Water flows in a streamline motion through a horizontal pipe of circular cross-section as shown in the figure. The pressure difference of water between P and Q is $15\text{ Nm}^{-2}$. The area of cross-section at P and Q are $40\text{ cm}^2$ and $20\text{ cm}^2$, respectively. The rate of flow of water through the pipe, in $\text{cm}^3\text{s}^{-1}$, is :
[Take density of water $= 1000\text{ kg m}^{-3}$]
To find the rate of flow of water through the pipe, we will use the principle of continuity and Bernoulli's equation for fluid flow.
The equation of continuity states that the product of the cross-sectional area and the fluid velocity is constant along the pipe for incompressible fluid flow. Mathematically, this is expressed as:
\(A_P v_P = A_Q v_Q\)
Where:
According to Bernoulli’s principle, for streamline flow, the total mechanical energy is constant, i.e.,
\(P_P + \frac{1}{2} \rho v_P^2 = P_Q + \frac{1}{2} \rho v_Q^2\)
Given:
Rearrange the equation to find the relation between \(v_P\) and \(v_Q\):
\(\frac{1}{2} \rho v_P^2 - \frac{1}{2} \rho v_Q^2 = P_P - P_Q = 15\)
From the continuity equation, we know:
\(v_Q = \frac{A_P}{A_Q} \cdot v_P = 2v_P\)
Substitute \(v_Q = 2v_P\) into Bernoulli's equation:
\(\frac{1}{2} \times 1000 \times v_P^2 - \frac{1}{2} \times 1000 \times (2v_P)^2 = 15\)
Simplify and solve for \(v_P\):
\(500v_P^2 - 2000v_P^2 = 15 \implies -1500v_P^2 = 15\)
\(v_P^2 = \frac{15}{1500} = \frac{1}{100} \implies v_P = \frac{1}{10} \text{ m/s}\)
The rate of flow (Q) is given by:
\(Q = A_P \cdot v_P\)
\(Q = 40 \times 10^{-4} \times \frac{1}{10} = 400 \text{ cm}^3/\text{s}\)
Thus, the rate of flow of water through the pipe is 400 cm3/s.
What will be the approximate terminal velocity of a rain drop of diameter $ 1.8\times {10}^{-3}. \mathrm{m}$, when density of rain water $ \approx {10}^{3}{\mathrm{kgm}}^{-3}$ and the coefficient of viscosity of air $ \approx 1.8\times {10}^{-5}. \mathrm{N}-\mathrm{sm}$$ {}^{-2} ?$ (Neglect buoyancy of air)
Match List I with List II :
| List I | List II |
| A. Young's Modulus | I. $\frac{\Delta d}{\Delta L} ( \frac{L}{d})$ |
| B. Compressibility | II. $\frac{FL}{A(\Delta L)}$ |
| C. Bulk Modulus | III. $-\frac{1}{\Delta P} (\frac{\Delta V}{V})$ |
| D. Poisson's Ratio | IV. $-P(\frac{V}{\Delta V})$ |
Choose the correct answer from the options given below :
A wire of length 10 cm and diameter 0.5 mm is used in a bulb. The temperature of the wire is 1727°C and power radiated by the wire is 94.2 W. Its emissivity is $\frac{x}{8}$ where x =……
(Given $\sigma = 6.0 \times 10^{-8}$ W m$^{-2}$ K$^{-4}$, $\pi = 3.14$ and assume that the emissivity of wire material is same at all wavelength.)
A water drop of radius $1$ cm is broken into eight equal droplets. Surface tension of water is $0.075$ N $m^{-1}$. The gain in surface energy is __________$\times10^{-7}$ J.(Take $\pi = 3.14$)
| List - I | List - II |
| (A) Heat capacity of body | (I) $J kg^{-1}$ |
| (B) Specific heat capacity of body | (II) $J K^{-1}$ |
| (C) Latent heat | (III) $J kg^{-1} K^{-1}$ |
| (D) Thermal conductivity | (IV) $J m^{-1} K^{-1} s^{-1}$ |
What will be the approximate terminal velocity of a rain drop of diameter $ 1.8\times {10}^{-3}. \mathrm{m}$, when density of rain water $ \approx {10}^{3}{\mathrm{kgm}}^{-3}$ and the coefficient of viscosity of air $ \approx 1.8\times {10}^{-5}. \mathrm{N}-\mathrm{sm}$$ {}^{-2} ?$ (Neglect buoyancy of air)