The diagram shows a stationary sphere that is just fully submerged in a liquid. The radius of the sphere is \(r\) and the density of the liquid is \(\rho\). The acceleration due to free fall is \(g\).
The air exerts pressure \(P_A\) on the surface of the liquid.
What is the pressure at the lowest point of the sphere?
Correct Answer: \( \boxed{\mathrm{C}} \)
Pressure in a liquid is given by \(P=P_A+\rho gh\).
Since the sphere is just fully submerged, its highest point is at the liquid surface, so the lowest point is at a depth of \(2r\).
Hence, \(P=P_A+\rho g(2r)=P_A+2r\rho g\).
Therefore, the correct answer is (C).
On Earth, a solid object that is fully submerged in a liquid experiences an upthrust \(U_{\mathrm{E}}\).
On Mars, the same object, fully submerged in the same liquid, experiences an upthrust \(U_{\mathrm{M}}\).
The acceleration of free fall on Mars is \(3.7\,\mathrm{m\,s^{-2}}\).
Assume that the liquid’s density and the object’s volume have the same values on Earth and Mars.
What is the ratio \( \dfrac{U_{\mathrm{M}}}{U_{\mathrm{E}}} \)?
Correct Answer: \( \boxed{\mathrm{A}} \)
Upthrust is given by \(U=\rho Vg\).
Since the liquid density and displaced volume are unchanged,
\(\dfrac{U_{\mathrm{M}}}{U_{\mathrm{E}}}=\dfrac{g_{\mathrm{M}}}{g_{\mathrm{E}}}=\dfrac{3.7}{9.8}\approx0.38\).
Therefore, the correct answer is (A).
A uniform cylinder of weight \(25.0\,\mathrm{N}\) is suspended from a newton meter.
The cylinder is fully submerged in water, as shown. The reading on the newton meter is \(10.0\,\mathrm{N}\).
The water is replaced by a liquid with a density \(10\%\) greater than the density of water. The cylinder remains fully submerged.
What is the new reading on the newton meter?
Correct Answer: \( \boxed{\mathrm{B}} \)
The apparent weight is
\( W_{\mathrm{apparent}}=W-F_{\mathrm{b}} \)
Initially, the buoyant force is
\( F_{\mathrm{b}}=25.0-10.0=15.0\,\mathrm{N} \)
Since buoyant force is proportional to the liquid density, increasing the density by \(10\%\) increases the buoyant force to
\( F_{\mathrm{b,new}}=1.10\times15.0=16.5\,\mathrm{N} \)
The new newton meter reading is
\( W_{\mathrm{apparent}}=25.0-16.5=8.5\,\mathrm{N} \)
Therefore, the correct answer is (A).
