Problem set 4

1. For a system performing small oscillations, which of the following statement is correct?
1. The number of normal modes and the number of normal coordinates is equal
2. The number of normal modes is twice the number of normal coordinates
3. The number of normal modes is half the number of normal coordinates
4. There is no specific relationship between the number of normal modes and the number of normal coordinates
(A) The number of normal modes and the number of normal coordinates is equal.
2. For any process, the second law of thermodynamics requires that the change of entropy of the universe be
1. Positive only
2. Positive or zero
3. Zero only
4. Negative or zero
(B) because for reversible process entropy is zero and for irreversible process entropy is positive
3. A body of mass $M=m_1+m_2$ at rest splits into two parts of masses $m_1$ and $m_2$ by an internal explosion which generates a kinetic energy $E$. The speed of mass $m_2$ relative to mass $m_1$ is
1. $\sqrt{\frac{E}{m_1m_2}}$
2. $\sqrt{\frac{2E}{m_1m_2}}$
3. $\sqrt{\frac{EM}{m_1m_2}}$
4. $\sqrt{\frac{2EM}{m_1m_2}}$
After explosion let two parts moves with velocities $v_1$ and $v_2$. According to law of conservation of momentum, the total momentum after split equal to the total momentum before split. i.e. $$m_1v_1+m_2v_2=0M$$ Kinetic energy generated is given by $$E=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2-\frac{1}{2}M0^2$$ $$E=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$$ \begin{align*} 2E&=m_1v_1^2+m_2v_2^2\\ &=-m_2v_2v_1+m_2v_2^2\\ &=m_2\left(v_2^2-v_1v_2\right) \end{align*} \begin{align*} 2E&=m_1v_1^2+m_2v_2^2\\ &=m_1v_1^2-m_1v_1v_2\\ &=m_1\left(v_1^2-v_1v_2\right) \end{align*} $$\frac{2E}{m_2}=\left(v_2^2-v_1v_2\right)$$ $$\frac{2E}{m_1}=\left(v_1^2-v_1v_2\right)$$ Adding these two equations we get $$\frac{2EM}{m_1m_2}=\left(v_1^2+v_2^2-2v_1v_2\right)$$ $$\sqrt{\frac{2EM}{m_1m_2}}=\left(v_2-v_1\right)$$
4. The value of $$x=1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\dots} }}$$
1. $\sqrt{2}$
2. 1.6
3. $\sqrt{3}$
4. 0.8
\begin{align*} x&=1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\dots} }}\\ &=1+\frac{1}{x} \end{align*} $$x^2-x-1=0$$ $$x=\frac{1\pm\sqrt{1+4}}{2}=1.6$$
5. In Young's double slit experiment, if one of the following parameters ($\lambda$, $d$ and $D$) is increased in the same order keeping the other two same, then the fringe width
1. decreases, decreases, increases
2. decreases, increases, increases
3. increases, decreases, increases
4. increases, increases, decreases
Fringe width is the distance between two fringes of the same kind. The formula for fringe width is $\Delta y=\frac{\lambda D}{d}$ If we increase $\lambda$ the fringe width increases. If we increase $d$ the fringe width decreases. If we increase $D$ the fringe width increases. Hence answer is (C)
6. Ideal Atwood machine is nothing but an inextensible string of negligible mass going around the fixed pulley with masses $m_1$ and $m_2$ attached to the ends of the string. If $m_1>m_2$, then the magnitude of acceleration of mass $m_1$ is
1. $\frac{m_1g}{(m_1+m_2)}$
2. $\frac{m_2g}{(m_1+m_2)}$
3. $\frac{(m_1-m_2)g}{(m_1+m_2)}$
4. $g$
Let us take upward direction as positive and downward direction as negative. Force on mass $m_1$ is $$T-m_1g=-m_1a_1$$ Force on mass $m_2$ is $$T-m_2g=m_2a_2$$ $$m_1a_1=m_1g-T$$ $$m_2a_2=T-m_2g$$ $$m_1a_1+m_2a_2=m_1g-m_2g$$ But $a_1=a_2=a$ $$a=\frac{(m_1-m_2)g}{(m_1+m_2)}$$
7. A particle of mass $m$ is released from a large height. Resistive force is directly proportional to velocity $\bar v$ with $k$ as a constant of proportionality. Asymptotic value of the velocity of particle is
1. $\frac{g}{k}$
2. $\frac{k}{m}$
3. $\frac{mg}{k}$
4. $\frac{g}{km}$
Let us take downward direction as positive. Hence force acting on particle is given by $$F=ma=mg-kv$$ $$\frac{dv}{dt}=g-\frac{k}{m}v$$ Asymptotic value of $v$ is the terminal velocity $v_t$, which is constant. Hence, $\frac{dv_t}{dt}=0$ $$g-\frac{k}{m}v_t=0$$ $$v_t=\frac{mg}{k}$$
8. The momentum of an electron (rest mass $m_0$), which has the same kinetic energy as its rest mass energy, is
1. $m_0c$
2. $\sqrt{2}m_0c$
3. $\sqrt{3}m_0c$
4. $2m_0c$
Energy of electron is given by $$E=mc^2=\sqrt{p^2c^2+m_0^2c^4}$$. Also \begin{align*} E&=mc^2\\ &=KE+Rest~mass~energy\\ &=KE+m_0c^2=2m_0c^2 \end{align*} $$2m_0c^2=\sqrt{p^2c^2+m_0^2c^4}$$ $$4m_0^2c^4=p^2c^2+m_0^2c^4$$ $$p^2c^2=3m_0^2c^4$$ $$p=\sqrt{3}m_0c$$
9. A planet of mass $m$ moves around the in an elliptic orbit. If $L$ denotes the angular momentum of the planet, then the rate at which area is swept by the radial vector is
1. $\frac{L}{2m}$
2. $\frac{L}{m}$
3. $\frac{2L}{2}$
4. $\frac{\sqrt{2}L}{m}$
(A)$\frac{L}{2m}$
10. The matrix $\begin{pmatrix}8&x&0\\4&0&2\\12&6&0\end{pmatrix}$ will become singular if the value of $x$ is
    1. $4$
    2. $6$
    3. $8$
    4. $12$
    If the matrix is singular, its determinant is zero. $$det=-96+24x=0$$ Hence, $x=4$