In the circuit shown assume the diode to be ideal. When V_i increases from 2V to 6 V, the change in the current is (in mA)

• zero

• 20

• 80/3

• 40

42.

Consider two concentric spherical metal shells of radii r₁ and r₂ (r₂ > r₁). If the outer shell has a charge q and the inner one is grounded, the charge on the inner shell is

• $\frac{- {\mathrm{r}}_2}{{\mathrm{r}}_1} \mathrm{q}$

• zero

• $\frac{- {\mathrm{r}}_1}{{\mathrm{r}}_2} \mathrm{q}$

• - q

Four cells, each of emf E and internal resistance r, are connected in series across an external resistance R. By mistake one of the cells is connected in reverse. Then the current in the external circuit is

• $\frac{2\mathrm{E}}{4\mathrm{r} + \mathrm{R}}$

• $\frac{3\mathrm{E}}{4\mathrm{r} + \mathrm{R}}$

• $\frac{3\mathrm{E}}{3\mathrm{r} + \mathrm{R}}$

• $\frac{2\mathrm{E}}{3\mathrm{r} + \mathrm{R}}$

The energy of gamma (γ) ray photon is E_γ and that of an X-ray photon is E_x. If the visible light photon has an energy of E_v, then we can say that

• E_X > E_γ > E_v

• E_γ > E_v > E_X

• E_γ > E_X > E_v

• E_X > E_v > Eγ

A long conducting wire carrying a current I is bent at 120° (see figure). The magnetic field B at a point P on the right bisector of bending angle at a distance d from the bend is (µ₀ is the permeability of free space)

• $\frac{3{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{\pi d}}$

• $\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{\pi d}}$

• $\frac{{\mathrm{\mu }}_0\mathrm{I}}{\sqrt{3} \mathrm{\pi d}}$

• $\frac{\sqrt{3} {\mathrm{\mu }}_0\mathrm{I}}{2 \mathrm{\pi d}}$

A circuit consists of three batteries of emf E₁ = 1 V, E₂ = 2 V and E₃ = 3 V and internal resistances 1 Ω, 2Ω and 1 Ω respectively which are connected in parallel as shown in the figure. The potential difference between points P and Q is

• 1.0 V

• 2.0 V

• 2.2 V

• 3.0 V

Half of the space between the plates of a parallel-plate capacitor is filled with a dielectric material of dielectric constant K. The remaining half contains air as shown in the figure. The capacitor is now given a charge Q. Then

• electric field in the dielectric-filled region is higher than that in the air-filled region.

• on the two halves of the bottom plate the charge densities are unequal.

• charge on the half of the top plate above the air-filled pant is $\frac{\mathrm{Q}}{\mathrm{K} + 1}$

• capacitance of the capacitor shown above is $\left(1 + \mathrm{K}\right) \frac{{\mathrm{C}}_0}{2}$, where C₀ is the capacitance of the same capacitor with the dielectric removed.

A stream of electrons and protons are directed towards a narrow slit in a screen (see figure). The intervening region has a uniform electric field E (vertically downwards) and a uniform magnetic field B (out of the plane of the figure) as shown. Then

• electrons and protons with speed $\frac{\left|\mathrm{E}\right|}{\left|\mathrm{B}\right|}$ will pass through the slit

• protons with speed $\frac{\left|\mathrm{E}\right|}{\left|\mathrm{B}\right|}$ will pass through the slit, electron of the same speed will not

• neither electrons nor protons will go through the slit irrespective of their speed

• electrons will always be deflected upwards irrespective of their speed

The output Y ofthe logic circuit given below is

• $\overline{\mathrm{A}} + \mathrm{B}$

• $\overline{\mathrm{A}}$

• $\overline{\left(\overline{\mathrm{A}} + \mathrm{B}\right)} . \overline{\mathrm{A}}$

• $\overline{\left(\overline{\mathrm{A}} + \mathrm{B}\right)} . \mathrm{A}$

The ionization energy of hydrogen is 13.6 eV. The energy of the photon released when an electron jumps from the first excited state (n = 2) to the ground state of a hydrogen atom is

• 3.4 eV

• 4.53 eV

• 10.2 eV

• 13.6 eV