Solveeit Logo
Login

Question

Question: A six-legged insect of mass 3.0 × 10⁻³ gm can stand on the surface of water. If the six legs bear eq...

A six-legged insect of mass 3.0 × 10⁻³ gm can stand on the surface of water. If the six legs bear equal weight and the tip of each leg is spherical in shape with a radius of 2.0 × 10⁻⁵ m, what will be the angle of contact between the insect’s leg and the water? [Surface tension of water = 7.2 × 10⁻² N m⁻¹, g = 9.8 m s⁻²]

A

30°

B

45°

C

60°

D

90°

Answer

45°

Explanation

Solution

The weight of the insect is distributed equally among its six legs. The total mass of the insect is m=3.0×10−3m = 3.0 \times 10^{-3} gm =3.0×10−6= 3.0 \times 10^{-6} kg. The total weight is W=m×g=(3.0×10−6 kg)×(9.8 m s⁻²)=29.4×10−6W = m \times g = (3.0 \times 10^{-6} \text{ kg}) \times (9.8 \text{ m s⁻²}) = 29.4 \times 10^{-6} N. The weight supported by each leg is Wleg=W6=29.4×10−6 N6=4.9×10−6W_{leg} = \frac{W}{6} = \frac{29.4 \times 10^{-6} \text{ N}}{6} = 4.9 \times 10^{-6} N.

The upward force due to surface tension on each leg supports this weight. The surface tension force acts tangentially along the line of contact. For a spherical tip of radius rr, the radius of the contact circle RR is related to the angle of contact θ\theta (assumed to be the angle between the tangent to the solid surface and the horizontal) by R=rcos⁡θR = r \cos \theta. The upward force due to surface tension is given by the vertical component of the force acting along the circumference of contact: Fup=(γ×2πR)sin⁡θF_{up} = (\gamma \times 2\pi R) \sin \theta.

Substituting R=rcos⁡θR = r \cos \theta, we get: Fup=γ×(2πrcos⁡θ)×sin⁡θF_{up} = \gamma \times (2\pi r \cos \theta) \times \sin \theta Fup=πγr(2sin⁡θcos⁡θ)F_{up} = \pi \gamma r (2 \sin \theta \cos \theta) Fup=πγrsin⁡(2θ)F_{up} = \pi \gamma r \sin(2\theta)

For equilibrium, Wleg=FupW_{leg} = F_{up}: 4.9×10−6 N=π×(7.2×10−2 N m⁻¹)×(2.0×10−5 m)×sin⁡(2θ)4.9 \times 10^{-6} \text{ N} = \pi \times (7.2 \times 10^{-2} \text{ N m⁻¹}) \times (2.0 \times 10^{-5} \text{ m}) \times \sin(2\theta) 4.9×10−6=(14.4π×10−7)sin⁡(2θ)4.9 \times 10^{-6} = (14.4\pi \times 10^{-7}) \sin(2\theta) 4.9×10−6≈(4.524×10−6)sin⁡(2θ)4.9 \times 10^{-6} \approx (4.524 \times 10^{-6}) \sin(2\theta) sin⁡(2θ)=4.9×10−64.524×10−6≈1.083\sin(2\theta) = \frac{4.9 \times 10^{-6}}{4.524 \times 10^{-6}} \approx 1.083

Since sin⁡(2θ)\sin(2\theta) cannot be greater than 1, the given parameters indicate that the insect cannot stand on the water surface. However, if we assume there is a slight inaccuracy in the problem's values and that a solution exists, the calculated value of sin⁡(2θ)≈1.083\sin(2\theta) \approx 1.083 is very close to 1. If sin⁡(2θ)=1\sin(2\theta) = 1, then 2θ=90∘2\theta = 90^\circ, which gives θ=45∘\theta = 45^\circ. This is a common angle in such problems, suggesting it might be the intended answer despite the numerical inconsistency.