Latihan Soal UTBK SBMPTN Eps. 27 • TKA Fisika • Fluida Statis

Summary

The video works through three UTBK-style problems on static fluids, applying Archimedes’ principle, the density relation for floating objects, and pressure in a liquid beneath a loaded piston.

1. Comparing two objects in water

Two objects have equal mass. Object A is suspended in the water, while object B floats at the surface. The presenter evaluates four statements:

• Object A is denser than object B — true. A suspended object has the same density as the water, while a floating object has a lower density.
• The objects experience the same buoyant force — true. Each object is in equilibrium, so its buoyant force equals its weight. Their masses, and therefore their weights, are equal.
• The buoyant force on B equals B’s weight — true. This is the equilibrium condition for a floating object.
• Buoyant force depends on the object’s density — false. By Archimedes’ principle, buoyant force depends on the fluid’s density, the volume of fluid displaced (equivalently, the submerged volume), and gravitational acceleration: [ F_b = \rho_{\text{fluid}} V_{\text{submerged}} g. ]

Answer: The first three statements are correct.

2. Submerged-to-exposed volume ratio

A plastic block floats in a liquid. The block’s density is (1.04\ \text{g/cm}^3), and the liquid’s density is (1.3\ \text{g/cm}^3).

For a floating object, the fraction of its volume that is submerged is:

[ \frac{V_{\text{submerged}}}{V_{\text{total}}} = \frac{\rho_{\text{object}}}{\rho_{\text{liquid}}} = \frac{1.04}{1.3} = 0.8 = \frac{4}{5}. ]

The exposed fraction is the remaining (\frac{1}{5}) of the total volume. Therefore:

[ V_{\text{submerged}} : V_{\text{exposed}} = 4 : 1. ]

Answer: (4:1).

3. Mass of a piston pressing on a liquid

A vessel is closed by a piston of mass (M) and area (A = 1.0\ \text{cm}^2 = 10^{-4}\ \text{m}^2). A (0.4\ \text{kg}) load sits on the piston. The liquid has a density of (1000\ \text{kg/m}^3), atmospheric pressure is (100{,}000\ \text{Pa}), and the pressure is (151{,}000\ \text{Pa}) at a depth of (0.10\ \text{m}).

The pressure at that depth is the sum of atmospheric pressure, hydrostatic pressure, and the pressure exerted by the piston and load:

[ P = p_0 + \rho g h + \frac{(M + m)g}{A}. ]

At (h = 0.10\ \text{m}), the hydrostatic contribution is (1000\ \text{Pa}). Subtracting the atmospheric and hydrostatic pressures from the stated pressure leaves (50{,}000\ \text{Pa}) due to the piston and load. This corresponds to a combined mass of (0.5\ \text{kg}). Since the load is (0.4\ \text{kg}):

[ M = 0.5 - 0.4 = 0.1\ \text{kg}. ]

The pressure stated at (0.30\ \text{m}) gives the same result: the additional depth adds (2000\ \text{Pa}), matching the increase from (151{,}000) to (153{,}000\ \text{Pa}).

Answer: The piston’s mass is (0.1\ \text{kg}).

Speakers and sources featured

• Speaker: An unnamed presenter from the Jendela Sains channel.
• Other speakers or sources: None identified in the subtitles.