($\pi = 22/7$)
This section provides a step-by-step calculation for the work done in increasing the diameter of a soap bubble, based on surface tension.
The work done ($W$) to change the surface area of a liquid film is related to surface tension ($T$) and the change in area ($\Delta A$). For a soap bubble, which has two surfaces (inner and outer), the formula is:
$W = 2 \times T \times \Delta A$
The surface area ($A$) of a sphere with radius ($r$) is given by $A = 4 \pi r^2$. Therefore, the change in area is $\Delta A = A_{final} - A_{initial} = 4 \pi r_{final}^2 - 4 \pi r_{initial}^2$.
The change in area depends on the squares of the radii:
$\Delta A = 4 \pi (r_2^2 - r_1^2)$
Substituting the values:
$\Delta A = 4 \times \frac{22}{7} \times \left( (0.07)^2 - (0.035)^2 \right) \text{ m}^2$
$\Delta A = \frac{88}{7} \times (0.0049 - 0.001225) \text{ m}^2$
$\Delta A = \frac{88}{7} \times 0.003675 \text{ m}^2$
$\Delta A = 0.0462 \text{ m}^2$
Use the work formula $W = 2 \times T \times \Delta A$:
$W = 2 \times (0.04 \text{ N/m}) \times (0.0462 \text{ m}^2)$
$W = 0.08 \times 0.0462 \text{ J}$
$W = 0.003696 \text{ J}$
Convert the work done to microJoules ($\mu J$):
$W = 0.003696 \times 10^6 \text{ }\mu\text{J} = 3696 \text{ }\mu\text{J}$
Equate the calculated work ($W$) with the given expression for work done:
$15000 - x = 3696$
Solve for $x$:
$x = 15000 - 3696$
$x = 11304$
The value of $x$ is 11304.
In the given figure the blocks $A$, $B$ and $C$ weigh 4 kg, 6 kg and 8 kg respectively. The co-efficient of sliding friction between any two surfaces is 0.5. The force $\vec{F}$ required to slide the block $C$ with constant speed is ______ N. (Use $g = 10 \text{ m/s}^2$)

A thin uniform rod ($X$) of mass $M$ and length $L$ is pivoted at a height $\left(\frac{L}{3}\right)$ as shown in the figure. The rod is allowed to fall from a vertical position and lie horizontally on the table. The angular velocity of this rod when it hits the table top, is __________.
($g = \text{gravitational acceleration}$)

In the given figure the blocks $A$, $B$ and $C$ weigh 4 kg, 6 kg and 8 kg respectively. The co-efficient of sliding friction between any two surfaces is 0.5. The force $\vec{F}$ required to slide the block $C$ with constant speed is ______ N. (Use $g = 10 \text{ m/s}^2$)
