Question

a string of length 0.5m carries a bob with a period 2πs. calculate angle of inination with the vertical and tension in the string

Answer 1

see the attachement,
  for equilibirium,
          mω²r = mgcosA
where, ω is angular velocity , m is the mass of bob , r is the radius of circular motion and A is the angle of inclination of bob with vertical.
  we know,
         ω = 2π/T , here T is time period 
so, m{2π/T}²r = mgcosA 
⇒ T = 2π√{r/gcosA}
   now, put r = 0.5m , T = 2π sec
 so, 2π = 2π√{0.5/10cosA} 
⇒1 = 1/20cosA ⇒ cosA = 1/20 
so, A = cos⁻¹(1/20) 
hence, angle of inclination = cos⁻¹(1/20)

Answer 2

This numerical in textbook is wrong because in this equaction we need mass,

Data=mass =0.2kg

period=root2

using fomula period of conical pendulam

T=2pi root lcostheta

g

the value of g is 9.8m/s^2

substitute this value

we get 5degreee 49minutes

for tension use T=mgsec theta

=0.2×9.8 yor perid for above answer

Note=check the answer in logarithm tables