$A$ particle of mass $m_1$ is moving with a velocity $v_1$ and another particle of mass $m_2$ is moving with a velocity $v_2$. Both of them have the same momentum,but their kinetic energies are $E_1$ and $E_2$ respectively. If $m_1 > m_2$,then:

$A$ body of mass $2 \, kg$ moving with a velocity of $3 \, m/s$ collides head-on with a body of mass $1 \, kg$ moving in the opposite direction with a velocity of $4 \, m/s$. After the collision,the two bodies stick together and move with a common velocity. What is the common velocity in $m/s$?

For a crystal,the angle of diffraction $(2 \theta)$ is $90^{\circ}$ and the second order line has a $d$ value of $2.28 \ \mathring{A}$. The wavelength (in $\mathring{A}$) of $X$-rays used for Bragg's diffraction is

Electrons ejected from the surface of a metal,when light of certain frequency is incident on it,are stopped fully by a retarding potential of $3 \ V$. Photoelectric effect in this metallic surface begins at a frequency $6 \times 10^{14} \ s^{-1}$. The frequency of the incident light in $s^{-1}$ is : [Planck's constant $= 6.4 \times 10^{-34} \ J \ s$,charge on the electron $= 1.6 \times 10^{-19} \ C$]

The work function of a metal is $4.2 \ eV$. Its threshold wavelength will be ............ $\mathring{A}$.

The amplitude of a wave represented by the displacement equation $y = \frac{1}{\sqrt{a}}\sin \omega t \pm \frac{1}{\sqrt{b}}\cos \omega t$ will be