Step 1: The equation is given as:
\[ \frac{24 + a + b}{8} = 4 \]
Step 2: Simplifying the equation:
\[ a + b = 8 \]
Step 3: The next equation is:
\[ 2 = \frac{4 + 1 + 1 + 0 + 1 + 9 + (a - 4)^2 + (b - 4)^2}{8} \] Simplifying: \[ 16 = 48 + a^2 + b^2 - 8a - 8b \]
Step 4: This leads to:
\[ a^2 + b^2 = 32 \]
Step 5: We also know:
\[ 32 = 2ab \] Solving for \( ab \): \[ ab = 16 \]
Step 6: From \( a + b = 8 \) and \( ab = 16 \), we find:
\[ a = 4, \quad b = 4 \]
Step 7: The mode is given as:
\[ \text{Mode} = 4 \]
Step 8: The mean deviation is calculated as:
\[ \text{Mean Deviation} = \frac{2 + 1 + 1 + 0 + 1 + 1 + 3 + 0 + 0}{8} = 1 \]
If the mean and the variance of 6, 4, a, 8, b, 12, 10, 13 are 9 and 9.25 respectively, then \(a + b + ab\) is equal to:
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\(^{-2}\).
