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Answer:
Step 1
Given: A tank is drained through an orifice from initial liquid level (h₀). The time(r), to drain the tank, depends on tank diameter (D), orifice diameter (d), acceleration due to gravity (g), liquid density (ρ) and liquid viscosity (μ).
r = f (h₀, D, d, g, ρ, μ)
Step 2
There are totally seven parameters r, h₀, D, d, g, ρ and μ
∴ n = 7
The primary variables are M, L, T
∴ m = 3
So, the number of repeating variables are, r = 3
Select the three repeating variables are ρ, d, g
Then according to Buckingham, there will be n-m (⇒ 7-3 = 4) dimensionless groups given by π₁, π₂, π₃ and π₄
So, The number of π terms = 4
Explanation:
See attached images for step 3 to 7
Answer:
Explanation:
First of all we shall calculate electric field near charge 2Q .
electric field due to charge Q = K x Q / (5 x 10⁻² )²
E₁ = KQ / 25 x 10⁻⁴ = KQ x 10⁴ / 25 . It is acting along positive x axis
E₁ = KQ x 10⁴ i / 25
Similarly electric field due to charge 3Q near 2Q
= 3KQ x 10⁴ i / 25 . It is acting along y-axis
E₂ = 3KQ x 10⁴ j / 25
Similarly electric field due to charge 4Q near 2Q
= 4KQ x 10⁴ j / (25 x 2 )
= 2 KQ x 10⁴ / 25 . It is acting acting along north east direction
unit vector in north east direction = ( i + j )/ √2
So E₃ can be represented by
E₃ = 2 KQ x 10⁴ ( i + j ) / 25 x √2
Total field = KQ x 10⁴ i / 25 + 3KQ x 10⁴ j / 25 + 2 KQ x 10⁴ ( i + j ) / ( 25 x √2 )
= KQ x 10⁴ [ i + 3 j + √2 i + √2 j ) / 25
= 400 KQ ( 2.414 i + 4.414 j ) N / C
Force on 2Q = Field x charge = 400 KQ ( 2.414 i + 4.414 j ) x 2Q N
= 800 KQ² ( 2.414 i + 4.414 j ) N
= 800 x 9 x 10⁹ x ( 2.5 x 10⁻⁶ )² x 2.414 x ( i + 2 j ) N
= 108.63 ( i + 2 j ) N .
Magnitude of this force
= 108.63 x √5
= 243 N approx .
