The excess pressure inside a soap bubble A in air is half the excess pressure inside another soap bubble B in air. If the volume of the bubble A is n times the volume of the bubble B, then the value of n is ________.

Numerical Answer Type:
Enter a numerical value Answer: 8 to 8 +4 marks

Solution & Explanation

### Related Formula Excess pressure inside a soap bubble in air (which has two liquid-gas interfaces) is given by: Delta P = frac4TR Rightarrow R propto frac1Delta P The volume V of a spherical bubble of radius R is: V = frac43pi R^3 Rightarrow V propto R^3 propto left(frac1Delta Pright)^3 ### Core Logic Given state: Delta P_A = frac12 Delta P_B Rightarrow fracDelta P_BDelta P_A = 2 ### Step 1: Calculate the ratio of radii fracR_AR_B = fracDelta P_BDelta P_A = 2 ### Step 2: Calculate volume scaling parameter (n) V_A = n V_B Rightarrow n = fracV_AV_B = left(fracR_AR_Bright)^3 = (2)^3 = 8 ### Pattern Recognition Volume scale factors depend on the cube of linear scale factors (V propto R^3). Since radius is inversely proportional to excess pressure, a halving of excess pressure leads to doubling of radius, scaling volume by 2^3 = 8. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Fluids

Reference Study Guides

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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