Units & Measurements

We have $[\omega ] = {T^{ - 1}}$ and

$[{\varepsilon _0}] = {{{{[QT]}^2}} \over {M{L^3}}}$

Therefore,

$\left[ {{1 \over {m{\varepsilon _0}}}} \right] = {{{L^3}} \over {{{[QT]}^2}}}$

Now, $\left[ {{{{e^2}} \over {m{\varepsilon _0}}}} \right] = {{{L^3}} \over {{{[T]}^2}}}$

Also, the number N is defined as number of particle in unit volume, that is, N = n/V.

$[N] = {1 \over {[V]}} = {1 \over {{L^3}}}$

Therefore,

$\left[ {{{N{e^2}} \over {m{\varepsilon _0}}}} \right] = {1 \over {{{[T]}^2}}}$

Therefore, the quantity $\sqrt {{{N{e^2}} \over {m{\varepsilon _0}}}} $ has the dimension of $\omega$.

Least count of screw gauge

$ = {{Pitch} \over {No.\,of\,divisions\,on\,the\,circular\,scale}}$

$ = {{0.5\,mm} \over {50}}$ = 0.01 mm

Diameter of ball, $D = MSR + CSR \times LC$

= 2.5 mm + (20)(0.01 mm) = 2.7 mm

Density, $\rho = {{Mass} \over {Volume}} = {M \over {{{4\pi } \over 3}{{\left( {{D \over 2}} \right)}^3}}}$

The relative error in the density is

${{\Delta \rho } \over \rho } = {{\Delta M} \over M} + {{3\Delta D} \over D}$

The relative percentage in the density is

${{\Delta \rho } \over \rho } \times 100 = \left[ {{{\Delta M} \over M} + {{3\Delta D} \over D}} \right] \times 100$

$ = 2\% + {{3 \times 0.01} \over {2.7}} \times 100 = 2\% + 1.11\% = 3.11\% $

In Vernier scale, 16 mm is divided into 20 parts, that is,

1 VSD = 16/20 mm.

Least count = 1 MSD $-$ 1 VSD

$LC = \left( {1 - {{16} \over {20}}} \right)$ mm = 0.22 mm

Relative error in measurement of time,

${{\Delta t} \over t} = {{1\,s} \over {40\,s}} = {1 \over {40}}$

Time period, $T = {{40\,s} \over {20}} = 2\,s$

Error in measurement of time period,

$\Delta T = T \times {{\Delta t} \over t} = 2\,s \times {1 \over {40}} = 0.05\,s$

The time period of simple pendulum is

$T = 2\pi \sqrt {{l \over g}} $

or, ${T^2} = {{4{\pi ^2}l} \over g}$

or, $g = {{4{\pi ^2}l} \over {{T^2}}}$

$\therefore$ ${{\Delta g} \over g} = {{2\Delta T} \over T} = 2 \times {1 \over {40}} = {1 \over {20}}$ ($\because$ ${{\Delta T} \over T} = {{\Delta t} \over t}$)

Percentage error in determination of g is

${{\Delta g} \over g} \times 100 = {1 \over {20}} \times 100 = 5\% $

Time period (T) $ = 2\pi \sqrt {{l \over g}} $

or ${t \over n} = 2\pi \sqrt {{l \over g}} $

$\therefore$ $g = {{(4{\pi ^2})({n^2})l} \over {{t^2}}}$

% error in $g = {{\Delta g} \over g} \times 100$

$ = \left( {{{\Delta l} \over l} \times {{2\Delta l} \over l}} \right) \times 100$

${E_I} = \left( {{{0.1} \over {64}} + {{2 \times 0.1} \over {128}}} \right) \times 100 = 0.3125\% $

${E_{II}} = \left( {{{0.1} \over {64}} + {{2 \times 0.1} \over {64}}} \right) \times 100 = 0.46875\% $

${E_{III}} = \left( {{{0.1} \over {20}} + {{2 \times 0.1} \over {36}}} \right) \times 100 = 1.055\% $

Hence, $\mathrm{E_I}$ is minimum.

Young's modulus $(Y)=\frac{\text { Stress }}{\text { Strain }}=\frac{4 m g l}{\pi d^{2} \Delta l}$

$\begin{gathered} \mathrm{Y}=\frac{4 m g l}{\pi d^{2} l} \\\\ =\frac{4 \times 1 \times 9.8 \times 2}{3.14 \times(0.4)^{2} \times(0.8) \times 10^{-6} \times 10^{-3}} \\\\ =2 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2} \end{gathered} $

Where, $\quad$ Mass $(m)=1 \mathrm{~kg}$

$\begin{aligned}g & =9.8 \mathrm{~m} / \mathrm{s}^{2} \\\\ \text { Length of wire } & =2 \mathrm{~m} \\\\ \pi & =3.14 \\\\ \text { Diameter }(d) & =0.4 \mathrm{~mm} \times 10^{-3} \mathrm{~m} \end{aligned}$

Extension in the length of wire

$\begin{aligned}(\Delta l) & =0.8 \mathrm{~mm} \\\\ & =0.8 \times 10^{-3} \mathrm{~m} \\\\ \frac{\Delta y}{y} & =\frac{2 \Delta d}{d}+\frac{\Delta l}{l} \\\\ & =\frac{2 \times(0.01)}{0.4}+\frac{(0.05)}{0.8} \\\\ & =0.1125\end{aligned}$

$\begin{aligned}\Rightarrow \Delta \mathrm{Y} & =(\mathrm{Y}) \times 0.1125 \\\\ & =2 \times 10^{11} \times 0.1125 \\\\ & =0.225 \times 10^{11} \\\\ & \simeq 0.2 \times 10^{11} \\\\ \text { (b) }\left(2 \pm 0.2 \times 10^{11} \mathrm{~N/}\right. & \left.\mathrm{m}^{2}\right) \end{aligned}$

Given:

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

Errors are in $l$ and $T$.

• $ l = 1 \text{ m} $

• $ \Delta l $ is given.

• The student records total time for $n$ oscillations.

So, time for one oscillation:

$ T = \frac{T_n}{n} $

Step 1: Relate fractional errors

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

$ \frac{\Delta g}{g} = \frac{\Delta l}{l} + 2 \frac{\Delta T}{T} $

But note $T$ is the period of one oscillation, and the student measures total time $T_n$ for $n$ oscillations.

So,

$ T = \frac{T_n}{n} \implies \frac{\Delta T}{T} = \frac{\Delta T_n}{T_n} $

Determine uncertainty in $T_n$.

Step 2: Error in measured total time $ T_n $

Total error in timing for $n$ oscillations (using a stopwatch of least count $\Delta T$):

$ \Delta T_n = \sqrt{(\Delta T)^2 + (0.1)^2} $

since least count and human error combine.

So,

$ \frac{\Delta T}{T} = \frac{\sqrt{(\Delta T)^2 + (0.1)^2}}{T_n} $

Step 3: Estimate the time for one oscillation

For $l = 1\,\text{m}$,

$ T = 2\pi \sqrt{\frac{1}{9.8}} \approx 2.0\,\text{s} $

Thus $T \approx 2\,\text{s}$.

Hence $T_n = nT = 2n\,\text{s}$.

Step 4: Compute fractional errors

$ \frac{\Delta g}{g} = \frac{\Delta l}{l} + 2 \frac{\Delta T_n}{T_n} $

Now compute for each case.

Option A

$ \Delta l = 5\text{ mm} = 0.005\text{ m}, \; \Delta T = 0.2\text{ s}, n = 10 $

$ \frac{\Delta l}{l} = 0.005 $

$ T_n = 2n = 20\,\text{s} $

$ \Delta T_n = \sqrt{(0.2)^2 + (0.1)^2} = \sqrt{0.05} \approx 0.224 $

$ \frac{\Delta T_n}{T_n} = 0.224/20 = 0.0112 $

$ \frac{\Delta g}{g} = 0.005 + 2(0.0112) = 0.0274 $

Option B

$ \Delta l = 0.005,\ \Delta T = 0.2,\ n=20 $

$ T_n = 40\text{ s} $

$ \Delta T_n = 0.224 $

$ \frac{\Delta T_n}{T_n} = 0.224/40 = 0.0056 $

$ \frac{\Delta g}{g} = 0.005 + 2(0.0056)=0.0162 $

Option C

$ \Delta l = 0.005, \; \Delta T = 0.1, \; n=10 $

$ \Delta T_n = \sqrt{(0.1)^2+(0.1)^2}=\sqrt{0.02}=0.141 $

$ T_n = 20\,\text{s} $

$ \frac{\Delta T_n}{T_n}=0.141/20 = 0.00705 $

$ \frac{\Delta g}{g}=0.005 + 2(0.00705)=0.0191 $

Option D

$ \Delta l = 1\text{ mm} = 0.001\,\text{m},\ \Delta T=0.1,\ n=50 $

$ \Delta T_n = 0.141 $

$ T_n = 100\,\text{s} $

$ \frac{\Delta T_n}{T_n} = 0.141 / 100 = 0.00141 $

$ \frac{\Delta g}{g} = 0.001 + 2(0.00141) = 0.00382 $

✅ Option D has the smallest fractional error, i.e. the most accurate measurement of $ g $.

✅ Final Answer: Option D

One division on main scale is $p=0.5 \mathrm{~mm}$ (pitch) and one division on circular scale (Least Count) is

$ \mathrm{LC}=p / N=0.5 / 50=0.01 \mathrm{~mm} . $

From the figure (i), zero error corresponding to five circular scale divisions is

$ e=5 \times \mathrm{LC}=5 \times 0.01=0.05 \mathrm{~mm} $

From the figure (ii), measurement by screw gauge corresponding to two main scale divisions and 25 circular scale divisions is

$ \begin{aligned} D_{\text {measured }} & =2 \times p+25 \times \mathrm{LC} \\ & =2 \times 0.5+25 \times 0.01=1.25 \mathrm{~mm} . \end{aligned} $

The diameter of the ball is equal to its measured value minus zero error of the screw gauge i.e.,

$ D_{\text {actual }}=D_{\text {measured }}-e=1.25-0.05=1.20 \mathrm{~mm} . $

Option A

Pressure, Young’s modulus, and Stress

All three have the same dimensional formula:

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

✅ All same dimensions.

Option B

EMF, Potential difference, and Electric potential

All represent potential difference (energy per unit charge):

$ M L^{2} T^{-3} I^{-1} $

✅ All same dimensions.

Option C

Heat, Work done, and Energy

All represent forms of energy:

$ M L^{2} T^{-2} $

✅ All same dimensions.

Option D

Dipole moment, Electric flux, Electric field

Let’s write the dimensional formulas:

• Electric dipole moment (p) = charge × distance

  $ [p] = Q L = I T L $

• Electric flux (Φ) = $ E \times A $

  Electric field $ E $ has $ M L T^{-3} I^{-1} $, so

  $ [Φ] = M L^{3} T^{-3} I^{-1} $

• Electric field (E)

  $ M L T^{-3} I^{-1} $

⛔ These three clearly have different dimensions.

✅ Answer: Option D

Explanation:

Only Dipole moment, Electric flux, and Electric field have different dimensional formulas; all other sets have same dimensions.

1 division of main scale = 1 mm

Number of division in vernier scale = 10

Mass of cube = 2.736 g

Least count of vernier callipers = (1 main scale division) $-$ (1 vernier scale division)

$\begin{aligned} & =1 \mathrm{~mm}-\frac{9}{10} \mathrm{~mm} \\ & =1 \mathrm{~mm}-0.9 \mathrm{~mm} \\ & =0.1 \mathrm{~mm} \end{aligned}$

Length of the cube as measured by vernier callipers

$\begin{aligned} & =10 \mathrm{~mm}+(1 \times 0.1 \mathrm{~mm}) \\ & =10 \mathrm{~mm}+0.1 \mathrm{~mm} \\ & =10.1 \mathrm{~mm} \\ & =1.01 \mathrm{~cm} \end{aligned}$

density of the material of the cube $=\frac{\text { mass }}{\text { volume }}$

$\begin{aligned} \rho & =\frac{2.736 \mathrm{~g}}{(1.01)^{3} \mathrm{~cm}^{3}} \\ & =2.66 \mathrm{~g} / \mathrm{cm}^{3} \\ \rho \text { in SI } & =2.66 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3} . \end{aligned}$

To determine which pairs have equivalent dimensions, we examine the respective dimensional formulas.

For work (W) specified as $ \overrightarrow{F} \cdot \overrightarrow{s} $:

$ W = F s \cos \theta $

The dimensions are:

$ W = [MLT^{-2}][L] = [ML^2T^{-2}] $

For torque ($\overrightarrow{\tau}$) given as $ \overrightarrow{r} \times \overrightarrow{F} $:

$ \tau = rF \sin \theta $

The dimensions are:

$ \tau = [L] [MLT^{-2}] = [ML^2T^{-2}] $

Therefore, the dimensions of torque and work are identical.