A boy is hanging from a horizontal branch of a tree. The tension in the arms will be maximum when the angle between the arms is

Question

A boy is hanging from a horizontal branch of a tree. The tension in the arms will be maximum when the angle between the arms is (A)  0°  (B)  60°  (C)  90°  (D)  120°

Tardigrade Admin · Accepted Answer

Different forces acting on the boy are shown in figure.

In equilibrium

2 T cos θ = m g

T = \frac{m g}{2 c o s θ}

Tension will be maximum,
when value of

2 cos θ

is

1

If

θ = 6 0^{\circ}

Then,

2 cos θ = 2 cos 6 0^{\circ}

= 2 \times \frac{1}{2} = 1

So, angle between the arms

= 2 \times 60 = 12 0^{\circ}

Q. A boy is hanging from a horizontal branch of a tree. The tension in the arms will be maximum when the angle between the arms is

$0^{\circ}$

$6 0^{\circ}$

$9 0^{\circ}$

$12 0^{\circ}$

Q. A boy is hanging from a horizontal branch of a tree. The tension in the arms will be maximum when the angle between the arms is

0∘

60∘

90∘

120∘

Different forces acting on the boy are shown in figure. In equilibrium 2Tcosθ=mg T=2cosθmg​ Tension will be maximum, when value of 2cosθ is 1 If θ=60∘ Then, 2cosθ=2cos60∘ =2×21​=1 So, angle between the arms =2×60=120∘

$0^{\circ}$

$6 0^{\circ}$

$9 0^{\circ}$

$12 0^{\circ}$