Verified Solution • 2025–26 Session moderate

QAIPMT-2009-Ph-2179: NEET 2009 Physics - Class 12 AIPMT-2009-Ph-2179 Solution: In the nuclear decay given below \(_Z^A X _{Z+1}^A Y…

In the nuclear decay given below ZAX→Z+1AY→Z−1A−4B→Z−1A−4B_Z^A X \rightarrow _{Z+1}^A Y \rightarrow _{Z-1}^{A-4} B \rightarrow _{Z-1}^{A-4} B the particles emitted in the sequence are γβα\gamma \beta \alpha βγα\beta \gamma \alpha αβγ\alpha \beta \gamma βαγ\beta \alpha \gamma

Multiple Choice Options

  • A ε2R2+(Lω−1ω)2R{{{\varepsilon ^2}\sqrt {{R^2} + {{\left( {L\omega - {1 \over {\omega }}} \right)}^2}} } \over R} R _ 2 R 2 + ( L _ _ _ 1 _ ) 2 _ _
  • B ε2R2+(Lω−1Cω)2R{{{\varepsilon ^2}\sqrt {{R^2} + {{\left( {L\omega - {1 \over {C\omega }}} \right)}^2}} } \over R} R _ 2 R 2 + ( L _ _ C _ 1 _ ) 2 _ _
  • C ε2RR2+(L−1Cω)2{{{\varepsilon ^2}R} \over {\sqrt {{R^2} + {{\left( {L - {1 \over {C\omega }}} \right)}^2}} }} R 2 + ( L _ C _ 1 _ ) 2 _ _ 2 R _
  • D ε2R[R2+(Lω−1Cω)2]{{{\varepsilon ^2}R} \over {\left[ {{R^2} + {{\left( {L\omega - {1 \over {C\omega }}} \right)}^2}} \right]}} [ R 2 + ( L _ _ C _ 1 _ ) 2 ] _ 2 R _

Step-by-Step Verified Solution

Power dissipated, P = E rms I rms cos _ cos _ = R Z But I rms = E r m s Z _ P = E r m s 2 . R Z 2 Also we know, Z = R 2 + ( X L _ X C ) 2 So, P = E r m s 2 . R R 2 + ( X L _ X C ) 2 = _ 2 R [ R 2 + ( L _ _ 1 C _ ) 2 ]

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