Problem set 9

1. The position vector $\bar r=x\hat i+y\hat j+z\hat k$, $\bar\nabla.\left(r^2\bar r\right)$ is given by:
1. 0
2. $5r^2$
3. $r^2$
4. $3r^2$
$r^2=x^2+y^2+z^2$ $${\scriptscriptstyle\bar\nabla.\left(r^2\bar r\right)=\left(\hat i\frac{\partial}{\partial x}+\hat j\frac{\partial}{\partial y}+\hat k\frac{\partial}{\partial x}\right)\cdot\left(\left(x^2+y^2+z^2\right)\left(x\hat i+y\hat j+z\hat k\right)\right)}$$ $$\bar\nabla.\left(r^2\bar r\right)=5r^2$$
2. The eigenvalues of the matrix: $\begin{pmatrix}1&2&3\\0&4&7\\0&0&3\end{pmatrix}$ are:
   1. 1, 4, 3
   2. 3, 7, 3
   3. 7, 3, 2
   4. 1, 2, 3
   In upper or lower triangular matrix, eigenvalues are the diagonal elements, Hence (A)1, 4, 3
3. Two solid spheres of radius $R$ and mass $M$ each are connected by a thin rod of negligible mass. The distance between the centres is $4R$. The moment of inertia about an axis passing through the centre of symmetry and perpendicular to the line joining the spheres is
1. $\frac{11}{5}MR^2$
2. $\frac{22}{5}MR^2$
3. $\frac{44}{5}MR^2$
4. $\frac{88}{5}MR^2$

MI of system= MI of sphere 1 about an axis through centre of symmetry +MI of sphere 2 about an axis through centre of symmetry

According to theorem of parallel axes MI of sphere 1 about an axis through centre of symmetry = MI of sphere 1 about an axis through its centre and parallel to axis through centre of symmetry + M X square of distance between axes. $$I_1=\frac{2}{5}MR^2+M(2R)^2=\frac{22}{5}MR^2$$ Similarly $$I_2=\frac{2}{5}MR^2+M(2R)^2=\frac{22}{5}MR^2$$ $$I=\frac{44}{5}MR^2$$
4. All solutions of the equation $e^z=-3$ are
1. $z=\ln \pi\ln 3,~n=\pm1,\pm2,\dots$
2. ${\scriptstyle z=\ln 3+i(2n+1)\pi,~n=0,\pm1,\pm2,\dots}$
3. ${\scriptstyle z=\ln 3+i~2n\pi,~n=0,\pm1,\pm2,\dots}$
4. $z=i3n\pi,~n=\pm1,\pm2,\dots$
Finding all solutions means to find all values of $z$ which will satify $e^z=-3$ \begin{align*} e^z&=-3\\ &=3(-1)\\ &={\scriptstyle3\left[\cos{(2n+1)\pi}+i\sin{(2n+1)\pi}\right]}\\ &=3e^{i(2n+1)\pi} \end{align*} Taking log on both sides we get $$z=\ln 3+i(2n+1)\pi~~~n=0,\pm1,\pm2,\dots$$
5. The solution of $\frac{dy}{dx}-y=e^{\lambda x}$ is :
1. $e^{-\lambda x}$
2. $\frac{1}{\lambda-1}e^{\lambda x}$
3. $e^{\lambda x}$
4. $\frac{1}{\lambda}e^{-\lambda x}$
Just by inspection or substituting $y=\frac{1}{\lambda-1}e^{\lambda x}$ in above equation one can verify that it is the solution. OR
The solution of $\frac{dy}{dx}+p(x)y=q(x)$ is given by $$y=\frac{\int u(x)q(x)dx+C}{u(x)}$$ where, $u(x)=exp\left(\int p(x)dx\right)$
Here $p(x)=-1$ and $q(x)=e^{\lambda x}$ $$u(x)=exp\left(-\int dx\right)=e^{-x}$$ \begin{align*} y&=\frac{\int e^{-x}e^{\lambda x}dx+C}{e^{-x}}\\ &=\frac{ e^{(\lambda-1) x}dx+C}{(\lambda-1)e^{-x}}\\ &=\frac{1}{\lambda-1}e^{\lambda x} \end{align*}