(Consider the density of water = $1000 \text{ kg m}^{-3}$, $1 \text{ atm} = 1 \times 10^5 \text{ Pa}$ and gravitational acceleration $g = 10 \text{ m/s}^2$)
Submarine Depth Calculation: Pressure Analysis
The problem asks for the maximum depth a submarine can reach based on its tolerance to absolute pressure.
Key Information Provided
- Maximum Absolute Pressure ($P_{abs\_max}$) the submarine can withstand: $100 \text{ atm}$
- Density of water ($\rho$): $1000 \text{ kg m}^{-3}$
- Atmospheric pressure conversion: $1 \text{ atm} = 1 \times 10^5 \text{ Pa}$
- Gravitational acceleration ($g$): $10 \text{ m/s}^2$
- Assume standard atmospheric pressure at the surface ($P_{atm}$): $1 \text{ atm} = 1 \times 10^5 \text{ Pa}$
Calculating Maximum Depth
The absolute pressure ($P_{abs}$) at a certain depth ($h$) in water is given by the formula:
$P_{abs} = P_{atm} + \rho g h$First, convert the maximum withstandable pressure to Pascals:
$P_{abs\_max} = 100 \text{ atm} \times (1 \times 10^5 \text{ Pa / atm}) = 1 \times 10^7 \text{ Pa}$Now, set the absolute pressure equal to the maximum withstandable pressure and solve for depth ($h$):
$1 \times 10^7 \text{ Pa} = (1 \times 10^5 \text{ Pa}) + (1000 \text{ kg m}^{-3}) \times (10 \text{ m/s}^2) \times h$Subtract the atmospheric pressure from the maximum absolute pressure:
$P_{water} = P_{abs\_max} - P_{atm}$ $P_{water} = (1 \times 10^7 \text{ Pa}) - (1 \times 10^5 \text{ Pa})$ $P_{water} = 10000000 \text{ Pa} - 100000 \text{ Pa} = 9900000 \text{ Pa}$This remaining pressure is due to the water column. Use the formula $P_{water} = \rho g h$ to find $h$:
$9900000 \text{ Pa} = (1000 \text{ kg m}^{-3}) \times (10 \text{ m/s}^2) \times h$ $9900000 = 10000 \times h$Solve for $h$:
$h = \frac{9900000}{10000}$ $h = 990 \text{ m}$Therefore, the submarine can go $990 \text{ m}$ deep.
Similar Questions
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 :- In the measurement of viscosity of liquids using terminal velocity experiment, spherical balls of same radius but having different densities are used. The variation of the terminal velocity ($v$) with the ratio of density of spherical ball ($\sigma$) to density of the liquid ($\rho$), is best represented by :
- The temperature of a metallic sphere of radius R is increased by a small amount $\Delta T$. If the linear coefficient of thermal expansion of the metal is $\alpha$, the approximate increase in the volume of the sphere is :
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 $= 1000\text{ kg m}^{-3}$]
Important Questions from Physical and Thermal Properties of Bulk Matter
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 $\pi = 3.14$)
- Two water drops of radius 'r' each coalesce to form a single large drop. If 'T' is the surface tension, then the surface energy released in this process is:
- Pressure of an ideal gas, contained in a closed vessel, is increased by $0.4\%$ when heated by $1^\circ C$. Its initial temperature must be:
A gas is kept in a container having walls which are thermally non-conducting. Initially the gas has a volume of $800 \ cm^3$ and temperature $27^{\circ}C$. The change in temperature when the gas is adiabatically compressed to $200 \ cm^3$ is:
(Take $\gamma = 1.5$; $\gamma$ is the ratio of specific heats at constant pressure and at constant volume)
