Units & Measurement and 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 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\% $.

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

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