Question

Time period depends on Gravitational constant (G), Planck constant (h) and speed of light (c), then T is proportional to
A. $\dfrac{{{G^{{1 {\left/ {\vphantom {1 2}} \right. } 2}}}{h^{{1 {\left/ {\vphantom {1 2}} \right. } 2}}}}}{{{c^{{3 {\left/ {\vphantom {3 2}} \right. } 2}}}}}$
B. $\dfrac{{{G^{{1 {\left/ {\vphantom {1 2}} \right. } 2}}}h}}{{{c^{{5 {\left/ {\vphantom {5 2}} \right. } 2}}}}}$
C. $\dfrac{{G{h^{{1 {\left/ {\vphantom {1 2}} \right. } 2}}}}}{{{c^{{5 {\left/ {\vphantom {5 2}} \right. } 2}}}}}$
D. $\dfrac{{{G^{{1 {\left/ {\vphantom {1 2}} \right. } 2}}}{h^{{1 {\left/ {\vphantom {1 2}} \right. } 2}}}}}{{{c^{{5 {\left/ {\vphantom {5 2}} \right. } 2}}}}}$

Answer 1

We will assume the relation of the time period in terms of gravitational constant, Planck constant (h), and speed of light (c) as per the problem statement. We will use the dimensional analysis to find out the final expression for the time period.

Complete step by step answer:
We know that the dimensional formula for gravitational constant (G) can be written as:
$\left[ G \right] = \left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right]$
We also know that the dimensional formula of Planck’s constant (h) can be written as:
$\left[ h \right] = \left[ {M{L^2}{T^{ - 1}}} \right]$
The dimensional formula of the speed of light is given as below:
$\left[ c \right] = \left[ {L{T^{ - 1}}} \right]$
It is given that the time period (T) depends on the gravitational constant (G), Planck’s constant (k), and speed of light (c), so we can write the dimension of the time period as below.
$\left[ T \right] = {\left[ G \right]^x}{\left[ h \right]^y}{\left[ c \right]^z}$……(1)
Here x, y, and z are unknown constants.
On substituting $\left[ {M{L^2}{T^{ - 1}}} \right]$ for $\left[ h \right]$, $\left[ {L{T^{ - 1}}} \right]$ for $\left[ c \right]$ and $\left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right]$ for $\left[ G \right]$ in the above expression, we get:
$\left[ T \right] = {\left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right]^x}{\left[ {M{L^2}{T^{ - 1}}} \right]^y}{\left[ {L{T^{ - 1}}} \right]^z}$
On rearranging the above expression, we can write:
$\left[ T \right] = \left[ {{M^{ - x + y}}{L^{3x + 2y + z}}{T^{2x - y - z}}} \right]$
Again rearranging it, we get:
$\left[ {{M^0}{L^0}T} \right] = \left[ {{M^{ - x + y}}{L^{3x + 2y + z}}{T^{2x - y - z}}} \right]$……(2)
On comparing the power of mass M, we get:

\- x + y = 0\\\ x = y $$……(3) On comparing the power of length L, we get: $$3x + 2y + z = 0$$……(4) On comparing the power of time T, we get: $$ - 2x - y - z = 1$$……(5) Using equation (3), equation (4), and equation (5), we get the values of x, y, and z as below:

x = \dfrac{1}{2}\\
y = \dfrac{1}{2}\\
z = - \dfrac{5}{2}

On substituting $$\dfrac{1}{2}$$ for x, $$\dfrac{1}{2}$$ for y, and $$ - \dfrac{5}{2}$$ for z in equation (1), we get: $$\left[ T \right] = {\left[ G \right]^{{1 {\left/ {\vphantom {1 2}} \right. } 2}}}{\left[ h \right]^{{1 {\left/ {\vphantom {1 2}} \right. } 2}}}{\left[ c \right]^{{{ - 5} {\left/ {\vphantom {{ - 5} 2}} \right. } 2}}}$$ We can write the above expression in proportionality form as below. $$T = \dfrac{{{G^{{1 {\left/ {\vphantom {1 2}} \right. } 2}}}{h^{{1 {\left/ {\vphantom {1 2}} \right. } 2}}}}}{{{c^{{5 {\left/ {\vphantom {5 2}} \right. } 2}}}}}$$ **Therefore, the time period is proportional to $$\dfrac{{{G^{{1 {\left/ {\vphantom {1 2}} \right. } 2}}}{h^{{1 {\left/ {\vphantom {1 2}} \right. } 2}}}}}{{{c^{{5 {\left/ {\vphantom {5 2}} \right. } 2}}}}}$$, and option (D) is correct.** **Note:** We can remember the unit of gravitational constant (G) is $${{\rm{m}}^3}{\rm{k}}{{\rm{g}}^{ - 1}}{{\rm{s}}^{ - 2}}$$, the unit of Planck’s constant (h) is $${\rm{kg}} \cdot {{\rm{m}}^2}{{\rm{s}}^{ - 1}}$$ , and the unit of speed of light is $${{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}$$. Using the base units of all the quantities, we can find the dimensional formula for them.