Glycerine of density 1.25 × 103 kg m-3 is flowing through the conical section of pipe. The area of cross-section of the pipe at its ends is 10 cm2 and 5 cm2 and pressure drop across its length is 3 Nm-2. The rate of flow of glycerine through the pipe is x x10-5 m3 s-1. The value of x is____.
Correct Answer: 4
Solution and Explanation
To solve the problem, we need to find the rate of flow of glycerin through a conical pipe given the density, cross-sectional areas, and pressure drop. Let's break this down step by step.
Step 1: Understand the given data
- Density of glycerin, ρ = 1.25 × 103 kg/m3
- Area at one end of the pipe, A1 = 10 cm2 = 10 × 10-4 m2 = 1 × 10-3 m2
- Area at the other end of the pipe, A2 = 5 cm2 = 5 × 10-4 m2
- Pressure drop, ΔP = 3 N/m2
Step 2: Apply the continuity equation
Using the principle of conservation of mass, we can apply the continuity equation:
A1v1 = A2v2
From this, we can express v2 in terms of v1:
v2 = (A1/A2)v1 = (10 × 10-4 / 5 × 10-4)v1 = 2v1
Step 3: Apply Bernoulli's equation
According to Bernoulli's equation, we have:
P1 + (1/2)ρv12 = P2 + (1/2)ρv22
Rearranging gives us:
P1 - P2 = (1/2)ρ(v22 - v12)
Substituting ΔP = P1 - P2 = 3 N/m2 and v2 = 2v1:
3 = (1/2)ρ((2v1)2 - v12)
3 = (1/2)ρ(4v12 - v12) = (1/2)ρ(3v12)
3 = (3/2)ρv12
Step 4: Solve for v1
Substituting the value of ρ:
3 = (3/2)(1.25 × 103)v12
3 = (3 × 1.25 × 103 / 2)v12
v12 = (3 × 2) / (3 × 1.25 × 103) = 2 / (1.25 × 103)
v12 = 2 / 1250 = 1 / 625 (m/s)2
v1 = √(1/625) = 1/25 = 0.04 m/s
Step 5: Calculate the rate of flow
The rate of flow Q is given by:
Q = A1v1
Substituting A1 = 1 × 10-3 m2 and v1 = 0.04 m/s:
Q = (1 × 10-3)(0.04) = 4 × 10-5 m3/s
Step 6: Express in terms of x
The rate of flow is given as x × 10-5 m3/s:
Q = 4 × 10-5 ⇒ x = 4
Final Answer
The value of x is 4.
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
Top JEE Main Pressure Questions
Two vessels A and B are of the same size and are at the same temperature. A contains 1 g of hydrogen and B contains 1 g of oxygen. \(P_A\) and \(P_B\) are the pressures of the gases in A and B respectively, then \(\frac{P_A}{P_B}\) is:
- In a container, 1 g of hydrogen and 1 g of oxygen are taken. Find the ratio of hydrogen pressure to oxygen pressure.
- \(T-V\) graph is given for two different pressures \(P_1\) and \(P_2\), Then :
- Mercury is filled in a tube of radius \(2 \, \text{cm}\) up to a height of \(30 \, \text{cm}\). The force exerted by mercury on the bottom of the tube is ___N.
(Given, atmospheric pressure = \(10^5 \, \text{N/m}^2\), density of mercury = \(1.36 \times 10^4 \, \text{kg/m}^3\), \(g = 10 \, \text{m/s}^2\), \(\pi = \frac{22}{7}\)) - Pressure inside a soap bubble is greater than the pressure outside by an amount;(given : R = Radius of bubble, S = Surface tension of bubble)
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.