JEE Units & Measurements Questions

Terminal velocity of a sphere: $$v_0 = \frac{2r^2(\sigma - \rho)g}{9\eta}$$, so $$\eta = \frac{2r^2(\sigma - \rho)g}{9v_0}$$.

Ignoring errors in $$\sigma, \rho, g$$:

$$\frac{\Delta\eta}{\eta} = 2\frac{\Delta r}{r} + \frac{\Delta v_0}{v_0}$$

(Since $$\eta \propto r^2/v_0$$, the maximum relative error is $$2\frac{\Delta r}{r} + \frac{\Delta v_0}{v_0}$$.)

The answer is Option 1: $$\frac{2\Delta r}{r} + \frac{\Delta v_0}{v_0}$$.

Measured value using Vernier Calipers is given by $$\text{Total Reading} = \text{MSR} + (\text{VSR} \times \text{LC}) - \text{Zero Error}$$.

From Fig. 1 (Zero Error and Least Count):

$$\text{Zero Error} = 0$$

$$1\text{ MSD} = \frac{1\text{ cm}}{10} = 0.1\text{ cm}$$

$$10\text{ VSD} = 9\text{ MSD} \implies 1\text{ VSD} = 0.09\text{ cm}$$

$$\text{LC} = 1\text{ MSD} - 1\text{ VSD} = 0.1 - 0.09 = 0.01\text{ cm}$$

From Fig. 2 (Measurement):

$$\text{MSR} = 0.1\text{ cm}$$

$$\text{VSR} = 3$$

$$\text{Value} = 0.1 + (3 \times 0.01) - 0 = 0.13\text{ cm}$$

Angular impulse:

Angular impulse = change in angular momentum

$$L=r\times p=r\cdot mv$$

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

So, (A) matches with (III)

Latent heat:

Latent heat = heat per unit mass

$$L=\frac{Q}{m}$$

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

$$[L]=\frac{ML^2T^{-2}}{M}=[L^2T^{-2}]$$

So, (B) matches with (I)

Electrical resistivity:

$$R=\rho\frac{L}{A}\Rightarrow\rho=R\frac{A}{L}$$

Now,

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

$$[V]=[ML^2T^{-3}A^{-1}]$$

$$[R]=\frac{ML^2T^{-3}A^{-1}}{A}=[ML^2T^{-3}A^{-2}]$$

So,

$$[\rho]=[ML^2T^{-3}A^{-2}]\cdot\frac{L^2}{L}=[ML^3T^{-3}A^{-2}]$$

So, (C) matches with (IV)

Electromotive force (EMF):

EMF has same dimensions as potential difference

$$[EMF]=[ML^2T^{-3}A^{-1}]$$

So, (D) matches with (II)

Final matching:

(A)−(III), (B)−(I), (C)−(IV), (D)−(II)

Solution :

Dimensions of momentum :

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

Given :

Charge,

$$[q] = AT$$

Current,

$$[I] = A$$

Permeability of vacuum :

From force per unit length between two current carrying wires,

$$\frac{\mu_0 I^2}{r}$$

has dimensions of force.

Therefore,

$$[\mu_0] = \frac{[F][r]}{[I]^2}$$

$$= \frac{(MLT^{-2})(L)}{A^2}$$

$$= MLT^{-2}A^{-2}$$

Now consider quantity :

$$\mu_0 qI$$

Its dimensions are :

$$[\mu_0 qI] = (MLT^{-2}A^{-2})(AT)(A)$$

$$= MLT^{-1}$$

which is the dimension of momentum.

Final Answer :

$$\mu_0 qI$$

Young’s modulus:

$$Y=\frac{\text{stress}}{\text{strain}}$$

Strain is dimensionless, so

$$[Y]=[\text{stress}]=\frac{\text{Force}}{\text{Area}}=\frac{MLT^{-2}}{L^2}=[ML^{-1}T^{-2}]$$

So, (A) matches with (II)

Torque:

$$\tau=r\times F$$

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

So, (B) matches with (IV)

Coefficient of viscosity:

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

So, (C) matches with (I)

Gravitational constant:

From Newton’s law:

$$F=\frac{GMm}{r^2}\Rightarrow G=\frac{Fr^2}{Mm}$$

$$[G]=\frac{MLT^{-2}\cdot L^2}{M^2}=[M^{-1}L^3T^{-2}]$$

So, (D) matches with (III)

Final matching:

(A)−(II), (B)−(IV), (C)−(I), (D)−(III)

The four physical quantities are gravitational constant $$G$$, gravitational potential energy $$U$$, gravitational potential $$V$$ and acceleration due to gravity $$g$$. We first write the defining equations for each quantity and then extract their dimensional formulas.

Case A : Gravitational constant  $$G$$
Newton’s law gives the gravitational force between two masses $$m_1$$ and $$m_2$$ separated by distance $$r$$ as
$$F = \dfrac{G\,m_1\,m_2}{r^{2}}$$
Re-arranging,
$$G = \dfrac{F\,r^{2}}{m_1\,m_2}$$
Force has dimensions $$[MLT^{-2}]$$. Hence
$$[G] = \dfrac{[MLT^{-2}]\,L^{2}}{M\,M} = [M^{-1}L^{3}T^{-2}]$$

Case B : Gravitational potential energy  $$U$$
Potential energy near Earth’s surface is $$U = mgh$$.
Mass $$m$$ has $$[M]$$, $$g$$ has $$[LT^{-2}]$$ and height $$h$$ has $$[L]$$. Therefore
$$[U] = [M]\,[LT^{-2}]\,[L] = [ML^{2}T^{-2}]$$

Case C : Gravitational potential  $$V$$
Potential is energy per unit mass: $$V = \dfrac{U}{m}$$.
Divide the dimensions of $$U$$ obtained above by $$[M]$$:
$$[V] = \dfrac{[ML^{2}T^{-2}]}{[M]} = [L^{2}T^{-2}]$$

Case D : Acceleration due to gravity  $$g$$
Acceleration is rate of change of velocity: $$g = \dfrac{dv}{dt}$$.
Velocity $$v$$ has $$[LT^{-1}]$$, so acceleration has
$$[g] = \dfrac{[LT^{-1}]}{[T]} = [LT^{-2}]$$

Now compare with LIST-II:
I. $$[LT^{-2}]$$  II. $$[L^{2}T^{-2}]$$  III. $$[ML^{2}T^{-2}]$$  IV. $$[M^{-1}L^{3}T^{-2}]$$

Matching each quantity with its dimensional formula:
A. $$G$$ ⟶ IV  B. $$U$$ ⟶ III  C. $$V$$ ⟶ II  D. $$g$$ ⟶ I

Thus the correct set is A-IV, B-III, C-II, D-I, which corresponds to Option A.

The quantity is $$Q = X^{-2}\,Y^{-\,\frac{3}{2}}\,Z^{\frac{2}{5}}$$.

For a physical quantity of the form $$Q = X^{a}\,Y^{b}\,Z^{c}$$, the maximum fractional (relative) error is obtained by adding the absolute contributions of each factor:
$$\frac{\Delta Q}{Q}\Big|_{\max}=|a|\;\frac{\Delta X}{X}+|b|\;\frac{\Delta Y}{Y}+|c|\;\frac{\Delta Z}{Z}$$.

Here $$a=-2,\;b=-\frac{3}{2},\;c=\frac{2}{5}$$.
Their absolute values are $$|a|=2,\;|b|=\frac{3}{2}=1.5,\;|c|=\frac{2}{5}=0.4$$.

The given fractional errors are
$$\frac{\Delta X}{X}=0.1,\qquad \frac{\Delta Y}{Y}=0.2,\qquad \frac{\Delta Z}{Z}=0.5$$.

Substituting these values:
$$\frac{\Delta Q}{Q}\Big|_{\max}=2(0.1)+1.5(0.2)+0.4(0.5)$$

Calculate each term:
$$2(0.1)=0.2,\qquad 1.5(0.2)=0.3,\qquad 0.4(0.5)=0.2$$

Add them:
$$0.2+0.3+0.2 = 0.7$$

Therefore, the maximum fractional error in $$Q$$ is $$0.7$$.

Option C is correct.

Let us analyze each statement about the Moving Coil Galvanometer (MCG):

Statement A: The torsional constant has dimensions $$[ML^{2}T^{-2}]$$.

The restoring torque is $$\tau = k\theta$$, where $$k$$ is the torsional constant and $$\theta$$ is dimensionless (in radians).

So $$[k] = [\tau] = [ML^{2}T^{-2}]$$. Statement A is correct.

Statement B: Increasing the current sensitivity may not necessarily increase the voltage sensitivity.

Current sensitivity: $$I_s = \frac{NAB}{k}$$

Voltage sensitivity: $$V_s = \frac{I_s}{R} = \frac{NAB}{kR}$$

If we increase $$N$$ to increase current sensitivity, the resistance $$R$$ also increases (since $$R \propto N$$). So voltage sensitivity may not increase. Statement B is correct.

Statement C: If $$N$$ is doubled to $$2N$$, voltage sensitivity doubles.

When $$N$$ is doubled, $$R$$ also roughly doubles. So $$V_s = \frac{NAB}{kR}$$. Since both $$N$$ and $$R$$ double, $$V_s$$ remains approximately the same. Statement C is incorrect.

Statement D: MCG can be converted into an ammeter by introducing a shunt resistance of large value in parallel.

To convert a galvanometer into an ammeter, a small (low) resistance shunt is connected in parallel, not a large resistance. Statement D is incorrect.

Statement E: Current sensitivity depends inversely on number of turns.

Current sensitivity $$I_s = \frac{NAB}{k}$$, which is directly proportional to $$N$$, not inversely. Statement E is incorrect.

Therefore, only statements A and B are correct.

The correct answer is Option 4: A, B Only.

The dimensional formula of a physical quantity is obtained from the defining equation of that quantity in terms of the fundamental mechanical quantities $$M$$ (mass), $$L$$ (length) and $$T$$ (time).

Case (A) : Mass density

Mass density $$\rho$$ is mass per unit volume:
$$\rho = \frac{\text{mass}}{\text{volume}}$$.
Mass has dimension $$[M]$$ and volume has dimension $$[L^3]$$, so

$$[\rho] = \frac{[M]}{[L^3]} = [M L^{-3} T^{0}]$$.
Thus (A) corresponds to (IV).

Case (B) : Impulse

Impulse $$J$$ is defined as force multiplied by the time interval:
$$J = F \, \Delta t$$.
Force has dimension $$[M L T^{-2}]$$ and time has dimension $$[T]$$, so

$$[J] = [M L T^{-2}] \,[T] = [M L T^{-1}]$$.
Thus (B) corresponds to (II).

Case (C) : Power

Power $$P$$ is work done per unit time:
    $$P = \frac{W}{t}$$.
Work (or energy) has dimension $$[M L^{2} T^{-2}]$$ and time has dimension $$[T]$$, hence

$$[P] = \frac{[M L^{2} T^{-2}]}{[T]} = [M L^{2} T^{-3}]$$.
Thus (C) corresponds to (I).

Case (D) : Moment of inertia

For a point mass $$m$$ at a distance $$r$$ from the axis, the moment of inertia $$I$$ is:
    $$I = m r^{2}$$.
Mass has dimension $$[M]$$ and distance squared has dimension $$[L^{2}]$$, so

$$[I] = [M] [L^{2}] = [M L^{2} T^{0}]$$.
Thus (D) corresponds to (III).

Collecting the matches:

(A) → (IV), (B) → (II), (C) → ( I ), (D) → (III).

This set of matches is given in Option C. Hence the correct answer is Option C.

Find the dimensions of $$B/\mu_0$$, where $$B$$ is the magnetic field and $$\mu_0$$ is the permeability of free space.

From the Lorentz force equation $$F = qvB$$, we can write $$B = \frac{F}{qv}$$.

$$[B] = \frac{[F]}{[q][v]} = \frac{MLT^{-2}}{(AT)(LT^{-1})} = \frac{MLT^{-2}}{ALT^{-1} \cdot T} = \frac{M}{AT^2} = MT^{-2}A^{-1}$$

From the relation $$B = \mu_0 nI$$ (magnetic field inside a solenoid), where $$n$$ is the number of turns per unit length and $$I$$ is the current:

$$[\mu_0] = \frac{[B]}{[n][I]} = \frac{MT^{-2}A^{-1}}{L^{-1} \cdot A} = MLT^{-2}A^{-2}$$

$$\left[\frac{B}{\mu_0}\right] = \frac{MT^{-2}A^{-1}}{MLT^{-2}A^{-2}} = \frac{M}{M} \cdot \frac{T^{-2}}{T^{-2}} \cdot \frac{A^{-1}}{A^{-2}} \cdot \frac{1}{L} = L^{-1}A$$

Physical significance: $$B/\mu_0$$ has the same dimensions as magnetic field intensity $$H$$, which has dimensions of $$L^{-1}A$$ (equivalent to A/m in SI units).

The correct answer is Option (3): $$L^{-1}A$$.

The main scale of the travelling microscope has $$300$$ equal divisions spread over a length of $$15 \text{ cm}$$.

Length of one main scale division (MSD) is therefore
$$\text{MSD} = \frac{15 \text{ cm}}{300} = 0.05 \text{ cm}$$

The vernier scale has $$25$$ divisions that coincide with $$24$$ divisions of the main scale.
Thus, length of the entire vernier scale is
$$25 \text{ VSD} = 24 \text{ MSD}$$

Using $$\text{MSD} = 0.05 \text{ cm}$$,
$$25 \text{ VSD} = 24 \times 0.05 \text{ cm} = 1.2 \text{ cm}$$

Length of one vernier scale division (VSD) is then
$$\text{VSD} = \frac{1.2 \text{ cm}}{25} = 0.048 \text{ cm}$$

Formula for least count (LC) of a vernier instrument:
$$\text{LC} = \text{MSD} - \text{VSD}$$

Substituting the values,
$$\text{LC} = 0.05 \text{ cm} - 0.048 \text{ cm} = 0.002 \text{ cm}$$

Hence, the least count of the travelling microscope is $$0.002 \text{ cm}$$.

Option B is correct.

We need to evaluate two statements about vernier callipers.

Statement I: "In a vernier callipers, one vernier scale division is always smaller than one main scale division."

In a standard vernier callipers, n vernier scale divisions (VSD) coincide with (n-1) main scale divisions (MSD). So:

$$n \times VSD = (n-1) \times MSD$$

$$VSD = \frac{(n-1)}{n} \times MSD$$

This means one VSD is smaller than one MSD. However, the statement says "always." In some vernier callipers, n VSD = (n+1) MSD, which would make VSD larger than MSD. So the word "always" makes Statement I false, as there exist vernier callipers where VSD > MSD.

Statement II: "The vernier constant is given by one main scale division multiplied by the number of vernier scale divisions."

The vernier constant (least count) is given by:

$$VC = 1 \text{ MSD} - 1 \text{ VSD} = MSD - \frac{(n-1)}{n} MSD = \frac{MSD}{n}$$

This equals one MSD divided by (not multiplied by) the number of vernier scale divisions. So Statement II is also false.

Both statements are false.

The correct answer is Option 3: Both Statement I and Statement II are false.

We need to find the dimensions of the product ABC, where $$v = At^2 + \frac{Bt}{C+t}$$.

Since $$C + t$$ must be dimensionally consistent, $$[C] = [t] = T$$.

From the first term $$At^2$$: $$[A][T^2] = [v] = LT^{-1}$$, so $$[A] = LT^{-3}$$.

From the second term $$\frac{Bt}{C+t}$$: for large $$t$$, this approaches $$B$$, so $$[B] = [v] = LT^{-1}$$.

(More rigorously: $$\frac{[B][T]}{[T]} = LT^{-1}$$, so $$[B] = LT^{-1}$$.)

Calculate $$[ABC]$$:

$$[ABC] = [A][B][C] = (LT^{-3})(LT^{-1})(T) = L^2T^{-3} = M^0L^2T^{-3}$$

The correct answer is Option (4): $$[M^0L^2T^{-3}]$$.

The dimensional formula for capacitance in farads (F) can be derived using the definition of capacitance. Capacitance $$C$$ is given by the formula:

$$ C = \frac{Q}{V} $$

where $$Q$$ is the charge and $$V$$ is the potential difference.

Charge $$Q$$ has the dimension $$[C]$$, so:

$$ [Q] = [C] $$

Potential difference $$V$$ is defined as work done per unit charge:

$$ V = \frac{W}{Q} $$

where $$W$$ is work done. Work done $$W$$ is force multiplied by displacement:

$$ W = F \cdot d $$

Force $$F$$ is mass times acceleration:

$$ F = m \cdot a $$

Acceleration $$a$$ has dimension $$[LT^{-2}]$$, so force $$F$$ has dimension:

$$ [F] = [M] \cdot [LT^{-2}] = [MLT^{-2}] $$

Displacement $$d$$ has dimension $$[L]$$, so work done $$W$$ has dimension:

$$ [W] = [F] \cdot [d] = [MLT^{-2}] \cdot [L] = [ML^2T^{-2}] $$

Therefore, potential difference $$V$$ has dimension:

$$ [V] = \frac{[W]}{[Q]} = \frac{[ML^2T^{-2}]}{[C]} = [ML^2T^{-2}C^{-1}] $$

Now, capacitance $$C$$ is:

$$ [C] = \frac{[Q]}{[V]} = \frac{[C]}{[ML^2T^{-2}C^{-1}]} $$

Simplifying the expression:

$$ [C] = [C] \times \frac{1}{[ML^2T^{-2}C^{-1}]} = [C] \times [M^{-1}L^{-2}T^{2}C] = [C^{1} \cdot C^{1}] \times [M^{-1}L^{-2}T^{2}] = [C^{2}M^{-1}L^{-2}T^{2}] $$

Thus, the dimensional formula for capacitance is $$[C^{2}M^{-1}L^{-2}T^{2}]$$.

Comparing with the options:

• A. $$[F]=[C^{2}M^{-1}L^{-2}T^{2}]$$
• B. $$[F]=[C^{2}M^{-2}L^{2}T^{2}]$$
• C. $$[F]=[CM^{-2}L^{-2}T^{-2}]$$
• D. $$[F]=[CM^{-1}L^{-2}T^{2}]$$

Option A matches the derived dimensional formula.

Therefore, the correct answer is A.

The dimensional analysis will be carried out for each physical quantity in LIST-I and then compared with the four alternatives given in LIST-II.

The average translational energy of one molecule in an ideal gas is $$E = \tfrac{3}{2}\,k_B\,T$$.
Hence $$k_B = \dfrac{E}{T}$$.
Energy has dimension $$ML^{2}T^{-2}$$ and temperature contributes the factor $$K$$ in the denominator.
Therefore $$[k_B] = \dfrac{ML^{2}T^{-2}}{K} = ML^{2}T^{-2}K^{-1}$$, which corresponds to LIST-II entry III.

Newton’s law of viscous flow states $$\tau = \eta \,\dfrac{dv}{dy}$$, where $$\tau$$ is shear stress.
Shear stress $$\tau = \dfrac{\text{Force}}{\text{Area}}$$ has dimension $$\dfrac{MLT^{-2}}{L^{2}} = ML^{-1}T^{-2}$$.
Velocity gradient $$\dfrac{dv}{dy}$$ has dimension $$T^{-1}$$.
Thus $$[\eta] = \dfrac{ML^{-1}T^{-2}}{T^{-1}} = ML^{-1}T^{-1}$$, matching LIST-II entry IV.

The photon energy is $$E = h\nu$$, and frequency $$\nu$$ has dimension $$T^{-1}$$.
Hence $$h = \dfrac{E}{\nu}$$.
Taking $$[E] = ML^{2}T^{-2}$$ and dividing by $$T^{-1}$$ gives $$[h] = ML^{2}T^{-1}$$, i.e. LIST-II entry I.

Fourier’s law of heat conduction is $$\dfrac{Q}{t} = -K\,A\,\dfrac{\Delta T}{L}$$.
Heat conducted per unit time $$\dfrac{Q}{t}$$ is power with dimension $$ML^{2}T^{-3}$$.
Area $$A$$ contributes $$L^{2}$$ and temperature gradient $$\dfrac{\Delta T}{L}$$ contributes $$K\,L^{-1}$$ in the denominator.
Therefore

$$[K] = \dfrac{ML^{2}T^{-3}}{L^{2}\,K\,L^{-1}} = \dfrac{ML^{2}T^{-3}}{L^{1}K} = MLT^{-3}K^{-1}$$, corresponding to LIST-II entry II.

Collecting the results:
A → III, B → IV, C → I, D → II

This is exactly the pairing given in Option A.

Hence, the correct answer is Option A.

We need to find the pair of physical quantities that do NOT have the same dimensions.

Check each option:

(1) Pressure and Young's modulus: Both have dimensions $$[ML^{-1}T^{-2}]$$. Same. ✓

(2) Surface tension and Impulse:

Surface tension = Force/Length = $$[MLT^{-2}]/[L] = [MT^{-2}]$$
Impulse = Force × Time = $$[MLT^{-2}][T] = [MLT^{-1}]$$
These are different. ✗

(3) Torque and Energy: Both have dimensions $$[ML^2T^{-2}]$$. Same. ✓

(4) Angular momentum and Planck's constant: Both have dimensions $$[ML^2T^{-1}]$$. Same. ✓

The correct answer is Option (2): Surface tension and impulse.

We first write one standard defining relation for each physical quantity and then deduce its dimensional formula.

Case A: Coefficient of viscosity ($$\eta$$)
Shear stress $$\tau = \eta \dfrac{dv}{dy}\quad -(1)$$
  • Dimension of shear stress $$\tau$$ is the same as pressure: $$[M\,L^{-1}\,T^{-2}]$$.
  • Velocity gradient $$\dfrac{dv}{dy}$$ has dimension $$T^{-1}$$.
Therefore,
$$\eta = \dfrac{\tau}{dv/dy} \; \Rightarrow \; [M\,L^{-1}\,T^{-2}] \; / \; [T^{-1}] = [M\,L^{-1}\,T^{-1}]$$.
Hence $$\eta$$ corresponds to List-II (IV).

Case B: Intensity of wave ($$I$$)
Intensity $$I = \dfrac{\text{Power}}{\text{Area}} \quad -(2)$$
  • Power $$P = \dfrac{\text{Energy}}{\text{time}}$$. Energy has dimension $$[M\,L^{2}\,T^{-2}]$$, so
$$P = [M\,L^{2}\,T^{-3}]$$.
  • Area has dimension $$[L^{2}]$$.
Therefore,
$$I = \dfrac{[M\,L^{2}\,T^{-3}]}{[L^{2}]} = [M\,L^{0}\,T^{-3}]$$.
Hence $$I$$ corresponds to List-II (I).

Case C: Pressure gradient ($$\nabla P$$)
Pressure gradient means pressure per unit length:
$$\nabla P = \dfrac{P}{L} \quad -(3)$$
  • Pressure $$P$$ has dimension $$[M\,L^{-1}\,T^{-2}]$$.
  • Length $$L$$ has dimension $$[L]$$.
Therefore,
$$\nabla P = \dfrac{[M\,L^{-1}\,T^{-2}]}{[L]} = [M\,L^{-2}\,T^{-2}]$$.
Hence $$\nabla P$$ corresponds to List-II (II).

Case D: Compressibility ($$\beta$$)
Compressibility is the reciprocal of bulk modulus (pressure):
$$\beta = \dfrac{1}{P} \quad -(4)$$
Since the dimension of pressure is $$[M\,L^{-1}\,T^{-2}]$$, we get
$$\beta = [M^{-1}\,L\,T^{2}]$$.
Hence $$\beta$$ corresponds to List-II (III).

Final matching
(A) $$\eta$$  →  (IV) [ML$$^{-1}$$T$$^{-1}$$]
(B) $$I$$  →  (I) [ML$$^{0}$$T$$^{-3}$$]
(C) $$\nabla P$$  →  (II) [ML$$^{-2}$$T$$^{-2}$$]
(D) $$\beta$$  →  (III) [M$$^{-1}$$LT$$^{2}$$]

Thus the correct set is: (A)-(IV), (B)-(I), (C)-(II), (D)-(III), which is given in Option B.

Answer : Option B

Magnetic induction B:

SI unit is tesla, and in CGS it is gauss.

So, (A) matches with (III)

Magnetic intensity H:

$$H=\frac{B}{\mu}$$

Unit:

$$A/m$$

So, (B) matches with (IV)

Magnetic flux $$\Phi$$:

$$\Phi=B\cdot A$$

Unit:

$$\text{weber}$$

So, (C) matches with (II)

Magnetic moment:

$$\mu=I\cdot A$$

Unit:

$$\text{Ampere}\cdot\text{meter}^2$$

So, (D) matches with (I)

Final matching:

(A)−(III), (B)−(IV), (C)−(II), (D)−(I)

Given $$E(t) = \alpha^3 e^{-\beta t}$$ with $$\beta = 0.3$$ s⁻¹ and relative errors $$\Delta\alpha/\alpha = 1.2\%$$, $$\Delta t/t = 1.6\%$$, we want to find the maximum percentage error in $$E$$ at $$t = 5\,$$s}.

We begin by taking the natural logarithm of $$E$$ to prepare for error propagation:

$$\ln E = 3\ln\alpha - \beta t$$

By differentiating this relation, the relative error in $$E$$ becomes

$$\frac{\Delta E}{E} = 3\frac{\Delta\alpha}{\alpha} + \beta \Delta t$$

Since $$\Delta t/t = 1.6\%$$ at $$t = 5\,$$s}, we calculate

$$\Delta t = 0.016 \times 5 = 0.08\ \text{s}$$

Substituting $$\Delta\alpha/\alpha = 1.2\%$$ and the computed $$\Delta t$$ into the expression for $$\Delta E/E$$ and converting to a percentage gives

$$\frac{\Delta E}{E} \times 100 = 3 \times 1.2\% + \beta \times \Delta t \times 100$$

$$= 3.6\% + 0.3 \times 0.08 \times 100$$

$$= 3.6\% + 2.4\%$$

$$= 6\%$$

Therefore, the maximum percentage error in $$E$$ at $$t=5\,$$s} is 6\%, which corresponds to Option 1.

Permeability of free space:

Using relation:

$$B=\mu_0\frac{I}{L}$$

$$\mu_0=\frac{BL}{I}$$

Substitute dimensions:

$$[B]=[MT^{-2}A^{-1}]$$

$$[\mu_0]=\frac{(MT^{-2}A^{-1})\cdot L}{A}=[MLT^{-2}A^{-2}]$$

So, Permeability → (III)

Magnetic field:

From Lorentz force,

$$F=qvB$$

$$B=\frac{F}{qv}$$

Now substitute dimensions:

$$[q]=[AT]$$

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

$$[B]=\frac{MLT^{-2}}{(AT)(LT^{-1})}=[MT^{-2}A^{-1}]$$

So, Magnetic field → (II)

Magnetic moment:

Magnetic moment is given by:

$$μ=IA$$

$$[\mu]=[A]\cdot[L^2]=[L^2A]$$

So, Magnetic moment → (IV)

Torsional constant (torsion constant) is defined from the relation:

$$C\theta\ =\tau$$

where $$τ=torque$$ and $$\theta$$ = angular displacement (dimensionless).

So,

$$C=\frac{\tau}{\theta}$$

Since θ\thetaθ is dimensionless:

$$[C]=[\tau]$$

Now, torque:

$$\tau=r\times F$$

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

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

So Torsional Constant →(I)

Final matching:

(A)−(III), (B)−(II), (C)−(IV), (D)−(I)

We need to find the dimension of the ratio $$\frac{\text{Modulus of Elasticity}}{\text{Torque}}$$ and determine the value of $$c$$ given that $$b = -3$$.

To begin, we write the dimension of Modulus of Elasticity ($$E$$). Modulus of elasticity (Young's modulus) is defined as stress/strain. Since strain is dimensionless:

$$ [E] = [\text{Stress}] = \frac{[\text{Force}]}{[\text{Area}]} = \frac{[M L T^{-2}]}{[L^2]} = [M L^{-1} T^{-2}] $$

Next, we write the dimension of Torque ($$\tau$$). Torque is given by Force $$\times$$ distance (lever arm):

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

We then compute the dimension of $$E/\tau$$:

$$ \left[\frac{E}{\tau}\right] = \frac{[M L^{-1} T^{-2}]}{[M L^2 T^{-2}]} $$

Dividing each base dimension:

$$ M: M^1 / M^1 = M^0 $$

$$ L: L^{-1} / L^2 = L^{-1-2} = L^{-3} $$

$$ T: T^{-2} / T^{-2} = T^0 $$

Therefore,

$$ \left[\frac{E}{\tau}\right] = [M^0 L^{-3} T^0] $$

Comparing this result with $$[M^a L^b T^c]$$ shows that $$a = 0$$, $$b = -3$$, and $$c = 0$$. This confirms the given value $$b = -3$$ and yields the value of $$c = 0$$.

The answer is 0.

A rectangular metallic sheet has length $$L = 5$$ mm and breadth $$B = 2.5$$ mm, and the screw gauge used has pitch = 0.75 mm with 15 divisions on the circular scale.

We first find the least count (LC) of the screw gauge.

$$ LC = \frac{\text{Pitch}}{\text{Number of divisions}} = \frac{0.75}{15} = 0.05 \text{ mm} $$

Next, we calculate the area of the sheet.

$$ A = L \times B = 5 \times 2.5 = 12.5 \text{ mm}^2 $$

We then determine the maximum fractional error in this area.

$$ \frac{\Delta A}{A} = \frac{\Delta L}{L} + \frac{\Delta B}{B} $$

Here, we take $$\Delta L = \Delta B = LC = 0.05$$ mm (since the maximum absolute error equals the least count).

$$ \frac{\Delta A}{A} = \frac{0.05}{5} + \frac{0.05}{2.5} = 0.01 + 0.02 = 0.03 = \frac{3}{100} $$

So $$\frac{x}{100} = \frac{3}{100}$$, giving $$x = 3$$.

Hence the answer is 3.

Screw gauge: LC = 0.01 mm. Pitch increased by 75%, divisions reduced by 50%. Find new LC.

LC = pitch/divisions. If pitch becomes 1.75p and divisions become 0.5d:

New LC = 1.75p/(0.5d) = 3.5(p/d) = 3.5 × 0.01 = 0.035 mm

In units of 0.001 mm: 35

The answer is 35.

For $$Q = \frac{ab^4}{cd}$$, the percentage error is given by

$$ \frac{\Delta Q}{Q} \times 100 = \left(\frac{\Delta a}{a} + 4\frac{\Delta b}{b} + \frac{\Delta c}{c} + \frac{\Delta d}{d}\right) \times 100 $$

Using the provided data, we calculate each relative error:

$$ \frac{\Delta a}{a} = \frac{3}{60} = 0.05 $$

$$ \frac{\Delta b}{b} = \frac{0.1}{20} = 0.005 $$

$$ \frac{\Delta c}{c} = \frac{0.2}{40} = 0.005 $$

$$ \frac{\Delta d}{d} = \frac{0.1}{50} = 0.002 $$

Substituting these results into the percentage error formula yields

$$ = (0.05 + 4 \times 0.005 + 0.005 + 0.002) \times 100 $$

$$ = (0.05 + 0.02 + 0.005 + 0.002) \times 100 $$

$$ = 0.077 \times 100 = 7.7\% $$

Since the percentage error can be expressed in the form $$\frac{x}{1000}$$, we set

$$ 7.7 = \frac{x}{1000} $$

$$ x = 7700 $$

The answer is 7700.

Boat speed downstream = 27+9 = 36 km/h = 10 m/s. Ball thrown at 10 m/s upward.

Time of flight = 2u/g = 2×10/10 = 2s. Range = horizontal speed × time = 10×2 = 20m = 2000 cm.

The answer is 2000.

Use dimensional analysis.

Assume the relation:

$$T\propto G^aM^br^c$$

Write dimensions:

$$[T]=T$$

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

$$[M]=M$$

$$[r]=L$$

Substitute:

$$T=(M^{-1}L^3T^{-2})^a\cdot M^b\cdot L^c$$

$$=M^{-a+b}\cdot L^{3a+c}\cdot T^{-2a}$$

Now compare powers with LHS$$[T^1]$$:

For mass:

$$-a+b=0$$

For length:

$$3a+c=0$$

For time:

$$-2a=1\Rightarrow a=-\frac{1}{2}$$

Substitute aaa:

$$b=a=-\frac{1}{2}$$

$$3a+c=0\Rightarrow c=\frac{3}{2}$$

Final relation:

$$T\propto G^{-1/2}M^{-1/2}r^{3/2}$$

$$T\propto\sqrt{\frac{r^3}{GM}}$$

Correct expression:

$$T=k\sqrt{\frac{r^3}{GM}}$$

$$T^2 = \frac{k r^3}{GM}$$

We need to find the dimensions of $$\epsilon_0 E^2$$, where $$\epsilon_0$$ is the permittivity of free space and $$E$$ is the electric field.

The electrostatic energy density stored in an electric field is $$u = \frac{1}{2}\epsilon_0 E^2$$. Since $$\frac{1}{2}$$ is dimensionless, $$\epsilon_0 E^2$$ has the same dimensions as energy density.

Energy density is defined as energy per unit volume, so

$$[u] = \frac{[E]}{[V]} = \frac{ML^2T^{-2}}{L^3} = ML^{-1}T^{-2}$$

Alternatively, we can verify this result by considering the dimensions of $$\epsilon_0$$ and $$E$$ separately. From Coulomb’s law, $$F = \frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r^2}$$, so

$$\epsilon_0 = \frac{q^2}{Fr^2} \implies [\epsilon_0] = \frac{A^2T^2}{MLT^{-2}\cdot L^2} = M^{-1}L^{-3}T^4A^2$$

The electric field has dimensions

$$[E] = \frac{[F]}{[q]} = \frac{MLT^{-2}}{AT} = MLT^{-3}A^{-1}$$

Therefore,

$$[\epsilon_0 E^2] = M^{-1}L^{-3}T^4A^2 \times M^2L^2T^{-6}A^{-2} = ML^{-1}T^{-2}$$

This confirms that $$[\epsilon_0 E^2] = ML^{-1}T^{-2}$$. The correct answer is Option (4): $$[ML^{-1}T^{-2}]$$.

Given $$m = kc^P G^{-1/2} h^{1/2}$$. Using dimensional analysis:

$$[m] = M$$, $$[c] = LT^{-1}$$, $$[G] = M^{-1}L^3T^{-2}$$, $$[h] = ML^2T^{-1}$$.

$$M = (LT^{-1})^P (M^{-1}L^3T^{-2})^{-1/2} (ML^2T^{-1})^{1/2}$$

$$= L^P T^{-P} \cdot M^{1/2}L^{-3/2}T^{1} \cdot M^{1/2}L^{1}T^{-1/2}$$

$$= M^{1} L^{P-3/2+1} T^{-P+1-1/2}$$

Matching M: $$1 = 1$$ ✓

Matching L: $$0 = P - 1/2 \Rightarrow P = 1/2$$

Matching T: $$0 = -P + 1/2 \Rightarrow P = 1/2$$ ✓

The answer is Option (1): $$\boxed{\frac{1}{2}}$$.

We need to find the percentage error in measuring the resistance of a wire, given percentage errors in length and diameter measurements.

The resistance of a wire is given by $$R = \frac{\rho L}{A}$$, where $$\rho$$ is the resistivity, $$L$$ is the length, and $$A$$ is the cross-sectional area. For a circular wire with diameter $$d$$, $$A = \frac{\pi d^2}{4}$$. Therefore, $$R = \frac{\rho L}{\frac{\pi d^2}{4}} = \frac{4\rho L}{\pi d^2}$$.

To determine the propagation of error for $$R = \frac{4\rho L}{\pi d^2}$$ (with $$\rho$$ treated as a constant), we take the natural logarithm: $$\ln R = \ln(4\rho/\pi) + \ln L - 2\ln d$$. Differentiating yields $$\frac{\Delta R}{R} = \frac{\Delta L}{L} + 2\frac{\Delta d}{d}$$, where absolute values are taken to represent maximum error.

Given that $$\frac{\Delta L}{L}\times100 = 0.1\%$$ and $$\frac{\Delta d}{d}\times100 = 0.1\%$$, it follows that $$\frac{\Delta R}{R}\times100 = 0.1\% + 2\times0.1\% = 0.3\%$$.

The percentage error in measuring the resistance is 0.3%. Hence the correct answer is Option (2): 0.3%.

The lens formula relates the focal length f to the object distance u and the image distance v as $$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$.

Differentiating this expression gives

$$ -\frac{df}{f^2} = -\frac{dv}{v^2} + \frac{du}{u^2} .$$

Considering maximum errors in u and v leads to

$$ \frac{|\Delta f|}{f^2} = \frac{\Delta v}{v^2} + \frac{\Delta u}{u^2} ,$$

which can be rearranged to

$$ \Delta f = f^2\left(\frac{\Delta u}{u^2} + \frac{\Delta v}{v^2}\right) .$$

Therefore, the correct answer is Option 3: $$f^2\left[\frac{\Delta u}{u^2} + \frac{\Delta v}{v^2}\right].$$

The order of magnitude of a number expressed in scientific notation as $$a \times 10^b$$ (where $$1 \leq a < 10$$) is defined based on the value of $$a$$. According to the convention used in this context:

If $$a \leq 5$$, then the order of magnitude is $$b$$.
If $$a > 5$$, then the order of magnitude is $$b+1$$.

Therefore, $$b$$ is the order of magnitude when $$a \leq 5$$.

Now, evaluating the options:

Option A states that $$b$$ is the order of magnitude for $$a \geq 5$$. This is incorrect because when $$a \geq 5$$, the order of magnitude is $$b$$ only if $$a \leq 5$$, but for $$a > 5$$, it is $$b+1$$.

Option B states that $$b$$ is the order of magnitude for $$a \leq 5$$. This is correct, as per the convention.

Option C states that $$a$$ is the order of magnitude for $$b \leq 5$$. This is incorrect because the order of magnitude is an exponent (a power of 10), not the coefficient $$a$$.

Option D states that $$b$$ is the order of magnitude for $$5 < a \leq 10$$. This is incorrect because when $$a > 5$$, the order of magnitude is $$b+1$$, not $$b$$.

Thus, the correct option is B.

1001:

All non-zero digits are significant, and zeros between non-zero digits are also significant.

So, 1001 has 4 significant figures.

(A) → (II)

010.1:

Leading zero is not significant. The zero between digits and decimal part is significant.

Digits: 1, 0, 1 → total 3 significant figures.

(B) → (I)

100.100:

Zeros between non-zero digits and trailing zeros after decimal are significant.

Digits: 1, 0, 0, 1, 0, 0 → total 6 significant figures.

(C) → (IV)

0.0010010:

Leading zeros are not significant. Zeros between digits and trailing zero after decimal are significant.

Digits: 1, 0, 0, 1, 0 → total 5 significant figures.

(D) → (III)

Final matching:

(A)−(II), (B)−(I), (C)−(IV), (D)−(III)

Coefficient of viscosity:

From Newton’s law of viscosity:

$$F=\eta A\frac{dv}{dx}$$

$$\eta=\frac{F}{A}\cdot\frac{dx}{dv}$$

$$[\eta]=\frac{MLT^{-2}}{L^2}\cdot\frac{L}{LT^{-1}}=[ML^{-1}T^{-1}]$$

So, (A) matches with (III)

Surface tension:

$$\text{Surface tension}=\frac{\text{Force}}{\text{length}}$$

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

So, (B) matches with (IV)

Angular momentum:

$$L=r\times p=r\cdot mv$$

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

So, (C) matches with (II)

Rotational kinetic energy:

$$E=\frac{1}{2}I\omega^2$$

$$[I]=ML^2,\quad[\omega]=T^{-1}$$

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

So, (D) matches with (I)

Final matching:

(A)−(III), (B)−(IV), (C)−(II), (D)−(I)

Latent heat is defined as the heat energy per unit mass required for a phase change:

$$L = \frac{Q}{m}$$

The dimension of heat energy $$Q$$ is $$[ML^2T^{-2}]$$, and mass $$m$$ has dimension $$[M]$$.

Therefore, $$[L] = \frac{[ML^2T^{-2}]}{[M]} = [M^0L^2T^{-2}]$$.

The correct answer is Option 2: $$[M^0L^2T^{-2}]$$.

Find percentage error in spring constant $$k$$ given 2% error in time and 1% error in mass.

Relation between k, m, and T.

$$T = 2\pi\sqrt{\frac{m}{k}}$$, so $$k = \frac{4\pi^2 m}{T^2}$$.

Error propagation.

$$\frac{\Delta k}{k} = \frac{\Delta m}{m} + 2\frac{\Delta T}{T} = 1\% + 2(2\%) = 1\% + 4\% = 5\%$$

Note: we take absolute values of errors for maximum percentage error regardless of sign.

The correct answer is Option A: 5%.

The relationship between temperature changes on Celsius and Fahrenheit scales is given by $$ \Delta F = \frac{9}{5} \Delta C $$. If the change in Celsius is $$\Delta C = 40°C$$, substituting this into the equation yields

$$ \Delta F = \frac{9}{5} \times 40 = 72°F $$. Therefore, the correct answer is Option 3: 72°F.

LC = 1/(20N) cm. MSD = 1mm = 1/10 cm. LC = MSD - VSD. VSD = MSD - LC = 1/10 - 1/(20N) = (2N-1)/(20N) cm. N×VSD = x×MSD. x = N×VSD/MSD = N×(2N-1)/(20N)÷(1/10) = (2N-1)/2.

Option (3).

$$E=\frac{B-x^2}{At}$$. $$[B]=[x^2]=L^2$$. $$[A]=[B]/([E][t])=L^2/(ML^2T^{-2}\cdot T)=L^2/(ML^2T^{-1})=M^{-1}T$$.

$$[AB]=M^{-1}T\cdot L^2=L^2M^{-1}T^1$$.

The answer is Option (2): $$L^2M^{-1}T^1$$.

Torque:

Torque is given by force × distance.

$$\tau=r\times F$$

Dimensions of force:

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

So,

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

So, Torque → (IV)

Magnetic field:

From Lorentz force,

$$F=qvB$$

$$B=\frac{F}{qv}$$

Now substitute dimensions:

$$[q]=[AT]$$

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

$$[B]=\frac{MLT^{-2}}{(AT)(LT^{-1})}=[MT^{-2}A^{-1}]$$

So, Magnetic field → (III)

Magnetic moment:

Magnetic moment is given by:

$$μ=IA$$

$$[\mu]=[A]\cdot[L^2]=[L^2A]$$

So, Magnetic moment → (II)

Permeability of free space:

Using relation:

$$B=\mu_0\frac{I}{L}$$

$$\mu_0=\frac{BL}{I}$$

Substitute dimensions:

$$[B]=[MT^{-2}A^{-1}]$$

$$[\mu_0]=\frac{(MT^{-2}A^{-1})\cdot L}{A}=[MLT^{-2}A^{-2}]$$

So, Permeability → (I)

Final matching:

(A)−(IV), (B)−(III), (C)−(II), (D)−(I)

We need to find the dimensional formula of angular impulse.

Angular impulse is the product of torque and time, giving Angular impulse = $$\tau \times t$$.

Torque ($$\tau$$) is force times the lever arm (perpendicular distance). The dimension of force is $$[M L T^{-2}]$$ and that of the lever arm is $$[L]$$, so $$[\tau] = [M L T^{-2}] \times [L] = [M L^2 T^{-2}]$$.

Thus the dimension of angular impulse is $$[\text{Angular impulse}] = [\tau] \times [t] = [M L^2 T^{-2}] \times [T] = [M L^2 T^{-1}]$$.

The correct answer is Option D: $$[M L^2 T^{-1}]$$.

The equation is: $$y = 2a\sin\left(\frac{2\pi nt}{\lambda}\right)\cos\left(\frac{2\pi x}{\lambda}\right)$$

For the equation to be dimensionally consistent:

- $$\frac{2\pi x}{\lambda}$$ must be dimensionless, so $$[\lambda] = [L]$$ and $$[x] = [L]$$

- $$\frac{2\pi nt}{\lambda}$$ must be dimensionless, so $$[nt/\lambda]$$ is dimensionless

- This gives $$[n] = [\lambda/t] = [L/T] = [LT^{-1}]$$ (velocity dimension)

Now checking each option:

Option 1: Dimensions of $$n/\lambda = [LT^{-1}]/[L] = [T^{-1}]$$, NOT $$[T]$$. This is NOT correct. ✓ (This is the wrong statement)

Option 2: Dimensions of n is $$[LT^{-1}]$$. Correct.

Option 3: Dimensions of x is [L]. Correct.

Option 4: Dimensions of nt = $$[LT^{-1}][T] = [L]$$. Correct.

The correct answer is Option 1: The dimensions of $$n/\lambda$$ is [T] (this is NOT correct since it should be $$[T^{-1}]$$).

$$a/V^2$$ has units of pressure: $$[a]=PV^2=ML^{-1}T^{-2}\cdot L^6=ML^5T^{-2}$$. $$[b]=V=L^3$$.

$$[a/b]=ML^5T^{-2}/L^3=ML^2T^{-2}$$.

The answer is Option (3): $$ML^2T^{-2}$$.

We need to find the percentage error in Young's modulus $$Y = 49000\frac{M}{l}$$.

For $$Y = k\frac{M}{l}$$, taking the logarithm gives $$\ln Y = \ln k + \ln M - \ln l$$, and differentiating yields $$\frac{\Delta Y}{Y} = \frac{\Delta M}{M} + \frac{\Delta l}{l}$$.

From the graph paper, the smallest scale division along the load axis is 5 g, so $$\Delta M = 5$$ g, and along the extension axis it is 0.02 cm, giving $$\Delta l = 0.02$$ cm.

$$\frac{\Delta Y}{Y} = \frac{5}{500} + \frac{0.02}{2} = 0.01 + 0.01 = 0.02 = 2\%$$

The correct answer is Option (2): 2%.

Find the least count of a Vernier calliper where 10 MSD = 11 VSD and each MSD = 5 units.

MSD stands for Main Scale Division and VSD for Vernier Scale Division. Since 10 main-scale divisions span the same length as 11 vernier-scale divisions, we have $$ 10 \times 5 = 11 \times \text{(1 VSD)} $$. It follows that $$ 1 \text{ VSD} = \frac{50}{11} \text{ units} $$. The least count (LC) of a Vernier instrument is the difference between one main-scale division and one vernier-scale division, so $$ LC = 1 \text{ MSD} - 1 \text{ VSD} = 5 - \frac{50}{11} = \frac{55 - 50}{11} = \frac{5}{11} $$. Therefore, the correct answer is Option D: $$\frac{5}{11}$$.

We are given a vernier callipers where 20 divisions on the vernier scale coincide with 19 divisions on the main scale, and the least count is 0.1 mm. We need to find the value of one main scale division (MSD).

The given condition is $$20 \text{ VSD} = 19 \text{ MSD}$$ which implies $$1 \text{ VSD} = \frac{19}{20} \text{ MSD}$$.

The least count (LC) is defined as the difference between one main scale division and one vernier scale division: $$\text{LC} = 1 \text{ MSD} - 1 \text{ VSD}$$.

Substituting the expression for VSD from the previous relation gives $$\text{LC} = 1 \text{ MSD} - \frac{19}{20} \text{ MSD} = \text{MSD} \left(1 - \frac{19}{20}\right) = \frac{1}{20} \text{ MSD}$$.

Since the least count is given as $$\text{LC} = 0.1$$ mm, one can write $$0.1 = \frac{1}{20} \times \text{MSD}$$ which yields $$\text{MSD} = 0.1 \times 20 = 2 \text{ mm}$$.

Therefore, one main scale division is equal to 2 mm. The correct answer is Option (2): 2.

We need to evaluate the Assertion and Reason about Vernier callipers.

Assertion (A): In a Vernier calliper, if positive zero error exists, the reading taken will be more than the actual reading.

This is true. Positive zero error means the zero of the vernier scale is to the right of the zero of the main scale when the jaws are closed. This adds an extra amount to every reading, making measured values larger than actual values. The correction requires subtracting the zero error.

Reason (R): The zero error in a Vernier calliper might have happened due to manufacturing defect or due to rough handling.

This is true. Manufacturing imperfections or damage from rough use can cause the jaws to not align properly at zero.

However, R merely explains the cause of zero error in general, not why positive zero error specifically leads to higher readings. R does not explain the mechanism described in A.

The correct answer is Option 2: Both (A) and (R) are correct but (R) is not the correct explanation of (A).

A) Radius of curvature of concave surface
A spherometer is specifically designed to measure radius of curvature (both concave and convex).
So this can be measured.

B) Specific rotation of liquids
This is an optical property measured using a polarimeter, not a spherometer.
Spherometer only measures mechanical dimensions (height, curvature).
So this cannot be measured.

C) Thickness of thin plates
Yes, a spherometer can measure very small thickness by comparing readings.
So this can be measured.

D) Radius of curvature of convex surface
Again, one of the main uses of a spherometer.
So this can be measured.

Final conclusion:
Only option B is not measurable using a spherometer.

Answer: Specific rotation of liquids

50 Vernier divisions = 49 main scale divisions.

1 VSD = $$\frac{49}{50}$$ MSD.

Least count = 1 MSD - 1 VSD = $$1 - \frac{49}{50} = \frac{1}{50}$$ MSD = $$\frac{0.5}{50}$$ mm = $$0.01$$ mm.

The answer is Option (4): $$\boxed{0.01 \text{ mm}}$$.

We need to find how many main scale divisions are in 1 cm for a vernier caliper.

First, the least count LC is defined as one main scale division (MSD) minus one vernier scale division (VSD), and since 50 VSD equal 49 MSD, we have 1 VSD = 49/50 MSD. Moreover, the zero error equals the number of coinciding VSD multiplied by the least count.

Expressing the least count in terms of MSD gives $$\text{LC} = 1\,\text{MSD} - 1\,\text{VSD} = 1\,\text{MSD} - \frac{49}{50}\,\text{MSD} = \frac{1}{50}\,\text{MSD}$$

Since the zero of the vernier has shifted left (indicating a positive zero error) and the fourth division on the vernier coincides with a main scale division, the zero error is four times the least count: $$\text{Zero error} = 4 \times \text{LC} = 4 \times \frac{1}{50}\,\text{MSD} = \frac{4}{50}\,\text{MSD}$$

We are told this zero error measures 0.04 mm, so $$\frac{4}{50}\,\text{MSD} = 0.04\,\text{mm} \implies \text{MSD} = 0.04 \times \frac{50}{4} = 0.04 \times 12.5 = 0.5\,\text{mm}$$

Hence, the number of main scale divisions in 1 cm (10 mm) is $$\frac{10\,\text{mm}}{0.5\,\text{mm}} = 20$$

The correct answer is Option (3): 20.

One main scale division (MSD) = $$m$$ units.

The $$n^{th}$$ division of the main scale coincides with the $$(n+1)^{th}$$ division of the vernier scale.

This means: $$n$$ MSD = $$(n+1)$$ VSD (vernier scale divisions).

Therefore: 1 VSD = $$\frac{n}{n+1}$$ MSD = $$\frac{nm}{n+1}$$ units.

Least count = 1 MSD - 1 VSD = $$m - \frac{nm}{n+1} = m\left(1 - \frac{n}{n+1}\right) = m \cdot \frac{1}{n+1} = \frac{m}{n+1}$$.

The correct answer is Option 3: $$\frac{m}{n+1}$$.

We need to find the density of a sphere measured with a vernier caliper.

9 main scale divisions = 10 vernier scale divisions. Smallest main scale division = 1 mm.

$$\text{LC} = 1 \text{ MSD} - 1 \text{ VSD} = 1 - \frac{9}{10} = 0.1 \text{ mm} = 0.01 \text{ cm}$$

Main scale reading = 2 cm. Vernier coincidence = 2nd division.

$$d = \text{MSR} + \text{VSR} \times \text{LC} = 2 + 2 \times 0.01 = 2.02 \text{ cm}$$

Radius: $$r = 1.01$$ cm.

$$V = \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times (1.01)^3$$

$$(1.01)^3 \approx 1.0303$$

$$V = \frac{4}{3} \times 3.1416 \times 1.0303 \approx 4.32 \text{ cm}^3$$

$$\rho = \frac{m}{V} = \frac{8.635}{4.32} \approx 2.0 \text{ g/cm}^3$$

The correct answer is Option (1): 2.0 g/cm$$^3$$.

We need to find the accuracy (percentage error) in the measurement of $$g$$ using a simple pendulum.

Length: $$l = 20$$ cm, $$\Delta l = 2$$ mm = 0.2 cm

Time for 50 oscillations: $$t = 40$$ s, $$\Delta t = 1$$ s

From $$T = 2\pi\sqrt{l/g}$$, we get: $$g = \frac{4\pi^2 l}{T^2}$$

Since $$T = t/50$$: $$g = \frac{4\pi^2 l \times 2500}{t^2} = \frac{10000\pi^2 l}{t^2}$$

Taking logarithmic differentiation:

$$\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T} = \frac{\Delta l}{l} + 2\frac{\Delta t}{t}$$

(Since $$T = t/50$$, $$\Delta T/T = \Delta t/t$$.)

$$\frac{\Delta g}{g} = \frac{0.2}{20} + 2 \times \frac{1}{40} = 0.01 + 0.05 = 0.06 = 6\%$$

Therefore $$N = 6$$.

The correct answer is Option 3: 6.

We need to find the diameter of a wire measured using a screw gauge that has a positive zero error. Since the least count of a screw gauge is given by $$\text{LC} = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}} = \frac{1 \text{ mm}}{100} = 0.01 \text{ mm}$$, we have determined the smallest measurable length increment.

The problem states that with no measuring quantity between the jaws, the zero of the circular scale lies 5 divisions below the reference line. From this we infer a positive zero error, which is calculated as $$\text{Zero error} = +5 \times \text{LC} = +5 \times 0.01 = +0.05 \text{ mm}$$. Since a positive zero error indicates that the instrument reads more than the actual value, we must subtract it from the measured reading.

Next, the observed reading consists of a main scale reading and a circular scale reading. For the main scale reading, 4 linear scale divisions are visible, and since the pitch is 1 mm, each linear division equals 1 mm. Hence $$\text{MSR} = 4 \times 1 = 4 \text{ mm}$$. The circular scale reading corresponds to the 60th division coinciding with the reference line, giving $$\text{CSR} = 60 \times \text{LC} = 60 \times 0.01 = 0.60 \text{ mm}$$. Altogether, the observed reading is $$\text{Observed reading} = \text{MSR} + \text{CSR} = 4 + 0.60 = 4.60 \text{ mm}$$.

Finally, applying the zero error correction yields $$\text{Corrected reading} = \text{Observed reading} - \text{Zero error} = 4.60 - 0.05 = 4.55 \text{ mm}$$. The correct answer is Option (3): 4.55 mm.

We need to find the diameter of a wire measured using a screw gauge. Recall that the screw gauge reading is given by: $$ \text{Reading} = \text{MSR} + \text{CSR} \times \text{Least Count} $$ where MSR = Main Scale Reading, CSR = Circular Scale Reading, and Least Count = Pitch / Number of divisions on circular scale.

Since the pitch is 1 mm and the circular scale has 100 divisions, the least count is: $$ LC = \frac{\text{Pitch}}{\text{Number of divisions}} = \frac{1 \text{ mm}}{100} = 0.01 \text{ mm} $$. Substituting in the formula gives the diameter: $$ d = 1 + 42 \times 0.01 = 1 + 0.42 = 1.42 \text{ mm} $$. Expressing 1.42 mm in fractional form: $$ d = 1.42 = \frac{142}{100} = \frac{71}{50} \text{ mm} $$. Therefore x = 71. The correct answer is Option (3): 71.

Let angle between $$\vec{A}$$ and $$\vec{B}$$ be $$\theta$$. Resultant $$\vec{R} = \vec{A} + \vec{B}$$ is perpendicular to $$\vec{A}$$.

$$\vec{R} \cdot \vec{A} = 0$$: $$A^2 + AB\cos\theta = 0 \implies \cos\theta = -A/B$$.

$$|\vec{R}| = B/2$$: $$A^2 + B^2 + 2AB\cos\theta = B^2/4$$.

$$A^2 + B^2 - 2A^2 = B^2/4 \implies B^2 - A^2 = B^2/4 \implies A^2 = 3B^2/4$$.

$$\cos\theta = -\frac{A}{B} = -\frac{\sqrt{3}}{2} \implies \theta = 150°$$.

The answer is $$\boxed{150}$$°.

The volume of a right circular cone in terms of the measured base diameter $$d$$ and height $$h$$ is

$$V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi\left(\frac{d}{2}\right)^{2}h=\frac{\pi}{12}\,d^{2}h$$

If a quantity $$Q$$ depends on the variables $$x,\,y,\,z$$ as $$Q=x^{m}y^{n}z^{p}$$, the maximum fractional (relative) error in $$Q$$ is

$$\frac{\Delta Q}{Q}=m\frac{\Delta x}{x}+n\frac{\Delta y}{y}+p\frac{\Delta z}{z}$$

Comparing $$V=\dfrac{\pi}{12}d^{2}h$$ with the general form, we have $$m=2$$ for $$d$$ and $$n=1$$ for $$h$$ (the constant $$\pi/12$$ does not affect errors).

Therefore, the maximum fractional error in volume is

$$\frac{\Delta V}{V}=2\frac{\Delta d}{d}+1\frac{\Delta h}{h}$$ $$-(1)$$

The measurements are:

Diameter $$d = 20.0\text{ cm}$$     Height $$h = 20.0\text{ cm}$$

The least count of the scale is $$2\text{ mm}=0.2\text{ cm}$$. In error calculations we take the full least count as the maximum absolute error, so

$$\Delta d = 0.2\text{ cm}, \qquad \Delta h = 0.2\text{ cm}$$

Compute the fractional errors:

$$\frac{\Delta d}{d}=\frac{0.2}{20.0}=0.01$$   (that is, $$1\%$$)

$$\frac{\Delta h}{h}=\frac{0.2}{20.0}=0.01$$   (that is, $$1\%$$)

Substitute into $$(1)$$:

$$\frac{\Delta V}{V}=2(0.01)+0.01=0.03$$

Converting to percentage:

Maximum percentage error $$=0.03\times100\% = 3\%$$

Hence, the maximum percentage error in determining the volume of the cone is 3 %.

In Young’s double slit experiment, intensity at a point is given by:

$$I = I_{\max}\cos^2\left(\frac{\phi}{2}\right)$$

When $$I = \frac{I_{\max}}{4}$$, we have

$$\cos^2\left(\frac{\phi}{2}\right) = \frac{1}{4} \implies \cos\left(\frac{\phi}{2}\right) = \frac{1}{2} \implies \frac{\phi}{2} = \frac{\pi}{3} \implies \phi = \frac{2\pi}{3}$$

The phase difference is related to the path difference by $$\phi = \frac{2\pi}{\lambda}\Delta$$, where $$\Delta = \frac{yd}{D}$$.

Thus,

$$\frac{2\pi}{3} = \frac{2\pi}{\lambda} \cdot \frac{yd}{D}$$

Rearranging to solve for $$y$$ gives

$$y = \frac{\lambda D}{3d} = \frac{600 \times 10^{-9} \times 1}{3 \times 10^{-3}} = \frac{600 \times 10^{-9}}{3 \times 10^{-3}} = 200 \times 10^{-6} \text{ m} = 200 \text{ }\mu\text{m}$$

Therefore, the answer is $$\boxed{200}$$ μm.

We need to find the dimensions of $$\sqrt{uG}$$ where $$u$$ is energy density and $$G$$ is the gravitational constant.

Dimensions of energy density $$u$$.

$$ [u] = \frac{\text{Energy}}{\text{Volume}} = \frac{[ML^2T^{-2}]}{[L^3]} = [ML^{-1}T^{-2}] $$

Dimensions of gravitational constant $$G$$.

From $$F = \frac{Gm_1m_2}{r^2}$$:

$$ [G] = \frac{[F][r^2]}{[m^2]} = \frac{[MLT^{-2}][L^2]}{[M^2]} = [M^{-1}L^3T^{-2}] $$

Dimensions of $$\sqrt{uG}$$.

$$ [uG] = [ML^{-1}T^{-2}] \times [M^{-1}L^3T^{-2}] = [M^0L^2T^{-4}] = [L^2T^{-4}] $$

$$ [\sqrt{uG}] = [LT^{-2}] $$

Identify the physical quantity.

$$[LT^{-2}]$$ is the dimension of acceleration, which is force per unit mass.

The correct answer is Option (4): Force per unit mass.

This question tests knowledge of the SI definition of the candela, the base unit of luminous intensity.

The Official SI Definition of the Candela:

The candela is defined as the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency $$540 \times 10^{12}$$ hertz and that has a radiant intensity in that direction of $$\frac{1}{683}$$ watt per steradian.

Why these specific values?

- The frequency $$540 \times 10^{12}$$ Hz corresponds to green light with a wavelength of approximately 555 nm. This is chosen because the human eye is most sensitive to light at this wavelength (the peak of the photopic luminosity function).

- The factor $$\frac{1}{683}$$ W/sr is chosen to maintain consistency with the historical definition of the candela, ensuring backward compatibility with previous measurement standards.

Matching with the question format:

The question states the frequency is $$A \times 10^{12}$$ Hz, so $$A = 540$$.

The question states the radiant intensity is $$\frac{1}{B}$$ W/sr, so $$B = 683$$.

The correct answer is Option 1: 540 and 683.

We need to determine how Young's modulus changes with temperature.

What is Young’s modulus? Young’s modulus ($$Y$$) is a measure of the stiffness of a material defined as the ratio of stress to strain within the elastic limit: $$ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} $$. It quantifies how much a material resists deformation under an applied force.

Effect of temperature on interatomic bonds: Young’s modulus depends on the strength of interatomic bonds, and as temperature increases, atoms vibrate with greater amplitude about their equilibrium positions. This increased vibration causes thermal expansion, increasing the average interatomic distance. Consequently, the interatomic forces become weaker because the atoms are, on average, further apart and lie on a flatter part of the potential energy curve. Since the restoring force per unit displacement decreases, the material becomes less stiff.

Therefore, Young’s modulus decreases with increasing temperature.

The correct answer is Option B: decreases.

For a planet revolving around the sun in a circular orbit of radius aaa:

The gravitational force provides the centripetal force.

$$\frac{GMm}{a^2}=\frac{mv^2}{a}$$

From this, we get the orbital velocity:

$$v^2=\frac{GM}{a}$$

Now kinetic energy of the planet is:

$$K=\frac{1}{2}mv^2=\frac{1}{2}m\cdot\frac{GM}{a}=\frac{GMm}{2a}$$

So, (A) matches with (II)

Gravitational potential energy of the sun-planet system is:

$$U=-\frac{GMm}{a}$$

So, (B) matches with (I)

Total mechanical energy is:

$$E=K+U=\frac{GMm}{2a}-\frac{GMm}{a}=-\frac{GMm}{2a}​$$​

So, (C) matches with (IV)

Escape energy per unit mass at the surface of a planet is:

$$\ \frac{\ 1}{2}​v_e^2​=\ \frac{\ GM}{r}$$

$$v_e=\sqrt{\frac{2GM}{r}}$$

Energy per unit mass:

$$v_e^2=\frac{GM}{r}$$

So, (D) matches with (III)

Thus, the correct matching is:

(A)−(II), (B)−(I), (C)−(IV), (D)−(III)

We need to find the SI units of $$A$$ and $$B$$ in the electric field expression $$\vec{E} = \frac{A}{x^2}\hat{i} + \frac{B}{y^3}\hat{j}$$.

Finding the unit of A:

The term $$\frac{A}{x^2}$$ represents the x-component of electric field.

$$[A] = [E] \times [x^2] = \text{N C}^{-1} \times \text{m}^2 = \text{N m}^2 \text{ C}^{-1}$$

Finding the unit of B:

The term $$\frac{B}{y^3}$$ represents the y-component of electric field.

$$[B] = [E] \times [y^3] = \text{N C}^{-1} \times \text{m}^3 = \text{N m}^3 \text{ C}^{-1}$$

Therefore, $$A$$ has SI unit N m$$^2$$ C$$^{-1}$$ and $$B$$ has SI unit N m$$^3$$ C$$^{-1}$$.

The correct answer is Option 2.

Statement I: Astronomical unit (Au), Parsec (Pc) and Light year (ly) are all units used for measuring astronomical distances. This is correct.

Statement II: Au < Parsec (Pc) < ly

The actual values are:

$$1 \text{ AU} \approx 1.496 \times 10^{11}$$ m

$$1 \text{ ly} \approx 9.461 \times 10^{15}$$ m

$$1 \text{ Pc} \approx 3.086 \times 10^{16}$$ m

So the correct order is: $$\text{Au} < \text{ly} < \text{Pc}$$

Statement II says Au < Pc < ly, which places Parsec before Light year. This is incorrect since 1 Pc is actually larger than 1 ly.

Therefore, Statement I is correct but Statement II is incorrect, which is Option B.

We need to find the dimensions of $$A$$ and $$B$$ in the equation $$(x - At)^2 + \left(y - \frac{t}{B}\right)^2 = a^2$$, where the dimension of $$t$$ is $$[T^{-1}]$$.

Principle of Dimensional Homogeneity: Each term in an equation must have the same dimensions.

Finding dimension of A:

The term $$At$$ must have the same dimension as $$x$$, which is $$[L]$$.

$$[At] = [L]$$

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

$$[A] = [LT]$$

Finding dimension of B:

The term $$\frac{t}{B}$$ must have the same dimension as $$y$$, which is $$[L]$$.

$$\left[\frac{t}{B}\right] = [L]$$

$$\frac{[T^{-1}]}{[B]} = [L]$$

$$[B] = \frac{[T^{-1}]}{[L]} = [L^{-1}T^{-1}]$$

Therefore, $$A = [LT]$$ and $$B = [L^{-1}T^{-1}]$$.

The correct answer is Option 2.

We have two resistances $$R_1 = (10 \pm 0.5)\;\Omega$$ and $$R_2 = (15 \pm 0.5)\;\Omega$$ connected in parallel, and we need to find the percentage error in the equivalent resistance.

For resistances in parallel, the equivalent resistance is:

$$R_{eq} = \frac{R_1 R_2}{R_1 + R_2} = \frac{10 \times 15}{10 + 15} = \frac{150}{25} = 6\;\Omega$$

Now, to find the error propagation, we start from $$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$$. Differentiating both sides and taking absolute values (since errors always add), the maximum absolute error in $$R_{eq}$$ is:

$$\Delta R_{eq} = R_{eq}^2 \left(\frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2}\right)$$

This follows because differentiating $$R_{eq}^{-1}$$ with respect to $$R_1$$ gives $$-R_1^{-2}$$, and similarly for $$R_2$$. Substituting the values with $$R_{eq}^2 = 36$$:

$$\Delta R_{eq} = 36 \left(\frac{0.5}{100} + \frac{0.5}{225}\right) = 36 \left(0.005 + 0.002222\right) = 36 \times 0.007222 = 0.26\;\Omega$$

So the percentage error is:

$$\%\;\text{error} = \frac{\Delta R_{eq}}{R_{eq}} \times 100 = \frac{0.26}{6} \times 100 = 4.33\%$$

Hence, the correct answer is Option 4.

Solution :

Refractive index of glass slab is :

$$\mu = \frac{\text{Real thickness}}{\text{Apparent thickness}}$$

Given :

$$t_r = 5.25\text{ mm}$$

$$t_a = 5.00\text{ mm}$$

Therefore,

$$\mu = \frac{5.25}{5.00}$$

$$= 1.05$$

Main scale :

$$20\text{ divisions} = 1\text{ cm}$$

Hence,

$$1\text{ MSD} = \frac{1}{20}\text{ cm}$$

$$= 0.05\text{ cm}$$

$$= 0.5\text{ mm}$$

Given,

$$50\text{ VSD} = 49\text{ MSD}$$

Least count :

$$LC = 1\text{ MSD} - 1\text{ VSD}$$

$$= \frac{1}{50}\text{ MSD}$$

$$= \frac{0.5}{50}\text{ mm}$$

$$= 0.01\text{ mm}$$

Fractional error in refractive index :

$$\frac{\Delta \mu}{\mu} = \frac{\Delta t_r}{t_r} + \frac{\Delta t_a}{t_a}$$

Substituting values :

$$\frac{\Delta \mu}{1.05} = \frac{0.01}{5.25} + \frac{0.01}{5.00}$$

$$\approx 0.0039$$

Therefore,

$$\Delta \mu = 1.05 \times 0.0039$$

$$\approx 0.0041$$

$$= 4.1 \times 10^{-3}$$

Given uncertainty is of the form :

$$\frac{x}{10} \times 10^{-3}$$

Comparing,

$$\frac{x}{10} = 4.1$$

$$x = 41$$

Final Answer :

$$41$$

Zero error: +0.06 cm (positive, since zero of vernier is to the right of main scale zero, with 6th division coinciding).

Observed reading: MSR = 3.2 cm, VSR = 4 × 0.01 = 0.04 cm. Total observed = 3.24 cm.

Corrected reading = 3.24 - 0.06 = 3.18 cm.

The correct answer is Option 2: 3.18 cm.

A screw gauge has 100 circular scale divisions and a pitch of 0.5 mm, meaning the main scale moves 0.5 mm per complete rotation. The least count is $$\frac{0.5}{100} = 0.005$$ mm.

Because the zero of the circular scale lies 6 divisions below the reference line, there is a positive zero error of $$+6 \times 0.005 = +0.03$$ mm.

When measuring the wire, the main scale reading is 4 divisions $$\times$$ 0.5 mm = 2.00 mm, and the circular scale reading is 46 $$\times$$ 0.005 = 0.23 mm. Thus the observed reading is 2.00 + 0.23 = 2.23 mm, and the corrected reading is 2.23 - 0.03 = 2.20 mm, which can also be written as $$= 220 \times 10^{-2}$$ mm.

Hence, the diameter of the wire is $$\boxed{220} \times 10^{-2}$$ mm.

The kinetic energy is given by $$K = \frac{1}{2}mv^2$$.

$$K = \frac{1}{2} \times 5 \times (20)^2 = \frac{1}{2} \times 5 \times 400 = 1000 \text{ J}$$

For error calculation, we use the formula for relative error in kinetic energy:

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

$$\frac{\Delta K}{K} = \frac{0.5}{5} + 2 \times \frac{0.4}{20}$$

$$\frac{\Delta K}{K} = 0.1 + 0.04 = 0.14$$

$$\Delta K = 0.14 \times 1000 = 140 \text{ J}$$

Therefore, kinetic energy = $$(1000 \pm 140)$$ J.

We need to check both the Assertion and the Reason.

Checking Assertion A: Dimension of P × t

Pressure $$P$$ has dimensions: $$[P] = [ML^{-1}T^{-2}]$$

Time $$t$$ has dimensions: $$[t] = [T]$$

So, $$[P \times t] = [ML^{-1}T^{-2}] \times [T] = [ML^{-1}T^{-1}]$$

Dimension of coefficient of viscosity ($$\eta$$):

From the formula: $$\eta = \frac{F}{A \times \text{velocity gradient}}$$

Velocity gradient = $$\frac{dv}{dx}$$ has dimensions $$[T^{-1}]$$

So, $$[\eta] = \frac{[MLT^{-2}]}{[L^2][T^{-1}]} = \frac{[MLT^{-2}]}{[L^2T^{-1}]} = [ML^{-1}T^{-1}]$$

Since $$[P \times t] = [ML^{-1}T^{-1}] = [\eta]$$, the Assertion is TRUE.

Checking Reason R:

The Reason states: Coefficient of viscosity = $$\frac{\text{Force}}{\text{Velocity gradient}}$$

The correct formula is: $$F = \eta A \frac{dv}{dx}$$, which gives $$\eta = \frac{F}{A \times \frac{dv}{dx}}$$

The Reason is missing the area $$A$$ in the denominator. So the Reason is FALSE.

Since Assertion A is true but Reason R is false, the correct answer is Option C.

We are given that the efficiency of a Carnot engine is expressed as $$\eta = \frac{\alpha\beta}{\sin\theta} \log_e\frac{\beta x}{kT}$$. Since efficiency $$\eta$$ is dimensionless and $$\sin\theta$$ is also dimensionless, the product $$\alpha\beta$$ must be dimensionless. Also, the argument of the logarithm $$\frac{\beta x}{kT}$$ must be dimensionless.

Since $$\alpha\beta$$ is dimensionless, $$\alpha$$ and $$\beta$$ must have reciprocal dimensions: $$[\alpha] = [\beta]^{-1}$$.

Now, from the condition that $$\frac{\beta x}{kT}$$ is dimensionless, we get $$[\beta] = \frac{[kT]}{[x]}$$. The Boltzmann constant $$k$$ has dimensions of energy per temperature, i.e., $$[k] = M L^2 T^{-2} K^{-1}$$. So $$[kT] = M L^2 T^{-2}$$ (energy). Since $$x$$ has dimensions of length $$[L]$$, we get $$[\beta] = \frac{M L^2 T^{-2}}{L} = M L T^{-2}$$, which is the dimension of force.

So the dimensions of $$\beta$$ are the same as that of force. This makes Option A correct.

Now, since $$[\alpha] = [\beta]^{-1} = M^{-1} L^{-1} T^{2}$$, let us check $$[\alpha^{-1} x]$$. We have $$[\alpha^{-1}] = M L T^{-2}$$ and $$[x] = L$$, so $$[\alpha^{-1} x] = M L^2 T^{-2}$$, which is the dimension of energy. This makes Option B correct.

For Option C, since $$\eta$$ is dimensionless and $$\sin\theta$$ is dimensionless, $$\eta^{-1}\sin\theta$$ is dimensionless. We already established that $$\alpha\beta$$ is dimensionless. So $$[\eta^{-1}\sin\theta] = [\alpha\beta]$$, both being dimensionless. This makes Option C correct.

For Option D, we found $$[\alpha] = M^{-1} L^{-1} T^{2}$$ and $$[\beta] = M L T^{-2}$$. These are clearly not the same dimensions (in fact they are reciprocals of each other). So Option D is incorrect.

Hence, the correct answer is Option D.

We need to check whether the expression $$T = K\sqrt{\dfrac{\rho r^3}{S^{3/2}}}$$ is dimensionally correct, where $$\rho$$ is density, $$r$$ is radius, $$S$$ is surface tension, and $$K$$ is dimensionless.

The dimensions of each quantity are: density $$[\rho] = ML^{-3}$$, radius $$[r] = L$$, and surface tension $$[S] = MT^{-2}$$ (force per unit length).

Now, the expression inside the square root has dimensions:

$$\dfrac{[\rho][r]^3}{[S]^{3/2}} = \dfrac{ML^{-3} \cdot L^3}{(MT^{-2})^{3/2}} = \dfrac{M}{M^{3/2}T^{-3}} = \dfrac{T^3}{M^{1/2}}$$

Taking the square root: $$\sqrt{\dfrac{T^3}{M^{1/2}}} = \dfrac{T^{3/2}}{M^{1/4}}$$

The dimension of the R.H.S. is $$\dfrac{T^{3/2}}{M^{1/4}}$$, which is NOT equal to $$[T] = T$$.

So, the Assertion (A) is false — the given expression is NOT dimensionally correct.

The Reason (R) states that using dimensional analysis, the R.H.S. has different dimensions than the time period. Since we showed that the R.H.S. has dimensions $$T^{3/2}M^{-1/4}$$ instead of $$T$$, the Reason is true.

Hence, the correct answer is Option D: (A) is false but (R) is true.

We need to identify which pair of physical quantities have the same dimensions.

For option A, velocity gradient and decay constant are considered. Velocity gradient is defined as the rate of change of velocity with respect to distance: $$\text{Velocity gradient} = \frac{dv}{dx}$$ Its dimensions are: $$\frac{[LT^{-1}]}{[L]} = [T^{-1}]$$ Decay constant $$\lambda$$ appears in the radioactive decay law $$N = N_0 e^{-\lambda t}$$ and since $$\lambda t$$ must be dimensionless, $$[\lambda] = [T^{-1}]$$ Both have dimensions $$[T^{-1}]$$ and therefore match.

For option B, angular frequency $$\omega$$ has dimensions $$[T^{-1}]$$, while angular momentum $$L = I\omega$$ has dimensions $$[ML^2T^{-1}]$$. These are different.

For option C, wave number $$k = 1/\lambda$$ has dimensions $$[L^{-1}]$$, while Avogadro number $$N_A$$ has dimensions $$[\text{mol}^{-1}]$$. These are different.

For option D, Wien's displacement constant $$b$$ has dimensions $$[LK]$$ (from $$\lambda_{max} T = b$$), while Stefan's constant $$\sigma$$ has dimensions $$[MT^{-3}K^{-4}]$$ (from $$P = \sigma A T^4$$). These are different.

Therefore, the correct answer is Option A: Velocity gradient and decay constant both have dimensions $$[T^{-1}]$$.

We need to identify the pair of physical quantities that have different dimensions.

Option A: Wave number and Rydberg's constant

Wave number $$k = \frac{1}{\lambda}$$, so its dimension is $$[L^{-1}]$$.

Rydberg's constant $$R$$ also has dimension $$[L^{-1}]$$ (it appears in $$\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$).

Both have the same dimensions.

Option B: Stress and Coefficient of elasticity

Stress $$= \frac{F}{A}$$, dimension $$= [ML^{-1}T^{-2}]$$.

Coefficient of elasticity (Young's modulus) $$= \frac{\text{Stress}}{\text{Strain}}$$. Since strain is dimensionless, its dimension is also $$[ML^{-1}T^{-2}]$$.

Both have the same dimensions.

Option C: Coercivity and Magnetisation

Both coercivity and magnetisation have the dimension of magnetic field intensity $$H$$, which is $$[AL^{-1}]$$.

Both have the same dimensions.

Option D: Specific heat capacity and Latent heat

Specific heat capacity $$c = \frac{Q}{m \Delta T}$$, dimension $$= \frac{[ML^2T^{-2}]}{[M][K]} = [L^2T^{-2}K^{-1}]$$.

Latent heat $$L = \frac{Q}{m}$$, dimension $$= \frac{[ML^2T^{-2}]}{[M]} = [L^2T^{-2}]$$.

These have different dimensions — specific heat capacity has an extra factor of $$K^{-1}$$.

Therefore, the correct answer is Option D.

Torque:

$$\tau=r\times F$$

Unit:

$$\text{N}\cdot\text{m}$$

So, (A) matches with (III)

Stress:

$$\text{Stress}=\frac{\text{Force}}{\text{Area}}$$

$$=\frac{\text{N}}{\text{m}^2}=\text{N m}^{-2}$$

So, (B) matches with (IV)

Latent heat:

Latent heat is heat per unit mass:

$$L=\frac{Q}{m}$$

$$=\frac{\text{J}}{\text{kg}}$$

So, (C) matches with (II)

Power:

$$P=\frac{\text{Work}}{\text{Time}}=\frac{\text{J}}{\text{s}}=\text{W}$$

Also,

$$\text{W}=\text{N m s}^{-1}$$

So, (D) matches with (I)

Final matching:

(A)−(III), (B)−(IV), (C)−(II), (D)−(I)

We need to find the dimensions of mutual inductance.

The emf induced in a coil is related to mutual inductance by:

$$\varepsilon = M \frac{dI}{dt}$$

So: $$M = \frac{\varepsilon \cdot dt}{dI}$$

The dimension of emf ($$\varepsilon$$) is the same as voltage:

$$[\varepsilon] = [V] = ML^2T^{-3}A^{-1}$$

The dimension of $$\frac{dt}{dI}$$ is:

$$\left[\frac{dt}{dI}\right] = \frac{T}{A} = TA^{-1}$$

Therefore, the dimension of mutual inductance is:

$$[M] = ML^2T^{-3}A^{-1} \times TA^{-1} = ML^2T^{-2}A^{-2}$$

Hence, the correct answer is Option B.

We have a travelling microscope whose main scale has 20 divisions per cm. This means each main scale division (MSD) is $$\dfrac{1}{20}$$ cm $$= 0.05$$ cm.

We are told that 25 Vernier scale divisions (VSD) coincide with 24 main scale divisions. Therefore, $$25 \text{ VSD} = 24 \text{ MSD}$$, which gives $$1 \text{ VSD} = \dfrac{24}{25} \text{ MSD}$$.

The least count (LC) of a Vernier instrument is the difference between one main scale division and one Vernier scale division: $$\text{LC} = 1\text{ MSD} - 1\text{ VSD} = 1\text{ MSD} - \dfrac{24}{25}\text{ MSD} = \dfrac{1}{25}\text{ MSD}$$.

Now substituting the value of one MSD: $$\text{LC} = \dfrac{1}{25} \times 0.05 \text{ cm} = \dfrac{0.05}{25} = 0.002 \text{ cm}$$.

Hence, the correct answer is Option C.

We are given a screw gauge with pitch = 0.5 mm and 50 divisions on the circular scale.

Step 1: Find the Least Count (LC)

$$\text{LC} = \dfrac{\text{Pitch}}{\text{Number of divisions}} = \dfrac{0.5}{50} = 0.01 \text{ mm}$$

Step 2: Find the diameter of the wire

$$d = \text{MSR} + \text{CSR} \times \text{LC} = 1.5 + 7 \times 0.01 = 1.57 \text{ mm}$$

Step 3: Calculate the curved surface area

The curved surface area of a cylinder is:

$$A = \pi d L$$

where $$d = 1.57 \text{ mm} = 0.157 \text{ cm}$$ and $$L = 6.8 \text{ cm}$$.

$$A = \pi \times 0.157 \times 6.8 = \pi \times 1.0676 = 3.3543 \text{ cm}^2$$

Step 4: Apply significant figures

The diameter (1.57 mm) has 3 significant figures, and the length (6.8 cm) has 2 significant figures. The result should be rounded to 2 significant figures:

$$A \approx 3.4 \text{ cm}^2$$

Therefore, the correct answer is Option B.

Given that 10 divisions of the Vernier scale equal 9 divisions of the main scale, and 1 main scale division (MSD) = 1 mm.

Since each Vernier scale division (VSD) is $$\tfrac{9}{10}$$ of an MSD, it follows that $$1 \text{ VSD} = \tfrac{9}{10} \text{ MSD} = 0.9 \text{ mm}.$$ Therefore the least count (LC) is the difference between one MSD and one VSD: $$\text{LC} = 1 \text{ MSD} - 1 \text{ VSD} = 1 - 0.9 = 0.1 \text{ mm}.$$

The zero of the Vernier scale is shifted to the left of the main scale zero, indicating a negative zero error. Because the 4th Vernier division coincides with the main scale, one finds $$\text{Zero error} = -(4 \times \text{LC}) = -(4 \times 0.1) = -0.4 \text{ mm}.$$

The main scale reading (MSR) is 30 mm (since the Vernier zero lies between 30 and 31), and the 6th Vernier division coincides, giving the Vernier scale reading (VSR). Hence $$\text{Observed reading} = \text{MSR} + \text{VSR} \times \text{LC} = 30 + 6 \times 0.1 = 30.6 \text{ mm}.$$

Subtracting the zero error from the observed reading yields the corrected measurement: $$\text{Corrected reading} = \text{Observed reading} - \text{Zero error} = 30.6 - (-0.4) = 30.6 + 0.4 = 31.0 \text{ mm} = 3.10 \text{ cm}.$$ Thus, the diameter of the spherical body is $$3.10 \text{ cm}.$$

The correct answer is Option C.

We need to identify which pair of physical quantities have the same dimensions.

For Option A, electric displacement $$\vec{D}$$ and surface charge density are considered.

From Gauss's law in dielectrics: $$\vec{D} = \varepsilon_0 \vec{E} + \vec{P}$$, and for a surface, $$\oint \vec{D} \cdot d\vec{A} = Q_{free}$$.

Therefore, $$[D] = \dfrac{[Q]}{[A]} = \dfrac{C}{m^2}$$

Surface charge density: $$[\sigma] = \dfrac{[Q]}{[A]} = \dfrac{C}{m^2}$$

Both have the same dimensions $$[AT L^{-2}]$$. This is correct.

For Option B, displacement current and the electric field are compared.

Displacement current has dimensions of current: $$[A]$$

Electric field has dimensions: $$[MLT^{-3}A^{-1}]$$

These are not the same.

For Option C, current density and surface charge density are compared.

Current density: $$[J] = \dfrac{[A]}{[L^2]} = [AL^{-2}]$$

Surface charge density: $$[\sigma] = \dfrac{[AT]}{[L^2]} = [ATL^{-2}]$$

These differ by a factor of $$[T]$$, so they are not the same.

For Option D, electric potential and energy are compared.

Electric potential: $$[ML^2T^{-3}A^{-1}]$$

Energy: $$[ML^2T^{-2}]$$

These are not the same.

Hence, the correct answer is Option A.

We need to find the least count of a travelling microscope.

Main scale: 40 divisions in 1 cm

50 Vernier scale divisions = 49 main scale divisions

$$1 \text{ MSD} = \frac{1 \text{ cm}}{40} = \frac{1}{40}$$ cm = $$0.025$$ cm

$$50 \text{ VSD} = 49 \text{ MSD}$$

$$1 \text{ VSD} = \frac{49}{50} \text{ MSD}$$

$$\text{LC} = 1 \text{ MSD} - 1 \text{ VSD} = 1 \text{ MSD} - \frac{49}{50} \text{ MSD} = \frac{1}{50} \text{ MSD}$$

$$= \frac{1}{50} \times 0.025 \text{ cm} = 0.0005 \text{ cm}$$

$$= 0.0005 \times 10^{-2} \text{ m} = 5 \times 10^{-6}$$ m

Comparing with $$\_\_\_ \times 10^{-6}$$ m, the answer is $$5$$.

The least count is $$5 \times 10^{-6}$$ m.

We need to find the vernier constant of a vernier callipers where each cm on the main scale is divided into 20 equal parts, and the 10th vernier scale division coincides with the 9th main scale division.

First, the value of one Main Scale Division (MSD) is determined by dividing 1 cm into 20 equal parts:

$$1 \text{ MSD} = \frac{1 \text{ cm}}{20} = \frac{10 \text{ mm}}{20} = 0.5 \text{ mm}$$

Next, since 10 vernier scale divisions coincide with 9 main scale divisions, the value of one Vernier Scale Division (VSD) is

$$1 \text{ VSD} = \frac{9}{10} \times 1 \text{ MSD} = \frac{9}{10} \times 0.5 = 0.45 \text{ mm}$$

The vernier constant (also called the least count) is the difference between one MSD and one VSD:

$$\text{VC} = 1 \text{ MSD} - 1 \text{ VSD}$$

$$\text{VC} = 0.5 - 0.45 = 0.05 \text{ mm}$$

Expressing this value in the required form gives

$$\text{VC} = 0.05 \text{ mm} = 5 \times 10^{-2} \text{ mm}$$

Therefore, the value of the vernier constant is 5 $$\times 10^{-2}$$ mm.

We are given a Vernier callipers where 1 main scale division (MSD) = $$1 \text{ mm}$$ and 10 Vernier scale divisions (VSD) = 9 MSD.

Since 1 VSD = $$\dfrac{9}{10}$$ MSD = 0.9 mm, the least count (LC) is the difference between one main scale division and one Vernier scale division:

$$\text{Least count (LC)} = 1 \text{ MSD} - 1 \text{ VSD} = 1 - 0.9 = 0.1 \text{ mm} = 0.01 \text{ cm}$$

When the jaws touch, the zero of the Vernier lies to the right of the zero of the main scale, and the 4th Vernier division coincides with a main scale division, indicating a positive zero error of

$$\text{Zero error} = +4 \t\times \text{LC} = +4 \t\times 0.01 = +0.04 \text{ cm}$$

In taking the measurement, the zero of the Vernier lies between $$4.1 \text{ cm}$$ and $$4.2 \text{ cm}$$ on the main scale, so the main scale reading (MSR) is:

$$\text{MSR} = 4.1 \text{ cm}$$

At the same time, the 6th Vernier division coincides with a main scale division, giving a Vernier scale reading of

$$\text{Vernier Scale Reading} = 6 \t\times \text{LC} = 6 \t\times 0.01 = 0.06 \text{ cm}$$

Therefore, the observed reading is the sum of MSR and VSR:

$$\text{Observed reading} = \text{MSR} + \text{VSR} = 4.1 + 0.06 = 4.16 \text{ cm}$$

Applying the zero correction by subtracting the zero error from the observed reading gives the correct reading:

$$\text{Correct reading} = \text{Observed reading} - \text{Zero error}$$

$$= 4.16 - 0.04 = 4.12 \text{ cm}$$

This result can be expressed in the required form as

$$4.12 \text{ cm} = 412 \t\times 10^{-2} \text{ cm}$$

The diameter of the bob is $$412 \t\times 10^{-2} \text{ cm}$$.

We need to find the corrected diameter measured using Vernier callipers. Vernier constant (least count) = 0.1 mm = 0.01 cm, coinciding Vernier division = 5, Vernier scale reading = 5 × 0.01 = 0.05 cm.

Observed reading = Main scale reading + Vernier scale reading $$= 1.7 + 0.05 = 1.75 \text{ cm}$$.

Zero error = −0.05 cm (negative zero error). Corrected reading = Observed reading − Zero error $$= 1.75 - (-0.05) = 1.75 + 0.05 = 1.80 \text{ cm}$$.

$$1.80 \text{ cm} = 180 \times 10^{-2} \text{ cm}$$. The answer is 180 $$\times 10^{-2}$$ cm.

The distance between successive main-scale divisions is given as $$0.5\text{ mm}$$.
For one complete rotation of the circular scale the sleeve advances by two such divisions, so the pitch (advance per rotation) is

$$\text{Pitch}=2\times 0.5\text{ mm}=1.0\text{ mm}$$

The circular scale has $$100$$ equal divisions; therefore the least count (smallest readable length) is

$$\text{Least count (LC)}=\frac{\text{Pitch}}{\text{No. of circular divisions}}=\frac{1.0\text{ mm}}{100}=0.01\text{ mm}$$

Zero reading (with jaws in contact)
Main-scale reading = $$0$$ division   ( = $$0\text{ mm}$$ )
Circular-scale reading = $$4$$ divisions

Observed length with no wire = $$4 \times \text{LC}=4\times 0.01=0.04\text{ mm}$$

Since the screw gauge reads $$0.04\text{ mm}$$ when the true separation is $$0$$, it possesses a positive zero error of $$+0.04\text{ mm}$$.
Hence the zero correction (to be applied to every reading) is

$$\text{Zero correction}=-0.04\text{ mm}$$

Attempt 1 (with wire)
Main-scale reading = $$4$$ divisions $$\Rightarrow 4\times 0.5 =2.00\text{ mm}$$
Circular-scale reading = $$20$$ divisions $$\Rightarrow 20\times 0.01 =0.20\text{ mm}$$

Observed diameter $$d_1 = 2.00+0.20 = 2.20\text{ mm}$$
Corrected diameter $$d_1^{\prime}=2.20-0.04 = 2.16\text{ mm}$$

Attempt 2 (with wire)
Main-scale reading = $$4$$ divisions $$\Rightarrow 2.00\text{ mm}$$
Circular-scale reading = $$16$$ divisions $$\Rightarrow 0.16\text{ mm}$$

Observed diameter $$d_2 = 2.00+0.16 = 2.16\text{ mm}$$
Corrected diameter $$d_2^{\prime}=2.16-0.04 = 2.12\text{ mm}$$

Mean diameter

$$\bar d=\frac{d_1^{\prime}+d_2^{\prime}}{2}=\frac{2.16+2.12}{2}=2.14\text{ mm}$$

Uncertainty in diameter
The two corrected readings differ by $$0.04\text{ mm}$$, so the spread about the mean is $$\pm 0.02\text{ mm}$$. Therefore

$$d = 2.14 \pm 0.02 \text{ mm}$$

Cross-sectional area
For a circular wire $$A=\dfrac{\pi d^{2}}{4}$$.

$$A=\frac{\pi (2.14\text{ mm})^{2}}{4} =\pi \times 1.1449\text{ mm}^2 \;\approx\; \pi(1.14\text{ mm}^2)$$

Uncertainty in area
Fractional error: $$\dfrac{\Delta d}{d}= \dfrac{0.02}{2.14}=0.00935$$.
Since $$A\propto d^{2}$$, fractional error in area is twice this value:

$$\frac{\Delta A}{A}=2\times 0.00935=0.0187$$

Absolute error: $$\Delta A = 0.0187 \times 1.14\text{ mm}^2 \approx 0.02\text{ mm}^2$$

Thus

$$A = \pi\bigl(1.14 \pm 0.02\bigr)\text{ mm}^2$$

Therefore, the measured values are

Diameter = $$2.14 \pm 0.02\text{ mm}$$,   Area = $$\pi(1.14 \pm 0.02)\text{ mm}^2$$

Hence the correct choice is
Option C which is: 2.14 ± 0.02 mm, $$\pi$$(1.14 ± 0.02) mm$$^2$$.

We have to find the fractional (relative) error in Young’s modulus, whose working formula for the experiment is $$Y=\frac{M g \, l^{3}}{4\, b\, d^{3}\, \delta}\, .$$

In error analysis, when a quantity is a product or quotient of measured quantities raised to powers, the fractional error in the result equals the algebraic sum of the absolute values of the fractional errors of each factor, each multiplied by its power. Stated formally, if $$Q = A^{p}\, B^{q}\, C^{r}\dots ,$$ then $$\frac{\Delta Q}{Q}=|p|\frac{\Delta A}{A}+|q|\frac{\Delta B}{B}+|r|\frac{\Delta C}{C}+\dots.$$

Comparing with $$Y=\dfrac{M g\, l^{3}}{4\, b\, d^{3}\, \delta},$$ we notice that $$g$$ and the numerical constant $$4$$ are exact, so they contribute no error. The powers of the remaining variables are:

$$$M^{+1},\quad l^{+3},\quad b^{-1},\quad d^{-3},\quad \delta^{-1}.$$$

Therefore,

$$\frac{\Delta Y}{Y}= \frac{\Delta M}{M}+3\frac{\Delta l}{l}+\frac{\Delta b}{b}+3\frac{\Delta d}{d}+\frac{\Delta\delta}{\delta}.$$

The absolute error in each measurement is taken as its least count. We now convert every quantity and its least count to the same unit (SI) and evaluate each fractional error.

Mass: observed $$$M = 2\ \text{kg},\quad \text{least count}=1\ \text{g}=0.001\ \text{kg}.$$$
$$\frac{\Delta M}{M}= \frac{0.001}{2}=0.0005.$$

Length of bar: observed $$$l = 1\ \text{m},\quad \text{least count}=1\ \text{mm}=0.001\ \text{m}.$$$
$$\frac{\Delta l}{l}= \frac{0.001}{1}=0.001.$$

Breadth: observed $$$b = 4\ \text{cm}=0.04\ \text{m},\quad \text{least count}=0.1\ \text{mm}=0.0001\ \text{m}.$$$
$$\frac{\Delta b}{b}= \frac{0.0001}{0.04}=0.0025.$$

Thickness: observed $$$d = 0.4\ \text{cm}=0.004\ \text{m},\quad \text{least count}=0.01\ \text{mm}=0.00001\ \text{m}.$$$
$$\frac{\Delta d}{d}= \frac{0.00001}{0.004}=0.0025.$$

Depression: observed $$$\delta = 5\ \text{mm}=0.005\ \text{m},\quad \text{least count}=0.01\ \text{mm}=0.00001\ \text{m}.$$$
$$\frac{\Delta \delta}{\delta}= \frac{0.00001}{0.005}=0.002.$$

Substituting all these fractional errors into the earlier relation, we get

$$$\begin{aligned} \frac{\Delta Y}{Y} &= 0.0005 \;+\; 3(0.001) \;+\; 0.0025 \;+\; 3(0.0025) \;+\; 0.002\\[4pt] &= 0.0005 + 0.003 + 0.0025 + 0.0075 + 0.002\\[4pt] &= 0.0155. \end{aligned}$$$

Thus the fractional error in the measurement of $$Y$$ is $$0.0155$$.

Hence, the correct answer is Option D.

Young's modulus is given by $$Y = \frac{FL}{\pi r^2 \Delta L}$$, where $$F$$ is the applied force, $$L$$ is the original length, $$r$$ is the radius, and $$\Delta L$$ is the elongation.

The fractional error in $$Y$$ is obtained by adding the fractional errors of each measured quantity. Since $$r$$ appears squared, its fractional error is doubled: $$\frac{\Delta Y}{Y} = \frac{\Delta F}{F} + \frac{\Delta L}{L} + 2\frac{\Delta r}{r} + \frac{\Delta(\Delta L)}{\Delta L}$$.

The force is $$F = mg$$, so $$\frac{\Delta F}{F} = \frac{\Delta m}{m} = \frac{1 \text{ g}}{1000 \text{ g}} = 0.001$$.

The fractional error in length is $$\frac{\Delta L}{L} = \frac{1 \text{ mm}}{1000 \text{ mm}} = 0.001$$.

The fractional error in radius is $$\frac{\Delta r}{r} = \frac{0.001 \text{ cm}}{0.2 \text{ cm}} = 0.005$$, so $$2\frac{\Delta r}{r} = 0.01$$.

The fractional error in elongation is $$\frac{\Delta(\Delta L)}{\Delta L} = \frac{0.001 \text{ cm}}{0.5 \text{ cm}} = 0.002$$.

Adding all contributions: $$\frac{\Delta Y}{Y} = 0.001 + 0.001 + 0.01 + 0.002 = 0.014 = 1.4\%$$.

Therefore, the fractional error in the value of Young's Modulus is $$1.4\%$$.

We start with the Rydberg constant $$R_H$$. In the Balmer‐type formula for spectral lines we write

$$\frac{1}{\lambda}=R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right).$$

Here the right-hand factor in brackets is dimensionless, so the left-hand side and $$R_H$$ must have the same dimension. Because $$\lambda$$ is a length, $$1/\lambda$$ has the dimension $$L^{-1}$$. Therefore, in SI units,

$$[R_H]=\text{metre}^{-1}=m^{-1},$$

which is entry (iii) in List-(II).

Now we look at Planck’s constant $$h$$. The fundamental relation is

$$E=h\nu,$$

where the energy $$E$$ has dimension $$ML^{2}T^{-2}$$ (joule) and frequency $$\nu$$ has dimension $$T^{-1}$$. Solving for $$h$$ we obtain

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

that is $$\text{kilogram metre}^{2}\ \text{second}^{-1}=kg\,m^{2}\,s^{-1}$$, which corresponds to entry (ii).

Next we examine the magnetic-field energy density, denoted here by $$\mu_B$$. For a magnetic field of magnitude $$B$$ we use the formula

$$u_B=\frac{B^{2}}{2\mu_0}.$$

Energy density is energy per unit volume, so its dimension is

$$\frac{\text{joule}}{\text{metre}^{3}}=\frac{ML^{2}T^{-2}}{L^{3}}=ML^{-1}T^{-2},$$

which in SI becomes $$kg\,m^{-1}\,s^{-2}$$. This matches entry (iv).

Finally, we consider the coefficient of viscosity $$\eta$$. In fluid mechanics the viscous force is given by

$$F=\eta\,A\,\frac{dv}{dy},$$

where $$F$$ is force, $$A$$ area, and $$dv/dy$$ the velocity gradient. Rearranging,

$$\eta=\frac{F}{A}\left(\frac{dv}{dy}\right)^{-1}.$$

Force has dimension $$MLT^{-2}$$, area has $$L^{2}$$, and the velocity gradient has $$T^{-1}$$. Substituting these we get

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

namely $$kg\,m^{-1}\,s^{-1}$$, which corresponds to entry (i).

Collecting our findings:

$$(a)\;R_H\ \longrightarrow\ (iii),\qquad (b)\;h\ \longrightarrow\ (ii),\qquad (c)\;\mu_B\ \longrightarrow\ (iv),\qquad (d)\;\eta\ \longrightarrow\ (i).$$

This exact pairing appears in Option C.

Hence, the correct answer is Option C.

To decide which relation is wrong we compare the dimensions on the left-hand side (LHS) with those on the right-hand side (RHS). If they differ, the equation is dimensionally incorrect.

First we tabulate the dimensions (in fundamental symbols $$M$$ for mass, $$L$$ for length, $$T$$ for time and $$A$$ for electric current).

$$\begin{aligned} \text{Work }(W)&:&\; \text{force}\times\text{distance}= (MLT^{-2})\,(L)=ML^{2}T^{-2}\\ \text{Torque }(\tau)&:&\; \text{force}\times\text{distance}=ML^{2}T^{-2}\\ \text{Pressure }(p)&:&\; \text{force}/\text{area}=ML^{-1}T^{-2}\\ \text{Surface tension }(s)&:&\; \text{force}/\text{length}=MT^{-2}\\ \text{Density }(\rho)&:&\; ML^{-3}\\ \text{Acceleration }(g)&:&\; LT^{-2}\\ \text{Viscosity }(\eta)&:&\; ML^{-1}T^{-1}\\ \text{Volume }(V)&:&\; L^{3}\\ \text{Current density }(J)&:&\; AL^{-2}\\ \text{Electric field }(E)&:&\; MLT^{-3}A^{-1}\\ \text{Permittivity }(\epsilon)&:&\; M^{-1}L^{-3}T^{4}A^{2} \end{aligned}$$ Length $$(a,r,h,L)$$ has dimension $$L$$, the angle $$\theta$$ is dimensionless.

Option A gives $$W=\tau\theta$$.

LHS $$\rightarrow$$ $$W:ML^{2}T^{-2}$$.

RHS $$\rightarrow$$ $$\tau\theta=(ML^{2}T^{-2})(1)=ML^{2}T^{-2}$$ because $$\theta$$ is dimensionless.

The two sides match, so Option A is dimensionally correct.

Option B gives $$V=\dfrac{\pi p a^{4}}{8\eta L}$$.

LHS $$\rightarrow$$ $$V:L^{3}$$.

RHS step by step:

We have $$p\,a^{4}$$ with dimensions $$\bigl(ML^{-1}T^{-2}\bigr)\,(L^{4}) = ML^{3}T^{-2}.$$

We divide by $$\eta L$$ whose dimensions are $$\bigl(ML^{-1}T^{-1}\bigr)\,(L)=MT^{-1}.$$

Therefore $$ \dfrac{p\,a^{4}}{\eta L}:\; \dfrac{ML^{3}T^{-2}}{MT^{-1}} =\dfrac{ML^{3}T^{-2}}{M}\,T^{1} =L^{3}T^{-1}. $$

So the RHS carries an extra factor $$T^{-1}$$ and has dimension $$L^{3}T^{-1}$$, whereas the LHS is only $$L^{3}$$. They differ, making this relation dimensionally inconsistent.

Option C gives $$h=\dfrac{2s\cos\theta}{\rho r g}$$.

LHS $$\rightarrow$$ $$h:L$$.

RHS numerator $$2s\cos\theta$$ has dimension $$MT^{-2}$$ (since $$\cos\theta$$ is dimensionless).

Denominator $$\rho r g$$ has dimension $$\bigl(ML^{-3}\bigr)\,(L)\,(LT^{-2})=ML^{-1}T^{-2}.$$ Hence $$ \dfrac{MT^{-2}}{ML^{-1}T^{-2}}=L, $$ matching the LHS. Option C is correct.

Option D gives $$J=\epsilon\dfrac{\partial E}{\partial t}$$.

LHS $$\rightarrow$$ $$J:AL^{-2}$$.

First write the time derivative: $$ \dfrac{\partial E}{\partial t} :\; \frac{MLT^{-3}A^{-1}}{T} = MLT^{-4}A^{-1}. $$

Now multiply by $$\epsilon$$: $$ \epsilon\dfrac{\partial E}{\partial t} =\bigl(M^{-1}L^{-3}T^{4}A^{2}\bigr)\bigl(MLT^{-4}A^{-1}\bigr) =L^{-2}A, $$ which is exactly $$AL^{-2}$$, so the dimensions tally. Option D is also correct.

Only Option B fails the dimensional test. Hence, the correct answer is Option B.

We have to find the dimensional formulas of the four physical quantities of List I and then compare them with the four possibilities given in List II. The dimensional symbols used are $$M$$ for mass, $$L$$ for length, $$T$$ for time and $$A$$ for electric current. After obtaining each formula we will match it with the corresponding item (i, ii, iii, iv) given in the problem.

Capacitance $$C$$
The defining relation is $$C=\dfrac{Q}{V}$$ where $$Q$$ is charge and $$V$$ is potential difference.
First, the dimension of charge is written from $$Q = I\,t,$$ so  $$[Q]=A\,T.$$ Next, the potential difference $$V$$ is work done per unit charge, $$V=\dfrac{W}{Q}.$$ Work (or energy) has the mechanical dimensions of force into distance, so $$[W]=M\,L^{2}\,T^{-2}.$$ Dividing by charge gives
$$[V]=\dfrac{M\,L^{2}\,T^{-2}}{A\,T}=M\,L^{2}\,T^{-3}\,A^{-1}.$$ Now substitute these in the formula for $$C$$:
$$[C]=\dfrac{A\,T}{M\,L^{2}\,T^{-3}\,A^{-1}} =A\,T\;M^{-1}\,L^{-2}\,T^{3}\,A^{1} =M^{-1}\,L^{-2}\,T^{4}\,A^{2}.$$ Thus capacitance corresponds to $$M^{-1}L^{-2}T^{4}A^{2},$$ which is item (iii) of List II.

Permittivity of free space $$\varepsilon_{0}$$
Coulomb’s law states $$F=\dfrac{1}{4\pi\varepsilon_{0}}\dfrac{q_{1}q_{2}}{r^{2}}.$$ Rearranging gives
$$\varepsilon_{0}=\dfrac{q_{1}q_{2}}{4\pi\,r^{2}\,F}.$$
Ignoring the numerical constant $$4\pi,$$ the dimensional expression is
$$[\varepsilon_{0}]=\dfrac{(A\,T)^{2}}{L^{2}\,(M\,L\,T^{-2})} =\dfrac{A^{2}\,T^{2}}{M\,L^{3}\,T^{-2}} =M^{-1}\,L^{-3}\,T^{4}\,A^{2}.$$ Hence $$\varepsilon_{0}$$ matches item (ii).

Permeability of free space $$\mu_{0}$$
For two long, straight, parallel conductors the magnetic force per unit length is
$$\dfrac{F}{l}=\dfrac{\mu_{0}}{2\pi}\dfrac{I_{1}I_{2}}{r}.$$
Rearranging yields $$\mu_{0}=\dfrac{F/l\;\,2\pi\,r}{I^{2}}.$$ Omitting the constant $$2\pi$$ we have
$$[\mu_{0}]=\dfrac{M\,T^{-2}\,L}{A^{2}} =M\,L\,T^{-2}\,A^{-2}.$$ Therefore $$\mu_{0}$$ corresponds to item (iv).

Electric field $$E$$
The electric field is force per charge: $$E=\dfrac{F}{Q}.$$ Using $$[F]=M\,L\,T^{-2}$$ and $$[Q]=A\,T,$$ we get
$$[E]=\dfrac{M\,L\,T^{-2}}{A\,T}=M\,L\,T^{-3}\,A^{-1}.$$ This equals item (i).

Collecting the results:
a. $$C$$ → (iii)    b. $$\varepsilon_{0}$$ → (ii)    c. $$\mu_{0}$$ → (iv)    d. $$E$$ → (i)

This sequence (a-iii, b-ii, c-iv, d-i) is exactly the pairing given in Option A.

Hence, the correct answer is Option A.

We need to find the dimensional formula for each quantity in List - I and match it with List - II.

Planck's constant $$h$$ has the dimension of energy multiplied by time. Since energy has dimensions $$[ML^2T^{-2}]$$, we get $$[h] = [ML^2T^{-2}] \times [T] = [ML^2T^{-1}]$$, which matches with (ii).

Kinetic energy $$E = \frac{1}{2}mv^2$$ has dimensions $$[M][LT^{-1}]^2 = [ML^2T^{-2}]$$, which matches with (iii).

Electric potential $$V = \frac{W}{q}$$, where work $$W$$ has dimensions $$[ML^2T^{-2}]$$ and charge $$q$$ has dimensions $$[IT]$$. Therefore $$[V] = \frac{[ML^2T^{-2}]}{[IT]} = [ML^2I^{-1}T^{-3}]$$, which matches with (iv).

Linear momentum $$P = mv$$ has dimensions $$[M][LT^{-1}] = [MLT^{-1}]$$, which matches with (i).

Therefore the correct matching is (a)→(ii), (b)→(iii), (c)→(iv), (d)→(i), which corresponds to Option (1).

We start by recalling that in dimensional analysis we express any physical quantity in terms of the fundamental dimensions: mass $$M$$, length $$L$$ and time $$T$$. The exponents of these symbols tell us how the quantity depends on each fundamental dimension.

For torque we use the definition “torque = force × perpendicular distance.” First we write the dimensional formula of force. From Newton’s second law, “force = mass × acceleration,” and acceleration is “length / time2.” Hence

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

Multiplying by distance (another factor of $$L$$) gives

$$[\text{Torque}] = M^1L^1T^{-2}\; \times \; L^1 = M^1L^{1+1}T^{-2} = ML^{2}T^{-2}.$$

So torque corresponds to $$ML^{2}T^{-2},$$ which is entry (iii).

For impulse we recall the relation “impulse = force × time.” Stating the dimensional formula once more, force is $$MLT^{-2}$$ and time is $$T^1$$. Hence

$$[\text{Impulse}] = (MLT^{-2}) \times (T^1) = ML\,T^{-2+1} = MLT^{-1}.$$

Thus impulse matches $$MLT^{-1},$$ which is entry (i).

For tension we simply note that tension is a kind of force (a pulling force in a string). Therefore its dimensional formula is identical to that of force:

$$[\text{Tension}] = MLT^{-2}.$$

This equals entry (iv).

For surface tension we remember the definition “surface tension = force per unit length.” We again insert the force dimension $$MLT^{-2}$$ and divide by length $$L^1$$:

$$[\text{Surface Tension}] = \dfrac{MLT^{-2}}{L} = MT^{-2}.$$

This matches entry (ii).

Collecting all the matches, we have

$$(a) \longrightarrow (iii), \quad (b) \longrightarrow (i), \quad (c) \longrightarrow (iv), \quad (d) \longrightarrow (ii).$$

Comparing with the options supplied, we see that Option B lists exactly this set of pairings.

Hence, the correct answer is Option B.

The given force is $$F = A \cos (B x) + C \sin (D t)$$.

Dimensions of force are $$[F] = [M \, L \, T^{-2}]$$.

Inside trigonometric functions the arguments are pure numbers (dimensionless). For the first term, $$B x$$ is dimensionless, therefore
$$[B] \,[x] = 1 \;\Longrightarrow\; [B] = [x]^{-1} = L^{-1}$$.

For the second term, $$D t$$ is dimensionless, therefore
$$[D] \,[t] = 1 \;\Longrightarrow\; [D] = [t]^{-1} = T^{-1}$$.

The quantities $$A$$ and $$C$$ multiply the trigonometric functions, so their dimensions must match the dimension of force: $$[A] = [C] = [F] = M \, L \, T^{-2}$$.

We need the dimensional formula of $$\dfrac{A D}{B}$$.
Write each factor with its dimensions:
$$[A] = M \, L \, T^{-2}, \quad [D] = T^{-1}, \quad [B] = L^{-1}$$.

Multiply and divide step by step:
$$[A]\, [D] = (M \, L \, T^{-2})(T^{-1}) = M \, L \, T^{-3}$$ $$-(1)$$
Now divide by $$[B]$$:
$$\dfrac{M \, L \, T^{-3}}{L^{-1}} = M \, L \, T^{-3} \times L^{1} = M \, L^{2} \, T^{-3}$$ $$-(2)$$.

Therefore, the dimensional formula of $$\dfrac{A D}{B}$$ is $$[M \, L^{2} \, T^{-3}]$$.

Option B is correct.

We recall the basic definitions for a screw gauge.

First, the pitch is the distance moved by the spindle on the main scale in one complete rotation of the circular scale. Algebraically,

$$\text{Pitch}=\frac{\text{Distance travelled on main scale}}{\text{Number of complete rotations}}.$$

In the given situation, the spindle advances $$5\ \text{mm}$$ on the main scale when the circular scale is rotated completely five times. Substituting these data, we have

$$\text{Pitch}=\frac{5\ \text{mm}}{5}=1\ \text{mm}.$$

Next, the least count (L.C.) of the screw gauge is obtained from the relation

$$\text{Least Count}=\frac{\text{Pitch}}{\text{Total divisions on the circular scale}}.$$

This formula is exactly what the Reason $$R$$ states.

We are told that the circular scale possesses $$50$$ total divisions. Substituting the numerical values,

$$\text{L.C.}=\frac{1\ \text{mm}}{50}=0.02\ \text{mm}.$$

To express this least count in centimetres we use the conversion $$1\ \text{mm}=0.1\ \text{cm}.$$ Hence,

$$0.02\ \text{mm}=0.02\times0.1\ \text{cm}=0.002\ \text{cm}.$$

So the calculated least count equals $$0.002\ \text{cm}$$ and not $$0.001\ \text{cm}$$ as claimed in Assertion $$A$$.

Therefore, Assertion $$A$$ is incorrect, whereas Reason $$R$$ is a correct and relevant statement.

Hence, the correct answer is Option A.

First, we recall the basic relations for a screw-gauge.

The pitch is the distance advanced by the spindle in one full rotation of the circular scale. It is given as $$0.5\ \text{mm}$$.

The least count (LC) is defined by the formula $$\text{LC} = \dfrac{\text{pitch}}{\text{number of circular-scale divisions}}.$$

We have $$\text{number of divisions} = 50,$$ so

$$\text{LC} = \dfrac{0.5\ \text{mm}}{50} = 0.01\ \text{mm}.$$

Now we determine the zero error. When the ratchet is fully closed (no object between the anvils), the reading should be zero, but the fifth division of the circular scale coincides with the reference line. Hence the instrument shows an apparent reading of

$$\text{zero error} = 5 \times \text{LC} = 5 \times 0.01\ \text{mm} = 0.05\ \text{mm}.$$

This reading is greater than the true value (which should be $$0$$), so it is a positive zero error. The zero correction is therefore negative:

$$\text{zero correction} = -0.05\ \text{mm}.$$

For the given observation we note the two parts of the apparent reading:

• Main-scale reading (MSR): $$5\ \text{mm}.$$

• Circular-scale reading (CSR): $$20$$ divisions.

The circular-scale contribution equals

$$\text{CSR} \times \text{LC} = 20 \times 0.01\ \text{mm} = 0.20\ \text{mm}.$$

So the observed reading becomes

$$\text{observed reading} = \text{MSR} + (\text{CSR} \times \text{LC}) = 5\ \text{mm} + 0.20\ \text{mm} = 5.20\ \text{mm}.$$

To obtain the true reading we apply the zero correction:

$$\text{true reading} = \text{observed reading} + \text{zero correction} = 5.20\ \text{mm} + (-0.05\ \text{mm}) = 5.15\ \text{mm}.$$

Hence, the correct answer is Option C.

We are given that one main scale division (MSD) of the vernier callipers is $$a$$ cm, and that $$n$$ divisions of the vernier scale coincide with $$(n-1)$$ divisions of the main scale.

From the given condition: $$n \times \text{VSD} = (n-1) \times \text{MSD}$$, which gives $$\text{VSD} = \frac{(n-1)}{n} \times a$$ cm.

The least count is $$\text{LC} = 1 \text{ MSD} - 1 \text{ VSD} = a - \frac{(n-1)}{n} \, a = a\left(1 - \frac{n-1}{n}\right) = a \times \frac{1}{n} = \frac{a}{n}$$ cm.

Converting to millimeters: $$\text{LC} = \frac{a}{n} \times 10 = \frac{10a}{n}$$ mm.

The pitch of the screw gauge is 1 mm and there are 100 divisions on the circular scale, so the least count is $$\text{LC} = \frac{1}{100} = 0.01$$ mm.

When nothing is placed between the jaws, the zero of the circular scale lies 8 divisions below the reference line. This means the zero error is positive: $$\text{Zero error} = +8 \times 0.01 = +0.08$$ mm.

When the wire is placed between the jaws, the first linear scale division is clearly visible, giving a main scale reading of 1 mm. The 72nd circular scale division coincides with the reference line, giving a circular scale reading of $$72 \times 0.01 = 0.72$$ mm.

The observed reading is $$1 + 0.72 = 1.72$$ mm.

The corrected reading (diameter of the wire) is $$\text{Observed reading} - \text{Zero error} = 1.72 - 0.08 = 1.64$$ mm.

The radius of the wire is $$\frac{1.64}{2} = 0.82$$ mm.

The correct answer is Option (2): 0.82 mm.

The vernier reading is computed as: main scale reading + (vernier coincidence $$\times$$ least count) $$-$$ zero error. The main scale reading (MSR) is 8.5 cm since the zero of the vernier lies between 8.5 cm and 8.6 cm. The vernier coincidence is 6 and the least count is 0.01 cm.

The observed reading is $$8.5 + 6 \times 0.01 = 8.5 + 0.06 = 8.56$$ cm. Since the instrument has a positive zero error of 0.2 mm = 0.02 cm, the corrected reading is $$8.56 - 0.02 = 8.54$$ cm.

The correct answer is option 2: 8.54 cm.

For a simple pendulum executing small-angle oscillations the well-known time-period formula is

$$T \;=\;2\pi\sqrt{\dfrac{L}{g}},$$

where $$L$$ is the length of the pendulum and $$g$$ is the local acceleration due to gravity.

Because $$L$$ is not given directly, we first express it in terms of the measured quantities $$T$$ and $$g$$. Rearranging the above equation, we obtain

$$L \;=\;\dfrac{gT^{2}}{4\pi^{2}}.$$

Next, we write the expression for the mechanical energy that has to be supplied to the pendulum so that it oscillates with a (small) angular amplitude $$\theta_{0}$$. The total energy in one cycle equals the maximum potential energy at the extreme position:

$$E \;=\;mgh,$$

with $$h$$ being the vertical rise of the bob. For small angles $$h\approx\dfrac{L\theta_{0}^{2}}{2}$$, therefore

$$E \;=\;mg\left(\dfrac{L\theta_{0}^{2}}{2}\right) \;=\;\dfrac{1}{2}\,m\,g\,L\,\theta_{0}^{2}.$$

Substituting $$L=\dfrac{gT^{2}}{4\pi^{2}}$$ from the earlier step, we get

$$E \;=\;\dfrac{1}{2}\,m\,g\left(\dfrac{gT^{2}}{4\pi^{2}}\right)\theta_{0}^{2} \;=\;\dfrac{m\theta_{0}^{2}}{8\pi^{2}}\;g^{2}T^{2}.$$

Thus the energy depends on the measurable quantities as

$$E \;\propto\;g^{2}\,T^{2}.$$

To find the percentage error in $$E$$ we use the rule for propagation of errors in a product of powers: if $$Q \propto a^{p}b^{q}$$, then $$\dfrac{\Delta Q}{Q}\;=\;|p|\dfrac{\Delta a}{a} + |q|\dfrac{\Delta b}{b}.$$

Here $$p=2$$ for $$g$$ and $$q=2$$ for $$T$$. The given percentage uncertainties are

$$\dfrac{\Delta g}{g}=4\% , \qquad \dfrac{\Delta T}{T}=3\%.$$

Therefore,

$$\dfrac{\Delta E}{E} =2\left(\dfrac{\Delta g}{g}\right) +2\left(\dfrac{\Delta T}{T}\right) =2(4\%) + 2(3\%) =8\% + 6\% =14\%. $$

So, the answer is $$14\%$$.

For a simple pendulum, $$g = \dfrac{4\pi^2 L}{T^2}$$ where $$T$$ is the time period. The percentage error in $$g$$ is:

$$\frac{\Delta g}{g} \times 100 = \frac{\Delta L}{L} \times 100 + 2\,\frac{\Delta T}{T} \times 100$$

where $$T = t_{\text{total}}/n$$, so $$\Delta T = \Delta t/n$$ and $$\dfrac{\Delta T}{T} = \dfrac{\Delta t}{t_{\text{total}}}$$.

The table gives: Student $$S_1$$: $$L = 1.0\,\text{m}$$, $$n = 10$$ oscillations, $$t_{\text{total}} = 20.0\,\text{s}$$; Student $$S_2$$: $$L = 0.5\,\text{m}$$, $$n = 20$$, $$t_{\text{total}} = 40.0\,\text{s}$$; Student $$S_3$$: $$L = 0.9\,\text{m}$$, $$n = 15$$, $$t_{\text{total}} = 45.0\,\text{s}$$. The least count of length is $$0.1\,\text{m}$$ and of time is $$0.1\,\text{s}$$.

For $$S_1$$: $$E_1 = \left(\dfrac{0.1}{1.0} + 2 \times \dfrac{0.1}{20.0}\right) \times 100 = (0.10 + 0.01) \times 100 = 11\%$$

For $$S_2$$: $$E_2 = \left(\dfrac{0.1}{0.5} + 2 \times \dfrac{0.1}{40.0}\right) \times 100 = (0.20 + 0.005) \times 100 = 20.5\%$$

For $$S_3$$: $$E_3 = \left(\dfrac{0.1}{0.9} + 2 \times \dfrac{0.1}{45.0}\right) \times 100 \approx (0.111 + 0.0044) \times 100 \approx 11.6\%$$

Comparing the three, $$E_1 = 11\% < E_3 \approx 11.6\% < E_2 = 20.5\%$$. The minimum percentage error is obtained by student number $$1$$.

We need to determine the magnitude of the downward negative lift $$F_L$$ acting on a racing car moving along a flat circular track.

1. Identify the Forces Acting on the Car

From the problem statement page, the car is moving in a horizontal circle of radius $$R$$ with a speed $$v$$. Let's analyze the forces acting in both vertical and horizontal directions:

• Vertical Direction:
  The forces keeping the car vertically balanced are:
  • Downward gravitational force: $$mg$$
  • Downward negative lift force (aerodynamic downforce): $$F_L$$
  • Upward normal reaction force from the track: $$N$$
  Since there is no vertical acceleration, the net vertical force is zero:

  $$N = mg + F_L \quad \text{--- (Equation 1)}$$

• Horizontal Direction:
  The centripetal force required to keep the car moving in a circle of radius $$R$$ at speed $$v$$ is provided entirely by the static friction ($$f_s$$) between the tires and the track:

  $$f_s = \frac{m v^2}{R}$$

2. Relate Friction to the Normal Force

To safely navigate the turn at speed $$v$$, the required centripetal friction must be provided by the maximum limiting static friction available ($$f_s = \mu_s N$$):

$$\mu_s N = \frac{m v^2}{R}$$

Solving for the normal force $$N$$ gives:

$$N = \frac{m v^2}{\mu_s R} \quad \text{--- (Equation 2)}$$

3. Solve for the Negative Lift ($$F_L$$)

Now, equate Equation 1 and Equation 2 to eliminate $$N$$:

$$mg + F_L = \frac{m v^2}{\mu_s R}$$

Isolate the negative lift force $$F_L$$:

$$F_L = \frac{m v^2}{\mu_s R} - mg$$

Factoring out the mass $$m$$ from both terms yields the final expression:

$$F_L = m \left( \frac{v^2}{\mu_s R} - g \right)$$

Conclusion

The magnitude of the negative lift acting downwards on the car is $$m \left( \frac{v^2}{\mu_s R} - g \right)$$, which corresponds to Option B.

We have to match each physical quantity in List - I with its dimensional formula in List - II. Let us determine the dimensions one by one, showing every algebraic step.

(a) Magnetic induction $$\mathbf B$$
The force on a moving charge is $$\mathbf F = q\,\mathbf v \times \mathbf B.$$
Using the basic dimensions $$$[\mathbf F] = ML T^{-2},\; [q] = AT,\; [\mathbf v] = LT^{-1},$$$ we write

$$$[\mathbf B] \;=\; \frac{[\mathbf F]}{[q]\,[\mathbf v]}\;=\; \frac{MLT^{-2}}{(AT)\,(LT^{-1})}.$$$

Simplifying the powers of $$L$$ and $$T$$:

$$[\mathbf B] = M T^{-2} A^{-1}.$$

This matches List-II item $$\text{(iii)}\; M T^{-2} A^{-1}.$$

(b) Magnetic flux $$\Phi_B$$
By definition $$\Phi_B = \mathbf B \cdot \text{Area}.$$ Hence

$$$[\Phi_B] = [\mathbf B]\,[\text{Area}] = \left(M T^{-2} A^{-1}\right)\left(L^{2}\right).$$$

Multiplying, we obtain

$$[\Phi_B] = M L^{2} T^{-2} A^{-1}.$$

This corresponds to List-II item $$\text{(i)}\; M L^{2} T^{-2} A^{-1}.$$

(c) Magnetic permeability $$\mu$$
The relation between magnetic induction and magnetic field strength is $$\mathbf B = \mu\,\mathbf H.$$ Rearranging gives $$\mu = \mathbf B / \mathbf H.$$

We already have $$[\mathbf B] = M T^{-2} A^{-1}.$$ Field strength $$\mathbf H$$ is current per unit length, so $$[\mathbf H] = A L^{-1}.$$

Therefore

$$$[\mu] = \frac{M T^{-2} A^{-1}}{A L^{-1}} = M L T^{-2} A^{-2}.$$$

This is exactly List-II item $$\text{(iv)}\; M L T^{-2} A^{-2}.$$

(d) Magnetization $$\mathbf M$$
Magnetization is magnetic moment per unit volume. Magnetic moment is current times area, so $$[\text{moment}] = A L^{2}.$$ Dividing by volume $$L^{3}$$ gives

$$[\mathbf M] = \frac{A L^{2}}{L^{3}} = A L^{-1}.$$

Since no mass term appears, we may write

$$[\mathbf M] = M^{0} L^{-1} A^{1}.$$

This equals List-II item $$\text{(ii)}\; M^{0} L^{-1} A.$$

Collecting the matches:

$$$ \begin{aligned} \text{a}\;&\to\;\text{(iii)},\quad \text{b}\;&\to\;\text{(i)},\quad \text{c}\;&\to\;\text{(iv)},\quad \text{d}\;&\to\;\text{(ii)}. \end{aligned} $$$

Option B lists exactly this correspondence. Hence, the correct answer is Option B.

We begin by recalling that a dimensionless quantity is one whose numerical value does not change when we change the fundamental units of measurement. In other words, if we write its dimensional formula in terms of the basic dimensions $$[M]$$ (mass), $$[L]$$ (length), $$[T]$$ (time), $$[I]$$ (electric current), and so on, every exponent must be zero, giving the overall dimensional formula $$[M^0L^0T^0I^0]$$, which we usually denote simply by $$1$$.

Now we examine each option one by one.

Option A : Power factor
The power factor is defined as $$\cos\phi$$, where $$\phi$$ is the phase angle between the alternating voltage and the alternating current. Since cosine is the ratio of two sides of a right-angled triangle in phasor representation, or more simply, the cosine of an angle, it is a pure number. Hence its dimensional formula is $$[M^0L^0T^0I^0]$$. Therefore, the power factor is dimensionless.

Option B : Quality factor
For an LCR resonant circuit, the quality factor $$Q$$ is defined by the relation $$ Q = \dfrac{\omega_0 L}{R} = \dfrac{1}{\omega_0 C R}, $$ where $$\omega_0$$ is the resonant angular frequency, $$L$$ is inductance, $$C$$ is capacitance, and $$R$$ is resistance. In every equivalent formula, the units in the numerator and denominator cancel out completely. This leaves the dimensional formula $$[M^0L^0T^0I^0]$$. Thus the quality factor is also dimensionless.

Option C : Permeability of free space $$\mu_0$$
The magnetic permeability of free space is defined from the relation $$ F = \dfrac{\mu_0 I_1 I_2 l}{2\pi r}, $$ or more commonly appears in the force law between two current-carrying conductors, in Ampère’s law, and in many electromagnetic formulas. From any such formula we can extract its dimensional formula. One very convenient starting point is the expression for the speed of light in vacuum: $$ c = \dfrac{1}{\sqrt{\mu_0 \varepsilon_0}}, $$ which gives $$ \mu_0 = \dfrac{1}{c^2 \varepsilon_0}. $$ Because $$\varepsilon_0$$ (permittivity of free space) has the dimensions $$ [\varepsilon_0] = [M^{-1}L^{-3}T^{4}I^{2}], $$ and the speed of light has dimensions $$ [c] = [LT^{-1}], $$ we substitute these into the above relation. We have $$ [\mu_0] = \dfrac{1}{[c]^2 \,[\varepsilon_0]} = \dfrac{1}{\bigl([L T^{-1}]\bigr)^2 \,[M^{-1}L^{-3}T^{4}I^{2}]} $$ $$\;\;\;= \dfrac{1}{[L^{2}T^{-2}]\,[M^{-1}L^{-3}T^{4}I^{2}]} $$ $$\;\;\;= [M^{1} L^{1} T^{-2} I^{-2}]. $$ The exponents here are not all zero, so $$\mu_0$$ is not dimensionless. Indeed, in SI units its unit is henry per metre (H m−1).

Option D : Relative magnetic permeability $$\mu_r$$
Relative permeability is defined as the ratio $$ \mu_r = \dfrac{\mu}{\mu_0}, $$ where $$\mu$$ is the absolute permeability of the medium. Because both $$\mu$$ and $$\mu_0$$ carry exactly the same dimensions, the ratio cancels those dimensions out completely. Therefore $$ [\mu_r] = \dfrac{[\mu]}{[\mu_0]} = 1, $$ so $$\mu_r$$ is dimensionless.

Gathering the conclusions:

Option A: dimensionless.
Option B: dimensionless.
Option C: has dimensions.
Option D: dimensionless.

Hence, the only quantity that is not dimensionless is the permeability of free space $$\mu_0$$, which corresponds to Option C (Option 3).

Hence, the correct answer is Option 3.

We need to determine the absolute value of the difference between the final circular scale readings observed by student A and student B when measuring the radius of a wire.

1. Calculate the Least Count (LC) of the Screw Gauges

From the problem page, both screw gauges have the same pitch and total number of circular divisions:

• Pitch = $$0.1\text{ cm}$$
• Number of circular divisions = $$100$$

The least count ($$\text{LC}$$) is calculated as:

$$\text{LC} = \frac{\text{Pitch}}{\text{Number of circular divisions}} = \frac{0.1\text{ cm}}{100} = 0.001\text{ cm}$$

2. Determine the Initial Zero Errors

Looking at the provided reference diagrams, we can determine the initial zero error for each student's instrument based on where the zero mark lies relative to the reference line:

• Screw Gauge A:
  The $$5^{\text{th}}$$ division coincides with the reference line when the studs are closed, meaning it has a positive zero error:

  $$\text{Zero Error}_A = +5 \times \text{LC} = +5 \times 0.001\text{ cm} = +0.005\text{ cm}$$

• Screw Gauge B:
  The $$92^{\text{nd}}$$ division coincides with the reference line. Since it is below the zero mark ($$100 - 92 = 8$$ divisions behind), it has a negative zero error:

  $$\text{Zero Error}_B = -(100 - 92) \times \text{LC} = -8 \times 0.001\text{ cm} = -0.008\text{ cm}$$

3. Relate Observed Readings to Actual Value

The true radius value is given as $$\text{True Reading} = 0.322\text{ cm}$$. The general formula for a measurement is:

$$\text{True Reading} = \text{Observed Reading} - \text{Zero Error}$$

$$\text{True Reading} = (\text{Main Scale Reading} + n \times \text{LC}) - \text{Zero Error}$$

Where $$n$$ represents the final circular scale reading. Let's express this for both students:

• For Student A:
  

  $$0.322 = \text{MSR}_A + (n_A \times 0.001) - 0.005$$

  $$n_A \times 0.001 = 0.327 - \text{MSR}_A$$

• For Student B:
  

  $$0.322 = \text{MSR}_B + (n_B \times 0.001) - (-0.008)$$

  $$n_B \times 0.001 = 0.314 - \text{MSR}_B$$

Since both gauges have a pitch of $$0.1\text{ cm}$$, the pitch/main scale reading ($$\text{MSR}$$) must be identical for measuring the same object ($$\text{MSR}_A = \text{MSR}_B = 0.3\text{ cm}$$). Let's solve for the exact circular scale divisions ($n_A$$ and $$n_B$$):

$$n_A \times 0.001 = 0.327 - 0.3 = 0.027 \implies n_A = 27$$

$$n_B \times 0.001 = 0.314 - 0.3 = 0.014 \implies n_B = 14$$

4. Calculate the Absolute Difference

Finally, find the absolute difference between the two circular scale readings ($$n_A$$ and $$n_B$$):

$$\text{Difference} = |n_A - n_B| = |27 - 14| = 13$$

Conclusion

The absolute value of the difference between the final circular scale readings observed by the students is 13.

We have a vernier callipers in which 9 main-scale divisions are equal in length to 10 vernier-scale divisions. First we find the length of one vernier division. One main-scale division (MSD) is given to be $$1\ \text{mm}.$$ Therefore

$$ \text{Length of one vernier division (VSD)} = \frac{9\ \text{MSD}}{10} = \frac{9 \times 1\ \text{mm}}{10} = 0.9\ \text{mm}. $$

The least count (LC) of a vernier callipers is defined as

$$ \text{LC} = 1\ \text{MSD} - 1\ \text{VSD}. $$

Substituting the values, we get

$$ \text{LC} = 1\ \text{mm} - 0.9\ \text{mm} = 0.1\ \text{mm}. $$

Now we read the scale. The main-scale reading (MSR) given is $$10\ \text{mm}.$$ The 8th vernier division coincides with a main-scale division, so the vernier-scale reading (VSR) is obtained by multiplying the coincident division number with the least count:

$$ \text{VSR} = 8 \times \text{LC} = 8 \times 0.1\ \text{mm} = 0.8\ \text{mm}. $$

The measured diameter (before zero-error correction) is therefore

$$ d_{\text{meas}} = \text{MSR} + \text{VSR} = 10\ \text{mm} + 0.8\ \text{mm} = 10.8\ \text{mm}. $$

Writing this in centimetres, since $$10\ \text{mm} = 1\ \text{cm},$$ we have

$$ d_{\text{meas}} = 10.8\ \text{mm} = 1.08\ \text{cm}. $$

The instrument possesses a positive zero error of $$0.04\ \text{cm}.$$ For a positive zero error, the true reading is obtained by subtracting the zero error from the measured reading. Hence

$$ d_{\text{true}} = d_{\text{meas}} - (\text{zero error}) = 1.08\ \text{cm} - 0.04\ \text{cm} = 1.04\ \text{cm}. $$

The radius is half of the true diameter:

$$ r = \frac{d_{\text{true}}}{2} = \frac{1.04\ \text{cm}}{2} = 0.52\ \text{cm}. $$

Expressing $$0.52\ \text{cm}$$ in the form $$N \times 10^{-2}\ \text{cm}$$ gives

$$ 0.52\ \text{cm} = 52 \times 10^{-2}\ \text{cm}. $$

Hence, the answer is $$52.$$

The entropy is $$S = \alpha^2\beta\ln\!\left[\frac{\mu kR}{J\beta^2}+3\right]$$, with $$[S] = \mathrm{J\,K^{-1}}$$. The logarithm is dimensionless, so $$[\alpha^2\beta] = [S] = \mathrm{J\,K^{-1}}$$.

For the log argument to be dimensionless: $$\left[\frac{\mu kR}{J\beta^2}\right] = 1$$. Since $$\mu$$ is dimensionless (number of moles treated as a pure number), $$[k] = \mathrm{J\,K^{-1}}$$, $$[R] = \mathrm{J\,K^{-1}}$$ (same dimension as $$k$$ for this analysis), and $$J$$ is dimensionless, we get $$[\beta^2] = [kR] = \mathrm{J^2\,K^{-2}}$$, so $$[\beta] = \mathrm{J\,K^{-1}}$$.

From $$[\alpha^2\beta] = \mathrm{J\,K^{-1}}$$ and $$[\beta] = \mathrm{J\,K^{-1}}$$, we get $$[\alpha^2] = 1$$, meaning $$\alpha$$ is dimensionless. The mechanical equivalent $$J$$ is also dimensionless, so $$\alpha$$ and $$J$$ share the same (dimensionless) dimensions — option 1 is correct. Also $$[S] = [\beta] = [k] = \mathrm{J\,K^{-1}}$$ and $$[\mu R] = \mathrm{J\,K^{-1}}$$, so option 2 is correct. $$S$$ and $$\alpha$$ differ (option 3 correct). However, option 4 claims $$\alpha$$ and $$k$$ have the same dimensions: $$[\alpha] = 1$$ (dimensionless) while $$[k] = \mathrm{J\,K^{-1}}$$ — these are different. Therefore option 4 is the incorrect statement.

We have the physical quantity $$z$$ expressed as

$$z \;=\; \frac{a^{2}\,b^{\,2/3}}{\sqrt{c}\,d^{3}}$$

(the square-root sign acts only on $$c$$, while $$d^{3}$$ is outside the root).

In error analysis we use the general rule: if a quantity can be written in the form

$$z \;=\; a^{\,m}\,b^{\,n}\,c^{\,p}\,d^{\,q},$$

then the fractional (relative) error in $$z$$ is obtained from

$$\frac{\Delta z}{z} =\; |m|\,\frac{\Delta a}{a} +\; |n|\,\frac{\Delta b}{b} +\; |p|\,\frac{\Delta c}{c} +\; |q|\,\frac{\Delta d}{d}.$$

This formula is nothing but the differential form of the logarithmic derivative $$\ln z = m\ln a + n\ln b + p\ln c + q\ln d$$ with all magnitudes taken positive because every separate error contributes additively to the maximum possible error.

Comparing the given expression with the standard form we read off the exponents directly:

$$m = 2,\qquad n = \frac23,\qquad p = -\frac12,\qquad q = -3.$$

Only the absolute values of these exponents matter for the percentage error, so we write

$$|m| = 2,\qquad |n| = \frac23,\qquad |p| = \frac12,\qquad |q| = 3.$$

The stated percentage errors in the measured quantities are

$$\frac{\Delta a}{a}\times100 = 2\%,\; \frac{\Delta b}{b}\times100 = 1.5\%,\; \frac{\Delta c}{c}\times100 = 4\%,\; \frac{\Delta d}{d}\times100 = 2.5\%.$$

Substituting these numbers into the error-propagation formula gives the percentage error in $$z$$:

$$ \frac{\Delta z}{z}\times100 = 2\,(2\%) + \left(\frac23\right)(1.5\%) + \left(\frac12\right)(4\%) + 3\,(2.5\%). $$

Now we carry out every multiplication one by one:

$$ 2\,(2\%) = 4\%, \\ \left(\frac23\right)(1.5\%) = 1\%, \\ \left(\frac12\right)(4\%) = 2\%, \\ 3\,(2.5\%) = 7.5\%. $$

Adding all these individual contributions we obtain

$$ \frac{\Delta z}{z}\times100 = 4\% + 1\% + 2\% + 7.5\% = 14.5\%. $$

So the percentage error in the calculated value of $$z$$ is $$14.5\%$$.

Hence, the correct answer is Option D.

We are given a physical quantity

$$f = \sqrt{\dfrac{h\,c^{5}}{G}}$$

To find the dimension of $$f$$, we first write the dimensional symbols of each constant in terms of the fundamental quantities mass $$M$$, length $$L$$ and time $$T$$.

• Planck’s constant $$h$$ has the dimension of action. By definition,

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

• The speed of light $$c$$ is a speed, so

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

• The universal gravitational constant $$G$$ appears in Newton’s law $$F = G\dfrac{m_{1}m_{2}}{r^{2}}$$. Therefore, recalling that force has dimension $$M\,L\,T^{-2}$$, we can write

$$[G] = \dfrac{F\,r^{2}}{M^{2}} = \dfrac{(M\,L\,T^{-2})\,L^{2}}{M^{2}} = M^{-1}\,L^{3}\,T^{-2}.$$

Now we construct the dimension of the expression under the square root.

First, the fifth power of $$c$$:

$$[c^{5}] = (L\,T^{-1})^{5} = L^{5}\,T^{-5}.$$

Multiplying by $$h$$:

$$[h\,c^{5}] = (M\,L^{2}\,T^{-1})\,(L^{5}\,T^{-5}) = M\,L^{7}\,T^{-6}.$$

Next, we divide by $$G$$. Dividing by a quantity is equivalent to multiplying by its reciprocal, so we use

$$[G]^{-1} = (M^{-1}\,L^{3}\,T^{-2})^{-1} = M\,L^{-3}\,T^{2}.$$

Thus,

$$\left[\dfrac{h\,c^{5}}{G}\right] = (M\,L^{7}\,T^{-6})\,(M\,L^{-3}\,T^{2}) = M^{2}\,L^{4}\,T^{-4}.$$

Finally, we take the square root because $$f$$ is the square root of that product:

$$[f] = \sqrt{M^{2}\,L^{4}\,T^{-4}} = M^{1}\,L^{2}\,T^{-2}.$$

The dimension $$M\,L^{2}\,T^{-2}$$ is precisely the dimension of energy (or work), since work $$= \text{force}\times\text{distance} = (M\,L\,T^{-2})\times L$$.

Hence, the correct answer is Option 2.

We are told that the physical quantity $$x$$ is expressed as

$$x=\dfrac{I\,F\,v^{2}}{W\,L^{4}}$$

where the symbols have the following usual meanings:

$$I=$$ moment of inertia,   $$F=$$ force,   $$v=$$ velocity,   $$W=$$ work,   $$L=$$ length.

Our task is to obtain the dimensional formula of $$x$$ and then compare it with the dimensions of the physical quantities listed in the options.

First, we write down the standard dimensional formulas (using $$[M]$$ for mass, $$[L]$$ for length and $$[T]$$ for time):

• Moment of inertia $$I$$ is mass × length$$^{2}$$, so $$[I]=[M^{1}L^{2}T^{0}]$$.

• Force $$F$$ follows Newton’s second law $$F=ma$$, giving $$[F]=[M^{1}L^{1}T^{-2}]$$.

• Velocity $$v$$ is length per time, therefore $$[v]=[M^{0}L^{1}T^{-1}]$$. Consequently $$[v^{2}]=[M^{0}L^{2}T^{-2}]$$.

• Work $$W$$ equals force × displacement, hence $$[W]=[M^{1}L^{2}T^{-2}]$$.

• Length $$L$$ of course has $$[L]=[M^{0}L^{1}T^{0}]$$, so $$[L^{4}]=[M^{0}L^{4}T^{0}]$$.

Now we substitute each of these into the defining expression for $$x$$:

Numerator dimensions:

$$[I\,F\,v^{2}]=[I]\,[F]\,[v^{2}]= \big[M^{1}L^{2}T^{0}\big]\, \big[M^{1}L^{1}T^{-2}\big]\, \big[M^{0}L^{2}T^{-2}\big].$$

Multiplying the like powers of $$M,\,L,\,T$$ one by one, we obtain

• For mass: $$M^{1}\times M^{1}\times M^{0}=M^{2}$$,

• For length: $$L^{2}\times L^{1}\times L^{2}=L^{5}$$,

• For time: $$T^{0}\times T^{-2}\times T^{-2}=T^{-4}$$.

So, $$[I\,F\,v^{2}]=[M^{2}L^{5}T^{-4}]$$.

Denominator dimensions:

$$[W\,L^{4}]=[W]\,[L^{4}]= \big[M^{1}L^{2}T^{-2}\big]\, \big[M^{0}L^{4}T^{0}\big] = [M^{1}L^{6}T^{-2}].$$

We now divide the numerator by the denominator to get the dimensions of $$x$$:

$$[x]=\dfrac{[M^{2}L^{5}T^{-4}]}{[M^{1}L^{6}T^{-2}]}.$$

Carrying out the division term by term, we subtract the powers:

• Mass exponent: $$2-1=1 \;\Rightarrow\; M^{1},$$

• Length exponent: $$5-6=-1 \;\Rightarrow\; L^{-1},$$

• Time exponent: $$-4-(-2)=-2 \;\Rightarrow\; T^{-2}.$$

Hence, $$[x]=[M^{1}L^{-1}T^{-2}].$$

Finally, we match this result with the dimensions of the quantities in the options:

(A) Planck’s constant $$h$$ has $$[M^{1}L^{2}T^{-1}]$$ - does not match.

(B) Force constant (spring constant $$k=F/\Delta L$$) has $$[M^{1}T^{-2}]$$ - does not match.

(C) Energy density (energy per unit volume) is $$\dfrac{\text{energy}}{\text{volume}}$$.

The dimensional formula of energy is $$[M^{1}L^{2}T^{-2}]$$ and of volume is $$[L^{3}]$$. Therefore

$$\dfrac{[M^{1}L^{2}T^{-2}]}{[L^{3}]}=[M^{1}L^{-1}T^{-2}],$$

which exactly matches $$[x]$$.

(D) Coefficient of viscosity $$\eta$$ has $$[M^{1}L^{-1}T^{-1}]$$ - does not match.

So the only option with the same dimensional formula as $$x$$ is energy density.

Hence, the correct answer is Option C.

We begin with the physical definition of thermal conductivity, usually denoted by the symbol $$k$$ (some books use $$K$$). According to Fourier’s law of heat conduction, the rate at which heat flows through a slab is proportional to the area, the temperature gradient, and the thermal conductivity. In symbols, the law is written as

$$\frac{Q}{t}=k\,A\,\frac{\Delta T}{\Delta x}$$

where

$$Q$$ = total heat conducted (a form of energy),
$$t$$ = time taken for that conduction,
$$A$$ = cross-sectional area of the slab,
$$\Delta T$$ = temperature difference across the slab, and
$$\Delta x$$ = thickness of the slab.

Re-arranging this expression to isolate $$k$$, we get

$$k=\frac{Q\,\Delta x}{A\,\Delta T\,t}$$

To obtain the dimensional formula of $$k$$, we now replace every quantity by its basic dimensions.

First, we recall that heat $$Q$$ is a form of energy, and the dimensional formula for energy is the same as that for work: $$[Q]=ML^{2}T^{-2}$$.

The remaining quantities have the following dimensions:

$$[\Delta x]=L,\qquad [A]=L^{2},\qquad [\Delta T]=K,\qquad [t]=T.$$

Substituting each of these into the expression for $$k$$ gives

$$[k]=\frac{[Q]\,[\Delta x]}{[A]\,[\Delta T]\,[t]} =\frac{\left(ML^{2}T^{-2}\right)\,L}{L^{2}\,K\,T}$$

Now we simplify the powers of $$L$$ and $$T$$ step by step:

$$[k]=\frac{M\,L^{2}T^{-2}\,L}{L^{2}\,K\,T} =\frac{M\,L^{3}\,T^{-2}}{L^{2}\,K\,T}$$

Dividing the exponents of like dimensions in numerator and denominator, we obtain

$$[k]=M\,L^{3-2}\,T^{-2-1}\,K^{-1} =M\,L^{1}\,T^{-3}\,K^{-1}$$

This can be written compactly as

$$[k]=MLT^{-3}K^{-1}$$

Comparing this with the given options, we see that it matches Option D.

Hence, the correct answer is Option D.

The problem asks us to compare the algebraic sums $$A_i+B_i+C_i$$ of four different data-sets and then decide which of the four statements given in the options describes their order correctly.

Whenever we add measured quantities we must obey the rule of significant figures meant for addition and subtraction.

Rule (state first): While adding or subtracting, the final result is written only up to that number of digits after the decimal point which is possessed by the addend having the least number of digits after the decimal point.

Now we evaluate every sum one by one, showing all arithmetic steps and then applying the above rule.

Set (i): $$A_1=24.36,\;B_1=0.0724,\;C_1=256.2$$

Raw sum (ordinary addition, no rounding yet):

$$24.36 + 0.0724 = 24.4324$$

$$24.4324 + 256.2 = 280.6324$$

The three addends possess respectively 2, 4 and 1 decimal places; the least is 1. So we round the result to one place after the decimal:

$$280.6324 \longrightarrow 280.6$$

Set (ii): $$A_2=24.44,\;B_2=16.082,\;C_2=240.2$$

Raw sum:

$$24.44 + 16.082 = 40.522$$

$$40.522 + 240.2 = 280.722$$

Here the addends have 2, 3 and 1 decimal places; the least is again 1. So we round to one decimal place:

$$280.722 \longrightarrow 280.7$$

Set (iii): $$A_3=25.2,\;B_3=19.2812,\;C_3=236.183$$

Raw sum:

$$25.2 + 19.2812 = 44.4812$$

$$44.4812 + 236.183 = 280.6642$$

The numbers have 1, 4 and 3 decimal places; the least is 1. Rounding to one decimal place gives:

$$280.6642 \longrightarrow 280.7$$

Set (iv): $$A_4=25,\;B_4=236.191,\;C_4=19.5$$

Raw sum:

$$25 + 236.191 = 261.191$$

$$261.191 + 19.5 = 280.691$$

Now the addends have 0, 3 and 1 decimal places; the least is 0. Hence the answer must be rounded to no decimal places at all:

$$280.691 \longrightarrow 281$$

We therefore have the four correctly rounded totals

$$\begin{aligned} A_1+B_1+C_1 &= 280.6,\\ A_2+B_2+C_2 &= 280.7,\\ A_3+B_3+C_3 &= 280.7,\\ A_4+B_4+C_4 &= 281. \end{aligned}$$

Arranging them in ascending order of magnitude gives

$$280.6 \; < \; 280.7 = 280.7 \; < \; 281,$$

or, written with the original symbols,

$$A_1+B_1+C_1 \; < \; A_2+B_2+C_2 \;=\; A_3+B_3+C_3 \; < \; A_4+B_4+C_4.$$

This statement is exactly what Option C expresses.

Hence, the correct answer is Option C.

We wish to evaluate the dimensional formula of the expression $$\dfrac{B^{2}}{2\mu_{0}}$$. The numerical factor $$2$$ is dimensionless, so the task reduces to finding the dimensions of $$\dfrac{B^{2}}{\mu_{0}}$$.

We begin with the magnetic induction $$B$$. By definition, the magnetic force on a straight conductor of length $$l$$ carrying current $$I$$ in a magnetic field $$B$$ is given by the formula

$$F = B\,I\,l\,\sin\theta.$$

Taking the maximum case $$\sin\theta = 1$$, we have $$B = \dfrac{F}{I\,l}$$. We now substitute the dimensional symbols:

$$[F] = M\,L\,T^{-2},\qquad [I] = I,\qquad [l] = L.$$

Hence

$$[B] \;=\; \dfrac{M\,L\,T^{-2}}{I\,L}\;=\;M\,T^{-2}\,I^{-1}.$$

Next we consider the magnetic permeability of free space $$\mu_{0}$$. It is related to $$B$$ and the magnetic field strength $$H$$ by the relation $$B = \mu_{0}H$$. The field strength $$H$$ is defined as current per unit length, so

$$[H] \;=\; \dfrac{I}{L} \;=\; I\,L^{-1}.$$

Therefore,

$$[\mu_{0}] \;=\; \dfrac{[B]}{[H]} \;=\; \dfrac{M\,T^{-2}\,I^{-1}}{I\,L^{-1}} \;=\; M\,L\,T^{-2}\,I^{-2}.$$

We now square the dimension of $$B$$:

$$[B^{2}] \;=\; \bigl(M\,T^{-2}\,I^{-1}\bigr)^{2} \;=\; M^{2}\,T^{-4}\,I^{-2}.$$

Dividing this by $$[\mu_{0}]$$ gives

$$\left[\dfrac{B^{2}}{\mu_{0}}\right] \;=\; \dfrac{M^{2}\,T^{-4}\,I^{-2}}{M\,L\,T^{-2}\,I^{-2}} \;=\; M\,L^{-1}\,T^{-2}.$$

Thus the complete expression $$\dfrac{B^{2}}{2\mu_{0}}$$ has the same dimension, because the factor $$2$$ is dimensionless:

$$\left[\dfrac{B^{2}}{2\mu_{0}}\right] \;=\; M\,L^{-1}\,T^{-2}.$$

This corresponds to Option D.

Hence, the correct answer is Option D.

We know that in the usual MLT system the dimensional formula for mechanical energy (work) is $$[E]=[M\,L^{2}\,T^{-2}]$$, because work is force times distance and force itself is mass times acceleration.

Now the problem tells us to treat momentum $$P$$, area $$A$$ and time $$T$$ as the new fundamental quantities. Therefore we must rewrite the above $$[M\,L^{2}\,T^{-2}]$$ in terms of $$P$$, $$A$$ and $$T$$ only.

First we recall the standard dimensional expressions of the given fundamental quantities in the MLT system:

Momentum is defined as mass times velocity, so $$[P]=[M\,L\,T^{-1}].$$

Area is the product of two lengths, hence $$[A]=[L^{2}].$$

Time is already fundamental, so $$[T]=[T].$$

Let us now assume that the energy dimension can be built as a product of powers of these three fundamental quantities. We therefore write $$[E]=P^{x}\,A^{y}\,T^{z},$$ where $$x$$, $$y$$ and $$z$$ are the unknown exponents to be determined.

Next, we substitute the MLT expressions of $$P$$, $$A$$ and $$T$$ into this assumed form:

$$$ [E]=\bigl(M^{1}L^{1}T^{-1}\bigr)^{x}\; \bigl(L^{2}\bigr)^{y}\; \bigl(T^{1}\bigr)^{z}. $$$

Collecting the powers of each fundamental symbol $$M$$, $$L$$ and $$T$$ separately, we get

$$$ [E]=M^{\,x}\; L^{\,x+2y}\; T^{\,-x+z}. $$$

This combined exponent form must equal the original energy dimension $$M^{1}L^{2}T^{-2}$$, so we equate the corresponding powers term by term:

For mass $$M$$: $$$ x = 1. $$$

For length $$L$$: $$$ x + 2y = 2. $$$

For time $$T$$: $$$ -x + z = -2. $$$

We now solve these three simple linear equations step by step.

From the first equation we immediately have $$$ x = 1. $$$

Substituting $$x=1$$ into the second equation gives $$$ 1 + 2y = 2 \quad\Longrightarrow\quad 2y = 1 \quad\Longrightarrow\quad y = \frac{1}{2}. $$$

Substituting $$x=1$$ into the third equation gives $$$ -1 + z = -2 \quad\Longrightarrow\quad z = -1. $$$

Thus the required exponents are $$$ x = 1,\qquad y = \tfrac12,\qquad z = -1. $$$

Putting these values back into the assumed form $$P^{x}A^{y}T^{z}$$, we obtain

$$$ [E]=P^{1}\,A^{1/2}\,T^{-1}. $$$

Therefore the dimensional formula for energy, when momentum, area and time are chosen as fundamental quantities, is written as $$[PA^{1/2}T^{-1}].$$

Looking at the given options, this matches Option C.

Hence, the correct answer is Option C.

We know that Young’s modulus $$Y$$ is defined as the ratio of stress to strain. Because strain is a pure number (dimensionless), the dimension of $$Y$$ is the same as the dimension of stress.

First, let us recall that stress is force per unit area, so

$$\text{Stress}=\dfrac{\text{Force}}{\text{Area}}.$$

In the ordinary $$M\!L\!T$$ (mass-length-time) system the dimensional formula of force is $$M\,L\,T^{-2},$$ while the dimensional formula of area is $$L^{2}.$$ Therefore, the dimensional formula of Young’s modulus in the $$M\!L\!T$$ system becomes

$$[Y]=\dfrac{M\,L\,T^{-2}}{L^{2}}=M\,L^{-1}\,T^{-2}.$$

Now the question asks us to re-express this same quantity using a new set of fundamental units: speed $$V,$$ area $$A$$ and force $$F.$$ We write the unknown dimensional formula as

$$[Y]=F^{\,a}\,A^{\,b}\,V^{\,c},$$

where $$a,\,b,\,c$$ are the exponents we must determine.

Next we translate each new fundamental unit back into the familiar $$M\!L\!T$$ language:

$$[F]=M\,L\,T^{-2},\qquad [A]=L^{2},\qquad [V]=L\,T^{-1}.$$

Substituting these into $$F^{\,a}\,A^{\,b}\,V^{\,c}$$ gives

$$F^{\,a}\,A^{\,b}\,V^{\,c}=\left(M\,L\,T^{-2}\right)^{a}\left(L^{2}\right)^{b}\left(L\,T^{-1}\right)^{c}=M^{a}\,L^{\,a+2b+c}\,T^{-2a-c}.$$

This result must match the already known $$M\!L\!T$$ exponents of Young’s modulus, namely $$M^{1}\,L^{-1}\,T^{-2}.$$ Hence we equate exponents term by term:

Mass exponent: $$a=1.$$(1)

Length exponent: $$a+2b+c=-1.$$(2)

Time exponent: $$-2a-c=-2.$$(3)

Using equation (1) $$a=1$$ inside equation (3) gives

$$-2(1)-c=-2\;\;\Longrightarrow\;\;-2-c=-2\;\;\Longrightarrow\;\;c=0.$$

With $$a=1$$ and $$c=0,$$ substitute into equation (2):

$$1+2b+0=-1\;\;\Longrightarrow\;\;2b=-2\;\;\Longrightarrow\;\;b=-1.$$

Thus we have determined

$$a=1,\qquad b=-1,\qquad c=0.$$

Putting these values back, the dimensional formula of Young’s modulus in the $$F\!A\!V$$ system is

$$[Y]=F^{1}\,A^{-1}\,V^{0}=F\,A^{-1}.$$

This matches option D, which is written as $$FA^{-1}V^{0}.$$

Hence, the correct answer is Option D.

First we recall the symbols for the fundamental (base) dimensions that we will use throughout the solution:

$$$[M] \;{\rm for\; mass},\qquad [L] \;{\rm for\; length},\qquad [T] \;{\rm for\; time},\qquad [I] \;{\rm for\; electric\; current}.$$$

We now determine the dimensions of the two electromagnetic constants that occur in the question.

For the permeability of free space we start from the defining relation $$B \;=\; \mu_0\,H.$$ Here $$B$$ is the magnetic flux density and $$H$$ is the magnetic field intensity. In SI units $$B$$ (tesla) has the dimension $$$[B] \;=\; \frac{{\rm force}}{{\rm current}\times{\rm length}} \;=\; \frac{[M\,L\,T^{-2}]}{[I]\,[L]} \;=\; [M\,T^{-2}\,I^{-1}],$$$ and $$H$$ (ampere per metre) has the dimension $$[H]\;=\;[I\,L^{-1}].$$ Therefore $$$[\mu_0] \;=\; \frac{[B]}{[H]} \;=\; \frac{[M\,T^{-2}\,I^{-1}]}{[I\,L^{-1}]} \;=\; [M\,L\,T^{-2}\,I^{-2}].$$$

For the permittivity of free space we employ Coulomb’s law $$F \;=\; \frac{1}{4\pi\varepsilon_0}\,\frac{q_1q_2}{r^2},$$ so that $$\varepsilon_0 \;=\; \frac{q_1q_2}{4\pi\,F\,r^2}.$$ Ignoring numerical constants, its dimension is $$$[\varepsilon_0] \;=\; \frac{[Q]^2}{[M\,L\,T^{-2}]\,[L]^2} \;=\; \frac{[I^2\,T^2]}{[M\,L^3\,T^{-2}]} \;=\; [M^{-1}\,L^{-3}\,T^{4}\,I^{2}].$$$

With these preliminaries completed, we are ready to analyse each of the three given quantities.

1. Dimension of $$x=\dfrac{1}{\sqrt{\mu_0\varepsilon_0}}.$$

First multiply the dimensions of $$\mu_0$$ and $$\varepsilon_0$$:

$$$[\mu_0\varepsilon_0] \;=\;[M\,L\,T^{-2}\,I^{-2}]\times[M^{-1}\,L^{-3}\,T^{4}\,I^{2}] \;=\;L^{(1-3)}\,T^{(-2+4)}\,M^{(1-1)}\,I^{(-2+2)} \;=\;[L^{-2}\,T^{2}].$$$

The square root of this product therefore has the dimension

$$\sqrt{[\mu_0\varepsilon_0]} \;=\;[L^{-1}\,T^{1}].$$

Because $$x$$ is the reciprocal of that square root, we invert the above result:

$$[x]\;=\;\frac{1}{[L^{-1}\,T^{1}]}\;=\;[L^{1}\,T^{-1}].$$

This is exactly the dimension of speed (length per time).

2. Dimension of $$y=\dfrac{E}{B}.$$

We already know $$[B]=[M\,T^{-2}\,I^{-1}].$$ Let us now find $$[E]$$, the electric field intensity. By definition $$E \;=\; \frac{\text{force}}{\text{charge}},$$ so $$$[E] \;=\;\frac{[M\,L\,T^{-2}]}{[I\,T]} \;=\;[M\,L\,T^{-3}\,I^{-1}].$$$

Dividing $$E$$ by $$B$$ gives

$$$[y] \;=\;\frac{[M\,L\,T^{-3}\,I^{-1}]}{[M\,T^{-2}\,I^{-1}]} \;=\;[L\,T^{-1}].$$$

Thus $$y$$ also has the dimension of speed.

3. Dimension of $$z=\dfrac{l}{C\,R}.$$

The length $$l$$ has the obvious dimension $$[l]=[L].$$

For the capacitance $$C$$ we use the definition $$C \;=\; \frac{Q}{V},$$ and voltage (potential difference) satisfies $$$V \;=\; \frac{\text{work}}{\text{charge}} \;=\; \frac{[M\,L^{2}\,T^{-2}]}{[I\,T]} \;=\;[M\,L^{2}\,T^{-3}\,I^{-1}].$$$ Hence

$$$[C] \;=\;\frac{[I\,T]}{[M\,L^{2}\,T^{-3}\,I^{-1}]} \;=\;[M^{-1}\,L^{-2}\,T^{4}\,I^{2}].$$$

For resistance $$R$$, Ohm’s law gives $$R = V/I,$$ so

$$$[R] \;=\;\frac{[M\,L^{2}\,T^{-3}\,I^{-1}]}{[I]} \;=\;[M\,L^{2}\,T^{-3}\,I^{-2}].$$$

The product $$C\,R$$ therefore possesses the dimension

$$$[C\,R] \;=\;[M^{-1}\,L^{-2}\,T^{4}\,I^{2}] \times[M\,L^{2}\,T^{-3}\,I^{-2}] \;=\;L^{(-2+2)}\,T^{(4-3)}\,M^{(-1+1)}\,I^{(2-2)} \;=\;[T^{1}].$$$

Finally, dividing the length $$l$$ by this product yields

$$[z] \;=\;\frac{[L]}{[T]} \;=\;[L\,T^{-1}].$$

Thus $$z$$ again has the dimension of speed.

We have shown that

$$$[x]=[L\,T^{-1}],\qquad [y]=[L\,T^{-1}],\qquad [z]=[L\,T^{-1}].$$$

All three quantities share the identical dimension (that of velocity).

Hence, the correct answer is Option A.

$$\text{Pitch} = 0.5\text{ mm},\quad \text{Divisions} = 50$$

$$\text{LC} = \frac{0.5\text{ mm}}{50} = 0.01\text{ mm} = 10\ \mu\text{m}$$

Since the circular scale is 4 units ahead of the pitch scale marking prior to use, it registers a non-zero positive reading when the studs are in contact.

$$\text{Zero Error} = +4 \times \text{LC} = +40\ \mu\text{m}\quad \text{(Positive Error)}$$

$$\text{Least Count} = 0.01\text{ mm}$$

Since the measurement cannot be more precise than the instrument's capability, both the average value and the uncertainty (standard deviation) must be rounded off to two decimal places:

$$\text{Average diameter} = 5.5375\text{ mm} \approx 5.54\text{ mm}$$

$$\text{Uncertainty} = 0.07395\text{ mm} \approx 0.07\text{ mm}$$

$$\text{Diameter} = (5.54 \pm 0.07)\text{ mm}$$

First, we recall what the pitch of a screw is. The pitch is defined as the linear distance moved by the tip of the screw for one complete rotation of the circular head.

According to the data given, six complete rotations of the screw make the tip advance through a distance of $$3\; \text{mm}$$ on the main scale. Hence, the pitch can be found by dividing the total linear movement by the number of rotations:

$$\text{Pitch} \;=\; \frac{\text{Total linear advance}}{\text{Number of rotations}}$$

Substituting the numbers, we get

$$\text{Pitch} \;=\; \frac{3\;\text{mm}}{6}$$

Simplifying the fraction,

$$\text{Pitch} \;=\; 0.5\;\text{mm}$$

Now we state the formula for the least count (L.C.) of a screw-gauge. The least count is the smallest length that can be measured with the instrument and is given by

$$\text{Least Count} \;=\; \frac{\text{Pitch}}{\text{Number of divisions on the circular scale}}$$

We have already found the pitch to be $$0.5\;\text{mm}$$, and we are told that the circular scale possesses $$50$$ equal divisions. Substituting these values into the formula, we obtain

$$\text{Least Count} \;=\; \frac{0.5\;\text{mm}}{50}$$

Carrying out the division gives

$$\text{Least Count} \;=\; 0.01\;\text{mm}$$

Usually, we may also wish to express the least count in centimetres for comparison with the answers listed. Since $$1\;\text{mm} \;=\; 0.1\;\text{cm},$$ we multiply $$0.01\;\text{mm}$$ by $$0.1$$ to convert the unit:

$$0.01\;\text{mm} \;=\; 0.01 \times 0.1\;\text{cm} \;=\; 0.001\;\text{cm}$$

Looking at the options, the value $$0.001\;\text{cm}$$ matches Option A.

Hence, the correct answer is Option A.

We are told that the least count of the main scale of the vernier calipers is $$1\ \text{mm}$$. Converting into centimetres first, we note that $$1\ \text{mm}=0.1\ \text{cm}$$, so each main-scale division (MSD) is $$0.1\ \text{cm}$$.

Next, the statement “10 vernier scale divisions (VSD) coincide with 9 main scale divisions” gives us the actual size of one vernier division. If 10 VSD equal 9 MSD, then one VSD is

$$1\ \text{VSD}=\frac{9\ \text{MSD}}{10}= \frac{9\times 0.1\ \text{cm}}{10}=0.09\ \text{cm}.$$

The least count (LC) of a vernier is defined by the formula

$$\text{Least Count}=1\ \text{MSD}-1\ \text{VSD}.$$

Substituting the values just obtained, we have

$$\text{LC}=0.1\ \text{cm}-0.09\ \text{cm}=0.01\ \text{cm}.$$

This $$0.01\ \text{cm}$$ (or $$0.1\ \text{mm}$$) is the resolution with which measurements can be refined by the vernier scale.

Now we examine the zero error. When the jaws touch each other, the seventh vernier division (i.e., $$n=7$$) lines up with a main-scale division and, importantly, the zero of the vernier lies to the right of the zero of the main scale. Whenever the vernier zero is to the right, the instrument is giving a reading even though the true length is zero; hence the zero error is positive.

The magnitude of that error equals the number of the coinciding division multiplied by the least count:

$$\text{Zero error}=n\times\text{LC}=7\times 0.01\ \text{cm}=0.07\ \text{cm}.$$

Because the error is positive, the zero correction (what we must algebraically add to any observed reading) is the negative of this value, namely $$-0.07\ \text{cm}.$$

We now proceed to the actual measurement of the cylinder. The problem states that the zero of the vernier scale lies between $$3.1\ \text{cm}$$ and $$3.2\ \text{cm}.$$ Therefore the main-scale reading (MSR)—the value just to the left of the vernier zero—is

$$\text{MSR}=3.1\ \text{cm}.$$

Further, the 4$$^{\text{th}}$$ vernier division coincides with a main-scale mark, so

$$\text{Vernier scale reading (VSR)}=4\times\text{LC}=4\times 0.01\ \text{cm}=0.04\ \text{cm}.$$

Adding the two parts together gives the observed reading (OR):

$$\text{OR}=\text{MSR}+\text{VSR}=3.1\ \text{cm}+0.04\ \text{cm}=3.14\ \text{cm}.$$

Finally, we must apply the zero correction. Since the correction is $$-0.07\ \text{cm},$$ we obtain the true length (TL) as

$$\text{TL}=\text{OR}+\text{correction}=3.14\ \text{cm}+(-0.07\ \text{cm})=3.07\ \text{cm}.$$

Hence, the correct answer is Option C.

We are dealing with a screw gauge. The very first thing to note is its pitch and the number of divisions on its circular scale.

The statement gives us

$$\text{Pitch}=0.1\ \text{cm}$$

and

$$\text{Number of circular scale divisions}=50.$$

For any screw gauge the smallest length it can measure - that is, the least count - is given by the standard relation

$$\text{Least Count}=\frac{\text{Pitch}}{\text{Number of divisions on circular scale}}.$$

Substituting the given values, we obtain

$$\text{Least Count}= \frac{0.1\ \text{cm}}{50} = 0.002\ \text{cm}.$$

This value tells us that every reading taken with this screw gauge must differ from the next possible reading by an exact multiple of $$0.002\ \text{cm}.$$ In simpler words, any valid final answer has to end in the thousandths place such that the last three-digit block is a multiple of 2 in the third decimal.

Now we inspect the four options:

$$\begin{aligned} \text{Option A:}&\;2.121\ \text{cm}\\ \text{Option B:}&\;2.124\ \text{cm}\\ \text{Option C:}&\;2.125\ \text{cm}\\ \text{Option D:}&\;2.123\ \text{cm} \end{aligned}$$

We test each by subtracting the integer part and checking divisibility by the least count:

$$2.121-2.000 = 0.121\ \text{cm},\quad\frac{0.121}{0.002}=60.5\;(\text{not an integer}).$$

$$2.124-2.000 = 0.124\ \text{cm},\quad\frac{0.124}{0.002}=62\;(\text{exact integer}).$$

$$2.125-2.000 = 0.125\ \text{cm},\quad\frac{0.125}{0.002}=62.5\;(\text{not an integer}).$$

$$2.123-2.000 = 0.123\ \text{cm},\quad\frac{0.123}{0.002}=61.5\;(\text{not an integer}).$$

Only Option B, $$2.124\ \text{cm},$$ is an exact multiple of the least count and therefore respects the precision of the screw gauge.

Hence, the correct answer is Option B.

We have to determine the maximum percentage error in the density of a solid metal sphere when its mass and diameter are measured with known percentage (relative) errors.

First, we recall the basic definition of density.

Density $$\rho$$ is defined by the formula

$$\rho \;=\; \frac{\text{mass}}{\text{volume}} \;=\; \frac{m}{V}.$$

For a sphere whose diameter is $$d$$, the radius is $$r = \dfrac{d}{2}$$. The volume of a sphere is

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

Substituting $$r = \dfrac{d}{2}$$, we find

$$V \;=\; \frac{4}{3}\pi\left(\frac{d}{2}\right)^{3} \;=\; \frac{4}{3}\pi\frac{d^{3}}{8} \;=\; \frac{\pi d^{3}}{6}.$$

Now we write the density explicitly in terms of the mass $$m$$ and the diameter $$d$$.

$$\rho \;=\; \frac{m}{\dfrac{\pi d^{3}}{6}} \;=\; \frac{6m}{\pi d^{3}}.$$

Hence $$\rho$$ is proportional to $$m$$ and inversely proportional to $$d^{3}$$. In algebraic form we may write

$$\rho \;\propto\; m^{\;1}\,d^{-3}.$$

To find the maximum (worst-case) percentage error in $$\rho$$, we use the well-known rule for propagation of relative errors for products and quotients:

When a quantity $$Q$$ depends on measured quantities as $$Q = A^{\,\alpha}B^{\,\beta}$$, the maximum relative error in $$Q$$ is

$$\frac{\Delta Q}{Q} \;=\; |\alpha|\frac{\Delta A}{A} \;+\; |\beta|\frac{\Delta B}{B}.$$

Applying this rule to $$\rho \propto m^{1}d^{-3}$$, we obtain

$$\frac{\Delta\rho}{\rho} \;=\; 1\bigl(\frac{\Delta m}{m}\bigr) + 3\bigl(\frac{\Delta d}{d}\bigr).$$

The problem states that the percentage (relative) errors in measuring the mass and the diameter are respectively

$$\frac{\Delta m}{m} = 6.0\%, \qquad \frac{\Delta d}{d} = 1.5\%.$$

Substituting these values, we get

$$\frac{\Delta\rho}{\rho} \;=\; 1(6.0\%) + 3(1.5\%) \;=\; 6.0\% + 4.5\% \;=\; 10.5\%.$$

According to the wording of the question, this maximum error is written in the form

$$\left(\frac{x}{100}\right)\%.$$

Equating the two expressions for the percentage error, we have

$$\left(\frac{x}{100}\right)\% = 10.5\%.$$

Removing the percent sign (which merely indicates “per hundred”) and equating the numerical parts, we find

$$\frac{x}{100} = 10.5 \;\;\Longrightarrow\;\; x = 10.5 \times 100 = 1050.$$

So, the answer is $$1050$$.

We have the dimensional equation $$X = 5\,Y\,Z^{2}$$. The numeral 5 is dimensionless, so the dimensions of the physical quantities must satisfy

$$[X] = [Y]\,[Z]^{2}\,.$$

Hence, to find the dimensions of $$Y$$ we need $$[Y] = \dfrac{[X]}{[Z]^{2}}.$$

Step 1 - Dimensions of $$X$$ (Capacitance)
Capacitance $$C$$ is defined by the relation $$C = \dfrac{Q}{V}$$, where $$Q$$ is electric charge and $$V$$ is potential difference.

The dimension of charge is current × time: $$[Q] = A\,T.$$

Voltage is work per unit charge: $$V = \dfrac{W}{Q}.$$ Work (or energy) has the dimensions of $$M\,L^{2}\,T^{-2}$$, so

$$[V] = \dfrac{M\,L^{2}\,T^{-2}}{A\,T} = M\,L^{2}\,T^{-3}\,A^{-1}\,.$$

Therefore, the dimensional formula of capacitance is

$$[C] = \dfrac{A\,T}{M\,L^{2}\,T^{-3}\,A^{-1}} = M^{-1}\,L^{-2}\,T^{4}\,A^{2}\,.$$

Since $$X$$ has the dimensions of capacitance, we write

$$[X] = M^{-1}\,L^{-2}\,T^{4}\,A^{2}\,.$$

Step 2 - Dimensions of $$Z$$ (Magnetic field)
Magnetic field $$B$$ is defined through the Lorentz force $$F = q\,v\,B$$ on a moving charge. Rearranging, $$B = \dfrac{F}{q\,v}.$$

The force dimension is $$[F] = M\,L\,T^{-2}\,,$$ charge is $$[q] = A\,T\,,$$ and speed is $$[v] = L\,T^{-1}\,.$$ Therefore

$$[B] = \dfrac{M\,L\,T^{-2}}{A\,T \, \cdot \, L\,T^{-1}} = \dfrac{M\,L\,T^{-2}}{A\,L} = M\,T^{-2}\,A^{-1}\,.$$

This is the dimensional formula of a magnetic field (tesla). Thus

$$[Z] = M\,T^{-2}\,A^{-1}\,.$$

Step 3 - Square of $$Z$$ and its inverse

Squaring $$Z$$ gives

$$[Z]^{2} = (M\,T^{-2}\,A^{-1})^{2} = M^{2}\,T^{-4}\,A^{-2}\,.$$

The reciprocal of this quantity, which we need for $$\dfrac{1}{[Z]^{2}}$$, is

$$[Z]^{-2} = M^{-2}\,T^{4}\,A^{2}\,.$$

Step 4 - Dimensions of $$Y$$

Now substitute the results into $$[Y] = \dfrac{[X]}{[Z]^{2}}$$: