Which of the following equations is dimensionally correct?

If discharge rate is given by  then the dimensions of η by taking velocity (v), time (T) and mass (M) as fundamental units, are :- (where V = rate of flow of volume; P = pressure difference; r = radius; ℓ = length of tube)

A particle moves 4 √2 m in South–West direction, 12m in upward direction and then 1m in East direction. Find the net displacement.

x-component of a vector  is twice of its y-component and √2 times of its z-component. Find out the angle made by the vector from y-axis :-

In following table some unknown physical quantities and their dimensions are given :

Tick the correct possible relation(s) :-

Which of the following is/are the unit vector(s) perpendicular to 

Which one of the following is/are CORRECT option for vectors 

The power for the hovering helicopter depends on its linear size, the density of air and (g × density of the helicopter) as p ∝ (linear size)^x (density of air)^y (g × density of helicopter)^z where g is acceleration due to gravity.

Given :

[Power] = ML²T^–3

Linear Size] = L

[Density] = ML^–3

[g × density] = ML^–2T^–2 and z = 3/2

The value of y in above expression is

The power for the hovering helicopter depends on its linear size, the density of air and (g × density of the helicopter) as p ∝ (linear size)^x (density of air)^y (g × density of helicopter)^z where g is acceleration due to gravity.

Given :

[Power] = ML²T^–3

[Linear Size] = L

[Density] = ML^–3

[g × density] = ML^–2T^–2 and z = 3/2

The ratio of the power out put to hover a helicopter on the earth and on the imaginary planet is [Given g on the planet is one fourth of g on the earth, i.e. g_planet = g/4 and density of air on earth is same as density of air on the imaginary planet]