Question

A metal stick of mass $1$ kg and length 1 metre is held vertically with one end on the floor and is then allowed to fall. All values are in SI units. Take approximations.
$\begin{array}{cc}\text{Column-I}& \text{Column-II}\\ \text{(a) Initial mechanical energy}& \text{(p)}5.425\\ {\text{(b)(Angular velocity)}}^2\text{just before hitting the ground}& \text{(q) 29.429}\\ \text{(c) Linear velocity of the free end just before hitting the ground}& \text{(r) 4.9}\\ \text{(d) Moment of inertia}& \text{(s) 0.333}\end{array}$

• A. $a-p;b-q;c-r;d-p$
• B. $a-r;b-q;c-p;d-s$
• C. $a-q;b-r;c-p;d-s$
• D. $a-r;b-q;c-s;d-p$

Answer 1

The correct option is B $a-r;b-q;c-p;d-s$

$\left(a\right)\to \left(r\right);\left(b\right)\to \left(q\right);\left(c\right)\to \left(p\right);\left(d\right)\to \left(s\right)$
Moment of inertia = $\frac{1}{3}\times 1\times 1=0.33$
Mechanical energy = $1\times 9.8\times \frac{1}{2}=4.9$
By M.E. conservation
also, $\frac{1}{2}\times \frac{1}{3}\times 1\times {\omega }^2=4.9$
$\Rightarrow {\omega }^2=6\times 4.9=29.4\text{For free end}\Rightarrow v=\omega l=\omega =5.425$