The minimum speed ball should have inside a smooth vertical tube when released from a height of h=2cm as shown in the figure so that it may reach the top of the tube is (Take g=10m/s2)
A
0.9m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
1.9m/s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
2.9m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
1.26m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B1.9m/s To just complete the circle, velocity at topmost point should be zero. Let the datum or the reference line be the line from where the ball is projected. By the conservation of mechanical energy we get, 12mu2+0=12mv2top+mgΔH vtop=√u2−2g(2R−h) 0=u2−2g(2R−h) u=√2g(2R−h) u=√2×10(2×10×10−2−2×10−2) u=√20×0.18 u=1.9m/s