Water flows through a horizontal tube as shown in the figure. The difference in height between the water columns in vertical tubes is 5 cm and the area of cross-sections at A and B are 6text cm^2 and 3text cm^2 respectively. The rate of flow will be ____ textcm^3/s. (take g = 10 textm/s^2)
Bernoulli's Principle diagram for Q29 - JEE Main 2026 Morning
The figure illustrates a Venturi meter setup with water flowing through a variable cross-section tube.

Solution & Explanation

### Related Formula A_1 V_1 = A_2 V_2 P_1 + frac12rho V_1^2 = P_2 + frac12rho V_2^2 P_1 - P_2 = rho g h ### Core Logic From the continuity equation between A and B: A_AV_A = A_BV_B implies 6V_A = 3V_B implies V_B = 2V_A Applying Bernoulli's equation between A and B for a horizontal pipe: P_A + frac12rho V_A^2 = P_B + frac12rho V_B^2 P_A - P_B = frac12rho(V_B^2 - V_A^2) Since the height difference is h = 5text cm = 0.05text m, the pressure difference is rho gh. rho g times 0.05 = frac12rho( (2V_A)^2 - V_A^2 ) g times 0.05 = frac12(3V_A^2) ### Step 1: Calculate Velocity and Volume Flow Rate Solving for V_A: V_A = sqrtfrac2 times 10 times 0.053 = sqrtfrac13 = frac1sqrt3text m/s V_A = frac100sqrt3text cm/s Volume flow rate Q = A_A V_A: Q = 6text cm^2 times frac100sqrt3text cm/s = frac600sqrt3 = 200sqrt3text cm^3text/s ### Pattern Recognition In a horizontal Venturi meter, substituting V_2 = V_1 (A_1/A_2) directly into rho g h = frac12rho (V_2^2 - V_1^2) is the standard path. Working in CGS vs MKS units requires careful tracking; convert V_A to cm/s before multiplying by A_A in cm². ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Fluids

More Mechanical Properties of Fluids Previous-Year Questions — Page 5

Q52 jee_main_2024_30_january_evening Surface Energy of Drops
A big drop is formed by coalescing 1000 small identical drops of water. If mathrmE_1 be the total surface energy of 1000 small drops of water and mathrmE_2 be the surface energy of single big drop of water, the mathrmE_1:mathrmE_2 is x:1 where x =
Numerical Answer. Answer: 10 to 10

Solution

### Related Formula V_textinitial = V_textfinal E = S times A ### Core Logic When small drops coalesce to form a large drop, the total volume is conserved. 1000 times frac43 pi r^3 = frac43 pi R^3 R^3 = 1000 r^3 implies R = 10r ### Step 1: Calculate Surface Energies The total surface energy of 1000 small drops (mathrmE_1) is: mathrmE_1 = 1000 times 4pi r^2 times S The surface energy of the single big drop (mathrmE_2) is: mathrmE_2 = 4pi R^2 times S = 4pi (10r)^2 times S = 100 times 4pi r^2 times S ### Step 2: Find the Ratio fracmathrmE_1mathrmE_2 = frac1000100 = frac101 Thus, the ratio is 10:1, which means x = 10. ### Pattern Recognition When N droplets merge to form one big drop, the radius scales as R = N^1/3 r. The ratio of total initial surface energy to final surface energy is N^1/3. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Fluids
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. fracv2
  • B. fracv4
  • C. 4v
  • D. 2v

Solution

### Related Formula At terminal velocity, downward force equals upward drag (assuming negligible buoyancy): Mg = 6pi eta r v v = fracMg6pi eta r ### Core Logic Since the density of the medium is negligible, we ignore buoyant forces. The mass 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, g, and eta are all constants: v propto frac1r ### Step 2: Evaluating the Ratio For the second ball, r' = 2r. Therefore, the new terminal velocity v' is: v' = v times left(fracrr'right) = v times left(fracr2rright) = fracv2 ### Pattern Recognition Read the constraints carefully. Usually, questions keep material density uniform (v propto r^2). However, this specifically says "same mass". This shifts the formula dependency from v propto r^2 entirely to v propto 1/r because M acts as a constant numerator. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Fluids
Q46 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?
  • A. textGraph 1
  • B. textGraph 2
  • C. textGraph 3
  • D. textGraph 4

Solution

### Related Formula mg - F_B - F_v = ma F_v = 6pieta r v ### Core Logic
Viscosity And Terminal Velocity diagram for Q46 - JEE Main 2024 Morning
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 - F_B - F_v = m fracdvdt left(rho frac43pi r^3right)g - left(rho_L frac43pi r^3right)g - 6pieta rv = m fracdvdt Let frac4pi r^3 g(rho - rho_L)3m = K_1 and frac6pieta rm = K_2. fracdvdt = K_1 - K_2 v Integrating from t=0, v=0: int_0^v fracdvK_1 - K_2 v = int_0^t dt -frac1K_2 ln left(fracK_1 - K_2 vK_1right) = t v = fracK_1K_2 left( 1 - e^-K_2 t right) This is an exponential curve starting from the origin and asymptotically approaching the terminal velocity V_T = K_1 / K_2. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties Of Fluids

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