrow a=1$

So, on putting all these values,

$F=V \rho g$

by substituting the dimension of $[T] = [T]$

$[S] = [M{T^{ - 2}}],\,[r] = [L],\,[\rho ] = [M{L^{ - 3}}]$

and by comparing the power of both the sides

$x = - 1/2,\,y = 3/2,\,z = 1/2$

so $T \propto \sqrt {\rho {r^3}/S} \Rightarrow T = k\sqrt {\frac{{\rho {r^3}}}{S}} $

Putting the dimensions in the above relation

$[M{L^2}{T^{ - 1}}] = k{[L{T^{ - 1}}]^x}{[L{T^{ - 2}}]^y}{[ML{T^{ - 2}}]^z}$

$⇒$ $[M{L^2}{T^{ - 1}}] = k[{M^z}{L^{x + y + z}}{T^{ - x - 2y - 2z}}]$

Comparing the powers of $M,\,L$ and $T$

$z = 1$ …$(i)$

$x + y + z = 2$ …$(ii)$

$ - x - 2y - 2z = - 1$ …$(iii)$

On solving $(i)$, $(ii)$ and $(iii)$ $x = 3,\,y = - 2,\,z = 1$

So dimension of $L$ in terms of $v,\,A$ and $f$
$[L] = [F{v^3}{A^{ - 2}}]$

$G=\left[E^g d^b P^c\right]$

$E=\left[M^2 T^{-2}\right]$

$d=\left[ ML ^{-3}\right]$

$P=\left[ ML ^2 T ^{-3}\right]$

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

$\left[ M ^{-1} L ^3 T ^{-2}\right]=\left[ ML ^2 T ^{-2}\right]^a\left[ ML ^{-3}\right]^b\left[ ML ^2 T ^{-3}\right]^c$

$a+b+c=-1$

$2 a-3 b+2 c=3$

$-2 a-3 c=-2 \Rightarrow 2 a+3 c=2$

On solving.

$a=-2$

$b=-1$

$c=2$

So, $G=\left[E^{-2} d^{-1} P^2\right]$

$\left[ {M{L^2}{T^{ - 2}}} \right] = {\left[ {ML{T^{ - 2}}} \right]^a}{\left[ {L{T^{ - 2}}} \right]^b}{\left[ T \right]^c}$

$\left[ {M{L^2}{T^{ - 2}}} \right] = \left[ {{M^a}{L^{a + b}}{T^{ - 2a - 2b + c}}} \right]$

$\therefore $ $a = 1$, $a + b = 2$ $⇒$ $b = 1$

and $ - 2a - 2b + c = - 2\;\; \Rightarrow \;c = 2$

$\therefore $ $E = KFA{T^2}$.

Moment of a force = Force $ \times $ Distance = $[M{L^2}{T^{ - 2}}]$

$\left[ {{n_2}} \right] = \left[ {{n_1}} \right] $= number of particles per unit volume $= [L-3]$

$[{x_2}] = [{x_1}]$= positions

$D = \frac{{[n]\;\left[ {{x_2} - {x_1}} \right]}}{{\left[ {{n_2} - {n_1}} \right]}} = \frac{{\left[ {{L^{ - 2}}{T^{ - 1}}} \right] \times [L]}}{{[{L^{ - 3}}]}}$ $ =\left[ {{L^2}{T^{ - 1}}} \right]$

Moment of Force = Force $ \times $ Perpendicular distance

= $[ML{T^{ - 2}}]\,[L]\, = \,[M{L^2}{T^{ - 2}}]$

$\left[ {{M^0}L{T^0}} \right] = {\left[ {L{T^{ - 2}}} \right]^m}{\left[ T \right]^n} = \left[ {{L^m}{T^{ - 2m + n}}} \right]$

$\therefore $ $m = 1$ and $ - 2m + n = 0$  $⇒$  $n = 2$.

Potential gradient$ = \frac{V}{x} = \frac{{[M{L^2}{T^{ - 3}}{A^{ - 1}}]}}{{[L]}}$$ = [ML{T^{ - 3}}{A^{ - 1}}]$

Energy gradient $ = \frac{E}{x} = \frac{{[M{L^2}{T^2}]}}{{[L]}} = [ML{T^{ - 2}}]$

and pressure gradient$ = \frac{P}{x} = \frac{{[M{L^{ - 1}}{T^{ - 2}}]}}{{[L]}} = [M{L^{ - 2}}{T^{ - 2}}]$

Stress $ = \frac{{{\rm{Restoring force}}}}{{{\rm{Area}}}} = M{L^{ - 1}}{T^{ - 2}}$

$Q=m^a s^b \theta^c$

$\left[ ML ^2 T ^{-2}\right]=\left[ M ^2\right]\left[ L ^{2 b} T ^{-2 b} K ^{-b}\right]\left[ K ^{ C }\right]$

$\Rightarrow a=1$,

$2 b=2 \Rightarrow b=1$

$-b+c=0$

$\Rightarrow b=c \Rightarrow c=1$

$Q=m s \Delta T$

$\Rightarrow s=\frac{Q}{m \Delta T}$

$=\frac{\text { Force }}{\text { Area }}=$ Modulus of elasticity

$\frac{\phi}{R}=a m p-\sec =$ charge

$* \frac{L}{R}=$ time constant

$\therefore \,\,{\rm{Dimension}}\,{\rm{of}}\,{\rm{[}}B{\rm{]}} = \frac{{{\rm{[}}F{\rm{]}}}}{{{\rm{[}}I{\rm{]}}\,{\rm{[}}L{\rm{]}}}} = \frac{{[ML{T^{ - 2}}]}}{{[I]\,[L]}}$=$[M{T^{ - 2}}{I^{ - 1}}]$

$[P] = \frac{{{B^2}{l^2}}}{m} = \frac{{{{[M{T^{ - 2}}{I^{ - 1}}]}^2} \times [{L^2}]}}{{[M]}}$ $ = [M{L^2}{T^{ - 4}}{I^{ - 2}}]$

$=\frac{\left[\mathrm{ML}^{2} \mathrm{T}^{-2}\right]}{\left[\mathrm{L}^{2}\right][\mathrm{T}]}=\left[\mathrm{ML}^{\circ} \mathrm{T}^{-3}\right]$

Dimension of $Y =\frac{ M L ^{-1} T ^{-2}}{ M ^{0} L ^{0} T ^{0}}$$=\left[ M L ^{-1} T ^{-2}\right]$

$F=\frac{\alpha-t^2}{\beta v^2}$

Dimensionally, $\alpha=\left[ T ^2\right]$

$\left[M L T^{-2}\right]=\frac{\left[ T ^2\right]}{\beta\left[L^2 T^{-2}\right]}$

$\beta=\frac{ T ^2}{\left[ MLT ^{-2} \cdot L ^2 T ^{-2}\right]}$

$\Rightarrow \beta=\left[ M ^{-1} L ^{-3} T ^6\right]$

Dimensions of $\frac{\alpha}{\beta}=\frac{ T ^2}{ M ^{-1} L ^{-3} T ^6}=\left[ ML ^3 T ^{-4}\right]$

$[e] = \left[ {\frac{{M{L^2}{T^{ - 2}}}}{{AT}}} \right] = [M{L^2}{T^{ - 2}}{Q^{ - 1}}]$

So dimensions of energy $ = M{L^2}{T^{ - 2}}$

$Y$= Young's modulus and ${r_0}$ = Interatomic distance

i.e. dimension of resistivity is $[M{L^3}{T^{ - 1}}{Q^{ - 2}}]$

Solar constant $[S]=\frac{\text { Energy }}{\text { Area } \times \text { Time }}=\frac{M^2 T^{-2}}{L^2 T} \Rightarrow\left[M T T^{-3}\right]$

$\mu_{0} \varepsilon_{0}=\frac{1}{\mathrm{C}^{2}}=\frac{1}{(\mathrm{velocity})^{2}}$

$\therefore $ $[a] = [P]\;[{V^2}] = [M{L^{ - 1}}{T^{ - 2}}]\; \times [{L^6}]$ = $[M{L^5}{T^{ - 2}}]$

Here, dv/dt is rate of change of velocity, v is velocity and t is time.

So, dimension of $A=\left[L T^{-2}\right] /[T]=\left[L T^{-1}\right]$

$\therefore$ dimension of $\mathrm{A}=\left[\mathrm{L} T^{-3}\right]$

Dimension of $\mathrm{B}=\left[\mathrm{L} T^{-2}\right] /\left[\mathrm{L} T^{-1}\right]=\left[\mathrm{T}^{-1}\right]$

$\therefore$ dimension of $\mathrm{B}=\left[\mathrm{T}^{-1}\right]$

the argument, $\theta$ of cos or sin should be dimensionless.

therefore,

dimension of $\mathrm{Bx}=[M L T]$

$[B]\left[L^{\prime}\right]=[M L T]$

$[B]=\left[M L^{0} T\right]$

Similarly $[D]\left[T^{\prime}\right]=[M L T]$

$[D]=\left[M L T^{0}\right.$

dimension of $D/B=\frac{\left[M L T^{0}\right]}{\left[M L^{0} T\right]}$

$=\left[L^{1} T^{-1}\right]$

$ \Rightarrow {\nu ^2} = \frac{{{P^2}}}{{4{l^2}}}\left[ {\frac{F}{m}} \right]$

$\therefore m \propto \frac{F}{{{l^2}{\nu ^2}}}$

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

Angular momentum = $[M{L^2}{T^{ - 1}}]$

So mass and length have the same dimensions

Dimensions for surface tension $=\left[ MLT ^{-2}\right]$

Again $\left[ {\frac{{{v_0}}}{\alpha }} \right] = [L]$so $[{v_0}] = [L{T^{ - 1}}]$

$u=\frac{A \sqrt{x}}{x+B}$

By the principle of homogeneity, $x=B$ (dimensionally)

$\Rightarrow B=[L]$

$\text { and }\left[ ML ^2 T ^{-2}\right]=\frac{A L^{1 / 2}}{L}$

${\left[ ML ^2 T ^{-2}\right]=A L^{-1 / 2}}$

$A=\left[ ML ^{3 / 2} T ^{-2}\right]$

$x =$ distance from a fixed origin

$u =\frac{ A \sqrt{ x }}{ x ^2+ B }$

unit of $B$ is same as $x ^2$. Unit of $x ^2=\left[ L ^2\right]$

$B =\left[ L ^2\right]$

$u =\frac{\left[ x ^{1 / 2}\right] A }{ x ^2+ B }$

$A =\frac{ u \left( x ^2+ B \right)}{\sqrt{ x }}=\frac{ kgm }{ s ^2} \times \frac{ m ^2}{ m ^{1 / 2}}$

$A =\frac{ kgm }{ s ^2}$

$A =\left[ ML ^{7 / 2} T ^{-2}\right]$

Dimensions of $AB =\left[ ML ^{7 / 2} T ^{-2}\right]\left[ L ^2\right]$

Dimensions of $AB =\left[ ML ^{11 / 2} T ^{-2}\right]$

$[\mathrm{U}]=\frac{[\mathrm{A}][\mathrm{x}]^{1 / 2}}{[\mathrm{x}]^{2}}=\frac{[\mathrm{A}]}{[\mathrm{x}]^{3 / 2}} \Rightarrow[\mathrm{A}]=[\mathrm{u}][\mathrm{x}]^{3 / 2}$

$=\mathrm{MI}^{2} \mathrm{T}^{-2} \times \mathrm{L}^{3 / 2}$

$[\mathrm{A}]=\mathrm{ML}^{7 / 2} \mathrm{T}^{-2}$

$\left[\frac{\mathrm{A}}{\mathrm{B}}\right]=\frac{\mathrm{ML}^{7 / 2} \mathrm{T}^{-2}}{\mathrm{L}^{2}}$

$[\mathrm{A} / \mathrm{B}]=\mathrm{ML}^{3 / 2} \mathrm{T}^{-2}$

$A= force =$ $[ML{T^{ - 2}}]\;$  [B]=$\frac{{[A]}}{{[m]}} = \frac{{[ML{T^{ - 2}}]}}{{[M{L^{ - 1}}]}}$$ = [{L^2}{T^{ - 2}}]$

This is same dimension as that of latent heat.

By substituting the dimension of each quantity and comparing the powers in both sides we get $[L{T^{ - 1}}] = {[L{T^{ - 2}}]^p}{[L]^q}$

$ \Rightarrow $ $p + q = 1,\,\, - 2p = - 1,\,$

$\therefore \,\,p = \frac{1}{2},\,q = \frac{1}{2}$

Dimension of wavelength $\lambda$ is $L$

Dimension of acceleration due to gravity $g$ is $LT ^{-2}$

Dimension of Density of water $p$ is $ML ^{-3}$

Let $v \propto \lambda^{ a } g ^{ b } \rho^{ c }$

Using Dimensional method

$a+b-3 c=1$

$c =0$

$-2 b =-1$

$b =0.5$

Hence, $a =0.5$

$\therefore v \propto \sqrt{ g \lambda}$

$v ^{2} \propto g \lambda$

$T=M^{0} L^{-1}$

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

$d=\left[M L^{-3}\right]$

$E=\left[M L^{2} T^{-2}\right]$

$M^{0} \cdot T^{\prime}=\left[M L^{-1} T\right]^{a}\left[M L^{-3}\right]^{b}\left[M L^{2} T^{-2}\right]^{c}$

$a+b+c=0-(1)$

$-a-3 b+2 c=0-(2)$

$-2 a-3 b-2 c=0-(3)$

Solve $e q^{n}(1),(2) \Delta(3)$

$a=\frac{-5}{6}, b=\frac{1}{2}, c=\frac{1}{3}$

$\eta=\frac{[M^1L^1T^{-2}][M^0L^1T^0]}{ [M^0L^2T^0] [M^0L^1T^{-1}]} = [M^1L^{-1}T^{-1}]$

Power = $\frac{1}{{{\rm{focal length}}}} = [{L^{ - 1}}]$

All quantities have dimensions

$\left[\right.$ Henry] $=\frac{\left[M T^{-1} Q^{-1} L^{2}\right]}{\left[Q T^{-1}\right]}= M L^{2} Q^{-2}$

Work = $[\vec F.\vec d] = [M{L^2}{T^{ - 2}}]$

${S_t}$= distance travelled in $t^{th}$ second $=\frac{{{\rm{Distance}}}}{{{\rm{time}}}} = [L{T^{ - 1}}]$

$u$ = velocity = $[L{T^{ - 1}}]$ and $\frac{1}{2}a(2t - 1) = [L{T^{ - 1}}]$

As dimensions of each term in the given equation are same, hence equation is dimensionally correct also.

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

$a$ will have same unit as $t^{2} .$ So, $a=\left[T^{2}\right]$

$b=\left[\frac{a-t^{2}}{p x}\right]=\frac{\left[T^{2}\right]}{\left[M L^{-1} T^{-2} L\right]}=\left[M^{-1} L^{0} T^{4}\right]$

$\frac{a}{b}=\frac{\left[T^{2}\right]}{\left[M^{-1} L^{0} T^{4}\right]}=\left[M L^{0} T^{-2}\right]$

$\therefore \,\,\,1\,\,amu\, = \,1.66 \times \,{10^{ - 24}}\,g\,\,\,(\because \,\,\,1\,kg\, = \,{10^3}\,g)$

$\therefore \,1\,\,g\,\, = \,\frac{1}{{1.66 \times {{10}^{ - 24}}\,\,}}\,\,amu$

$\therefore \,1\,\,g\,\, = \,6.02\, \times \,{10^{23}}\,\,amu$

Surface area of cube = $6{a^2}$

according to problem $a^3 = 6a^2 $ $⇒$ $ a = 6$

$V = {a^3} = 216\,units$.

Imprecise $\to $ when variation in readings is more

${n_2} = 10{\left[ {\frac{m}{{{{10}^3}m}}} \right]^1}{\left[ {\frac{{\sec }}{{3600\sec }}} \right]^{ - 2}}$$ = 129600$

Time span of human life $=10^9 \,s$

Age of universe $=10^{17} \,s$

So, $\frac{\text { Age of universe }}{\text { Time of human }}=\frac{10^{17}}{10^9}=10^8$

If, $\frac{\text { Age of universe }}{100}=10^8$

$\Rightarrow \text { Age of universe }=10^{10}\,s$

Area of the image of the house formed on the screen $=1.55 m ^{2}$

$=1.55 \times 10^{4} cm ^{2}$

Arial magnification, $m_{a}=\frac{\text { Area of image }}{\text { Area of object }}=\frac{1.55}{1.75} \times 10^{4}$

$\therefore$ Linear magnifications, $m_{l}=\sqrt{m_{\alpha}}$ $=\sqrt{\frac{1.55}{1.75} \times 10^{4}}=94.11$

$ \Rightarrow g = \frac{{4{\pi ^2}l}}{{{T^2}}}$

$Here \,\% \,error \,in \,l = \frac{{1mm}}{{100cm}} \times 100 = \frac{{0.1}}{{100}} \times 100 = 0.1\%$  and $\%\, error\, in\, T =\frac{{0.1}}{{2 \times 100}} \times 100 = 0.05\% $  

$\% \,error\, in\, g = \% \,error\, in \,l + 2(\%\, error\, in \,T)$

$ = 0.1 + 2 \times 0.05= 0.2 \%$

Random error in $400$ obsrvation $= ?$

If observation is increased by '$n$' times then random error decreases by $\frac{1}{ n }$ times therefore random error $=\frac{x}{4}$

Hence, the answer is $\frac{x}{4}$.

To express maximum estimate of error, the time period should be written as $(2.00 \pm 0.05)\,sec$

Expressing it in percentage error, we have,

$S = 13.8 \pm \frac{{0.2}}{{13.8}} \times 100\% = 13.8 \pm 1.4\% $

and $t = 4.0 \pm \frac{{0.3}}{4} \times 100\% = 4 \pm 7.5\% $

$ = \frac{{13.8 \pm 1.4}}{{4 \pm 7.5}} = (3.45 \pm 0.3)\;m/s.$

$ = \frac{{0.2}}{{13.8}} \times 100 + \frac{{0.3}}{4} \times 100$

$= 1.44 + 7.5 = 8.94 \%$

Hence, percentage increase in area of square sheet $ = 2 \times 2\% $ $= 4\%$

So maximum error in a is given by

${\left( {\frac{{\Delta a}}{a} \times 100} \right)_{\max }} = \alpha \,.\,\frac{{\Delta b}}{b} \times 100 + \beta \,.\,\frac{{\Delta c}}{c} \times 100$

$ + \gamma \,.\,\frac{{\Delta d}}{d} \times 100 + \delta \,.\,\frac{{\Delta e}}{e} \times 100$

$ = \left( {\alpha {b_1} + \beta {c_1} + \gamma {d_1} + \delta {e_1}} \right)\% $

$ = \frac{5}{{100}} \times 100 + \frac{{0.2}}{{10}} \times 100$

$ = (5 + 2)\% $

$= 7\%$

Now $|\Delta {T_1}|\, = \,2.63 - 2.62 = 0.01$

$|\Delta {T_2}|\, = \,2.62 - 2.56 = 0.06$

$|\Delta {T_3}|\, = \,2.62 - 2.42 = 0.20$

$|\Delta {T_4}|\, = \,2.71 - 2.62 = 0.09$

$|\Delta {T_5}|\, = \,2.80 - 2.62 = 0.18$

Mean absolute error

$\Delta T = \frac{{|\Delta {T_1}| + |\Delta {T_2}| + |\Delta {T_3}| + |\Delta {T_4}| + |\Delta {T_5}|}}{5}$

$ = \frac{{0.54}}{5} = 0.108\, = 0.11sec$

Percentage error in volume

$\frac{{\Delta V}}{V} \times 100 = \frac{{2\Delta r}}{r} \times 100 + \frac{{\Delta l}}{l} \times 100$

$ = \left( {2 \times \frac{{0.01}}{{2.0}} \times 100 + \frac{{0.1}}{{5.0}} \times 100} \right)$

$ = (1 + 2)\% $ =$3\% $

So maximum permissible error in

$Y = \frac{{\Delta Y}}{Y} \times 100 $ $= \left( {\frac{{\Delta M}}{M} + \frac{{\Delta g}}{g} + \frac{{\Delta L}}{L} + \frac{{2\Delta D}}{D} + \frac{{\Delta l}}{l}} \right) \times 100$

$ = \left( {\frac{1}{{300}} + \frac{1}{{981}} + \frac{1}{{2820}} + 2 \times \frac{1}{{41}} + \frac{1}{{87}}} \right) \times 100$

$ = 0.065 \times 100 = 6.5\% $

$\therefore \,\,\,\frac{{\Delta H}}{H} \times 100 = \left( {\frac{{2\Delta I}}{I} + \frac{{\Delta R}}{R} + \frac{{\Delta t}}{t}} \right) \times 100$

$ = (2 \times 3 + 4 + 6)\% $ $ = 16\% $