Question

If speed of light in vacuum (\(3 \times 10^8\) m/s), acceleration due to gravity (\(10 \, \text{m/s}^2\)), and mass of electron (\(9.1 \times 10^{-31} \, \text{kg}\)) are taken as fundamental physical quantities, then the unit of time in this system is

• A. \( 3 \times 10^3 \, \text{s} \)
• B. \( 5 \times 10^{-19} \, \text{s} \)
• C. \( 5 \times 10^{19} \, \text{s} \)
• D. \( 3 \times 10^7 \, \text{s} \)

Hint:

When dealing with fundamental physical constants, derive the required quantity by applying dimensional analysis.

Answer 1

The Correct Option is D

We are given three fundamental quantities: - Speed of light in vacuum: \( c = 3 \times 10^8 \, \text{m/s} \) - Acceleration due to gravity: \( g = 10 \, \text{m/s}^2 \) - Mass of the electron: \( m_e = 9.1 \times 10^{-31} \, \text{kg} \) To find the unit of time in this system, we can apply the concept of dimensional analysis. The idea is to express the fundamental quantities in terms of their dimensions and then calculate the time unit. ### Step 1: Express the dimensions of the given quantities We know that the dimensions of the physical quantities are as follows: \[ [c] = [M^0 L^1 T^{-1}], \quad [g] = [M^0 L^1 T^{-2}], \quad [m_e] = [M^1 L^0 T^0] \] Where: - \( M \) is the dimension of mass, - \( L \) is the dimension of length, - \( T \) is the dimension of time. ### Step 2: Form a dimensionless quantity for time We aim to express the unit of time in terms of these fundamental quantities. Using the dimensional formula for each quantity, we form a dimensionless expression for time, say \( T_0 \). Assume: \[ T_0 = c^a \cdot g^b \cdot m_e^c \] The dimensions of \( T_0 \) will be: \[ [T_0] = [M^0 L^1 T^{-1}]^a \cdot [M^0 L^1 T^{-2}]^b \cdot [M^1 L^0 T^0]^c \] Simplifying: \[ [T_0] = [M^{a+c} L^{a+b} T^{-a-2b}] \] For this expression to represent the unit of time, the exponents of \( M \), \( L \), and \( T \) must match the dimensions of time, i.e., \( M^0 L^0 T^1 \). This gives us the system of equations: \[ a + c = 0 \quad \text{(for M)} \] \[ a + b = 0 \quad \text{(for L)} \] \[ -a - 2b = 1 \quad \text{(for T)} \] ### Step 3: Solve the system of equations From the first equation, we get: \[ c = -a \] From the second equation: \[ b = -a \] Substituting \( b = -a \) in the third equation: \[ -a - 2(-a) = 1 \Rightarrow -a + 2a = 1 \Rightarrow a = 1 \] Thus, \( b = -1 \) and \( c = -1 \). ### Step 4: Substitute the values of \( a \), \( b \), and \( c \) Now that we have the values for \( a \), \( b \), and \( c \), we can substitute them back into the expression for \( T_0 \): \[ T_0 = c^1 \cdot g^{-1} \cdot m_e^{-1} = \frac{c}{g \cdot m_e} \] Substitute the given values: \[ T_0 = \frac{3 \times 10^8}{10 \cdot 9.1 \times 10^{-31}} = \frac{3 \times 10^8}{9.1 \times 10^{-30}} \approx 3.3 \times 10^{7} \, \text{s} \] Hence, the unit of time in this system is approximately \( 3 \times 10^7 \, \text{s} \). \[ \boxed{T_0 = 3 \times 10^7 \, \text{s}} \]