Question

The minimum speed ball should have inside a smooth vertical tube when released from a height of $h=2\text{cm}$ as shown in the figure so that it may reach the top of the tube is
(Take $g=10m/s^2$)

• A. $0.9\text{m/s}$
• B. $1.9\text{m/s}$
• C. $2.9\text{m/s}$
• D. $1.26\text{m/s}$

Answer 1

The correct option is B $1.9\text{m/s}$

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,
$\frac{1}{2}m{u}^2+0=\frac{1}{2}m{v}_{top}^2+mg\Delta H$
$v_{top}=\sqrt{u^2-2g(2R-h)}$
$0=u^2-2g(2R-h)$
$u=\sqrt{2g(2R-h)}$
$u=\sqrt{2\times 10(2\times 10\times {10}^{-2}-2\times {10}^{-2})}$
$u=\sqrt{20\times 0.18}$
$u=1.9\text{m/s}$

Hence option B is the correct answer