A) \[a=1\], \[b=1,\,\,c=2\]
B) \[a=1\], \[b=2,\,\,c=1\]
C) \[a=\frac{5}{6}\], \[b=\frac{1}{2},c=\frac{1}{3}\]
D) \[a=-\frac{5}{6},b=\frac{1}{2},c=\frac{1}{3}\]
Correct Answer: D
Solution :
Key Idea: Every equation relating physical quantities should be in dimensional balance. Given, \[T\propto {{p}^{a}}{{d}^{b}}{{E}^{c}}\] Since, only similar quantities can be equated, therefore dimensions of the terms on both sides of the equation must be same. Hence, we have Dimensions of \[T=[{{M}^{0}}{{L}^{0}}{{T}^{1}}]\] Dimensions of pressure \[P=[M{{L}^{-1}}{{T}^{-2}}]\] Dimensions of density \[d=[M{{L}^{-3}}]\] Dimensions of energy \[E=[M{{L}^{2}}{{T}^{-2}}]\] \[\therefore \] We have \[[{{M}^{0}}{{L}^{0}}{{T}^{1}}]=k{{[M{{L}^{-1}}{{T}^{-2}}]}^{a}}{{[M{{L}^{-3}}]}^{b}}{{[M{{L}^{2}}{{T}^{-2}}]}^{c}}\] where, A: is a constant. Comparing dimensions of similar terms, we have \[[{{M}^{0}}{{L}^{0}}{{T}^{1}}]=k[{{M}^{a+b+c}}{{L}^{-a-3b+2c}}{{T}^{-2a-2c}}]\] Comparing powers of M, we have \[0=a+b+c\] ... (i) Comparing powers of£, we have \[0=-a-3b+2c\] ... (ii) Comparing powers of T, we have \[1=-2\,a-2c\] ... (iii) Solving Eqs. (i), (ii) and (iii), we have \[a=-\frac{5}{6},\,b=\frac{1}{2},\,c=\frac{1}{3}\]
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