Given:
Let \( x \) be the side length of the square cut from each corner of the square sheet.
\( V(x) = (30 - 2x)^2x = 4x^3 - 120x^2 + 900x \).
\( V'(x) = \frac{dV}{dx} = 12x^2 - 240x + 900 \).
Set \( V'(x) = 0 \):\( 0 = 12x^2 - 240x + 900 \).
Simplify:\( 0 = x^2 - 20x + 75 \).
Factorize:\( 0 = (x - 5)(x - 15) \).
So, \( x = 5 \) or \( x = 15 \).If \( x = 15 \), the length and breadth become \( 30 - 2(15) = 0 \), which is not possible. Therefore, \( x = 5 \).
\( S(x) = (30 - 2x)^2 + 4x(30 - 2x) \).
Substituting \( x = 5 \):\( S(5) = (30 - 2(5))^2 + 4(5)(30 - 2(5)) \).
Simplify:\( S(5) = (20)^2 + 20(20) = 400 + 400 = 800 \, \text{cm}^2 \).
Final Answer: The surface area of the open box is \( 800 \, \text{cm}^2 \).
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient 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.
(Given: $g = 10\,\text{m s}^{-2}$) 
Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).
