As we know that,
Dimensional formula of volume
=[M0L3T0] ...(i)
Since,
F=6πηrv ∴η∝ where, η is viscosity and F is force.
∴[η]==[ML−1T−1] So,
[]=| [ML−1T−2][L4] |
| [ML−1T−1][L1] |
=[M0L3T−1] ...(ii)
Since, Eq. (i) is not equal to Eq (ii), so option (a) is wrong.
Now, since formula of capillary rise in tube,
h= Dimensional formula of LHS part,
∴ [h] = [L]
Dimensional formula of RHS part
===[L] Hence,
h= is dimensionally correct.
So, option (b) will also be dimensionally correct.
In option (c),
J=ε ...(iii)
⇒J=ε Dimension of current density J is calculated as
Since,
J= ∴[J]== ⇒[J]=[AL−2] ...(iv)
Again, we know that
E=. ⇒εE=. =. ⇒[]== []=[AL−2] ...(v)
Form Eqs. (iv) and (v), we see that Eq. (iii) is dimensionally correct.
In option (d) W = τθ
and
τ=r×F So, dimensional formula of
[τ][θ]=[r][F]=[L][MLT−2]=[ML2T−2] =[W]