Problem set 69

1. A particle moves in two dimensions on the ellipse $x^2+4y^2=8$. At a particular instant it is at the point $(x,y)=(2,1)$ and the $x$-component of its velocity is 6 (in suitable units). Then the $y$-component of its velocity is
1. -3
2. -2
3. 1
4. 4

$$x^2+4y^2=8$$ differentiating with respect to $t$ $$2x\frac{dx}{dt}+8y\frac{dy}{dt}=0$$ At point $(x,y)=(2,1)$, $\frac{dx}{dt}=6$. Hence, $$2\times 2\times 6+8\times 1\times\frac{dy}{dt}=0$$ $$\frac{dy}{dt}=-3$$

Hence, answer is (A)

2. a particle of mass $m$ moves in the one-dimensional potential $V(x)=\frac{\alpha}{3}x^3+\frac{\beta}{4}x^4$ where $\alpha,\beta > 0$. One of the equilibrium points is $x=0$. The angular frequency of small oscillations about the other equilibrium point is
1. $\frac{2\alpha}{\sqrt{3m\beta}}$
2. $\frac{\alpha}{\sqrt{m\beta}}$
3. $\frac{\alpha}{\sqrt{12m\beta}}$
4. $\frac{\alpha}{\sqrt{24m\beta}}$

$$V(x)=\frac{\alpha}{3}x^3+\frac{\beta}{4}x^4$$ $$\frac{dV}{dx}=\alpha x^2+\beta x^3$$ At equilibrium points $\frac{dV}{dx}=0$, which gives $$\alpha x^2+\beta x^3=0$$ $$\Rightarrow x=0, x=-\frac{\alpha}{\beta}$$ For small oscillations, system performs SHM about equilibrium points. Hence, $$F=-kx\Rightarrow -\frac{dV}{dx}=-kx$$ $$\Rightarrow k=\frac{d^2V}{dx^2}$$ $$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{\frac{d^2V}{dx^2}}{m}}$$ $$\frac{d^2V}{dx^2}=2\alpha x+3\beta x^2$$ \begin{align*} \left.\frac{d^2V}{dx^2}\right|_{x=-\frac{\alpha}{\beta}}&=-\frac{2\alpha^2}{\beta}+\frac{3\alpha^2}{\beta}\\ &=\frac{\alpha^2}{\beta} \end{align*} $$\omega=\frac{\alpha}{\sqrt{m\beta}}$$

Hence, answer is (B)

3. A particle of unit mass moves in the $xy$-plane in such a way that $\dot x(t)=y(t)$ and $\dot y(t)=-x(t)$. We can conclude that it is in a conservative force-field which can be derived from the potential
1. $\frac{1}{2}(x^2+y^2)$
2. $\frac{1}{2}(x^2-y^2)$
3. $x+y$
4. $x-y$

For a conservative force field, we have, $$m\ddot{\vec r}=-\vec\nabla V$$ For $m=1$ $$\ddot{\vec r}=-\vec\nabla V$$ $$\ddot x\hat i+\ddot y\hat j=-\left(\frac{\partial V}{\partial x}\hat i+\frac{\partial V}{\partial y}\hat j\right)$$ Using $\dot x(t)=y(t)$ and $\dot y(t)=-x(t)$ in above equation we get $$ x\hat i+ y\hat j=\frac{\partial V}{\partial x}\hat i+\frac{\partial V}{\partial y}\hat j$$ Clearly, $V=\frac{1}{2}(x^2+y^2)$ satisfies above condition.

Hence, answer is (A)

4. The concentration of electrons, $n$, and holes, $p$, for an intrinsic semiconductor at a temperature $T$ can be expressed as $n=p=AT^{3/2}\exp{\left(-\frac{E_g}{2k_BT}\right)}$, where $E_g$ is the band gap and $A$ is a constant. If the mobility of both types of carriers is proportional to $T^{-3/2}$, then the log of the conductivity is a linear function of $T^{-1}$, with slope
1. $E_g/(2k_B)$
2. $E_g/k_B$
3. $-E_g/(2k_B)$
4. $-E_g/k_B$

Conductivity is given by $$\sigma=ne\mu$$ $$\sigma=AT^{3/2}\exp{\left(-\frac{E_g}{2k_BT}\right)}e\mu$$ But $\mu\propto T^{-3/2}$ $$\sigma=Be\exp{\left(-\frac{E_g}{2k_BT}\right)}$$ $$\ln\sigma=\ln{Be}-\frac{E_g}{2k_BT}$$ slope$=-\frac{E_g}{2k_B}$

Hence, answer is (C)

5. The rank-2 tensor $x_ix_j$, where $x_i$ are the Cartesian coordinates of the position vector in three dimensions, has 6 independent elements. Under rotation, these 6 elements decompose into irreducible sets (that is, the elements of each set transform only into linear combinations of elements in that set) containing
1. 4 and 2 elements
2. 5 and 1 elements
3. 3, 2 and 1 elements
4. 4, 1 and 1 elements

The rank-2 tensor $T_{ij}=x_ix_j$ can be decomposed into into irreducible representations in the following way $$T_{ij}=T_{ij}^{(0)}+T_{ij}^{(1)}+T_{ij}^{(2)}$$ where,

$T_{ij}^{(0)}$ is a tensor of rank 0 and has 1 independent component only.

$T_{ij}^{(1)}$ is a tensor of rank 1 and has 3 independent components.

$T_{ij}^{(2)}$ is a tensor of rank 2 and has 5 independent components.

Hence, answer is (B)