In the photoelectric effect, the number of photo-electrons emitted is proportional to:

1.	velocity of incident beam
2.	frequency of incident beam
3.	intensity of incident beam
4.	work function for cathode material

The de-Broglie wavelength of electrons falling on the target in an X-ray tube is 'A. The cut-off wavelength of the emitted X-ray is:

1.  ${\mathrm{\lambda }}_0=\frac{{\left(\mathrm{mc\lambda }\right)}^2}{\mathrm{h}}$

2.  ${\mathrm{\lambda }}_0=\frac{{\mathrm{m}}^2\mathrm{c\lambda }}{{\mathrm{h}}^2}$

3.  ${\mathrm{\lambda }}_0=\frac{2\mathrm{m} {\mathrm{c\lambda }}^2}{\mathrm{h}}$

4.  ${\mathrm{\lambda }}_0=\frac{2\mathrm{m} {\mathrm{c\lambda }}^2}{{\mathrm{h}}^2}$

If M₀ is the mass of an oxygen isotope ₈O¹⁷, M_P and M_n are the masses of a proton and a neutron, respectively, the nuclear binding energy of the isotope is:

1.  ${\mathrm{M}}_{\mathrm{o}}{\mathrm{c}}^2$

2.  $\left({\mathrm{M}}_{\mathrm{o}}-17{\mathrm{M}}_{\mathrm{c}}\right) {\mathrm{c}}^2$

3.  $\left({\mathrm{M}}_{\mathrm{o}}-8{\mathrm{M}}_{\mathrm{p}}\right) {\mathrm{c}}^2$

4.  $\left(8{\mathrm{M}}_{\mathrm{p}}+9{\mathrm{M}}_{\mathrm{n}}-{\mathrm{M}}_{\mathrm{o}}\right) {\mathrm{c}}^2$

A nucleus disintegrates into two nuclear parts which have their velocities in the ratio \(2:1\). The ratio of their nuclear size will be:
1. \( 2^{1 / 3}: 1 \)
2. \( 1: 3^{1 / 2} \)
3. \( 3^{1 / 2}: 1 \)
4. \( 1: 2^{1 / 3}\)

A rod Ab of length 1 is moving with ends remaining in contact with frictionless wall and floor. If at the instant, shown the velocity of end B is 2 m/s towards the negative direction of x. The speed of end A will be:
       

1.  $\sqrt{3}{\mathrm{ms}}^{-1}$

2.  $\frac{2}{\sqrt{3}}{\mathrm{ms}}^{-1}$

3.  $\sqrt{2}{\mathrm{ms}}^{-1}$

4.  $\frac{\sqrt{5}}{2}{\mathrm{ms}}^{-1}$