Bennett Mechanical Comprehension Practice Test 2025 - Free Practice Questions and Review Guide

Question: 1 / 400

If a liquid becomes twice denser and the depth of an immersed object halves, what happens to the liquid's pressure?

Quadruples

To understand how the pressure in the liquid changes when its density doubles and the depth of an immersed object halves, it's essential to grasp the relationship between pressure, depth, and density in a fluid.

Pressure in a liquid is calculated using 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 depth of the liquid.

When the liquid becomes twice as dense, the new density can be represented as \( 2\rho \). If the depth of the object is halved, the new depth is \( \frac{h}{2} \).

Now substituting these new values into the pressure formula gives us:

\[ P' = (2\rho) g \left(\frac{h}{2}\right) = \;\rho g h \]

From this, we see that the new pressure \( P' \) is equal to the original pressure \( P \) because the effects of doubling the density and halving the depth cancel each other out.

Thus, the pressure does not change, and the correct understanding is that the pressure remains constant

Get further explanation with Examzify DeepDiveBeta

Reduces by a factor of 4

Doesn't change

Doubles

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