Consider a completely full cylindrical water tank of height 1.6~m and cross-sectional area 0.5~m2 . It has a small hole…

Solution & Explanation

Related Formula

Bernoulli's Principle between the top surface (1) and the hole (2):

P₁ + (1)/(2)ρ v₁² + ρ g h₁ = P₂ + (1)/(2)ρ v₂² + ρ g h₂

Core Logic

Let the top surface be point 1, and the opening of the hole be point 2.

Height of tank, $H = 1.6~m$ $H = 1.6\mathrm{~m}$ (thus height of point 1, $h₁ = 1.6~m$ $h_1 = 1.6\mathrm{~m}$)
Height of hole, $h₂ = 90~cm = 0.9~m$ $h_2 = 90\mathrm{~cm} = 0.9\mathrm{~m}$ from the bottom.
Depth of the hole below top surface, $h = H - h₂ = 1.6 - 0.9 = 0.7~m$ .
Since area of hole is negligibly small compared to tank area, the velocity of fluid surface at the top is negligible ( $v₁ \approx 0$ $v_1 \approx 0$).
Pressure at point 1 (top surface):

P₁ = P₀ + (Mg)/(A)

where $M = 50~kg$ $M = 50\mathrm{~kg}$, $A = 0.5~m²$ $A = 0.5\mathrm{~m}^2$, $P₀$ $P_0$ is atmospheric pressure.

Pressure at point 2 (outside the hole):

$P₂ = P₀$ $P_2 = P_0$

Step 1: Applying Bernoulli's Equation

Substitute these values into Bernoulli's equation with reference datum at the hole ( $h₂ = 0$ $h_2 = 0$):

P₁ + ρ g h = P₂ + (1)/(2)ρ v₂²

(P₀ + (Mg)/(A)) + ρ g h = P₀ + (1)/(2)ρ v₂²

(Mg)/(A) + ρ g h = (1)/(2)ρ v₂²

Step 2: Substitution and Calculation

Substitute $M = 50~kg$ $M = 50\mathrm{~kg}$, $g = 10~m/s²$ $g = 10\mathrm{~m/s}^2$, $A = 0.5~m²$ $A = 0.5\mathrm{~m}^2$, $ρ = 10³~kg/m³$ $\rho = 10^3\mathrm{~kg/m}^3$, and $h = 0.7~m$ $h = 0.7\mathrm{~m}$:

(50 \times 10)/(0.5) + 10³ \times 10 \times 0.7 = (1)/(2) \times 10³ \times v₂²

1000 + 7000 = 500 \times v₂²

8000 = 500 \times v₂²

v₂² = 16 v₂ = 4~m/s

Cylindrical water tank pressure and velocity profile diagram for Q2

Pattern Recognition

Modified Torricelli's formula with top pressure overload:

v = \sqrt(2gh + (2Δ P)/(ρ))

Substitute $Δ P = (Mg)/(A) = (500)/(0.5) = 1000~Pa$ . This quickly yields: $v = \sqrt(2(10)(0.7) + (2(1000))/(1000)) = \sqrt(14 + 2) = 4~m/s$ . Keep this shortcut in mind for pressurized tanks!

Chapter Mix

Class 11 Physics: Mechanical Properties of Fluids

More Fluid Mechanics Previous-Year Questions — Page 7

Q47 jee_main_2024_31_jan_evening Viscosity and Terminal Velocity

A small spherical ball of radius

r

, falling through a viscous medium of negligible density has terminal velocity 'v'. Another ball of the same mass but of radius

2r

, falling through the same viscous medium will have terminal velocity:

A. $(v)/(2)$ $\frac{v}{2}$
B. $(v)/(4)$ $\frac{v}{4}$
C. $4v$ $4v$
D. $2v$ $2v$

Solution

Related Formula

At terminal velocity, downward force equals upward drag (assuming negligible buoyancy):

Mg = 6π η r v

v = (Mg)/(6π η r)

Core Logic

Since the density of the medium is negligible, we ignore buoyant forces. The mass $M$ $M$ of the ball is specified to remain the same in both cases, despite the change in radius (implying the material density of the second ball is lower).

Step 1: Setup Proportionality

Since $M$ $M$, $g$ $g$, and $η$ $\eta$ are all constants:

v \propto (1)/(r)

Step 2: Evaluating the Ratio

For the second ball, $r' = 2r$ $r' = 2r$. Therefore, the new terminal velocity $v'$ $v'$ is:

v' = v \times ((r)/(r')) = v \times ((r)/(2r)) = (v)/(2)

Pattern Recognition

Read the constraints carefully. Usually, questions keep material density uniform ( $v \propto r²$ $v \propto r^2$). However, this specifically says "same mass". This shifts the formula dependency from $v \propto r²$ $v \propto r^2$ entirely to $v \propto 1/r$ $v \propto 1/r$ because $M$ $M$ acts as a constant numerator.

Chapter Mix

Class 11 Physics: Mechanical Properties of Fluids

Q jee_main_2024_31_jan_morning Viscosity And Terminal Velocity

A small steel ball is dropped into a long cylinder containing glycerine. Which one of the following is the correct representation of the velocity time graph for the transit of the ball?

Solution

Related Formula

mg - FB - Fv = ma

Fv = 6πη r v

Core Logic

Viscosity And Terminal Velocity diagram for Q46 - JEE Main 2024 Morning

When dropped, three forces act on the ball: Gravity downwards, Buoyant force upwards, and Viscous drag upwards.

mg - FB - Fv = m (dv)/(dt)

(ρ (4)/(3)π r³)g - (ρL (4)/(3)π r³)g - 6πη rv = m (dv)/(dt)

Let $(4π r³ g(ρ - ρL))/(3m) = K₁$ and $(6πη r)/(m) = K₂$ $\frac{6\pi\eta r}{m} = K_2$.

(dv)/(dt) = K₁ - K₂ v

Integrating from $t=0, v=0$ $t=0, v=0$:

\int₀^v (dv)/(K₁ - K₂ v) = \int₀^t dt

-(1)/(K₂) ln ((K₁ - K₂ v)/(K₁)) = t

v = (K₁)/(K₂) ( 1 - e-K₂ t )

This is an exponential curve starting from the origin and asymptotically approaching the terminal velocity $VT = K₁ / K₂$ $V_T = K_1 / K_2$.

Chapter Mix

Class 11 Physics: Mechanical Properties Of Fluids

Solution & Explanation

Related Formula

Core Logic

Step 1: Applying Bernoulli's Equation

Step 2: Substitution and Calculation

Pattern Recognition

Chapter Mix

Reference Study Guides

More Fluid Mechanics Previous-Year Questions — Page 7

Solution

Related Formula

Core Logic

Step 1: Setup Proportionality

Step 2: Evaluating the Ratio

Pattern Recognition

Chapter Mix

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Related Formula

Core Logic

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Solution & Explanation

Related Formula

Core Logic

Step 1: Applying Bernoulli's Equation

Step 2: Substitution and Calculation

Pattern Recognition

Chapter Mix

Reference Study Guides

More Fluid Mechanics Previous-Year Questions — Page 7

Solution

Related Formula

Core Logic

Step 1: Setup Proportionality

Step 2: Evaluating the Ratio

Pattern Recognition

Chapter Mix

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Related Formula

Core Logic

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