Correct answer is (15)

In this problem, the net force on the cylinder is the tension $T$ in the rope, which is equal to the force applied to the rope:

$F=T=52.5\mathrm{N}$

The force causes the cylinder to accelerate with an angular acceleration $\alpha$, which is related to its linear acceleration $a$ and the radius of the cylinder $R$ by the equation:

$\alpha =\frac{a}{R}$

The linear acceleration $a$ of the cylinder can be found using the formula $F=ma$:

$ma=F=52.5\mathrm{N}$

where $m=5\mathrm{kg}$ is the mass of the cylinder. Solving for $a$, we get:

$a=\frac{F}{m}=\frac{52.5\mathrm{N}}{5\mathrm{kg}}=10.5\mathrm{m}/{\mathrm{s}}^2$

Substituting this value of $a$ into the equation for $\alpha$, we get:

$\alpha =\frac{a}{R}=\frac{10.5\mathrm{m}/{\mathrm{s}}^2}{0.7\mathrm{m}}=\boxed{{\displaystyle 15\mathrm{rad}/{\mathrm{s}}^2}}$

Therefore, the angular acceleration of the cylinder is $15\mathrm{rad}/{\mathrm{s}}^2$.

Alternate Method:

Let's first draw a free body diagram of the cylinder. The force $F$ applied to the rope creates a tension in the rope, which in turn exerts a force on the cylinder in the opposite direction. This force is given by:

$T=F$

where $T$ is the tension in the rope. The cylinder also experiences a torque due to the tension in the rope, which causes it to rotate. The torque is given by:

$\tau =TR$

where $R$ is the radius of the cylinder.

The net torque on the cylinder is equal to the product of the moment of inertia $I$ of the cylinder and its angular acceleration $\alpha$:

$\tau =I\alpha$

The moment of inertia of a hollow cylinder about its geometrical axis which is parallel to its length is given by:

$I=M{R}^2$

where $M$ is the mass of the cylinder.

Substituting the given values, we get:

$\tau =TR=I\alpha =M{R}^2\alpha$

Solving for $\alpha$, we get:

$\alpha =\frac{T}{MR}=\frac{F}{MR}=\frac{52.5\mathrm{N}}{5\mathrm{kg}\cdot 0.7\mathrm{m}}=\boxed{{\displaystyle 15\mathrm{rad}/{\mathrm{s}}^2}}$

Therefore, the angular acceleration of the cylinder is $15\mathrm{rad}/{\mathrm{s}}^2$.

Correct answer is (3)

Let the point where the force acts be A, the origin of the coordinate system (0, 0, 0), and let the point about which the torque is calculated be B (2, -3, 0). The force vector is given by $\overrightarrow{F}=-P\hat{k}$.

To find the torque, we first find the position vector of point A with respect to point B:

${\overrightarrow{r}}_{AB}={\overrightarrow{r}}_A-{\overrightarrow{r}}_B=(0-2)\hat{i}+(0-(-3))\hat{j}+(0-0)\hat{k}=-2\hat{i}+3\hat{j}$

To calculate the cross product, we can use the determinant method with a 3x3 matrix:

$\overrightarrow{\tau }={\overrightarrow{r}}_{AB}\times \overrightarrow{F}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -2& 3& 0\\ 0& 0& -P\end{array}\right|$

Now, we will calculate the cross product components by expanding the determinant along the first row:

1. ${\tau }_i=\hat{i}\left|\begin{array}{cc}3& 0\\ 0& -P\end{array}\right|=\hat{i}((3)(-P)-(0)(0))=-3P\hat{i}$

2. ${\tau }_j=-\hat{j}\left|\begin{array}{cc}-2& 0\\ 0& -P\end{array}\right|=-\hat{j}((-2)(-P)-(0)(0))=-2P\hat{j}$

(Notice the negative sign in front of the $\hat{j}$ term, as it comes from the expansion of the determinant.)

3. ${\tau }_k=\hat{k}\left|\begin{array}{cc}-2& 3\\ 0& 0\end{array}\right|=\hat{k}((-2)(0)-(3)(0))=0\hat{k}$

Now, combine the components to get the torque vector:

$\overrightarrow{\tau }=-3P\hat{i}-2P\hat{j}+0\hat{k}=-3P\hat{i}-2P\hat{j}$

Comparing this to the given torque vector $\overrightarrow{\tau }=P(a\hat{i}+b\hat{j})$, we find that:

$a=-3$

$b=-2$

Thus, the ratio $\frac{a}{b}=\frac{-3}{-2}=\frac{x}{2}$.

Therefore, $x=3$.

Correct Answer is Option (C)

$\overrightarrow{\tau }=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 2& 1\\ 5& 3& -7\end{array}\right|$

$\begin{array}{rl}& =\hat{i}(-14-3)+\hat{j}(5+14)+\hat{k}(6-10)\\ \\ & =-17\hat{i}+19\hat{j}-4\hat{k}\end{array}$