Here are the solutions to the questions: --- Question W: Derive the dimension of the volume of a cylinder using height (h), radius (r), and diameter (d). Step 1: Recall the formula for the volume of a cylinder. The volume of a cylinder ($V$) is given by: $$V = \pi r^2 h$$ Step 2: Identify the dimensions of each physical quantity. Radius ($r$) is a length, so its dimension is $[L]$. Height ($h$) is a length, so its dimension is $[L]$. Diameter ($d$) is also a length, so its dimension is $[L]$. $\pi$ is a dimensionless constant. Step 3: Substitute the dimensions into the volume formula. The dimension of $V$ is: $$[V] = [r]^2 [h]$$ $$[V] = [L]^2 [L]$$ $$[V] = [L^3]$$ Step 4: Express the dimension using diameter (d). Since the radius $r$ is half the diameter $d$ ($r = \frac{d}{2}$), its dimension is still $[L]$. Therefore, using diameter instead of radius does not change the overall dimension of volume. $$[V] = [d]^2 [h]$$ $$[V] = [L]^2 [L]$$ $$[V] = [L^3]$$ The dimension of the volume of a cylinder is $\boxed{[L^3]}$. --- Question 4: If a bob is attached to a string of length 80 cm and lifted at an angle of 45 degrees to cause a displacement. i) Find the speed of the moving bob. Assumption: We assume the question asks for the speed of the bob at the lowest point of its swing after being released from rest at an angle of $45^\circ$. We will use the principle of conservation of mechanical energy. Step 1: Convert the string length to meters. Length of the string, $L = 80 \, \text{cm} = 0.80 \, \text{m}$. Step 2: Calculate the vertical height ($h$) the bob falls. When the bob is lifted to an angle $\theta$, its vertical height from the pivot is $L \cos\theta$. When it is at the lowest point, its vertical height from the pivot is $L$. The change in vertical height (the height it falls) is: $$h = L - L \cos\theta = L(1 - \cos\theta)$$ Given $\theta = 45^\circ$: $$h = 0.80 \, \text{m} (1 - \cos 45^\circ)$$ $$h = 0.80 \, \text{m} \left(1 - \frac{\sqrt{2}}{2}\right)$$ $$h = 0.80 \, \text{m} (1 - 0.7071)$$ $$h = 0.80 \, \text{m} (0.2929)$$ $$h = 0.23432 \, \text{m}$$ Step 3: Apply the conservation of mechanical energy. At the initial position (lifted at $45^\circ$), the bob has potential energy ($PE_i$) and zero kinetic energy ($KE_i = 0$). At the lowest point, the bob has zero potential energy ($PE_f = 0$) and maximum kinetic energy ($KE_f$). $$PE_i + KE_i = PE_f + KE_f$$ $$mgh + 0 = 0 + \frac{1}{2}mv^2$$ Where $m$ is the mass of the bob, $g$ is the acceleration due to gravity ($9.8 \, \text{m/s}^2$), and $v$ is the speed at the lowest point. Step 4: Solve for the speed ($v$). $$mgh = \frac{1}{2}mv^2$$ Divide both sides by $m$: $$gh = \frac{1}{2}v^2$$ $$v^2 = 2gh$$ $$v = \sqrt{2gh}$$ Substitute the values for $g$ and $h$: $$v = \sqrt{2 \times 9.8 \, \text{m/s}^2 \times 0.23432 \, \text{m}}$$ $$v = \sqrt{4.592672 \, \text{m}^2/\text{s}^2}$$ $$v \approx 2.143 \, \text{m/s}$$ The speed of the moving bob at the lowest point is $\boxed{2.14 \, \text{m/s}}$. ii) Calculate X from the diagram shown. The diagram mentioned in this part of the question is not provided. Therefore, it is not possible to calculate X.