The question asks for the SI unit of pressure. To answer this, we need to recall the definition of pressure and its standard unit in the International System of Units (SI).
\[ P = \frac{F}{A} \] where \(P\) is pressure, \(F\) is force, and \(A\) is area.
Correct Option: A) Pascal is the SI unit of pressure.
The question asks to identify the instrument used for measuring atmospheric pressure. We need to recall the functions of the instruments listed in the options.
Correct Option: B) A barometer is an instrument used to measure atmospheric pressure. It works by balancing the weight of a column of mercury or by using a sealed metal chamber that expands and contracts with changes in pressure.
The question asks about the relationship between fluid pressure and depth. This is a fundamental concept in fluid mechanics, specifically hydrostatics. We need to recall the formula for pressure at a certain depth in a fluid and analyze its components.
A) Increases linearly with depth. As derived from the formula \(P = \rho g h\), pressure is directly proportional to depth, indicating a linear increase.
This question asks about the fundamental principle that governs why objects, specifically ships, float in water. We need to identify the physical law that describes buoyancy and displacement.
Correct Option: B) Archimedes' principle
To calculate the pressure exerted by a column of liquid, we use the formula \(P = \rho g h\), where \(P\) is the pressure, \(\rho\) is the density of the liquid, \(g\) is the acceleration due to gravity, and \(h\) is the height of the liquid column. It is crucial to ensure all units are consistent (SI units in this case) before performing the calculation.
The formula for pressure due to a liquid column is:
\[P = \rho g h\]The calculated value \(101,292.8 \text{ Pa}\) is closest to \(101,920 \text{ Pa}\). The slight difference might be due to rounding \(g\) to 9.81 m/s² or 10 m/s² in some contexts, or the option itself being an approximation. If we use \(g = 10 \text{ m/s}^2\) for a quick check, \(P = 13600 \times 10 \times 0.76 = 103360 \text{ Pa}\). If we use \(g = 9.81 \text{ m/s}^2\), \(P = 13600 \times 9.81 \times 0.76 = 101392.96 \text{ Pa}\). Option A is the closest.
Let's re-evaluate the question's given \(g = 98 \text{ m/s}^2\). If we strictly use \(g = 98 \text{ m/s}^2\), then:
\[P = 13,600 \times 98 \times 0.76\] \[P = 1,012,928 \text{ Pa}\]This value is not among the options. Therefore, it is highly probable that \(g\) was intended to be \(9.8 \text{ m/s}^2\).
Using \(g = 9.8 \text{ m/s}^2\), our result \(101,292.8 \text{ Pa}\) is very close to \(101,920 \text{ Pa}\). This discrepancy is common in competitive exams where options might be rounded or a slightly different value of \(g\) (like \(9.81 \text{ m/s}^2\)) is implicitly used. Given the options, \(101,920 \text{ Pa}\) is the most plausible answer.
A) \(101,920 \text{ Pa}\) is the closest value to the calculated pressure of \(101,292.8 \text{ Pa}\) when using \(g = 9.8 \text{ m/s}^2\). This is also approximately the standard atmospheric pressure, which is often measured by a 76 cm mercury column.
Atmospheric pressure is the force exerted by the weight of the air column above a given point. To understand why it changes with altitude, we need to consider how the amount and density of air change as one moves higher in the atmosphere.
Correct Option: B) Due to a decrease in air density
As altitude increases, the air column above a given point becomes shorter, and the air molecules are spread out over a larger volume due to reduced gravitational compression. This leads to a significant decrease in the number of air molecules per unit volume, meaning the air density decreases. Since atmospheric pressure is the weight of the air column, a decrease in air density directly results in a decrease in atmospheric pressure.