Question

A spehere of mass $0.2$ kg is attached to an inextensible string of length $0.5m$ whose upper end is fixed to the cielling . The sphere is made to describe a horizontal circle of radius $0.3m$ The speed of the sphere will be

Answer 1

$m=200g$
$l=130cm$

$130cm=1.3$ metre
So the time period of a pendulum doesn't depend on the mass of the Bob. It is equal to :
$\frac{2\pi \sqrt{l}}{\sqrt{g}}$
$(2pi\times rootl/rootg)$ where l is length and g is gravity. Which gives us:
$t2=4\pi 2\times 1.3/9.8$ (squaring both sides)
$\pi 2$ and $9.8$ are approx the same so we can cut it out, giving us:
$4\times 1.3=520s$
Therefore $t2=5.2$
$t=2.29s$

As per the second part of the question, the force on the string is simple mg = $200\times 10=2000$ gram m/s^2
Or $200gf$
Or $0.2kg\times 10=2N$