Correct answer is (A)

Y = k [F]^x [A]^y [V]^z

[M¹L¹T ^–2] = [MLT^–2]^x [L²]^y [LT^–1]^z

[M¹L¹T ^–2] = [M]^x [L]^x+2y+z[T]^–2x–z

Comparing power of M, L and T

x = 1 ……(1)

x + 2y + z = –1 ……(2)

–2x – z = –2 ……(3)

After solving

x = 1

y = –1

z = 0

$\therefore$ Y = FA^–1V⁰

Correct answer is (B)

V₀ $\propto$ h^Pc^QG^RI^S

[V₀] = [M¹L²T^–3A^–1]

[c] = [L¹T^–1]

[h] = [M¹L²T^–1]

[G] = [M^–1L³T^–2]

[I] = [A]

$\therefore$ [M¹L²T^–3A^–1] = [M^P–R L^2P+Q+3R T^{–P–Q–2R} A^S]

Comparing dimension of M, L, T, A, we get

P – R = 1 ; 2P + Q + 3R = 2

– P – Q – 2R = – 3 ; S = – 1

$\Rightarrow$ P = 0, Q = 5, R = –1, S = –1

$\therefore$ V₀ $\propto$ h⁰ c⁵ G^-1 A^-1

Correct answer is (A)

We know,

surface tension (S) = $\frac{F}{L}$ = $\frac{\left[ML{T}^{-2}\right]}{\left[L\right]}$

$\therefore$ [S] = $\left[M{T}^{-2}\right]$

Moment of inertia (I) = mr²

$\therefore$ [I] = $\left[M{L}^2\right]$

Planck's constant (h) = $\frac{E}{f}$ = Et

$\therefore$ [h] = $\left[M{L}^2{T}^{-1}\right]$

Also linear momentum (p) = mv = $\left[ML{T}^{-1}\right]$

Now we have to express p in terms of s, I and h.

$\therefore$ Let, [P] = [S^a I^b h^c]

$\Rightarrow$ $\left[ML{T}^{-1}\right]$ = $\left[M{T}^{-2}\right]$^a $\left[M{L}^2\right]$^b $\left[M{L}^2{T}^{-1}\right]$^c

$\Rightarrow$ $\left[ML{T}^{-1}\right]$ = [ M^{a + b + c} L^{2b +2c} T^{- 2a - c} ]

By comparing the dimensions of both sides, we get

a + b + c = 1 .........(1)

2b +2c = 1 ..............(2)

- 2a - c = -1 ...................(3)

By solving those three equations we get,

a = $\frac{1}{2}$

b = $\frac{1}{2}$

c = 0

$\therefore$ linear momentum [p] = [S^1/2I^1/2h⁰]

Correct answer is (C)

We know,

Young's modulus (Y) = $\frac{\frac{F}{A}}{\frac{\mathrm{\Delta }l}{l}}$

$\therefore$ [Y] = $\frac{\left[ML{T}^{-2}\right]}{\left[L^2\right]}$ = [ ML^-1T^-2]

Let [Y] = [V]^x [A]^y [F]^z

$\therefore$ [ ML^-1T^-2] =

[LT^-1]^x [LT^-2]^y [MLT^-2]^z

$\Rightarrow$ [ ML^-1T^-2] =

[ M^z L^{x + y + z} T^{-x -2y - 2z}

For dimensional balance, the dimension on both sides should be same.

So, z = 1

x + y + z = -1

$\Rightarrow$ x + y = -2 ........(1)

and -x -2y - 2z = -2

$\Rightarrow$ x + 2y = 0 ...........(2)

By solving those two equations we get,

x = -4 and y = 2

$\therefore$ [Y] = V^$-$4A²F¹

Correct answer is (C)

Let t $\propto$ G^x h^y c^z

$\therefore$   [t] = [G]^x [h]^y [c]^z    . . . . . (1)

We know,

F = $\frac{G{M}^2}{R^2}$

$\Rightarrow$  G = $\frac{F{R}^2}{M^2}$

$\therefore$  [G] = $\frac{\left[ML{T}^{-2}\right]\left[L^2\right]}{\left[M^2\right]}$

[G] = $[M^{-1}{L}^3{T}^{-2}]$

Also,

E = hf

$\therefore$  [h] = $\frac{[E]}{[F]}$

= $\frac{\left[M{L}^2{T}^{-2}\right]}{[T^{-1}]}$

= $\left[M{L}^2{T}^{-1}\right]$

[C] = $\left[M^{o}L{T}^{-1}\right]$

From equation (1) we get,

$\left[M^o{L}^o{T}^1\right]={\left[M^{-1}{L}^3{T}^{-2}\right]}^x{\left[M{L}^2{T}^{-1}\right]}^y{\left[M^{o}L{T}^{-1}\right]}^z$

$\left[M^o{L}^o{T}^1\right]=\left[M^{-x+y}{L}^{3x+2yz}{T}^{-2x-y-z}\right]$

By comparing the power of M, L, T

$-$ x + y = 0

$\Rightarrow$  x = y

3x + 2y + z = 0

$\Rightarrow$  5x + z = 0 . . . . . (2)

$-$ 2x $-$ y $-$ z = 1

$\Rightarrow$  $-$ 3x $-$ z = 1  . . . . (3)

By solving (2) and (3), we get,

x = $\frac{1}{2}$ = y and z = $-$ $\frac{5}{2}$

$\therefore$  t $\propto$ $\sqrt{\frac{Gh}{C^5}}$