1. Young’s modulus of elasticity \( Y \) is expressed in terms of three derived quantities, namely, the gravitational constant \( G \), Planck’s constant \( h \) and the speed of light \( c \), as \( Y = c^\alpha h^\beta G^\gamma \). Which of the following is the correct option?
[JEE (Advanced) 2023]
(A) \( \alpha = 7, \beta = -1, \gamma = -2 \)
(B) \( \alpha = -7, \beta = -1, \gamma = -2 \)
(C) \( \alpha = 7, \beta = -1, \gamma = 2 \)
(D) \( \alpha = -7, \beta = 1, \gamma = -2 \)
Solution:
To find the correct values for \( \alpha \), \( \beta \), and \( \gamma \) in the expression for Young’s modulus \( Y = c^\alpha h^\beta G^\gamma \), we use dimensional analysis.
1. Identify Dimensions of Given Quantities
- Young’s Modulus (\( Y \)): Since \( Y = \frac{\text{Stress}}{\text{Strain}} \) and strain is dimensionless,
\( [Y] = [\text{Pressure}] = [M^1 L^{-1} T^{-2}] \). - Speed of light (\( c \)): \( [c] = [L T^{-1}] \).
- Planck’s constant (\( h \)): Since \( E = h\nu \),
\( [h] = [E][T] = [M L^2 T^{-2}][T] = [M^1 L^2 T^{-1}] \). - Gravitational constant (\( G \)): From \( F = \frac{G m_1 m_2}{r^2} \),
\( [G] = \frac{[F][L^2]}{[M^2]} = \frac{[M L T^{-2}][L^2]}{[M^2]} = [M^{-1} L^3 T^{-2}] \).
2. Set Up the Dimensional Equation
\[
[Y] = [c]^\alpha [h]^\beta [G]^\gamma
\]
\[
[M^1 L^{-1} T^{-2}] = [L T^{-1}]^\alpha [M^1 L^2 T^{-1}]^\beta [M^{-1} L^3 T^{-2}]^\gamma
\]
Grouping terms:
\[
[M^1 L^{-1} T^{-2}] = M^{(\beta – \gamma)} \cdot L^{(\alpha + 2\beta + 3\gamma)} \cdot T^{(-\alpha – \beta – 2\gamma)}
\]
3. Solve for \( \alpha, \beta, \gamma \)
Equating powers of \( M, L, T \):
- \( 1 = \beta – \gamma \)…………..Eq. 1
- \( -1 = \alpha + 2\beta + 3\gamma \) …………Eq. 2
- \( -2 = -\alpha – \beta – 2\gamma \)……………Eq. 3
Adding equations (2) and (3) correctly:
$$(-1) + (-2) = (\alpha + 2\beta + 3\gamma) + (-\alpha – \beta – 2\gamma)$$ $$-3 = \beta + \gamma$$
Now we have a system of two simple equations for $\beta$ and $\gamma$:
* $\beta – \gamma = 1$
* $\beta + \gamma = -3$
Solving for $\beta$ and $\gamma$:
* Adding them: $2\beta = -2 \implies \mathbf{\beta = -1}$
* Subtracting them: $-2\gamma = 4 \implies \mathbf{\gamma = -2}$
Solving for $\alpha$ using equation (3):
$$2 = \alpha + \beta + 2\gamma$$ $$2 = \alpha + (-1) + 2(-2)$$ $$2 = \alpha – 1 – 4$$ $$2 = \alpha – 5 \implies \mathbf{\alpha = 7}$$
The values are $\alpha = 7$, $\beta = -1$, and $\gamma = -2$.
Ans: (A) \( \alpha = 7, \beta = -1, \gamma = -2 \)
2. In a particular system of units, a physical quantity can be expressed in terms of the electric charge \( e \), electron mass \( m_e \), Planck’s constant \( h \), and Coulomb’s constant \( k = \dfrac{1}{4\pi \varepsilon_0} \). In terms of these physical constants, the dimension of the magnetic field is
\[
[B] = [e]^\alpha [m_e]^\beta [h]^\gamma [k]^\delta
\]
The value of \( \alpha + \beta + \gamma + \delta \) is ________.
[JEE(Advanced) 2022]
Solution:
Step 1. Dimension of magnetic field \( B \):
Magnetic field is force per unit charge per unit velocity:
\[
[B] = \frac{F}{qv} = \frac{M L T^{-2}}{Q \cdot L T^{-1}} = M T^{-1} Q^{-1}
\]
\[
[B] = M^1 L^0 T^{-1} Q^{-1}
\]
Step 2. Dimensions of constants:
- \( e \): charge → \( [e] = Q \)
- \( m_e \): mass → \( [m_e] = M \)
- \( h \): Planck’s constant → \( [h] = M L^2 T^{-1} \)
- \( k \): Coulomb’s constant → \( [k] = M L^3 T^{-2} Q^{-2} \)
Step 3. General form:
\[
[B] = [e]^\alpha [m_e]^\beta [h]^\gamma [k]^\delta
\]
\[
[B] = Q^\alpha \cdot M^\beta \cdot (M L^2 T^{-1})^\gamma \cdot (M L^3 T^{-2} Q^{-2})^\delta
\]
Step 4. Collect exponents:
- Mass exponent: \( \beta + \gamma + \delta \)
- Length exponent: \( 2\gamma + 3\delta \)
- Time exponent: \( -\gamma – 2\delta \)
- Charge exponent: \( \alpha – 2\delta \)
\[
[B] = M^{\beta + \gamma + \delta} \, L^{2\gamma + 3\delta} \, T^{-(\gamma + 2\delta)} \, Q^{\alpha – 2\delta}
\]
Step 5. Match with required dimension \( M^1 L^0 T^{-1} Q^{-1} \):
- Mass: \( \beta + \gamma + \delta = 1 \)
- Length: \( 2\gamma + 3\delta = 0 \)
- Time: \( -(\gamma + 2\delta) = -1 \Rightarrow \gamma + 2\delta = 1 \)
- Charge: \( \alpha – 2\delta = -1 \)
Step 6. Solve equations:
From length:
\[
2\gamma + 3\delta = 0 \Rightarrow \gamma = -\frac{3}{2}\delta
\]
Substitute into time:
\[
-\frac{3}{2}\delta + 2\delta = 1 \Rightarrow \frac{1}{2}\delta = 1 \Rightarrow \delta = 2
\]
Then,
\[
\gamma = -\frac{3}{2}(2) = -3
\]
From mass:
\[
\beta + \gamma + \delta = 1 \Rightarrow \beta – 3 + 2 = 1 \Rightarrow \beta = 2
\]
From charge:
\[
\alpha – 2\delta = -1 \Rightarrow \alpha – 4 = -1 \Rightarrow \alpha = 3
\]
Step 7. Add exponents:
\[
\alpha + \beta + \gamma + \delta = 3 + 2 – 3 + 2 = 4
\]