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Six-point charges are kept at the vertices of a regular hexagon of side L  and centre O as shown in the figure.  Given that K = $\frac{1}{4 \pi \epsilon_{0}} \frac{q}{L^{2}}$, which of the following statements(s) is(are) correct.

Answer:

Explanation:

Resultant of 2K and 2K (at $120^{0}C)$ is also 2K towards 4K. Therefore , net electric field is 6K

 (b)   $V_{0}=\frac{1}{4 \pi \epsilon_{0}}$

      =$ \frac{1}{4 \pi \epsilon_{0} L} (q_{A}+.....+q_{E})$

=0

 Because     $q_{A}+q_{B}+q_{C}+q_{D}+q_{E}+q_{F}=0$

 (c) only line PR , potential  is same (=0)

The $\beta$ -decay process, discovered around 1900, is basically the decay of a neutron (n). In the laboratory, a proton (p) and an electron ($e^{-}$) are observed as the decay products of the neutron. Therefore. considering the decay of a neutron as a two-body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant. But experimentally, it was observed that the electron kinetic energy has a continuous spectrum. Considering a three-body decay process, i,e., $n \rightarrow p+e^{-}+v_{e'}^{-}$, around
1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino ($\overline{v_{e}}$) to be massless and possessing negligible energy., and the neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is
0.8 x 106 eV. The kinetic energy carried by the proton is only the recoil energy.

What is the maximum energy of the anti-neutrino?

Answer:

Explanation:

Maximum kinetic energy of anti-neutrino is nearly ($0.8 \times 10^{6} )$eV

The $\beta$ -decay process, discovered around 1900, is basically the decay of a neutron (n). In the laboratory, a proton (p)
and an electron ($e^{-}$) are observed as the decay products of the neutron. Therefore. considering the decay of a neutron as a
two-body decay process, it was predicted theoretically that the kinetic energy of the
electron should be a constant. But experimentally, it was observed that the electron kinetic energy has a continuous
spectrum. Considering a three-body decay process, i,e., $n \rightarrow p+e^{-}+v_{e'}^{-}$, around
1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino ($\overline{v_{e}}$) to be massless and possessing negligible energy., and the neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is
0.8 x 10⁶ eV. The kinetic energy carried by the proton is only the recoil energy.

If the anti-neutrino had a mass of $3eV/c^{2}$ (where c is the speed of light) instead of zero mass, what should 'be the range of the kinetic energy K, of the electron?

Answer:

Explanation:

Maximum kinetic energy of anti-neutrino is nearly ($0.8 \times 10^{6} )$eV

The general motion of a rigid body can be considered to be a combination of (i) a motion of its centre of mass about an axis, and (ii) its motion about an instantaneous axis passing through the centre of mass. These axes need not be stationary. Consider, for example, a thin uniform disc welded (rigidly fixed) horizontally at its rim to a massless stick, as shown in the figure. When the disc-stick system is rotated about the origin on a horizontal frictionless plane with angular speed ω, the motion at any instant can be taken as a combination of (i) a rotation of the centre of mass of the disc about the z-axis, and (ii) a rotation of the disc through an instantaneous .vertlcal axis passing through its centre of mass (as is seen from the changed orientation of poinis P and Q). Both these motions have the same angular speed $\omega$ in this case.

Now consider two similar systems as shown in the figure: Case (a) the disc with its face vertical and parallel to x-z plane; Case (b) the disc with its face making an angle of $45^{0}$ with x-y plane and its horizontal diameter parallel to x-axis. In both cases, the disc is welded at point P, and the systems are rotated with constant angular speed to about the z-axis.

Which of the following statemenrs about the instantaneous axis (passing
through the centre of mass) is correct?

Answer:

Explanation:

(i) Every particle of the disc rotating in a horizontal circle.
(ii) Actual velocity of any particle horizontal.
(iii) Magnitude of the velocity of any particle is  $v=r \omega$ where r is the perpendicular distance of that particle from actual axis of rotation

(z-axis)
(iv) When it is broken into two parts then the actual velocity of any particle is
the resultant of two velocities

 $V_{1}=r_{1}\omega_{1}$  and $v_{2}=r_{2}\omega_{2}$

Here,

$r_{1}=$perpendicular distance of the centre of mass from z-axis.

$\omega_{1}$=angular speed of rotation of centre of mass from z-axis.

$r_{2}$= distance of the particle from centre of mass and

$\omega_{2}$=angular speed of rotation of the disc about the axis passing through centre of mass. (v) Net v wiil be horizontal, if $v_{1}$ and $v_{2}$ both are horizontal. Further, v, is

already horizontal, because the centre of mass is rotating about vertical z-axis.
To make v₂, aiso horizontal, second axis should also be vertical. 

The general motion of a rigid body can be considered to be a combination of (i) a motion of its centre of mass about an axis, and (ii) its motion about an instantaneous axis passing through the centre of mass. These axes need not be stationary. Consider, for example, a thin uniform disc welded (rigidly fixed) horizontally at its rim to a massless stick, as shown in the figure. When the disc-stick system is rotated about the origin on a horizontal frictionless plane with angular speed ω, the motion at any instant can be taken as a combination of (i) a rotation of the centre of mass of the disc about the z-axis, and (ii) a rotation of the disc through an instantaneous .vertlcal axis passing through its centre of mass (as is seen from the changed orientation of points P and Q). Both these motions have the same angular speed $\omega$ in this case.

Now consider two similar systems as shown in the figure: Case (a) the disc with its face vertical and parallel to x-z plane; Case (b) the disc with its face making an angle of $45^{0}$ with the x-y plane and its horizontal diameter parallel to the x-axis. In both cases, the disc is welded at point P, and the systems are rotated with constant angular speed to about the z-axis.

Which of the following statements regarding the angular speed about the instantaneous axis (passing through the centre of mass) is correct?

Answer:

Explanation:

(i) Every particle of the disc rotating in a horizontal circle.
(ii) Actual velocity of any particle horizontal.
(iii) Magnitude of the velocity of any particle is  $v=r \omega$ where r is the perpendicular distance of that particle from the actual axis of rotation (z-axis)

(iv) When it is broken into two parts then the actual velocity of any particle is the resultant of two velocities

 $v_{1}=r_{1}\omega_{1}$  and $v_{2}=r_{2}\omega_{2}$

Here,

$r_{1}=$perpendicular distance of the centre of mass from z-axis.

$\omega_{1}$=angular speed of rotation of centre of mass from z-axis.

$r_{2}$= distance of the particle from centre of mass and

$\omega_{2}$=angular speed of rotation of the disc about the axis passing through the centre of mass. (v) Net v will be horizontal, if $v_{1}$ and $v_{2}$ both are horizontal. Further, v, is already horizontal, because the centre of mass is rotating about the vertical z-axis.

To make v₂, also horizontal, the second axis should also be vertical. 

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