Fluid MechanicsHard
Question
A uniform rod OB of length 1m, cross-sectional area 0.012 m2 and relative density 2.0 is free to rotate about O in vertical plane. The rod is held with a horizontal string AB which can withstand a maximum tension of 45 N. The rod and string system is kept in water as shown in figure. The maximum value of angle α which the rod can make with vertical without breaking the string is
Options
A.45o
B.37o
C.53o
D.60o
Solution
F.B.D. of rod NM+ dk F.B.D. :
W = (0.012) (1) (2 × 103) (10) = 240 N
Fb = (0.012) (1) (103) (10) = 120 N
Torque about O
(For equilibrium)
(240 − 120)
= 45 (cos α) ⇒ = 
= 
⇒ α = 37o
W = (0.012) (1) (2 × 103) (10) = 240 N
Fb = (0.012) (1) (103) (10) = 120 N
Torque about O
(For equilibrium)
(240 − 120)
= 45 (cos α) ⇒ = = ⇒ α = 37oCreate a free account to view solution
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