If the earth were considered a perfect sphere and its radius was reduced to half without any change in its mass, what would be its period of rotation?
Answers
Explanation:
Correct option is
Correct option isAngular momentum is conserved hence,
Correct option isAngular momentum is conserved hence,L
Correct option isAngular momentum is conserved hence,L i
Correct option isAngular momentum is conserved hence,L i
Correct option isAngular momentum is conserved hence,L i =L
Correct option isAngular momentum is conserved hence,L i =L f
Correct option isAngular momentum is conserved hence,L i =L f
Correct option isAngular momentum is conserved hence,L i =L f
Correct option isAngular momentum is conserved hence,L i =L f I
Correct option isAngular momentum is conserved hence,L i =L f I I
Correct option isAngular momentum is conserved hence,L i =L f I I
Correct option isAngular momentum is conserved hence,L i =L f I I ω
Correct option isAngular momentum is conserved hence,L i =L f I I ω i
Correct option isAngular momentum is conserved hence,L i =L f I I ω i
Correct option isAngular momentum is conserved hence,L i =L f I I ω i =I
Correct option isAngular momentum is conserved hence,L i =L f I I ω i =I f
Correct option isAngular momentum is conserved hence,L i =L f I I ω i =I f
Correct option isAngular momentum is conserved hence,L i =L f I I ω i =I f ω
Correct option isAngular momentum is conserved hence,L i =L f I I ω i =I f ω f
Correct option isAngular momentum is conserved hence,L i =L f I I ω i =I f ω f
Correct option isAngular momentum is conserved hence,L i =L f I I ω i =I f ω f
Correct option isAngular momentum is conserved hence,L i =L f I I ω i =I f ω f 5
2
mr
mr 2
mr 2 ω
mr 2 ω i
mr 2 ω i
mr 2 ω i =
5
52
52
52 m(
52 m( 2
52 m( 2r
52 m( 2r
52 m( 2r )
52 m( 2r ) 2
52 m( 2r ) 2 ω
52 m( 2r ) 2 ω f
52 m( 2r ) 2 ω f
52 m( 2r ) 2 ω f
52 m( 2r ) 2 ω f ∴ω
52 m( 2r ) 2 ω f ∴ω f
52 m( 2r ) 2 ω f ∴ω f
52 m( 2r ) 2 ω f ∴ω f =4ω
52 m( 2r ) 2 ω f ∴ω f =4ω i
52 m( 2r ) 2 ω f ∴ω f =4ω i
52 m( 2r ) 2 ω f ∴ω f =4ω i
52 m( 2r ) 2 ω f ∴ω f =4ω i or,
52 m( 2r ) 2 ω f ∴ω f =4ω i or, T
52 m( 2r ) 2 ω f ∴ω f =4ω i or, T f
52 m( 2r ) 2 ω f ∴ω f =4ω i or, T f
52 m( 2r ) 2 ω f ∴ω f =4ω i or, T f
52 m( 2r ) 2 ω f ∴ω f =4ω i or, T f 2π
52 m( 2r ) 2 ω f ∴ω f =4ω i or, T f 2π
52 m( 2r ) 2 ω f ∴ω f =4ω i or, T f 2π =4×
52 m( 2r ) 2 ω f ∴ω f =4ω i or, T f 2π =4× T
52 m( 2r ) 2 ω f ∴ω f =4ω i or, T f 2π =4× T i
52 m( 2r ) 2 ω f ∴ω f =4ω i or, T f 2π =4× T i
52 m( 2r ) 2 ω f ∴ω f =4ω i or, T f 2π =4× T i
52 m( 2r ) 2 ω f ∴ω f =4ω i or, T f 2π =4× T i 2π
∴T
T f
T f
T f =
T f = 4
T f = 4T
T f = 4T i
T f = 4T i
T f = 4T i
T f = 4T i
T f = 4T i =
T f = 4T i = 4
T f = 4T i = 424
T f = 4T i = 424
T f = 4T i = 424 =6hours
Answer:
Heya Here's the required Answer mate
- The period of rotation will be 6 hrs
Hope it helps