Problem set 19

1. A $2\times2$ matrix $A$ has eigenvalues $e^{i\pi/5}$ and $e^{i\pi/6}$. The smallest value of $n$ such that $A^n = I$ is
1. 20
2. 30
3. 60
4. 120
As $e^{i\pi/5}$ and $e^{i\pi/6}$ are eigenvalues of matrix $A$, eigenvalues of matrix of matrix $A^n$ will be $e^{in\pi/5}$ and $e^{in\pi/6}$. However, $A^n$ is identity matrix whose eignevalues will be 1 and 1, the diagonal elements. Hence, $e^{in\pi/5}=1$ and $e^{ni\pi/6}=1$ $$e^{in\pi/5}=\cos{(n\pi/5)}+i\sin{(n\pi/5)}=1$$ $$\Rightarrow \cos{(n\pi/5)}=1 \text{ and } sin{(n\pi/5)}=0$$ Hence, $n\pi/5=2m\pi$, where $m=0,1,2,\dots$, hence, $n=10m$, where $m=0,1,2,\dots$ Hence, $n=0, 10,20, 30,40,50,60\dots$ $$e^{in\pi/6}=\cos{(n\pi/6)}+i\sin{(n\pi/6)}=1$$ $$\Rightarrow \cos{(n\pi/6)}=1 \text{ and } sin{(n\pi/6)}=0$$ Hence, $n\pi/6=2m\pi$, where $m=0,1,2,\dots$, hence, $n=12m$, where $m=0,1,2,\dots$ Hence, $n=0, 12,24, 36,48,60,72\dots$
The common minimum value of $n$ is 60. Hence, answer is (C)
2. In a series of five Cricket matches, one of the captains calls "Heads" every time when the toss is taken. The probability that he will win 3 times and lose 2 times is
1. 1/8
2. 5/8
3. 3/16
4. 5/16
Here, the coin is tossed five times. In a single toss probability getting head is $p=\frac{1}{2}$ and probability of getting tail is $q=\frac{1}{2}$. In a sequences of $n=5$ tosses probability of getting $m=3$ heads and $n-m=2$ tails is given by $$p^mq^{(n-m)}=\left(\frac{1}{2}\right)^m\left(\frac{1}{2}\right)^{n-m}$$ However, total combinations of $m$ times head up and $(n-m)$ times tail up are given by $$\frac{n!}{m!(n-m)!}$$ Hence, probability of getting $m$ heads is \begin{align*} P&=\frac{n!}{m!(n-m)!}\left(\frac{1}{2}\right)^m\left(\frac{1}{2}\right)^{n-m}\\ &=\frac{n!}{m!(n-m)!}\left(\frac{1}{2}\right)^n\\ &=\frac{5!}{3!(5-3)!}\left(\frac{1}{2}\right)^5\\ &=10\left(\frac{1}{2}\right)^5=\frac{5}{16} \end{align*} Hence, answer is (D).
3. The unit normal vector at the point $\left(\frac{a}{\sqrt{3}},\frac{b}{\sqrt{3}},\frac{c}{\sqrt{3}}\right)$ on the surface of the ellipsoid $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$
1. $\frac{bc\hat i+ca\hat j+ab\hat k}{\sqrt{a^2+b^2+c^2}}$
2. $\frac{a\hat i+b\hat j+c\hat k}{\sqrt{a^2+b^2+c^2}}$
3. $\frac{b\hat i+c\hat j+a\hat k}{\sqrt{a^2+b^2+c^2}}$
4. $\frac{\hat i+\hat j+\hat k}{\sqrt{3}}$
The unit normal is defined as $\hat n=\frac{\nabla f}{|\nabla f|}$ $$\nabla f=2\left(\frac{x}{a^2}\hat i+\frac{y}{b^2}\hat j+\frac{z}{c^2}\hat k\right)$$ \begin{align*} &\left(\nabla f\right)_{\left(\frac{a}{\sqrt{3}},\frac{b}{\sqrt{3}},\frac{c}{\sqrt{3}}\right)}\\ &=\frac{2}{\sqrt{3}}\left(\frac{\hat i}{a^2}+\frac{\hat j}{b^2}+\frac{\hat k}{c^2}\right) \end{align*} \begin{align*} &\left(|\nabla f|\right)_{\left(\frac{a}{\sqrt{3}},\frac{b}{\sqrt{3}},\frac{c}{\sqrt{3}}\right)}\\ &=\frac{2}{\sqrt{3}}\sqrt{\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)}\\ &=\frac{2}{\sqrt{3}}\sqrt{\left(\frac{b^2c^2+a^2c^2+b^2a^2}{a^2b^2c^2}\right)} \end{align*} $$\frac{\nabla f}{|\nabla f|}=\frac{bc\:\hat i+ac\:\hat j+ab\:\hat k}{\sqrt{b^2c^2+a^2c^2+b^2a^2}}$$
4. A solid cylinder of height $H$, radius $R$ and density $\rho$, floats vertically on the surface of a liquid of density $\rho_0$. The cylinder will be set into oscillatory motion when a small instantaneous downward force is applied. The frequency of oscillation is
1. $\frac{\rho g}{\rho_0 H}$
2. $\frac{\rho}{\rho_0}\sqrt{\frac{g}{H}}$
3. $\sqrt{\frac{\rho g}{\rho_0 H}}$
4. $\sqrt{\frac{\rho_0g}{\rho H}}$
Let the cylinder of height $H$ and area of cross section $A$ is floating on the water vertically. Let some part of the cylinder is submerged inside the water. Hence, volume $V$ of water will get displaced. Hence, on a floating object there are two forces acting viz. 1) weight $W$ acting in the downward direction and 2) buoyant force or up thrust force $F_b$ acting in upward direction. When the floating body is at rest, these two forces are equal. i.e. $$-W=Fb$$ This is the equilibrium position of the cylinder. When we push the cylinder inside the water by applying downward force $F$, it get submerged by an extra distance $x$ inside the water. Hence, there is change in up thrust force. The change in up thrust force is equal to extra weight of the liquid displaced by the body. The extra volume of the liquid displaced is equal to extra volume of the cylinder submerged in the liquid, given by $$\Delta V=Ax$$ Hence, the change in up thrust force is given by $$\Delta F_b=\text{mass of liquid displaced}\times g=Ax\rho_0g$$ When we keep the body in the displaced position at rest then we have, $$-F=\Delta F_b=Ax\rho_0g$$ This the restoring force acting on the body. When we release the body it starts moving upward with acceleration $a$. Hence, $$F=ma=m\frac{d^2x}{dt^2}=-Ax\rho_0g$$ is the equation of motion of the object. Here, $m=AH\rho$ is the mass of cylinder. \begin{align*} \frac{d^2x}{dt^2}&=-\frac{A\rho_0g}{m}x &=-\frac{A\rho_0g}{AH\rho}x\\ &=-\frac{\rho_0g}{H\rho}x \end{align*} This equation is similar to the equation of simple harmonic motion, $\frac{d^2x}{dt^2}=-\omega^2 x$. Hence, we have, $$\omega=\sqrt{\frac{\rho_0g}{H\rho}}$$ Hence, period of oscillation is $$T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{H\rho}{\rho_0g}}$$ Frequency, of oscillation is $$n=\frac{1}{2\pi}\sqrt{\frac{\rho_0g}{H\rho}}$$
5. Three particles of equal mass $m$ are connected by two identical massless springs of stiffness constant $k$ as shown in the figure:
   
   If $x_1$ ,$x_2$ and $x_3$ denote the horizontal displacements of the masses from their respective equilibrium positions, the potential energy of the system is
1. $\frac{1}{2}\left[x_1^2+x_2^2+x_3^2\right]$
2. $\frac{1}{2}\left[x_1^2+x_2^2+x_3^2-x_2(x_1+x_3)\right]$
3. $\frac{1}{2}\left[x_1^2+2x_2^2+x_3^2+2x_2(x_1+x_3)\right]$
4. $\frac{1}{2}\left[x_1^2+2x_2^2+x_3^2-2x_2(x_1+x_3)\right]$
Let us take origin of the coordinate system on left side mass. Then extension in left string $=(x_2-x_1)$ and extension in right string $=(x_3-x_2)$ \begin{align*} V&=\frac{1}{2}(x_2-x_1)^2+\frac{1}{2}(x_3-x_2)^2\\ &={\scriptstyle\frac{1}{2}\left[x_1^2+2x_2^2+x_3^2-2x_2(x_1+x_3)\right]} \end{align*} Hence, answer is (D)