THE TRUE UNIT: TRIADIC ROTATIONAL UNITS OF EQUIVALENCE (TRUE) AND THE THIRD
FORM OF REALITY: GIMMEL; APPLYING THE CONVEYANCE EQUATION (PART 12)
The true quantum unit of mass/energy, as defined above, is very
useful in dimensional extrapolation processes and as the basic measurement unit
of phenomenological distinctions in the calculus of distinctions. It is the
smallest possible measurable discrete quantity of the universal substance of
reality. Every elementary particle is therefore composed of a whole number of
these true quantum units of the universal substance. Quantum mechanical
phenomena that defy explanation in terms of classical physics concepts, are
explicable if they are symmetrical vortical structures spinning at near
lightspeed angular velocities in the mathematically required nine dimensional
domain of quantized reality.
The electron is measurable as one
single true quantum unit of mass/energy equivalence in the 3S1t dimensional
domain of observable reality, but as we shall see, the electron is not
identical with one true quantum unit. We have found that it must be much more
to exist as part of a stable atom. All other stable nonradiating subatomic
entities are measurable in multiples of these subquantal units also. These are
units of measurement, not subquantal entities existing as independent
phenomena. Until impacting on a receptor in an irreversible way, gimmel, the
substance of these units, is a massless, energyless third substance which is
required for stable atomic and subatomic structure.
When we choose to measure the substance of a quantum distinction,
the effects of spinning in the three planes of space register as inertia or
mass, and spin in the timelike dimensional planes manifests as energy because
time is nonexistent without movement, and any movement of mass relative to an
observer is measured by that observer as kinetic energy. Spinning in the
additional planes of reality containing
the space and time domains, requires a third form of the stuff of reality, in
addition to, but not registering as, either mass or energy, to complete the
minimum quantum volume required for the stability of that distinct object.
Because this third form of the stuff of reality is unknown in
current science, we need an appropriate symbol to represent it. Every letter in
the English and Greek alphabets has been used, some for multiple subjects, as a
symbol for something in math and science, so we have gone to possibly the
historically oldest maintained alphabet, Hebrew, at an estimated 3100 years,
but likely older. [1]
We have represented that potential third form of reality here with the third
letter of the Hebrew alphabet, ג (Gimmel),
and we will call this unitary measure of the three forms of reality the Triadic Rotational
Unit of Equivalence, or TRUE
Unit.
The mix of the three equivalent forms of the substance of
reality, (mass, energy, and gimmel) needed to maintain symmetric stability, present in any given 3S1t
measurement, can be determined by a symmetric threedimensional conveyance
equation: We found above that the smallest set of integer values that satisfies
the threedimensional form of the conveyance equation is the set 3, 4, 5 and 6.
So the Diophantine equation 3^{3}
+ 4^{3 }+ 5^{3}= 6^{3} describes the addition of
three volumes with integer radii 3, 4, and 5 to form a symmetric volume with
the integer radius r = 6.
When n = m = 3, the Conveyance Equation Σ^{n}_{i=1} (X_{n})^{m} = Z^{m}
yields:
(X_{1})^{3} + (X_{2})^{3 }+
(X_{3})^{3}= Z^{3}
The
integer solutions of this Diophantine equation, the conveyance equation with in
TRUE units represent the possible combinations of three symmetric vortical
distinctions forming a fourth threedimensional symmetric vortical distinction.
The primary level of symmetric stability – quarks and the conveyance
equation
Because of Planck’s discovery that energy only occurs in integer
multiples of a very small quantum, and Einstein’s discovery of the equivalence
of matter and energy, (E = mc^{2}) we know that the substance of the
universe is quantized. With the appropriate integer
values for X_{1}, X_{2},
X_{3}, and Z, in TRUE units, the threedimensional
conveyance equation (X_{1})^{3}
+ (X_{2})^{3 }+ (X_{3})^{3}= Z^{3} represents
the stable combination of three quarks to form a Proton or Neutron. There are
many integer solutions for this equation and historically, methods for solving
it were first developed by Leonhard Euler ^{99}.
Applying mathematics empirically
Our approach is empirical mathematical testing: We start with the
smallest integer solution of this Conveyance Equation, 3^{3} + 4^{3 }+ 5^{3}= 6^{3}, and
see if it can describe the combination of mass/energy and gimmel consistent
with particle collider data.
In order to test the mathematical
hypothesis that the combination of the volume and content of three quarks to
form protons and neutrons can be adequately described using the Diophantine
conveyance equations, we can start by using the simplest 3D conveyance equation
solution of 3^{3} + 4^{3 }+
5^{3}= 6^{3}. If this equation doesn’t fit the empirical
data, we need to establish what does work.
When we use the smallest integer
solution, 3^{3} + 4^{3 }+
5^{3}= 6^{3}, to the 3D conveyance equation to attempt to
find the appropriate values of ג for the Proton, we obtain negative values
for ג for the first upquark and the downquark and zero for the
second upquark. It is conceivable that some quarks may contain no ג
units, but negative values are a problem. They cannot be allowed because a negative
number of total ג units would
produce an entity with fewer total observable TRUE units in 3S1t than the sum
of mass/energy units of that entity, violating the conservation of mass and
energy, destroying the particle’s equilibrium and identity.
We now compare two tables showing
hypothesized TRUE and gimmel in the proton and then the neutron. We apply a
trial and error approach, knowing that we need positive integers and ultimately
quantal volumetric figures, where the cube roots are integral. For consistency in a
quantized reality, charge has also been normalized in these tables.
In Table 2P1, we attempt to use the
smallest integer solution of the conveyance equation to describe the
combination of two upquarks and one downquarks in a proton, but some of the quarks have negative ג units.
In Table 2N1, we attempt to use the
smallest integer solution of the conveyance equation to describe the
combination of one upquark and two downquarks in a neutron, all of the quarks have negative ג units.
This means the data in Table 2P1 and
2N1 for the proton and neutron are empirically incorrect: This is impossible.
The table numbering is complex here [2]
Table 12AP1: Trial Combination of Two UpQuarks and One
DownQuark, i.e. The Proton, applying minimal TRUE Units
Particle

Charge^{*}

Mass/Energy

ג

Total TRUE Units

MREV^{**[3]}

u_{1}

+ 2

4

1

3

27

u_{2}

+ 2

4

0

4

64

d

 1

9

4

5

125

Total

+ 3

17

5

12

216=6^{3}

^{ }
And the neutron:
Table 12BN1: Trial Combination of One UpQuark and Two
DownQuarks in TRUE Units as in the neutron (N^{0})
Particle

Charge

Mass/Energy

ג

Total TRUE
Units

MREV

u

+ 2

4

1

3

27

d_{1}

 1

9

5

4

64

d_{2}

 1

9

4

5

125

Totals

0

22

10

12

216=6^{3}

In conformance with Bohr’s solution of the EPR paradox (the
Copenhagen interpretation of quantum mechanics ^{100}), newly formed
elementary entities do not exist as localized particles in 3S1t until a 3S1t
measurement or observation is made. We propose that this is only possible if all TRUE units are undetectable in
3S1t, before observation and measurement. This means that they exist in the
substrate underlying all dimensional domains and will manifest as either
mass/energy, or ג units, to exhibit
the logical patterns of the substrate in observable symmetrically stable 3S1t
forms. In this way, the encompassing substrate, the additional five plus
dimensions of the ninedimensional structure of reality, organizes the 3S1t
world that we experience through the physical senses and their extensions into
discrete forms.
The mathematical distribution of TRUE units cannot result in the
appearance of negative ג units in
the internal structure of an entity. A triadic entity with negative total ג units is not possible because a
negative number of total ג units
would violate the conservation of mass and energy, destroying the particle’s
equilibrium and identity. Why? Because analogous to the axiom ‘nature abhors a
vacuum’, a result of the second law of thermodynamics, just as the electrons of
an incomplete shell rush around the entire volume of the shell trying to fill
it, negative ג units would cause
TRUE units of the mass/energy of the particle to fill the void and the
measurable mass/energy of the particle would no longer be that of a proton or
neutron, and conservation of mass/energy in 3S1t would be violated because the
measured mass/energy equivalence would be changed and the proton or neutron
would become unstable.
Attempting to use the smallest integer solution, (3, 4, 5, 6) of
the Conveyance Equation to find the appropriate values of ג for both the proton and neutron, we obtain negative total ג unit values. This would change the
particle’s measurable mass/energy identity and violate conservation of mass and
energy, so this solution of the conveyance equation will not work and we
continue to look for an appropriate solution. The next numerically smallest
integer solution for the Conveyance Equation is 1^{3} + 6^{3 }+ 8^{3}= 9^{3}, but,
using it also results in negative values of gimmel.
Therefore, the smallest integer solution of the conveyance equation
that produces no negative values of ג
and also no zeroes for the Proton is 6^{3}
+ 8^{3 }+ 10^{3}= 12^{3}.
Using this solution, we have the
electrically and symmetrically stable Proton. This would mean if we adequate
figures for the Neutron (and the Electron) then our calculations would be
viable for symmetrical, stable particles. [4]
Table 12AP2: The Proton (P^{+}) Solution
Particle*

Charge

Mass/Energy

ג

Total
TRUE Units

MREV

u_{1}

+ 2

4

2

6

216

u_{2}

+ 2

4

4

8

512

d_{1}

 1

9

1

10

1,000

Total

+ 3

17

7

24

1728=12^{3}

Nature, reflecting the patterns of the dimensional substrate,
does not have to rely upon random particle encounters to build complex
structural forms. Compound structures are formed within the mathematical
organization of the Conveyance Equation, and useful building blocks have a
significant level of stability in 3S1t for protons to combine with other
compound particles and create structures sufficiently complex to support life.
To see how other structures arise from quarks, protons and electrons, we need to
know how protons, neutrons and electrons relate to the Conveyance Equation: (X_{1})^{3} + (X_{2})^{3
}+ (X_{3})^{3}= Z^{3}. If the total number of
TRUE units in the proton is equal to the integer X_{1}, the number of TRUE units in the neutron = X_{2}, the number of TRUE units
in the electron = X_{3},
then the resulting compound entity, will be stable in the 3S1T domain of
physical observations.
We know that the 24 TRUEunit Proton must combine with an
electron to form a Hydrogen atom, and with other protons, electrons and
neutrons to form the other elements. In order to find the smallest solution of
the conveyance equation that can include the 24 TRUE units of the proton, we
may start by trying the solutions we’ve used so far.
24 is a multiple of 2, 3, 4, 6, and 8, any one of which can be a
factor of X_{1} in the
conveyance equation solutions we’ve used so far. Up to this point we’ve only
used the first two of the smallest primitive integer solutions of the equation:
3^{3} + 4^{3 }+ 5^{3 }=
6^{3} and 1^{3} + 6^{3
}+ 8^{3 }= 9^{3}. (A primitive Diophantine solution is
defined as one without a common factor in all terms.) We have also tried to use
6^{3} + 8^{3 }+ 10^{3}=
12^{3}, an integer solution obtained by multiplying all of the
terms of the smallest primitive solution by 2. The first 36 integer solutions
of the conveyance equation (X_{1})^{3}
+ (X_{2})^{3 }+ (X_{3})^{3 }= Z^{3} are
listed below in ascending order. Primitive solutions are in bold in Table 3.
Table 12C: The First 36 Conveyance Equation Integer Solutions
for n=m=3.
3^{3} + 4^{3} + 5^{3}
= 6^{3}

1^{3} + 6^{3} + 8^{3 }=
9^{3}

6^{3}
+ 8^{3} + 10^{3} = 12^{3}

2^{3}+
12^{3} + 16^{3} = 18^{3}

3^{3} + 10^{3} + 18^{3
}= 19^{3}

7^{3} + 14^{3} + 17^{3
}= 20^{3}

12^{3}
+ 16^{3} + 20^{3 }= 24^{3}

4^{3} + 17^{3} + 22^{3}
= 25^{3}

3^{3}
+ 18^{3} + 24^{3 }= 27^{3}

18^{3} + 19^{3} + 21^{3
}= 28^{3}

11^{3} + 15^{3} + 27^{3}
= 29^{3}

15^{3}
+ 20^{3} + 25^{3} = 30^{3}

4^{3}
+ 24^{3} + 32^{3} = 36^{3}

18^{3}
+ 24^{3} + 30^{3} = 36^{3}

2^{3} + 17^{3} + 40^{3}
= 41^{3}

6^{3} + 32^{3} + 33^{3}
= 41^{3}

16^{3} + 23^{3} + 41^{3}
= 44^{3}

5^{3}
+ 30^{3} + 40^{3} = 45^{3}

3^{3}
+ 36^{3} + 37^{3} = 46^{3}

27^{3}
+ 30^{3} + 37^{3} = 46^{3}

24^{3}
+ 32^{3} + 40^{3} = 48^{3}

8^{3}
+ 34^{3} + 44^{3} = 50^{3}

29^{3} + 34^{3} + 44^{3}
= 53^{3}

12^{3} + 19^{3} + 53^{3}
= 54^{3}

36^{3} + 38^{3} + 42^{3}
= 56^{3}

15^{3}
+ 42^{3} + 49^{3} = 58^{3}

21^{3}
+ 42^{3} + 51^{3 }= 60^{3}

30^{3}
+ 40^{3} + 50^{3} = 60^{3}

7^{3}
+ 42^{3} + 56^{3} = 63^{3}

22^{3} + 51^{3} + 54^{3}
= 67^{3}

36^{3} + 38^{3} + 61^{3}
= 69^{3}

7^{3} + 54^{3} + 57^{3}
= 70^{3}

14^{3} + 23^{3} + 70^{3}
= 71^{3}

34^{3}
+ 39^{3} + 65^{3} = 72^{3}

38^{3}
+ 43^{3} + 66^{3} = 75^{3}

31^{3}
+ 33^{3} + 72^{3} = 76^{3}

The numbers appearing in the totals in the tables describing
quarks, protons, neutrons and atoms are the smallest possible nonnegative
integers consistent with the empirical data and the requirement for symmetry
that the sum of the three totals cubed must equal an integer cubed. Thus, we
can calculate the number of ג units involved, and the totals of TRUE
units required by the conveyance equation to yield results consistent with
empirical particle collider data. Note that the TRUE units in these tables,
consistent with 3S1t observation, are measurements of threedimensional
objects in multiples of the unitary linear measure of their volumes, and their
minimal rotational equivalence volumes (MREV), listed in the last column, are
equal to the TRUE unit values cubed.
As indicated, negative values for ג cannot occur because of the conservation
of mass and energy as negatives would destroy the mass/energy/ ג balance and turn the
quarks into unstable combinations which would decay quickly. Note that unstable
quarks, e.g. top, charm or bottom quarks, will likely fall into specific
unstable series of conveyance Diophantine equations. But this is a subject for
further research. For now, we must find the smallest unique conveyance equation solution for each combination of
subatomic particles. Nature is parsimonious, and we must never make a
mathematical description or demonstration any more complicated than it has to
be. The correct unique solution can be found for each triadic subatomic
particle by starting with the smallest integer solution of the conveyance
equation and moving up the integer scale by trial and error, until no negative
values are obtained. Also, a solution with the total for any term equal to zero
cannot be allowed, because, in that case, there would be no solution as the
resulting Diophantine equation and the Fermat inequality would apply. Using the
solution 6^{3} + 8^{3 }+ 10^{3}= 12^{3}, the
first attempt to find the TRUE unit configuration of the neutron is shown
below:
Table 12BN2: The Neutron (N^{0}) Solution
Trial Combination of One UpQuark and Two DownQuarks
in TRUE Units
Particle

Charge

Mass/Energy

ג

Total TRUE
Units

MREV

u_{1}

+ 2

4

2

6

216

d_{1}

 1

9

1

8

512

d_{2}

 1

9

1

10

1000

Totals

0

22

2

24

1728=12^{3}

Since this solution still produces a negative value of ג for d1,
we must move to the next larger solution to represent the Neutron. The smallest
unique Conveyance Equation solution with no negative or zero values of ג for
the stable Neutron is 93 + 123 + 153= 183
These TRUE unit numbers give us a stable neutron; but now we have
another problem: None of the solutions with a term equal to 24 have a second
term equal to 36. Nor do any of the solutions listed have two terms with the
ratio 24/36 =2/3. This is a problem because it means that atoms with equal
numbers of protons and neutrons could not be stable because they would not
satisfy any of the solutions of the conveyance equation, and we know that the
element Helium, and other elements are stable combinations with equal numbers
of protons and neutrons.
Table 12BN3 Trial of Quark Combinations for the Neutron (N^{0})
Particle

Charge

Mass/Energy

ג

Total TRUE
Units

MREV

u_{3}

+ 2

4

5

9

729

d_{2}

 1

9

3

12

1,728

d_{3}

 1

9

6

15

3,375

Totals

0

22

14

36

5,832=18^{3}

We now apply the stable proton and neutron to the smallest
element with both neutrons (hydrogen does not have a neutron) and protons. To
describe a stable neutron, proton, electron combination, the conveyance
equation solution would have to be either 4^{3} + 24^{3} + 32^{3}
= 36^{3}, 18^{3} + 24^{3} + 30^{3} = 36^{3},
or some other combination of the integers 24 and 36. For example: looking at
the TRUEunits analysis of Helium, with protons consisting of 24 TRUE units and
neutrons consisting of 36 TRUE units, we have:
Table 12DHe1: Attempt to Construct a Helium Atom with P^{+ }=
24 and N^{0} = 36
Particle

Charge

Mass/Energy

ג

Total
TRUE Units

MREV

2e

 6

2

78

80^{}

512,000

2P^{+ }

+ 6

34

14

48

110,592

2N^{0}

0

44

28

72

373,248

Totals

0

80

120

200

995,840=(99.
861…)^{3}

The number of TRUE units making up the electron is unknown at
this point. This value was chosen because it is the integer value that produced
a total MREV nearest to a cube, as it must be for a stable Helium atom. So
these figures for protons or neutrons or electrons must be incorrect with us
applying the derived figures: We have found that the smallest integer value in
TRUE units that can satisfy the conveyance equation to produce a stable proton
is 24, and the smallest integer value in TRUE units that can produce a stable
neutron is 36. But, if the proton consists of 24 TRUE units and the neutron
consists of 36 TRUE units, or multiples of these integers, atoms with equal
numbers of protons and neutrons, like Helium, cannot combine to satisfy the
conveyance equation. This would contradict the empirical fact that stable
Helium atoms do exist, so, following the law of parsimony, i.e. using the
smallest possible integers, we have to seek another integer solution of the
conveyance equation for the neutron.
Table 12BN4 The trial that works of Quark Combinations for the
Neutron N^{0 }
Particle

Charge

Mass/Energy

ג

Total TRUE
Units

MREV

u_{3}

+ 2

4

3

7

343

d_{2}

 1

9

5

14

2,744

d_{3}

 1

9

8

17

4,913

Totals

0

22

16

38

8,000=20^{3}

Next, we need to see if this quark combination for the neutron
combined with protons and electrons will yield stable atomic structures. Using
these values for P^{+} and N^{0}, the first integer
solution of the conveyance equation containing the values X_{1 }= 24 and X_{2
}= 38, or multiples of them, is obtained by multiplying both sides of
the primitive solution 12^{3} +
19^{3} + 53^{3} = 54^{3} by 2, yielding the integer
solution 24^{3} + 38^{3}
+ 106^{3} = 108^{3}.
Note that we have different kinds of quarks with different ratios
of mass/energy to gimmel: There are three different kinds (or colors) of
upquarks u1, u2, u3 with u3 in the neutron being different from the u1 and u2
in the proton. Similarly, d1 in the down quark of the proton, is different from
the d2 and d3 in the neutron. Therefore, each up quark and each down quark is
triadic. They logically come in threes fitting the integer solutions to the
conveyance equation.
Table 12DHe2: Helium Atom with P^{+ }= 24 and N^{0}
= 38
Particle

Charge

Mass/Energy

ג

Total TRUE
Units

MREV

2e

 6

2

210

212^{*}^{}

9,528,128

2P^{+ }

+ 6

34

14

48

110,592

2N^{0}

0

44

32

76

438,976

Totals

0

80

256

336

10,077,696=216^{3}

With the TRUE units determined for protons and neutrons, the
Helium atom is stable only if the total number of TRUE units for the electron
is 106.
Besides the TRUE units that appear as mass/energy in given
elementary particles, because of the embedded nature (dimensional tethering) of
dimensional domains in TDVP, there must be a minimum number of ג units
associated with each particle for stability. Consistent with up and downquark
decay from the strange quark, the stabilization requirement of an integer solution
for the conveyance equation, and the additional TRUE units of ג
needed for particle stability, the following Table 4A describes the electron,
proton and neutron in TRUE units, with up quarks composed of a total of 24 TRUE
units, down quarks composed of a total of 38 TRUE units and electrons composed
of a total of 106 TRUE units. 106^{3}+24^{3}+38^{3}=108^{3}
It
therefore represents the normalized mass/energy, minimum ג
and total volumes for stable electrons, protons and neutrons, the building
blocks of the physical universe.
Whether mass, energy or gimmel (ג),
upon measurement, each TRUE unit of the substance of reality occupies
the same volume, i.e. the minimal volume for an elementary particle as a
spinning object, as required by relativity and defined in TDVP as the basic
unit of volume is consistently the same for any electrons (106 with 105
gimmel), protons (24 with 7 gimmel) and neutrons (38 with 16 gimmel).
Each TRUE unit is capable of contributing to the structure of
physical reality as m, E or ג to form a stable
particle, according to the logical pattern in the substrate reflected in the
Conveyance Equation, and the relative volume of each particle (in the three
dimensions of space) is equal to the total number of TRUE units cubed times the
shape factor.
Table 12E1: The Building Blocks of the Elements in TRUE Units
Particle

Charge

Mass/
Energy

ג

Total
TRUE Units

Volume

e

 3

1

105

106

1,191,016

P^{+}

+ 3

17

7

24

13,824

N^{0}

0

22

16

38

54,872

As noted
before, the shape factor of any regular form always cancels out of the
conveyance equation. (As demonstrated above for the sphere, the shape factor, 4/3π, occurs in all terms of the
equation, and thus can be cancelled by dividing both sides of the equation by 4/3π.) Thus the same equation is
obtained regardless of the shape of the particles, as long as the shape and
substance is the same for all three particles). For this reason, the righthand column in these tables
contains cubed integer amounts
representing the Minimum Relative
Equivalence Volume (MREV) for each particle making up the combination of
subatomic particles.
The TRUE unit values for these elementary particles are uniquely
determined by conditions necessary for the existence of a stable universe. The
values for up and downquarks are the necessary values for the proton and
neutron, as determined above, and the number of ג units and the total TRUE
units for the electron are determined by calculating the ג units necessary to form stable atoms like the Helium atom. They
also determine the smallest possible stable atoms, Hydrogen H1, Deuterium H2
and Tritium H3, as shown below.
Atoms are semistable
structures composed of electrons, protons and neutrons. They are not as stable
as protons and neutrons, but they are generally more stable than molecules.
Some molecules, like H_{2}O, are more stable than others ostensibly
because of higher gimmel content, but all of the factors that contribute to
stability must be considered, especially symmetry.
[1] Hebrew is the oldest continuously
enduring language and regarded as the “holy language”. As this third substance
has a postulated possibly mystical significance, the name gimmel, as the third
letter of the Hebrew alphabet, may be appropriate.
[2] The numbering here as a
convenience. It involves the part e.g. Part 12 and the first table so 12A. But
in the instances of testing it has a suffix. So here Table 12A – P1 has the –P1
referring to the first in the test sequence of Protons so P1. Because this
might not work out, the next would be Table 12A P2. This allows convenience
for those observing the mathematical test sequence only.
[3] Minimum Rotational Equivalent Volume
(MREV): This is a term we apply so we can reflect cubes as required in quantal
volumes.
[4] Upquarks are designated u, and
downquarks as d: u_{1 }and u_{2} in the proton, have the same
number of TRUE units of mass and energy, and therefore will register as
upquarks in the collider data, but have different numbers of TRUE units of
equivalent volume participating as ג
to produce the volumetrically symmetric, and therefore stable, (and also vary
in spin proportions of 0.5) We could refer to u_{1 }and u_{2}
using another method of particle description commonly employed in physics,
namely distinction by color, as in chromodynamics theory (QCD). We would have
little difficulty, e.g., saying that because the stable quarks in the proton
come in threes and they could be referred to as ‘green’ for u_{1 }and
‘yellow’ for u_{2} which have the same mass and energy in collider data
but have different third substance gimmel values and are therefore different in
the combination. With this scheme, it is clearly indicated that stable quarks
are in fact triadic, occurring only in threes in the proton. The d_{1}
for the downquark could be another color, e.g., ‘orange’. The converse applies
to the neutron, which is still triadic with three stable quarks but this time
what is referred to as 2 downquarks would be the d2 and d3 and the colors
could be “blue” and “red” but again reflecting the massenergy collider data of
downquarks, plus say a “purple” for u3, the third upquark.
The state class 6th, 7th, 8th, and 9th standard of high school level Bengali medium, English medium, Hindi medium, Urdu medium, and other medium students of WBBSE have been announced. Every year, West Bengal Madhyamik Model Paper 2022 the WB Kolkata board releases the WB Madhyamik Shiksha Model Question Bank 2022 with suggested answers for all general and vocational course students in grades VI to IX, and this year was no exception.
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