A thin rod MN , free to rotate in the vertical plane about the fixed end N , is held horizontal. When the end M is released the speed of this end, when the rod makes an angle $\alpha$ with the horizontal, will be proportional to: [Figure]
(1) $\sqrt { \cos \alpha }$
(2) $\cos \alpha$
(3) $\sin \alpha$
(4) $\sqrt { \sin \alpha }$

A pendulum bob has a speed of $3\mathrm{~m~s}^{-1}$ at its lowest position. The pendulum is 50 cm long. The speed of bob, when the length makes an angle of $60^{\circ}$ to the vertical will be ($g = 10\mathrm{~m~s}^{-2}$) \_\_\_\_ $\mathrm{m~s}^{-1}$.

A pendulum is suspended by a string of length 250 cm. The mass of the bob of the pendulum is 200 g. The bob is pulled aside until the string is at $60^\circ$ with vertical as shown in the figure. After releasing the bob, the maximum velocity attained by the bob will be $\_\_\_\_$ $\mathrm{m\,s^{-1}}$. (if $g = 10\mathrm{~m\,s^{-2}}$)

In the given figure, the block of mass $m$ is dropped from the point $|A|$. The expression for kinetic energy of block when it reaches point $|B|$ is
(1) $mgy_0$
(2) $\frac{1}{2}mgy_0^2$
(3) $\frac{1}{2}mgy^2$
(4) $mg(y - y_0)$

A bob of mass $m$ is suspended by a light string of length $L$. It is imparted a minimum horizontal velocity at the lowest point $A$ such that it just completes half circle reaching the top most position $B$. The ratio of kinetic energies $\frac { ( \text { K.E. } ) _ { A } } { ( \text { K.E. } ) _ { B } }$ is:
(1) $3 : 2$
(2) $5 : 1$
(3) $2 : 5$
(4) $1 : 5$

A particle is placed at the point A of a frictionless track ABC as shown in figure. It is gently pushed towards right. The speed of the particle when it reaches the point $B$ is: (Take $g = 10 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ ).
(1) $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$
(2) $\sqrt { 10 } \mathrm {~m} \mathrm {~s} ^ { - 1 }$
(3) $2 \sqrt { 10 } \mathrm {~m} \mathrm {~s} ^ { - 1 }$
(4) $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$

A body of mass $m$ connected to a massless and unstretchable string goes in a vertical circle of radius $R$ under gravity $g$. The other end of the string is fixed at the center of circle. If velocity at top of circular path is $n\sqrt{gR}$, where $n \geq 1$, then ratio of kinetic energy of the body at bottom to that at top of the circle is:
(1) $\frac{n^2}{n^2 + 4}$
(2) $\frac{n^2 + 4}{n^2}$
(3) $\frac{n+4}{n}$
(4) $\frac{n}{n+4}$

Q4. A body of $m \mathrm {~kg}$ slides from rest along the curve of vertical circle from point $A$ to $B$ in friction less path. The [Figure] velocity of the body at $B$ is: (given, $R = 14 \mathrm {~m} , g = 10 \mathrm {~m} / \mathrm { s } ^ { 2 }$ and $\sqrt { 2 } = 1.4$ )
(1) $16.7 \mathrm {~m} / \mathrm { s }$
(2) $19.8 \mathrm {~m} / \mathrm { s }$
(3) $10.6 \mathrm {~m} / \mathrm { s }$
(4) $21.9 \mathrm {~m} / \mathrm { s }$