Units & Measurement

Time, $t=6 \min 40 \mathrm{~s}$

$ \begin{aligned} & =360+40 \\ & =400 \mathrm{~s} \end{aligned} $

Distance in new system $d=v t$

$ =1 \times 400=400 $

The side of the cube is

$ a = 9.0\ \text{cm} = 9.0 \times 10^{-2}\ \text{m} $

The volume of the cube is

$ V = a^3 $

$ V = (9.0 \times 10^{-2})^3\ \text{m}^3 $

$ V = 7.29 \times 10^{-4}\ \text{m}^3 $

Density is

$ \rho = \frac{\text{mass}}{\text{volume}} $

$ \rho = \frac{5.580}{7.29 \times 10^{-4}} $

$ \rho = 7.654 \times 10^3\ \text{kg m}^{-3} $

Now we apply significant figures.

Mass $5.580\ \text{kg}$ has $4$ significant figures.

Side $9.0\ \text{cm}$ has $2$ significant figures.

Since density involves multiplication/division, the final answer should have the least number of significant figures, which is $2$.

So,

$ 7.654 \times 10^3 \approx 7.7 \times 10^3\ \text{kg m}^{-3} $

Hence,

$ X = 7.7 $

Correct option: D.

Least count of a vernier calliper

The least count (L.C.) is the smallest measurement a vernier calliper can read. It is given by:

$ \text{L.C.} = 1\,\text{MSD} - 1\,\text{VSD} \quad ...(i) $

It is given that 20 vernier scale divisions (VSD) coincide with 16 main scale divisions (MSD). Each MSD is 1 mm long.

So,

$ 20\,\text{VSD} = 16\,\text{MSD} $

Dividing both sides by 20, we get

$ 1\,\text{VSD} = \frac{16}{20}\,\text{MSD} $

Substitute this value in equation (i):

$ \text{L.C.} = 1\,\text{MSD} - \frac{16}{20}\,\text{MSD} $

Simplifying,

$ \text{L.C.} = \frac{4}{20}\,\text{MSD} $

Since 1 MSD = 1 mm,

$ \text{L.C.} = \frac{1}{5}\,\text{mm} = 0.2\,\text{mm} $

In centimetre, the least count is:

$ \text{L.C.} = 0.02\,\text{cm} $

$ \begin{aligned} & \sigma_s=\mathrm{MT}^{-3} \mathrm{~K}^{-4} \\ & k_B=\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-2} \mathrm{~K}^{-1} \\ & b=\mathrm{M}^0 \mathrm{LT}^0 \mathrm{~K} \\ & \sigma_s k_B^{-1} b=\left[\mathrm{MT}^{-3} \mathrm{~K}^{-4}\right]\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^2 \mathrm{~K}\right]\left[\mathrm{M}^0 \mathrm{LT}^0 \mathrm{~K}\right] \\ & =\mathrm{T}^{-1} \mathrm{~K}^{-2} \mathrm{~L}^{-1} \\ & =\mathrm{L}^{-1} \mathrm{~T}^{-1} \mathrm{~K}^{-2} \end{aligned} $

⇒ There is negative zero error

⇒ Zero error $=-(10-4) \times$ L.C. $\ldots$ (i)

From … (i),

Zero error $=-(10-4) \times 0.01=-0.06 \mathrm{~cm}$

⇒ Zero correction $=-$ Zero error

    $=-(-0.06)$

    $=0.06 \mathrm{~cm}$

⇒ Correct reading $=R^{\prime}$

$R^{\prime}=R+$ Zero correction

     $=1+0.06$

     $=1.06 \mathrm{~cm}$

$\therefore \quad$ No option is correct

The relationship for $ T $ is given by:

$ T \propto S^\alpha A^\beta \rho^\gamma R^\delta $

To find constants $ \alpha $, $ \beta $, $ \gamma $, and $ \delta $, analyze the dimensional formulae:

Dimension analysis for Time $ T $:

$ \text{M}^0 \text{L}^0 \text{T}^1 = \left(\text{M} \text{T}^{-2}\right)^\alpha \left(\text{L}^2\right)^\beta \left(\text{M} \text{L}^{-3}\right)^\gamma \text{L}^\delta $

Equate dimensions:

$ \text{M}^0 \text{L}^0 \text{T}^1 = \text{M}^{\alpha + \gamma} \text{L}^{2\beta - 3\gamma + \delta} \text{T}^{-2\alpha} $

From comparing dimensions:

For mass (M): $\alpha + \gamma = 0$

For length (L): $2\beta - 3\gamma + \delta = 0$

For time (T): $ -2\alpha = 1 \Rightarrow \alpha = -\frac{1}{2} $

Solving equations:

Using $\alpha = -\frac{1}{2}$,

$ \alpha + \gamma = 0 \Rightarrow -\frac{1}{2} + \gamma = 0 \Rightarrow \gamma = \frac{1}{2} $

Substituting these into the equation for length:

$ 2\beta - 3\left(\frac{1}{2}\right) + \delta = 0 $

Assume $\beta = -1$ to solve for $\delta$:

$ 2(-1) - \frac{3}{2} + \delta = 0 \quad \Rightarrow \quad \delta = \frac{7}{2} $

By this analysis, the values are $\alpha = -\frac{1}{2}$, $\beta = -1$, $\gamma = \frac{1}{2}$, and $\delta = \frac{7}{2}$.

The physical quantity $ P $ is defined in terms of four observations $ a, b, c, $ and $ d $ by the formula:

$ P = \frac{a^3 b^2}{c \sqrt{d}} $

To find the percentage error in $ P $, we use the formula for maximum percentage error. The general rule for percentage errors when a quantity is a product or a quotient is to add the percentage errors, while for powers, multiply the error by the power.

The percentage error for $ P $ is calculated using:

$ \text{Maximum } \% \text{ error in } P = 3 \left( \frac{\Delta a}{a} \times 100 \right) + 2 \left( \frac{\Delta b}{b} \times 100 \right) + \left( \frac{\Delta c}{c} \times 100 \right) + \frac{1}{2} \left( \frac{\Delta d}{d} \times 100 \right) $

Given the percentage errors:

$ a $ has a percentage error of $ 1\% $

$ b $ has a percentage error of $ 3\% $

$ c $ has a percentage error of $ 2\% $

$ d $ has a percentage error of $ 4\% $

Substituting these into the formula:

$ = 3 \times (1) + 2 \times (3) + (2) + \frac{1}{2} \times (4) $

$ = 3 + 6 + 2 + 2 $

$ = 13\% $

Thus, the percentage error in the quantity $ P $ is $ 13\% $.

To measure the diameter of a spherical object with Vernier calipers, consider the following details:

Vernier Scale and Main Scale Relationship: 10 Vernier Scale Divisions (V.S.D.) are equivalent to 9 Main Scale Divisions (M.S.D.).

Least Value on Main Scale: 0.1 cm

Zero Error: The zero marking on the Vernier Scale (V.S.) is at 0.1 cm when the caliper jaws are closed.

Given measurements:

Main Scale Reading: $ M = 5 \, \text{cm} $

Coinciding Vernier Division: 8

Calculations:

Determining the Least Count:

$ \text{Least Count} = \text{1 MSD} - \text{1 VSD} $

$ = 1 \times 0.1 \, \text{cm} - \frac{9}{10} \times 0.1 \, \text{cm} $

$ = 0.1 \, \text{cm} - 0.09 \, \text{cm} = 0.01 \, \text{cm} $

Accounting for Zero Error:

Zero Error = +0.1 cm

Vernier Scale Reading Calculation:

$ = 8 \times 0.01 \, \text{cm} = 0.08 \, \text{cm} $

Final Diameter Measurement After Zero Error Correction:

$ \text{Measured Diameter} = \text{Main Scale Reading} + \text{Vernier Reading} - \text{Zero Error} $

$ = 5 \, \text{cm} + 0.08 \, \text{cm} - 0.1 \, \text{cm} $

$ = 4.98 \, \text{cm} $

Thus, the corrected measurement for the diameter is 4.98 cm.

To determine the resistance $R$ using Ohm's law, we use the formula:

$R = \frac{V}{I}$

Given:

$V = (200 \pm 4)$ volts

$I = (20 \pm 0.2)$ amperes

First, calculate the nominal value of the resistance:

$R = \frac{200}{20} = 10 \, \Omega$

Next, we need to calculate the uncertainty in the resistance. For division, the relative uncertainties add up. The relative uncertainty for voltage is:

$\frac{\Delta V}{V} = \frac{4}{200} = 0.02$

and for current:

$\frac{\Delta I}{I} = \frac{0.2}{20} = 0.01$

Adding these relative uncertainties gives us the relative uncertainty for the resistance:

$\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I} = 0.02 + 0.01 = 0.03$

Now, we calculate the absolute uncertainty for the resistance:

$\Delta R = R \times 0.03 = 10 \times 0.03 = 0.3 \, \Omega$

Therefore, the resistance with its uncertainty is:

$R = 10 \pm 0.3 \, \Omega$

Thus, the correct answer is:

Option B $(10 \pm 0.3) \Omega$

To determine the diameter of the wire using the screw gauge, we need to consider both the pitch scale reading and the circular scale reading. The formula to calculate the diameter is:

$ D = P + (C \times L) $

Where:

• $D$ is the diameter of the wire.
• $P$ is the pitch scale reading.
• $C$ is the circular scale reading.
• $L$ is the least count of the screw gauge.

Given:

• Pitch of the screw gauge, $P = 1 \mathrm{~mm}$
• Number of divisions on the circular scale = 100
• Pitch scale reading, $P = 1 \mathrm{~mm}$
• The circular scale reads $63^{\text{rd}}$ division.

First, we calculate the least count of the screw gauge:

$ L = \frac{ \text{Pitch} }{ \text{Number of divisions} } = \frac{ 1 \mathrm{~mm} }{ 100 } = 0.01 \mathrm{~mm} $

Next, we substitute the values into the formula for the diameter:

$ D = 1 \mathrm{~mm} + (63 \times 0.01 \mathrm{~mm}) $

Calculate the product:

$ 63 \times 0.01 \mathrm{~mm} = 0.63 \mathrm{~mm} $

Now, add the pitch scale reading and the circular scale reading:

$ D = 1 \mathrm{~mm} + 0.63 \mathrm{~mm} = 1.63 \mathrm{~mm} $

Convert $1.63 \mathrm{~mm}$ to centimeters:

$ 1.63 \mathrm{~mm} = 0.163 \mathrm{~cm} $

Therefore, the diameter of the wire is:

Option B: $0.163 \mathrm{~cm}$

To determine the dimensions of $\frac{A^2}{B}$, we start with the given potential energy expression:

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

The dimensions of potential energy ($V$) are $[ML^2T^{-2}]$. So, we need to consider the dimensions of the other terms in the expression to match these dimensions. Let's break it down step-by-step.

For the term $\sqrt{x}$, where $x$ represents distance:

$\sqrt{x}$ has dimensions $[L^{1/2}]$

Since both terms in the denominator $\sqrt{x} + B$ have to be of the same dimensions for them to be added together, the dimensions of $B$ must also be $[L^{1/2}]$.

Now the expression becomes:

$V = \frac{A x^2}{L^{1/2}}$

Thus, the dimensions must be:

$[V] = \left[\frac{A x^2}{L^{1/2}}\right]$

Substitute the dimensions of $V$, $x$, and $L$:

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

Simplifying the right-hand side:

$[ML^2T^{-2}] = [A L^{3/2}]$

Therefore, the dimensions of $A$ must be:

$[A] = [ML^2T^{-2}] \cdot [L^{-3/2}]$

$[A] = [ML^{1/2} T^{-2}]$

We have dimensions of $A$ and $B$. Now we need to find the dimensions of $\frac{A^2}{B}$:

$[A^2] = [ML^{1/2} T^{-2}]^2$

$[A^2] = [M^2 L T^{-4}]$

Divide by the dimensions of $B$:

$\left[\frac{A^2}{B}\right] = \left[\frac{M^2 L T^{-4}}{L^{1/2}}\right]$

$\left[\frac{A^2}{B}\right] = [M^2 L^{1/2} T^{-4}]$

Thus, the dimensions of $\frac{A^2}{B}$ are:

Option C: $\left[M^2 L^{1/2} T^{-4}\right]$

V.C = MSD $-$ VSD ..... (1)

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

$\mathrm{VSD}=\left(\frac{N}{N+1}\right) \mathrm{MSD}$ ..... (2)

From (1) and (2)

$\begin{aligned} & V . C=(M S D)-\frac{N}{N+1}(M S D) \\ & =M S D\left(1-\frac{N}{N+1}\right)=\frac{M S D}{N+1} \\ & =\frac{0.01}{N+1}=\frac{1}{100(N+1)} \end{aligned}$

Solid angle $d \Omega=\frac{d A}{r^2}$ has dimensions $[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0]$

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

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

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

The force acting on a particle at a given time $ t $ is defined as $ F = \alpha t^2 + \beta t $, where $ \alpha $ and $ \beta $ are constants. We need to determine which factor is dimensionless if $ \alpha $ and $ \beta $ are constants.

Using the principle of homogeneity of dimensions:

$ \begin{aligned} & [F] = [\alpha t^2] = [\beta t] \\ & [\alpha] = \frac{[F]}{[t^2]} \quad \text{and} \quad [\beta] = \frac{[F]}{[t]} \\ & \therefore \quad [\alpha][t] = [\beta] \\ & \therefore \quad \frac{\alpha t}{\beta} = \text{dimensionless} \end{aligned} $

$\text { } \begin{aligned} \text { Mean diameter } & =\frac{\mathrm{d}_1+\mathrm{d}_2+\mathrm{d}_3+\mathrm{d}_4+\mathrm{d}_5}{5} \\ & =\frac{3.33+3.32+3.34+3.33+3.32}{5} \\ & =3.328 \approx 3.33 \end{aligned}$

Angular momentum $=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]$

Coeff of thermal conductivity $=\left[\mathrm{MLT}^{-3} \mathrm{~K}^{-1}\right]$

Torque $=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]$

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

So, gravitational constant has power of $(-1)$ of $\mathrm{M}$.

Error arise due to unpredictable fluctuation in temperature and voltage supply are $\rightarrow$ random errors.

$\rho=\frac{M}{V}$

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

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

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

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

$=1.6 \%$

Pressure $[P] = \left[ {{F \over A}} \right] = \left[ {{{ML{T^{ - 2}}} \over {{L^2}}}} \right] = [M{L^{ - 1}}{T^{ - 2}}]$

And, dimensions of Young's modulus $[Y] = \left[ {{{Stress} \over {Strain}}} \right] = [M{L^{ - 1}}{T^{ - 2}}]$

[Force] $ = [ML{T^{ - 2}}]$

[Momentum] $ = [ML{T^{ - 1}}]$

[Coefficient of viscosity] $ = [M{L^{ - 1}}{T^{ - 1}}]$

Given $g = {{4{p^2}L} \over {{T^2}}}$

Fractional error in value of g

${{\Delta g} \over g} = {{\Delta L} \over L} + 2{{\Delta T} \over T}$

$ \Rightarrow {{\Delta g} \over g} \times 100 = {{\Delta L} \over L} \times 100 + 2{{\Delta T} \over T} \times 100$

$ \Rightarrow {{\Delta g} \over g} \times 100 = {{0.1} \over {10}} \times 100 + 2 \times {1 \over {100}} \times 100 = 3\% $

Hence, percentage error in measurement of g is 3%

Dimensional formula of magnetic permeability is [MLT^$-$2A^$-$2]

Plane angle = ${{Arc} \over {Radius}} = {{[L]} \over {[L]}}\buildrel {} \over \longrightarrow $ Unit = Radian

Solid angle = ${{Area} \over {{{(Radius)}^2}}}\buildrel {} \over \longrightarrow $ Unit = Steradian

$ = {{{L^2}} \over {{L^2}}} = [{M^0}{L^0}{T^0}]$

$\therefore$ Both have units but no dimensions.

Area = Length $\times$ Breadth

= 55.3 $\times$ 25 m²

= 1382.5 m²

= 14 $\times$ 10² m² (Rounding off of two significant figures)

Mean of the five observations,

$\mu=\frac{80.0+80.5+81.0+81.5+82}{5}=\frac{405.0}{5}=81$

$\therefore$ Mean error

$=\frac{|80-\mu|+|80.5-\mu|+|81.0-\mu|+|81.5-\mu| +|82-\mu|}{5}$

$=\frac{|80-81|+|80.5-81|+|81.0-81| +|81.5-81|+|82-81|}{5}$

$\begin{gathered} =\frac{1+0.5+0+0.5+1}{5}=\frac{3}{5}=0.6 \\ \therefore \quad \text { Mean % error }=\frac{0.6}{81} \times 100 \%=0.74 \% \end{gathered}$

According to question, $[X]=[C]=\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^2 \mathrm{Q}^2\right]$ and $[Z]=[B]=\left[\mathrm{MT}^{-1} \mathrm{Q}^{-1}\right]$

$\begin{aligned} \therefore \quad & {[Y]=\frac{[X]}{\left[Z^2\right]} } \\ & {[Y]=\frac{\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^2 \mathrm{Q}^2\right]}{\left[\mathrm{M}^2 \mathrm{~T}^{-2} \mathrm{Q}^{-2}\right]}=\left[\mathrm{M}^{-3} \mathrm{~L}^{-2} \mathrm{~T}^4 \mathrm{Q}^4\right] } \end{aligned}$