Problem set 81

1. The Gauss hypergeometric function $F(a,b,c;z)$, defined by the Taylor series expansion around $z=0$ as $${\scriptstyle F(a,b,c;z)=\sum\limits_{n=0}^\infty\frac{a(a+1)\cdots(a+n-1)b(b+1)\cdots(b+n-1)}{c(c+1)\cdots(c+n-1)n!}z^n}$$ satisfies the recursion relation
1. ${\scriptstyle \frac{d}{dz}F(a,b,c;z)=\frac{c}{ab}F(a-1,b-1,c-1;z)}$
2. ${\scriptstyle \frac{d}{dz}F(a,b,c;z)=\frac{c}{ab}F(a+1,b+1,c+1;z)}$
3. ${\scriptstyle \frac{d}{dz}F(a,b,c;z)=\frac{ab}{c}F(a-1,b-1,c-1;z)}$
4. ${\scriptstyle \frac{d}{dz}F(a,b,c;z)=\frac{ab}{c}F(a+1,b+1,c+1;z)}$

$${\scriptstyle \frac{d}{dz}F(a,b,c;z)=\sum\limits_{n=1}^\infty n\frac{a(a+1)\cdots(a+n-1)b(b+1)\cdots(b+n-1)}{c(c+1)\cdots(c+n-1)n!}z^{n-1}}$$ $${\scriptstyle \frac{d}{dz}F(a,b,c;z)=\sum\limits_{n=1}^\infty\frac{a(a+1)\cdots(a+n-1)b(b+1)\cdots(b+n-1)}{c(c+1)\cdots(c+n-1)(n-1)!}z^{n-1}}$$ $${\scriptstyle \frac{d}{dz}F(a,b,c;z)=\sum\limits_{n=0}^\infty\frac{a(a+1)\cdots(a+n)b(b+1)\cdots(b+n)}{c(c+1)\cdots(c+n)(n)!}z^{n}}$$ $${\scriptstyle \frac{d}{dz}F(a,b,c;z)=\frac{ab}{c}\sum\limits_{n=0}^\infty\frac{(a+1)\cdots(a+n)(b+1)\cdots(b+n)}{(c+1)\cdots(c+n)(n)!}z^{n}}$$ $${\scriptstyle \frac{d}{dz}F(a,b,c;z)=\frac{ab}{c}\frac{d}{dz}F(a+1,b+1,c+1;z)}$$

Hence, answer is (D)

2. Let $(x,y)$ and $(x',y')$ be the coordinate systems used by the observers $O$ and $O'$, respectively. Observer moves with a velocity $v= \beta c$ along their common positive $x$-axis. If $x_+=x+ct$ and $x_-=x-ct$ are the linear combinations of the coordinates, the Lorentz transformation relating $O$ and $O'$ takes the form
1. $x_+'=\frac{x_--\beta x_+}{\sqrt{1-\beta^2}}$ and $x_-'=\frac{x_+-\beta x_-}{\sqrt{1-\beta^2}}$
2. $x_+'=\sqrt{\frac{1+\beta}{1-\beta}}x_+$ and $x_-'=\sqrt{\frac{1-\beta}{1+\beta}}x_-$
3. $x_+'=\frac{x_+-\beta x_-}{\sqrt{1-\beta^2}}$ and $x_-'=\frac{x_--\beta x_+}{\sqrt{1-\beta^2}}$
4. $x_+'=\sqrt{\frac{1-\beta}{1+\beta}}x_+$ and $x_-'=\sqrt{\frac{1+\beta}{1-\beta}}x_-$

According to Lorentz transformation $$x'=\gamma(x-vt)$$ $$y'=y$$ $$z'=z$$ $$t'=\gamma(t-xv/c^2)$$ where, $\gamma=\frac{1}{\sqrt{1-v^2/c^2}}$. Hence, for $v= \beta c$, $\gamma=\frac{1}{\sqrt{1-\beta^2}}$ \begin{align*} x_+'&=x'+ct'\\ &=\gamma(x-\beta ct)+c\gamma(t-x\beta/c)\\ &=\gamma\left\{x-\beta ct+ct-x\beta\right\}\\ &=\gamma\left\{x_+-\beta (x+ct)\right\}\\ &=\gamma\left\{1-\beta \right\}x_+\\ &=\sqrt{\frac{1-\beta}{1+\beta}}x_+ \end{align*} \begin{align*} x_-'&=x'-ct'\\ &=\gamma(x-\beta ct)-c\gamma(t-x\beta/c)\\ &=\gamma\left\{x-\beta ct-ct+x\beta\right\}\\ &=\gamma\left\{x_-+\beta (x-ct)\right\}\\ &=\gamma\left\{1+\beta \right\}x_-\\ &=\sqrt{\frac{1+\beta}{1-\beta}}x_- \end{align*}

Hence, answer is (D)

3. A particle of mass $m$ is constrained to move in a vertical plane along a trajectory given by $x=A\cos\theta$, $y=A\sin\theta$, where $A$ is constant.
1. The Lagrangian of the particle is
1. $\frac{1}{2}mA^2\dot\theta^2-mgA\cos\theta$
2. $\frac{1}{2}mA^2\dot\theta^2-mgA\sin\theta$
3. $\frac{1}{2}mA^2\dot\theta^2$
4. $\frac{1}{2}mA^2\dot\theta^2+mgA\cos\theta$
2. The equation of motion of particle is
1. $\ddot\theta-\frac{g}{A}\cos\theta=0$
2. $\ddot\theta+\frac{g}{A}\sin\theta=0$
3. $\ddot\theta=0$
4. $\ddot\theta-\frac{g}{A}\sin\theta=0$

1. $$L=\frac{1}{2}mA^2\dot\theta^2-mgA\cos\theta$$

   Hence, answer is (A)

2. $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot\theta}\right)-\frac{\partial L}{\partial\theta}=0$$ gives $$\ddot\theta-\frac{g}{A}\sin\theta=0$$

   Hence, answer is (D)

4. The $x$- and $z$-components of a static magnetic field in a region are $B_x=B_0(x^2-y^2)$ and $B_z=0$, respectively. Which of the following solutions for its $y$-component is consistent with the Maxwell equations?
1. $B_y=B_0xy$
2. $B_y=-2B_0xy$
3. $B_y=B_0(x^2-y^2)$
4. $B_y=B_0(\frac{1}{3}x^3-xy^2)$

$\vec\nabla\cdot\vec B=0$ gives $$2B_0x+\frac{\partial B_y}{\partial y}=0$$ $$\frac{\partial B_y}{\partial y}=-2B_0x$$ $$B_y=-2B_0xy$$

Hence, answer is (B)

5. A magnetic field $\vec B$ is $B\hat z$ in the region $x > 0$ and zero elsewhere. A rectangular loop, in the $xy$-plane, of sides $l$ (along the $x$-direction) and $h$ (along the $y$-direction) is inserted into the $x > 0$ region from the $x < 0$ region at a constant velocity $\vec v =v \hat x$. Which of the following values of $l$ and $h$ will generate the largest EMF?
1. $l=8$, $h=3$
2. $l=4$, $h=6$
3. $l=6$, $h=4$
4. $l=12$, $h=2$

Let us consider an area element $\vec A=hx\hat z$ passing through magnetic field $\vec B=B\hat z$. Magnetic flux passing through area element is $$\Phi=\vec B\cdot\vec A=Bhx$$ Rate of change of magnetic flux $$\frac{d\Phi}{dt}=Bh\frac{dx}{dt}=Bhv$$ $$\epsilon=-\frac{d\Phi}{dt}=-Bhv$$ Hence, EMF $\epsilon$ will be largest for largest value of $h$

Hence, answer is (B)

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)