Centre of mass (Point Mass) Questions in English

(B) The position of the centre of mass $(R_{cm})$ for a two-particle system is given by the formula $R_{cm} = \frac{m_1r_1 + m_2r_2}{m_1 + m_2}$.
For two masses $m$ and $M$ separated by a distance $d$,the centre of mass lies on the line joining them.
Specifically,the distance of the centre of mass from the mass $M$ is $r_M = \frac{m \cdot d}{M + m}$ and from the mass $m$ is $r_m = \frac{M \cdot d}{M + m}$.
Since $M > m$,it follows that $r_M < r_m$.
Therefore,the centre of mass is always located closer to the heavier mass,which is $M$.

(C) Given masses are $m_1 = 200 \ gm = 0.2 \ kg$ and $m_2 = 500 \ gm = 0.5 \ kg$.
Velocities are $\vec{v}_1 = 10 \hat{i} \ m/s$ and $\vec{v}_2 = (3 \hat{i} + 5 \hat{j}) \ m/s$.
The velocity of the centre of mass $\vec{v}_{cm}$ is given by the formula:
$\vec{v}_{cm} = \frac{m_1 \vec{v}_1 + m_2 \vec{v}_2}{m_1 + m_2}$
Substituting the values:
$\vec{v}_{cm} = \frac{0.2(10 \hat{i}) + 0.5(3 \hat{i} + 5 \hat{j})}{0.2 + 0.5}$
$\vec{v}_{cm} = \frac{2 \hat{i} + 1.5 \hat{i} + 2.5 \hat{j}}{0.7}$
$\vec{v}_{cm} = \frac{3.5 \hat{i} + 2.5 \hat{j}}{0.7}$
$\vec{v}_{cm} = \frac{3.5}{0.7} \hat{i} + \frac{2.5}{0.7} \hat{j}$
$\vec{v}_{cm} = 5 \hat{i} + \frac{25}{7} \hat{j} \ m/s$.

(B) Let the masses be placed at the vertices of the parallelogram. Let mass $m$ be at $(0, 0)$. Since the side length is $a$ and one angle is $60^o$,the mass $4m$ is at $(a, 0)$. The mass $2m$ is at $(a \cos 60^o, a \sin 60^o) = (a/2, \sqrt{3}a/2)$. The mass $3m$ is at $(a + a/2, \sqrt{3}a/2) = (3a/2, \sqrt{3}a/2)$.
The $x$-coordinate of the centre of mass is $x_{cm} = \frac{m(0) + 4m(a) + 2m(a/2) + 3m(3a/2)}{m + 2m + 3m + 4m} = \frac{0 + 4ma + ma + 4.5ma}{10m} = \frac{9.5ma}{10m} = 0.95a$.
The $y$-coordinate of the centre of mass is $y_{cm} = \frac{m(0) + 4m(0) + 2m(\sqrt{3}a/2) + 3m(\sqrt{3}a/2)}{10m} = \frac{5\sqrt{3}ma/2}{10m} = \frac{\sqrt{3}a}{4}$.
Thus,the centre of mass is at $(0.95a, \frac{\sqrt{3}}{4}a)$.

(C) The coordinates of the centre of mass $(X_{cm}, Y_{cm})$ for a system of particles are given by:
$X_{cm} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3}$
$Y_{cm} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3}$
Given $m_1 = m_2 = m_3 = m$ and positions $(x_1, y_1) = (1, 1), (x_2, y_2) = (2, 2), (x_3, y_3) = (3, 3)$.
$X_{cm} = \frac{m(1) + m(2) + m(3)}{m + m + m} = \frac{6m}{3m} = 2$
$Y_{cm} = \frac{m(1) + m(2) + m(3)}{m + m + m} = \frac{6m}{3m} = 2$
Thus,the centre of mass is at $(2, 2)$.

(A) The coordinates of the four spheres are $(1, 1)$,$(-1, 1)$,$(-1, -1)$,and $(1, -1)$.
Since all spheres have the same mass $m$,the coordinates of the centre of mass $(X_{cm}, Y_{cm})$ are given by:
$X_{cm} = \frac{m(1) + m(-1) + m(-1) + m(1)}{m + m + m + m} = \frac{0}{4m} = 0$
$Y_{cm} = \frac{m(1) + m(1) + m(-1) + m(-1)}{m + m + m + m} = \frac{0}{4m} = 0$
Thus,the centre of mass is at $(0, 0)$.

(C) Let the position of mass $m$ be at the origin $(x_1 = 0)$ and the position of mass $M$ be at $(x_2 = L)$ along the $x$-axis.
The formula for the center of mass $X_{cm}$ of a two-particle system is given by:
$X_{cm} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2}$
Substituting the values:
$X_{cm} = \frac{m(0) + M(L)}{m + M}$
$X_{cm} = \frac{ML}{m + M} = L\left( \frac{M}{m + M} \right)$
Thus,the distance of the center of mass from $m$ is $L\left( \frac{M}{m + M} \right)$.

(B) Let the radius of each sphere be $r$. Since the spheres are identical and touching,the distance between the centres of adjacent spheres is $2r$.
Let the positions of the centres be $x_K = 0$,$x_L = 2r$,and $x_M = 4r$.
The mass of each sphere is $m = 1 \ kg$.
The centre of mass $X_{cm}$ is given by $X_{cm} = \frac{m x_K + m x_L + m x_M}{m + m + m} = \frac{0 + 2r + 4r}{3} = \frac{6r}{3} = 2r$.
Since $KL = 2r$ and $KM = 4r$,we have $X_{cm} = KL = \frac{KM}{2}$.
Alternatively,$X_{cm} = \frac{KL + KM}{3} = \frac{2r + 4r}{3} = 2r$.
Thus,the distance of the centre of mass from $K$ is $\frac{KL + KM}{3}$.

(C) Let the carbon atom be at the origin $(0, 0)$ and the oxygen atom be placed at $(1.1, 0)$ on the $x$-axis.
Given: $m_1 = 12 \ a.m.u.$ (carbon),$m_2 = 16 \ a.m.u.$ (oxygen).
Position vectors are $\vec{r}_1 = 0\hat{i}$ and $\vec{r}_2 = 1.1\hat{i}$.
The center of mass $R_{cm}$ is given by:
$R_{cm} = \frac{m_1 r_1 + m_2 r_2}{m_1 + m_2}$
$R_{cm} = \frac{12(0) + 16(1.1)}{12 + 16}$
$R_{cm} = \frac{17.6}{28} \ \mathring{A}$
$R_{cm} \approx 0.63 \ \mathring{A}$ from the carbon atom.

(B) Let corner $A$ of the square $ABCD$ be at the origin $(0,0)$. The mass $8\,kg$ is placed at $A$. The diagonal of the square is $d = a\sqrt{2} = 80\,cm$,where $a$ is the side length.
$a = \frac{80}{\sqrt{2}} = 40\sqrt{2}\,cm$.
The masses are: $m_A = 8\,kg$,$m_B = 2\,kg$,$m_C = 4\,kg$,$m_D = 2\,kg$.
The coordinates of the corners are: $A(0,0)$,$B(a,0)$,$C(a,a)$,$D(0,a)$.
The coordinates of the centre of mass $(X_{CM}, Y_{CM})$ are given by:
$X_{CM} = \frac{m_A x_A + m_B x_B + m_C x_C + m_D x_D}{m_A + m_B + m_C + m_D} = \frac{8(0) + 2(a) + 4(a) + 2(0)}{8 + 2 + 4 + 2} = \frac{6a}{16} = \frac{3a}{8}$.
$Y_{CM} = \frac{m_A y_A + m_B y_B + m_C y_C + m_D y_D}{m_A + m_B + m_C + m_D} = \frac{8(0) + 2(0) + 4(a) + 2(a)}{8 + 2 + 4 + 2} = \frac{6a}{16} = \frac{3a}{8}$.
Substituting $a = 40\sqrt{2}\,cm$:
$X_{CM} = \frac{3(40\sqrt{2})}{8} = 15\sqrt{2}\,cm$.
$Y_{CM} = \frac{3(40\sqrt{2})}{8} = 15\sqrt{2}\,cm$.
The distance of the centre of mass from $A(0,0)$ is $d = \sqrt{X_{CM}^2 + Y_{CM}^2} = \sqrt{(15\sqrt{2})^2 + (15\sqrt{2})^2} = \sqrt{450 + 450} = \sqrt{900} = 30\,cm$.

(C) Given masses are $m_1 = 7 \, g$,$m_2 = 4 \, g$,and $m_3 = 10 \, g$.
Their position vectors are $\vec{r}_1 = (1\hat{i} + 5\hat{j} - 3\hat{k}) \, cm$,$\vec{r}_2 = (2\hat{i} + 5\hat{j} + 7\hat{k}) \, cm$,and $\vec{r}_3 = (3\hat{i} + 3\hat{j} - 1\hat{k}) \, cm$.
The position of the centre of mass $\vec{R}_{cm}$ is given by $\vec{R}_{cm} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2 + m_3\vec{r}_3}{m_1 + m_2 + m_3}$.
Substituting the values:
$\vec{R}_{cm} = \frac{7(1\hat{i} + 5\hat{j} - 3\hat{k}) + 4(2\hat{i} + 5\hat{j} + 7\hat{k}) + 10(3\hat{i} + 3\hat{j} - 1\hat{k})}{7 + 4 + 10}$
$\vec{R}_{cm} = \frac{(7\hat{i} + 35\hat{j} - 21\hat{k}) + (8\hat{i} + 20\hat{j} + 28\hat{k}) + (30\hat{i} + 30\hat{j} - 10\hat{k})}{21}$
$\vec{R}_{cm} = \frac{(7+8+30)\hat{i} + (35+20+30)\hat{j} + (-21+28-10)\hat{k}}{21}$
$\vec{R}_{cm} = \frac{45\hat{i} + 85\hat{j} - 3\hat{k}}{21} = \frac{45}{21}\hat{i} + \frac{85}{21}\hat{j} - \frac{3}{21}\hat{k}$
$\vec{R}_{cm} = \frac{15}{7}\hat{i} + \frac{85}{21}\hat{j} - \frac{1}{7}\hat{k}$.
Thus,the coordinates are $\left( \frac{15}{7}, \frac{85}{21}, -\frac{1}{7} \right) \, cm$.

(C) The system is in equilibrium when supported at its center of mass $(COM)$.
The masses are $m_i = i$ (in grams) and their positions are $x_i = i$ (in cm) for $i = 1, 2, \dots, 100$.
The center of mass is given by:
$COM = \frac{\sum_{i=1}^{100} m_i x_i}{\sum_{i=1}^{100} m_i} = \frac{\sum_{i=1}^{100} i^2}{\sum_{i=1}^{100} i}$
Using the summation formulas:
$\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$ and $\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$
For $n = 100$:
$COM = \frac{\frac{100(101)(201)}{6}}{\frac{100(101)}{2}} = \frac{201}{3} = 67\,cm$.
Thus,the system should be supported at the $67\,cm$ mark.

(B) The masses of the atoms are $m_1 = 1 \ amu$ (for $H$) and $m_2 = 35.5 \ amu$ (for $Cl$). The interatomic distance is $L = 1 \ \mathring A = 10^{-10} \ m$.
Let $x$ be the distance of $H$ from the center of mass ($C$.$M$.). Then the distance of $Cl$ from the $C$.$M$. is $(L - x)$.
Using the definition of the center of mass: $m_1 x = m_2 (L - x) \Rightarrow 1 \cdot x = 35.5 \cdot (1 - x)$.
$x = 35.5 - 35.5x \Rightarrow 36.5x = 35.5 \Rightarrow x = \frac{35.5}{36.5} \ \mathring A \approx 0.9726 \ \mathring A$.
The distance of $Cl$ from the $C$.$M$. is $(L - x) = 1 - 0.9726 = 0.0274 \ \mathring A$.
The moment of inertia $I$ is given by $I = m_1 x^2 + m_2 (L - x)^2$.
Using $1 \ amu = 1.67 \times 10^{-27} \ kg$:
$I = (1 \times 1.67 \times 10^{-27}) \times (0.9726 \times 10^{-10})^2 + (35.5 \times 1.67 \times 10^{-27}) \times (0.0274 \times 10^{-10})^2$.
$I = 1.67 \times 10^{-27} \times 10^{-20} \times [0.9726^2 + 35.5 \times 0.0274^2]$.
$I = 1.67 \times 10^{-47} \times [0.946 + 35.5 \times 0.00075] \approx 1.67 \times 10^{-47} \times [0.946 + 0.0266] \approx 1.67 \times 10^{-47} \times 0.9726 \approx 1.62 \times 10^{-47} \ kg \ m^2$.
This is approximately $1.61 \times 10^{-47} \ kg \ m^2$.

(B) Since the net external force on the system is zero (or the forces are internal to the system),the center of mass of the system remains at rest if it was initially at rest.
They will collide at their center of mass.
The position of the center of mass is given by $x_{cm} = \frac{M x_A + m x_B}{M + m}$.
Since $M > m$,the center of mass is closer to the heavier mass $A$.
Therefore,they will collide near $A$.

(C) Let the square plate be placed in the $XY$-plane with its center at the origin $O(0,0)$.
Let the four corners be located at $(a, a), (-a, a), (-a, -a),$ and $(a, -a)$.
Let the squares $1, 2, 3,$ and $4$ be at these corners respectively.
When squares $1$ and $3$ (diagonally opposite) are removed,the remaining system is symmetric about the line passing through the centers of squares $2$ and $4$.
This line corresponds to the diagonal $Y = -X$ (or $OX'$ axis depending on the coordinate orientation).
Due to the symmetry of the remaining mass distribution about this diagonal,the center of mass must lie on this line of symmetry.
Therefore,the center of mass will be located on the axis $OX'$.

(C) The velocity of the center of mass $\vec{V}_{cm}$ is given by the formula:
$\vec{V}_{cm} = \frac{m_1 \vec{v}_1 + m_2 \vec{v}_2}{m_1 + m_2}$
Given: $m_1 = 200 \ g$,$m_2 = 500 \ g$,$\vec{v}_1 = 10 \ \hat{i} \ m/s$,$\vec{v}_2 = 3 \ \hat{i} + 5 \ \hat{j} \ m/s$.
Substituting the values:
$\vec{V}_{cm} = \frac{200(10 \ \hat{i}) + 500(3 \ \hat{i} + 5 \ \hat{j})}{200 + 500}$
$\vec{V}_{cm} = \frac{2000 \ \hat{i} + 1500 \ \hat{i} + 2500 \ \hat{j}}{700}$
$\vec{V}_{cm} = \frac{3500 \ \hat{i} + 2500 \ \hat{j}}{700}$
$\vec{V}_{cm} = 5 \ \hat{i} + \frac{25}{7} \ \hat{j} \ m/s$.

(B) Let the masses be $m_1 = 1 \, g, m_2 = 2 \, g, m_3 = 3 \, g$ and $m_4 = 4 \, g$.
Since the center of mass of the first three particles is at the origin $(0, 0, 0)$,their combined position vector is $\vec{R}_{123} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2 + m_3\vec{r}_3}{m_1 + m_2 + m_3} = (0, 0, 0)$,which implies $m_1\vec{r}_1 + m_2\vec{r}_2 + m_3\vec{r}_3 = 0$.
The position vector of the fourth particle is $\vec{r}_4 = \alpha(\hat{i} + 2\hat{j} + 3\hat{k})$.
The center of mass of the new system is $\vec{R}_{cm} = (1, 2, 3)$.
The formula for the center of mass is $\vec{R}_{cm} = \frac{(m_1\vec{r}_1 + m_2\vec{r}_2 + m_3\vec{r}_3) + m_4\vec{r}_4}{m_1 + m_2 + m_3 + m_4}$.
Substituting the known values: $(1, 2, 3) = \frac{0 + 4\alpha(\hat{i} + 2\hat{j} + 3\hat{k})}{1 + 2 + 3 + 4} = \frac{4\alpha(\hat{i} + 2\hat{j} + 3\hat{k})}{10}$.
Comparing the $x$-components: $1 = \frac{4\alpha}{10} \implies 10 = 4\alpha \implies \alpha = \frac{10}{4} = 2.5 = 5/2$.
Similarly,for $y$-components: $2 = \frac{8\alpha}{10} \implies 20 = 8\alpha \implies \alpha = 2.5 = 5/2$.
Thus,the value of $\alpha$ is $5/2$.

(B) Let vertex $A$ be at the origin $(0, 0)$. The side length of the square $s$ is related to the diagonal $d = 80 \, cm$ by $s = d / \sqrt{2} = 80 / \sqrt{2} = 40\sqrt{2} \, cm$.
The coordinates of the vertices are: $A(0, 0)$,$B(40\sqrt{2}, 0)$,$C(40\sqrt{2}, 40\sqrt{2})$,and $D(0, 40\sqrt{2})$.
The masses are $m_A = 8 \, kg, m_B = 2 \, kg, m_C = 4 \, kg, m_D = 2 \, kg$. Total mass $M = 8 + 2 + 4 + 2 = 16 \, kg$.
The $x$-coordinate of the center of mass is $X_{cm} = \frac{m_A x_A + m_B x_B + m_C x_C + m_D x_D}{M} = \frac{8(0) + 2(40\sqrt{2}) + 4(40\sqrt{2}) + 2(0)}{16} = \frac{240\sqrt{2}}{16} = 15\sqrt{2} \, cm$.
The $y$-coordinate of the center of mass is $Y_{cm} = \frac{m_A y_A + m_B y_B + m_C y_C + m_D y_D}{M} = \frac{8(0) + 2(0) + 4(40\sqrt{2}) + 2(40\sqrt{2})}{16} = \frac{240\sqrt{2}}{16} = 15\sqrt{2} \, cm$.
The distance from $A(0, 0)$ is $R = \sqrt{X_{cm}^2 + Y_{cm}^2} = \sqrt{(15\sqrt{2})^2 + (15\sqrt{2})^2} = \sqrt{450 + 450} = \sqrt{900} = 30 \, cm$.

(A) The formula for the position vector of the center of mass is $\vec{r}_{cm} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2}{m_1 + m_2}$.
Given $m_1 = 1 \ kg$,$\vec{r}_1 = \hat{i} + 2\hat{j} + \hat{k}$ and $m_2 = 3 \ kg$,$\vec{r}_2 = -3\hat{i} - 2\hat{j} + \hat{k}$.
Substituting these values:
$\vec{r}_{cm} = \frac{1(\hat{i} + 2\hat{j} + \hat{k}) + 3(-3\hat{i} - 2\hat{j} + \hat{k})}{1 + 3}$
$\vec{r}_{cm} = \frac{\hat{i} + 2\hat{j} + \hat{k} - 9\hat{i} - 6\hat{j} + 3\hat{k}}{4}$
$\vec{r}_{cm} = \frac{-8\hat{i} - 4\hat{j} + 4\hat{k}}{4}$
$\vec{r}_{cm} = -2\hat{i} - \hat{j} + \hat{k}$.

(B) Let the two particles be placed on the $x$-axis. Let mass $m_1$ be at the origin $(0, 0)$ and mass $m_2$ be at $(L, 0)$.
The formula for the center of mass $x_{CM}$ is given by:
$x_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$
Substituting the values $x_1 = 0$ and $x_2 = L$:
$x_{CM} = \frac{m_1(0) + m_2(L)}{m_1 + m_2}$
$x_{CM} = \frac{m_2 L}{m_1 + m_2}$
Since the particles are on the $x$-axis,$y_{CM} = 0$ and $z_{CM} = 0$.

(B) Let the position of point $P$ be at the origin $(0, 0)$.
Since the spheres are identical and arranged along the $X$-axis,the positions of the centers of mass are $P = 0$,$Q = PQ$,and $R = PR$.
The mass of each sphere is $m = 1 \ kg$.
The center of mass $x_{cm}$ of the system is given by:
$x_{cm} = \frac{m_1x_1 + m_2x_2 + m_3x_3}{m_1 + m_2 + m_3}$
$x_{cm} = \frac{(1 \times 0) + (1 \times PQ) + (1 \times PR)}{1 + 1 + 1}$
$x_{cm} = \frac{PQ + PR}{3}$

(A) Let the three equal masses be $m_1 = m_2 = m_3 = m$.
The coordinates are $(x_1, y_1) = (0,0)$,$(x_2, y_2) = (a,0)$,and $(x_3, y_3) = \left( \frac{a}{2}, \frac{a\sqrt{3}}{2} \right)$.
The $x$-coordinate of the center of mass is given by:
$x_{CM} = \frac{m_1x_1 + m_2x_2 + m_3x_3}{m_1 + m_2 + m_3} = \frac{m(0) + m(a) + m(\frac{a}{2})}{m + m + m} = \frac{m(a + \frac{a}{2})}{3m} = \frac{\frac{3a}{2}}{3} = \frac{a}{2}$.
The $y$-coordinate of the center of mass is given by:
$y_{CM} = \frac{m_1y_1 + m_2y_2 + m_3y_3}{m_1 + m_2 + m_3} = \frac{m(0) + m(0) + m(\frac{a\sqrt{3}}{2})}{m + m + m} = \frac{m(\frac{a\sqrt{3}}{2})}{3m} = \frac{a\sqrt{3}}{6}$.
Thus,the coordinates of the center of mass are $\left( \frac{a}{2}, \frac{a\sqrt{3}}{6} \right)$.

(D) The centre of mass of the metre stick is at its midpoint,which is at a distance of $L/2 = 0.5 \ m$ from the pivot.
When the stick is displaced by an angle $\theta = 60^o$,the vertical height $h$ by which the centre of mass rises is given by:
$h = \frac{L}{2} - \frac{L}{2} \cos \theta$
Given $L = 1 \ m$ and $\theta = 60^o$:
$h = 0.5 - 0.5 \cos 60^o = 0.5 - 0.5 \times 0.5 = 0.5 - 0.25 = 0.25 \ m$
The increase in potential energy $\Delta U$ is given by:
$\Delta U = mgh$
Given mass $m = 400 \ g = 0.4 \ kg$ and taking $g = 10 \ m/s^2$:
$\Delta U = 0.4 \times 10 \times 0.25 = 4 \times 0.25 = 1 \ J$.

(A) Before the explosion,the particle is moving along the $x$-axis,so its initial velocity component in the $y$-direction is zero $(v_{y,i} = 0)$.
Since there are no external forces acting in the $y$-direction,the position of the center of mass $(y_{cm})$ remains constant at $0$ (assuming the initial $y$-coordinate was $0$).
Using the formula for the center of mass: $y_{cm} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2}$.
Given $m_1 = m/4$,$y_1 = 15 \ cm$,$m_2 = 3m/4$,and $y_{cm} = 0$.
$0 = \frac{(m/4)(15) + (3m/4)(y_2)}{m/4 + 3m/4}$.
$0 = \frac{m}{4} (15 + 3y_2) / m$.
$15 + 3y_2 = 0$.
$3y_2 = -15$.
$y_2 = -5 \ cm$.

(D) Since the three spheres are identical and their centers form an equilateral triangle,the system possesses rotational symmetry about the axis passing through the centroid of the triangle.
The center of mass of a system of identical particles is located at the geometric center (centroid) of the configuration.
Therefore,the center of mass of the system of three spheres is located at the centroid of the equilateral triangle formed by their centers.

(A) Let the position of the center $P$ be at the origin $(0, 0)$.
Since the spheres are identical and touching,the distance between the centers of adjacent spheres is equal to the diameter of a sphere,say $d$.
Thus,the coordinates of the centers are:
$P = (0, 0)$
$Q = (d, 0)$
$R = (2d, 0)$
Note that $PQ = d$ and $PR = 2d$.
The $x$-coordinate of the center of mass $(x_{cm})$ is given by:
$x_{cm} = \frac{m_P x_P + m_Q x_Q + m_R x_R}{m_P + m_Q + m_R}$
Since $m_P = m_Q = m_R = 1 \ kg$:
$x_{cm} = \frac{1(0) + 1(d) + 1(2d)}{1 + 1 + 1} = \frac{3d}{3} = d$
Since $PQ = d$,the distance of the center of mass from $P$ is $PQ$.
However,looking at the options provided,the expression for the center of mass in terms of distances is:
$x_{cm} = \frac{0 + PQ + PR}{3} = \frac{PQ + PR}{3}$.

(A) The center of mass $x$ from the first mass $m_1$ is given by the formula: $x = \frac{m_2 d}{m_1 + m_2}$.
Here,$m_1 = 1.5 \ g$,$m_2 = 2.5 \ g$,and $d = 16 \ cm$.
Substituting the values:
$x = \frac{2.5 \times 16}{1.5 + 2.5}$
$x = \frac{40}{4}$
$x = 10 \ cm$.

(C) Let the left edge of the bottom book (book $1$) be at $x = 0$. The length of each book is $L$.
For book $1$,the center of mass is at $x_1 = L/2$.
For book $2$,the left edge is at $x = L/4$,so the center of mass is at $x_2 = L/4 + L/2 = 3L/4$.
For book $3$,the left edge is at $x = L/4 + L/4 = L/2$,so the center of mass is at $x_3 = L/2 + L/2 = L$.
For book $4$,the left edge is at $x = L/4 + L/4 + L/4 = 3L/4$,so the center of mass is at $x_4 = 3L/4 + L/2 = 5L/4$.
Since all books have equal mass $m$,the $x$-coordinate of the center of mass of the system is:
$x_{cm} = \frac{m(x_1 + x_2 + x_3 + x_4)}{4m} = \frac{L/2 + 3L/4 + L + 5L/4}{4}$
$x_{cm} = \frac{(2L + 3L + 4L + 5L)/4}{4} = \frac{14L/4}{4} = \frac{14L}{16} = \frac{7L}{8} = 0.875\ L$.

(C) Let the three rods be $OA$,$OB$,and $AB$,each of mass $m$ and length $a$.
$1$. The center of mass $(CM)$ of rod $OA$ (along the x-axis from $(0,0)$ to $(a,0)$) is at $(a/2, 0)$.
$2$. The center of mass $(CM)$ of rod $OB$ (along the y-axis from $(0,0)$ to $(0,a)$) is at $(0, a/2)$.
$3$. The center of mass $(CM)$ of rod $AB$ (connecting $(a,0)$ and $(0,a)$) is at the midpoint of the line segment,which is $(a/2, a/2)$.
Now,calculate the coordinates of the center of mass of the system:
$X_{cm} = \frac{m(a/2) + m(0) + m(a/2)}{m + m + m} = \frac{ma}{3m} = \frac{a}{3}$
$Y_{cm} = \frac{m(0) + m(a/2) + m(a/2)}{m + m + m} = \frac{ma}{3m} = \frac{a}{3}$
Thus,the coordinates of the center of mass are $(a/3, a/3)$.

(C) Let the rod be placed along the $x$-axis with one end at the origin. Since the rod lies on the $x$-axis,the $y$ and $z$ coordinates of the center of mass are zero,i.e.,$y_{CM} = 0$ and $z_{CM} = 0$.
Consider a small element of length $dx$ at a distance $x$ from the origin. The mass of this small element is $dm = \lambda dx = (A + Bx)dx$.
The $x$-coordinate of the center of mass is given by:
$x_{CM} = \frac{\int_{0}^{L} x dm}{\int_{0}^{L} dm} = \frac{\int_{0}^{L} x(A + Bx) dx}{\int_{0}^{L} (A + Bx) dx}$
Evaluating the integrals:
Numerator: $\int_{0}^{L} (Ax + Bx^2) dx = [\frac{Ax^2}{2} + \frac{Bx^3}{3}]_{0}^{L} = \frac{AL^2}{2} + \frac{BL^3}{3} = \frac{3AL^2 + 2BL^3}{6} = \frac{L^2(3A + 2BL)}{6}$
Denominator: $\int_{0}^{L} (A + Bx) dx = [Ax + \frac{Bx^2}{2}]_{0}^{L} = AL + \frac{BL^2}{2} = \frac{2AL + BL^2}{2} = \frac{L(2A + BL)}{2}$
Therefore,$x_{CM} = \frac{L^2(3A + 2BL)}{6} \times \frac{2}{L(2A + BL)} = \frac{L(3A + 2BL)}{3(2A + BL)}$.

(A) For an object to undergo pure translational motion without rotation,the external force must be applied at its center of mass $(CM)$.
Let the mass of the horizontal rod $AB$ of length $l$ be $m$. Then the mass of the vertical rod $CD$ of length $2l$ will be $2m$.
The center of mass of rod $AB$ is at point $D$ (midpoint of $AB$),and the center of mass of rod $CD$ is at its midpoint,which is at a distance $l$ from $D$ and $l$ from $C$.
Let point $C$ be the origin $(0,0)$.
Coordinates of the center of mass of rod $CD$ $(m_1 = 2m)$: $(0, l)$.
Coordinates of the center of mass of rod $AB$ $(m_2 = m)$: $(0, 2l)$.
The $y$-coordinate of the center of mass of the system is given by:
$Y_{cm} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2}$
$Y_{cm} = \frac{(2m)(l) + (m)(2l)}{2m + m}$
$Y_{cm} = \frac{2ml + 2ml}{3m} = \frac{4ml}{3m} = \frac{4}{3}l$
Thus,the force must be applied at a distance of $\frac{4}{3}l$ from point $C$.

(D) Let the square be in the $XY$-plane with $P(0, a), Q(a, a), R(a, 0)$ and $S(0, 0)$.
The masses are $m_P = 1 \ kg, m_Q = 1 \ kg, m_R = 2 \ kg, m_S = 2 \ kg$.
The center of mass of $P$ and $S$ is at $O_1$ on the side $PS$ at a distance $y_1 = \frac{m_P \cdot a + m_S \cdot 0}{m_P + m_S} = \frac{1 \cdot a + 2 \cdot 0}{1 + 2} = \frac{a}{3}$ from $S$.
The center of mass of $Q$ and $R$ is at $O_2$ on the side $QR$ at a distance $y_2 = \frac{m_Q \cdot a + m_R \cdot 0}{m_Q + m_R} = \frac{1 \cdot a + 2 \cdot 0}{1 + 2} = \frac{a}{3}$ from $R$.
The total mass of the system is $M = 1 + 1 + 2 + 2 = 6 \ kg$.
The center of mass of the whole system lies on the line joining $O_1$ and $O_2$ at a distance $x_{cm} = \frac{(m_P+m_S) \cdot 0 + (m_Q+m_R) \cdot a}{M} = \frac{3 \cdot 0 + 3 \cdot a}{6} = \frac{a}{2}$ from the $Y$-axis.
Thus,the center of mass is at $(\frac{a}{2}, \frac{a}{3})$.
Comparing distances from corners: The center of mass is closest to $R$ and $S$ (distance $\sqrt{(a/2)^2 + (a/3)^2}$) and farthest from $P$ and $Q$ (distance $\sqrt{(a/2)^2 + (2a/3)^2}$).

(B) The linear mass density is given by $\lambda = \frac{dm}{dx}$,so $dm = \lambda \, dx$.
The position of the center of mass $x_{cm}$ is given by:
$x_{cm} = \frac{\int x \, dm}{\int dm}$
Substituting $\lambda = 2 + x$ and the limits from $0$ to $3$:
$x_{cm} = \frac{\int_{0}^{3} x(2 + x) \, dx}{\int_{0}^{3} (2 + x) \, dx}$
$x_{cm} = \frac{\int_{0}^{3} (2x + x^2) \, dx}{\int_{0}^{3} (2 + x) \, dx}$
Evaluating the integrals:
Numerator: $[x^2 + \frac{x^3}{3}]_{0}^{3} = (3^2 + \frac{3^3}{3}) = 9 + 9 = 18$
Denominator: $[2x + \frac{x^2}{2}]_{0}^{3} = (2(3) + \frac{3^2}{2}) = 6 + 4.5 = 10.5 = \frac{21}{2}$
$x_{cm} = \frac{18}{21/2} = \frac{36}{21} = \frac{12}{7} \ m$

(D) Let the mass of a wire of length $l$ be $m$.
For segment $AB$ (length $l$): Mass $m_1 = m$,Center of mass $\vec{r}_1 = \left( \frac{l}{2}, 0 \right)$.
For segment $BC$ (length $2l$): Mass $m_2 = 2m$,Center of mass $\vec{r}_2 = (0, l)$.
For segment $CD$ (length $l$): Mass $m_3 = m$,Center of mass $\vec{r}_3 = \left( \frac{l}{2}, 2l \right)$.
The coordinates of the center of mass $(X_{cm}, Y_{cm})$ are given by:
$X_{cm} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3} = \frac{m(\frac{l}{2}) + 2m(0) + m(\frac{l}{2})}{m + 2m + m} = \frac{m(l)}{4m} = \frac{l}{4}$.
$Y_{cm} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3} = \frac{m(0) + 2m(l) + m(2l)}{m + 2m + m} = \frac{4ml}{4m} = l$.
Thus,the center of mass is $\left( \frac{l}{4}, l \right)$.

(D) The center of mass $x_{cm}$ is given by the formula:
$x_{cm} = \frac{\int x dm}{\int dm} = \frac{\int_{0}^{L} x \cdot k(x/L)^n dx}{\int_{0}^{L} k(x/L)^n dx} = \frac{\int_{0}^{L} x^{n+1} dx}{\int_{0}^{L} x^n dx}$
Evaluating the integrals:
$x_{cm} = \frac{L^{n+2} / (n+2)}{L^{n+1} / (n+1)} = L \left( \frac{n+1}{n+2} \right)$
Analyzing the behavior:
$1$. When $n = 0$,$x_{cm} = L(1/2) = L/2$.
$2$. As $n \to \infty$,$x_{cm} = L \left( \frac{1 + 1/n}{1 + 2/n} \right) \to L$.
$3$. The derivative $\frac{dx_{cm}}{dn} = L \left( \frac{(n+2) - (n+1)}{(n+2)^2} \right) = \frac{L}{(n+2)^2} > 0$,which means $x_{cm}$ is an increasing function of $n$.
Thus,the graph starts at $L/2$ for $n=0$ and asymptotically approaches $L$ as $n$ increases. This corresponds to the graph shown in option $D$.

(D) Let the positions of the particles be:
$m$ at $(0, 0)$
$2m$ at $(a, 0)$
$3m$ at $(a, a)$
$4m$ at $(0, a)$
The $x$-coordinate of the center of mass is given by:
$X_{CM} = \frac{m(0) + 2m(a) + 3m(a) + 4m(0)}{m + 2m + 3m + 4m} = \frac{5ma}{10m} = \frac{a}{2}$
The $y$-coordinate of the center of mass is given by:
$Y_{CM} = \frac{m(0) + 2m(0) + 3m(a) + 4m(a)}{m + 2m + 3m + 4m} = \frac{7ma}{10m} = \frac{7}{10}a$
Thus,the coordinates of the center of mass are $\left( \frac{a}{2}, \frac{7}{10}a \right)$.

(A) Let the mass per unit area of the given objects be $\sigma$.
The mass of the square plate is $m_1 = \sigma l^2$.
The mass of the circular disc is $m_2 = \sigma \pi \left( \frac{l}{2} \right)^2 = \sigma l^2 \left( \frac{\pi}{4} \right)$.
Since $m_1 > m_2$ (as $\pi/4 \approx 0.785 < 1$),the center of mass of the system will be closer to the heavier object,which is the square plate. Therefore,the center of mass lies inside the square plate.

(C) For a uniform triangular lamina,the center of mass is located at the centroid of the triangle.
If the vertices of the right-angled triangle are at $(0, 0)$,$(b, 0)$,and $(0, h)$,the coordinates of the center of mass $(x_{cm}, y_{cm})$ are given by the average of the coordinates of the vertices:
$x_{cm} = \frac{x_1 + x_2 + x_3}{3} = \frac{0 + b + 0}{3} = \frac{b}{3}$
$y_{cm} = \frac{y_1 + y_2 + y_3}{3} = \frac{0 + 0 + h}{3} = \frac{h}{3}$
Therefore,the coordinates of the center of mass are $\left( \frac{b}{3}, \frac{h}{3} \right)$.

(B) Let the mass per unit length be $\lambda = kx$,where $k$ is a constant.
Consider a small element of length $dx$ at a distance $x$ from the end $x=0$.
The mass of this element is $dm = \lambda dx = kx dx$.
The center of mass $x_{cm}$ is given by the formula:
$x_{cm} = \frac{\int x dm}{\int dm}$
Substituting the values:
$x_{cm} = \frac{\int_0^3 x (kx dx)}{\int_0^3 kx dx} = \frac{\int_0^3 x^2 dx}{\int_0^3 x dx}$
Evaluating the integrals:
$x_{cm} = \frac{[x^3/3]_0^3}{[x^2/2]_0^3} = \frac{27/3}{9/2} = \frac{9}{4.5} = 2 \ m$.

(B) The length of the rod is the distance between the two masses:
$L = |\vec{r}_2 - \vec{r}_1| = \sqrt{(4 - 2)^2 + (2 - 5)^2} = \sqrt{2^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13} \ m$.
The center of mass $\vec{r}_{CM}$ is given by:
$\vec{r}_{CM} = \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2}{m_1 + m_2} = \frac{3(2 \hat{i} + 5 \hat{j}) + 2(4 \hat{i} + 2 \hat{j})}{3 + 2}$
$\vec{r}_{CM} = \frac{(6 \hat{i} + 15 \hat{j}) + (8 \hat{i} + 4 \hat{j})}{5} = \frac{14 \hat{i} + 19 \hat{j}}{5} = \left( \frac{14}{5} \hat{i} + \frac{19}{5} \hat{j} \right) \ m$.

(C) Let the mass of the $H$ atom be $m_1 = 1 \ \text{unit}$ and the mass of the $Cl$ atom be $m_2 = 35.5 \ \text{units}$.
Place the $H$ atom at the origin $(0, 0)$. Then the position of the $H$ atom is $r_1 = 0$ and the position of the $Cl$ atom is $r_2 = 1.27 \ \mathring A$.
The center of mass $R_{cm}$ is given by the formula:
$R_{cm} = \frac{m_1 r_1 + m_2 r_2}{m_1 + m_2}$
Substituting the values:
$R_{cm} = \frac{(1)(0) + (35.5)(1.27)}{1 + 35.5}$
$R_{cm} = \frac{35.5 \times 1.27}{36.5}$
$R_{cm} \approx 1.24 \ \mathring A$
Thus,the center of mass is at a distance of $1.24 \ \mathring A$ from the $H$ atom.

(A) The total mass of the system is $M = 6m + m + m + m + m = 10m$.
The $y$-coordinates of the centers of mass of the individual parts are:
Outer circle: $y_1 = 0$,mass $m_1 = 6m$
Left inner circle: $y_2 = a$,mass $m_2 = m$
Right inner circle: $y_3 = a$,mass $m_3 = m$
Horizontal line segment: $y_4 = -a$,mass $m_4 = m$
The $y$-coordinate of the center of mass is given by:
$Y_{cm} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3 + m_4 y_4}{M}$
$Y_{cm} = \frac{(6m)(0) + (m)(a) + (m)(a) + (m)(-a)}{10m}$
$Y_{cm} = \frac{0 + ma + ma - ma}{10m} = \frac{ma}{10m} = \frac{a}{10}$

(B) Let the total mass of the first three particles be $M = 10 + 20 + 30 = 60 \ kg$. The center of mass of these particles is at $\vec{r}_{cm} = (0, 0, 0)$.
Let the mass of the fourth particle be $m = 40 \ kg$ and its position be $\vec{r} = (x, y, z)$.
The new center of mass of the system is given as $\vec{r}'_{cm} = (3, 3, 3)$.
The formula for the center of mass of the combined system is $\vec{r}'_{cm} = \frac{M\vec{r}_{cm} + m\vec{r}}{M + m}$.
Substituting the values: $(3, 3, 3) = \frac{60(0, 0, 0) + 40(x, y, z)}{60 + 40}$.
$(3, 3, 3) = \frac{40(x, y, z)}{100} = 0.4(x, y, z)$.
Comparing the coordinates: $3 = 0.4x \Rightarrow x = \frac{3}{0.4} = 7.5 \ m$.
Similarly,$y = 7.5 \ m$ and $z = 7.5 \ m$.
Thus,the position of the fourth particle is $(7.5, 7.5, 7.5) \ m$.

(A) Given:
$m_1 = 1 \ kg$,$\vec{r}_1 = \hat{i} + 2\hat{j} + \hat{k}$
$m_2 = 3 \ kg$,$\vec{r}_2 = -3\hat{i} - 2\hat{j} + \hat{k}$
The position vector of the centre of mass is given by:
$\vec{r}_{CM} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2}{m_1 + m_2}$
Substituting the values:
$\vec{r}_{CM} = \frac{1(\hat{i} + 2\hat{j} + \hat{k}) + 3(-3\hat{i} - 2\hat{j} + \hat{k})}{1 + 3}$
$\vec{r}_{CM} = \frac{\hat{i} + 2\hat{j} + \hat{k} - 9\hat{i} - 6\hat{j} + 3\hat{k}}{4}$
$\vec{r}_{CM} = \frac{-8\hat{i} - 4\hat{j} + 4\hat{k}}{4}$
$\vec{r}_{CM} = -2\hat{i} - \hat{j} + \hat{k}$

(A) The distance of the center of mass $(X_{CM})$ of a system of particles from the origin is given by the formula:
$X_{CM} = \frac{m_1x_1 + m_2x_2 + m_3x_3}{m_1 + m_2 + m_3}$
Given:
$m_1 = 300 \, g, x_1 = 0 \, cm$
$m_2 = 500 \, g, x_2 = 40 \, cm$
$m_3 = 400 \, g, x_3 = 70 \, cm$
Substituting the values:
$X_{CM} = \frac{300 \times 0 + 500 \times 40 + 400 \times 70}{300 + 500 + 400}$
$X_{CM} = \frac{0 + 20000 + 28000}{1200}$
$X_{CM} = \frac{48000}{1200}$
$X_{CM} = 40 \, cm$

(B) The center of mass of a system of two particles is located on the line joining them. The position of the center of mass $R_{cm}$ is given by $R_{cm} = \frac{m_1r_1 + m_2r_2}{m_1 + m_2}$.
For two masses $m$ and $M$ separated by a distance $d$,if we place $M$ at the origin $(0,0)$ and $m$ at $(d,0)$,the center of mass is at $x = \frac{M(0) + m(d)}{M + m} = \frac{md}{M + m}$.
Since $M > m$,the center of mass is closer to the heavier mass $M$. Therefore,the center of mass is located towards $M$.

(C) Let the mass of the hydrogen atom be $m_1 = 1$ and the mass of the chlorine atom be $m_2 = 35.5$.
Place the hydrogen atom at the origin $(0, 0)$,so its position is $\vec{r}_1 = 0$.
The chlorine atom is at a distance of $1.27 \mathring{A}$ from the hydrogen atom along the $x$-axis,so its position is $\vec{r}_2 = 1.27 \hat{i} \mathring{A}$.
The position of the center of mass $\vec{R}_{cm}$ is given by the formula:
$\vec{R}_{cm} = \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2}{m_1 + m_2}$
Substituting the values:
$\vec{R}_{cm} = \frac{1 \times 0 + 35.5 \times 1.27 \hat{i}}{1 + 35.5}$
$\vec{R}_{cm} = \frac{35.5 \times 1.27}{36.5} \hat{i}$
$\vec{R}_{cm} \approx 0.9726 \times 1.27 \hat{i} \approx 1.235 \hat{i} \mathring{A}$
Rounding to two decimal places,the position is approximately $1.24 \mathring{A}$ from the hydrogen atom.

(C) Let the carbon atom be at the origin $(0, 0)$. The position of the carbon atom is $x_1 = 0$ with mass $m_1 = 12 \, a.m.u.$
The position of the oxygen atom is $x_2 = 1.1 \, \mathring A$ with mass $m_2 = 16 \, a.m.u.$
The center of mass $X_{cm}$ is given by:
$X_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$
$X_{cm} = \frac{12 \times 0 + 16 \times 1.1}{12 + 16}$
$X_{cm} = \frac{17.6}{28} \, \mathring A$
$X_{cm} \approx 0.6285 \, \mathring A \approx 0.63 \, \mathring A$
Thus,the center of mass is at a distance of $0.63 \, \mathring A$ from the carbon atom.

(A) The velocity of the center of mass $\vec{v}_{cm}$ is given by the formula: $\vec{v}_{cm} = \frac{m_1 \vec{v}_1 + m_2 \vec{v}_2 + m_3 \vec{v}_3}{m_1 + m_2 + m_3}$
Given: $m_1 = 20 \, g$,$m_2 = 30 \, g$,$m_3 = 50 \, g$ and $\vec{v}_1 = 10 \, \hat{i} \, cm/s$,$\vec{v}_2 = 10 \, \hat{j} \, cm/s$,$\vec{v}_3 = 10 \, \hat{k} \, cm/s$.
Total mass $M = 20 + 30 + 50 = 100 \, g$.
Substituting the values: $\vec{v}_{cm} = \frac{20(10 \, \hat{i}) + 30(10 \, \hat{j}) + 50(10 \, \hat{k})}{100}$
$\vec{v}_{cm} = \frac{200 \, \hat{i} + 300 \, \hat{j} + 500 \, \hat{k}}{100}$
$\vec{v}_{cm} = 2 \, \hat{i} + 3 \, \hat{j} + 5 \, \hat{k}$

(B) Let the masses at the vertices be $m_1 = m, m_2 = 2m, m_3 = 3m, m_4 = 4m$.
Based on the geometry of the rhombus in the $x-y$ plane:
$r_1 = (0, 0)$
$r_2 = (a \cos 60^o, a \sin 60^o) = (a/2, a\sqrt{3}/2)$
$r_3 = (a + a \cos 60^o, a \sin 60^o) = (3a/2, a\sqrt{3}/2)$
$r_4 = (a, 0)$
The center of mass $(X_{cm}, Y_{cm})$ is given by:
$X_{cm} = \frac{m_1x_1 + m_2x_2 + m_3x_3 + m_4x_4}{m_1 + m_2 + m_3 + m_4} = \frac{m(0) + 2m(a/2) + 3m(3a/2) + 4m(a)}{10m} = \frac{0 + am + 4.5am + 4am}{10m} = \frac{9.5am}{10m} = 0.95a$
$Y_{cm} = \frac{m_1y_1 + m_2y_2 + m_3y_3 + m_4y_4}{m_1 + m_2 + m_3 + m_4} = \frac{m(0) + 2m(a\sqrt{3}/2) + 3m(a\sqrt{3}/2) + 4m(0)}{10m} = \frac{am\sqrt{3} + 1.5am\sqrt{3}}{10m} = \frac{2.5a\sqrt{3}}{10} = \frac{\sqrt{3}}{4}a$
Thus,the center of mass is $(0.95a, \frac{\sqrt{3}}{4}a)$.

(B) Let the side length of the square be $a$. The diagonal is $d = a\sqrt{2} = 80 \ cm$,so $a = 40\sqrt{2} \ cm$.
Place $A$ at $(0, 0)$,$B$ at $(a, 0)$,$C$ at $(a, a)$,and $D$ at $(0, a)$.
The masses are $m_A = 8 \ kg, m_B = 2 \ kg, m_C = 4 \ kg, m_D = 2 \ kg$.
The coordinates of the center of mass $(X, Y)$ are:
$X = \frac{m_A x_A + m_B x_B + m_C x_C + m_D x_D}{m_A + m_B + m_C + m_D} = \frac{8(0) + 2(a) + 4(a) + 2(0)}{8 + 2 + 4 + 2} = \frac{6a}{16} = \frac{3a}{8}$.
$Y = \frac{m_A y_A + m_B y_B + m_C y_C + m_D y_D}{m_A + m_B + m_C + m_D} = \frac{8(0) + 2(0) + 4(a) + 2(a)}{16} = \frac{6a}{16} = \frac{3a}{8}$.
Substituting $a = 40\sqrt{2} \ cm$:
$X = \frac{3(40\sqrt{2})}{8} = 15\sqrt{2} \ cm$ and $Y = 15\sqrt{2} \ cm$.
The distance from $A(0, 0)$ is $r = \sqrt{X^2 + Y^2} = \sqrt{(15\sqrt{2})^2 + (15\sqrt{2})^2} = \sqrt{450 + 450} = \sqrt{900} = 30 \ cm$.