Q.Torricelli's barometer used mercury. Pascal duplicated it using French wine of density $984\ \text{kg m}^{-3}$. Determine the height of the wine column for normal atmospheric pressure.

Showing the 12 most recent of 21 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.An iceberg floats on sea water of density $\rho=1.03\times10^3\ kg\,m^{-3}$. The percentage of total volume of ice above the surface of water is about (Density of ice $\sigma=0.92\times10^3\ kg\,m^{-3}$) (A) 10.7% (B) 12.5% (C) 9.8% (D) 11.3%

   ›Reveal solutionSolution

   Floating equilibrium requires the weight of the iceberg to equal the buoyant force from the displaced seawater — this directly gives the submerged fraction as the density ratio, and the rest is the visible fraction above water. Answer: (A) 10.7%.

   Concept and Intuition

   When an object floats, it displaces just enough fluid so the buoyant force (weight of displaced fluid) equals its own weight. If $V$ is the iceberg's total volume and $V_{sub}$ the submerged part, then:

   $\sigma V g = \rho V_{sub}g \quad\Rightarrow\quad \frac{V_{sub}}{V}=\frac{\sigma}{\rho}.$

   Since ice is less dense than seawater, this ratio is less than 1 — most of an iceberg is submerged (the famous "tip of the iceberg" only shows the small fraction above water).

   Step-by-Step Solution

   1. Submerged fraction: $\frac{V_{sub}}{V}=\frac{\sigma}{\rho}=\frac{0.92\times10^3}{1.03\times10^3}=\frac{0.92}{1.03}=0.8932.$ …

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2. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.When a body floats in water and another liquid separately, it floats with one third of its volume outside water and $\dfrac{3}{4}$ of its volume outside another liquid. The density of the liquid is (A) $9.4\ \mathrm{g\,cc^{-1}}$ (B) $4\ \mathrm{g\,cc^{-1}}$ (C) $\dfrac{8}{3}\ \mathrm{g\,cc^{-1}}$ (D) $\dfrac{3}{8}\ \mathrm{g\,cc^{-1}}$

   ›Reveal solutionSolution

   This tests the law of flotation (weight of body = weight of displaced liquid). Using the fraction submerged in each liquid gives the body's density from water, then the unknown liquid's density from the second condition.

   Concept and Intuition

   When a body floats in equilibrium, its weight equals the weight of the liquid it displaces (Archimedes' principle applied to flotation): $\rho_b V g = \rho_{liquid} V_{sub}\, g$, i.e. $\rho_b V = \rho_{liquid} V_{sub}$. If a fraction $f$ of the volume is outside the liquid, then the submerged fraction is $(1-f)$, and

   $\rho_b = (1-f)\,\rho_{liquid}.$

   Applying this once with water (whose density we know, $1\ \text{g/cc}$) pins down $\rho_b$. Applying it a second time with the unknown liquid, now that $\rho_b$ is known, gives the unknown liquid's density.

   Step-by-Step Solution

   1. In water: fraction outside $=1/3$, so submerged fraction $=2/3$.

   $\rho_b = \frac{2}{3}\times \rho_{water} = \frac{2}{3}\times 1 = \frac{2}{3}\ \text{g/cc}.$

   2. In the other liquid: fraction outside $=3/4$, so submerged fraction $=1/4$.

   $\rho_b = \frac{1}{4}\times \rho_{L}.$

   3. Equate the two expressions for $\rho_b$ (same body, same mass/volume): …

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3. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.A vessel contains oil of density $0.8\ gcm^{-3}$ over mercury of density $13.6\ gcm^{-3}$. A homogenous sphere floats with half of its volume immersed in mercury and the other half in oil. Then the density of the material of the sphere is (A) $3.3\ gcm^{-3}$ (B) $6.4\ gcm^{-3}$ (C) $7.2\ gcm^{-3}$ (D) $12.8\ gcm^{-3}$

   ›Reveal solutionSolution

   Tests Archimedes' principle for a body floating at the interface of two immiscible liquids. Each half of the sphere's volume displaces one liquid, and equilibrium gives $\rho_{sphere} = 7.2\ g/cm^3$, the simple average of the two liquid densities (since the two immersed volumes are equal).

   Concept and Intuition

   When a solid floats in equilibrium (whether fully or partly submerged, or split between two liquids), Newton's first law for the floating body requires the net upward buoyant force to exactly balance its weight. Here the sphere straddles the oil-mercury interface with exactly half its volume in each liquid. Each half displaces its own liquid and contributes its own buoyant force — we simply add the two buoyant contributions and set the sum equal to the total weight of the sphere.

   Step-by-Step Solution

   1. Let the sphere have total volume $V$ and density $\rho_s$. Its weight is $\rho_s V g$.
   2. Half the volume ($V/2$) is immersed in oil (density $0.8\ g/cm^3$), contributing buoyant force $\rho_{oil}\cdot\frac{V}{2}\cdot g$.
   3. The other half ($V/2$) is immersed in mercury (density $13.6\ g/cm^3$), contributing buoyant force $\rho_{Hg}\cdot\frac{V}{2}\cdot g$.
   4. Equilibrium (floating, net force zero): …

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4. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.In a hydraulic lift, if the radius of the smaller piston is 5 cm, and the radius of the larger piston is 50 cm, then the weight that the larger piston can support when a force of 250 N is applied to the smaller piston is (A) 50 kN (B) 100 kN (C) 40 kN (D) 25 kN

   ›Reveal solutionSolution

   Hydraulic lift multiplies force by the ratio of piston areas (radius squared); the larger piston supports 25 kN.

   Concept and Intuition

   Pascal's law says pressure applied to an enclosed incompressible fluid is transmitted undiminished in all directions. Since pressure $=F/A$ must be equal on both pistons, a small force on a small piston produces a large force on a large piston — the mechanical advantage is the ratio of the piston areas, i.e. the square of the radius ratio.

   Step-by-Step Solution

   1. Pascal's law: $\dfrac{F_1}{A_1}=\dfrac{F_2}{A_2}$, with $A\propto r^2$.
   2. $F_2 = F_1\left(\dfrac{r_2}{r_1}\right)^2 = 250\times\left(\dfrac{50}{5}\right)^2$. …

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5. AP EAPCET 2024Set ap-2024-05-17-FN1 markMCQ
   Q.A U-tube is partially filled with water. Oil which does not mix with water is next poured into one-side of U-tube until entire water rises by 25 cm on the other side. If the density of oil is $0.8\ \mathrm{g\,cm^{-3}}$, the oil level will stand higher than the water level by (A) 6.25 cm (B) 12.50 cm (C) 31.75 cm (D) 63.50 cm

   ›Reveal solutionSolution

   This tests pressure balance in a U-tube with two immiscible liquids; the answer is (B) 12.50 cm.

   Concept and Intuition

   When oil is poured into one arm of a U-tube containing water, it cannot mix, so it floats as a layer on top of the water in that arm and pushes the water down. Because the tube has the same cross-section throughout, the volume of water pushed down in the oil arm must reappear as a rise in the plain-water arm — so the drop of the oil–water interface equals the rise of the water surface in the other arm. Once the system is in equilibrium, the pressure at any given horizontal level, measured through connected fluid, must be equal on both arms. This is the standard "connected liquids" pressure-balance idea.

   Step-by-Step Solution

   1. Let the original common water level (before oil was added) be the datum, height $0$.
   2. Water rises by $h=25$ cm on the plain-water arm ⇒ its free surface is at $+h=+25$ cm.
   3. By volume conservation (equal areas), the oil–water interface in the oil arm falls by the same amount, to $-h=-25$ cm.
   4. Take the horizontal level of this interface ($-h$) as the reference for pressure balance.
      • Via the plain-water arm: the pressure at this level equals atmospheric pressure plus the pressure of the water column from $-h$ up to the free surface at $+h$, i.e. a column of height $2h=50$ cm of water: $P=P_{atm}+\rho_{water}g(2h)$.
      • Via the oil arm: the pressure at the same level equals atmospheric plus the oil column above it, of height $H$ (unknown, up to the oil's free surface): $P=P_{atm}+\rho_{oil}gH$. …

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6. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.Two cylindrical vessels A and B of different areas of cross-section kept on same horizontal plane are filled with water to the same height. If the volume of water in vessel A is 3 times the volume of water in vessel B, then the ratio of the pressures at the bottom of the vessels A and B is (A) 1 : 1 (B) 1 : 3 (C) 1 : 9 (D) 1 : 6

   ›Reveal solutionSolution

   This tests whether you know that hydrostatic pressure depends only on depth, not on the shape/volume of the container.

   Concept and Intuition

   The pressure at the bottom of a liquid column comes purely from the weight of the liquid sitting directly above a unit area, which is captured by $P = \rho g h$. This formula has no term for cross-sectional area or total volume — it is the classic "hydrostatic paradox": a wide short vessel and a narrow tall vessel filled to the same height exert the same pressure at the bottom, even though they hold very different volumes of liquid, because the extra volume in the wide vessel is exactly compensated by the extra area over which that weight is spread.

   Step-by-Step Solution

   1. Let vessel A have base area $A_1$ and vessel B have base area $A_2$, both filled to the same height $h$.
   2. Volume in A: $V_A = A_1 h$. Volume in B: $V_B = A_2 h$. Given $V_A = 3V_B \Rightarrow A_1 = 3A_2$ — this tells us about the areas, not the pressures. …

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7. AP EAPCET 2023Set ap-2023-05-22-FN1 markMCQ
   Q.A liquid of density $800\ \mathrm{kgm^{-3}}$ is used instead of mercury, in the Torricelli barometer. The height of the liquid column is (Density of mercury $= 13.6\times10^{3}\ \mathrm{kgm^{-3}}$ and atmospheric pressure = 76 cm of Hg) (A) $13.2\ \mathrm{m}$ (B) $12.9\ \mathrm{m}$ (C) $13.6\ \mathrm{m}$ (D) $800\ \mathrm{m}$

   ›Reveal solutionSolution

   A Torricelli barometer works because the atmospheric pressure balances the weight of a liquid column with a vacuum above it; replacing mercury with a lighter liquid makes the column proportionally taller.

   Concept and Intuition

   In a barometer, the space above the liquid in the closed tube is (nearly) vacuum, so the pressure at the base of the column, $\rho g h$, must equal the atmospheric pressure pressing on the open reservoir. This balance condition is independent of which liquid is used — only $\rho h$ must stay the same (since $g$ and $P_{atm}$ are fixed). A less dense liquid therefore needs a taller column to produce the same pressure.

   Step-by-Step Solution

   1. For mercury: $P_{atm}=\rho_{Hg}\,g\,h_{Hg}$, with $h_{Hg}=76\ \text{cm}=0.76\ \text m$ and $\rho_{Hg}=13.6\times10^{3}\ \mathrm{kg\,m^{-3}}$.
   2. For the new liquid: $P_{atm}=\rho_{liq}\,g\,h_{liq}$, with $\rho_{liq}=800\ \mathrm{kg\,m^{-3}}$. …

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8. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.The gauge pressure at a depth of 50 m in a sea is (Density of sea water is 1025 kg m$^{-3}$ and g = 10 ms$^{-2}$) (A) 1025 Pa (B) 512500 Pa (C) 20000 Pa (D) 15000 Pa

   ›Reveal solutionSolution

   Direct substitution into $P=\rho g h$ gives the gauge pressure at the given depth: 512500 Pa.

   Concept and Intuition

   Gauge pressure (pressure in excess of atmospheric) at a depth $h$ in a fluid of density $\rho$ is $P=\rho g h$, arising from the weight of the fluid column above that point.

   Step-by-Step Solution

   1. Given: $\rho = 1025\,\text{kg m}^{-3}$, $g=10\,\text{ms}^{-2}$, $h=50\,\text{m}$.
   2. $P = \rho g h = 1025\times10\times50$.
   3. Compute: $1025\times10 = 10250$; $10250\times50 = 512500\,\text{Pa}$.

   Common Mistakes …

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9. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
   Q.A tank of oil has height of 4 m and density of $850\ \mathrm{kg\,m}^{-3}$. The gauge pressure at the bottom of the tank is (1 atm = $10^5$ Pa, Acceleration due to gravity = $10\ \mathrm{ms}^{-2}$) (A) 34 kPa (B) 384 kPa (C) 284 kPa (D) 200 kPa

   ›Reveal solutionSolution

   This tests the direct hydrostatic pressure formula $P=\rho gh$, giving gauge pressure (pressure in excess of atmospheric) at the base of a liquid column.

   Concept and Intuition

   Gauge pressure is the pressure due to the fluid column alone, excluding atmospheric pressure (which acts equally on any exposed surface and is usually already "zeroed out" by the measuring instrument). For a fluid of density $\rho$ at depth $h$, this is simply $\rho gh$ — it does not depend on the shape or cross-section of the tank, only on the vertical height of fluid above the point.

   Step-by-Step Solution

   1. Given: height of oil column $h=4\ \mathrm m$, density $\rho=850\ \mathrm{kg\,m^{-3}}$, $g=10\ \mathrm{ms^{-2}}$.
   2. Gauge pressure at the bottom: $P_{gauge}=\rho gh=850\times10\times4$.
   3. $850\times10=8500$; $8500\times4=34{,}000\ \mathrm{Pa}$.
   4. Convert to kPa: $34{,}000\ \mathrm{Pa}=34\ \mathrm{kPa}$. …

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10. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
    Q.A swimming pool has depth of 3 meters. The pressure at the bottom of the pool due to water alone is (Density of water is $1000\ \mathrm{kg\,m^{-3}}$, Acceleration due to gravity = $10\ \mathrm{ms^{-2}}$) (A) $10^4\ \mathrm{Pa}$ (B) $3\times10^3\ \mathrm{Pa}$ (C) $29\times10^3\ \mathrm{Pa}$ (D) $30\times10^3\ \mathrm{Pa}$

    ›Reveal solutionSolution

    The hydrostatic pressure due to water alone at depth $h$ is $\rho gh$, giving $30\times10^3$ Pa.

    Concept and Intuition

    The pressure at a depth $h$ in a static fluid (due to the fluid column alone, ignoring atmospheric pressure which the question explicitly excludes by saying "due to water alone") is $P=\rho g h$.

    Step-by-Step Solution

    1. $\rho = 1000\ \mathrm{kg\,m^{-3}}$, $g=10\ \mathrm{ms^{-2}}$, $h=3\ \mathrm{m}$. …

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11. AP EAPCET 2022Set ap-2022-07-11-FN1 markMCQ
    Q.A container is filled to a height of 20 cm with water. A 30 cm thick layer of oil with specific gravity 0.8 floats on the top of water. If the density of water is $1000\ \mathrm{kg\,m^{-3}}$ and atmospheric pressure is $1\times10^5$ Pa, then the total pressure at the bottom of the container is (Acceleration due to gravity $=10\ \mathrm{m\,s^{-2}}$) (A) $1.044\times10^5$ Pa (B) $1.24\times10^5$ Pa (C) $1.062\times10^5$ Pa (D) $1.15\times10^5$ Pa

    ›Reveal solutionSolution

    Total pressure at the bottom is the atmospheric pressure plus the weight-per-area contributions of each fluid layer stacked above: $P = P_{atm} + \rho_{oil} g h_{oil} + \rho_{water} g h_{water} = 1.044\times10^5\ \mathrm{Pa}$.

    Concept and Intuition

    Pressure in a fluid at rest increases with depth due to the weight of fluid above. When layers of different fluids are stacked, the total pressure at the bottom is simply the sum of the atmospheric pressure at the top surface plus each layer's own $\rho g h$ contribution (heavier/lighter fluids each contribute according to their own density and thickness).

    Step-by-Step Solution

    1. Oil layer: density $= 0.8 \times 1000 = 800\ \mathrm{kg\,m^{-3}}$, thickness $h_{oil}=0.30$ m. Contribution: $\rho_{oil} g h_{oil} = 800 \times 10 \times 0.30 = 2400\ \mathrm{Pa}$.
    2. Water layer: density $=1000\ \mathrm{kg\,m^{-3}}$, thickness $h_{water}=0.20$ m. …

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12. AP EAPCET 2022Set ap-2022-07-12-FN1 markMCQ
    Q.An object hangs from a spring balance. The balance shows 40 N when the object is in air, 30 N when the object is immersed in water, and 34 N when the object is immersed in another liquid. If the density of water is $1000\ \mathrm{kg\,m^{-3}}$, then the density of liquid is (Acceleration due to gravity = $10\ \mathrm{m\,s^{-2}}$) (A) $500\ \mathrm{kg\,m^{-3}}$ (B) $800\ \mathrm{kg\,m^{-3}}$ (C) $750\ \mathrm{kg\,m^{-3}}$ (D) $600\ \mathrm{kg\,m^{-3}}$

    ›Reveal solutionSolution

    The loss of apparent weight in each fluid equals the buoyant force $\rho_{fluid}Vg$; comparing the two losses (with the same $V$ and $g$) gives the unknown liquid's density directly. Answer: $600\ \mathrm{kg\,m^{-3}}$.

    Concept and Intuition

    Archimedes' principle says the apparent weight loss when submerged equals the weight of fluid displaced, $\rho_{fluid} V g$, where $V$ is the object's volume (same in both fluids). Taking the ratio of weight losses in water and in the unknown liquid eliminates the unknown $V$, leaving a simple density ratio.

    Step-by-Step Solution

    1. Weight in air: $40\,$N (true weight of the object).
    2. Weight in water: $30\,$N, so buoyant force in water $= 40-30=10\,$N $=\rho_w V g$.
    3. Weight in liquid: $34\,$N, so buoyant force in liquid $= 40-34=6\,$N $=\rho_L V g$. …

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