Units & Measurement and Dimensions

$ \begin{aligned} &\text { Area of plate, } A=l^2\\ &\begin{aligned} & \therefore \text { Pressure, } p=\frac{F}{A}=\frac{F}{l^2} \\ & \therefore \frac{\Delta p}{p} \times 100 \%=\frac{\Delta F}{F} \times 100 \%+2 \frac{\Delta l}{l} \times 10 \% \\ & =3 \%+2 \times 1 \%=5 \% \end{aligned} \end{aligned} $

Given, reading of viscosity (in mPa.s) $2.62,2.68,2.58,2.57,2.54$, $2.55$

Mean

$ \begin{aligned} & =\frac{2.62+2.68+2.58+2.57+2.54+2.55}{6} \\ & =\frac{15.54}{6}=2.59(\mathrm{mPa} . \mathrm{s}) \end{aligned} $

Absolute Errors

$ \begin{aligned} & 12.62-2.591=0.03 \\ & 12.68-2.591=0.09 \\ & 12.58-2.591=0.01 \\ & 12.57-2.591=0.02 \\ & 12.54-2.591=0.05 \\ & 12.55-2.591=0.04 \end{aligned} $

The mean absolute error

$ \begin{aligned} & =\frac{0.03+0.09+0.01+0.02+0.05+0.04}{6} \\ & =\frac{0.24}{6}=0.04[\mathrm{mPa} . \mathrm{s}] \end{aligned} $

To find the length of the remaining rod, we subtract the length of the cut piece from the original length of the rod.

Original length = 43.4 m

Cut length = 3.532 m

Remaining length = Original length - Cut length

Remaining length = 43.4 - 3.532

Let's perform the subtraction:

43.400 (adding zeros to align decimal places for calculation)

• 3.532

39.868

Now, we need to consider the rules for significant figures in subtraction. When adding or subtracting measurements, the result should be rounded to the same number of decimal places as the measurement with the fewest decimal places.

• The original length (43.4 m) has one decimal place.

• The cut length (3.532 m) has three decimal places.

The measurement with the fewest decimal places is 43.4 m, which has one decimal place. Therefore, our final answer must be rounded to one decimal place.

Our calculated exact result is 39.868.

To round 39.868 to one decimal place, we look at the second decimal place, which is 6. Since 6 is 5 or greater, we round up the first decimal place (8).

So, 39.868 rounded to one decimal place is 39.9.

Comparing this with the given options:

Option A: 39.9

Option B: 39.8

Option C: 39.868

Option D: 39.87

The correct answer is 39.9 m.

The final answer is $\boxed{\text{39.9}}$

The measured length is 830600 mm.

Because there is no decimal point in this number, the zeros at the end (the last two zeros) are not counted as significant figures.

So, only the digits 8, 3, 0, and 6 are significant.

Therefore, the number of significant figures in 830600 mm is 4.

We know that the formula for resonance angular frequency is:

$ \omega = \frac{1}{\sqrt{LC}} $

So, $ \omega = (LC)^{-1/2} $

This means the dimensional formula for $(LC)^{-1/2}$ is the same as the dimensional formula for $\omega$.

The dimensional formula for angular frequency $\omega$ is time inverse: $[\mathrm{T}^{-1}]$.

In full form, this is $[\mathrm{M}^0\,\mathrm{L}^0\,\mathrm{T}^{-1}]$.

According to principle of homogenity, physical quality can only be added or subtracted if they have the same dimensions.

$\frac{A-C}{B}$ is correct only when $A$ and $C$ must have some dimensions.

Number of significant figure in $0.03240=4$

Leading zeros are not significant, while trailing zeros after the decimal point are significant. Therefore, the significant figures are 3, 2, 4 and trailing zero.

$ \begin{aligned} {\left[\sqrt{\frac{K}{T}}\right] } & =\sqrt{\frac{\left[\mathrm{ML}^2 \mathrm{~T}^2\right]}{\left[\mathrm{MT}^{-2}\right]}}=\sqrt{\left[\mathrm{L}^2\right]}=[\mathrm{L}] \\ & =\text { Dimension of distance } \end{aligned} $

Force $=\frac{\alpha}{\text { density }+\beta^3}$

According to principle of homogeneity,

$ \begin{array}{ll} & {\left[\beta^3\right]=[\text { density }]} \\ & {\left[\beta^3\right]=\left[M L^{-3} T^0\right]} \\ \Rightarrow \quad & {[\beta]=\left[M^{\frac{1}{3}} L^{-1} T^0\right]} \end{array} $

$ \begin{aligned} \text { And }[\alpha] & =[\text { density }] \times[\text { force }] \\ & =\left[\mathrm{ML}^{-3} \mathrm{~T}^0\right] \times\left[\mathrm{MLT}^{-2}\right] \\ & =\left[\mathrm{M}^2 \mathrm{~L}^{-2} \mathrm{~T}^{-2}\right] \end{aligned} $

Given:

Error in surface area $ S $ = $ 1.2\% $

We need to find the percentage error in volume $ V $ of a sphere.

Relations:

For a sphere of radius $ r $:

$ S = 4\pi r^2 $

$ V = \frac{4}{3}\pi r^3 $

Step 1: Express relative errors

For small errors,

$ \frac{dS}{S} = 2\frac{dr}{r} $

$ \frac{dV}{V} = 3\frac{dr}{r} $

Step 2: Connect the two

From the first equation:

$ \frac{dr}{r} = \frac{1}{2}\frac{dS}{S} $

Substitute into the second:

$ \frac{dV}{V} = 3 \times \frac{1}{2}\frac{dS}{S} = \frac{3}{2}\frac{dS}{S} $

Step 3: Substitute given error

$ \text{Error in }V = \frac{3}{2} \times 1.2\% = 1.8\% $

✅ Final Answer: Option B) 1.8%

Let’s restate the given problem carefully:

The velocity of a particle is given by

$ v = a t + \frac{b}{t + c} $

We need to find the dimensions of $a$, $b$, and $c$.

Step 1. Dimensional formula of velocity

$ [v] = [LT^{-1}] $

Step 2. Dimensional analysis of first term $a t$

In the expression $v = a t + \frac{b}{t+c}$, both terms on the right-hand side must have the same dimensions as velocity.

The first term: $a t $

$ [a][t] = [v] = [LT^{-1}] $

$ [a] = \frac{[LT^{-1}]}{[T]} = [LT^{-2}] $

So,

$ [a] = [LT^{-2}] $

Step 3. The second term $\frac{b}{t + c}$

This term must also have the dimensions of velocity:

$ \left[\frac{b}{t + c}\right] = [LT^{-1}] $

Now, $t + c$ is a sum — you can only add quantities with the same dimensions, so:

$ [c] = [t] = [T] $

Then,

$ [b] = [LT^{-1}] \cdot [t + c] = [LT^{-1}] \cdot [T] = [L] $

So,

$ [b] = [L] $

✅ Final Dimensions:

$ [a] = [LT^{-2}], \quad [b] = [L], \quad [c] = [T] $

✅ Correct Option:

Option A: $ \left[LT^{-2}\right], [L], [T] $

Stress $=\frac{F}{A}$

$ \Rightarrow[\text { Stress }]=\frac{[\mathrm{F}]}{[\mathrm{A}]}=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{L}^2\right]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right] $

Linear momentum, $p=m v$

$ \begin{aligned} {[p] } & =[m][v] \\ & =[\mathrm{M}]\left[\mathrm{LT}^{-1}\right]=\left[\mathrm{MLT}^{-1}\right] \end{aligned} $

Significant digit in simplification of $\frac{0.501}{0.05}(0.312-0.03)$

In the subtraction of $0.312-0.03$.

The number 0.03 has two digits after decimal.

So, $0.312-0.03=0.282$

But only upto two decimal places it can be written as 0.28 .

Now, $\frac{0.501}{0.05}$ the term 0.05 has two digits after decimal

$ \frac{0.501}{0.05}=10.02 \approx 10 $

Now, $10.02 \times 0.28=2.8056$

But this number should be rounded of to 2.8 .

So, number of significant digit is 2 .

Let's determine the dimensional formula of Planck’s constant (h).

We know from quantum theory:

$ E = h\nu $

where

$E$ = energy = $[ML^2T^{-2}]$

$\nu$ = frequency = $[T^{-1}]$

So,

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

✅ Therefore, the dimensional formula of Planck’s constant is

$ \boxed{[ML^2T^{-1}]} $

Correct option: C

The uncertainty in the difference is the sum of the uncertainties.

i.e. $(0.5+0.5)=1.0^{\circ} \mathrm{C}$

The difference in the central values

$ =44-22=22^{\circ} \mathrm{C} $

Thus, temperature difference

$ =22^{\circ} \mathrm{C} \pm 1^{\circ} \mathrm{C} $

$ \begin{aligned} &\text { Young's modulus }\\ &\begin{aligned} & \begin{aligned} {[Y]=\frac{[\text { Stress }]}{[\text { Strain }]} } & =\frac{\left[\frac{F}{A}\right]}{\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0\right]} \\ =\left[\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2}\right] & =\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right] \\ {[\text { Energy density }] } & =\left[\frac{E}{V}\right]=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]}{\left[\mathrm{L}^3\right]} \\ & =\left[\mathrm{ML}^{-1} \mathrm{~T}^2\right] \end{aligned} \end{aligned} \end{aligned} $

The statement that is not correct is:

Option A: All conserved quantities are necessarily scalars.

In physics, conserved quantities are those that remain constant in time under certain conditions. While some conserved quantities, like mass or energy, can be scalars, angular momentum, and linear momentum are examples of vector quantities that are also conserved. Thus, not all conserved quantities are scalars.

To understand this further, consider:

Energy and Mass are examples of scalar conserved quantities.

Linear Momentum can be written as a vector $\vec{p}$, where $\vec{p} = m \vec{v}$, with $m$ being mass and $\vec{v}$ the velocity vector.

Angular Momentum is given by a vector $\vec{L} = \vec{r} \times \vec{p}$, which is a cross product of position vector $\vec{r}$ and momentum vector $\vec{p}$.

The remaining options are correct:

Option B: The laws of nature do not change with time, as they are universal and consistent over time.

Option C: The laws of nature are the same everywhere in the universe, reflecting the principle of universality.

Option D: The law of gravitation is the same both on the Moon and the Earth, which illustrates the universality of gravitational law as described by Newton's law of universal gravitation.

Given,

Internal diameter,

$ d_{1}=(5.73 \pm 0.01) \mathrm{cm} $

External diameter,

$ d_{2}=(6.01 \pm 0.01) \mathrm{cm} $

Internal radius, $ r_{1}=\frac{d_{1}}{2} $

External radius, $ r_{2}=\frac{d_{2}}{2} $

Actual thickness, $ t=\frac{1}{2}\left(d_{2}-d_{1}\right) $

$ t=\frac{1}{2}(0.28)=0.14 \mathrm{~cm} $

Now, error in thickness,

We know from the properties of error that total error in substraction operation is addition of each error.

$ \begin{aligned} \Delta t & =\Delta t_{1}+\Delta t_{2} \\\\ \Delta t_{1} & =0.01 \mathrm{~cm}, \Delta t_{2}=0.01 \mathrm{~cm} \\\\ \Delta t & =0.02 \mathrm{~cm} \end{aligned} $

Hence, $ t=(0.14 \pm 0.02) \mathrm{cm} $

Given, $I=(10 \pm 0.5) \mathrm{A}$

Voltage $=(100 \pm 6) V$

We know that, $V=I R$

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

$ \begin{aligned} \Rightarrow \quad \frac{\Delta R}{R} & =\frac{\Delta V}{V}+\frac{\Delta I}{I} \\\\ & =\frac{6}{100}+\frac{0.5}{10} \\\\ \frac{\Delta R}{R} & =\frac{11}{100} \\\\ \text { \% error } & =\frac{\Delta R}{R} \times 100 \% \\\\ & =\frac{11}{100} \times 100 \%=11 \% \end{aligned} $

Given the equation of motion,

$ m \frac{d^2 x}{d t^2}+b \frac{d x}{d t}+k x=0 $

We need to find the dimensional formula of $\frac{b}{\sqrt{k m}}$.

Here,

$ \begin{aligned} {[b] } & =[\mathrm{M}][\mathrm{T}]^{-1} \\ {[k] } & =[\mathrm{M}][\mathrm{T}]^{-2} \\ {[m] } & =[\mathrm{M}] \\ \frac{b}{\sqrt{k m}} & =\frac{[\mathrm{M}][\mathrm{T}]^{-1}}{\sqrt{[\mathrm{M}][\mathrm{T}]^{-2} \cdot[\mathrm{M}]}} \\ & =\frac{[\mathrm{M}][\mathrm{T}]^{-1}}{\sqrt{[\mathrm{M}]^2[\mathrm{~T}]^{-2}}}=\frac{[\mathrm{M}][\mathrm{T}]^{-1}}{[\mathrm{M}][\mathrm{T}]^{-1}} \\ & =\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0\right] \end{aligned} $

So, the dimensional formula of $\frac{b}{\sqrt{\mathrm{~km}}}$ is $\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0\right]$.

The SI unit of permeability μ is henry per metre, i.e.

$[μ]=\frac{\rm H}{\rm m}\,. $

Now let’s check each option:

Henry metre⁻¹

$\frac{\rm H}{\rm m}$

⇒ correct unit.

Weber ampere⁻¹ metre⁻¹

$\frac{\rm Wb}{\rm A\,m} = \frac{\rm (V·s)}{\rm A\,m} = \frac{\rm H}{\rm m}$

⇒ equivalent to H/m.

Ohm second metre⁻¹

$\frac{\rm Ω·s}{\rm m} = \frac{\rm H}{\rm m}$

(since 1 H = 1 Ω·s)

⇒ also H/m.

Volt second metre⁻¹

$\frac{\rm V·s}{\rm m} = \frac{\rm Wb}{\rm m}$

which is missing the “per ampere” factor, so it is not H/m.

Answer: Option D.

In the given equation, $(p + \frac{a}{V^2})(V-b) = RT$, we need to find the dimensions of $a$.

Since a pressure can only be added to another pressure, this implies that $\frac{a}{V^2}$ must have the same dimensions as pressure $p$.

$ \frac{a}{V^2} = p $

Solving for $a$, we get:

$ a = p \times V^2 $

The dimensions of pressure $p$ are $\left[M L^{-1} T^{-2}\right]$.

The dimensions of volume $V$ are $\left[L^3\right]$.

Therefore, substituting the dimensions into the equation for $a$:

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

$ [a] = \left[M L^{-1} T^{-2}\right] \times \left[L^6\right] $

Simplifying, we find:

$ [a] = \left[M L^{5} T^{-2}\right] $

Thus, the dimensions of $a$ are $\left[M L^5 T^{-2}\right]$.

Given:

The length of the side of the cube, $ L = 1.2 \times 10^{-2} \, \mathrm{m} $.

To find the volume of the cube, we use the formula:

$ \text{Volume of the cube} = L^3 = (1.2 \times 10^{-2})^3 $

Calculating the volume:

$ = 1.728 \times 10^{-6} \, \mathrm{m}^3 $

Since the length $ L $ is given with 2 significant figures, the calculated volume should also be expressed in 2 significant figures. Therefore, the volume of the cube is:

$ 1.7 \times 10^{-6} \, \mathrm{m}^3 $

Given the equation:

$ F = a \sqrt{x} + b t^2 $

where:

$ F $ is force with the SI unit of $\text{kg} \cdot \text{m/s}^2$,

$ x $ represents distance in meters ($ \text{m} $),

$ t $ is time in seconds ($ \text{s} $).

From the principle of dimensional consistency, equating similar terms gives us:

$ \text{kg} \cdot \text{m/s}^2 = a \sqrt{\text{m}} + b t^2 $

Dimensional Analysis

Analyzing $ a \sqrt{x} $:

$ a \sqrt{x} $ must have the same dimensions as force, i.e., $\text{kg} \cdot \text{m/s}^2$.

For $ a \sqrt{\text{m}} = \text{kg} \cdot \text{m/s}^2 $,

$ a(\text{m}^{1/2}) = \text{kg} \cdot \text{m/s}^2 $

Solving for $ a $, we find:

$ a = \text{kg} \cdot \frac{\text{m}^{1/2}}{\text{s}^2} $

Analyzing $ b t^2 $:

$ b t^2 $ also has the dimension of force.

For $ b(\text{s}^2) = \text{kg} \cdot \text{m/s}^2 $,

$ b = \text{kg} \cdot \text{m/s}^4 $

Finding the Ratio $\frac{a}{b}$:

To derive the dimensions of $\frac{a}{b}$:

$ \frac{a}{b} = \frac{\text{kg} \cdot \frac{\text{m}^{1/2}}{\text{s}^2}}{\text{kg} \cdot \text{m/s}^4} $

Simplifying, we get:

$ = \frac{\frac{1}{\text{m}^{1/2}} \cdot \text{s}^2}{1} $

The dimensional formula for $\frac{a}{b}$:

$ \left[\text{M}^0 \text{L}^{-1/2} \text{T}^{2}\right] $

The dimensional formulas for energy $ E $, mass $ M $, angular momentum $ L $, and the gravitational constant $ G $ are given as follows:

$ \begin{aligned} & [E] = \left[\mathrm{M L}^2 \mathrm{T}^{-2}\right] \\ & [M] = [\mathrm{M}] \\ & [L] = \left[\mathrm{M L}^2 \mathrm{T}^{-1}\right] \\ & [G] = \left[\mathrm{M}^{-1} \mathrm{L}^3 \mathrm{T}^{-2}\right] \end{aligned} $

To find the dimensions of $\frac{E L^2}{M^5 G^2}$, we proceed as follows:

$ \begin{aligned} \Rightarrow \frac{\left[\mathrm{M L}^2 \mathrm{T}^{-2}\right]\left[\mathrm{M L}^2 \mathrm{T}^{-1}\right]^2}{\left[\mathrm{M}^5\right]\left[\mathrm{M}^{-1} \mathrm{L}^3 \mathrm{T}^{-2}\right]^2} \end{aligned} $

Simplifying further:

$ \begin{aligned} &= \frac{\left[\mathrm{M L}^2 \mathrm{T}^{-2}\right]\left[\mathrm{M}^2 \mathrm{L}^4 \mathrm{T}^{-2}\right]}{\left[\mathrm{M}^5\right]\left[\mathrm{M}^{-2} \mathrm{L}^6 \mathrm{T}^{-4}\right]} \\ &= \frac{\left[\mathrm{M}^3 \mathrm{L}^6 \mathrm{T}^{-4}\right]}{\left[\mathrm{M}^3 \mathrm{L}^6 \mathrm{T}^{-4}\right]} \\ &= \left[\mathrm{M}^0 \mathrm{L}^0 \mathrm{T}^0\right] \end{aligned} $

This shows that the expression $\frac{E L^2}{M^5 G^2}$ is dimensionless. A dimensionless quantity is equivalent to an angle in terms of its dimensional representation.

The kinetic energy $ K $ is given by:

$ K = \frac{1}{2} m v^2 $

To calculate the percentage error in the measurement of kinetic energy, we apply the rule of propagation of errors. According to this rule, the relative error in multiplication is the sum of the relative errors of the quantities involved. For kinetic energy:

$ \frac{\Delta K}{K} \times 100\% = \frac{\Delta m}{m} \times 100\% + 2 \frac{\Delta v}{v} \times 100\% $

Given that the percentage error in mass ($ \Delta m/m $) is $ 3\% $ and the percentage error in velocity ($ \Delta v/v $) is $ 4\% $, we can substitute these values into the formula:

$ \frac{\Delta K}{K} \times 100\% = 3\% + 2 \times 4\% $

$ = 3\% + 8\% $

$ = 11\% $

Thus, the percentage error in the measurement of kinetic energy is $ 11\% $.

Given, $X=\frac{2 k^3 l^2}{m \sqrt{n}}$ error percentage $k=1 \%, l=2 \%, m=3 \%$, $n=4 \%$

Total percentage error in $X$

$ \begin{aligned} \frac{\Delta X}{X} \times 100= & 3\left(\frac{\Delta k}{k}\right) \times 100+2\left(\frac{\Delta l}{l}\right) \times 100 \\ & +\left(\frac{\Delta m}{m} \times 100\right)+\frac{1}{2}\left(\frac{\Delta n}{n}\right) \times 100 \end{aligned} $

$ \begin{aligned} \Rightarrow \quad \frac{\Delta X}{X} \times 100= & 3 \times 1 \%+2 \times 2 \% +3 \%+\frac{1}{2} \times 4 \% \\ = & 3+4+3+2=12 \% \end{aligned} $

Given,

Figure $=0.079000$

Significant figures is the measured value of digits which are accurate. The rules for counting significant figures,

(i) All non zero digit are significant.

(ii) Zero placed left of a digit are never significant.

(iii) Zero placed after a digit is significant.

Applying these rules on figure given above.

Significant Figure $=5$

Given, efficiency of heat engine,

$ \eta=\frac{\alpha \beta}{\sin \theta} \cdot \log \frac{\beta x}{K T} $

$\eta$ is a dimensionless variable. Therefore $\alpha \beta$ and $\frac{\beta x}{K T}$ must have no dimensions.

Since, $\alpha, \beta$ are non-zero constants, thus $\beta$ cannot have demensions of force.

Similarly $\alpha^{-1} x$ also cannot have dimensions of energy.

$[\alpha \beta]=$ dimensionless and are constants. Therefore, both $\alpha$ and $\beta$ have same dimensions.

Given, $v_2=\frac{n}{m^2} v_1 ; a_2=\frac{a_1}{m n}$

Let initial velocities $u_1$ and $u_2$ be zero. Therefore,

$ \begin{aligned} \frac{v_1}{v_2} & =\frac{a_1 t_1}{a_2 t_2} \\ \Rightarrow \frac{v_1}{\frac{n}{m^2} v_1} & =\frac{a_1 t_1}{\frac{a_1}{n m} t_2} \Rightarrow \frac{m^2}{n}=n m \frac{t_1}{t_2} \\ \Rightarrow \quad t_2 & =\left(\frac{n^2}{m}\right) t_1 \end{aligned} $

From second equation of motion we have,

$ \begin{aligned} s_2 & =u_2 t_2+\frac{1}{2} a_2 t_2^2=\frac{1}{2}\left(\frac{a_1}{n m}\right)\left(t_2\right)^2 \\ & =\frac{1}{2} \frac{a_1}{n m} \frac{n^4}{m^2} t_1^2=\frac{n^3}{m^3} \frac{1}{2} a_1 t_1^2=\left(\frac{n}{m}\right)^3 S_1 \end{aligned} $

Hence, option (a) is correct.

In the given number, $3.78 \times 10^{22} \, \mathrm{kg}$, we need to determine the number of significant figures.

Significant figures are the digits in a number that contribute to its precision. They include all non-zero digits, any zeros between significant digits, and any trailing zeros in a decimal number. Here, the number is expressed in scientific notation. The mantissa, $3.78$, determines the significant figures.

The digits "3", "7", and "8" in $3.78$ are all non-zero digits.

Therefore, the number of significant figures in $3.78 \times 10^{22} \, \mathrm{kg}$ is 3.

Let's analyze the dimensions of each pair:

$\textbf{Current Density vs. Charge Density}$

Current density (usually denoted as $\vec{J}$) is the current per unit area:

$[\vec{J}] = \frac{[I]}{L^2}$

Charge density (commonly referring to volume charge density, $\rho$) is the charge per unit volume:

$[\rho] = \frac{[Q]}{L^3}$

Since charge $Q$ has dimensions $[I \cdot T]$, charge density becomes:

$[\rho] = \frac{[I \cdot T]}{L^3}$

These two are not equivalent because $[\vec{J}] = [I L^{-2}]$ and $[\rho] = [I T L^{-3}]$ have different dimensions.

$\textbf{Angular Momentum vs. Linear Momentum}$

Linear momentum $\vec{p}$ has dimensions:

$[\vec{p}] = MLT^{-1}$

Angular momentum $\vec{L}$ is given by $\vec{r} \times \vec{p}$, so its dimensions are:

$[\vec{L}] = L \times MLT^{-1} = ML^2T^{-1}$

They differ by a factor of length and are therefore not the same.

$\textbf{Spring Constant vs. Surface Energy}$

The spring constant $k$ in Hooke's law (where $F=kx$) has dimensions:

$[k] = \frac{[F]}{[L]} = \frac{MLT^{-2}}{L} = MT^{-2}$

Surface energy is energy per unit area. Energy has dimensions:

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

Dividing by area $L^2$ gives:

$ \frac{ML^2T^{-2}}{L^2} = MT^{-2}$

These both have the same dimensions, $MT^{-2}$.

$\textbf{Force vs. Torque}$

Force has dimensions:

$[F] = MLT^{-2}$

Torque is the cross product of a lever arm (length) and force:

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

They differ by an extra factor of length.

Thus, the pair with the same dimensions is:

$\boxed{\text{Option C: Spring constant and surface energy.}}$

Velocity of light, $[c]=\left[L T^{-1}\right]$

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

Planck's constant, $[h]=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]$

Let dimensions of density $d$ in new system are $d=c^a, G^b, h^c$

Then expressing both sides in terms of dimensions of fundamental units,

we get

$ \begin{aligned} {\left[\mathrm{ML}^{-3}\right] } & =\left[\mathrm{LT}^{-1}\right]^a\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]^b\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]^c \\ \Rightarrow\left[\mathrm{ML}^{-3}\right] & =\left[\mathrm{M}^{-b+c} \mathrm{~L}^{a+3 b+2 c} \mathrm{~T}^{-a-2 b-c}\right] \end{aligned} $

Comparing both sides we get,

$ \begin{aligned} -b+c & =1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ a+3 b+2 c & =-3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

and $-a-2 b-c=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

Solving these, we get

$ a=5, b=-2, c=-1 $

So, density can be expressed as

$ d=c^5 G^{-2} h^{-1} $

A → II, Boltzmann constant has dimensions of energy divided iemperature, the same as entropy. B → III Energy density (energy per unit volume) of stretched wire is given as

$ u=\frac{1}{2} \times Y \times \sigma^2 $

where, $Y=$ Young's modulus, $\sigma=$ strain.

Thus, energy density of a wire is directly proportional to Young's modulus.

$\mathrm{C} \rightarrow$ IV. According to Bohr's Model, angular momentum ( $L$ ) of moving electron is equal to integral multiple $h / 2 \pi$.

i.e. $L=n \frac{h}{2 \pi}$ where, $n=1,2,3,4 \ldots \ldots \ldots$ where, $h$ is Planck's constant. $\mathrm{D} \rightarrow \mathrm{I}$, Dimensional formula of decay constant and angular velocity is same as $\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^{-1}\right]$.

Given, $S=\frac{\sqrt{a b}}{c^3 d^4}$

Percentage error in $S$ is given as, $\frac{\Delta s}{s} \times 100=\frac{1}{2} \times \frac{\Delta a}{a} \times 100+\frac{1}{2} \times \frac{\Delta b}{b} \times 100$

$ \begin{aligned} &\begin{aligned} & +3 \times \frac{\Delta c}{c} \times 100+4 \times \frac{\Delta d}{d} \times 100 \\ = & \frac{1}{2} \times 2 \%+\frac{1}{2} \times 1 \%+3 \times 1 \%+4 \times 1 \% \\ = & 1 \%+\frac{1}{2} \%+3 \%+4 \% \\ = & 8.5 \% \end{aligned}\\ &\text { Which is nearest to } 9 \% \text {. } \end{aligned} $

$ \begin{aligned} &\text { On multiplying } R \text { and } C \text {, we get }\\ &\begin{gathered} R C=\frac{V}{I} \times \frac{q}{V} \\ {\left[\because R=\frac{V}{I} \text { Ohm's law and } q=C V\right]} \end{gathered} \end{aligned} $

$ \begin{array}{rlrl} \Rightarrow & & R C & =\frac{q}{I} \\ \text { Also, } & I & =\frac{q}{t} \\ \therefore & & R C & =\frac{q}{q / t}=t \end{array} $

$\therefore R C$ will have dimensional formula of time.

Option (a) is incorrect because all basic laws of physics are universal. option (b) is correct because it follows Noether's first theorem which states that any differentiable symmetry of the action of a physical system has a corresponding conservation law. option (c) is incorrect as there are only 4 fundamental forces in nature. Option (d) is incorrect as there are examples where physics came out from technology, for example measurement of time (the technology of clocks) enabled Galileo to understand the physics of velocity and acceleration quite differently from Aristotle, giving rise to Galilean / Newtonian physics. Measurement of charge, in the technology of electricity, gave rise to millikans oil drop experiment and the discovery of the electron.

We know, for a formula to be correct only when dimensions of LHS and dimensions of RHS must be equal.

$ E=\frac{2 E_0 L}{L_0} \Rightarrow \frac{E}{E_0}=\frac{2 L}{L_0} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

⇒ Both LHS and RHS of Eq. (i), are dimensionless. Hence, the formula is dimensionally correct.

• In $E=E_0 e^{-\frac{2 L}{L_0}}$, clearly the exponent is dimensionless because, $L$ and $L_0$ will have same dimension and dimensions of $E$ and $E_0$ will be equal, since both $E$ and $E_0$ will have dimensions of energy. Thus, this formula is dimensionally correct.

• In $E=L e^{-L / E_0}$, clearly the power of exponential $\frac{-L}{E_0}$ will have dimensions and hence the power of exponent is not dimensionless, hence this equation is dimensionally incorrect.

• In $E=2\left(\frac{E_0}{L_0}\right) \times e^{-L / L_0}$, here $\frac{E_0}{L_0}$ will have different dimensions from $E$, hence this formula is also dimensionally incorrect.

Energy of the system as a function of time, $E(t)=\alpha t-\beta t^3$.

Where, $\alpha$ and $\beta$ are constant.

According to principle of homogenity, dimension of $\alpha t=$ dimension of energy i.e

$\begin{array}{rlrl} \Rightarrow & {[\alpha][t]=[E]} \\ \Rightarrow & {[\alpha][\mathrm{T}]} =\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right] \\ \Rightarrow & {[\alpha]} =\left[\mathrm{ML}^2 \mathrm{~T}^{-3}\right] \end{array}$

Again, dimension of $\beta t^3=$ dimension of energy

$\begin{array}{rr} \Rightarrow & {[\beta]\left[t^3\right]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]} \\ \Rightarrow & {[\beta]\left[\mathrm{T}^3\right]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]} \\ \Rightarrow & {[\beta]=\left[\mathrm{ML}^2 \mathrm{~T}^{-5}\right]} \end{array}$

In SI units, we know that:

$ \begin{aligned} \text{Work} &= \text{Force} \times \text{Displacement} \\ &= \text{Mass} \times \text{Acceleration} \times \text{Displacement} \\ &= \mathrm{kg} \times \mathrm{m} \, \mathrm{s}^{-2} \times \mathrm{m} \\ &= \mathrm{kg} \cdot \mathrm{m}^2 \cdot \mathrm{s}^{-2} \end{aligned} $

Therefore, the SI unit of work is $\mathrm{kg} \cdot \mathrm{m}^2 \cdot \mathrm{s}^{-2}$, which is equivalent to a Joule.

Given,

$\begin{aligned} & A=0.001204 \mathrm{~m} \\ & B=43120000 \mathrm{~m}, \text { and } C=1.200 \mathrm{~m} \end{aligned}$

We know that, zeroes at the right hand side of any number are non significant, hence number of significant figure in $B, N_B=4$. Zeroes after decimal point and before any number are not significant. Therefore, no. of significant figure in $A, N_A=4$.

Zeroes at the R.H.S adjacent to number after the decimal point are significant.

$\therefore$ No. of significant figure in $C, N_C=4$

$\therefore \quad N_A=N_B=N_C=4$

United State declared 2005 as International year of Physics, because that was the 100th anniversary of Albert Einstein.

As we know that,

1$\mathop A\limits^o$ = 10$^{-10}$ m

and 1 mm = 10$^{-3}$ m

$\therefore$ 1$\mathop A\limits^o$ = 10$^{-7}$ mm

As we know that,

Stress $=$ Force $/$ Area

In terms of dimensional formula,

$[\text { Stress }]=\frac{\left[\mathrm{ML} \mathrm{T}^{-2}\right]}{\left[\mathrm{L}^2\right]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]$

Given, speed of ripple of water $=v$

Surface tension $=\sigma$

Density $=\rho$

Wavelength $=\lambda$

As we know that, $\sigma=\frac{\text { Force }}{\text { Length }}$

and dimensional formula of $\sigma = {{[ML{T^{ - 2}}]} \over {[L]}} = [M{L^0}{T^{ - 2}}]$

and dimensional formula of $\rho, v$ and $\lambda$ are $[M{L^{ - 3}}{T^0}],[{M^0}L{T^{ - 1}}]$ and $\left[\mathrm{M}^0 \mathrm{LT}^0\right]$.

Let $\quad[\sigma]=[\rho]^a[v]^b[\lambda]^c$

$\begin{aligned} & \Rightarrow \quad\left[\mathrm{ML}^0 \mathrm{~T}^{-2}\right]=\left[\mathrm{ML}^{-3} \mathrm{~T}^0\right]^a\left[\mathrm{M}^0 \mathrm{LT}^{-1}\right]^b\left[\mathrm{M}^0 \mathrm{LT}^0\right]^c \\ & =\left[\mathrm{M}^a \mathrm{~L}^{-3 a+b+c} \mathrm{~T}^{-b}\right] \\ & \Rightarrow \quad a=1 \Rightarrow b=2 \\ & -3 a+b+c=0 \Rightarrow-3+2+c=0 \\ & c=1 \\ & \therefore \quad[\sigma]=[\rho]^1[v]^2[\lambda]^1 \\ & \text { Hence, } \quad \sigma \propto \rho v^2 \lambda \\ & \Rightarrow \quad v^2 \propto \frac{\sigma}{\rho \lambda} \\ & \end{aligned}$

Angular momentum for a point mass about an axis with velocity component of $v$ perpendicular to the radial direction is given by Angular momentum, $L=m v r$ Dimension formula of angular momentum $=[\mathrm{M}]\left[\mathrm{LT}^{-1}\right][\mathrm{L}]=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]$

According to the question, given that 29 main scale division (MSD) $=30$ vernier scale division (VSD)

$ \begin{aligned} & \Rightarrow \quad 29 \mathrm{MSD}=30 \mathrm{VSD} \\ & \Rightarrow \quad 1 \mathrm{VSD}=\frac{29}{30} \times 1 \mathrm{MSD} \\ & \Rightarrow \quad 1 \mathrm{VSD}=\frac{29}{30}\left(0.5^{\circ}\right) \\ & \quad \quad\left[\because 1 \mathrm{MSD}=0.5^{\circ}\right] \\ & \Rightarrow \quad 1 \mathrm{VSD}=\left(\frac{29}{60}\right)^{\circ} \end{aligned} $

∴ Least count of vernier calliperse is given as

$ \begin{aligned} \mathrm{LC} & =1 \mathrm{MSD}-1 \mathrm{VSD} \\ & =0.5^{\circ}-\left(\frac{29}{60}\right)^{\circ}=\left(\frac{30-29}{60}\right)^{\circ}=\left(\frac{1}{60}\right)^{\circ} \\ & =1 \mathrm{~min}\left[\because 1 \mathrm{~min}=\frac{1}{60} \text { degree }\right] \end{aligned} $

$ \begin{aligned} &\text { Mean mass of the ball is }\\ &\begin{aligned} \bar{M} & =\frac{m_1+m_2+m_3+m_4+m_5}{5} \\ & =\frac{2.61+2.58+2.40+2.73+2.80}{5} \\ \bar{M} & =2.624 \mathrm{~g} \simeq 2.62 \mathrm{~g} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { ∴ Mean absolute error, }\\ &\begin{aligned} \Delta \bar{M} & =\frac{\begin{array}{r} |2.62-2.61|+|2.62-2.58| \\ +|2.62-2.40|+|2.62-2.73| \\ +|2.62-2.80| \end{array}}{5} \\ & =\frac{0.01+0.04+0.22+0.11+0.18}{5} \\ & =\frac{0.56}{5}=0.112 \simeq 0.11 \mathrm{~g} \end{aligned} \end{aligned} $

Given, time-period of oscillation $T$ is

$ T \propto p^\alpha S^\beta E^\gamma \text { or } T=k p^\alpha S^\beta E^\gamma $

Now, subtituting dimensions of $T, p, S$ and $E$, we have

$ \left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^1\right]=k\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]^\alpha\left[\mathrm{ML}^{-3}\right]^\beta\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]^\gamma $

Equating powers of similar terms, we have

$ \begin{array}{r} \alpha+\beta+\gamma=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ -\alpha-3 \beta+2 \gamma=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ -2 \alpha-2 \gamma=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{array} $

Solving, we get $\alpha=\frac{-5}{6}, \beta=\frac{1}{2}, \gamma=\frac{1}{3}$