JEE PYQs MCQs

Hint

(d)

\(
\begin{aligned}
F & =\frac{q}{V} \\
& =\frac{q^2}{q V} \\
& =\frac{q^2}{\text { Work Done }}
\end{aligned}
\)
\(
\begin{aligned}
{[F] } & =\left[\frac{q^2}{\text { Force x Dis }}\right] \\
& =\left[\frac{C^2}{M L T^{-2} L}\right]=C^2 M^{-1} L^{-2} T^2
\end{aligned}
\)

Hint

(a) Step 1: Determine the dimensions of A, B, C, and D
The principle of dimensional homogeneity requires every term in the position equation \(x(t)=A \sin t+B \cos ^2 t+C t^2+D\) to have the dimension of length, \(L\). The dimension of time is \(T\).
Dimension of \(D\) must be length: \([D]=L\).
Dimension of \(C t^2\) must be length: \(\left[C t^2\right]=[C] T^2=L\), so \([C]=L T^{-2}\).
Dimension of \(\boldsymbol{A}\) sin \(\boldsymbol{t}\) must be length. Trigonometric functions are dimensionless:
\(
[A \sin t]=[A] \cdot 1=L, \text { so }[A]=L .
\)
Dimension of \(B \cos ^2 t\) must be length. Trigonometric functions are dimensionless:
\(
\left[B \cos ^2 t\right]=[B] \cdot 1=L, \text { so }[B]=L .
\)
Step 2: Calculate the dimension of the expression \(\frac{A B C}{D}\)
Substitute the determined dimensions into the expression:
\(
\begin{gathered}
{\left[\frac{A B C}{D}\right]=\frac{[A][B][C]}{[D]}} \\
{\left[\frac{A B C}{D}\right]=\frac{L \cdot L \cdot\left(L T^{-2}\right)}{L}} \\
{\left[\frac{A B C}{D}\right]=\frac{L^3 T^{-2}}{L}} \\
{\left[\frac{A B C}{D}\right]=L^2 T^{-2}}
\end{gathered}
\)

Hint

(d) This is determined by dimensional analysis:
Electric flux \((\phi)\) has units of \(\mathrm{V} \cdot \mathrm{m}\) or \(\mathrm{N} \cdot \mathrm{m}^2 / \mathrm{C}\).
Surface charge density \((\sigma)\) has units of \(\mathrm{C} / \mathrm{m}^2\).
Linear charge density ( \(\lambda\) ) has units of \(\mathrm{C} / \mathrm{m}\).
For the equation \(\phi=\alpha \sigma+\beta \lambda\) to be dimensionally consistent, both terms on the right must have the same units as \(\phi\), and thus the same units as each other: units of \((\alpha \sigma)=\) units of \((\beta \lambda)\).
Therefore, the units of the ratio \(\left(\frac{\alpha}{\beta}\right)\) must be units of \(\left(\frac{\lambda}{\sigma}\right)\) :
\(
\frac{\mathrm{C} / \mathrm{m}}{\mathrm{C} / \mathrm{m}^2}=\frac{\mathrm{C}}{\mathrm{~m}} \times \frac{\mathrm{m}^2}{\mathrm{C}}=\mathrm{m} \text { (meters). }
\)
Meters are the unit of length or displacement.

Hint

(b) Surface tension and impulse.
Explanation
Surface tension has dimensions of force per unit length: \(\left[\mathrm{M} \mathrm{L}^0 \mathrm{~T}^{-2}\right]\) or \(\left[\mathrm{M} \mathrm{T}^{-2}\right]\).
Impulse has dimensions of force multiplied by time, which is the same as linear momentum: \(\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-1}\right]\).
Since the dimensions \(\left[\mathrm{M} \mathrm{T}^{-2}\right]\) and \(\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-1}\right]\) are different, this pair does not have the same dimensions.

Why other options are incorrect
(a) Pressure and Young’s modulus: Both have the same dimension of stress (force per unit area): \(\left[\mathrm{M} \mathrm{L}^{-1} \mathrm{~T}^{-2}\right]\).
(c) Torque and energy: Both have the same dimension of force multiplied by distance: \(\left[\mathrm{M} \mathrm{L}^2 \mathrm{~T}^{-2}\right]\).
(d) Angular momentum and Planck’s constant: Both have the same dimension of energy multiplied by time: \(\left[\mathrm{M} \mathrm{L}^2 \mathrm{~T}^{-1}\right]\).

Hint

(b)

(A) Magnetic induction is measured in (III) Gauss (CGS unit) or Tesla (SI unit).
(B) Magnetic intensity is measured in (IV) Ampere meter (more accurately, Ampere per meter, A/m, in SI units).
(C) Magnetic flux is measured in (II) Weber (Wb).
(D) Magnetic moment: Magnetic moment is measured in Ampere-meter², which corresponds to (I). 

Hint

(c) Step 1: Determine the dimensions of B and \(\mu_0\)
The dimension of the magnetic field, \(B\), is \([B]=\mathrm{MT}^{-2} \mathrm{~A}^{-1}\).
The dimension of the permeability of free space, \(\mu_0\), is \(\left[\mu_0\right]=\mathrm{MLT}^{-2} \mathrm{~A}^{-2}\).
Step 2: Calculate the dimensions of the ratio \(\left(\boldsymbol{B} / \boldsymbol{\mu}_0\right)\)
To find the dimension of the ratio, divide the dimension of \(B\) by the dimension of \(\mu_0\) :
\(
\left[\frac{B}{\mu_0}\right]=\frac{[B]}{\left[\mu_0\right]}=\frac{\mathrm{MT}^{-2} \mathrm{~A}^{-1}}{\mathrm{MLT}^{-2} \mathrm{~A}^{-2}}=\mathrm{L}^{-1} \mathrm{~A}
\)

Hint

(d) Permeability of free space (A):
The dimensional formula is \(\left[M L T^{-2} A^{-2}\right]\).
Thus, (A) matches with (III).
Magnetic field (B):
The dimensional formula is \(\left[M T^{-2} A^{-1}\right]\).
Thus, (B) matches with (II).
Magnetic moment (C):
The dimensional formula is \(\left[L^2 A\right]\).
Thus, (C) matches with (IV).
Torsional constant (D):
The dimensional formula is \(\left[M L^2 T^{-2}\right]\).
Thus, (D) matches with (I).

Hint

(d) Angular Impulse is the product of torque and time, with dimensions of angular momentum: \(\left[\mathbf{M ~ L}^{\mathbf{2}} \mathbf{T}^{-\mathbf{1}}\right] (\mathrm{I})\).

Latent Heat is energy per unit mass: Energy has dimensions [ \(\mathrm{M} \mathrm{L}^2 T^{-2}\) ]. Dividing by mass \([\mathrm{M}]\) gives \(\left[\mathrm{M}^{\circ} \mathrm{L}^2 \mathrm{~T}^{-2}\right](\mathrm{II})\).

Electrical Resistivity has the dimensions of \(\left[\mathrm{M} \mathrm{L}^3 \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right] (\mathrm{III})\)

Electromotive Force (EMF) is potential difference (voltage), which has the dimensions of \(\left[\mathbf{M ~ L}^{\mathbf{2}} \mathbf{T}^{-\mathbf{3}} \mathbf{A}^{-\mathbf{1}}\right] (\mathrm{IV})\).

Hint

(d) Step 1: Analyze the dimensional homogeneity
The principle of dimensional homogeneity requires every term in the equation \(v=A t^2+\frac{B t}{C+t}\) to have the same dimension as velocity \([v]=\left[\mathrm{L}^1 \mathrm{~T}^{-1}\right]\).
For the terms in the denominator \(C+t\) to be added, they must have the same dimension. Therefore, the dimension of \(C\) must be the same as the dimension of time \([t]=\left[\mathrm{T}^1\right]\).
\(
[C]=[T]
\)
Step 2: Determine the dimension of \(B\)
The entire second term \(\frac{B t}{C+t}\) must have the dimension of velocity.
\(
\left[\frac{B t}{C+t}\right]=[v] \Longrightarrow \frac{[B][t]}{[C+t]}=\left[\mathrm{L}^1 \mathrm{~T}^{-1}\right]
\)
Since \([C+t]=[T]\),
\(
\frac{[B][\mathrm{T}]}{[\mathrm{T}]}=\left[\mathrm{L}^1 \mathrm{~T}^{-1}\right] \Longrightarrow[B]=\left[\mathrm{L}^1 \mathrm{~T}^{-1}\right]
\)
Step 3: Determine the dimension of A
The first term \(\boldsymbol{A} \boldsymbol{t}^{\mathbf{2}}\) must have the dimension of velocity.
\(
\begin{gathered}
{\left[A t^2\right]=[v] \Longrightarrow[A]\left[t^2\right]=\left[\mathrm{L}^1 \mathrm{~T}^{-1}\right]} \\
{[A]\left[\mathrm{T}^2\right]=\left[\mathrm{L}^1 \mathrm{~T}^{-1}\right] \Longrightarrow[A]=\left[\mathrm{L}^1 \mathrm{~T}^{-1}\right]\left[\mathrm{T}^{-2}\right]} \\
{[A]=\left[\mathrm{L}^1 \mathrm{~T}^{-3}\right]}
\end{gathered}
\)
Step 4: Calculate the dimension of ABC
Multiply the dimensions found for \(\mathrm{A}, \mathrm{B}\), and C to find the dimension of their product.
\(
\begin{gathered}
{[A B C]=[A][B][C]=\left(\left[\mathrm{L}^1 \mathrm{~T}^{-3}\right]\right)\left(\left[\mathrm{L}^1 \mathrm{~T}^{-1}\right]\right)\left(\left[\mathrm{T}^1\right]\right)} \\
{[A B C]=\left[\mathrm{L}^{1+1} \mathrm{~T}^{-3-1+1}\right]=\left[\mathrm{L}^2 \mathrm{~T}^{-3}\right]}
\end{gathered}
\)
Including the mass dimension \(\left[\boldsymbol{M}^0\right]\) for completeness:
\(
[A B C]=\left[\mathrm{M}^0 \mathrm{~L}^2 \mathrm{~T}^{-3}\right]
\)

Hint

(a) (A) Young’s Modulus \(\rightarrow(\mathrm{II}) \mathrm{ML}^{-1} \mathrm{~T}^{-2}\)
(B) Torque \(\rightarrow(\mathrm{IV}) \mathrm{ML}^2 \mathrm{~T}^{-2}\)
(C) Coefficient of Viscosity \(\rightarrow(\mathrm{I}) \mathrm{ML}^{-1} \mathrm{~T}^{-1}\)
(D) Gravitational Constant \(\rightarrow(\mathrm{III}) \mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\)

Hint

(c)

\(
\begin{aligned}
& \text { Fractional error }=2 \frac{\Delta X}{X}+\frac{3}{2} \frac{\Delta Y}{Y}+\frac{2}{5} \frac{\Delta Z}{Z} \\
& =2(0.1)+\frac{3}{2}(0.2)+\frac{2}{5}(0.5) \\
& =0.2+0.3+0.2=0.7
\end{aligned}
\)

Hint

(c) Step 1: Determine the dimensions of \(\mu_0\) and \(\epsilon_0\)
The dimensional formula for vacuum permeability ( \(\mu_0\) ) in terms of mass ( M ), length ( L ), time ( T ), and current ( A ) is:
\(
\left[\mu_0\right]=\left[M L T^{-2} A^{-2}\right]
\)
The dimensional formula for vacuum permittivity ( \(\epsilon_0\) ) is:
\(
\left[\epsilon_0\right]=\left[M^{-1} L^{-3} T^4 A^2\right]
\)
Step 2: Calculate the dimension of the expression
The dimension of the ratio \(\frac{\mu_0}{\varepsilon_0}\) is:
\(
\begin{aligned}
& {\left[\frac{\mu_0}{\varepsilon_0}\right] }=\frac{\left[M L T^{-2} A^{-2}\right]}{\left[M^{-1} L^{-3} T^4 A^2\right]} \\
& {\left[\frac{\mu_0}{\varepsilon_0}\right]=\left[M^{1-(-1)} L^{1-(-3)} T^{-2-4} A^{-2-2}\right] } \\
& {\left[\frac{\mu_0}{\varepsilon_0}\right] }=\left[M^2 L^4 T^{-6} A^{-4}\right]
\end{aligned}
\)
\(
\begin{aligned}
&\text { The dimension of } \sqrt{\frac{\mu_0}{\epsilon_0}} \text { is the square root of this result: }\\
&\begin{gathered}
{\left[\sqrt{\frac{\mu_0}{\epsilon_0}}\right]=\left[M^2 L^4 T^{-6} A^{-4}\right]^{1 / 2}} \\
{\left[\sqrt{\frac{\mu_0}{\epsilon_0}}\right]=\left[M L^2 T^{-3} A^{-2}\right]}
\end{gathered}
\end{aligned}
\)
The dimensions of the options are:
Voltage: \(\left[M L^2 T^{-3} A^{-1}\right]\)
Inductance: \(\left[M L^2 T^{-2} A^{-2}\right]\)
Resistance: \(\left[M L^2 T^{-3} A^{-2}\right]\)
Capacitance: \(\left[M^{-1} L^{-2} T^4 A^2\right]\)
The dimension of \(\sqrt{\frac{\mu_0}{\epsilon_0}}\) matches the dimension of Resistance. This quantity is also known as the characteristic impedance of free space, often denoted by \(\mathbf{Z}_{\mathbf{0}}\).

Hint

(d) Mass density: Mass per unit volume \(\left(\frac{\mathrm{kg}}{\mathrm{m}^3}\right)\), dimensions \(\left[\mathrm{ML}^{-3} \mathrm{~T}^0\right]\).
Impulse: Force multiplied by time or change in momentum ( \(\mathrm{kg} \cdot \frac{\mathrm{m}}{\mathrm{s}}\) ), dimensions \(\left[\mathrm{MLT}^{-1}\right]\).
Power: Work done per unit time \(\left(\frac{\mathrm{J}}{\mathrm{S}}\right.\) or W\()\), dimensions \(\left[\mathrm{ML}^2 \mathrm{~T}^{-3}\right]\).
Moment of inertia: Mass multiplied by the square of the distance \(\left(\mathrm{kg} \cdot \mathrm{m}^2\right)\), dimensions \(\left[\mathrm{ML}^2 \mathrm{~T}^0\right]\).

Hint

(b) Dimensional Analysis
Electric Dipole Moment ( \(p\) ): Defined as charge ( \(q\) ) times distance ( \(d\) ).
Dimensions: \([\mathbf{A} \cdot \mathbf{T} \cdot \mathbf{L}]\) or \(\left[\mathbf{L}^1 \mathbf{T}^1 \mathbf{A}^1\right]\).
Magnetic Dipole Moment \((m)\) : Defined as current \((I)\) times area \((A)\).
Dimensions: \(\left[\mathbf{A} \cdot \mathbf{L}^{\mathbf{2}}\right]\) or \(\left[\mathbf{L}^{\mathbf{2}} \mathbf{A}^{\mathbf{1}}\right]\).
Ratio \(\left(\frac{p}{m}\right)\) : The ratio of electric to magnetic dipole moment.
Dimensions: \(\frac{\left[\mathrm{L}^1 \mathrm{~T}^1 \mathrm{~A}^1\right]}{\left[\mathrm{L}^2 \mathrm{~A}^1\right]}=\left[\mathrm{L}^{-1} \mathrm{~T}^1 \mathrm{~A}^0\right]=\left[\mathrm{M}^0 \mathrm{~L}^{-1} \mathrm{~T}^1 \mathrm{~A}^0\right]\).
Comparing the resulting dimension \(\left[\mathrm{M}^0 \mathrm{~L}^{-1} \mathrm{~T}^1 \mathrm{~A}^0\right]\) with the general form \(\left[\mathrm{M}^{\mathrm{P}} \mathrm{L}^{\mathrm{Q}} \mathrm{T}^R A^{\mathrm{S}}\right]\) :
\(P=0\)
\(Q=-1\)

Hint

(a) Step 1: Determine the value of one main scale division (MSD)
The main scale has 300 divisions equal to 15 cm. The value of one MSD is calculated as the total length divided by the number of divisions.
\(
1 \mathrm{MSD}=\frac{15 \mathrm{~cm}}{300}=0.05 \mathrm{~cm}
\)
Step 2: Determine the value of one vernier scale division (VSD)
The vernier scale has 25 divisions (VSD) equal to 24 divisions of the main scale (MSD). The value of one VSD is \(\frac{24}{25}\) times the value of one MSD.
\(
1 \mathrm{VSD}=\frac{24}{25} \times 1 \mathrm{MSD}=\frac{24}{25} \times 0.05 \mathrm{~cm}=0.048 \mathrm{~cm}
\)
Step 3: Calculate the least count (LC)
The least count of the microscope is the difference between one main scale division and one vernier scale division.
\(
L C=1 \mathrm{MSD}-1 \mathrm{VSD}=0.05 \mathrm{~cm}-0.048 \mathrm{~cm}=0.002 \mathrm{~cm}
\)
The least count (LC) of the travelling microscope is \(\mathbf{0 . 0 0 2 ~ c m}\).

Hint

(d) Step 1: Determine the dimensional formula for electric flux \(\boldsymbol{(} \boldsymbol{\Phi}_{\boldsymbol{E}}{)}\)
The dimensional formula for electric flux is derived from the formula \(\boldsymbol{\Phi}_{\boldsymbol{E}} \boldsymbol{=} \boldsymbol{E} \cdot \boldsymbol{A}\), where \(E\) is the electric field and \(A\) is the area. The dimension for the electric field is \(\left[M^1 L^1 T^{-3} A^{-1}\right]\) and for area is \(\left[L^2\right]\).
\(
\left[\Phi_E\right]=\left[M^1 L^1 T^{-3} A^{-1}\right] \times\left[L^2\right]=\left[M^1 L^3 T^{-3} A^{-1}\right]
\)
Step 2: Determine the dimensional formula for magnetic flux \(\boldsymbol{(} \boldsymbol{\Phi}_{\boldsymbol{B}}{)}\)
The dimensional formula for magnetic flux is derived from the formula \(\boldsymbol{\Phi}_{\boldsymbol{B}}=\boldsymbol{B} \cdot \boldsymbol{A}\), where \(\boldsymbol{B}\) is the magnetic field and \(\boldsymbol{A}\) is the area. The dimension for the magnetic field is \(\left[M^1 L^0 T^{-2} A^{-1}\right]\) (often written as \(\left[M^1 T^{-2} A^{-1}\right]\) as per search snippets, derived from \(F=q v B\) ) and for area is \(\left[L^2\right]\).
\(
\left[\Phi_B\right]=\left[M^1 T^{-2} A^{-1}\right] \times\left[L^2\right]=\left[M^1 L^2 T^{-2} A^{-1}\right]
\)
Step 3: Calculate the dimensional formula for the ratio of electric flux to magnetic flux
The ratio of electric flux to magnetic flux is \(\frac{\Phi_E}{\Phi_B}\).
\(
\begin{gathered}
\frac{\left[\Phi_E\right]}{\left[\Phi_B\right]}=\frac{\left[M^1 L^3 T^{-3} A^{-1}\right]}{\left[M^1 L^2 T^{-2} A^{-1}\right]} \\
=\left[M^{1-1} L^{3-2} T^{-3-(-2)} A^{-1-(-1)}\right] \\
=\left[M^0 L^1 T^{-1} A^0\right]
\end{gathered}
\)
The resulting dimension is \(\left[L^1 T^{-1}\right]\), which is the dimension of velocity.
The dimensional formula is in the form \(M^P L^Q T^R A^S\). Comparing this with the calculated dimension \(\left[M^0 L^1 T^{-1} A^0\right]\) :
The values of \(Q\) and \(R\) are 1 and -1 , respectively.

Hint

(d) (P) Boltzmann constant
\(
\frac{\text { Energy }}{\text { Temperature }}=\frac{M L^2 T^{-2}}{K}=\left[M L^2 T^{-2} K^{-1}\right]
\)
(Q) Coefficient of viscosity \((\eta)\) :
\(
\begin{aligned}
& \eta=\frac{F \delta x}{A \Delta V} \\
& {[\eta]=\frac{\left[M L T^{-2}\right][L]}{\left[L^2\right]\left[L T^{-1}\right]}=\left[M L^{-1} T^{-1}\right]}
\end{aligned}
\)
(R) Plank constant (h):
\(
E=h v,[h]=\frac{\left[M L^2 T^{-2}\right]}{\left[T^{-1}\right]}=\left[M L^2 T^{-1}\right]
\)
(S) Thermal conductivity
\(
\begin{aligned}
& K=\frac{\Delta Q l}{\Delta t(A \theta)} \\
& {[K]=\frac{\left[M L^2 T^{-2}\right][L]}{[T]\left[L^2\right][K]}=M L T^{-3} K^{-1}}
\end{aligned}
\)

Hint

(d) Step 1: Sum the masses
First, we calculate the arithmetic sum of the three given masses: \(435.42 \mathrm{~g}, 226.3 \mathrm{~g}\), and 0.125 g.
\(
435.42 \mathrm{~g}+226.3 \mathrm{~g}+0.125 \mathrm{~g}=661.845 \mathrm{~g}
\)
Step 2: Determine significant figures rule for addition
The rule for addition (and subtraction) with significant figures requires that the result be rounded to the same number of decimal places as the measurement with the fewest decimal places.
435.42 g has two decimal places.
226.3 g has one decimal place.
0.125 g has three decimal places.
The sum must be rounded to the tenths place (one decimal place).
Step 3: Round the sum to the correct decimal place
We round the calculated sum of 661.845 g to the tenths place:
The digit in the hundredths place is 4 , which is less than 5 , so we keep the tenths digit as 8.
The correctly rounded sum is 661.8 g.

Hint

(b) Step 1: Relate the constants to a known physical quantity
The permeability of free space ( \(\mu_0\) ) and permittivity of free space ( \(\epsilon_0\) ) are related to the speed of light in a vacuum ( \(c\) ) by the equation:
\(
c=\frac{1}{\sqrt{\mu_0 \epsilon_0}}
\)
Step 2: Express the given expression in terms of the physical quantity
The expression to evaluate is \(\left(\frac{1}{\mu_0 \epsilon_0}\right)\). Squaring the relationship from Step 1 gives:
\(
\frac{1}{\mu_0 \epsilon_0}=c^2
\)
Step 3: Determine the dimension of the resulting expression
The dimension of speed (c) is length per unit time, denoted as \(\left[L T^{-1}\right]\) or \(L / T\). The dimension of \(c^2\) is the square of the dimension of speed:
\(
\left[c^2\right]=[c]^2=\left[L T^{-1}\right]^2=\left[L^2 T^{-2}\right]
\)
This is equivalent to \(\mathrm{L}^2 / \mathrm{T}^2\).

Hint

(d) To determine which quantity has the dimensions of momentum, we analyze the dimensions of each option in terms of mass \((\mathrm{M})\), length \((\mathrm{L})\), time \((\mathrm{T})\), and current ( \(A\) or \(I\)):
Momentum ( \(p\) ): Mass × Velocity, so the dimension is \(\left[M L T^{-1}\right] .\)
Charge ( \(q\) ): Current \(\times\) Time, so the dimension is \([A T]\).
Current (\(I\)): Dimension is [latex]A[/latex].
Permeability of vacuum \(\left(\mu_0\right)\) : Dimension is \(\left[M L T^{-2} A^{-2}\right]\)
Let’s check the dimensions of option (d):
\(
\begin{gathered}
{\left[q \mu_0 I\right]=[q]\left[\mu_0\right][I]=[A T]\left[M L T^{-2} A^{-2}\right][A]} \\
{\left[q \mu_0 I\right]=\left[A T M L T^{-2} A^{-2} A\right]} \\
{\left[q \mu_0 I\right]=\left[M L T^{-1}\right]}
\end{gathered}
\)
This matches the dimension of momentum.

Hint

(c) Here are the dimensional formulas:
Coefficient of viscosity, \(\mu\)
\(
[\mu]=\frac{\text { pressure } \quad \times \text { time }}{=}=\underline{\left[M L^{-1} T^{-2}\right][T]}=\left[M L^{-1} T^{-1}\right] \quad \longrightarrow(\mathrm{IV})
\)
Intensity of a wave, \(I\)
\(
[I]=\frac{\text { power }}{\text { area }}=\frac{\left[M L^2 T^{-3}\right]}{\left[L^2\right]}=\left[M L^0 T^{-3}\right] \quad \longrightarrow(\mathrm{I})
\)
Pressure gradient, \(\mathrm{dP} / \mathrm{dx}\)
\(
\left[\frac{d P}{d x}\right]=\frac{\left[M L^{-1} T^{-2}\right]}{[L]}=\left[M L^{-2} T^{-2}\right] \quad \longrightarrow(\mathrm{II})
\)
Compressibility, \([\kappa]\)
\(
[\kappa]=\frac{1}{[\text { pressure }]}=\left[M^{-1} L T^2\right] \quad \longrightarrow(\mathrm{III})
\)

Hint

(d) Step 1: Determine the number of significant figures in each value
32.3 has 3 significant figures.
1125 has 4 significant figures.
27.4 has 3 significant figures.
The number with the fewest significant figures is 3 .
Step 2: Perform the calculation and round the result
The mathematical calculation is \(y=\frac{32.3 \times 1125}{27.4} \approx 1326.186 \ldots\)
According to the rules for multiplication and division with significant figures, the result must be rounded to the same number of significant figures as the value with the least number of significant figures (which is 3 ).
Rounding 1326.186… to 3 significant figures yields 1330. The first three significant digits are 1,3 , and 2 . Since the fourth digit is 6 (which is 5 or greater), we round up the third digit (2) to 3, and use a zero as a placeholder to maintain the magnitude.
The reported value of \(\boldsymbol{y}\) is \(\mathbf{y = 1 3 3 0}\).

Hint

(a) Step 1: Determine the maximum relative error formula
The energy is given by \(E=\alpha^3 \mathrm{e}^{-\beta t}\). Using logarithmic differentiation to find the relative error and considering the maximum possible error accumulation (absolute values), we find the maximum relative error formula:
\(
\frac{\Delta E}{E}=3 \frac{\Delta \alpha}{\alpha}+\beta \Delta t
\)
Step 2: Calculate the values of the individual terms
The given percentage errors are \(\frac{\Delta \alpha}{\alpha} \times 100 \%=1.2 \%\) (or \(\frac{\Delta \alpha}{\alpha}=0.012\) ) and \(\frac{\Delta t}{t} \times 100 \%=1.6 \%\). At \(t=5 \mathrm{~s}\), the absolute error in time \(\Delta t\) is:
\(
\Delta t=\frac{1.6 \%}{100 \%} \times t=0.016 \times 5 \mathrm{~s}=0.08 \mathrm{~s}
\)
The value of \(\beta\) is \(0.3 \mathrm{~s}^{-1}\).
Step 3: Substitute values and calculate the maximum percentage error
Substitute the values into the maximum relative error formula:
\(
\begin{gathered}
\frac{\Delta E}{E}=3 \times 0.012+0.3 \mathrm{~s}^{-1} \times 0.08 \mathrm{~s} \\
\frac{\Delta E}{E}=0.036+0.024=0.06
\end{gathered}
\)
To express this as a percentage error:
\(
\frac{\Delta E}{E} \times 100 \%=0.06 \times 100 \%=6 \%
\)
The maximum percentage error in the energy at \(t=5 \mathrm{~s}\) is \(6 \%\).

Hint

(c) Formula for density ( \(\rho\) ) of a wire: \(\rho=\frac{m}{V}=\frac{m}{\pi r^2 l}\) where \(m\) is mass, \(r\) is radius, and \(l\) is length.
Formula for maximum percentage error: The total percentage error is the sum of the percentage errors of the individual measurements, with the percentage error of the radius multiplied by 2 (due to the \(\boldsymbol{r}^2\) term).
% error in \(\rho=(\%\) error in \(m)+(2 \times \%\) error in \(r)+(\%\) error in \(l)\)
Calculate individual percentage errors:
% error in \(m=\frac{\Delta m}{m} \times 100=\frac{0.003}{0.60} \times 100=0.5 \%\)
% error in \(r=\frac{\Delta r}{r} \times 100=\frac{0.01}{0.50} \times 100=2 \%\)
% error in \(l=\frac{\Delta l}{l} \times 100=\frac{0.05}{10.00} \times 100=0.5 \%\)
Calculate total percentage error in density:
% error in \(\rho=0.5 \%+(2 \times 2 \%)+0.5 \%=0.5 \%+4 \%+0.5 \%=5 \%\)

Hint

(b) Statement I is true; a vernier scale division must be smaller than a main scale division for the calliper to function. Statement II is false; the vernier constant is calculated as the value of one main scale division minus the value of one vernier scale division, or by dividing the value of one main scale division by the total number of vernier scale divisions.

Hint

(c) Step 1: Isolate Planck’s constant (h)
The given de Broglie wavelength formula is \(\lambda=h / \sqrt{2 m E}\). Rearranging the equation to solve for Planck’s constant ( \(h\) ) gives \(h=\lambda \sqrt{2 m E}\).
Step 2: Determine the dimensions of each variable
The dimensions for the relevant variables are:
Wavelength ( \(\lambda\) ): [L]
Mass ( \(m\) ): [M]
Energy (E): \(\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]\)
The number 2 is a dimensionless constant.
Step 3: Substitute and simplify dimensions
Substituting the dimensions into the equation for \(h\) :
\(
\begin{gathered}
h=[\mathrm{L}] \sqrt{[\mathrm{M}]\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]} \\
h=[\mathrm{L}] \sqrt{\left[\mathrm{M}^2 \mathrm{~L}^2 \mathrm{~T}^{-2}\right]} \\
h=[\mathrm{L}]\left[\mathrm{MLT}^{-1}\right] \\
h=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]
\end{gathered}
\)

Hint

(c) The latent heat, \(L\), is defined as the amount of heat energy ( \(Q\) ) required to change the state of a unit mass ( \(m\) ) of a substance, given by the formula \(Q=m L\), which can be rearranged to \(L=Q / m\).
The dimensional formula for energy (heat) is \(\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]\).
The dimensional formula for mass is [M].
The dimensional formula for latent heat is calculated as the ratio of the dimensions of energy to mass:
\(
[L]=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]}{[\mathrm{M}]}=\left[\mathrm{M}^0 \mathrm{~L}^2 \mathrm{~T}^{-2}\right]
\)

Hint

(b) Step 1: Establish the relationship between MSD and VSD
The problem implies that the length of \(n\) main scale divisions (MSD) is equal to the length of ( \(n+1\) ) vernier scale divisions (VSD) in the design of the caliper.
\(
n \times \mathrm{MSD}=(n+1) \times \mathrm{VSD}
\)
We are given that one MSD equals \(\boldsymbol{m}\) units. We can express the VSD in terms of MSD (or \(m\) ):
\(
\mathrm{VSD}=\frac{n}{(n+1)} \times \mathrm{MSD}=\frac{n}{(n+1)} \times m
\)
Step 2: Calculate the Least Count (LC)
The least count (LC) of a vernier caliper is the difference between one main scale division and one vernier scale division:
\(
\mathrm{LC}=\mathrm{MSD}-\mathrm{VSD}
\)
Substitute the expression for VSD from Step 1 into the LC formula:
\(
\begin{aligned}
\mathrm{LC} & =\mathrm{MSD}-\frac{n}{(n+1)} \times \mathrm{MSD} \\
\mathrm{LC} & =\mathrm{MSD} \times\left(1-\frac{n}{(n+1)}\right) \\
\mathrm{LC} & =\mathrm{MSD} \times\left(\frac{(n+1)-n}{(n+1)}\right) \\
\mathrm{LC} & =\mathrm{MSD} \times\left(\frac{1}{(n+1)}\right)
\end{aligned}
\)
Finally, substitute \(\mathrm{MSD}=m\) units into the equation:
\(
\mathrm{LC}=\frac{m}{(n+1)}
\)

Hint

(a) Step 1: Identify the relevant physical relationship
The quantity \(\frac{1}{2} \epsilon_0 \mathrm{E}^2\) represents the energy density ( \(U\) ) stored in an electric field.
Step 2: Determine the dimensions of energy density
Energy density is defined as energy per unit volume. The dimensions for energy (work) are \(\left[M L^2 T^{-2}\right]\), and for volume are \(\left[L^3\right]\).
The dimensions of energy density are:
\(
U=\frac{\text { Energy }}{\text { Volume }}=\frac{\left[M L^2 T^{-2}\right]}{\left[L^3\right]}=\left[M L^{-1} T^{-2}\right]
\)
Since \(\frac{1}{2}\) is a dimensionless constant, the dimensions of \(\epsilon_{\mathrm{o}} \mathrm{E}^2\) are the same as the dimensions of energy density, \(\left[M L^{-1} T^{-2}\right]\).

Hint

(c) Step 1: Calculate the Least Count
The least count (LC) of the screw gauge is calculated using the formula:
\(
\begin{gathered}
\mathrm{LC}=\frac{\text { Pitch }}{\text { Number of divisions on circular scale }} \\
\mathrm{LC}=\frac{1 \mathrm{~mm}}{100}=0.01 \mathrm{~mm}
\end{gathered}
\)
Step 2: Determine the Zero Error
The zero of the circular scale lies 5 divisions below the reference line, indicating a positive zero error.
\(
\begin{gathered}
\text { Zero Error }(\mathrm{ZE})=+(\text { Number of divisions below }) \times \mathrm{LC} \\
\mathrm{ZE}=+5 \times 0.01 \mathrm{~mm}=+0.05 \mathrm{~mm}
\end{gathered}
\)
Step 3: Calculate the Observed Diameter
The observed diameter is the sum of the linear scale reading (LSR) and the circular scale reading (CSR) multiplied by the least count.
\(
\begin{gathered}
\text { Observed Reading }(\mathrm{OR})=\mathrm{LSR}+(\mathrm{CSR} \times \mathrm{LC}) \\
\mathrm{OR}=4 \mathrm{~mm}+(60 \times 0.01 \mathrm{~mm})=4 \mathrm{~mm}+0.60 \mathrm{~mm}=4.60 \mathrm{~mm}
\end{gathered}
\)
Step 4: Calculate the True Diameter
The true diameter is obtained by subtracting the zero error from the observed reading.
True Diameter (TD) = OR – ZE
\(
\mathrm{TD}=4.60 \mathrm{~mm}-0.05 \mathrm{~mm}=4.55 \mathrm{~mm}
\)

Hint

(d) Step 1: Define variables and formulas
The least count (LC) of a vernier caliper is given by the formula:
\(
\mathrm{LC}=\mathrm{MSD}-\mathrm{VSD}
\)
where MSD is the value of one main scale division and VSD is the value of one vernier scale division.
The problem provides:
\(\mathrm{LC}=\frac{1}{20 N} \mathrm{~cm}\)
\(\mathrm{MSD}=1 \mathrm{~mm}=0.1 \mathrm{~cm}=\frac{1}{10} \mathrm{~cm}\)
Step 2: Calculate the value of one VSD
Using the LC formula and given values:
\(
\begin{gathered}
\frac{1}{20 N} \mathrm{~cm}=\frac{1}{10} \mathrm{~cm}-\mathrm{VSD} \\
\mathrm{VSD}=\frac{1}{10} \mathrm{~cm}-\frac{1}{20 N} \mathrm{~cm} \\
\mathrm{VSD}=\left(\frac{2 N-1}{20 N}\right) \mathrm{cm}
\end{gathered}
\)
Step 3: Determine the coinciding divisions
The problem asks for the number of main scale divisions ( \(\boldsymbol{M}\) ) that coincide with \(\boldsymbol{N}\) divisions of the vernier scale. This relationship is defined by:
\(
M \times \mathrm{MSD}=N \times \mathrm{VSD}
\)
Substitute the values for MSD and VSD:
\(
M \times\left(\frac{1}{10} \mathrm{~cm}\right)=N \times\left(\frac{2 N-1}{20 N}\right) \mathrm{cm}
\)
Step 4: Solve for M
Simplify the equation to find the value of \(M\) :
\(
\begin{aligned}
\frac{M}{10} & =\frac{N(2 N-1)}{20 N} \\
\frac{M}{10} & =\frac{2 N-1}{20} \\
M & =\frac{2 N-1}{20} \times 10 \\
M & =\frac{2 N-1}{2}
\end{aligned}
\)

Hint

(b) In standard scientific notation, a quantity is written as
\(
N=a \times 10^b,
\)
where
\(1 \leq a<10\) (the mantissa),
\(b\) is an integer (the exponent).
The order of magnitude of \(N\) is the exponent \(b\) (provided the number is written in proper scientific notation).
So the exponent \(b\) is the order of magnitude whenever \(a\) lies in the valid range \(1 \leq a<10\).
Now check the options:
A: ” \(a\) is order of magnitude for \(b \leq 5\).” → Incorrect. The order of magnitude is not \(a\).
B: “b is order of magnitude for \(a \leq 5\).” → Only partially true; should be \(1 \leq a<10\), not just \(a \leq 5\).
C: ” \(b\) is order of magnitude for \(a \geq 5\).” → Also only partially true.
D: “b is order of magnitude for \(5<a \leq 10\).” → Again only partially true.
Correct statement (not present exactly):
\(b\) is the order of magnitude for any a satisfying \(1 \leq a<10\).

Hint

(b) Step 1: Calculate the least count (LC)
The value of one main scale division (MSD) is 1 mm. Given that 10 vernier scale divisions (VSD) equal 9 main scale divisions, \(1 \mathrm{VSD}=\frac{9}{10} \mathrm{MSD}=0.9 \mathrm{~mm}\). The least count (LC) is the difference between one MSD and one VSD:
\(
\mathrm{LC}=1 \mathrm{MSD}-1 \mathrm{VSD}=1 \mathrm{~mm}-0.9 \mathrm{~mm}=0.1 \mathrm{~mm}
\)
In centimeters, \(\mathrm{LC}=0.01 \mathrm{~cm}\).
Step 2: Determine the diameter
The total reading (diameter, \(d\) ) is calculated using the formula:
Reading \(=\mathrm{MSR}+(\mathrm{VSR} \times \mathrm{LC})\).
The main scale reading (MSR) is given as 2 cm. The vernier scale reading (VSR) is 2 (the second division coincides).
Diameter \((d)=2 \mathrm{~cm}+(2 \times 0.01 \mathrm{~cm})=2 \mathrm{~cm}+0.02 \mathrm{~cm}=2.02 \mathrm{~cm}\)
Step 3: Calculate the radius
The radius \((r)\) is half of the diameter:
\(
r=\frac{d}{2}=\frac{2.02 \mathrm{~cm}}{2}=1.01 \mathrm{~cm}
\)
Step 4: Calculate the volume
The volume ( \(V\) ) of a sphere is given by the formula \(V=\frac{4}{3} \pi r^3\) :
\(
V=\frac{4}{3} \times \pi \times(1.01 \mathrm{~cm})^3 \approx 4.316 \mathrm{~cm}^3
\)
Step 5: Calculate the density
The mass \((m)\) is 8.635 g . Density \((\rho)\) is mass per unit volume:
\(
\rho=\frac{m}{V}=\frac{8.635 \mathrm{~g}}{4.316 \mathrm{~cm}^3} \approx 2.0 \mathrm{~g} / \mathrm{cm}^3
\)

Hint

(d) The least count (LC) is calculated from the zero error and the coinciding division when the jaws are closed.
Zero of the vernier scale is left, so the zero error is positive.
Zero error \(=\) Coinciding Vernier Division × LC.
\(
0.04 \mathrm{~mm}=4 \times \mathrm{LC} \Longrightarrow \mathrm{LC}=0.01 \mathrm{~mm} .
\)
The relationship \(50 \mathrm{VSD}=49 \mathrm{MSD}\) means \(\mathrm{LC}=1 \mathrm{MSD}-1 \mathrm{VSD}=1 / 50 \mathrm{MSD}\).
From the \(\mathrm{LC}, 0.01 \mathrm{~mm}=1 / 50 \mathrm{MSD} \Longrightarrow 1 \mathrm{MSD}=0.5 \mathrm{~mm}\).
Since \(1 \mathrm{~cm}=10 \mathrm{~mm}\), the number of MSD in 1 cm is \(10 \mathrm{~mm} / 0.5 \mathrm{~mm}=20\) divisions.

Hint

(d) Step 1: Calculate the least count
The value of one Main Scale Division (MSD) is determined from the information that there are 20 divisions in 1 cm , so \(1 \mathrm{MSD}=\frac{1}{20} \mathrm{~cm}=0.05 \mathrm{~cm}\).
The least count (LC) is given by the formula LC \(=1 \mathrm{MSD}-1 \mathrm{VSD}\), or LC \(=\frac{\text { Value of one MSD }}{\text { Number of VSD }}\). Given 50 Vernier Scale Divisions (VSD) \(=49\) MSD, the formula is:
\(
\mathrm{LC}=\frac{1 \mathrm{MSD}}{50}=\frac{0.05 \mathrm{~cm}}{50}=0.001 \mathrm{~cm}
\)
The actual position reading (R) for each observation is calculated using the formula \(\mathrm{R}=\mathrm{MSR}+(\mathrm{VC} \times \mathrm{LC}):\)
For the mark on paper:
\(
\mathrm{R}_1=8.45 \mathrm{~cm}+(26 \times 0.001 \mathrm{~cm})=8.45 \mathrm{~cm}+0.026 \mathrm{~cm}=8.476 \mathrm{~cm}
\)
For the mark seen through the slab:
\(
\mathrm{R}_2=7.12 \mathrm{~cm}+(41 \times 0.001 \mathrm{~cm})=7.12 \mathrm{~cm}+0.041 \mathrm{~cm}=7.161 \mathrm{~cm}
\)
For the powder particle on the top surface:
\(
\mathrm{R}_3=4.05 \mathrm{~cm}+(1 \times 0.001 \mathrm{~cm})=4.05 \mathrm{~cm}+0.001 \mathrm{~cm}=4.051 \mathrm{~cm}
\)
Step 3: Determine real depth and apparent depth
Assuming the microscope reading increases downwards, the real and apparent depths are differences between the positions:
Real Depth is the distance between the top surface and the actual position of the mark at the bottom:
\(
\text { Real Depth }=\mathrm{R}_1-\mathrm{R}_3=8.476 \mathrm{~cm}-4.051 \mathrm{~cm}=4.425 \mathrm{~cm}
\)
Apparent Depth is the distance between the top surface and the apparent position of the mark seen through the slab:
\(
\text { Apparent Depth }=\mathrm{R}_2-\mathrm{R}_3=7.161 \mathrm{~cm}-4.051 \mathrm{~cm}=3.110 \mathrm{~cm}
\)
Step 4: Calculate refractive index
The refractive index \((\eta)\) is the ratio of the real depth to the apparent depth:
\(
\eta=\frac{\text { Real Depth }}{\text { Apparent Depth }}=\frac{4.425 \mathrm{~cm}}{3.110 \mathrm{~cm}} \approx 1.42
\)

Hint

(a) Step 1: Establish the relationship between variables
The relationship between the spring constant \((k)\), mass \((m)\), and the time period of oscillation \((t)\) is given by the formula for a simple harmonic oscillator:
\(
t=2 \pi \sqrt{\frac{m}{k}}
\)
Step 2: Rearrange the formula for \(k\)
To determine the error in \(k\), we must express \(k\) as a function of the measured variables \(m\) and \(t\) :
\(
k=\frac{4 \pi^2 m}{t^2}
\)
For our equation \(k=\left(4 \pi^2\right) \cdot m^1 \cdot t^{-2}\), the constant term has no error. The percentage error in \(k\left(\% E_k\right)\) is:
\(
\begin{gathered}
\% E_k=|1|\left(\% E_m\right)+|-2|\left(\% E_t\right) \\
\% E_k=\left(\% E_m\right)+2\left(\% E_t\right)
\end{gathered}
\)
Step 3: Calculate the final percentage error
Given a \(1 \%\) error in mass \(\left(\% E_m=1 \%\right)\) and a \(2 \%\) error in time \(\left(\% E_t=2 \%\right)\), the total percentage error in \(k\) is:
\(
\begin{gathered}
\% E_k=1 \%+2(2 \%) \\
\% E_k=1 \%+4 \% \\
\% E_k=5 \%
\end{gathered}
\)

Hint

(d) A-IV, B-III, C-II, D-I

Permeability of free space: \(\left[M L T^{-2} A^{-2}\right]\)
Magnetic field: \(\left[M T^{-2} A^{-1}\right]\)
Magnetic moment: \(\left[L^2 A\right]\)
Torsional constant: \(\left[M L^2 T^{-2}\right]\)

Hint

(b) Step 1: Calculate the Least Count
The least count (LC) of the screw gauge is calculated using the formula:
\(
\mathrm{LC}=\frac{\text { Pitch }}{\text { Number of divisions on circular scale }}
\)
Substituting the given values, Pitch \(=1 \mathrm{~mm}\) and Number of divisions \(=100\) :
\(
\mathrm{LC}=\frac{1 \mathrm{~mm}}{100}=0.01 \mathrm{~mm}
\)
Step 2: Calculate the Total Reading (Diameter)
The total reading (TR), which is the diameter of the wire, is calculated using the formula:
\(
\mathrm{TR}=\text { Main Scale Reading }+ \text { (Circular Scale Reading } \text { × } \text { LC })
\)
Substituting the given values, Main Scale Reading \(=1 \mathrm{~mm}\) and Circular Scale Reading = 42:
\(
\mathrm{TR}=1 \mathrm{~mm}+(42 \times 0.01 \mathrm{~mm})=1 \mathrm{~mm}+0.42 \mathrm{~mm}=1.42 \mathrm{~mm}
\)
Step 3: Determine the value of \(x\)
The diameter of the wire is given in the form \(\frac{x}{50} \mathrm{~mm}\). We equate our calculated diameter to this expression to find the value of \(x\) :
\(
1.42 \mathrm{~mm}=\frac{x}{50} \mathrm{~mm}
\)
Solving for \(x\) :
\(
x=1.42 \times 50=71
\)

Hint

(d) Step 1: Determine the dimension of \(b\)
The term \(\boldsymbol{b}\) is subtracted from the volume \(\boldsymbol{V}\) in the equation ( \(\boldsymbol{V}-\boldsymbol{b}\) ). According to the principle of dimensional homogeneity, terms that are subtracted from each other must have the same dimensions. The dimension of volume is \(\mathrm{L}^3\).
Therefore, the dimension of \(b\) is \([\mathbf{b}]=\left[\mathrm{L}^3\right]\).
Step 2: Determine the dimension of \(a\)
The term \(\frac{a}{V^2}\) is added to pressure \(P\) in the equation \(\left(P+\frac{a}{V^2}\right)\). Thus, their dimensions must be equal. The dimension of pressure \(\boldsymbol{P}\) (Force/Area) is \([\mathrm{P}]=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\). The dimension of volume squared \(V^2\) is \(\left[\mathrm{V}^2\right]=\left[\mathrm{L}^6\right]\).
Setting the dimensions equal:
\(
[P]=\left[\frac{a}{V^2}\right] \Longrightarrow[a]=[P]\left[V^2\right]
\)
Substituting the dimensions:
\(
[\mathrm{a}]=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]\left[\mathrm{L}^6\right]=\left[\mathrm{ML}^5 \mathrm{~T}^{-2}\right]
\)
Step 3: Determine the dimension of \(\mathrm{ab}^{-1}\)
To find the dimension of \(a b^{-1}\), we divide the dimension of \(a\) by the dimension of \(b\) :
\(
\begin{gathered}
{\left[\mathrm{ab}^{-1}\right]=\frac{[\mathrm{a}]}{[\mathrm{b}]}=\frac{\left[\mathrm{ML}^5 \mathrm{~T}^{-2}\right]}{\left[\mathrm{L}^3\right]}} \\
{\left[\mathrm{ab}^{-1}\right]=\left[\mathrm{ML}^{5-3} \mathrm{~T}^{-2}\right]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]}
\end{gathered}
\)

Hint

(b) Step 1: Understand the Vernier Caliper Principle
The least count (LC) of a vernier caliper is defined as the difference between one main scale division (MSD) and one vernier scale division (VSD). The formula for the least count is given by LC \(=\) MSD – VSD. The relationship between the scales is given by \(N \times\) VSD \(=(N-1) \times\) MSD, where \(N\) is the number of divisions on the vernier scale that coincide with ( \(N-1\) ) main scale divisions. From this, we can derive an alternative formula: \(\mathrm{LC}=\frac{\mathbf{M S D}}{\boldsymbol{N}}\).
Step 2: Calculate the Main Scale Division Value
Using the formula relating least count, main scale division, and the number of vernier divisions:
\(
\mathrm{MSD}=\mathrm{LC} \times N
\)
We are given that \(\mathrm{LC}=0.1 \mathrm{~mm}\) and \(N=20\) divisions. Substituting these values into the equation:
\(
\begin{gathered}
\mathrm{MSD}=0.1 \mathrm{~mm} \times 20 \\
\mathrm{MSD}=2.0 \mathrm{~mm}
\end{gathered}
\)

Hint

(b) Step 1: Calculate the arithmetic mean
The arithmetic mean \(({\bar{T}})\) is calculated by summing the four readings and dividing by the number of readings. The sum is \(4.62 \mathrm{~s}+4.632 \mathrm{~s}+4.6 \mathrm{~s}+4.64 \mathrm{~s}=18.492 \mathrm{~s}\).
The mean is:
\(
\bar{T}=\frac{18.492 \mathrm{~s}}{4}=4.623 \mathrm{~s}
\)
Step 2: Apply significant figure rules
The rule for reporting the mean of measured values requires the result to be rounded to the least precise decimal place of the original measurements.
4.62 s is precise to the hundredths place.
4.632 s is precise to the thousandths place.
4.6 s is precise to the tenths place.
4.64 s is precise to the hundredths place.
The least precise measurement is 4.6 s, which is precise to the tenths place. Therefore, the calculated mean of 4.623 s must be rounded to the tenths place.
Rounding 4.623 s to the nearest tenth gives 4.6 s.

Hint

(c) Step 1: Determine Dimensions of Physical Quantities
We first establish the dimensions for each variable:
Time period: \([T]=[T]\)
Radius: \([r]=[L]\)
Mass: \([M]=[M]\)
The dimension of the gravitational constant, \(G\), derived from \(F=G \frac{M^2}{r^2}\), is
\(
[G]=[\mathrm{M}]^{-1}[\mathrm{~L}]^3[\mathrm{~T}]^{-2} .
\)
Step 2: Check Dimensional Homogeneity for Each Option
The Left Hand Side (LHS) of all options has the dimension \(\left[\boldsymbol{T}^{\mathbf{2}}\right]=[\boldsymbol{T}]^{\mathbf{2}}\). We check the Right Hand Side (RHS) dimensions:
(A) \([\mathrm{RHS}]=\frac{[\mathrm{L}]}{[\mathrm{M}]^{-1}[\mathrm{~L}]^3[\mathrm{~T}]^{-2}[\mathrm{M}]^2}=[\mathrm{L}]^{-2}[\mathrm{M}]^{-1}[\mathrm{~T}]^2 \neq[\mathrm{T}]^2\)
(B) \([\mathrm{RHS}]=[\mathrm{L}]^3 \neq[\mathrm{T}]^2\)
(C) \([\) RHS \(]=\frac{[L]^3}{[M]^{-1}[L]^3[T]^{-2}[M]}=\frac{[L]^3}{[L]^3[T]^{-2}}=[T]^2\)
(D) \([\) RHS \(]=\frac{[L]^2}{[M]^{-1}[L]^3[T]^{-2}[M]}=[L]^{-1}[T]^2 \neq[T]^2\)

Hint

(d) Step 1: Analyze the dimensions of the trigonometric arguments
The arguments of trigonometric functions must be dimensionless. In the given equation \(y=2 a \sin \left(\frac{2 \pi n t}{\lambda}\right) \cos \left(\frac{2 \pi x}{\lambda}\right)\), the arguments are \(\frac{2 \pi n t}{\lambda}\) and \(\frac{2 \pi x}{\lambda}\).
We require:
\(
\begin{aligned}
& {\left[\frac{n t}{\lambda}\right]=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0\right]} \\
& {\left[\frac{x}{\lambda}\right]=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0\right]}
\end{aligned}
\)
Step 2: Determine the dimension of \(\boldsymbol{n}\)
From the second argument, we know \([x]=[L]\) and \([\lambda]=[L]\). Thus \(\left[\frac{x}{\lambda}\right]=[L / L]\), which is dimensionless, confirming statement C is correct.
From the first argument, we substitute the known dimensions:
\(
\begin{aligned}
& \frac{[n][t]}{[\lambda]}=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0\right] \\
& \frac{[n][\mathrm{T}]}{[\mathrm{L}]}=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0\right]
\end{aligned}
\)
Solving for \([n]\) :
\(
[n]=\frac{[\mathrm{L}]}{[\mathrm{T}]}=\left[\mathrm{LT}^{-1}\right]
\)
This confirms statement B is correct. The variable \(n\) has the dimensions of velocity.
Step 3: Evaluate the dimensions of the expressions in the options
Using the dimension \([n]=\left[\mathrm{LT}^{-1}\right]\), we evaluate the dimensions of the expressions in options A and D :
For option A:
\(
[n t]=[n][t]=\left[\mathrm{LT}^{-1}\right][\mathrm{T}]=[\mathrm{L}]
\)
This confirms statement \(A\) is correct.
For option D:
\(
\left[\frac{n}{\lambda}\right]=\frac{[n]}{[\lambda]}=\frac{\left[\mathrm{LT}^{-1}\right]}{[\mathrm{L}]}=\left[\mathrm{T}^{-1}\right]
\)
The dimension of \(n / \lambda\) is \(\left[\mathrm{T}^{-1}\right]\), which is the dimension of frequency, not \([\mathrm{T}]\) (time).
The statement that is NOT correct is D, as the dimension of \(n / \lambda\) is \(\left[\mathrm{T}^{-1}\right]\), not \([\mathrm{T}]\).

Hint

(a) The correct match between the numbers in List-I and their significant figures in List-II is:
(A) 1001 – (II) 4
(B) 010.1 – (I) 3
(C) 100.100 – (IV) 6
(D) 0.0010010 – (III) 5
Explanation of Significant Figures:
(A) 1001: All non-zero digits are significant, and any zeros between non-zero digits are also significant. Thus, all four digits (1, 0, 0, 1) are significant.
(B) 010.1: Leading zeros (zeros to the left of the first non-zero digit) are not significant. The digits 1,0 , and 1 (after the leading zero) are significant (the zero is between non-zero digits). The number is effectively 10.1, which has 3 significant figures.
(C) 100.100: Zeros between non-zero digits are significant. Trailing zeros to the right of the decimal point are also significant as they indicate precision. Therefore, all six digits are significant.
(D) 0.0010010 : The leading zeros ( 0.00 ) are not significant placeholders. The subsequent digits \(1,0,0,1\) are significant (zeros between non-zero digits). The final trailing zero is to the right of the decimal point and follows non-zero digits, so it is significant. This gives a total of 5 significant figures.

Hint

(a) Step 1: Define variables and relationship
The problem states that 10 main scale divisions (MSD) coincide with 11 Vernier scale divisions (VSD), and each MSD has a value of 5 units [calculated]. This relationship can be expressed as:
\(
10 \times \mathrm{MSD}=11 \times \mathrm{VSD}
\)
We know the value of 1 MSD : \(\mathrm{MSD}=5\) units.
Step 2: Calculate the value of one Vernier scale division (VSD)
Substitute the value of the MSD into the relationship to find the VSD value:
\(
\begin{gathered}
10 \times 5 \text { units }=11 \times \mathrm{VSD} \\
\mathrm{VSD}=\frac{50}{11} \text { units }
\end{gathered}
\)
Step 3: Calculate the least count (LC)
The least count of a Vernier calliper is the difference between one main scale division and one Vernier scale division:
\(
\mathrm{LC}=\mathrm{MSD}-\mathrm{VSD}
\)
Substitute the known values of MSD and VSD into the formula:
\(
\begin{gathered}
\mathrm{LC}=5 \text { units }-\frac{50}{11} \text { units } \\
\mathrm{LC}=\frac{55}{11} \text { units }-\frac{50}{11} \text { units } \\
\mathrm{LC}=\frac{5}{11} \text { units }
\end{gathered}
\)
The least count of the instrument is \(\frac{\mathbf{5}}{\mathbf{1 1}}\) units.

Hint

(c) Definition: Angular impulse equals the product of torque \((\tau)\) and the time interval ( \(\boldsymbol{\Delta} \boldsymbol{t}\) ) over which it acts (Angular Impulse \(\boldsymbol{=} \boldsymbol{\tau} \boldsymbol{\Delta} \boldsymbol{t}\) ). It is also equal to the change in angular momentum.

Derivation:
Torque has a dimensional formula of \(\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]\) (force × distance).
Time has a dimensional formula of [T].
Multiplying these gives: \(\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right] \times[\mathrm{T}]=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]\).
Equivalence: The dimensional formula \(\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]\) is the same as that for angular momentum and Planck’s constant.

Hint

(d) Step 1: Resistivity Formula
The relationship between resistance \((R)\), length \((l)\), radius \((r)\), and resistivity \((\rho)\) of a wire is given by the formula \(R=\frac{\rho l}{\pi r^2}\). Rearranging to solve for resistivity gives:
\(
\rho=\frac{\pi R r^2}{l}
\)
Step 2: Error Propagation Formula
To find the maximum possible fractional error in the calculated resistivity, we sum the individual fractional errors of the measured quantities. The formula for the fractional error in \(\rho\) is:
\(
\frac{\Delta \rho}{\rho}=\frac{\Delta R}{R}+2 \frac{\Delta r}{r}+\frac{\Delta l}{l}
\)
The percentage error is \(100 \times \frac{\Delta \rho}{\rho}\).
Step 3: Calculation
We substitute the given values and their errors into the formula:
\(\frac{\Delta R}{R}=\frac{10}{100}=0.1\)
\(\frac{\Delta r}{r}=\frac{0.05}{0.35}=\frac{1}{7}\)
\(\frac{\Delta l}{l}=\frac{0.2}{15}=\frac{1}{75}\)
The fractional error in resistivity is:
\(
\frac{\Delta \rho}{\rho}=0.1+2\left(\frac{1}{7}\right)+\frac{1}{75}=\frac{1}{10}+\frac{2}{7}+\frac{1}{75}=\frac{419}{1050}
\)
The percentage error is:
\(
\text { Percentage Error }=100 \times \frac{419}{1050} \approx 39.9 \%
\)

Hint

(b) Step 1: Determine the dimension of \(B\)
According to the principle of dimensional homogeneity, all terms in a sum or difference must have the same dimension. In the expression \(\left(B-x^2\right)\), both \(B\) and \(x^2\) must have the same dimension as length squared, \(\left[L^2\right]\).
\(
[B]=\left[x^2\right]=L^2
\)
Step 2: Determine the dimension of \(A\)
The given equation is \(E=\frac{B-x^2}{A t}\). Rearranging for \(A\) gives \(A=\frac{B-x^2}{E t}\).
The dimensions are \([\boldsymbol{E}]=\boldsymbol{M L}^2 \boldsymbol{T}^{-2},[\boldsymbol{t}]=\boldsymbol{T}\), and \(\left[\boldsymbol{B}-\boldsymbol{x}^2\right]=\boldsymbol{L}^2\).
\(
\begin{gathered}
{[A]=\frac{\left[B-x^2\right]}{[E][t]}} \\
{[A]=\frac{L^2}{\left(M L^2 T^{-2}\right)(T)}} \\
{[A]=\frac{L^2}{M L^2 T^{-1}}} \\
{[A]=M^{-1} T^1}
\end{gathered}
\)
Step 3: Determine the dimension of \(A B\)
The dimension of the product \(A B\) is the product of their individual dimensions.
\(
\begin{gathered}
{[A B]=[A][B]} \\
{[A B]=\left(M^{-1} T^1\right)\left(L^2\right)} \\
{[A B]=L^2 M^{-1} T^1}
\end{gathered}
\)

Hint

(a)

\(
\begin{aligned}
&\begin{aligned}
& \mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{~g}}} \\
& \mathrm{~g}=\frac{4 \pi^2 \ell}{\mathrm{~T}^2} \\
& \frac{\Delta \mathrm{~g}}{\mathrm{~g}}=\frac{\Delta \ell}{\ell}+\frac{2 \Delta \mathrm{~T}}{\mathrm{~T}} \\
& =\frac{0.2}{20}+2\left(\frac{1}{40}\right) \\
& =\frac{1.2}{20}
\end{aligned}\\
&\text { Percentage change }=\frac{1.2}{20} \times 100=6 \%
\end{aligned}
\)

Hint

(d)

\(
\begin{aligned}
& \mathrm{R}=\frac{\rho \mathrm{L}}{\pi \frac{\mathrm{~d}^2}{4}} \\
& \frac{\Delta \mathrm{R}}{\mathrm{R}}=\frac{\Delta \mathrm{L}}{\mathrm{~L}}+\frac{2 \Delta \mathrm{~d}}{\mathrm{~d}} \\
& \frac{\Delta \mathrm{~L}}{\mathrm{~L}}=0.1 \% \text { and } \frac{\Delta \mathrm{d}}{\mathrm{~d}}=0.1 \% \\
& \frac{\Delta \mathrm{R}}{\mathrm{R}}=0.3 \%
\end{aligned}
\)

Hint

(b) To determine the dimensions of \(\frac{b^2}{a}\), let’s start by identifying the dimensions of each term in the equation \(F=a x^2+b t^{\frac{1}{2}}\), where \(F\) represents force, \(x\) represents distance, and \(t\) represents time.
The dimension of force \((F)\) is given by \(\left[\mathrm{MLT}^{-2}\right]\), where \(M\) is mass, \(L\) is length, and \(T\) is time.
The term \(a x^2\) has the same dimension as force, so:
\(
[a]=\frac{[F]}{[x]^2}=\frac{M L T^{-2}}{L^2}=M L^{-1} T^{-2}
\)
The term \(b t^{\frac{1}{2}}\) also has the same dimension as force, which gives:
\(
[b]=\frac{[F]}{[t]^{\frac{1}{2}}}=\frac{M L T^{-2}}{T^{\frac{1}{2}}}=M L T^{-\frac{5}{2}}
\)
Now, to find \(\frac{b^2}{a}\), we substitute the dimensions of \(b\) and \(a\) :
\(
\left[\frac{b^2}{a}\right]=\frac{\left(M^2 L^2 T^{-5}\right)}{\left(M L^{-1} T^{-2}\right)}=\left[M^1 L^3 T^{-3}\right]
\)
Therefore, the dimensions of \(\frac{b^2}{a}\) are \(\left[\mathrm{ML}^3 \mathrm{~T}^{-3}\right]\), which corresponds to mass times length cubed per time cubed.

Hint

(a) Step 1: Identify given values and formula
The value of one main scale division is given as \(S=0.5 \mathrm{~mm}\). The number of Vernier divisions ( \(n\) ) equal to \(n-1\) main scale divisions is 50. The formula for the Vernier constant (LC) is:
\(
L C=\frac{S}{n}
\)
Step 2: Calculate the Vernier constant
Substitute the given values into the formula:
\(
\begin{aligned}
& L C=\frac{0.5 \mathrm{~mm}}{50} \\
& L C=0.01 \mathrm{~mm}
\end{aligned}
\)
The Vernier constant of the traveling microscope is \(\mathbf{0 . 0 1 ~ m m}\).

Alternate: Step 1: Calculate Vernier Constant Formula
The relationship between the main scale division (MSD) and the Vernier scale division (VSD) is given as 50 Vernier divisions equal to 49 main scale divisions, which means \(50 \times \mathrm{VSD}=49 \times \mathrm{MSD}\). Therefore, one Vernier scale division is \(\mathrm{VSD}=\frac{49}{50} \times \mathrm{MSD}\).
The Vernier constant (VC) or least count is the difference between one main scale division and one Vernier scale division:
\(
\mathrm{VC}=\mathrm{MSD}-\mathrm{VSD}
\)
Substituting the expression for VSD:
\(
\mathrm{VC}=\mathrm{MSD}-\frac{49}{50} \times \mathrm{MSD}=\mathrm{MSD} \times\left(1-\frac{49}{50}\right)=\mathrm{MSD} \times\left(\frac{50-49}{50}\right)=\frac{\mathrm{MSD}}{50}
\)
Step 2: Substitute values and calculate
The smallest reading of the main scale (MSD) is given as \(\mathbf{0 . 5 ~ m m}\). Using the derived formula for the Vernier constant:
\(
\begin{aligned}
& \mathrm{VC}=\frac{0.5 \mathrm{~mm}}{50} \\
& \mathrm{VC}=0.01 \mathrm{~mm}
\end{aligned}
\)
The value can also be expressed in centimeters:
\(
\mathrm{VC}=0.01 \mathrm{~mm} \times \frac{1 \mathrm{~cm}}{10 \mathrm{~mm}}=0.001 \mathrm{~cm}
\)
The Vernier constant of the traveling microscope is \(\mathbf{0 . 0 1 ~ m m}\).

Hint

(c)

\(
\begin{aligned}
&\begin{aligned}
& \mathrm{m}=\mathrm{kc}^{\mathrm{P}} \mathrm{G}^{-1 / 2} \mathrm{~h}^{1 / 2} \\
& \mathrm{M}^1 \mathrm{~L}^0 \mathrm{~T}^0=\left[\mathrm{LT}^{-1}\right]^{\mathrm{P}}\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]^{-1 / 2}\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]^{1 / 2}
\end{aligned}\\
&\text { By comparing } \mathrm{P}=1 / 2
\end{aligned}
\)

Hint

(d)

\(
\begin{aligned}
& F=\eta A \frac{d v}{d y} \\
& {\left[M L T^{-2}\right]=\eta\left[L^2\right]\left[T^{-1}\right]} \\
& \eta=\left[M L^{-1} T^{-1}\right] \\
& S \cdot T=\frac{F}{\ell}=\frac{\left[M L T^{-2}\right]}{[L]}=\left[M L^0 T^{-2}\right] \\
& L=m v r=\left[M L^2 T^{-1}\right] \\
& K \cdot E=\frac{1}{2} I \omega^2=\left[M L^2 T^{-2}\right]
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
& \mathrm{Q}=\frac{\mathrm{a}^4 \mathrm{~b}^3}{\mathrm{c}^2} \\
& \frac{\Delta \mathrm{Q}}{\mathrm{Q}}=4 \frac{\Delta \mathrm{a}}{\mathrm{a}}+3 \frac{\Delta \mathrm{~b}}{\mathrm{~b}}+2 \frac{\Delta \mathrm{c}}{\mathrm{c}} \\
& \frac{\Delta \mathrm{Q}}{\mathrm{Q}} \times 100=4\left(\frac{\Delta \mathrm{a}}{\mathrm{a}} \times 100\right)+3\left(\frac{\Delta \mathrm{~b}}{\mathrm{~b}} \times 100\right)+2\left(\frac{\Delta \mathrm{c}}{\mathrm{c}} \times 100\right) \\
& \% \text { error in } \mathrm{Q}=4 \times 3 \%+3 \times 4 \%+2 \times 5 \% \\
& =12 \%+12 \%+10 \% \\
& =34 \%
\end{aligned}
\)

Hint

(d)

\(
\begin{aligned}
&\mathrm{R}=\frac{\mathrm{V}}{1}\\
&\text { According to error analysis }\\
&\begin{aligned}
\frac{\mathrm{dR}}{\mathrm{R}} & =\frac{\mathrm{dV}}{\mathrm{~V}}+\frac{\mathrm{dI}}{\mathrm{I}} \\
\frac{\mathrm{dR}}{\mathrm{R}} & =\frac{5}{200}+\frac{0.2}{20} \\
\frac{\mathrm{dR}}{\mathrm{R}} & =\frac{7}{200}
\end{aligned}\\
&\% \text { error } \frac{\mathrm{dR}}{\mathrm{R}} \times 100=\frac{7}{200} \times 100=3.5 \%
\end{aligned}
\)

Hint

(b) Both (A) and (R) are correct, but (R) is not the correct explanation of (A).
Assertion (A) is correct: A positive zero error means the instrument reads a positive value when the jaws are closed, so any measurement taken will be more than the actual reading by that positive value.
Reason ( R ) is correct: Zero errors can indeed occur due to manufacturing defects or rough handling of the instrument, which affects the alignment of the scales.
Reason (R) does not explain (A): Reason (R) describes the cause of the error, while Assertion (A) describes the effect of the error on the measurement. The cause (R) does not explain the physical principle (how the positive offset leads to an overestimated measurement) described in (A).

Hint

(a) 

\(
\begin{aligned}
&\begin{aligned}
& {[\mathrm{h}]=\mathrm{ML}^2 \mathrm{~T}^{-1}} \\
& {[\mathrm{~L}]=\mathrm{ML}^2 \mathrm{~T}^{-1}} \\
& {[\mathrm{P}]=\mathrm{MLT}^{-1}} \\
& {[\tau]=\mathrm{ML}^2 \mathrm{~T}^{-2}}
\end{aligned}\\
&\text { (Here } \mathrm{h} \text { is Planck’s constant, } \mathrm{L} \text { is angular momentum, } \mathrm{P} \text { is linear momentum and } \tau \text { is moment of force) }
\end{aligned}
\)

Hint

(d)

\(
v=\lambda^{\mathrm{a}} \mathrm{~g}^{\mathrm{b}} \rho^{\mathrm{c}}
\)
using dimension formula
\(
\begin{aligned}
& \Rightarrow\left[\mathrm{M}^0 \mathrm{~L}^1 \mathrm{~T}^{-1}\right]=\left[\mathrm{L}^1\right]^{\mathrm{a}}\left[\mathrm{~L}^1 \mathrm{~T}^{-2}\right]^{\mathrm{b}}\left[\mathrm{M}^1 \mathrm{~L}^{-3}\right]^{\mathrm{c}} \\
& \Rightarrow\left[\mathrm{M}^0 \mathrm{~L}^1 \mathrm{~T}^{-1}\right]=\left[\mathrm{M}^{\mathrm{c}} \mathrm{~L}^{\mathrm{a}+\mathrm{b}-\mathrm{c}} \mathrm{~T}^{-2 \mathrm{~b}}\right] \\
& \therefore \mathrm{c}=0, \mathrm{a}+\mathrm{b}-3 \mathrm{c}=1,-2 \mathrm{~b}=-1 \Rightarrow \mathrm{~b}=\frac{1}{2}
\end{aligned}
\)
Now \(\mathrm{a}+\mathrm{b}-3 \mathrm{c}=1\)
\(
\begin{aligned}
& \Rightarrow \mathrm{a}+\frac{1}{2}-0=1 \\
& \Rightarrow \mathrm{a}=\frac{1}{2} \\
& \therefore \mathrm{a}=\frac{1}{2}, \mathrm{~b}=\frac{1}{2}, \mathrm{c}=0
\end{aligned}
\)

Hint

(b) Given that, \(\left[X+\frac{a}{Y^2}\right][Y-b]=\mathrm{R} T\)
\(\therefore X\) and \(\frac{a}{Y^2}\) have the same dimensions and \(Y\) and \(b\) have the same dimensions, let’s analyze the dimensions of \(\frac{a}{b}\).
Since \(X\) represents pressure, it has dimensions of \(\left[M L^{-1} T^{-2}\right]\).
Since \(X\) and \(\frac{a}{Y^2}\) have the same dimensions, we have:
\(
\left[\frac{a}{Y^2}\right]=\left[M L^{-1} T^{-2}\right]
\)
Then, the dimensions of \(a\) are:
\(
[a]=\left[M L^{-1} T^{-2}\right]\left[Y^2\right]=\left[M L^5 T^{-2}\right]
\)
Now, since \(Y\) and \(b\) have the same dimensions and \(Y\) represents volume, we have:
\(
[b]=\left[L^3\right]
\)
Now, let’s find the dimensions of the ratio \(\frac{a}{b}\) :
\(
\frac{[a]}{[b]}=\frac{\left[M L^5 T^{-2}\right]}{\left[L^3\right]}=\left[M L^2 T^{-2}\right]
\)
Indeed, the dimensions of \(\frac{a}{b}\) are \(\left[M L^2 T^{-2}\right]\), which corresponds to the dimensions of energy.

Hint

(a) Step 1: Calculate the nominal kinetic energy
The kinetic energy \((K E)\) is calculated using the formula \(K E=\frac{1}{2} m v^2\).
The nominal values are \(m=5 \mathrm{~kg}\) and \(v=20 \mathrm{~m} / \mathrm{s}\).
\(
K E=\frac{1}{2} \times 5 \mathrm{~kg} \times(20 \mathrm{~m} / \mathrm{s})^2=\frac{1}{2} \times 5 \times 400 \mathrm{~J}=1000 \mathrm{~J}
\)
Step 2: Calculate the uncertainty in kinetic energy
To find the maximum possible uncertainty ( \(\Delta K E\) ), we use the formula for propagation of errors for multiplication and powers:
\(
\frac{\Delta K E}{K E}=\frac{\Delta m}{m}+2 \frac{\Delta v}{v}
\)
Given uncertainties are \(\Delta m=0.5 \mathrm{~kg}\) and \(\Delta v=0.4 \mathrm{~m} / \mathrm{s}\).
\(
\begin{gathered}
\frac{\Delta K E}{1000 \mathrm{~J}}=\frac{0.5 \mathrm{~kg}}{5 \mathrm{~kg}}+2 \frac{0.4 \mathrm{~m} / \mathrm{s}}{20 \mathrm{~m} / \mathrm{s}} \\
\frac{\Delta K E}{1000 \mathrm{~J}}=0.1+2 \times 0.02=0.1+0.04=0.14
\end{gathered}
\)
Now, we solve for \(\triangle K E\) :
\(
\Delta K E=0.14 \times 1000 \mathrm{~J}=140 \mathrm{~J}
\)
The kinetic energy of the body is \((1000 \pm 140) \mathrm{J}\).

Hint

(c) Step 1: Analyze Spring Constant
The spring constant ( \(\boldsymbol{k}\) ) is derived from Hooke’s Law, \(\boldsymbol{F}=\boldsymbol{k} \boldsymbol{x}\). The dimensional formula for force is \(\left[\mathrm{MLT}^{-2}\right]\), and for displacement is \([\mathrm{L}]\).
The dimensional formula of spring constant, \(k\) :
\(
[\mathrm{k}]=\frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{L}]}=\left[\mathrm{MT}^{-2}\right]
\)
So, A matches with II.
Step 2: Analyze Angular Speed
Angular speed ( \(\omega\) ) is the rate of change of angular displacement with respect to time. Angular displacement is dimensionless. The dimensional formula for angular speed is the reciprocal of the dimensional formula for time:
\(
[\omega]=\left[\mathrm{T}^{-1}\right]
\)
So, B matches with I.
Step 3: Analyze Angular Momentum
Angular momentum ( \(L\) ) can be defined as the product of moment of inertia ( \(I\) ) and angular speed \((\omega), \boldsymbol{L}=\boldsymbol{I} \omega\). The dimensional formula for moment of inertia is \(\left[\mathbf{M L}^2\right]\), and for angular speed is \(\left[\mathrm{T}^{-1}\right]\).
The dimensional formula of angular momentum, \(L\) :
\(
[\mathrm{L}]=[I][\omega]=\left[\mathrm{ML}^2\right]\left[\mathrm{T}^{-1}\right]=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]
\)
So, C matches with IV.
Step 4: Analyze Moment of Inertia
The moment of inertia ( \(I\) ) depends on the mass ( \(m\) ) of the object and its distribution around the axis of rotation (distance \(r\) ). It is proportional to \(m r^2\).
The dimensional formula of moment of inertia, \(\boldsymbol{I}\) :
\(
[\mathrm{I}]=[\mathrm{M}]\left[\mathrm{L}^2\right]=\left[\mathrm{ML}^2\right]
\)
So, D matches with III.

Hint

(d) The fundamental dimensions for force (F), velocity (V), time (T), and density ( \(\rho\) ) in terms of Mass (M), Length (L), and Time (T) are:
\([\rho]=M L^{-3}\)
\([F]=M L T T^{-2}\)
\([V]=L T^{-1}\)
\([T]=T\)
Step 2: Formulate Equations
We express density as a function of \(\mathrm{F}, \mathrm{V}\), and T with unknown exponents \(a, b\), and \(c\) :
\(
[\rho]=[F]^a[V]^b[T]^c
\)
Substituting the M, L, T dimensions into the equation:
\(
\begin{aligned}
& M L^{-3} T^0=\left(M L T^{-2}\right)^a\left(L T^{-1}\right)^b(T)^c \\
& M L^{-3} T^0=M^a L^{a+b} T^{-2 a-b+c}
\end{aligned}
\)
Equating the exponents for M, L, and T gives a system of linear equations:
For M: \(a=1\)
For L: \(a+b=-3\)
For T: \(-2 a-b+c=0\)
Step 3: Solve for Exponents
Solving the system of equations:
From the M equation, \(a=1\).
Substitute \(a=1\) into the L equation: \(1+b=-3 \Longrightarrow b=-4\).
Substitute \(a=1\) and \(b=-4\) into the T equation:
\(
-2(1)-(-4)+c=0 \Longrightarrow-2+4+c=0 \Longrightarrow c=-2 .
\)
The exponents are \(a=1, b=-4\), and \(c=-2\).
The dimensional formula for density in terms of \(\mathrm{F}, \mathrm{V}\), and T is \(\mathrm{FV}^{-4} \mathrm{~T}^{-2}\).

Hint

(d)

Statement I is correct: The Astronomical Unit (AU), Parsec (Pc), and Light-year (ly) are all standard units used by astronomers to measure the vast distances in space.

Statement II is incorrect: The correct order of these units from smallest to largest is \(\mathbf{A U}<\mathbf{l y}<\mathbf{P c}\).
1 AU is the Earth-Sun distance (approx. \(1.5 \times 10^{11} \mathrm{~m}\) ).
1 ly is approximately \(63,240 \mathrm{AU}\) (approx. \(9.46 \times 10^{15} \mathrm{~m}\) ).
1 Pc is approximately 206,265 AU and 3.26 ly (approx. \(3.09 \times 10^{16} \mathrm{~m}\) ).

Hint

(a)

Zero Error: The zero of the vernier scale is to the right, indicating a positive zero error.
Zero Error \(=\) Coinciding Vernier Division × Least Count \(=6 \times 0.1 \mathrm{~mm}=+0.6 \mathrm{~mm}=\) +0.06 cm.

Observed Reading:
Main Scale Reading (MSR) = 3.2 cm (the mark just before the vernier zero).
Vernier Scale Reading (VSR) \(=\) Coinciding Vernier Division × Least Count \(= 4 \times 0.1 \mathrm{~mm}=0.4 \mathrm{~mm}=0.04 \mathrm{~cm}\).
Total Observed Reading \(=\mathrm{MSR}+\mathrm{VSR}=3.2 \mathrm{~cm}+0.04 \mathrm{~cm}=3.24 \mathrm{~cm}\).

Corrected Diameter (Actual Reading): The true measurement is the observed reading minus the zero error.
Actual Reading = Total Observed Reading – Zero Error = \(3.24 \mathrm{~cm}-0.06 \mathrm{~cm}=3.18 \mathrm{~cm}\).

Hint

(b) The percentage error in a quantity that is a product or quotient of other quantities is given by the sum of the percentage errors in those quantities, each multiplied by the power to which it is raised in the expression for the quantity.
Given the physical quantity \(P\) as
\(
P=\frac{a^2 b^3}{c \sqrt{d}}
\)
The percentage error in P , denoted as \(\Delta P / P\), is given by:
\(
\frac{\Delta P}{P}=2\left(\frac{\Delta a}{a}\right)+3\left(\frac{\Delta b}{b}\right)+\left(\frac{\Delta c}{c}\right)+\frac{1}{2}\left(\frac{\Delta d}{d}\right)
\)
Substituting the given percentage errors for \(\mathrm{a}, \mathrm{b}, \mathrm{c}\), and d :
\(
\frac{\Delta P}{P}=2(0.01)+3(0.02)+0.03+\frac{1}{2}(0.04)=0.02+0.06+0.03+0.02=0.13
\)
Therefore, the percentage error in P is \(13 \%\).

Hint

(b) Let’s analyze the SI units of each quantity from List I:
Torque: Torque \((\tau)\) is given by the cross product of the radius \((r)\) and the force \((F)\). Therefore, its Sl units are Newton meter \((\mathrm{Nm})\), which translates to \(\mathrm{ML}^2 \mathrm{~T}^{-2}\) in fundamental units.
Stress: Stress is force per unit area. The Sl unit for force is the Newton \((\mathrm{N})\) and for area is meter squared \(\left(\mathrm{m}^2\right)\). Therefore, the Sl unit for stress is Pascal (Pa), which translates to \(\mathrm{ML}^{-1} \mathrm{~T}^{-2}\) in fundamental units.
Pressure gradient: The pressure gradient is the rate of increase or decrease in pressure. It has units of pressure per distance. In SI units, that’s Pascal per meter \((\mathrm{Pa} / \mathrm{m})\), which translates to \(\mathrm{ML}^{-2} \mathrm{~T}^{-2}\) in fundamental units.
Coefficient of viscosity: This is a measure of a fluid’s resistance to shear or flow, and its Sl units are the Pascal second (Pa.s), which translates to \(\mathrm{ML}^{-1} \mathrm{~T}^{-1}\) in fundamental units.
Therefore, the correct matches are:
A – II, B – IV, C – I, D – III

Hint

(c) The density of a cylindrical wire is given by the formula:
\(
\rho=\frac{m}{V}=\frac{m}{\pi r^2 l}
\)
where \(m\) is the mass, \(r\) is the radius, and \(l\) is the length.
The relative error in a calculated quantity is the sum of the relative errors in the quantities it depends on. For the density, this is given by:
\(
\frac{\Delta \rho}{\rho}=\frac{\Delta m}{m}+2 \frac{\Delta r}{r}+\frac{\Delta l}{l}
\)
Given that \(\Delta m=0.01 \mathrm{~g}, m=0.4 \mathrm{~g}, \Delta r=0.03 \mathrm{~mm}, r=6 \mathrm{~mm}, \Delta l=0.04 \mathrm{~cm}\), and \(l=8 \mathrm{~cm}\), we can substitute these values into the formula to find the relative error in the density:
\(
\frac{\Delta \rho}{\rho}=\frac{0.01}{0.4}+2 \frac{0.03}{6}+\frac{0.04}{8}=0.025+0.01+0.005=0.04
\)
So the relative error in the density is 0.04 , or \(4 \%\).

Hint

(c) In the problem, we are given two resistances, \(R_1\) and \(R_2\), each with a certain measurement error, \(\Delta R_1\) and \(\Delta R_2\). These resistances are connected in parallel, and we are asked to find the percentage error in the equivalent resistance of this combination.
The formula for the equivalent resistance \(R\) of two resistors \(R_1\) and \(R_2\) in parallel is:
\(
\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}
\)
We want to find the percentage error in \(R\), which is given by \((\Delta R / R) \times 100 \%\).
In order to find \(\Delta R / R\), we differentiate both sides of the above equation with respect to \(R, R_1\), and \(R_2\). This gives us:
\(
\frac{\Delta R}{R^2}=\frac{\Delta R_1}{R_1^2}+\frac{\Delta R_2}{R_2^2}
\)
We can then solve this equation for \(\Delta R / R\) :
\(
\frac{\Delta R}{R}=\left(\frac{\Delta R_1}{R_1^2}+\frac{\Delta R_2}{R_2^2}\right) R
\)
Substituting the given values, \(R_1=10 \Omega, R_2=15 \Omega, \Delta R_1=\Delta R_2=0.5 \Omega\), and \(R=R_1 R_2 /\left(R_1+R_2\right)=6 \Omega\), we get:
\(
\frac{\Delta R}{R}=\left(\frac{0.5}{100}+\frac{0.5}{225}\right) \times 6=\frac{13}{300}
\)
Finally, to convert this to a percentage, we multiply by 100 , giving:
\(
\frac{\Delta R}{R} \times 100=\frac{13}{3}=4.33 \%
\)
This tells us that the percentage error in the equivalent resistance of the two resistances in parallel is \(4.33 \%\).

Hint

(c)

\(
\begin{aligned}
& {[\mathrm{M}]=[\mathrm{G}]^{\mathrm{x}}[\mathrm{~h}]^{\mathrm{y}}[\mathrm{c}]^{\mathrm{z}}} \\
& {[\mathrm{M}]=\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]^x\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]^y\left[\mathrm{LT}^{-1}\right]^z} \\
& {\left[\mathrm{M}^1 \mathrm{~L}^0 \mathrm{~T}^0\right]=\left[\mathrm{M}^{-x+y}\right]\left[\mathrm{L}^{3 x+2 y+z}\right]\left[\mathrm{T}^{-2 x-y-z}\right]} \\
& \mathrm{y}-\mathrm{x}=1 \ldots \ldots(1) \\
& 3 \mathrm{x}+2 \mathrm{y}+\mathrm{z}=0 \ldots \ldots(2) \\
& -2 \mathrm{x}-\mathrm{y}-\mathrm{z}=0 \ldots \ldots \ldots(3) \\
& \text { On solving, } \mathrm{x}=-\frac{1}{2}, \mathrm{y}=\frac{1}{2}, \mathrm{z}=\frac{1}{2} \\
& \text { So } \mathrm{m}=\sqrt{\frac{\mathrm{hc}}{\mathrm{G}}}
\end{aligned}
\)

Hint

(b) 

\(
\begin{aligned}
& \begin{array}{l}
\vec{L}=\vec{r} \times \vec{p} \Rightarrow[\mathrm{~L}]=\left[\mathrm{M}^0 \mathrm{~L}^1 \mathrm{~T}^0\right]\left[\mathrm{M}^1 \mathrm{~L}^1 \mathrm{~T}^{-1}\right] \\
=\left[\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-1}\right]
\end{array} \\
& \begin{aligned}
\vec{\tau}=\vec{r} \times \vec{F} \Rightarrow[\tau] & =\left[\mathrm{L}^1\right]\left[\mathrm{MLT}^{-2}\right] \\
& =\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]
\end{aligned} \\
& \text { Stress } \equiv \text { Pressure }=\frac{F}{A} \Rightarrow[\text { Stress }]=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right] \\
& \text { Pressure Gradient }=\frac{d P}{d x} \Rightarrow[\text { Pressure Gradient }]=\left[\mathrm{ML}^{-2} \mathrm{~T}^{-2}\right]
\end{aligned}
\)

Hint

(a) Torque (A): Defined as force times distance \((\tau=F \cdot r)\). The unit of force is Newtons (N) or \(\mathrm{kg} \mathrm{m} \mathrm{s}^{-2}\), so the unit of torque is N m or \(\mathrm{kg} \mathrm{m}^2 \mathrm{~s}^{-2}\).

Energy density (B): Defined as energy per unit volume. The unit of energy (Joule) is \(\mathrm{kg} \mathrm{m}^2 \mathrm{~s}^{-2}\). The unit of volume is \(\mathrm{m}^3\). So, the unit of energy density is \(\left(\mathrm{kg} \mathrm{m}^2 \mathrm{~s}^{-2}\right) / \mathrm{m}^3 =\mathrm{kg} \mathrm{m}^{-1} \mathrm{~s}^{-2}\).

Pressure gradient (C): Defined as the change in pressure per unit distance ( \(\boldsymbol{\Delta} \boldsymbol{P} \boldsymbol{/} \boldsymbol{\Delta} \boldsymbol{x}\) ). The unit of pressure (Pascal) is \(\mathrm{N} / \mathrm{m}^2\) or \(\mathrm{kg} \mathrm{m}^{-1} \mathrm{~s}^{-2}\). So, the unit of pressure gradient is \(\left(\mathrm{kg} \mathrm{m}^{-1} \mathrm{~s}^{-2}\right) / \mathrm{m}=\mathrm{kg} \mathrm{m}^{-2} \mathrm{~s}^{-2}\).

Impulse (D): Defined as force times time ( \(J=F \cdot \Delta t\) ) or change in momentum. The unit of force is \(\mathrm{kg} \mathrm{m} \mathrm{s}^{-2}\), so the unit of impulse is \(\mathrm{kg} \mathrm{m} \mathrm{s}^{-2} \cdot \mathrm{~s}=\mathrm{kg} \mathrm{m} \mathrm{s}^{-1}\). This is the same as the unit of momentum (mass ⋅ velocity \(=\mathrm{kg} \cdot \mathrm{m} / \mathrm{s}\) ).
Therefore, the correct matching sequence is A-IV, B-I, C-III, D-II.

Hint

(c) Here, \(At\) is distance, so dimensions of
\(
[A t]=[x]=[L]
\)
Given. The dimensions of t is \(\left[\mathrm{T}^{-1}\right]\)
\(
\begin{aligned}
& {\left[A \times \mathrm{T}^{-1}\right]=[\mathrm{L}] \Rightarrow[A]=[\mathrm{LT}]} \\
& {\left[\frac{t}{B}\right]=[y]=[L]} \\
& \Rightarrow \frac{\mathrm{T}^{-1}}{[B]}=[L] \Rightarrow B=\left[\mathrm{L}^{-1} \mathrm{~T}^{-1}\right]
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
& \text { Pressure gradient }=\frac{d p}{d x}=\frac{\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]}{[\mathrm{L}]} \\
& =\left[\mathrm{M}^1 \mathrm{~L}^{-2} \mathrm{~T}^{-2}\right] \\
& \text { Energy density }=\frac{\text { energy }}{\text { volume }}=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]}{\left[\mathrm{L}^3\right]} \\
& =\left[\mathrm{M}^1 \mathrm{~L}^{-1} \mathrm{~T}^{-2}\right] \\
& \text { Electric field }=\frac{\text { Force }}{\text { ch arge }}=\frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{A.T}]} \\
& =\left[\mathrm{M}^1 \mathrm{~L}^1 \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right] \\
& \text { Latent heat }=\frac{\text { heat }}{\text { mass }}=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]}{[\mathrm{M}]} \\
& =\left[\mathrm{M}^0 \mathrm{~L}^2 \mathrm{~T}^{-2}\right]
\end{aligned}
\)

Hint

(d)

\(
\begin{aligned}
&\begin{aligned}
& \mathrm{Y}=\frac{\text { Stress }}{\text { Strain }}=\frac{\mathrm{F} / \mathrm{A}}{\Delta \ell / \ell}=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{L}^2\right]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right] \\
& \mathrm{F}=6 \pi \eta \mathrm{rv} \Rightarrow \eta=\frac{\mathrm{F}}{6 \pi \mathrm{rv}} \\
& {[\eta]=\frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{L}]\left[\mathrm{LT}^{-1}\right]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]} \\
& \mathrm{E}=\mathrm{h} v \Rightarrow \mathrm{~h}=\frac{\mathrm{E}}{v}=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]}{\left[\mathrm{T}^{-1}\right]}=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]
\end{aligned}\\
&\text { Work function has same dimension as that of energy, so }[\phi]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]
\end{aligned}
\)

Hint

(b) (A) Planck’s constant
\(
\begin{aligned}
& \mathrm{h} v=\mathrm{E} \\
& \mathrm{~h}=\frac{\mathrm{E}}{v}=\frac{\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-2}}{\mathrm{~T}^{-1}}=\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-1} (III)
\end{aligned}
\)
(B) \(\mathrm{E}=\mathrm{qV}\)
\(
V=\frac{E}{q}=\frac{M^1 L^2 T^{-2}}{A^1 T^1}=M^1 L^2 T^{-3} A^{-1} (IV)
\)
(C) \(\phi(\) work function \()=\) energy
\(
=\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-2} (I)
\)
(D) Momentum \((p) = F.t\)
\(
\begin{aligned}
& =\mathrm{M}^1 \mathrm{~L}^1 \mathrm{~T}^{-2} \mathrm{~T}^1 \\
& =\mathrm{M}^1 \mathrm{~L}^1 \mathrm{~T}^{-1} \quad (II)
\end{aligned}
\) 

Hint

(b) Torque → Nm
Stress \(\rightarrow N / \mathrm{m}^2\)
Latent heat → J/kg
Power → Nm/s