($\pi = 22/7$)
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.
Suppose there is a uniform circular disc of mass $M \text{ kg}$ and radius $r \text{ m}$ shown in figure. The shaded regions are cut out from the disc. The moment of inertia of the remainder about the axis $A$ of the disc is given by $\frac{x}{256} M r^2$. The value of $x$ is ________.

A block of mass M is attached to a wire. The upper part of wire is made of steel and lower part of wire of copper (as shown in figure). Both parts have the same area of cross section. Neglecting the mass of wires, the ratio of longitudinal strain developed in steel and copper wires is :
(Given, Young's modulus of steel = $2 \ \times \ 10^{11} \ N/m^2$ and Young's modulus of copper = $1.5 \ \times \ 10^{11} \ N/m^2$)

A fly wheel of mass $10$ kg and radius $50$ cm is rotating at a rate of$360$ rpm. Assuming mass to be concentrated at the rim, the constant retarding torque required to stop the rotation of wheel in $6$ rotations will be x Nm. The value of x is ____________.
A body moving horizontally on a smooth surface with a speed of $20$ $ms^{-1}$ splits into two parts and continue to move in the same direction. The masses of two parts are in the ratio of $1:2$. The smaller part moves with a speed of $30$ $ms^{-1}$. The fractional change is the kinetic energy is given by $\frac{1}{x}$. The value of x is__________
A thin uniform bar of length L and mass 8 m lies on a smooth horizontal table. Two point masses m and 2 m are moving in the same horizontal plane from opposite sides of the bar with speeds 2v and v respectively. The masses stick to the bar after collision at a distance $\frac{L}{3}$ and $\frac{L}{6}$ respectively from the centre of the bar. If the bar starts rotating about its center of mass as a result of collision, the angular speed of the bar will be :

As shown in the figure, forces of $10^5$ N each are applied in opposite directions, on the upper and lower faces of a cube of side 10 cm, shifting the upper face parallel to itself by 0.5 cm. If the side of another cube of the same material is 20 cm, then under similar conditions as above, the displacement will be :

Suppose there is a uniform circular disc of mass $M \text{ kg}$ and radius $r \text{ m}$ shown in figure. The shaded regions are cut out from the disc. The moment of inertia of the remainder about the axis $A$ of the disc is given by $\frac{x}{256} M r^2$. The value of $x$ is ________.
