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JEE Mains · Physics · STD 11 - 1. units,dimensions and measurement

यदि द्रव्यमान, लम्बाई और समय के स्थान पर समय \(( T )\), वेग \(( C )\) तथा कोणीय संवेग \(( h )\) को मूलभूत राशियाँ मान लें तो द्रव्यमान की विमा को इन राशियों के रूप में निम्न तरीके से लिखेंगे

  1. A \(\left[ M \right] = \left[ {{T^{ - 1}}\,{C^{ - 2}}\,h} \right]\)
  2. B \(\left[ M \right] = \left[ {{T^{ - 1}}\,{C^2}\,h} \right]\)
  3. C \(\left[ M \right] = \left[ {{T^{ - 1}}\,{C^{ - 2}}\,{h^{ - 1}}} \right]\)
  4. D \(\left[ M \right] = \left[ {T\,{C^{ - 2}}\,h} \right]\)
Verified Solution

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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