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If momentum (p), area (A) and time (T) are taken to be fundamental quantities, then energy has the dimensional formula.

Answer» <html><body><p>`(pA^(-1)T^(1)]`<br/>`[p^(2)AT]`<br/>`[pA^(-(1)/(2))T]`<br/>`[pA^((1)/(2)T)]`<br/></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Given, fundamental quantities are momentum (p), <a href="https://interviewquestions.tuteehub.com/tag/area-13372" style="font-weight:bold;" target="_blank" title="Click to know more about AREA">AREA</a> (A) and time (T). <br/>We can write energy E as <br/> `E <a href="https://interviewquestions.tuteehub.com/tag/prop-607409" style="font-weight:bold;" target="_blank" title="Click to know more about PROP">PROP</a> p^(A)A^(b)T^(c)` <br/> `E= kp^(a)A^(A)T^(c)` <br/> `:. [E]=[p]^(a)[A]^(A)[T]^(c)` <br/> where <a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a> is <a href="https://interviewquestions.tuteehub.com/tag/dimensionless-954191" style="font-weight:bold;" target="_blank" title="Click to know more about DIMENSIONLESS">DIMENSIONLESS</a> constant of proportionality <br/> Dimensions of `E=[E]=[M^(1)L^(2)T^(-2)] and p]=[M^(1)L^(1)T^(-1)]` <br/> `[A]=[L^(2)]` <br/> `[T]=[T^(1)]` <br/> Putting all the dimensions, we get <br/> `M^(1)L^(2)T^(-2)=[M^(1)L^(1)T^(-1)]^(a)[L^(2)][T]^(c)` <br/> `:. M^(1)L^(2)T^(-2)=M^(a)L^(2b+b)T^(-a+c)` <br/> By comparison of power <br/> `a=1, 2b+a=2` <br/> `:.2b+1=2` <br/> `b=(1)/(2)` <br/> `-a+c=-2` <br/> `c=-2+a=-2+1=-1` <br/> Thus`|E|=[pA^((1)/(2))T^(-1)]`</body></html>


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