JEE Mains · Physics · STD 11 - 1. units,dimensions and measurement
Time \((T)\), velocity \((C)\) and angular momentum \((h)\) are chosen as fundamental quantities instead of mass, length and time. In terms of these, the dimensions of mass would be
- A \(\left[ M \right] = \left[ {{T^{ - 1}}\,{C^{ - 2}}\,h} \right]\)
- B \(\left[ M \right] = \left[ {{T^{ - 1}}\,{C^2}\,h} \right]\)
- C \(\left[ M \right] = \left[ {{T^{ - 1}}\,{C^{ - 2}}\,{h^{ - 1}}} \right]\)
- D \(\left[ M \right] = \left[ {T\,{C^{ - 2}}\,h} \right]\)
Answer & Solution
Correct Answer
(A) \(\left[ M \right] = \left[ {{T^{ - 1}}\,{C^{ - 2}}\,h} \right]\)
Step-by-step Solution
Detailed explanation
Let mass related as \(M \propto \,{T^x}{C^y}{h^z}\) \({M^1}{L^0}{T^0} = {\left( T \right)^x}{\left( {{L^1}{T^{ - 1}}} \right)^y}{\left( {{M^1}{L^2}{T^{ - 1}}} \right)^z}\) \({M^1}{L^0}{T^0} = {M^z}{L^{y + 2z}} + {T^{x - y - z}}\) \(z = 1\) \(y + 2z = 0\) \(x - y - z = 0\)…
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