1. The dimension of \(\dfrac{B^2}{2\mu_0}\), where \(B\) is the magnetic field and \(\mu_0\) is the magnetic permeability of vacuum, is:
1. \([ML^{-1}T^{-2}]\)
2. \([ML^2T^{-1}]\)
3. \([MLT^{-2}]\)
4. \([ML^2T^{-2}]\)
Solution:
The quantity \(\dfrac{B^2}{2\mu_0}\) represents magnetic energy stored per unit volume (Energy Density).
Energy Density \(=\dfrac{\text{Energy}}{\text{Volume}}\)
\[
\left[\frac{B^2}{\mu_0}\right] = \frac{ML^2T^{-2}}{L^3} = \boxed{ML^{-1}T^{-2}}
\]
Answer: 1. \([ML^{-1}T^{-2}]\)
2. The dimension of stopping potential \(V_0\) in the photoelectric effect in units of Planck’s constant \(h\), speed of light \(c\), gravitational constant \(G\) and Ampere \(A\) is:
1. \(h^2G^{3/2}c^{1/3}A^{-1}\)
2. \(h^{-2/3}c^{-1/3}G^{4/3}A^{-1}\)
3. \(h^{1/3}G^{2/3}c^{1/3}A^{-1}\)
4. \(h^{0}c^{5}G^{-1}A^{-1}\)
Solution:
The stopping potential \( V_0 \) is the minimum negative (retarding) voltage applied to the collector plate that is just sufficient to stop the most energetic photoelectrons from reaching it, thereby reducing the photoelectric current to zero.
At this stage, the work done by the electric field equals the maximum kinetic energy:
\[
K_{\text{max}} = eV_0
\]
- \( K_{\text{max}} \): Maximum kinetic energy
- \( e \): Elementary charge \( (1.602 \times 10^{-19} \, \text{C}) \)
- \( V_0 \): Stopping potential
Let \(V_0 = h^x\, c^y\, G^z\, A^w\).
Dimensions of stopping potential \(= \dfrac{ML^2T^{-2}}{AT}\)
Known:
\([h]=ML^2T^{-1},\quad [c]=LT^{-1},\quad [G]=M^{-1}L^3T^{-2}\)
Equations:
Comparing powers of \(M, L, T, A\) on both sides:
\[
w = -1,\quad x = 0,\quad y = 5,\quad z = -1
\]
Result:
\(V_0 = h^0\,c^5\,G^{-1}\,A^{-1}\)
3. A quantity \(f\) is given by \(f = \sqrt{\dfrac{hc^5}{G}}\), where \(c\) is the speed of light, \(G\) is the universal gravitational constant and \(h\) is Planck’s constant. The dimension of \(f\) is that of:
1. Momentum
2. Area
3. Energy
4. Volume
Solution:
Known:
\([h]=ML^2T^{-1},\quad [c]=LT^{-1},\quad [G]=M^{-1}L^3T^{-2}\)
\[
[f] = \sqrt{\frac{ML^2T^{-1}\cdot L^5T^{-5}}{M^{-1}L^3T^{-2}}}
= \sqrt{\frac{M^2L^4T^{-4}}{1}}
= M^1L^2T^{-2}
\]
Compare:
\([\text{Energy}] = ML^2T^{-2}\)
Answer: (3) Energy
4. If speed \(V\), area \(A\) and force \(F\) are chosen as fundamental units, then the dimension of Young’s Modulus will be:
1. \(FA^{-1}V^{0}\)
2. \(FA^2V^{-1}\)
3. \(FA^2V^{-3}\)
4. \(FA^2V^{-2}\)
Solution:
Dimension: Young’s Modulus \(Y\) has dimension \([ML^{-1}T^{-2}]\)
Let:
\(Y = F^x A^y V^z\)
\[
ML^{-1}T^{-2} = [MLT^{-2}]^x \cdot [L^2]^y \cdot [LT^{-1}]^z
\]
\[
= M^x\, L^{x+2y+z}\, T^{-2x-z}
\]
Comparing powers of \(M, L, T\): \(x=1,\; y=-1,\; z=0\)
Answer: (1) \(FA^{-1}V^0\)
5. If momentum \(P\), area \(A\) and time \(T\) are taken to be the fundamental quantities, then the dimensional formula for energy is:
1. \([PA^{-1}T^{-2}]\)
2. \([P^{1/2}A^2T^{-1}]\)
3. \([P^2AT^{-2}]\)
4. \([P^{1/2}AT^{-1}]\)
Solution:
Let:
\([E] = P^x A^y T^z\)
\[
ML^2T^{-2} = [MLT^{-1}]^x\,[L^2]^y\,[T]^z
= M^x\,L^{x+2y}\,T^{-x+z}
\]
Solve:
From \(M\): \(x=1\)
From \(L\): \(1+2y=2 \Rightarrow y=\frac{1}{2}\)
From \(T\): \(-1+z=-2 \Rightarrow z=-1\)
\[
[E] = \left[P^1\,A^{1/2}\,T^{-1}\right]
\]
6. Amount of solar energy received on the earth’s surface per unit area per unit time is defined as the solar constant. The dimension of solar constant is:
1. \([ML^2T^{-2}]\)
2. \([MLT^{-2}]\)
3. \([M^2L^0T^{-1}]\)
4. \([ML^0T^{-3}]\)
Solution:
To find the dimension of the solar constant \( S \), we use its definition:
\[
S = \frac{\text{Energy}}{\text{Area} \times \text{Time}}
\]
- Energy (\( E \)): \( [M L^2 T^{-2}] \)
- Area (\( A \)): \( [L^2] \)
- Time (\( t \)): \( [T] \)
Calculation:
Substitute the dimensions into the formula:
\[
[S] = \frac{[M L^2 T^{-2}]}{[L^2] \times [T]}
\]
Cancel \( L^2 \) and combine powers of \( T \):
\[
[S] = [M L^{(2-2)} T^{(-2-1)}]
\]
\[
[S] = [M L^0 T^{-3}]
\]
7. A quantity \(x\) is given by \(\dfrac{IFv^2}{WL^4}\) in terms of moment of inertia \(I\), force \(F\), velocity \(v\), work \(W\) and length \(L\). The dimensional formula for \(x\) is the same as that of:
1. Planck’s constant
2. Force constant
3. Coefficient of viscosity
4. Energy density
Solution:
Known:
\([I]=ML^2,\quad [F]=MLT^{-2},\quad [v]=LT^{-1},\quad [W]=ML^2T^{-2},\quad [L]=L\)
\[
[x] = \frac{ML^2 \cdot MLT^{-2} \cdot L^2T^{-2}}{ML^2T^{-2} \cdot L^4}
= \frac{M^2L^5T^{-4}}{ML^6T^{-2}}
= ML^{-1}T^{-2}
\]
Compare:
Energy Density \(= \dfrac{E}{V} = \dfrac{ML^2T^{-2}}{L^3} = ML^{-1}T^{-2}\)
Answer: (4) Energy Density
8. The quantities \(x\), \(y\) and \(z\) are defined as:
\[
x = \frac{1}{\sqrt{\mu_0\varepsilon_0}},\qquad
y = \frac{E}{B},\qquad
z = \frac{l}{RC}
\]
where \(C\) = capacitance, \(R\) = resistance, \(l\) = length, \(E\) = electric field, \(B\) = magnetic field, \(\varepsilon_0,\mu_0\) = free-space permittivity and permeability. Which of the following is true?
1. Only \(x\) and \(y\) have the same dimension
2. \(x\), \(y\) and \(z\) have the same dimension
3. Only \(x\) and \(z\) have the same dimension
4. Only \(y\) and \(z\) have the same dimension
Solution:
\(x\):
\(x = \dfrac{1}{\sqrt{\mu_0\varepsilon_0}} = c\) ⇒ speed of light
\([x] = LT^{-1}\)
\(y\):
\(y = \dfrac{E}{B}\) ⇒ ratio gives speed
\([y] = LT^{-1}\)
\(z\):
\(z = \dfrac{l}{RC}\), where \(RC = \tau\) (time constant)
\([z] = \dfrac{L}{T} = LT^{-1}\)
\[
[x] = [y] = [z] = LT^{-1}
\]
Answer: (2) All three have the same dimension
9. Given a charge \( q \), current \( I \) and permeability of vacuum \( \mu_0 \). Which of the following quantities has the dimension of momentum?
1. \( \dfrac{qI}{\mu_0} \)
2. \( q\mu_0 I \)
3. \( q^2 \mu_0 I \)
4. \( \dfrac{q\mu_0}{I} \)
Solution:
Dimension of momentum:
\[
[p] = [M L T^{-1}]
\]
Now, write dimensions of given quantities:
- Charge \( q \): \( [Q] \)
- Current \( I \): \( [Q T^{-1}] \Rightarrow [Q] = [I T] \)
- Permeability \( \mu_0 \):
\[
[\mu_0] = [M L T^{-2} I^{-2}]
\]
Check Option (4):
\[
\frac{q\mu_0}{I}
\]
Substitute dimensions:
\[
= \frac{[Q] \cdot [M L T^{-2} I^{-2}]}{[I]}
\]
Replace \( Q = I T \):
\[
= \frac{(I T)\cdot M L T^{-2} I^{-2}}{I}
\]
\[
= M L \cdot T^{-1} \cdot I^{(1-2-1)}
\]
\[
= M L T^{-1}
\]
This matches the dimension of momentum.
Correct Answer: (4) \( \dfrac{q\mu_0}{I} \)
10. If \(\mu_0\) and \(\varepsilon_0\) are the permeability and permittivity of free space respectively, then the dimension of \(\dfrac{1}{\mu_0\varepsilon_0}\) is:
1. \(L/T^2\)
2. \(L^2/T^2\)
3. \(T^2/L\)
4. \(T^2/L^2\)
Solution:
Key Relation:
Speed of light \(c = \dfrac{1}{\sqrt{\mu_0\varepsilon_0}}\)
\[
\frac{1}{\mu_0\varepsilon_0} = c^2 \Rightarrow \left[\frac{1}{\mu_0\varepsilon_0}\right] = L^2T^{-2}
\]
Answer: (2) \(L^2/T^2\)
11. The equation for a real gas is given by
\[
\left(P + \frac{a}{V^2}\right)(V – b) = RT
\]
where \(P,\,V,\,T,\,R\) are pressure, volume, temperature and gas constant. The dimension of \(ab^{-2}\) is equivalent to that of:
1. Planck’s constant
2. Compressibility
3. Strain
4. Energy density
Solution:
Dimension of \(a\):
Since \(\dfrac{a}{V^2}\) has dimensions of pressure \([P] = ML^{-1}T^{-2}\):
\[
[a] = [P]\cdot[V^2] = ML^{-1}T^{-2}\cdot L^6 = ML^5T^{-2}
\]
Dimension of \(b\):
\([b] = [V] = L^3\)
\[
[ab^{-2}] = ML^5T^{-2}\cdot L^{-6} = \boxed{ML^{-1}T^{-2}}
\]
Compare:
Energy Density \(= \dfrac{ML^2T^{-2}}{L^3} = ML^{-1}T^{-2}\)
Answer: (4) Energy Density
12. Match List–I with List–II and choose the correct answer:
| List–I (Physical Quantity) | List–II (Dimension) |
|---|---|
| (A) Coefficient of viscosity | (I) [ML0T-3] |
| (B) Intensity of wave | (II) [ML-2T-2] |
| (C) Pressure gradient | (III) [M-1L1T2] |
| (D) Compressibility | (IV) [ML-1T-1] |
Options:
(1) A–I, B–IV, C–III, D–II
(2) A–IV, B–I, C–II, D–III
(3) A–IV, B–II, C–I, D–III
(4) A–II, B–III, C–IV, D–I
Answer: (2) A–IV, B–I, C–II, D–III
Explanation:
(A) Coefficient of Viscosity (\( \eta \)):
From Newton’s law of viscosity:
\[
F = \eta A \frac{dv}{dx}
\]
\[
\eta = \frac{F}{A \cdot \frac{dv}{dx}}
= \frac{[M L T^{-2}]}{[L^2][T^{-1}]}
= [M L^{-1} T^{-1}]
\]
(B) Intensity of Wave (\( I \)):
\[
I = \frac{E}{A t}
\]
\[
I = \frac{[M L^2 T^{-2}]}{[L^2][T]}
= [M L^0 T^{-3}]
\]
(C) Pressure Gradient \( \left(\frac{dP}{dx}\right) \):
\[
\frac{dP}{dx} = \frac{\text{Pressure}}{\text{Length}}
= \frac{[M L^{-1} T^{-2}]}{[L]}
= [M L^{-2} T^{-2}]
\]
(D) Compressibility (\( \beta \)):
\[
\beta = \frac{1}{B} = \frac{1}{\text{Pressure}}
= \frac{1}{[M L^{-1} T^{-2}]}
= [M^{-1} L T^2]
\]
Q. A physical quantity \( C \) is related to four other quantities \( p \), \( q \), \( r \), and \( s \) as follows:
\[
C = \frac{p q^2 r^3}{\sqrt{s}}
\]
13. The percentage errors in the measurement of \( p \), \( q \), \( r \), and \( s \) are \( 1\% \), \( 2\% \), \( 3\% \), and \( 2\% \), respectively. The percentage error in the measurement of \( C \) will be _________ %.
Solution
1. General Error Formula
For a quantity of the form:
\[
C = \frac{p^a q^b r^c}{s^d}
\]
The maximum percentage error is:
\[
\frac{\Delta C}{C} \times 100
= a\left(\frac{\Delta p}{p} \times 100\right)
+ b\left(\frac{\Delta q}{q} \times 100\right)
+ c\left(\frac{\Delta r}{r} \times 100\right)
+ d\left(\frac{\Delta s}{s} \times 100\right)
\]
2. Applying the Exponents
From the given expression:
\[
C = \frac{p q^2 r^3}{\sqrt{s}} = p^1 q^2 r^3 s^{-1/2}
\]
So,
\[
a = 1,\quad b = 2,\quad c = 3,\quad d = \frac{1}{2}
\]
\[
\% \text{ error in } C
= 1(\% \text{ error in } p)
+ 2(\% \text{ error in } q)
+ 3(\% \text{ error in } r)
+ \frac{1}{2}(\% \text{ error in } s)
\]
3. Substituting Values
\( \% \text{ error in } p = 1\% \)
\( \% \text{ error in } q = 2\% \)
\( \% \text{ error in } r = 3\% \)
\( \% \text{ error in } s = 2\% \)
\[
\% \text{ error in } C = 1(1\%) + 2(2\%) + 3(3\%) + \frac{1}{2}(2\%)
\]
\[
= 1\% + 4\% + 9\% + 1\%
\]
\[
= 15\%
\]
14. Three identical spheres of mass \(m\) are placed at the vertices of an equilateral triangle of length \(a\). When released, they interact only through gravitational force and collide after time \(T = 4\) seconds. If the sides of the triangle are increased to \(2a\) and the masses are made \(2m\), they will collide after:
1. 4 s
2. 8 s
3. 16 s
4. 2 s
Solution:
To derive the collision time \( T \) for three identical spheres of mass \( m \) placed at the vertices of an equilateral triangle of side \( a \), we use dimensional analysis.
Governing Parameters:
- Initial separation \( a \): \( [L] \)
- Mass \( m \): \( [M] \)
- Gravitational constant \( G \): \( [M^{-1} L^3 T^{-2}] \)
Dimensional Analysis:
Assume:
\[
T \propto a^x m^y G^z
\]
\[
[T] = [L]^x [M]^y [M^{-1}L^3T^{-2}]^z
\]
\[
[M^0 L^0 T^1] = [M^{y-z} L^{x+3z} T^{-2z}]
\]
Equating powers:
- Time: \( -2z = 1 \Rightarrow z = -\frac{1}{2} \)
- Mass: \( y – z = 0 \Rightarrow y = -\frac{1}{2} \)
- Length: \( x + 3z = 0 \Rightarrow x = \frac{3}{2} \)
\[
T \propto a^{3/2} m^{-1/2} G^{-1/2}
\]
\[
T \propto \sqrt{\frac{a^3}{Gm}}
\]
- Case 1: \( a, m \Rightarrow T_1 = 4\,\text{s} \)
- Case 2: \( 2a, 2m \Rightarrow T_2 = ? \)
\[
\frac{T_2}{T_1} = \frac{\sqrt{(2a)^3/(2m)}}{\sqrt{a^3/m}}
= \sqrt{\frac{8a^3}{2m} \cdot \frac{m}{a^3}}
= \sqrt{4} = 2
\]
\[
T_2 = 2 \times 4 = 8\,\text{s}
\]
15. Assertion–Reason
Statement (I): The dimensions of Planck’s constant and angular momentum are the same.
Statement (II): In Bohr’s model, electrons revolve around the nucleus only in those orbits for which angular momentum is an integral multiple of Planck’s constant.
1. Both Statement I and II are correct
2. Statement I is incorrect but II is correct
3. Statement I is correct but II is incorrect
4. Both Statement I and II are incorrect
Solution:
Statement I:
From \(E = hf\):
\[
[h] = \frac{[E]}{[f]} = \frac{ML^2T^{-2}}{T^{-1}} = ML^2T^{-1}
\]
Angular Momentum:
\[
[L] = [mvr] = M \cdot LT^{-1} \cdot L = ML^2T^{-1}
\]
Hence, both have same dimensions ✓
Statement II:
Bohr’s quantisation condition:
\[
L = \frac{nh}{2\pi}, \quad n = 1,2,3,\ldots
\]
Angular momentum is an integral multiple of \(\dfrac{h}{2\pi}\), not \(h\) → Statement II is incorrect.
Answer: Option (3) — Statement I correct, Statement II incorrect