(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
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
The strain-stress plot for materials $A, B, C$ and $D$ is shown in the figure. Which material has the largest Young's modulus ?
- Eight mercury drops, each of radius $r$, coalesce to form a bigger drop. The surface energy released in this process is ___________. ($\text{S}$ is the surface tension of mercury).
- An ideal gas at pressure $P$ and temperature $T$ is expanding such that $P T^{3} = \text{constant}$. The coefficient of volume expansion of the gas is ___________.
- The surface tension of a soap solution is $3.5 \times 10^{-2}~\text{N/m}$. The work required to increase the radius of a soap bubble from $1~\text{cm}$ to $2~\text{cm}$ is $\alpha \times 10^{-6}~\text{J}$. The value of $\alpha$ is________.
($\pi = 22/7$) - The temperature of a metal strip having coefficient of linear expansion $\alpha$ is increased from $T_1$ to $T_2$ resulting in increase of its length by $\Delta L_1$. The temperature is further increased from $T_2$ to $T_3$ such that the increase in its length is $\Delta L_2$.
Given $T_3 + T_1 = 2T_2$ and $T_2 - T_1 = \Delta T$, the value of $\Delta L_2$ is _______.
