Physics, asked by Anonymous, 1 year ago

Answer the question which is given in attachment.

Attachments:

QGP: N is number of particles per unit area per unit time

N1 and N2 are number of particles per unit volume

Z1 and Z2 are positions at Z-axis. [Their unit is metre]

Now try the question

Answers

Answered by QGP
5
Here we first need to understand what each term means:

N is the Number of particles per unit Area per unit Time.
Now, number of particles can be taken as unitless, or we can give the name "mole", which after all has no dimensions.

So, Dimension would be:

 [ \, N \, ] = \frac{1}{[ \, Area \, \times \, Time\, ]} \\ \\ \\ \implies [ \, N \, ] = \frac{1}{[ \, L^2 \, ] \, \, [\, T \, ] } \\ \\ \\ \implies [ \, N \, ] = [ \, L^{-2} \, T^{-1} \, ]



N_1 and N_2 are Number of particles per unit volume

Their dimensions would be:

 [\, N_1 \, ] = \frac{1}{[\, Volume \, ]} \\ \\ \\ \implies [ \, N_1 \, ] = \frac{1}{[ \, L^3 \, ]} \\ \\ \\ \implies [ \, N_1 \, ] = [ \, N_2 \, ] = [ \, L^{-3} \, ]



Z_1 and Z_2 are positions at Z-axis. They simply have the unit metre and have the dimension of [ L ] 




Now, we can find the dimensions of D.
N = -D \frac{(N_2-N_1)}{(Z_2-Z_1)} \\ \\ \\ \implies [ \, N \, ] = [ \, D \, ] \, \frac{[\, N_1 \, ]}{[ \, Z_1 \, ]} \\ \\ \\ \implies [ \, L^{-2} \, T^{-1} \, ] = [ \, D \, ] \, \frac{[ \, L^{-3} \, ]}{[\, L \, ]} \\ \\ \\ \implies [\, L^{-2} \, T^{-1} \, ] = [ \, D \, ] \, \, [\, L^{-4} \, ] \\ \\ \\ \implies [ \, D \, ] = \frac{[\, L^{-2} \, T^{-1} \, ]}{[ \, L^{-4} \, ]} \\ \\ \\ \\ \implies \boxed{[ \, D \, ] = [ \, L^2 \, T^{-1} \, ]} \\ \\ \\ \\ \implies \boxed{[ \, D \, ] = [ \, M^0 \, L^2 \, T^{-1} \, ]}


Thus, the answer is Option (3).




Hope it helps
Purva
Brainly Community


Anonymous: Thank u so much!
QGP: You are welcome :))
Answered by yashgandhi74
0

Here we first need to understand what each term means:

NN is the Number of particles per unit Area per unit Time.

Now, number of particles can be taken as unitless, or we can give the name "mole", which after all has no dimensions.

So, Dimension would be:

\begin{lgathered}[ \, N \, ] = \frac{1}{[ \, Area \, \times \, Time\, ]} \\ \\ \\ \implies [ \, N \, ] = \frac{1}{[ \, L^2 \, ] \, \, [\, T \, ] } \\ \\ \\ \implies [ \, N \, ] = [ \, L^{-2} \, T^{-1} \, ]\end{lgathered}

[N]=

[Area×Time]

1

⟹[N]=

[L

2

][T]

1

⟹[N]=[L

−2

T

−1

]

N_1N

1

and N_2N

2

are Number of particles per unit volume

Their dimensions would be:

\begin{lgathered}[\, N_1 \, ] = \frac{1}{[\, Volume \, ]} \\ \\ \\ \implies [ \, N_1 \, ] = \frac{1}{[ \, L^3 \, ]} \\ \\ \\ \implies [ \, N_1 \, ] = [ \, N_2 \, ] = [ \, L^{-3} \, ]\end{lgathered}

[N

1

]=

[Volume]

1

⟹[N

1

]=

[L

3

]

1

⟹[N

1

]=[N

2

]=[L

−3

]

Z_1Z

1

and Z_2Z

2

are positions at Z-axis. They simply have the unit metre and have the dimension of [ L ] .

Now, we can find the dimensions of D.

\begin{lgathered}N = -D \frac{(N_2-N_1)}{(Z_2-Z_1)} \\ \\ \\ \implies [ \, N \, ] = [ \, D \, ] \, \frac{[\, N_1 \, ]}{[ \, Z_1 \, ]} \\ \\ \\ \implies [ \, L^{-2} \, T^{-1} \, ] = [ \, D \, ] \, \frac{[ \, L^{-3} \, ]}{[\, L \, ]} \\ \\ \\ \implies [\, L^{-2} \, T^{-1} \, ] = [ \, D \, ] \, \, [\, L^{-4} \, ] \\ \\ \\ \implies [ \, D \, ] = \frac{[\, L^{-2} \, T^{-1} \, ]}{[ \, L^{-4} \, ]} \\ \\ \\ \\ \implies \boxed{[ \, D \, ] = [ \, L^2 \, T^{-1} \, ]} \\ \\ \\ \\ \implies \boxed{[ \, D \, ] = [ \, M^0 \, L^2 \, T^{-1} \, ]}\end{lgathered}

N=−D

(Z

2

−Z

1

)

(N

2

−N

1

)

⟹[N]=[D]

[Z

1

]

[N

1

]

⟹[L

−2

T

−1

]=[D]

[L]

[L

−3

]

⟹[L

−2

T

−1

]=[D][L

−4

]

⟹[D]=

[L

−4

]

[L

−2

T

−1

]

[D]=[L

2

T

−1

]

[D]=[M

0

L

2

T

−1

]

Thus, the answer is Option (3).

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