Last Updated On [03-11-2023]
A particle of mass 20 g is released with an initial velocity 5 m/s along the curve from the point A, as shown in the figure. The point A is at height h from point B. The particle slides along the frictionless surface. When the particle reaches point B, its angular momentum about O will be : (Take g= 10 m/s2)
(a)
8kg - m2/s
(b)
6kg - m2/s
(c)
3kg - m2/s
(d)
2kg - m2/s
A particle of mass 20 g is released with an initial velocity 5 m/s along the curve from the point A, as shown in the figure. The point A is at height h from point B. The particle slides along the frictionless surface. When the particle reaches point B, its angular momentum about O will be : (Take g= 10 m/s2)
(a)
8kg - m2/s
(b)
6kg - m2/s
(c)
3kg - m2/s
(d)
2kg - m2/s
Question ID - 50725 :-
A particle of mass 20 g is released with an initial velocity 5 m/s along the curve from the point A, as shown in the figure. The point A is at height h from point B. The particle slides along the frictionless surface. When the particle reaches point B, its angular momentum about O will be : (Take g= 10 m/s2)
(a)
8kg - m2/s
(b)
6kg - m2/s
(c)
3kg - m2/s
(d)
2kg - m2/s
|
A particle of mass 20 g is released with an initial velocity 5 m/s along the curve from the point A, as shown in the figure. The point A is at height h from point B. The particle slides along the frictionless surface. When the particle reaches point B, its angular momentum about O will be : (Take g= 10 m/s2)
|
|||
|
(a) |
8kg - m2/s |
(b) |
6kg - m2/s |
|
(c) |
3kg - m2/s |
(d) |
2kg - m2/s |
Work Energy Theorem from A to B
mgh =
m
− 
mv2
2gh = 
−
2×10×10= 
−52
Angular momentum about 0
L0= m
r
= 20×10−3×20
L0 = 6 kg.m2/s
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