Units and Measurements

A vernier scale is designed such that its divisions are slightly smaller than the main scale divisions. This difference allows precise measurements by aligning a particular vernier division with the main scale.
The vernier constant (least count) is actually given by:
Least Count = Value of One Main Scale Division-Value of One Vernier Scale Division
Thus, the correct option is: 2.

Magnetic field B :
The dimensional formula for B is derived from the Lorentz force law:

\(
F=q v B \quad \Rightarrow \quad B=\frac{F}{q v}
\)

1. Force (F) has the dimensional formula:

\(
[F]=\left[\mathrm{MLT}^{-2}\right]
\)

2. Charge (q) has the dimensional formula:

\(
[q]=[\mathrm{AT}]
\)

3. Velocity (v) has the dimensional formula:

\(
[v]=\left[\mathrm{LT}^{-1}\right]
\)

Thus, the dimensional formula for B is:

\(
[B]=\frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{AT}]\left[\mathrm{LT}^{-1}\right]}=\frac{[\mathrm{M}]\left[\mathrm{T}^{-2}\right]}{[\mathrm{A}]}=[\mathrm{M}]\left[\mathrm{A}^{-1}\right]\left[\mathrm{T}^{-2}\right]
\)

Magnetic permeability ( \(\mu_0\) ):
The dimensional formula of \(\mu_0\) is:

\(
\left[\mu_0\right]=[\mathrm{M}][\mathrm{L}]\left[\mathrm{T}^{-2}\right]\left[\mathrm{A}^{-2}\right]
\)

Dimensions of \(\frac{B}{\mu_0}\) :
Using the formula:

\(
\frac{B}{\mu_0}=\frac{[\mathrm{M}]\left[\mathrm{A}^{-1}\right]\left[\mathrm{T}^{-2}\right]}{[\mathrm{M}][\mathrm{L}]\left[\mathrm{T}^{-2}\right]\left[\mathrm{A}^{-2}\right]}
\)

Cancel the terms:

\(
\frac{B}{\mu_0}=[\mathrm{A}]\left[\mathrm{L}^{-1}\right]
\)

The displacement $x(t)$ has the dimension $[L]$. Each term in the equation must also have the same dimension $[L]$:

• $A\sin(t)$: $\sin(t)$ is dimensionless, so $A$ has $[L]$.
• $B\cos^2(t)$: $\cos^2(t)$ is dimensionless, so $B$ has $[L]$.
• $Ct^2$: $t^2$ has dimensions $[T^2]$, so $C$ must have $[L T^{-2}]$.
• $D$: As part of the displacement, $D$ has $[L]$.

Thus, the dimensions of $\frac{ABC}{D}$​ are:

$$\frac{[L] \cdot [L] \cdot [L T^{-2}]}{[L]} = [L^2 T^{-2}]$$

Therefore, the correct answer is $\mathbf{(b)}$.

\begin{equation}
\begin{aligned}
&\begin{aligned}
& {[P]=\left[\frac{a}{v^2}\right]} \\
& \Rightarrow[a]=\left[P V^2\right] \\
& {[a]=\left[M L^{-1} T^{-2} L^6\right]=\left[M L^5 T^{-2}\right]}
\end{aligned}\\
&\text { Now, }[v]=[b]\\
&\begin{aligned}
& {[b]=\left[L^3\right]} \\
& \therefore\left[a b^{-1}\right]=\left[M L^5 T^{-2} L^{-3}\right]=\left[M L^2 T^{-2}\right]
\end{aligned}
\end{aligned}
\end{equation}

Capaciatnce is given by

$$\begin{gathered} \mathrm{C}\hspace{0.17em}= \frac{\mathrm{Q}}{\mathrm{V}} \\ \mathrm{C} =\frac{\mathrm{Q}\times \mathrm{Q}}{\mathrm{W}} \\ \mathrm{Dimensions} \mathrm{of} \mathrm{C} = \frac{\left[{\mathrm{C}}^2\right]}{\left[{\mathrm{ML}}^2{\mathrm{T}}^{-2}\right]}=[{\mathrm{M}}^{-1} {\mathrm{L}}^{-2} {\mathrm{T}}^2 {\mathrm{C}}^2] \end{gathered}$$

$$\begin{gathered} \tau =C\theta \\ c=\left[M{L}^2{T}^{-2}\right] \end{gathered}$$

$$\begin{gathered} M=NIA \\ M=\left[L^{2}A\right] \end{gathered}$$

$$\begin{gathered} F=qVB \\ \left[ML{T}^{-2}\right]=\left[AT\right]\left[L{T}^{-1}\right]B \\ B=\left[M{T}^{-2}{A}^{-1}\right] \end{gathered}$$

$$\begin{gathered} B =\frac{{\mu }_{0}i}{2r} \\ {\mu }_0 =\frac{\left[M{T}^{-2}{A}^{-1}\right]\left[L\right]}{\left[A\right]} \end{gathered}$$

${\mu }_0=\left[ML{T}^{-2}{A}^{-2}\right]$

\begin{equation}
\begin{aligned}
& \text { } E=\frac{Q}{4 \pi \varepsilon_0 R^2} \Rightarrow \varepsilon_0=\frac{Q}{4 \pi R^2 E} \\
& \text { Now, } \varepsilon_0 E^2=\frac{Q}{4 \pi R^2 E} \cdot E^2=\frac{Q}{4 \pi R^2} \cdot E \\
& {\left[\varepsilon_0 E^2\right]=\left[\frac{Q E}{R^2}\right]=\frac{[Q][E]}{\left[R^2\right]}=\frac{[Q]}{\left[R^2\right][Q][R]}[W]} \\
& =\frac{[W]}{\left[R^3\right]}=\frac{M L^2 T^{-2}}{L^3}=M L^{-1} T^{-2}
\end{aligned}
\end{equation}

$\mathrm{v}={\mathrm{At}}^2+\frac{\mathrm{Bt}}{\mathrm{C}+\mathrm{t}}$

Dimensions of C is same as the dimensions of t.

$\left[\mathrm{C}\right] = \left[\mathrm{T}\right]$

$$\begin{gathered} \mathrm{Dimensions} \mathrm{of} \mathrm{Bt} \mathrm{is} \mathrm{same} \mathrm{as} \mathrm{the} \mathrm{dimensions} \mathrm{of} \mathrm{v} \mathrm{t} \\ \mathrm{So}, \left[\mathrm{B}\right] = \left[{\mathrm{LT}}^{-1}\right] \\ \mathrm{Dimensions} \mathrm{of} {\mathrm{At}}^2 \mathrm{is} \mathrm{same} \mathrm{as} \mathrm{dimensions} \mathrm{of} \mathrm{v} \\ \mathrm{So}, \left[\mathrm{A}\right] = \left[{\mathrm{LT}}^{-3}\right] \\ \mathrm{Thus}, \mathrm{the} \mathrm{dimensions} \mathrm{of} \left[\mathrm{ABC}\right] = \left[{\mathrm{LT}}^{-3}\right]\left[{\mathrm{LT}}^{-1}\right]\left[\mathrm{T}\right] = \left[{\mathrm{L}}^2{\mathrm{T}}^{-3}\right] \end{gathered}$$

Sum of number by considering significant digit sum =4.6+4.6+4.6+4.6=18.4

Arithmetic Mean = $\frac{sum}{4} = \frac{18.4}{4} = 4.6$

The density of the wire is calculated using the formula:

$$\begin{gathered} \mathrm{\rho }=\mathrm{m}/\mathrm{V} \\ \mathrm{where},\mathrm{V}={\mathrm{\pi r}}^2\mathrm{l} \end{gathered}$$

$\mathrm{\rho } = \frac{\mathrm{m}}{{\mathrm{\pi r}}^2\mathrm{l}}$

Now, the percentage error in density is given by the sum of the relative percentage errors of the individual measurements:

$$\begin{gathered} \frac{∆\mathrm{\rho }}{\mathrm{\rho }}\times 100 = \frac{∆\mathrm{m}}{m}\times 100 + 2\times \frac{∆\mathrm{r}}{r}\times 100 + \frac{∆\mathrm{l}}{l}\times 100 \\ \frac{∆\mathrm{\rho }}{\mathrm{\rho }}\times 100 = \left(\frac{0.003}{0.6}\times 100\right) + 2\times \left(\frac{0.01}{0.5}\times 100\right) + \left(\frac{0.05}{10}\times 100\right) \\ \frac{∆\mathrm{\rho }}{\mathrm{\rho }}\times 100 =0.5 + 4 + 0.5 = 5 \% \end{gathered}$$

$$\begin{gathered} \left[L{T}^{-1}\right]=\left[A\right]\left[T^2\right]=\frac{\left[B\right]\left[T\right]}{\left[C\right]+\left[T\right]} \\ \left[C\right]=\left[T\right] \\ \left[A\right]=\left[L{T}^{-3}\right] \\ \left[B\right]=\left[L{T}^{-1}\right] \\ \left[ABC\right]=\left[L^2{T}^{-3}\right] \end{gathered}$$

\begin{equation}
\begin{aligned}
& \text { Given: } 10 \mathrm{MSD}=11 \mathrm{VSD} \\
& \Rightarrow 1 \mathrm{VSD}=\frac{10}{11} \mathrm{MSD} \\
& \text { As, } \mathrm{LC}=1 \mathrm{MSD}-1 \mathrm{VSD} \\
& =1 \mathrm{MSD}-\frac{10}{11} \mathrm{MSD}=\frac{1 \mathrm{MSD}}{11} \\
& =\frac{5}{11} \text { units }
\end{aligned}
\end{equation}

We analyze the dimensions of each pair of physical quantities to determine which have the same dimensions.

• Angular momentum ($L$):

  $$L = mvr = \text{mass} \times \text{velocity} \times \text{distance}$$ $$= [M] [LT^{-1}] [L] = [M L^2 T^{-1}]$$

• Planck's constant ($h$):
  Planck’s constant is defined in the equation:

  $$E = h f$$

  where $E$ is energy and $f$ is frequency.
  Since energy has dimension $[M L^2 T^{-2}]$ and frequency $[T^{-1}]$, we get:

  $$[h] = \frac{[E]}{[f]} = \frac{[M L^2 T^{-2}]}{[T^{-1}]} = [M L^2 T^{-1}]$$

Both have the same dimensions: $[M L^2 T^{-1}]$.

• Torque ($\tau$):

  $$\tau = r \times F = [L] \times [M L T^{-2}] = [M L^2 T^{-2}]$$

• Moment of inertia ($I$):

  $$I = m r^2 = [M] [L^2] = [M L^2]$$

Different dimensions.

• Impulse ($J$):

  $$J = \text{Force} \times \text{time} = [M L T^{-2}] [T] = [M L T^{-1}]$$

• Surface tension ($T$):

  $$T = \frac{\text{Force}}{\text{Length}} = \frac{[M L T^{-2}]}{[L]} = [M T^{-2}]$$

Different dimensions.

• Momentum ($p$):

  $$p = m v = [M] [L T^{-1}] = [M L T^{-1}]$$

• Work Done ($W$):

  $$W = \text{Force} \times \text{Distance} = [M L T^{-2}] [L] = [M L^2 T^{-2}]$$

Different dimensions.

$$\boxed{A} \quad \text{Angular momentum and Planck's constant}$$

We are given the equation:

$$Q = \frac{a b^4}{c d}$$​

To determine the percentage error $Q$, we use the error propagation rule:

$$\frac{\Delta Q}{Q} = \frac{\Delta a}{a} + 4 \frac{\Delta b}{b} + \frac{\Delta c}{c} + \frac{\Delta d}{d}$$​

• Percentage error in a = 2%
• Percentage error in b = 1%
• Percentage error in c = 2%
• Percentage error in d = 1%

Now, substituting these values into the equation:

$$\text{Percentage error in } Q = 2 + 4(1) + 2 + 1$$ $$= 2 + 4 + 2 + 1 = 9\%$$

$$\boxed{C} \quad 9\%$$C​

Problem Statement: Find the dimensional formula for $Z=\frac{S^2\sigma }{\kappa }$.

Solution:

1. Determine Dimensions of Constituent Quantities:

• Seebeck Coefficient (S): Defined as EMF per unit temperature.$\left[S\right]=\frac{\left[V\right]}{\left[\text{Δ}T\right]}=\frac{\left[W\right]/\left[Q\right]}{\left[K\right]}=\frac{\left[M{L}^2{T}^{-2}\right]/\left[IT\right]}{\left[K\right]}=\left[M{L}^2{T}^{-3}{I}^{-1}{K}^{-1}\right]$
• Electrical Conductivity ($\sigma$): Inverse of resistivity ($\rho$). $R=\rho \frac{L}{A}\left[\rho \right]=\left[R\right]\frac{\left[A\right]}{\left[L\right]}=\frac{\left[V\right]}{\left[I\right]}\left[L\right]=\left[M{L}^2{T}^{-3}{I}^{-2}\left]\left[L\right]=\right[M{L}^3{T}^{-3}{I}^{-2}\right]$$\left[\sigma \right]=\frac{1}{\left[\rho \right]}=\left[M^{-1}{L}^{-3}{T}^3{I}^2\right]$
• Thermal Conductivity ($\kappa$): From Power flow equation $P=\frac{\kappa A\text{Δ}T}{L}$.$\left[\kappa \right]=\frac{\left[P\right]\left[L\right]}{\left[A\right]\left[\text{Δ}T\right]}=\frac{\left[M{L}^2{T}^{-3}\right]\left[L\right]}{\left[L^2\right]\left[K\right]}=\left[ML{T}^{-3}{K}^{-1}\right]$

2. Calculate Dimensions of $Z$: $Z=\frac{S^2\sigma }{\kappa }$

First, calculate $\left[S^2\right]$: $\left[S^2\left]=\right[M{L}^2{T}^{-3}{I}^{-1}{K}^{-1}{]}^2=\right[M^2{L}^4{T}^{-6}{I}^{-2}{K}^{-2}]$

Multiply by $\left[\sigma \right]$: $\left[S^2\sigma \left]=\right[M^2{L}^4{T}^{-6}{I}^{-2}{K}^{-2}\left]\times \right[M^{-1}{L}^{-3}{T}^3{I}^2\left]=\right[M^1{L}^1{T}^{-3}{I}^0{K}^{-2}\right]$

Divide by $\left[\kappa \right]$: $\left[Z\right]=\frac{\left[ML{T}^{-3}{K}^{-2}\right]}{\left[ML{T}^{-3}{K}^{-1}\right]}$

$\left[Z\right]=\left[M^{1-1}{L}^{1-1}{T}^{-3-\left(-3\right)}{I}^0{K}^{-2-\left(-1\right)}\right]$ $\left[Z\right]=\left[M^0{L}^0{T}^0{I}^0{K}^{-1}\right]$

Correct Option: (B) $\left[M^0{L}^0{T}^0{I}^0{K}^{-1}\right]$

$\text{ (A) } \left[Y\right]=\frac{F}{A\left(\frac{\Delta \ell }{\ell }\right)} =M{L}^{-1}{T}^{-2} \left(\frac{\Delta \ell }{\ell }\right. =\text{ Dimensionless })$

(B) Torque

$\left(\overrightarrow{\tau }\right)=\overrightarrow{r}\times \overrightarrow{F}=L\times ML{T}^{-2}=M{L}^2{T}^{-2}$

(C) Coefficient of viscosity $\Rightarrow \frac{F}{A·\frac{dv}{dt}}$

$$\begin{gathered} \eta \to Pa\text{ sec. } \\ \left[\eta \right]=\frac{ML{T}^{-2}}{L^2}\times T=M{L}^{-1}{T}^{-1} \end{gathered}$$

$$\begin{gathered} \text{ (D) Gravitational constant ( }G\text{ ) } \\ \left[G\right]=\frac{F·r^2}{m_1{m}_2}=\frac{ML{T}^{-2}\times L^2}{M^2}=M^{-1}{L}^3{T}^{-2} \end{gathered}$$

Dimensions of surface tension is [MT^-2]

Dimensions of Impulse is [MLT^-1]

Magnetic field B :
The dimensional formula for B is derived from the Lorentz force law:

\(
F=q v B \quad \Rightarrow \quad B=\frac{F}{q v}
\)

1. Force (F) has the dimensional formula:

\(
[F]=\left[\mathrm{MLT}^{-2}\right]
\)

2. Charge (q) has the dimensional formula:

\(
[q]=[\mathrm{AT}]
\)

3. Velocity (v) has the dimensional formula:

\(
[v]=\left[\mathrm{LT}^{-1}\right]
\)

Thus, the dimensional formula for B is:

\(
[B]=\frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{AT}]\left[\mathrm{LT}^{-1}\right]}=\frac{[\mathrm{M}]\left[\mathrm{T}^{-2}\right]}{[\mathrm{A}]}=[\mathrm{M}]\left[\mathrm{A}^{-1}\right]\left[\mathrm{T}^{-2}\right]
\)

Magnetic permeability ( \(\mu_0\) ):
The dimensional formula of \(\mu_0\) is:

\(
\left[\mu_0\right]=[\mathrm{M}][\mathrm{L}]\left[\mathrm{T}^{-2}\right]\left[\mathrm{A}^{-2}\right]
\)

Dimensions of \(\frac{B}{\mu_0}\) :
Using the formula:

\(
\frac{B}{\mu_0}=\frac{[\mathrm{M}]\left[\mathrm{A}^{-1}\right]\left[\mathrm{T}^{-2}\right]}{[\mathrm{M}][\mathrm{L}]\left[\mathrm{T}^{-2}\right]\left[\mathrm{A}^{-2}\right]}
\)

Cancel the terms:

\(
\frac{B}{\mu_0}=[\mathrm{A}]\left[\mathrm{L}^{-1}\right]
\)

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

The maximum percentage error in the measurement of the density of a wire

$$\begin{gathered} \frac{\Delta \rho }{\rho }=\frac{\Delta M}{M}+2\frac{\Delta r}{r}+\frac{\Delta l}{l} \\ \Rightarrow \frac{\Delta \rho }{\rho }=\left(\frac{0.003}{0.6}+\frac{2\times 0.01}{0.5}+\frac{0.05}{10}\right)100 \end{gathered}$$

Total error =0.5 %+4 %+0.5 %=5 %

$\text{ (A) } \left[Y\right]=\frac{F}{A\left(\frac{\Delta \ell }{\ell }\right)} =M{L}^{-1}{T}^{-2} \left(\frac{\Delta \ell }{\ell }\right. =\text{ Dimensionless })$

(B) Torque

$\left(\overrightarrow{\tau }\right)=\overrightarrow{r}\times \overrightarrow{F}=L\times ML{T}^{-2}=M{L}^2{T}^{-2}$

(C) Coefficient of viscosity $\Rightarrow \frac{F}{A·\frac{dv}{dt}}$

$$\begin{gathered} \eta \to Pa\text{ sec. } \\ \left[\eta \right]=\frac{ML{T}^{-2}}{L^2}\times T=M{L}^{-1}{T}^{-1} \end{gathered}$$

$$\begin{gathered} \text{ (D) Gravitational constant ( }G\text{ ) } \\ \left[G\right]=\frac{F·r^2}{m_1{m}_2}=\frac{ML{T}^{-2}\times L^2}{M^2}=M^{-1}{L}^3{T}^{-2} \end{gathered}$$

The formula for percentage error propagation in a function is:

Given:

1. For $x^3$: The percentage error for $x^3$ is:

   $$\Delta E_{\% \, x} = 3 \cdot \Delta x_{\%}$$​

   Substituting $\Delta x_{\%} = 1.2\%$

   $$\Delta E_{\% \, x} = 3 \cdot 1.2 = 3.6\%$$

2. For $e^{-\beta t}$: The percentage error due to $t$t comes from the exponential term. Using:

   $$\Delta E_{\% \, t} = \beta \cdot t \cdot \Delta t_{\%}$$​

   Substituting $\beta = 0.3 \, \text{s}^{-1}, t = 5 \, \text{s}, \Delta t_{\%} = 1.6\%$

   $$\Delta E_{\% \, t} = 0.3 \cdot 5 \cdot 1.6 = 2.4\%$$

The total percentage error in $E$E is:

Substituting the values:

Correct answer: (d) 6%

\begin{aligned}
& \text { (d) } 50 \mathrm{~V}+\mathrm{S}=49 \mathrm{~S}+\mathrm{S} \\
& \mathrm{~S}=50(\mathrm{~S}-\mathrm{V}) \\
& 0.5=50(\mathrm{~S}-\mathrm{V})
\end{aligned}
The Vernier constant of travelling microscope is $S-V = \frac{0.5}{50} = \frac{1}{100}= 0.01mm$

$$\begin{gathered} y=\frac{32.3\times 1125}{27.4} \\ y=1326.186 \end{gathered}$$

Significant Figures Rule:
The given numbers have 3 significant digits.
The result must also be rounded to 3 significant digits.
Rounding 1326.186 to 3 significant digits:
$1326.186\to 1330$ (since 1326 rounds up to 1330 )
the correct answer is 1330 .

\begin{equation}
\begin{aligned}
& \text { } E=\frac{Q}{4 \pi \varepsilon_0 R^2} \Rightarrow \varepsilon_0=\frac{Q}{4 \pi R^2 E} \\
& \text { Now, } \varepsilon_0 E^2=\frac{Q}{4 \pi R^2 E} \cdot E^2=\frac{Q}{4 \pi R^2} \cdot E \\
& {\left[\varepsilon_0 E^2\right]=\left[\frac{Q E}{R^2}\right]=\frac{[Q][E]}{\left[R^2\right]}=\frac{[Q]}{\left[R^2\right][Q][R]}[W]} \\
& =\frac{[W]}{\left[R^3\right]}=\frac{M L^2 T^{-2}}{L^3}=M L^{-1} T^{-2}
\end{aligned}
\end{equation}

Magnetic field B :
The dimensional formula for B is derived from the Lorentz force law:

\(
F=q v B \quad \Rightarrow \quad B=\frac{F}{q v}
\)

1. Force (F) has the dimensional formula:

\(
[F]=\left[\mathrm{MLT}^{-2}\right]
\)

2. Charge (q) has the dimensional formula:

\(
[q]=[\mathrm{AT}]
\)

3. Velocity (v) has the dimensional formula:

\(
[v]=\left[\mathrm{LT}^{-1}\right]
\)

Thus, the dimensional formula for B is:

\(
[B]=\frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{AT}]\left[\mathrm{LT}^{-1}\right]}=\frac{[\mathrm{M}]\left[\mathrm{T}^{-2}\right]}{[\mathrm{A}]}=[\mathrm{M}]\left[\mathrm{A}^{-1}\right]\left[\mathrm{T}^{-2}\right]
\)

Magnetic permeability ( \(\mu_0\) ):
The dimensional formula of \(\mu_0\) is:

\(
\left[\mu_0\right]=[\mathrm{M}][\mathrm{L}]\left[\mathrm{T}^{-2}\right]\left[\mathrm{A}^{-2}\right]
\)

Dimensions of \(\frac{B}{\mu_0}\) :
Using the formula:

\(
\frac{B}{\mu_0}=\frac{[\mathrm{M}]\left[\mathrm{A}^{-1}\right]\left[\mathrm{T}^{-2}\right]}{[\mathrm{M}][\mathrm{L}]\left[\mathrm{T}^{-2}\right]\left[\mathrm{A}^{-2}\right]}
\)

Cancel the terms:

\(
\frac{B}{\mu_0}=[\mathrm{A}]\left[\mathrm{L}^{-1}\right]
\)

$\left[\frac{B}{{\mu }_0}\right]=\left[ni\right]=L^{-1}{A}^1$

The dimensions of $(\beta/\alpha)$ are derived by ensuring dimensional consistency. The term $\phi = \alpha\lambda$ gives the dimensions of $\alpha$α as $[ML^4T^{-4}A^{-2}]$,and $\phi = \beta\sigma$ gives the dimensions of $\beta$ as $[ML^5T^{-4}A^{-2}]$. The ratio $(\beta/\alpha)$ simplifies to $[L]$, which corresponds to displacement as it is a measure of length.

$x\left(t\right)=A\sin \left(t\right)+{Bcos}^2\left(t\right)+C{t}^2+D$

Here, A, B, C, and D are constants, and t is time.

$$\begin{gathered} \left[A\right]=L, \left[B\right]=L, \left[C\right]=\frac{L}{T^2},\left[D\right]=L \\ \left[\frac{ABC}{D}\right]=\left[\frac{L\times L\times L{T}^{-2}}{L}\right]=\left[L^2{T}^{-2}\right] \end{gathered}$$

$x\left(t\right)=A\sin \left(t\right)+{Bcos}^2\left(t\right)+C{t}^2+D$

Here, A, B, C, and D are constants, and t is time.

$$\begin{gathered} \left[A\right]=L, \left[B\right]=L, \left[C\right]=\frac{L}{T^2},\left[D\right]=L \\ \left[\frac{ABC}{D}\right]=\left[\frac{L\times L\times L{T}^{-2}}{L}\right]=\left[L^2{T}^{-2}\right] \end{gathered}$$

\begin{aligned}
& L C=\frac{\text { pitch }}{\text { No. of CSD }}=\left(\frac{1}{100}\right) \\
& =0.01 \mathrm{~mm}
\end{aligned}
$$\begin{gathered} Diameter = MSR +LC\times CSD \\ Diameter = 1+(0.01)\times 42 mm \end{gathered}$$

therefore, $\frac{x}{50} =1.42$

x=71

\begin{equation}
\frac{\Delta Q}{Q}=4 \frac{\Delta a}{a}+3 \frac{\Delta b}{b}+2 \frac{\Delta c}{c}
\end{equation}

\begin{aligned}
& \frac{\Delta Q}{Q} \times 100=4\left(\frac{\Delta a}{a} \times 100\right)+ \\
& 3\left(\frac{\Delta b}{b} \times 100\right)+2\left(\frac{\Delta c}{c} \times 100\right)
\end{aligned}

% error in Q $= 4\times 3\%+3\times 4\%+2\times 5\% = 34\%$

Dimension formula for [B] = $L^2$

$A = \frac{x^2}{Et} = \frac{L^2}{TM{L}^2{T}^{-2}} = \frac{1}{M{T}^{-1}}$

Dimension formula for [A] = $M^{-1}T$

dimension formula for [B] = $L^2{M}^{-1}T$

Vernier Constant = MSD - VSD

$\begin{array}{rl}& =MSD-\frac{N}{N+1}MSD\\ & =\frac{1}{N+1}MSD\\ & =\frac{1}{N+1}\left(0.01\right)\mathrm{cm}\\ & =\frac{1}{100(N+1)}\end{array}$

$$\begin{gathered} \tau =C\theta \\ c=\left[M{L}^2{T}^{-2}\right] \end{gathered}$$

$$\begin{gathered} M=NIA \\ M=\left[L^{2}A\right] \end{gathered}$$

$$\begin{gathered} F=qVB \\ \left[ML{T}^{-2}\right]=\left[AT\right]\left[L{T}^{-1}\right]B \\ B=\left[M{T}^{-2}{A}^{-1}\right] \end{gathered}$$

$$\begin{gathered} B =\frac{{\mu }_{0}i}{2r} \\ {\mu }_0 =\frac{\left[M{T}^{-2}{A}^{-1}\right]\left[L\right]}{\left[A\right]} \end{gathered}$$

${\mu }_0=\left[ML{T}^{-2}{A}^{-2}\right]$

\begin{equation}
\begin{aligned}
& \text { Least count of screw gauge }=\frac{\text { Pitch }}{\text { No. of divisions on circular scale }} \\
& \Rightarrow \text { Least count }=\frac{1}{100}=0.01 \mathrm{~mm} \\
& \text { diameter of wire}=\mathrm{MSR}+\mathrm{CSR} \times \text { L.C. } \\
& =1 \mathrm{~mm}+(63)(0.01) \mathrm{mm} \\
& =1.63 \mathrm{~mm} \\
& =0.163 \mathrm{~cm}
\end{aligned}
\end{equation}

$$\begin{gathered} E\left(t\right)={\alpha }^3{e}^{-\beta t} \\ dE\left(t\right)=3{\alpha }^2{e}^{-\beta t}d\alpha -\beta {\alpha }^3{e}^{-\beta t}dt \\ \frac{dE\left(t\right)}{E\left(t\right)}=\frac{3{\alpha }^2{e}^{-\beta t}d\alpha }{E\left(t\right)}+\frac{\beta {\alpha }^3{e}^{-\beta t}dt}{E\left(t\right)} \\ \frac{\Delta E}{E}=\frac{3\Delta \alpha }{\alpha }+\beta \Delta t \\ \frac{\Delta t}{t}\times \frac{100}{E}=1.6 \\ \frac{\Delta t}{5}=0.016\Rightarrow \Delta t=0.08 \end{gathered}$$

$$\begin{gathered} \frac{\Delta E}{E}=3\times 0.012+0.3\times 0.08 \\ \frac{\Delta E}{E}=0.036+0.024\Rightarrow \frac{\Delta E}{E}=0.06 \\ \%E=6\% \end{gathered}$$

\begin{equation}
\begin{aligned}
&\frac{1}{f}=\frac{1}{v}-\frac{1}{u} \Rightarrow f^{-1}=v^{-1}-u^{-1}\\
&\text { differentiating w.r.t } f\\
&\begin{aligned}
& \Rightarrow-f^{-2} d f=-v^{-2} d v-u^{-2} d u \\
& \Rightarrow \frac{d f}{f^2}=\frac{d v}{v^2}+\frac{d u}{u^2} \\
& \Rightarrow d f=f^2\left[\frac{d v}{v^2}+\frac{d u}{u^2}\right]
\end{aligned}
\end{aligned}
\end{equation}

Capacitance is defined as:

$C=\frac{Q}{V}$

- Charge Q has dimensional formula:

$\left[Q\right]=\left[C\right]$

- Voltage V is energy per charge:

$V=\frac{W}{Q}=\frac{M{L}^2{T}^{-2}}{C}=M{L}^2{T}^{-2}{C}^{-1}$

Now, substituting in $C=\frac{Q}{V}$ :

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

We are given the equation:

$$Q = \frac{a b^4}{c d}$$​

To determine the percentage error $Q$, we use the error propagation rule:

$$\frac{\Delta Q}{Q} = \frac{\Delta a}{a} + 4 \frac{\Delta b}{b} + \frac{\Delta c}{c} + \frac{\Delta d}{d}$$​

• Percentage error in a = 2%
• Percentage error in b = 1%
• Percentage error in c = 2%
• Percentage error in d = 1%

Now, substituting these values into the equation:

$$\text{Percentage error in } Q = 2 + 4(1) + 2 + 1$$ $$= 2 + 4 + 2 + 1 = 9\%$$

$$\boxed{C} \quad 9\%$$C​