Problem set 80

1. A dielectric sphere of radius $R$ carries polarization $\vec P = kr^2\hat r$, where $r$ is the distance from the centre and $k$ is a constant. In the spherical polar coordinate system, $\hat r$, $\hat \theta$ and $\hat \phi$ are the unit vectors.
1. The bound volume charge density inside the sphere at a distance $r$ from the centre is
1. $-4kR$
2. $-4kr$
3. $-4kr^2$
4. $-4kr^3$
2. The electric field inside the sphere at a distanced $d$ from the centre is
1. $\frac{-kd^2}{\epsilon_0}\hat r$
2. $\frac{-kR^2}{\epsilon_0}\hat r$
3. $\frac{-kd^2}{\epsilon_0}\hat\theta$
4. $\frac{-kR^2}{\epsilon_0}\hat\theta$

1. The unit vector $\hat n$ on the surface of the sphere is equal to the radial unit vector. The bound surface charge is equal to $\sigma_b=\vec P\cdot\hat n|_{r=R}$

   The bound volume charge is equal to $$\begin{align*} \rho_b&=-\vec\nabla\cdot\vec P\\ &=-\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\:kr^2\right)\\ &=-4kr \end{align*}$$

   Hence, answer is (B)

2. The electric field inside the sphere at a distanced $d$ from the centre is $$\begin{align*} \vec E_{volume}&=\frac{1}{4\pi\epsilon_0}\frac{\frac{4}{3}\pi d^3\rho_b}{d^2}\hat r\\ &=\frac{1}{4\pi\epsilon_0}\frac{\frac{4}{3}\pi d^3(-4kd)}{d^2}\hat r\\ &=-\frac{4kd^2}{3\epsilon_0}\hat r \end{align*}$$

   Hence, answer is ()

2. Let $X$ and $Y$ be two independent random variables, each of which follow a normal distribution with the same standard deviation $\sigma$, but with means $+\mu$ and $-\mu$, respectively. Then the sum follows a
1. distribution with two peaks at $\pm\mu$ and mean $0$ and standard deviation $\sigma\sqrt{2}$
2. normal distribution with mean 0 and standard deviation $2\sigma$
3. distribution with two peaks at $\pm\mu$ and mean 0 and standard deviation $2\sigma$
4. normal distribution with mean 0 and standard deviation $\sigma\sqrt{2}$

The convolution of two normal densities with means $\mu_1$ and $\mu_2$ and variances $\sigma_1$ and $\sigma_2$ is again a normal density, with mean $\mu_1+\mu_2$ and variance $\sigma_1^2+\sigma_2^2$. $$\mu_t=-\mu+\mu=0$$ $$\sigma^2_t=\sigma^2+\sigma^2=2\sigma^2$$ $$\sigma_t=\sqrt{2}\sigma$$

Hence, answer is (D)

3. Using dimensional analysis, Planck defined a characteristic temperature $T_p$ from powers of the gravitational constant $G$, Planck’s constant $h$, Boltzmann constant $k_B$ and the speed of light $c$ in vacuum. The expression for $T_p$ is proportional to
1. $\sqrt{\frac{hc^5}{k_B^2G}}$
2. $\sqrt{\frac{hc^3}{k_B^2G}}$
3. $\sqrt{\frac{G}{hc^4k_B^2}}$
4. $\sqrt{\frac{hk_B^2}{Gc^3}}$

The Planck temperature is defined as: $$T_p=\frac{m_pc^2}{k_B}$$ where, $m_p$ is Plank mass.

Now, let $$m_p=c^{n_1}G^{n_2}\hbar^{n_3}$$ Using dimensional analysis we have $${\scriptstyle M^1L^0T^0=\left[M^0L^{n_1}T^{-n_1}\right]\left[M^{-n_2}L^{3n_2}T^{-2n_2}\right]\left[M^{n_3}L^{2n_3}T^{-n_3}\right]}$$ $$n_{3}-n_{2}=1$$ $$n_{1}+3n_{2}+2n_{3}=0$$ $$-n_{1}-2n_{2}-n_{3}=0$$ $$\Rightarrow n_{1}=1/2,n_{2}=-1/2,n_{3}=1/2$$ $$m_p=\sqrt{\frac{c\hbar}{G}}$$ $$T_p=\sqrt{\frac{\hbar c^5}{k_B^2G}}$$

Hence, answer is (A)

4. A ball of mass $m$, initially at rest, is dropped from a height of 5 meters. If the coefficient of restitution is 0.9, the speed of the ball just before it hits the floor the second time is approximately (take $g = 9.8\: m/s^2$)
1. 9.80 m/s
2. 9.10 m/s
3. 8.91 m/s
4. 7.02 m/s

For an object bouncing off a stationary object, such as a floor, coefficient of restitution, $e=\sqrt{\frac{h'}{h}}$ $$h'=e^2h=4.05\:m$$ $$v=\sqrt{2gh}$$ $$v=\sqrt{2\times 9.8\times4.05}=8.91\:m/s$$

Hence, answer is (C)