The correct answer is (A) : 2.8 × 10–4 J
\(r′ = \frac{r}{4}\)
⇒ ΔE = T(ΔS)
= T × 4π(nr′2 –r2), n = 64
= T × 4π × (4 – 1)r2
⇒ ΔE = 0.075 × 4 × 3.142 (3) × 10–4 J
= 2.8 × 10–4 J
Two soap bubbles of radius 2 cm and 4 cm, respectively, are in contact with each other. The radius of curvature of the common surface, in cm, is _______________.
Let $A = \begin{bmatrix} \cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta \end{bmatrix}$. If for some $\theta \in (0, \pi)$, $A^2 = A^T$, then the sum of the diagonal elements of the matrix $(A + I)^3 + (A - I)^3 - 6A$ is equal to
Let $A = \{ z \in \mathbb{C} : |z - 2 - i| = 3 \}$, $B = \{ z \in \mathbb{C} : \text{Re}(z - iz) = 2 \}$, and $S = A \cap B$. Then $\sum_{z \in S} |z|^2$ is equal to
Let $C$ be the circle $x^2 + (y - 1)^2 = 2$, $E_1$ and $E_2$ be two ellipses whose centres lie at the origin and major axes lie on the $x$-axis and $y$-axis respectively. Let the straight line $x + y = 3$ touch the curves $C$, $E_1$, and $E_2$ at $P(x_1, y_1)$, $Q(x_2, y_2)$, and $R(x_3, y_3)$ respectively. Given that $P$ is the mid-point of the line segment $QR$ and $PQ = \frac{2\sqrt{2}}{3}$, the value of $9(x_1 y_1 + x_2 y_2 + x_3 y_3)$ is equal to
Let $A$ and $B$ be two distinct points on the line $L: \frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2}$. Both $A$ and $B$ are at a distance $2\sqrt{17}$ from the foot of perpendicular drawn from the point $(1, 2, 3)$ on the line $L$. If $O$ is the origin, then $\overrightarrow{OA} \cdot \overrightarrow{OB}$ is equal to:
The amount of energy required to increase the liquid's surface area by one unit area is known as surface tension. In other words, it is a property of the liquid surface to resist force.
Surface tension is defined as,
The ratio of the surface force F to the length L along which the force acts.
Mathematically, the surface tension formula can be expressed as follows:
T=F/L
Where,
Read More: Detergents and Surface Tension
The SI unit of Surface Tension is Newton per Meter or N/m.