Problems in Physics(Part—1)

 

Q1. Planck’s constant has dimensions ………, fill the blank

Sol. Planck's constant (h) has dimensions of energy × time, which is equivalent to action.

In terms of fundamental dimensions (Mass [M], Length [L], Time [T]), Planck’s constant has the dimensional formula:

[h]=[M][L]²[T]⁻¹

Explanation:

Planck’s constant relates the energy (E) of a photon to its frequency (ν) via the equation:

E=hν

• Energy (E) has dimensions [M][L]²[T]⁻² (same as work or force × distance).
• Frequency (ν) has dimensions [T]⁻¹.

Thus, solving for h:

[h]=[E]/[ν]=[M][L]²[T]⁻²/[T]⁻¹=[M][L]²[T]⁻¹

Summary:

The blank can be filled as:
"Planck’s constant has dimensions [M][L]²[T]⁻¹."

Or, in words:
"Planck’s constant has dimensions of energy multiplied by time (or action)."

Q2. In the formula X = 3YZ² X^, and Z have dimensions of capacitance and magnetic induction, respectively. The dimensions of Y in MKSQ system are ……….

 

Sol. We are given the equation:

X=3YZ²

where:

• Z has the dimensions of magnetic induction (B).
• X has the dimensions of capacitance (C).

We need to find the dimensions of Y in the MKSQ system (Meter-Kilogram-Second-Quantum Charge).

Step 1: Recall the dimensions of X (capacitance) and Z (magnetic induction)

1. Capacitance (C) has the dimensional formula:

[C]=[M]⁻¹[L]⁻²[T]²[Q]²

(since capacitance C=Q/V, and voltage V=W/Q, where W is work).

2. Magnetic induction (B) has the dimensional formula:

[B]=[M][T] ⁻¹[Q]⁻¹

(Since F=QvB, and force [F]=[M][L][T] ⁻².

Step 2: Rewrite the equation in terms of dimensions

Given X=3YZ², the dimensions must satisfy:

[X]=[Y][Z]²

Substitute the known dimensions:

[M]⁻¹[L]⁻²[T]²[Q]²=[Y]⋅([M][T]⁻¹[Q]⁻¹)²

Step 3: Solve for [Y]

Simplify [Z]²:

[Z]²=[M]²[T]⁻²[Q]⁻²

Now, isolate [Y]:

[Y]=[X]/[Z]²=[M]⁻¹[L]⁻²[T]²[Q]² /[M]²[T]⁻²[Q]⁻²

Combine exponents:

[Y]=[M]⁻¹⁻²[L]⁻²[T]²⁻⁽⁻²⁾ [Q]²⁻⁽⁻²⁾

[Y]=[M]⁻³[L]⁻²[T]⁴[Q]⁴

Final Answer:

The dimensions of Y in the MKSQ system are:

[M]⁻³[L]⁻²[T]⁴[Q]⁴

Interpretation:

This means Y has dimensions of inverse cubic mass × inverse square length × time⁴ × charge⁴. Such combinations appear in certain electromagnetic or quantum mechanical relations.

Q3. The dimensions of electrical conductivity is ……….

 

Sol. The dimensions of electrical conductivity (σ) can be derived from its definition.

Key Formula:

Electrical conductivity (σ) is defined as:

σ=J/E

where:

• J = current density ([J]=[I][L] ⁻²,
• E = electric field ([E]=[V][L] ⁻¹.

Step-by-Step Derivation:

1. Current density (J) has dimensions:

[J]=[I][L] ⁻²

(Since current I is charge per time [Q][T] ⁻¹, and area is [L]².

2. Electric field (E) has dimensions:

[E]=[V]/[L]=[M][L]²[T]⁻³[I]⁻¹/[L]=[M][L][T] ⁻³[I]⁻¹

(Since voltage [V]=[W]/[Q]=[M][L]²[T]⁻² /[I][T] =[M][L]²[T]⁻³[I]⁻¹.

3. Conductivity (σ=J/E):

[σ]=[I][L] ⁻² /[M][L][T] ⁻³[I]⁻¹=[I]²[L]⁻³[M]⁻¹[T]³

Since [I]=[Q]/[T]⁻¹, we can express in terms of charge [Q]:

[σ]=[Q]²[T]⁻²[L]⁻³[M]⁻¹[T]³=[M]⁻¹[L]⁻³[T][Q]²

Final Dimensions:

In the MKSQ system (Meter-Kilogram-Second-Quantum Charge), the dimensions of electrical conductivity are:

[M]⁻¹[L]⁻³[T][Q]²

Equivalent Forms:

• In terms of current ([I]) instead of charge ([Q]):

[σ]=[M]⁻¹[L]⁻³[T]³[I]²

• In SI base units, it simplifies to Siemens per meter (S/m), where 1 S = Ω⁻¹.

Verification:

From Ohm's Law (R=V/I) and resistivity (ρ=RA/L​), conductivity (σ=1/ρ​) should match the derived result.

Thus, the correct dimensional formula is confirmed.

Q4. The equation of state of a real gas is given by (1997)

 

(P+a/V²) (V-b) =RT

 

where p, V and T are pressure, volume and temperature, respectively and R is the universal gas constant. The dimensions of the constant a in the above equation is…….

 

Sol. To determine the dimensions of the constant aa in the given real gas equation of state:

Given Equation:

(p+aV²) (V−b) =RT

where:

• p = pressure,
• V = volume,
• T = temperature,
• R = universal gas constant,
• a, b = van der Waals constants.

Step 1: Identify the dimensions of each term

1. Pressure (p):

[p]=[M][L] ⁻¹[T]⁻²

(since pressure = force/area, and force = mass × acceleration).

2. Volume (V):

[V]=[L]³

3. Universal gas constant (R):

[R]=[M][L]²[T]⁻²[K]⁻¹

(from the ideal gas equation PV=nRT).

4. Temperature (T):

[T]=[K]]

(Kelvin, a base unit in thermodynamics).

Step 2: Analyse the term aV²​

The term (p+aV²) implies that a/V² must have the same dimensions as pressure (p) for dimensional consistency.

Thus:

[a/V²] =[p]=[M][L] ⁻¹[T]⁻²

Step 3: Solve for [a]

[a]=[p][V]²=[M][L] ⁻¹[T]⁻²×[L]⁶=[M][L]⁵[T]⁻²

Final Dimensions of a:

[M][L]⁵[T]⁻²

Verification:

In the van der Waals equation, aa accounts for intermolecular forces and is known to have these dimensions. This matches standard results in thermodynamics.

Summary:

The constant aa has dimensions of:

• Mass ([M])
• Length to the fifth power ([L]⁵)
• Time to the inverse square ([T]⁻²)

This ensures the equation is dimensionally consistent.

Q5. In the formula X = 3Y Z², X and Z have dimensions of capacitance and magnetic induction, respectively. What are the dimensions of Y in MKSQ system? (1995)

 

(a) [M^–3 L^–1 T ³Q⁴] (b) [M^–3 L ^–2 T⁴ Q ⁴]

 

(c) [M ^-2 L^-2 T ⁴Q⁴] (d) [M^-3 L^-2 T⁴ Q]

 

Sol. We have the formula:

X=3YZ²

Given:

• X has dimensions of capacitance.
• Z has dimensions of magnetic induction.
• We need to find the dimensions of Y in the MKSQ system (Meter, Kilogram, Second, Coulomb).

First, let's recall the dimensions of capacitance and magnetic induction in the MKSQ system.

Dimensions of Capacitance (X)

Capacitance (C) is given by:

C=Q/V

where:

• Q is charge (dimension [Q]),
• V is voltage (potential difference).

Voltage (V) is work per unit charge:

V=W/Q

Work (W) is force times distance:

W=F⋅d

Force (F) is mass times acceleration:

F=m⋅a

So, putting it all together:

[V]=[W]/[Q]=[F][d]/[Q]=[M][L][T] ⁻²[L]/[Q]=[M][L]²[T]⁻²[Q]⁻¹

Therefore, capacitance:

[C]=[Q]/[V]=[Q]/[M][L]²[T]⁻²[Q]⁻¹=[M]⁻¹[L]⁻²[T]²[Q]²

So, the dimensions of X (capacitance) are:

[X]=[M]⁻¹[L]⁻²[T]²[Q]²

Dimensions of Magnetic Induction (Z)

Magnetic induction (B) is related to the force on a moving charge:

F=qvB

where:

• F is force ([M][L][T] ⁻²,
• q is charge ([Q]),
• v is velocity ([L][T] ⁻¹).

So:

[B]=[F]/[q][v] =[M][L][T] ⁻² /[Q][L][T] ⁻¹=[M][T] ⁻¹[Q]⁻¹

Therefore, the dimensions of Z (magnetic induction) are:

[Z]=[M][T] ⁻¹[Q]⁻¹

Expressing Y in Terms of X and Z

Given:

X=3YZ²

The constant 3 is dimensionless, so:

[X]=[Y][Z]²

Solving for [Y]:

[Y]=[X]/[Z]²

Now, compute [Z]²:

[Z]²=([M][T] ⁻¹[Q]⁻¹)²=[M]²[T]⁻²[Q]⁻²

Now, divide [X] by [Z]²:

[Y]=[M]⁻¹[L]⁻²[T]²[Q]² /[M]²[T]⁻²[Q]⁻²

When dividing, subtract the exponents:

• For [M]: −1−2=−3
• For [L]: −2−0=−2
• For [T]: 2−(−2) =4
• For [Q]: 2−(−2) =4

So:

[Y]=[M]⁻³[L]⁻²[T]⁴[Q]⁴

Matching with the Given Options

Now, let's look at the options:

(a) [M⁻³L⁻¹T³Q⁴]
(b) [M⁻³L⁻²T⁴Q⁴]
(c) [M⁻²L⁻²T⁴Q⁴]
(d) [MLTQ]

Our derived dimensions for Y are [M]⁻³[L]⁻²[T]⁴[Q]⁴, which matches option (b).

Verifying the Answer

Just to ensure I didn't make a mistake, let's quickly verify:

Given:

[Y]=[M]⁻³[L]⁻²[T]⁴[Q]⁴

Now, let's reconstruct [X] from [Y] and [Z]²:

[X]=[Y][Z]²=[M]⁻³[L]⁻²[T]⁴[Q]⁴×[M]²[T]⁻²[Q]⁻²

Adding exponents:

• [M]: −3+2=−1
• [L]: −2+0=−2
• [T]: 4−2=2
• [Q]: 4−2=2

So:

[X]=[M]⁻¹[L]⁻²[T]²[Q]²

Which matches the dimensions of capacitance we derived earlier. So, our answer is correct.

Final Answer

The correct option is:
(b) [M⁻³L⁻²T⁴Q⁴]