Units & Measurements

S is emf per unit temperature ⇒

$[S]=\frac{[V]}{[K]}=\frac{M\,L^2\,T^{-3}\,I^{-1}}{K} =M\,L^2\,T^{-3}\,I^{-1}\,K^{-1}.$

σ is electrical conductivity ⇒

$[σ]=\frac{[J]}{[E]} =\frac{I\,L^{-2}}{M\,L\,T^{-3}\,I^{-1}} =M^{-1}\,L^{-3}\,T^3\,I^2.$

κ is thermal conductivity ⇒

$[κ]=\frac{\dot Q/(A)}{∇T} =\frac{M\,L^2\,T^{-3}/L^2}{K\,L^{-1}} =M\,L\,T^{-3}\,K^{-1}.$

Now

$[Z] =[S]^2\,[σ]\,[κ]^{-1} =(M\,L^2\,T^{-3}\,I^{-1}\,K^{-1})^2 \times M^{-1}\,L^{-3}\,T^3\,I^2 \times (M\,L\,T^{-3}\,K^{-1})^{-1}$

Simplifying exponents gives

$[Z]=M^0\,L^0\,T^0\,I^0\,K^{-1}.$

Answer: Option B.

For vernier scale,

$ \begin{aligned} & 10 \mathrm{div}=7 \mathrm{~mm} \\ & 1 \mathrm{div}=0.7 \mathrm{~mm} \end{aligned} $

Reading: main scale $=1 \mathrm{~mm}$

VS 1 marking matches with MS div

So VS reads $(1-0.7 \mathrm{~mm})=0.3 \mathrm{~mm}$

$ \begin{aligned} \text { So total } & =(1+0.3) \mathrm{mm} \\ & =1.3 \mathrm{~mm} \\ & =0.13 \mathrm{~cm} \text { Ans. } \end{aligned} $

To find the values of $(\alpha, \beta, \gamma, \delta)$ for which the expression $e^\alpha \varepsilon_0{ }^\beta h^\gamma c^\delta$ is dimensionless, we need to ensure that the product of the quantities has no overall physical dimensions. This means the dimensions of charge ($e$), permittivity ($\varepsilon_0$), Planck's constant ($h$), and speed of light ($c$) need to cancel each other out.

Let's look at the dimensions of each of the physical quantities:

Electronic charge ($e$): The dimension of charge in terms of fundamental dimensions is:

$[e] = [A \cdot T]$

where $A$ is the dimension of current and $T$ is the dimension of time.

Permittivity of free space ($\varepsilon_0$): The dimension is given by:

$[\varepsilon_0] = [\text{M}^{-1} \text{L}^{-3} \text{T}^4 \text{A}^2]$

Planck's constant ($h$): The dimension is:

$[h] = [\text{M} \text{L}^2 \text{T}^{-1}]$

Speed of light ($c$): The dimension is:

$[c] = [\text{L} \text{T}^{-1}]$

Now, we write the expression for the dimensionless quantity and set its dimensions to zero:

$[e^\alpha \varepsilon_0{ }^\beta h^\gamma c^\delta] = [A^\alpha T^\alpha] [\text{M}^{-\beta} \text{L}^{-3\beta} \text{T}^{4\beta} \text{A}^{2\beta}] [\text{M}^\gamma \text{L}^{2\gamma} \text{T}^{-\gamma}] [\text{L}^\delta \text{T}^{-\delta}]$

Combining the dimensions, we get:

$[e^\alpha \varepsilon_0{ }^\beta h^\gamma c^\delta] = [\text{M}^{-\beta + \gamma} \text{L}^{-3\beta + 2\gamma + \delta} \text{T}^{\alpha + 4\beta - \gamma - \delta} \text{A}^{\alpha + 2\beta}]$

For the quantity to be dimensionless, the exponents of $\text{M}$, $\text{L}$, $\text{T}$, and $\text{A}$ must all be zero:

• $-\beta + \gamma = 0$
• $-3\beta + 2\gamma + \delta = 0$
• $\alpha + 4\beta - \gamma - \delta = 0$
• $\alpha + 2\beta = 0$

Solving the above set of equations, we start with the fourth equation:

$\alpha + 2\beta = 0 \Rightarrow \alpha = -2\beta$

Substituting $\alpha = -2\beta$ into the third equation, we get:

$-2\beta + 4\beta - \gamma - \delta = 0 \Rightarrow 2\beta - \gamma - \delta = 0 \Rightarrow \gamma + \delta = 2\beta$

From the first equation:

$-\beta + \gamma = 0 \Rightarrow \gamma = \beta$

Then, substituting $\gamma = \beta$ into $\gamma + \delta = 2\beta$:

$\beta + \delta = 2\beta \Rightarrow \delta = \beta$

Thus, we have:

$\alpha = -2\beta$

$\gamma = \beta$

$\delta = \beta$

Let $\beta = -n$ where $n$ is a non-zero integer. Then, we can write the values as:

$\alpha = -2(-n) = 2n$

$\beta = -n$

$\gamma = -n$

$\delta = -n$

Hence, the tuple $(\alpha, \beta, \gamma, \delta)$ is given by:

$ (2n, -n, -n, -n) $

So, the correct option is Option A: $(2n,-n,-n,-n)$

Given the equation, $ Y = c^\alpha h^\beta G^\gamma $

where Y represents Young's modulus, c the speed of light, h is Planck's constant, G is the gravitational constant, and α, β, and γ are powers to be determined.

We start by expressing the quantities involved in terms of their fundamental dimensions: Mass (M), Length (L), and Time (T). The equation in terms of dimensions is :

$ [M^1 L^{-1} T^{-2}] = [M^0 L^1 T^{-1}]^\alpha [M^1 L^2 T^{-1}]^\beta [M^{-1} L^3 T^{-2}]^\gamma $

Next, by equating the powers of the dimensions M, L, and T on both sides of the equation, we form three separate equations:

1. Equating the powers of M, we get : $ \beta - \gamma = 1 $

2. Equating the powers of L, we get : $ \alpha + 2\beta + 3\gamma = -1 $

3. Equating the powers of T, we get : $ -\alpha - \beta - 2\gamma = -2 $

Now we can solve this system of equations.

From the first equation, we get $ \beta = \gamma + 1 $.

Substituting $ \beta $ from the first equation into the second and third equations, we get :

For the second equation, substituting $ \beta $ we get : $ \alpha + 2(\gamma + 1) + 3\gamma = -1 $, which simplifies to $ \alpha = -5\gamma - 3 $.

For the third equation, substituting $ \beta $ we get : $ -\alpha - (\gamma + 1) - 2\gamma = -2 $, which simplifies to $ \alpha = -3\gamma - 1 $.

Setting the two equations for $ \alpha $ equal to each other gives : $ -5\gamma - 3 = -3\gamma - 1 $, which simplifies to $ \gamma = -2 $.

Substituting $ \gamma = -2 $ back into the equations for $ \alpha $ and $ \beta $ we find $ \alpha = 7 $ and $ \beta = -1 $.

So, the solution to the system is $ \alpha = 7, \beta = -1, \gamma = -2 $, which corresponds to Option A.

$ \begin{aligned} & \text {LC }=\frac{0.1}{100}=0.001 \mathrm{~mm} \\\\ & \text {Zero error }=4 \times 0.001=0.004 \mathrm{~mm} \\\\ & \text {Reading } 1=0.5 \times 4+20 \times 0.001-0.004=2.16 \mathrm{~mm} \\\\ & \text {Reading } 2=0.5 \times 4+16 \times 0.001-0.004=2.12 \mathrm{~mm} \\\\ & \text {Mean value }=2.14 \mathrm{~mm} \\\\ & \text {Mean absolute error }=\frac{0.02+0.02}{2}=0.02 \\\\ & \text {Diameter }=2.14 \pm 0.02 \\\\ & \text {Area }=\frac{\pi}{4} \mathrm{~d}^2 \end{aligned} $

Least count of Vernier calipers (L.C) = $\left( {1 - {9 \over {10}}} \right)0.1$ = 0.01 cm

We know that main scale reading (MSR) is the first reading on the main scale immediately to the left of the zero of the Vernier scale. But there are no marks on the main scale before zero of the Vernier scale. We claim that MSR = −0.1 cm. The Vernier scale reading is VSR = 6 and the least count is LC = 0.01 cm.

Substitute these values to get,

Zero Error = MSR + VSR $ \times $ LC

= -0.1 + 6 $ \times $ 0.01 = -0.04 cm

Now in the second figure, the reading from main scale is 3.1 cm will be added to 1^st matching division of vernier so,

Reading = 3.1 cm + 1 $ \times $ L.C

= 3.1 cm + 0.01

= 3.11 cm

So correct diameter of the sphere

= 3.11 - (Zero Error)

= (3.11 + 0.04) cm

= 3.15 cm

Given, ratio $r = {{1 - a} \over {1 + a}}$ ..... (1)

Given, error in measurement of a is $\Delta$a ($\Delta$a/a << 1)

Taking natural log of Eq. (1), we get

$\ln r = \ln \left( {{{1 - a} \over {1 + a}}} \right)$

$ \Rightarrow \ln r = \ln (1 - a) - \ln (1 + a)$

${{\Delta r} \over r} = {{\Delta a} \over {1 - a}} + {{\Delta a} \over {1 + a}} = {{(1 + a)\Delta a + (1 - a)\Delta a} \over {(1 - a)(1 + a)}}$

$ \Rightarrow {{\Delta r} \over r} = {{2\Delta a} \over {(1 - a)(1 + a)}}$

$ \Rightarrow \Delta r = r\,.\,{{2\Delta a} \over {(1 - a)(1 + a)}}$

Substituting value of r, we get

$\Delta r = {{1 - a} \over {1 + a}}{{2\Delta a} \over {(1 - a)(1 + a)}} = {{2\Delta a} \over {{{(1 + a)}^2}}}$

Therefore, error $\Delta$r in determining r is ${{2\Delta a} \over {{{(1 + a)}^2}}}$

The law of radioactive decay gives number of undecayed nuclei N at time t as $N = {N_0}{e^{ - \lambda t}}$, where N₀ is number of nuclei at t = 0 and $\lambda$ is the decay constant. The number of nuclei decayed till time t is given by ${N_d} = {N_0} - N$. Thus, the decay constant is given by

$\lambda = {1 \over t}\ln {{{N_0}} \over N} = {1 \over t}\ln {{{N_0}} \over {{N_0} - {N_d}}}$.

Since, N₀ = 3000 and t = 1 s are given without any error, we assume that error in $\lambda$ is due to error in N_d only. Hence, we can write above equation as

$\lambda + \Delta \lambda = {1 \over t}\ln {{{N_0}} \over {{N_0} - ({N_d} + \Delta {N_d})}}$

$ = {1 \over t}\ln {{{N_0}} \over {({N_0} - {N_d})(1 - \Delta {N_d}/({N_0} - {N_d}))}}$

$ = {1 \over t}\ln {{{N_0}} \over {{N_0} - {N_d}}} - {1 \over t}\ln \left( {1 - {{\Delta {N_d}} \over {{N_0} - {N_d}}}} \right)$

$ \approx \lambda + {1 \over t}{{\Delta {N_d}} \over {{N_0} - {N_d}}}$,

which gives $\Delta \lambda = {{\Delta {N_d}} \over {t({N_0} - {N_d})}}$. Substitute $\Delta$N_d = 40, t = 1 s, and N_d = 1000 to get $\Delta$$\lambda$ = 0.02.

Let the time taken by stone to reach bottom of will be t₁ and time taken by sound to reach the observer be t₂ then

${t_1} = \sqrt {{{2L} \over g}} $ .... (1_)

and ${t_2} = {L \over {{v_s}}}$ .... (2)

Where L is depth of well

g is acceleration due to gravity

v_s is velocity of sound

Total time taken, $T = {t_1} + {t_2} = {{\sqrt {2L} } \over g} + {L \over {{v_s}}}$ ..... (3)

If $\delta$L is error in depth of well measured and $\delta$T is error is time measured then

$T + \delta T = \sqrt {{{2(L + \delta L)} \over g}} + {{(L + \delta L)} \over {{v_s}}}$

$T + \delta T = \sqrt {{{2L} \over g}\left( {1 + {{\delta L} \over L}} \right)} + {L \over {{v_s}}}\left( {1 + {{\delta L} \over L}} \right)$

expanding ${\left( {1 + {{\delta L} \over L}} \right)^{1/2}}$ using binomial approximation

$T + \delta T = \sqrt {{{2L} \over g}} \left( {1 + {1 \over 2}{{\delta L} \over L}} \right) + {L \over {{V_s}}}\left( {1 + {{\delta L} \over L}} \right)$

$ = \sqrt {{{2L} \over g}} + \sqrt {{{2L} \over g}} \left( {{1 \over 2}{{\delta L} \over L}} \right) + {L \over {{v_s}}} + {L \over {{v_s}}}{{\delta L} \over L}$

$ = \sqrt {{{2L} \over g}} + {L \over {{v_s}}} + \left( {{1 \over 2}\sqrt {{{2L} \over g}} + {L \over {{v_s}}}} \right){{\delta L} \over L}$

given, L = 20 m, g = 10 m/s, v_s = 300 m/s and using equation (3) : $\sqrt {{{2L} \over g}} + {L \over {{v_s}}} = T$, above equation becomes

$T + \delta T = T + \left( {{1 \over 2}\sqrt {{{2 \times 20} \over {10}}} + {{20} \over {300}}} \right){{\delta L} \over L}$

$ \Rightarrow \delta T = \left( {{1 \over 2}\sqrt 4 + {1 \over {15}}} \right){{\delta L} \over L} = \left( {1 + {1 \over {15}}} \right){{\delta L} \over L} = \left( {{{16} \over {15}}} \right){{\delta L} \over L}$

$ \Rightarrow {{\delta L} \over L} = \delta T\left( {{{15} \over {16}}} \right)$

It is given that $\delta$T = 0.01 sec, therefore,

$\left( {{{\delta L} \over L}} \right) \times 100\% = {{15} \over {16}} \times {1 \over {100}} \times 100\% = {{15} \over {16}}\% $

error $ \simeq 1\% $

Answer (B)

In both calipers C₁ and C₂, 1 cm is divided into 10 equal divisions on the main scale. Thus, 1 division on the main scale is equal to x_m1 = x_m2 = 1 cm/10 = 0.1 cm.

In calipers C₁, 10 equal divisions on the Vernier scale are equal to 9 main scale divisions.
Thus, 1 division on the Vernier scale of C₁ is equal to x_v1 = 9x_m1/10 = 0.09 cm.

In calipers C₂, 10 equal divisions on the Vernier scale are equal to 11 main scale divisions.
Thus, 1 division on the Vernier scale of C₂ is equal to x_v2 = 11x_m2/10 = 0.11 cm.


Let main scale reading be MSR and v^th division of the Vernier scale coincides with m^th division of the main scale (m is counted beyond MSR). The value measured by this calipers is

X = MSR + x = MSR + mx_m $-$ vx_v .........(1)

In calipers C₁, MSR₁ = 2.8 cm, m₁ = 7 and v₁ = 7 and in calipers C₂, MSR₂ = 2.8 cm, m₂ = 8 and v₂ = 7. Substitute these values in equation (1) to get

X₁ = MSR₁ + m₁x_m1 $-$ v₁x_v1

= 2.8 + 7(0.1) $-$ 7(0.09) = 2.87 cm.

X₂ = MSR₂ + m₂x_m2 $-$ v₂x_v2

= 2.8 + 8(0.1) $-$ 7(0.11) = 2.83 cm.

Let's break down the dimensional analysis of each physical quantity:

P. Boltzmann Constant ($k_B$)

The Boltzmann constant relates energy to temperature. Its dimensional formula is derived from the relationship: $E = k_B T$.

Energy ($E$) has dimensions of [ML²T^-2]. Temperature ($T$) has dimensions of [K]. Therefore:

$[k_B] = \frac{[E]}{[T]} = \frac{[ML^2T^{-2}]}{[K]} = [ML^2T^{-2}K^{-1}]$

Q. Coefficient of Viscosity ($\eta$)

Viscosity is a measure of a fluid's resistance to flow. It is defined as the ratio of shear stress to shear rate. Shear stress has dimensions of [ML^-1T^-2] (force per unit area), and shear rate has dimensions of [T^-1] (velocity gradient). Therefore:

$[\eta] = \frac{[ML^{-1}T^{-2}]}{[T^{-1}]} = [ML^{-1}T^{-1}]$

R. Plank Constant ($h$)

Planck's constant relates energy to frequency. Its dimensional formula is derived from the relationship: $E = h\nu$.

Energy ($E$) has dimensions of [ML²T^-2]. Frequency ($\nu$) has dimensions of [T^-1]. Therefore:

$[h] = \frac{[E]}{[\nu]} = \frac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}]$

S. Thermal Conductivity ($k$)

Thermal conductivity is a measure of a material's ability to conduct heat. It is defined as the rate of heat transfer per unit area per unit temperature gradient. Heat transfer rate has dimensions of [ML²T^-3]. Area has dimensions of [L²], and temperature gradient has dimensions of [K/L]. Therefore:

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

Matching the Dimensions:

Based on the dimensional analysis, we can match the quantities in List I with the dimensions in List II as follows:

List I	List II
P. Boltzmann Constant ($k_B$)	4. [ML2T-2K-1]
Q. Coefficient of Viscosity ($\eta$)	2. [ML-1T-1]
R. Plank Constant ($h$)	1. [ML2T-1]
S. Thermal conductivity ($k$)	3. [MLT-3K-1]

Therefore, the correct answer is Option C.

From the given data :

• One main scale division (MSD) is $ 5.15 \text{ cm} - 5.10 \text{ cm} = 0.05 \text{ cm} $.


• One Vernier scale division (VSD) is $ \frac{2.45 \text{ cm}}{50} = 0.049 \text{ cm} $.

The least count (LC) of the Vernier calipers is:

$ \text{LC} = 1 \text{ MSD} - 1 \text{ VSD} = 0.05 \text{ cm} - 0.049 \text{ cm} = 0.001 \text{ cm} $

Given the main scale reading (MSR) is 5.10 cm and the Vernier scale reading (VSR) is 24. The diameter $ D $ of the cylinder is calculated as:

$ \begin{aligned} D & = \text{MSR} + \text{VSR} \times \text{LC} \\\\ & = 5.10 \text{ cm} + 24 \times 0.001 \text{ cm} \\\\ & = 5.124 \text{ cm} \end{aligned} $

To determine Young's modulus using Searle's method, we use the formula:

$ Y = \frac{4MLg}{\pi l d^2} $

Given:
Length of wire, $L = 2 \text{ m}$
Diameter of wire, $d = 0.5 \text{ mm}$
Load, $M = 2.5 \text{ kg}$
Extension in length, $l = 0.25 \text{ mm}$

The quantities $d$ and $l$ are measured using a screw gauge and a micrometer, respectively, both having a pitch of 0.5 mm and 100 divisions on their circular scale. The least count, which is the smallest measurement that can be accurately read, is:

$ \text{Least count} = \frac{\text{Pitch}}{\text{Total number of divisions on the circular scale}} = \frac{0.5 \text{ mm}}{100} = 0.005 \text{ mm} $

Hence, the least count, $\Delta d$ and $\Delta l$, are both 0.005 mm.

Error in measurement of $l$ (extension length) is:

$ \frac{\Delta l}{l} = \frac{0.005}{0.25} = \frac{1}{50} $

Error in measurement of $d$ (diameter) is:

$ \frac{\Delta d}{d} = \frac{0.005}{0.5} = \frac{1}{100} $

Using the formula for Young's modulus, the combined error is:

$ Y = \frac{4MLg}{\pi ld^2} $

The relative error in $Y$ is given by:

$ \frac{\Delta Y}{Y} = \frac{\Delta l}{l} + 2 \frac{\Delta d}{d} $

As can be seen:

$ \frac{\Delta l}{l} = 2 \frac{\Delta d}{d} = \frac{1}{50} $

Therefore, the contributions to the error in the measurement of $Y$ from errors in measuring $l$ and $d$ are the same.

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

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

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

The given expression is

$P = {\alpha \over \beta }\exp \left( { - {{\alpha z} \over {k\theta }}} \right)$

Now,

$[z] = [{L^1}]$

$[\theta ] = [{K^1}]$

$[k] = [{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}]$

Since the terms which are raised to power in an exponential function are dimensionless, we get

$[\alpha] = \left[ {{{k\theta } \over z}} \right] = [{M^1}{L^1}{T^{ - 2}}]$

Also, $[P] =$ $\left[ {{\alpha \over \beta }} \right]$

$ \Rightarrow $ $\left[ \beta \right] = \left[ {{\alpha \over P}} \right]$

$ \Rightarrow $ $\left[ \beta \right] = \left[ {{{{M^1}{L^1}{T^{ - 2}}} \over {M{L^{ - 1}}{T^{ - 2}}}}} \right]$ = $[{M^0}{L^2}{T^0}]$

Answer (C)

We have

V = l³ = (1.2 $\times$ 10^$-$2)³ = 1.728 $\times$ 10^$-$6 m³

Since l has two significant figures, V should also have two significant figures; hence,

V = 1.7 $\times$ 10^$-$6 m³

Answer (A)

Given that $X = {\varepsilon _0}L\left( {{{\Delta V} \over {\Delta t}}} \right)$

Therefore, dimensional formula for X is

$[{M^{ - 1}}{L^{ - 3}}{T^4}{I^2}][L]{{[M{L^2}{T^{ - 3}}{I^{ - 1}}]} \over {[T]}} = [I]$

Thus, dimensional formula for X is same as that of current.

Answer (D)

Dimension of $\varepsilon $₀ = M^$-$1L^$-$3T⁴A²

Dimension of electric field E = MLT^$-$3A^$-$1

Now, dimension of

${1 \over 2}$$\varepsilon $₀E² = (M^$-$1L^$-$3T⁴A²)(MLT^$-$3A^$-$1)²

= M^$-$1L^$-$3T⁴A²M²L²T^$-$6A^$-$2 = ML^$-$1T^$-$2

Answer (C)

$ \begin{aligned} & l=10.5 \mathrm{~cm} \rightarrow 3 \mathrm{SF} \\ & b=0.05 \mathrm{~mm}=5 \times 10^{-3} \mathrm{~cm} \rightarrow 1 \mathrm{SF} \\ & h=6.0 \mu \mathrm{~m}=6.0 \times 10^{-4} \mathrm{~cm} \rightarrow 2 \mathrm{SF} \end{aligned} $

we know, $V=$ lbh

$ \begin{aligned}So\,\,\,\, v & =10.5 \times 5 \times 10^{-3} \times 6.0 \times 10^{-4} \\ & =315 \times 10^{-7} \mathrm{~cm}^3 \\ & =3.15 \times 10^{-5} \mathrm{~cm}^3 \end{aligned} $

The final answer should have the same number of significant figures as the factor with the fewest significant figures.

$ \text { So } V=3 \times 10^{-5} \mathrm{~cm}^3 \quad \rightarrow 1 \mathrm{SF} $

Dimension of electric field $[E] = [M{A^{ - 1}}L{T^{ - 3}}]$

Dimensions of magnetic field [B]

$[B] = [M{A^{ - 1}}L{T^{ - 2}}]$

Dimensions of magnetic permebility

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

$[S] = {{EB} \over {{\mu _0}}} = {{[M{A^{ - 1}}L{T^{ - 3}}][M{A^{ - 1}}{T^{ - 2}}]} \over {[M{A^{ - 2}}{T^{ - 2}}L]}} = [M{T^{ - 3}}]$

$[x] = {x^\alpha } \Rightarrow {p \over v} = time$

$[v] = {x^\beta }$

$[a] = {x^p} \Rightarrow {v \over a} = time$

$[p] = {x^q}$

$[F] = {x^r}$

$ \Rightarrow \left( {{{position} \over {speed}}} \right) = \left( {{{speed} \over {acceleration}}} \right) = \left( {{p \over F}} \right)$

$ \Rightarrow {{{x^\alpha }} \over {{x^\beta }}} = {{{x^\beta }} \over {{x^\alpha }}} = {{{x^q}} \over {{x^r}}}$

$ \Rightarrow {x^{\alpha - \beta }} = {x^{\beta - p}} = {x^{q - r}}$

$ \therefore $ $\alpha - \beta = \beta - p = q - r$

$\alpha + p = 2\beta $

$q + p = \beta + r$

$p + q - r = \beta $

$[M] = [Mass] = [{M^0}{L^0}{T^0}]$

[J] = [Angular momentum] = [ML²T^-1]

[L] = [Length]

Now, $[M{L^2}{T^{ - 1}}] = [{M^0}{L^0}{T^0}]$

$ \therefore $ [L²] = [T]

Power [P] = [MLT².LT^-1] = [ML²T^-3]

= [L²L^-6]

[P] = [L^-4\

Energy/Work [W] = [MLT^-2.L]

= [L²L^-4] = [L^-2]

Force [F] = [MLT^-2] = [L.L^-4] = [L^-3]

Linear momentum [p] = [MLT^-1] = [L.L^-2]

[p] = [L^-1]

The time period is measured for five successive measurements as follows :

T(s)	0.52	0.56	0.57	0.54	0.59
$\Delta $T	0.04	0	0.01	0.02	0.03

Therefore,

${T_{mean}} = {{0.52 + 0.56 + 0.57 + 0.54 + 0.59} \over 5}$

$ = {{2.78} \over 5} = 0.556 \approx 0.56$

and we know $\Delta$T(absolute error) = $\left| {{T_{mean}} - T} \right|$

Error in reading = $|{T_{mean}} - {T_1}| = 0.04$
$|{T_{mean}} - {T_2}| = 0.00$
$|{T_{mean}} - {T_3}| = 0.01$
$|{T_{mean}} - {T_4}| = 0.02$
$|{T_{mean}} - {T_5}| = 0.03$

$ \Rightarrow \Delta {T_{mean}} = {{0.04 + 0 + 0.01 + 0.02 + 0.03} \over 5} = 0.02$

Thus, the error in the measurement of T is

${{\Delta T} \over T} \times 100 = {{0.02} \over {0.56}} \times 100 = 3.57\% $ .... (1)

Hence, option (B) is correct.

The error in the measurement of r is

${{\Delta r} \over r} \times 100 = {1 \over {10}} \times 100 = 10\% $ ..... (2)

Hence, option (A) is correct.

Now, it is given that

$T = 2\pi \sqrt {{{7(R - r)} \over {5g}}} $

$ \Rightarrow {T^2} = 4{\pi ^2}\left( {{{7(R - r)} \over {5g}}} \right)$

$ \Rightarrow g = {{28{\pi ^2}} \over 5}\left( {{{R - r} \over {{T^2}}}} \right)$

Taking log and differentiating on both the sides of this equation, we get

$\ln g = \ln \left( {{{28{\pi ^2}} \over 5}} \right) + \ln (R - r) - 2\ln T$

$ \Rightarrow {{\Delta g} \over g} = \left( {{{\Delta R + \Delta r} \over {R - r}}} \right) + {{2\Delta T} \over T}$

$ = \left( {{{1 + 1} \over {50}}} \right) + 2 \times 0.0357$

$ \Rightarrow {{\Delta g} \over g} \times 100 = \left( {{1 \over {25}} + 2 \times 0.0357} \right) \times 100$

$ = (4 + 7.14)\% = 11.14\% = 11\% $

Therefore, the error in the determined value g is 11%.

Answer (A), (B), (D)

$[n] = [{L^{ - 3}}];[q] = [AT]$

$[\varepsilon ] = [{M^{ - 1}}{L^{ - 3}}{A^2}{T^4}]$

$[T] = [L]$

$[l] = [L]$

$[{k_B}] = [{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}]$

(a) RHS

$ = \sqrt {{{[{L^{ - 3}}{A^2}{T^2}]} \over {[{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}][{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}][K]}}} $

$ = \sqrt {{{[{L^{ - 3}}{A^2}{T^2}]} \over {[{L^{ - 1}}{A^2}{T^2}]}}} = \sqrt {[{L^{ - 2}}]} = [{L^{ - 1}}]$ Wrong

(b) RHS

$ = \sqrt {{{[{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}][{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}][K]} \over {[{L^{ - 3}}{A^2}{T^2}]}}} $

$ = \sqrt {{{[{L^{ - 1}}{A^2}{T^2}]} \over {[{L^{ - 3}}{A^2}{T^2}]}}} = L$ Correct

(c) RHS

$ = \sqrt {{{[{A^2}{T^2}]} \over {[{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}][{L^{ - 2}}][{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}][K]}}} $

$ = \sqrt {[{L^3}]} $ Wrong

(d) RHS

$ = \sqrt {{{[{A^2}{T^2}]} \over {[{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}][{L^{ - 1}}][{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}][K]}}} $

$ = \sqrt {{{[{A^2}{T^2}]} \over {[{L^{ - 2}}{T^2}{A^2}]}}} = [L]$ Correct

The dimensions of Planck's constant is $h = [{M^1}{L^3}{T^{ - 2}}]$

Speed of light is $c = [{L^1}{T^{ - 1}}]$

Gravitational constant is $G = [{M^{ - 1}}{L^3}{T^{ - 2}}]$

Let $L \propto {h^x}{c^y}{G^z}$.

$[L] = {[{M^1}{L^2}{T^{ - 1}}]^x}{[{L^1}{T^{ - 1}}]^y}{[{M^{ - 1}}{L^3}{T^{ - 2}}]^z}$

Comparing, we get

$\left. {\matrix{ \hfill {x - z = 0} \cr \hfill {2x + y + 3z = 1} \cr \hfill { - x - y - 2z = 0} \cr } } \right\}$

Solving, we get

$x = z$

$y + 5x = 1$

$ - y - 3x = 0$

or, $2x = 1$

$x = {1 \over 2} = z$

$y = - {3 \over 2}$

Therefore,

$L \propto {h^{1/2}}{C^{ - 3/2}}{G^{1/2}}$

$L \propto \sqrt h $

$L \propto \sqrt G $

Let $M \propto {h^a}{C^b}{G^c}$

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

Comparing, we get $a - c = 1$.

$\left. \matrix{ 2a + b + 3c = 0 \hfill \cr - a - b - 2c = 0 \hfill \cr} \right\}a + c = 0$

Therefore,

$2a = 1$

$a = {1 \over 2}$

$c = - {1 \over 2}$

$b = {1 \over 2}$

Therefore,

$M \propto {h^{1/2}}{c^{1/2}}{G^{ - 1/2}}$

$M \propto \sqrt c $

In given Vernier callipers, each 1 cm is equally divided into 8 main scale divisions (MSD). Thus, 1 MSD = ${1 \over 8}$ = 0.125 cm. Further, 4 main scale divisions coincide with 5 Vernier scale divisions (VSD) i.e., 4 MSD = 5 VSD. Thus, 1 VSD = 4/5 MSD = 0.8 $\times$ 0.125 = 0.1 cm. The least count of the Vernier callipers is given by

LC = 1 MSD $-$ 1 VSD = 0.125 $-$ 0.1 = 0.025 cm.

In screw gauge, let l be the distance between two adjacent divisions on the linear scale. The pitch p of the screw gauge is the distance travelled on the linear scale when it makes one complete rotation. Since circular scale moves by two divisions on the linear scale when it makes one complete rotation, we get p = 2l. The least count of the screw gauge is defined as ratio of the pitch to the number of divisions on the circular scale (n) i.e.,

$LC' = {p \over n} = {{2l} \over {100}} = {l \over {50}}$ ....... (1)

If p = 2 LC = 2(0.025) = 0.05 cm, then $l = {p \over 2} = 0.025$ cm. Substitute l in equation (1) to get the least count of the screw gauge

$LC' = {{0.025} \over {50}} = 5 \times {10^{ - 4}}$ cm = 0.005 mm.

If l = 2 LC = 2(0.025) = 0.05 cm then equation (1) gives

$LC' = {{0.05} \over {50}} = 1 \times {10^{ - 3}}$ cm = 0.01 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\% $

From coulombs law, we know that

$F = {1 \over {4\pi {\varepsilon _0}}}{{{q_1}{q_2}} \over {{r^2}}} \Rightarrow {\varepsilon _0} = {1 \over {4\pi }}{{{q_1}{q_2}} \over {{r^2}F}}$

Therefore, ${\varepsilon _0} = {{[{I^2}{T^2}]} \over {[{L^2}][ML{T^{ - 2}}]}} = {M^{ - 1}}{L^{ - 3}}{T^4}{I^2}$

Now, $c = {1 \over {\sqrt {{\varepsilon _0}{\mu _0}} }} \Rightarrow {c^2} = {1 \over {{\varepsilon _0}{\mu _0}}}$

Therefore, ${\mu _0} = {1 \over {{c^2}{\varepsilon _0}}} = {1 \over {{{[L{T^{ - 1}}]}^2}[{M^{ - 1}}{L^{ - 3}}{T^4}{I^2}]}} = [ML{T^{ - 2}}{I^{ - 2}}]$

Answer (B), (C)

Induction $L = {\phi \over i} = $ weber/Ampere

Also, $e = - L\left( {{{di} \over {dt}}} \right) \Rightarrow L = {{ - e} \over {(di/dt)}} = {{volt - {\mathop{\rm second}\nolimits} } \over {Ampere}}$

Also, $U = {1 \over 2}L{i^2} \Rightarrow L = {{2U} \over {{i^2}}} = {{Joule} \over {{{(Ampere)}^2}}}$

and $U = {1 \over 2}L{i^2} = {i^2}RT \Rightarrow L = RT = $ Ohm-second

Answer (A), (B), (C), (D)