JEE Main 31-Aug-2021 Shift 1 Solved Paper

As we know that,
Dimensional formula of volume =[M0L3T0] ...(i)
Since, F=6πηrv
∴η∝

F

rv

where, η is viscosity and F is force.
∴[η]=

[ML‌T−2]

[L.LT−1]

=[ML−1T−1]
So,[

πpa4

8ηL

]=

[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=

2s‌cos‌θ

ρgr

Dimensional formula of LHS part,
∴ [h] = [L]
Dimensional formula of RHS part
=

[s]

[ρ][g][r]

=

[MT−2]

[ML−3][LT−2][L]

=[L]
Hence, h=

2s‌cos‌θ

ρrg

is dimensionally correct.
So, option (b) will also be dimensionally correct.
In option (c),
J=ε

dE

∂t

...(iii)
⇒J=ε

E

t

Dimension of current density J is calculated as
Since, J=

l

A

∴[J]=

[l]

[A]

=

[A]

[L2]

⇒[J]=[AL−2] ...(iv)
Again, we know that
E=

1

4πε

.

q

r2

⇒εE=

1

4π

.

q

r2

⇒εE

t

=

1

4π

.

q

tr2

⇒[

εE

t

]=

[q]

[t][r2]

=

[AT]

[T][L2]

[

εE

t

]=[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]