Conceptual PCQs

Hint

(b) length, time and velocity.
Explanation
Fundamental quantities are defined as quantities that are independent of other quantities and cannot be derived from them. Velocity is a derived quantity, defined as length divided by time (velocity = length/time). Therefore, it depends on both length and time for its definition. A set of fundamental quantities must consist of quantities that are all mutually independent. Since velocity is dependent on length and time, the set containing all three cannot be a set of fundamental quantities.

Hint

(d) as the numerical value \(n\) is inversely proportional to the unit size \(u\).
A physical quantity \(Q\) remains constant regardless of the unit system used for measurement. The quantity is expressed as the product of a numerical value \(n\) and a unit \(u\), so \(Q=n u\).
If the same quantity is expressed in two different units, \(u_1\) and \(u_2\), with corresponding numerical values \(n_1\) and \(n_2\), the following relationship holds:
\(
n_1 u_1=n_2 u_2
\)
This can be rearranged to show the inverse proportionality:
\(
\frac{n_1}{n_2}=\frac{u_2}{u_1}
\)
This demonstrates that \(n\) is inversely proportional to the unit size \(u\), or \(n \propto \frac{1}{u}\). A larger unit size results in a smaller numerical value, and vice versa.

Hint

(d) Mass (M) may be represented in terms of length (L), time (T), and the quantity \(x\) if \(\boldsymbol{a} \neq \mathbf{0}\). This is only possible if the dimensional equation for \(\boldsymbol{x}\) can be rearranged to isolate M , which requires that M is present in the expression for \(x\).

Explanation: The quantity mass has the dimensional representation \(\mathbf{M}^1 \mathbf{L}^0 \mathbf{T}^0\).
Mass is a fundamental quantity, represented by the dimension \(\mathbf{M}\).
It does not inherently depend on length ( L ) or time ( T ) in the standard MLT dimensional system.
The exponents in the general formula \([x]=\mathrm{M}^a \mathrm{~L}^b \mathrm{~T}^c\) for mass are \(a=1, b=0\), and \(c=0\).

Hint

(c) Let’s start with the dimension of a physical quantity for which we will need to address the fundamental quantities. A fundamental quantity belongs to a class of physical quantities of its own kind only, each of them having no dependence with other fundamental quantities. These quantities are used to determine the dimensions of all other physical quantities. The fundamental quantities are, mass \([M]\), length \([L]\), time \([T]\), current \([I]\), luminous intensity \([c d]\), temperature \([K]\) and amount of substance \([\mathrm{mol}]\). These quantities are used to determine the dimensions of other physical quantities. How do we calculate these dimensions? Well, dimensions of a physical quantity are the powers to which the fundamental units are raised in order to represent that quantity. Let’s take velocity for an example. We know that velocity is the rate of displacement which is mathematically represented as \(v=\frac{s}{t}\), where \(s\) and \(t\) are total displacement and total time for that displacement respectively. The dimensions for velocity will be the dimensions of displacement (length) over dimensions of time. So the dimensions of velocity are \(\frac{[L]}{[T]}\) or \(\left[L T^{-1}\right]\). Does that ring any rings? We know that S.I. unit of velocity is \(m s^{-1}\) or length per time. This is exactly how we can use dimensions to determine the unit of a physical quantity. Another example, the dimensions for force are \(\left[M^1 L^1 T^{-2}\right]\) (Calculate this yourself). So the S.I. unit of force will be kg.m. \(\mathrm{s}^{-1}\) or Newton. Dimensionless quantities on the other hand are the quantities which do not possess any dimensions.

Example of such a quantity is, specific gravity (ratio of absolute density of an object to absolute density of water). As these quantities are often a pure ratio between similar physical quantities they do not possess dimension and hence no units. But when we look at the dimensionless quantity- angle, it has a unit assigned to it, the radians or degrees. So we cannot say that all dimensionless quantities are unit-less.

Hint

(a) never has a nonzero dimension.
Unitless (dimensionless) quantities are pure numbers (e.g., a ratio, a constant) and are represented by the dimension symbol 1 (or \(\mathbf{M}^0 \mathbf{L}^0 \mathbf{T}^0 \mathbf{A}^0 \ldots\) ).
A quantity has a nonzero dimension only if it can be expressed in terms of base units of measurement (like length, mass, or time).

Hint

(a) Step 1: Evaluate the integral
Substitute the rewritten expression back into the integral:
\(
\int \frac{d x}{\sqrt{a^2-(x-a)^2}}
\)
Using the standard integral form \(\int \frac{d u}{\sqrt{C^2-u^2}}=\sin ^{-1}\left(\frac{u}{C}\right)+\mathrm{C}\), with \(u=x-a\) and \(C=a\), the integral evaluates to:
\(
\sin ^{-1}\left(\frac{x-a}{a}\right)+\mathbf{C}=\sin ^{-1}\left(\frac{x}{a}-1\right)+\mathbf{C}
\)
Step 2: Compare results to find \(\boldsymbol{n}\)
We compare the result of our integration with the given right-hand side of the equation:
\(
\sin ^{-1}\left(\frac{x}{a}-1\right)=a^n \sin ^{-1}\left(\frac{x}{a}-1\right)
\)
For this equation to hold true, the coefficient \(a^n\) must equal 1. Assuming \(a \neq 0\), this requires that the value of \(n\) is 0.

Hint

(c, d ) We know that pressure is the amount of force acting on a unit area.
We can write the expression for pressure as,
\(
P=\frac{F}{A}
\)
Where \(P\) stands for the pressure, \(F\) stands for the force acting on a surface, and \(A\) stands for the area of the surface.
We know that force is the product of acceleration and mass. Hence we can write the dimensional formula for force as,
\(
F=\left[M L T^{-2}\right]
\)
The dimensional formula for the area can be written as,
\(
A=\left[L^2\right]
\)
Substituting the dimensional formulae for force and area in the expression of pressure, we get
\(
P=\frac{F}{A}=\frac{\left[M L T^{-2}\right]}{\left[L^2\right]}=\left[M L^{-1} T^{-2}\right]
\)
Since we have already calculated the dimensional formula for force while calculating for pressure, we can exclude force from the options.
Now let us calculate energy per unit volume.
We know that energy is the same as the work is done, hence we can consider energy as the product of force and displacement.
We can write energy as,
\(
E=F x
\)
Where \(E\) stands for the energy, \(F\) stands for the force, and \(x\) stands for the displacement. The dimensional formula for force can be written as,
\(
F=\left[M L T^{-2}\right]
\)
The dimensional formula for displacement is,
\(
x=[L]
\)
So, we can write the dimensional formula for energy as,
\(
E=F x=\left[M L T^{-2}\right][L]=\left[M L^2 T^{-2}\right]
\)
The energy per unit volume can be written as,
\(
\frac{E}{V}=\frac{\left[M L^2 T^{-2}\right]}{\left[L^3\right]}=\left[M L^{-1} T^{-2}\right]
\)
Linear momentum is the product of mass and velocity,
The linear momentum can be written as,
\(
P=m v
\)
The dimensional formula for momentum can be written as,
\(
P=m v=[M]\left[L T^{-1}\right]=\left[M L T^{-1}\right]
\)

Hint

(a), (b), and (d).
Explanation
(a) A dimensionally correct equation may be correct. This is true. The principle of homogeneity (same dimensions on both sides) is a necessary condition for an equation to be physically valid. A correct physical equation, like \(s=u t+\frac{1}{2} a t^2\), is always dimensionally correct.
(b) A dimensionally correct equation may be incorrect. This is also true. An equation can have the correct dimensions on both sides but still be numerically or physically incorrect, possibly due to a missing or wrong dimensionless constant. For example, \(s=u t+a t^2\) is dimensionally correct, but the correct factor for acceleration is \(\frac{1}{2}\).
(d) A dimensionally incorrect equation may be incorrect. This is true. If an equation is dimensionally incorrect, it violates the fundamental principle of homogeneity and therefore cannot be a correct physical equation.

Hint

(a), (b), and (c).

Explanation
(a) All quantities may be represented dimensionally in terms of the base quantities. This is correct because base quantities (like mass, length, time) form the foundation of dimensional analysis, and all physical quantities, whether base or derived, can be expressed in terms of these fundamental dimensions.
(b) A base quantity cannot be represented dimensionally in terms of the rest of the base quantities. This is correct because base quantities are fundamentally independent of each other. For example, mass cannot be expressed in terms of only length and time.
(c) The dimension of a base quantity in other base quantities is always zero. This is correct. When expressing a base quantity (e.g., length, L) in terms of other base quantities (e.g., mass, M , and time, T ), its dimensional formula is \(M^0 L^1 T^0\), meaning its dimension in mass and time is zero.

Why other options are incorrect
(d) The dimension of a derived quantity is never zero in any base quantity. This statement is incorrect. A derived quantity can have a dimension of zero in one or more base quantities. For instance, acceleration \(\left(L T^{-2}\right)\) has a dimension of zero in mass ( \(\boldsymbol{M}^0\) ). A dimensionless quantity like the Reynolds number has zero dimensions in all base quantities.

Hint

(b) Given, number of divisions on circular scale \(=100\)
Pitch \(=0.01 \mathrm{~cm}\)
\(
\begin{aligned}
\therefore \text { Least count }(\mathrm{LC}) & =\frac{\text { Pitch }}{\text { Number of divisions on circular scale }} \\
& =\frac{0.01}{100}=10^{-4} \mathrm{~cm}
\end{aligned}
\)

Hint

(c) The diameter ( \(d\) ) of a circle is directly proportional to its radius ( \(r\) ), as given by the formula \(d=2 r\). The relationship between the change (error) in diameter ( \(\Delta d\) ) and the change in radius ( \(\Delta r\) ) is also linear: \(\Delta d=2 \Delta r\).
The percentage error is defined as the ratio of the error to the actual value, multiplied by \(100 \%\). The percentage error in the diameter is:
\(
\% \text { Error }_d=\left(\frac{\Delta d}{d}\right) \times 100 \%
\)
Given that \(\%\) Error \(_d=4 \%\).
Substituting \(\Delta d=2 \Delta r\) and \(d=2 r\) into the formula for percentage error in diameter:
\(
\% \text { Error }_d=\left(\frac{2 \Delta r}{2 r}\right) \times 100 \%=\left(\frac{\Delta r}{r}\right) \times 100 \%
\)
The expression \(\left(\frac{\Delta r}{r}\right) \times 100 \%\) is the percentage error in the radius \(\left(\%\right.\) Error \(\left._r\right)\).
Thus, the percentage error in measuring the radius is equal to the percentage error in measuring the diameter.
\(
\% \text { Error }_r=\% \text { Error }_d=4 \%
\)

Hint

(d) Step 1: Determine the known values
The length of a single rod is given as \(L=(11.05 \pm 0.2) \mathrm{cm}\). This means the measured length is \(L_{\text {value }}=11.05 \mathrm{~cm}\) and the absolute uncertainty is \(\Delta L=0.2 \mathrm{~cm}\).
Step 2: Calculate the total length
When two identical rods are joined side by side, their lengths are added together. The total length \(\boldsymbol{L}_{\text {total }}\) is calculated as:
\(
L_{\text {total }}=L_{\text {value }}+L_{\text {value }}=11.05 \mathrm{~cm}+11.05 \mathrm{~cm}=22.10 \mathrm{~cm}
\)
Step 3: Calculate the total uncertainty
When two quantities with uncertainties are added, their absolute uncertainties are summed. The total uncertainty \(\boldsymbol{\Delta} \boldsymbol{L}_{\text {total }}\) is:
\(
\Delta L_{\text {total }}=\Delta L+\Delta L=0.2 \mathrm{~cm}+0.2 \mathrm{~cm}=0.4 \mathrm{~cm}
\)
Step 4: Express the final measurement with uncertainty
The net length of the system of rods is expressed in the format \(\left(L_{\text {total }} \pm \Delta L_{\text {total }}\right)\) :
\(
L_{\mathrm{net}}=(22.10 \pm 0.4) \mathrm{cm}
\)

Hint

(b) Error limits in velocity is \(\frac{\Delta v}{v}=\frac{\Delta d}{d}+\frac{\Delta t}{t}\) where \(\Delta d\) is the distance error and d is the total distance and \(\Delta t\) is the time error and t is the total time.
Given that,
Distance \(=(13.8+0.2)\) meter
Time \(=(4.0+0.3)\) second
Distance error \(\Delta d=0.2\) meter
Time error \(\Delta t=0.3\) second
Therefore velocity \(=\frac{\text { Distance }}{\text { Time }}=\frac{13.8}{4}=3.45 \frac{\mathrm{~m}}{\mathrm{~s}}\)
Error limits in velocity \(=\frac{\Delta v}{v}=\frac{\Delta d^4}{d}+\frac{\Delta t}{t}\)
\(
\begin{aligned}
& =\frac{0.2}{13.8}+\frac{0.3}{4.0} \\
& =0.089 \\
& \Rightarrow \frac{\Delta v}{v}=0.089
\end{aligned}
\)
Therefore change is velocity = \(\Delta v=0.089 \times 3.45=0.30\)
Hence Velocity within error limits \(=(3.45 \pm 0.3) \frac{m}{s}\)
\(
\text { Percentage error }=\frac{\Delta v}{v} \times 100=0.089 \times 100=8.9 \%
\)

Hint

\(
\begin{aligned}
&\text { (c) Volume of cuboid, }\\
&\begin{array}{rlrl}
& V=  l \times 2 l \times 3 l=6 l^3 \\
\therefore & & \frac{\Delta V}{V} \times 100=3\left(\frac{\Delta l}{l}\right)=3 \% & \left(\because \frac{\Delta l}{l} \times 100=1 \%\right)
\end{array}
\end{aligned}
\)

Hint

(a) 

\(
\begin{aligned}
& \text { Pressure }=\frac{\text { Force }}{\text { Area }} \therefore P=\frac{F}{L^2} \\
& \therefore \frac{\Delta P}{P}=\frac{\Delta F}{F}+2 \frac{\Delta L}{L}=\frac{4}{100}+2 \times \frac{2}{100} \\
& \therefore \frac{\Delta P}{P}=\frac{8}{100}=8 \%
\end{aligned}
\)

Hint

(b) \(H=I^2 R t\)
\(
\frac{\Delta H}{H}=2 \frac{\Delta I}{I}+ \frac{\Delta R}{R}+ \frac{\Delta t}{t}
\)
\(
\begin{gathered}
\frac{\Delta H}{H}=2(2 \%)+(1 \%)+(1 \%) \\
\frac{\Delta H}{H}=4 \%+1 \%+1 \% \\
\frac{\Delta H}{H}=6 \%
\end{gathered}
\)

Hint

(c) We know that, \(K=\frac{p^2}{2 m}\)
The error in measurement of momentum is \(+100 \%\).
Therefore, the actual momentum with error, \(p^{\prime}=p+p=2 p\)
∵ Kinetic energy with error,
\(
K^{\prime}=\frac{\left(p^{\prime}\right)^2}{2 m}=\frac{(2 p)^2}{2 m}
\)
\(
\begin{aligned}
K^{\prime} & =\frac{4 p^2}{2 m} \\
K^{\prime} & =4 K
\end{aligned}
\)
So, percentage change in kinetic energy,
\(
\begin{aligned}
\mathrm{KE} & =\left(\frac{K^{\prime}-K}{K}\right) \times 100=\left(\frac{4 K-K}{K}\right) \times 100 \\
& =\left(\frac{4-1}{1}\right) \times 100=300 \%
\end{aligned}
\)

Hint

(a) Radius of ball \(=5.2 \mathrm{~cm}\)
\(
\text { ume, } \begin{aligned}
V & =\frac{4}{3} \pi R^3 \\
\frac{\Delta V}{V} & =3\left(\frac{\Delta R}{R}\right) \\
\left(\frac{\Delta V}{V}\right) \times 100 & =3\left(\frac{0.2}{5.2}\right) \times 100 \simeq 11 \%
\end{aligned}
\)

Hint

(d) 

\(
\begin{aligned}
&\begin{gathered}
\frac{1}{\mathrm{R}}=\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_2} \\
\text { or, } \mathrm{R}^{-1}=\mathrm{R}_1^{-1}+\mathrm{R}_2^{-1}
\end{gathered}\\
&\text { On differentiating both sides, we get }\\
&\begin{aligned}
&\left(\frac{-1}{\mathrm{R}^2}\right) \mathrm{dR}=\left(\frac{-1}{\mathrm{R}_1^2}\right) \mathrm{dR}_1+\left(\frac{-1}{\mathrm{R}_2^2}\right) \mathrm{dR}_2 \\
& \Rightarrow \frac{\mathrm{dR}}{\mathrm{R}}=\left[\left(\frac{\mathrm{dR}_1}{\mathrm{R}_1^2}\right)+\left(\frac{\mathrm{dR}_2}{\mathrm{R}_2^2}\right)\right](\mathrm{R}) \\
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
&\begin{aligned}
& \frac{\mathrm{dR}}{\mathrm{R}}=\left[\frac{0.2}{5^2}+\frac{0.1}{10^2}\right](10 / 3) \\
&=(0.008+0.001)(10 / 3) \\
&=(0.009)(10 / 3) \\
&=(0.009)(10 / 3) \\
&=0.03 \\
& \therefore \frac{\mathrm{dR}}{\mathrm{R}}=0.03 \\
& \text { or } \frac{\mathrm{dR}}{\mathrm{R}} \times 100=0.03 \times 100 \\
& \text { or }\left(\frac{\mathrm{dR}}{\mathrm{R}} \times 100\right)=3
\end{aligned}\\
&\text { or, % error in } \mathrm{R} \text { is 3%. }
\end{aligned}
\)

Hint

(b) The number of significant figures in 0.06900 is 4 .
Leading zeros (0.0) are not significant.
The digits 6 and 9 are non-zero digits and are significant.
Trailing zeros to the right of a decimal point (6900) are significant.

Hint

(c) Step 1: Add the numbers arithmetically:
\(
436.32+227.2+0.301=663.821
\)
Step 2: Determine the least number of decimal places in the original numbers:
436.32 has two decimal places.
227.2 has one decimal place.
0.301 has three decimal places.
The least number of decimal places is one, from the number 227.2.
Step 3: Round the sum to the least number of decimal places:
The sum, 663.821 , must be rounded to one decimal place. Since the digit in the hundredths place (2) is less than 5, the tenths place digit (8) remains unchanged.
The final answer is 663.8.

Hint

(b) The physical quantity will be (b) pressure, if \(a=1, b=-1, c=-2\).
The dimensions of various physical quantities in terms of mass \((\mathrm{M})\), length \((\mathrm{L})\), and time (\(T\)) are as follows:
Force is defined as mass × acceleration. The dimensional formula is \(\left[\mathrm{M}^1 \mathrm{~L}^1 \mathrm{~T}^{-2}\right]\).
Pressure is defined as force per unit area. The dimensional formula is \(\left[\mathrm{M}^1 \mathrm{~L}^{-1} \mathrm{~T}^{-2}\right]\)
Velocity is defined as displacement per unit time. The dimensional formula is \(\left[\mathrm{M}^0 \mathrm{~L}^1 \mathrm{~T}^{-1}\right]\).
Acceleration is defined as velocity per unit time. The dimensional formula is \(\left[\mathrm{M}^0 \mathrm{~L}^1 \mathrm{~T}^{-2}\right]\).

Comparing these with the given general form \(\left[\mathrm{M}^a \mathrm{~L}^b \mathrm{~T}^c\right]\) :
For option (a), \(a=0, b=-1, c=-2\), which is not force.
For option (b), \(a=1, b=-1, c=-2\), which matches the dimensions of pressure.
For option (c), \(a=1, b=0, c=-1\), which is not velocity.
For option (d), \(a=1, b=1, c=-2\), which is not acceleration.

Hint

(c) Step 1: Apply the Principle of Homogeneity
According to the principle of homogeneity of dimensions, every term in a valid physical equation must have the same dimensions. The given equation is \(y=A t^2-B t^3\). The variable \(y\) has the dimension of length \([L]\), and \(t\) has the dimension of time \([T]\).
Step 2: Determine the Dimensions of \(B\)
For the equation to be dimensionally consistent, the dimension of the term \(\boldsymbol{B t}^3\) must be equal to the dimension of \(y\) :
\(
\left[B t^3\right]=[y]
\)
Substituting the known dimensions:
\(
[B]\left[T^3\right]=[L]
\)
Solving for the dimensions of \(\boldsymbol{B}\) :
\(
\begin{gathered}
{[B]=\frac{[L]}{\left[T^3\right]}} \\
{[B]=\left[L T^{-3}\right]}
\end{gathered}
\)

Hint

(d) The numbers 2.745 and 2.735 on rounding off to 3 significant figures will give 2.74 and 2.74

The rounding follows the “round half to even” or “banker’s rounding” convention, which is standard in many scientific contexts.
For 2.745 : The third significant figure is 4 (an even number). The next digit is 5 . When the digit to be dropped is exactly 5 , the preceding even digit is left unchanged, resulting in 2.74.
For 2.735 : The third significant figure is 3 (an odd number). The next digit is 5 . When the digit to be dropped is exactly 5 , the preceding odd digit is increased by one, resulting in 2.74.

Hint

(c) The density is calculated as mass/volume. The result must have the same number of significant figures as the measurement with the fewest significant figures.
Mass ( 4.237 g ) has four significant figures.
Volume \(\left(2.5 \mathrm{~cm}^3\right)\) has two significant figures.
The calculation gives approximately \(1.6948 \mathrm{~g} \mathrm{~cm}^{-3}\), which must be rounded to two significant figures.
The correct answer is (c) \(1.7 \mathrm{~g} \mathrm{~cm}^{-3}\).

Hint

(a) 5.00 mm.
Explanation
Precision refers to the level of detail in a measurement, which is determined by the smallest unit the measuring instrument can record. A smaller unit of measurement indicates a higher degree of precision.
All the given measurements are expressed to two decimal places, which might suggest similar precision in terms of significant figures, but the unit of measurement is the determining factor here.
We compare the units:
Millimeter (mm)
Centimeter (cm)
Meter (m)
Kilometer (km)
Millimeter is the smallest unit of length among these options ( \(1 \mathrm{~m}=1000 \mathrm{~mm}\); \(1 \mathrm{~cm}= 10 \mathrm{~mm} ; 1 \mathrm{~km}=1,000,000 \mathrm{~mm})\). A measurement in millimeters, therefore, can express the length to a finer detail than one in centimeters, meters, or kilometers.
The absolute error in 5.00 mm (commonly taken as 0.01 mm ) is the least compared to the others \((0.01 \mathrm{~cm}=0.1 \mathrm{~mm} ; 0.01 \mathrm{~m}=10 \mathrm{~mm} ; 0.01 \mathrm{~km}=10,000 \mathrm{~mm})\), making it the most precise measurement.
Why other options are incorrect
(b) 5.00 cm is less precise than (a) 5.00 mm because a centimeter is a larger unit than a millimeter \((1 \mathrm{~cm}=10 \mathrm{~mm})\), meaning the measurement is less detailed.
(c) 5.00 m is less precise as a meter is a much larger unit \((1 \mathrm{~m}=1000 \mathrm{~mm})\). The potential error range is much larger compared to a millimeter scale.
(d) 5.00 km is the least precise of all, as a kilometer is the largest unit provided (1 \(\mathrm{km}=1,000,000 \mathrm{~mm}\) ), indicating a measurement with the coarsest scale.

Hint

(a) Given, length, \(l=5 \mathrm{~cm}\)
Now, checking the errors with each options one by one, we get
\(
\begin{aligned}
& \Delta l_1=5-4.9=0.1 \mathrm{~cm} \\
& \Delta l_2=5-4.805=0.195 \mathrm{~cm} \\
& \Delta l_3=5.25-5=0.25 \mathrm{~cm} \\
& \Delta l_4=5.4-5=0.4 \mathrm{~cm}
\end{aligned}
\)
Error \(\Delta l_1\) is least.
Hence, 4.9 cm is most accurate.

Hint

(b) Step 1: Analyze dimensions of y and A
The equation for the oscillating particle’s displacement is given by \(\boldsymbol{y}=\boldsymbol{A} \sin (\boldsymbol{B x}+\boldsymbol{C t}+\boldsymbol{D})\). Displacement \(\boldsymbol{y}\) has the dimension of length, \([\boldsymbol{L}]\). Since the sine function is dimensionless, the amplitude \(\boldsymbol{A}\) must also have the dimension of length.
\(
[A]=[L]
\)
Step 2: Analyze dimensions of the sine argument
The argument of the sine function, \((\boldsymbol{B x}+\boldsymbol{C} t+\boldsymbol{D})\), must be dimensionless. For each term in the sum to be dimensionless, we must have:
\(
\begin{aligned}
& {[B x]=\left[M^0 L^0 T^0\right]=[1]} \\
& {[C t]=\left[M^0 L^0 T^0\right]=[1]} \\
& {[D]=\left[M^0 L^0 T^0\right]=[1]}
\end{aligned}
\)
Step 3: Determine dimensions of B, C, and D
Knowing that \([x]=[L]\) and \([t]=[T]\), we can determine the dimensions of \(B, C\), and \(D\)
\(
\begin{gathered}
{[B]=\frac{[1]}{[x]}=\frac{[1]}{[L]}=\left[L^{-1}\right]} \\
{[C]=\frac{[1]}{[t]}=\frac{[1]}{[T]}=\left[T^{-1}\right]} \\
{[D]=[1]}
\end{gathered}
\)
Step 4: Calculate the dimension of the product (ABCD)
The dimensional formula for the product \((A B C D)\) is the product of their individual dimensional formulas:
\(
\begin{aligned}
{[A B C D] } & =[A] \times[B] \times[C] \times[D] \\
{[A B C D] } & =[L] \times\left[L^{-1}\right] \times\left[T^{-1}\right] \times[1] \\
{[A B C D] } & =\left[L^{1-1} T^{-1}\right]=\left[L^0 T^{-1}\right]
\end{aligned}
\)
Which can be written as:
\(
[A B C D]=\left[M^0 L^0 T^{-1}\right]
\)

Hint

\(
\begin{aligned}
& \text { (c) Given, } Q=\frac{A^3 B^3}{C \sqrt{D}} \\
& \frac{\Delta Q}{Q}=3\left(\frac{\Delta A}{A}\right)+3\left(\frac{\Delta B}{B}\right)+\left(\frac{\Delta C}{C}\right)+\frac{1}{2}\left(\frac{\Delta D}{D}\right)
\end{aligned}
\)
\(
\begin{aligned}
& \text { Here, } \frac{\Delta A}{A} \times 100=2 \%, \frac{\Delta B}{B} \times 100=1 \% \\
& \quad \frac{\Delta C}{C} \times 100=3 \%, \frac{\Delta D}{D} \times 100=4 \% \\
& \therefore \quad \frac{\Delta Q}{Q} \times 100=(3 \times 2 \%)+(3 \times 1 \%)+(3 \%)+\frac{1}{2} \times(4 \%)= \pm 14 \%
\end{aligned}
\)

Hint

(a) Given, length, \(l=(16.2 \pm 0.1) \mathrm{cm}\)
Breadth, \(b=(10.1 \pm 0.1) \mathrm{cm}\)
Area, \(A=l \times b=(16.2 \mathrm{~cm}) \times(10.1 \mathrm{~cm})=163.62 \mathrm{~cm}^2\)
Rounding off to three significant digits, area, \(A=164 \mathrm{~cm}^2\)
\(
\begin{aligned}
\because \quad \frac{\Delta A}{A} & =\frac{\Delta l}{l}+\frac{\Delta b}{b}=\frac{0.1}{16.2}+\frac{0.1}{10.1}=\frac{1.01+1.62}{16.2 \times 10.1}=\frac{2.63}{163.62} \\
\Rightarrow \quad \Delta A & =A \times \frac{2.63}{163.62}=163.62 \times \frac{2.63}{163.62} \\
& =2.63 \mathrm{~cm}^2
\end{aligned}
\)
\(\Delta A=3 \mathrm{~cm}^2\) (By rounding off to one significant figure)
\(\therefore \quad\) Area, \(A=A \pm \Delta A=(164 \pm 3) \mathrm{cm}^2\)

Hint

(c) (a) Work \(=\) Force × Distance \(=\left[\mathrm{MLT}^{-2}\right][\mathrm{L}]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]\)
\(
\text { Torque }=\text { Force × Distance }=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]
\)
(b) Angular momentum \(=m v r=[\mathrm{M}]\left[\mathrm{LT}^{-1}\right][\mathrm{L}]=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]\)
\(
\text { Planck’s constant }=\frac{E}{\mathrm{v}}=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]}{\left[\mathrm{T}^{-1}\right]}=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]
\)
\(
\begin{aligned}
& \text { (c) } \text { Tension }=\text { Force }=\left[\mathrm{MLT}^{-2}\right] \\
& \text { Surface tension }=\frac{\text { Force }}{\text { Length }}=\frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{L}]}=\left[\mathrm{ML}^0 \mathrm{~T}^{-2}\right]
\end{aligned}
\)
\(
\begin{aligned}
& \text { (d) Impulse }=\text { Force × Time }=\left[\mathrm{MLT}^{-2}\right][\mathrm{T}]=\left[\mathrm{MLT}^{-1}\right] \\
& \text { Momentum }=\text { Mass } \text { × } \text { Velocity }=[\mathrm{M}]\left[\mathrm{LT}^{-1}\right]=\left[\mathrm{MLT}^{-1}\right]
\end{aligned}
\)

Hint

(a) Given, \(A=2.5 \mathrm{~ms}^{-1} \pm 0.5 \mathrm{~ms}^{-1}, B=0.10 \mathrm{~s} \pm 0.01 \mathrm{~s}\)
\(
\begin{aligned}
x & =A B=(2.5)(0.10)=0.25 \mathrm{~m} \\
\frac{\Delta x}{x} & =\frac{\Delta A}{A}+\frac{\Delta B}{B}=\frac{0.5}{2.5}+\frac{0.01}{0.10}=\frac{0.05+0.025}{0.25}=\frac{0.075}{0.25}
\end{aligned}
\)
\(\Delta x=0.075=0.08 \mathrm{~m}\) (rounding off to one significant figure)
\(
A B=(0.25 \pm 0.08) \mathrm{m}
\)

Hint

\(
\begin{aligned}
&\text { (c) Given, Young’s modulus, } Y=1.9 \times 10^{11} \mathrm{Nm}^{-2}\\
&\begin{aligned}
& 1 \mathrm{~N}=10^5 \text { dyne } \\
& \text { Hence, } \quad Y=19 \times 10^{11} \times 10^5 \text { dyne } / \mathrm{m}^2
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
&\text { We know that, } 1 \mathrm{~m}=100 \mathrm{~cm}\\
&\begin{aligned}
\therefore \quad Y & =19 \times 10^{11} \times 10^5 \text { dyne } /(100)^2 \mathrm{~cm}^2 \\
& =1.9 \times 10^{16-4} \text { dyne } / \mathrm{cm}^2 \\
Y & =1.9 \times 10^{12} \text { dyne } / \mathrm{cm}^2
\end{aligned}
\end{aligned}
\)

Hint

(c) Given, voltage, \(V=(100 \pm 5) \mathrm{V}\)
Current,
\(
I=(10 \pm 0.2) \mathrm{A}
\)
From Ohm’s law, \(V=I R \Rightarrow\) Resistance, \(R=\frac{V}{I}\)
\(
\begin{aligned}
&\text { Maximum percentage error in resistance, }\\
&\begin{aligned}
\left(\frac{\Delta R}{R} \times 100\right) & =\left(\frac{\Delta V}{V} \times 100\right)+\left(\frac{\Delta I}{I} \times 100\right) \\
& =\left(\frac{5}{100} \times 100\right)+\left(\frac{0.2}{10} \times 100\right)=5+2=7 \%
\end{aligned}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { (d) Density, } \quad \rho=\frac{m}{\pi r^2 L} \\
& \therefore \quad \frac{\Delta \rho}{\rho} \times 100=\left[\frac{\Delta m}{m}+\frac{2 \Delta r}{r}+\frac{\Delta L}{L}\right] \times 100
\end{aligned}
\)
After substituting the values, we get the maximum percentage error in density \(=4 \%\).

Hint

(c) Given \(x=(10.0 \pm 0.1), \quad 2 x=(20.0 \pm 0.2) y=(10.0 \pm 0.1), \quad 2 y=(20.0 \pm 0.2)\)
Thus \(2 x-2 y=(20.0-20.0) \pm(0.2+0.2)=(0 \pm 0.4)\)

Hint

(a) The dimensions of \(a\) are \(\left[\mathrm{ML}^5 \mathrm{~T}^{-2}\right]\). The constant \(a / V^2\) is added to the pressure \(p\), and quantities in a sum must have the same dimensions.
Dimensions of pressure \(p\) : \(\left[\mathbf{M L}^{-1} \mathbf{T}^{-2}\right]\) (force per unit area).
Dimensions of volume \(V:\left[\mathrm{L}^3\right]\).
For \(a / V^2\) to have the same dimensions as \(p\) :
Dimensions of \(a=\) (Dimensions of \(p) \times\left(\right.\) Dimensions of \(V^2\)
Dimensions of \(a=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right] \times\left[\mathrm{L}^3\right]^2\)
Dimensions of \(a=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right] \times\left[\mathrm{L}^6\right]\)
Dimensions of \(a=\left[\mathrm{ML}^5 \mathrm{~T}^{-2}\right]\)

Hint

\(
\begin{aligned}
&\text { (d) Area of strip }=l b\\
&\begin{aligned}
\therefore\left(\frac{\Delta A}{A}\right) \times 100 & =\left(\frac{\Delta l}{l}\right) \times 100+\left(\frac{\Delta b}{b}\right) \times 100 \\
& =\frac{0.1}{10} \times 100+\frac{0.01}{1} \times 100= \pm 2 \%
\end{aligned}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { (b) Volume of cylinder, } V=\pi r^2 L, r=\left(\frac{D}{2}\right) \\
& \begin{aligned}
\therefore \quad\left(\frac{\Delta V}{V}\right) \times 100 & =2\left(\frac{\Delta D}{D}\right) \times 100+\left(\frac{\Delta L}{L}\right) \times 100 \\
& =2\left(\frac{0.01}{4.0}\right) \times 100+\left(\frac{0.1}{5}\right) \times 100=2.5 \%
\end{aligned}
\end{aligned}
\)

Hint

(d) Given, \(A=1.0 \mathrm{~m} \pm 0.2 \mathrm{~m}, B=2.0 \mathrm{~m} \pm 0.2 \mathrm{~m}\)
Let \(Y=\sqrt{A B}=\sqrt{(1.0)(2.0)}=1.414 \mathrm{~m}\)
Rounding off to two significant digit \(Y=1.4 \mathrm{~m}\)
\(
\because \quad \frac{\Delta Y}{Y}=\frac{1}{2}\left[\frac{\Delta A}{A}+\frac{\Delta B}{B}\right]=\frac{1}{2}\left[\frac{0.2}{1.0}+\frac{0.2}{2.0}\right]=\frac{0.6}{2 \times 2.0}
\)
\(
\Rightarrow \quad \Delta Y=\frac{0.6 Y}{2 \times 2.0}=\frac{0.6 \times 1.4}{2 \times 2.0}=0.212
\)
Rounding off to one significant digit, \(\Delta Y=0.2 \mathrm{~m}\) Thus, correct value for \(\sqrt{A B}=Y+\Delta Y=1.4 \mathrm{~m} \pm 0.2 \mathrm{~m}\)

Hint

\(
\begin{aligned}
&\text { (a) In the given equation, } \frac{\alpha Z}{k \theta} \text { should be dimensionless. }\\
&\begin{aligned}
& \therefore \quad[\alpha]=\left[\frac{k \theta}{Z}\right] \Rightarrow[\alpha]=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~K}^{-1}\right][\mathrm{K}]}{[\mathrm{L}]}=\left[\mathrm{MLT}^{-2}\right] \\
& \text { and } \quad p=\frac{\alpha}{\beta} \Rightarrow[\beta]=\left[\frac{\alpha}{p}\right]=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]}=\left[\mathrm{M}^0 \mathrm{~L}^2 \mathrm{~T}^0\right]
\end{aligned}
\end{aligned}
\)

Hint

(a) The values of \(p, q\), and \(r\) are respectively \(-\mathbf{1} / \mathbf{2}, \mathbf{1} / \mathbf{2}, 5/2\).
Step 1: Determine the dimensions of each physical quantity
The dimensions of energy \(\boldsymbol{E}\), universal gravitational constant \(\boldsymbol{G}\), Planck’s constant \(\boldsymbol{h}\), and velocity of light \(c\) are:
Energy: \([E]=\left[M L^2 T^{-2}\right]\)
Gravitational constant: \([G]=\left[M^{-1} L^3 T^{-2}\right]\)
Planck’s constant: \([h]=\left[M L^2 T^{-1}\right]\)
Velocity of light: \([c]=\left[L T^{-1}\right]\)
Step 2: Formulate the dimensional equation
Substitute the dimensions into the given equation \(E=G^p h^q c^r\) :
\(
\left[M L^2 T^{-2}\right]=\left(\left[M^{-1} L^3 T^{-2}\right]\right)^p\left(\left[M L^2 T^{-1}\right]\right)^q\left(\left[L T^{-1}\right]\right)^r
\)
This simplifies to:
\(
\left[M^1 L^2 T^{-2}\right]=\left[M^{-p+q} L^{3 p+2 q+r} T^{-2 p-q-r}\right]
\)
Step 3: Set up and solve the system of linear equations
Equating the exponents for each fundamental dimension ( \(M, L, T\) ) yields a system of three linear equations:
For \(\mathrm{M}: 1=-p+q\)
For L: \(2=3 p+2 q+r\)
For \(\mathrm{T}:-2=-2 p-q-r\)
Solving this system of equations (by substitution or elimination) gives:
\(
\begin{gathered}
p=-1 / 2 \\
q=1 / 2 \\
r=5 / 2
\end{gathered}
\)

Hint

(d) The values of \(a, b\), and \(c\) are \(a=-5 / 6, b=1 / 2\), and \(c=1 / 3\).
Step 1: Define Dimensions
The period \(T\) is proportional to \(p^a d^b E^c\), so the dimensional equation is \([T]=[p]^a[d]^b[E]^c\). The dimensions for each variable are:
Period T: [T]
Pressure \(p:\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\)
Density \(d:\left[\mathbf{M L}^{-3}\right]\)
Energy \(E:\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]\)
Step 2: Formulate Dimensional Equations
Substitute the dimensions into the proportional relationship:
\(
[\mathrm{T}]^1\left[\mathrm{M}^0[\mathrm{~L}]^0=\left(\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\right)^a\left(\left[\mathrm{ML}^{-3}\right]\right)^b\left(\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]\right)^c\right.
\)
Equating the exponents for each fundamental dimension (M, L, T) yields a system of linear equations:
For M: \(a+b+c=0\)
For L: \(-a-3 b+2 c=0\)
For T: \(-2 a-2 c=1\)
Step 3: Solve the System of Equations
From the \(T\) equation, we find \(a+c=-1 / 2\).
Using the M equation, \(b=-(a+c)\), so \(b=-(-1 / 2)=1 / 2\).
Substitute \(b=1 / 2\) into the L equation:
\(
-a-3(1 / 2)+2 c=0 \Longrightarrow 2 c-a=3 / 2
\)
We also have \(c=-1 / 2-a\). Substitute this into the modified \(L\) equation:
\(
2(-1 / 2-a)-a=3 / 2 \Longrightarrow-1-2 a-a=3 / 2 \Longrightarrow-3 a=5 / 2 \Longrightarrow a=-5 / 6
\)
Finally, solve for \(c\) using \(a+c=-1 / 2\) :
\(
-5 / 6+c=-1 / 2 \Longrightarrow c=-1 / 2+5 / 6 \Longrightarrow c=-3 / 6+5 / 6=2 / 6=1 / 3
\)