When a charge +Q is placed at the centre of spherical cavity as shown in the figure, then charge induced on the inner surface of a shell is - Q and charge induced on the outer surface of a shell is + Q.

Therefore, surface charge density on inner surface of shell is \(\frac{-Q}{4 \pi R_{1}^{2}}\) and on outer surface of shell is \(\frac{+Q}{4 \pi R_{2}^{2}}\)

Yes, it changes as the distance becomes less

The electric fields find the atoms to netural entity.As it is known that, electrostatic fields are caused by excess charges.However there is no excess charge on the inner surface of an isolated conductor.Therefore, electrostatic  field inside a conductor is zero.

Two point charge system is taken from air to water keeping other variable(e.g distance, magnitude of charge) unchanged.so the only factor which may affect the interacting force is dielectric constant of medium.
Force acting between two point charges,
\(F=\frac { 1 }{ 4\pi \varepsilon _{ 0 }K } \frac { q_{ 1 }q_{ 2 } }{ r^{ 2 } } \ or \ F\propto \ \frac { 1 }{ K } \Rightarrow \frac { F_{ air } }{ F_{ medium } } =K\)
\(\Rightarrow \frac { 8 }{ F_{ water } } =80 \ \Rightarrow \ F_{ water }=\frac { 8 }{ 80 } =\frac { 1 }{ 10 } N\).

\(\tau =\rho E\sin { \theta ,\tau \ is \ minimum \ when \ \theta ={ 0 }^{ 0 } } \)

Current in the wire is steadily decreasing, so the induced current in rings 1 and 2 will flow in such a way that it opposes the decrease of current. So, it will flow in same direction. Now, from the figure, it is clear that the direction of induced current in
(i) ring 1 is clockwise
(ii) ring 2 is anti-clockwise

No, this is because the magnetic field due to the current in the coil (A and B) will be parallel to the plane of the other coil (B and A).Hence, the magnetic flux linked with the other coil will be zero and so no current will be induced in it.

When the switch is thrown from the OFF position (open circuit) to the ON position (closed circuit), then neither B nor A and the angle between B and A does not change.Thus, no change in magnetic flux linked with coil occur, hence no electromotive force is produced and consequently, no current will flow in the circuit.

In photoelectric effect,. we can observe that most electrons get scattered into the metal by absorbing a photon.
Thus, all the electrons that absorb a photon does not come out as photoelectron. Only a few comes out of metal whose energy becomes greater than the work functibn of metal.

The size of the dopant atom should be such that their presence in the pure semiconductor does not distort the semiconductor but easily contribute the charge carriers on forming covalent bonds with Si or Ge atoms, which are provided by group XIII or group XV elements.