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Question: If energy, \(E={{G}^{p}}{{h}^{q}}{{c}^{r}}\) where \(G\) is the universal gravitational constant,\(h...

If energy, E=GphqcrE={{G}^{p}}{{h}^{q}}{{c}^{r}} where GG is the universal gravitational constant,hh is the Planck’s constant and cc is the velocity of light, then the values of p,qp,q and rr respectively:
A.−12,12and52-\dfrac{1}{2},\dfrac{1}{2}and\dfrac{5}{2}
B.12,−12and−52\dfrac{1}{2},-\dfrac{1}{2}and\dfrac{-5}{2}
C.−12,12and32\dfrac{-1}{2},\dfrac{1}{2}and\dfrac{3}{2}
D.12,−12and−32\dfrac{1}{2},\dfrac{-1}{2}and\dfrac{-3}{2}

Explanation

Solution

Hint Write to both sides dimensions and compare both sides
dimension of E=E= dimension of Gphqcr{{G}^{p}}{{h}^{q}}{{c}^{r}}
Where G=G= Universal gravitational constant, h=h= Plank’s constant and c=c= Speed of light

Complete step-by-step solution : Dimension:- The dimension of a physical quantity are the power to which the fundamental units are raised in order to obtain the derived unit of that quantity.
To express the dimensions of physical quantities in mechanics, the length, mass and time are denoted by [L]\left[ L \right] [M]\left[ M \right] and [T]\left[ T \right] . The dimensions of physical quantities areaain length, bb in mass and cc in time, then the dimensions of that physical quantity shall be written in the following manner
[LaMbCC]\left[ {{L}^{a}}{{M}^{b}}{{C}^{C}} \right] [dimension formula]
Given
E=GphqcrE={{G}^{p}}{{h}^{q}}{{c}^{r}} ……………(1)
G=G= Gravitational constant
h=h= Plank’s constant
c=c= Speed of light
Dimension of G=[M−1L3T−2]G=\left[ {{M}^{-1}}{{L}^{3}}T-2 \right]
Dimension of h=[ML2T−1]h=\left[ M{{L}^{2}}{{T}^{-1}} \right]
Dimension of c=[LT−1]=[M0LT−1]c=\left[ L{{T}^{-1}} \right]=\left[ {{M}^{0}}L{{T}^{-1}} \right]
Dimension of E=[ML2T−2]E=\left[ M{{L}^{2}}{{T}^{-2}} \right]
Putt these values in equation (1)
[ML2T−2]=[M−1L−3T−2]p[ML2T−1]q[M0LT−1]r\left[ M{{L}^{2}}{{T}^{-2}} \right]={{\left[ {{M}^{-1}}{{L}^{-3}}{{T}^{-2}} \right]}^{p}}{{\left[ M{{L}^{2}}{{T}^{-1}} \right]}^{q}}{{\left[ {{M}^{0}}L{{T}^{-1}} \right]}^{r}}
[ML2T−2]=[M−p+q+0L3p+2q+rT−2p−q−r]\left[ M{{L}^{2}}{{T}^{-2}} \right]=\left[ {{M}^{-p+q+0}}{{L}^{3p+2q+r}}{{T}^{-2p-q-r}} \right]
Comparing the power of MM both sides
1=−p+q1=-p+q
−p+q=1-p+q=1 ………………….(2)
Comparing the power of LL both sides
2=3p+2q+r2=3p+2q+r
3p+2q+r=23p+2q+r=2 ………………..(3)
Comparing the power of TT both sides
−2=−2p−q−r −(2p+q+r)=−2 \begin{aligned} & -2=-2p-q-r \\\ & -\left( 2p+q+r \right)=-2 \\\ \end{aligned}
2p+q+r=22p+q+r=2 …………….(4)
Equation (3) substract from equation (4)
2p+q+r−3p−2q−r=2−2 −p−q=0 \begin{aligned} & 2p+q+r-3p-2q-r=2-2 \\\ & -p-q=0 \\\ \end{aligned}
p=−qp=-q…………..(5)
Put the value of pp in equation (1)
−p+q=1 −(−q)+q=1 q+q=1 2q=1 q=12 \begin{aligned} & -p+q=1 \\\ & -\left( -q \right)+q=1 \\\ & q+q=1 \\\ & 2q=1 \\\ & q=\dfrac{1}{2} \\\ \end{aligned}
From equation (5)
p=−q p=−12 \begin{aligned} & p=-q \\\ & p=-\dfrac{1}{2} \\\ \end{aligned}
Value of pp and qq put in equation (3)
3×(−12)+2×12+r=2 −32+1+r=2 −12+r=2 r=2+12 r=32 \begin{aligned} & 3\times \left( -\dfrac{1}{2} \right)+2\times \dfrac{1}{2}+r=2 \\\ & -\dfrac{3}{2}+1+r=2 \\\ & -\dfrac{1}{2}+r=2 \\\ & r=2+\dfrac{1}{2} \\\ & r=\dfrac{3}{2} \\\ \end{aligned}
Putting the value of p,qp,q and rr in equation (1)
E=G−1h12c32E={{G}^{-1}}{{h}^{\dfrac{1}{2}}}{{c}^{\dfrac{3}{2}}}

Note: Student thought that GG is a universal constant, so it has no dimension, but it has the dimension[M−1L3T−2]\left[ {{M}^{-1}}{{L}^{3}}{{T}^{-2}} \right]