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QAIPMT-2010-Ph-1974: NEET 2010 Physics - Class 12 AIPMT-2010-Ph-1974 Solution: A thin circular ring of mass M and radius r is rotating about…
A thin circular ring of mass M and radius r is rotating about its axis with constant angular velocity . Two objects each of mass m are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with angular velocity given by
Multiple Choice Options
- A 23.6 MeV
- B 2.2 MeV
- C 28.0 MeV
- D 30.2 MeV
Step-by-Step Verified Solution
1 H 2
+ 1 H 2 _ 2 He 4
+ _ E
The binding energy per nucleon of a deuteron = 1.1 MeV
_ Total binding energy = 2 _ 1.1 = 2.2 MeV
The binding energy per nucleon of a helium nuclei
= 7 MeV
_ Total binding energy = 4 _ 7 = 28 MeV
Hence, energy released
_ E = (28 � 2 _ 2.2) = 23.6 MeV