Q.A wooden block with a coin placed on its top floats in water in a container. The vertical distance $l$ is the depth of the block that lies submerged below the water surface, and $h$ is the height of the water level in the container (the height of the water column, measured from the bottom of the container up to the free water surface). After some time the coin slides off the block and falls into the water, sinking to the bottom of the container (the coin is denser than water). Considering the block's submerged depth $l$ and the water level $h$ afterwards, which of the following statements are correct? (More than one option may be correct.)

Showing the 12 most recent of 18 on this concept.

1. MHT-CET 2026Set pcm-2026-04-18-M1 markMCQ
   Q.A piston of cross-sectional area 2.5×10−22 is used in a hydraulic lift to exert a force of 250 on water. The cross-sectional area of the other piston which supports a car of mass 3000 is (𝑔=9.82) (A) 1.962 (B) 2.942 (C) 3.922 (D) 5.882

   ›Reveal solutionSolution

   The second piston's area is about 2.94×10⁻² m².

   By Pascal's law, pressure is transmitted equally: F1/A1 = F2/A2, so A2 = A1·(F2/F1).

   The car's weight is F2 = mg = 3000 kg × 9.8 m/s² = 29400 N. …

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2. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.The maximum length of water column that can stay without falling in a vertically held capillary tube of diameter 1 mm and open at both the ends is (Acceleration due to gravity $= 10~\mathrm{ms}^{-2}$ and surface tension of water $= 0.07~\mathrm{Nm}^{-1}$) (A) 1.4 cm (B) 2.8 cm (C) 5.6 cm (D) 0 cm

   ›Reveal solutionSolution

   The water column is held by surface tension at both open ends, so the maximum length is twice the rise in a single capillary — giving 5.6 cm.

   The key idea is that a capillary tube open at both ends can support a water column because surface tension acts at both the top and bottom menisci. Each meniscus exerts an upward force equal to the surface tension times the inner circumference. For the column not to fall, the total upward force from both menisci must balance the weight of the water column.

   Let’s work through it step by step.

   1. Force from a single meniscus In a capillary tube, the surface tension force acts along the inner circumference. For a tube of radius $r$, the force from one meniscus is

   $F_{\text{one}} = \sigma \times (2\pi r)$

   where $\sigma = 0.07~\mathrm{N/m}$ is the surface tension. Here the diameter is $1~\mathrm{mm}$, so radius $r = 0.5~\mathrm{mm} = 5 \times 10^{-4}~\mathrm{m}$.

   2. Total upward force Since the tube is open at both ends, water forms a concave meniscus at both the top and bottom. Both pull upward (the bottom meniscus also pulls up because the water is trying to climb the inner wall). So the total upward force is

   $F_{\text{up}} = 2 \times \sigma \times 2\pi r = 4\pi r \sigma$

   3. Weight of the water column The water column has height $h$ (the maximum length we want), cross-sectional area $\pi r^2$, density $\rho = 1000~\mathrm{kg/m^3}$, and $g = 10~\mathrm{m/s^2}$. Its weight is

   $W = \rho g \times (\pi r^2 h)$

   4. Balance condition For the column to just stay without falling, the upward force equals the weight:

   $4\pi r \sigma = \rho g \pi r^2 h$

   Cancel $\pi r$ from both sides:

   $4\sigma = \rho g r h$

   5. Solve for $h$ …

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3. COMEDK 2025Set 2025-M1 markMCQ
   Q.A wooden block floats with $\frac{3}{5}$ of its volume submerged in a tank of water. If a denser liquid is poured into the tank, the wooden block floats with half its volume in the liquid and the remaining half in water. The relative density of the liquid is: (A) $\frac{1}{2}$ (B) $\frac{1}{3}$ (C) $\frac{1}{6}$ (D) $\frac{1}{5}$

   ›Reveal solutionSolution

   Floating $\tfrac{3}{5}$ submerged gives block density $0.6$; floating half-in-liquid, half-in-water gives $0.6=\tfrac12(1+\rho_l)$, so $\rho_l=\tfrac15$. The correct option is (D).

   By Archimedes' principle a floating body's weight equals the weight of fluid it displaces. Taking water's relative density as $1$:

   1. Water only. With $\tfrac{3}{5}$ of the volume submerged,

   $\rho_b\,V=\rho_w\left(\tfrac{3}{5}V\right)\;\Longrightarrow\;\rho_b=\tfrac{3}{5}=0.6.$

   2. Two-liquid float. Half the volume displaces water, half displaces the added liquid $\rho_l$:

   $\rho_b\,V=\rho_w\left(\tfrac12V\right)+\rho_l\left(\tfrac12V\right)\;\Longrightarrow\;\rho_b=\tfrac12\left(\rho_w+\rho_l\right).$

   3. Solve for $\rho_l$ with $\rho_b=0.6,\ \rho_w=1$: …

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4. MHT-CET 2025Set pcm-2025-04-19-E1 markMCQ
   Q.A small metal sphere of density $\rho$ is dropped from height $h$ into a jar containing liquid of density $\sigma(\sigma>\rho)$. The maximum depth up to which the sphere sinks is (Neglect damping forces) (A) $\frac{\rho}{\rho-\sigma}$ (B) $\frac{\mathrm{h} \sigma}{(\rho-\sigma)}$ (C) $\frac{\sigma}{(\rho-\sigma)}$ (D) $\frac{\mathrm{h} \rho}{(\rho-\sigma)}$

   ›Reveal solutionSolution

   Depth $=\dfrac{h\rho}{\sigma-\rho}$ (option D form).

   The sphere enters the liquid with speed $v=\sqrt{2gh}$. Inside the liquid ($\sigma>\rho$) the net upward force is $(\sigma-\rho)Vg$, giving deceleration

   $a=\frac{(\sigma-\rho)Vg}{\rho V}=\frac{(\sigma-\rho)g}{\rho}.$

   Using $v^2=2ad$ at maximum depth (final speed zero): …

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5. MHT-CET 2025Set pcm-2025-04-23-E1 markMCQ
   Q.A solid cylinder of length $l$ and cross-sectional area $\frac{a}{5}$ is immersed such that it floats with its axis vertical at the liquid-liquid interface with length $l / 4$ in the denser liquid as shown in figure. The lower density liquid ( $\rho$ ) is open to atmosphere having pressure $\mathrm{P}_0$. The density d of solid cylinder is (A) $\frac{1}{2} \rho$ (B) $\frac{3}{2} \rho$ (C) $\frac{3}{4} p$ (D) $\rho$

   ›Reveal solutionSolution

   Balancing the cylinder's weight against the buoyancy from both liquids (upper $\rho$ over $3l/4$, lower $3\rho$ over $l/4$) gives $d=\dfrac{3}{2}\rho$.

   Let the cylinder (length $l$, cross-section $A=\tfrac{a}{5}$) float with $\tfrac{l}{4}$ in the denser lower liquid and the remaining $\tfrac{3l}{4}$ in the upper liquid of density $\rho$. From the figure the denser lower liquid has density $3\rho$.

   Equilibrium — weight equals total buoyant force:

   $d\,A\,l\,g=\rho\,A\left(\frac{3l}{4}\right)g+3\rho\,A\left(\frac{l}{4}\right)g.$

   Cancel $A\,g$ and divide by $l$: …

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6. TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
   Q.A uniform narrow tube of length one metre with one end closed contains 25 cm long mercury thread, which traps a column of air at the closed end. When the tube is held vertically with the open end up, the length of the air column near the closed end is 21 cm and when the tube is held horizontally, the length of the air column near the closed end is 'L'. If the atmospheric pressure is equal to the pressure of 75 cm of mercury, then L is (A) 35 cm (B) 28 cm (C) 7 cm (D) 14 cm

   ›Reveal solutionSolution

   Using Boyle’s law for the trapped air, the horizontal length L is found by equating the product of pressure and volume in the vertical and horizontal cases. The result is L = 28 cm.

   Concept & Intuition

   The key is that the air trapped in the tube obeys Boyle’s law (constant temperature), so $P_1 V_1 = P_2 V_2$. The mercury thread shifts, changing the air column’s length and thus its volume. The pressure on the trapped air is the sum of atmospheric pressure and the pressure due to the mercury column (when vertical). When horizontal, the mercury exerts no extra pressure along the tube, so the air pressure equals atmospheric pressure alone. By comparing the two vertical orientations (open end up) and the horizontal orientation, we can solve for L.

   Step-by-step solution

   1. Define the setup

      • Tube length = 100 cm, closed at one end.
      • Mercury thread length = 25 cm (constant).
      • Atmospheric pressure = 75 cm of Hg.
      • When vertical with open end up: air column length = 21 cm.
      • When horizontal: air column length = L (unknown).

   2. Pressure on the trapped air when vertical (open end up)

      The open end is at the top, so the mercury thread is above the trapped air. The pressure at the top of the mercury column is atmospheric (75 cm Hg). The mercury column of height 25 cm adds its own weight, so the pressure at the bottom of the mercury (just above the trapped air) is:

   $P_{\text{vertical}} = 75 + 25 = 100 \text{ cm Hg}.$

   This is the pressure of the trapped air.

   3. Volume of trapped air when vertical

      The tube has uniform cross-section area A. The air column length is 21 cm, so volume $V_{\text{vertical}} = 21A$.

   4. Pressure on the trapped air when horizontal

      When the tube is horizontal, the mercury thread does not add any hydrostatic pressure along the tube’s length. The trapped air is only opposed by atmospheric pressure at the open end. Hence:

      $P_{\text{horizontal}} = 75 \text{ cm Hg}.$ …

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7. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.A test tube of mass 6 g and uniform area of cross section $10\ \mathrm{cm^2}$ is floating in water vertically when 10 g of mercury is in the bottom. The tube is depressed by a small amount and then released. The time period of oscillation is (Acceleration due to gravity $= 10\ \mathrm{ms^{-2}}$) (A) $0.75\ \mathrm{s}$ (B) $0.5\ \mathrm{s}$ (C) $0.25\ \mathrm{s}$ (D) $0.85\ \mathrm{s}$

   ›Reveal solutionSolution

   The floating test tube (with mercury as ballast) oscillates like a hydrometer with restoring constant $k=\rho g A$; plugging in the numbers gives $T \approx 0.25\ \text{s}$.

   Concept and Intuition

   When a floating body is pushed down by a small extra depth $x$, it displaces an extra volume $Ax$ of liquid, so it feels an extra upward buoyant force $\rho g A x$ — exactly like a spring pushing back proportional to displacement. This makes the tube perform SHM with an effective spring constant $k = \rho g A$, where $\rho$ is the liquid's density and $A$ is the tube's cross-sectional area.

   Step-by-Step Solution

   1. Total oscillating mass: $m = 6\ \text{g (tube)} + 10\ \text{g (mercury)} = 16\ \text{g} = 0.016\ \text{kg}$.
   2. Cross-section: $A = 10\ \text{cm}^2 = 10\times10^{-4}\ \text{m}^2 = 1\times10^{-3}\ \text{m}^2$.
   3. Effective "spring constant" from buoyancy: $k = \rho g A = 1000 \times 10 \times 10^{-3} = 10\ \text{N/m}$. …

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8. MHT-CET 2024Set pcm-2024-05-11-E1 markMCQ
   Q.A piece of wood has length, breadth and height, ' $a$ ', ' $b$ ' and ' $c$ ' respectively. Its relative density, is ' $d$ '. It is floating in water such that the side ' $a$ ' is vertical. It is pushed down a little and released. The time period of S.H.M. executed by it is ($\mathrm{g}=$ acceleration due to gravity) (A) $2 \pi \sqrt{\frac{a b c}{g}}$ (B) $2 \pi \sqrt{\frac{\mathrm{bc}}{\mathrm{dg}}}$ (C) $2 \pi \sqrt{\frac{g}{d a}}$ (D) $2 \pi \sqrt{\frac{\mathrm{ad}}{\mathrm{g}}}$

   ›Reveal solutionSolution

   $T=2\pi\sqrt{ad/g}$. …

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9. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
   Q.A 20 g copper block is suspended by a vertical spring causing 1 cm elongation over the natural length of spring. If a beaker of water is placed below the block so that the copper block is completely immersed in the liquid, the elongation of the spring is (Density of copper $9000\ \mathrm{kg\,m^{-3}}$, Density of water $1000\ \mathrm{kg\,m^{-3}}$, $g = 10\ \mathrm{ms^{-2}}$) (A) 0.25 cm (B) 0.15 cm (C) 0.78 cm (D) 0.89 cm

   ›Reveal solutionSolution

   This tests how buoyancy reduces the effective weight supported by a spring, hence reducing its elongation. The new elongation is about 0.89 cm.

   Concept and Intuition

   A spring's elongation is proportional to the net downward force it must support. In air, that force is simply $mg$. When the block is submerged, Archimedes' principle says the water pushes up with a buoyant force equal to the weight of displaced water, $\rho_w V g$. The spring now only needs to balance the reduced net weight $mg-\rho_wVg$, so its elongation shrinks proportionally.

   Step-by-Step Solution

   1. In air: $kx_1 = mg$, with $x_1=1$ cm.
   2. Volume of copper block: $V=\dfrac{m}{\rho_c}$.
   3. Buoyant force when submerged: $F_B=\rho_w V g = \rho_w\dfrac{m}{\rho_c}g$.
   4. New spring force balance: $kx_2 = mg-\rho_w\dfrac{m}{\rho_c}g = mg\left(1-\dfrac{\rho_w}{\rho_c}\right)$. …

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10. COMEDK 2023Set 2023-E1 markMCQ
    Q.A spherical metal ball of density '$\rho$' and radius '$r$' is immersed in a liquid of density '$\sigma$'. When an electric field is applied in the upward direction the metal ball remains just suspended in the liquid. Then the expression for the charge on the metal ball is : (A) $q=\frac{\left[\frac{4}{3} \pi r^3 \rho g\right]}{\sigma E}$ (B) $q=\frac{\left[4 \pi r^3(\rho-\sigma) g\right]}{3 E}$ (C) $q=\frac{\left[4 \pi r^2 \rho g\right]}{\sigma E}$ (D) $q=\frac{\left[4 \pi r^2(\rho-\sigma) g\right]}{E}$

    ›Reveal solutionSolution

    (Note the answer must contain the DENSITY DIFFERENCE (rho - sigma) - buoyancy cannot be ignored - and must go as r^3, which rules out (A), (C) and (D).)

    Concept: equilibrium of a charged ball in a liquid - the upward electric force plus the upthrust balance the weight.

    Forces on the ball (volume V = (4/3) pi r^3):

    • weight (down): V rho g = (4/3) pi r^3 rho g
    • buoyancy / upthrust (up): V sigma g = (4/3) pi r^3 sigma g
    • electric force (up, given): qE

    'Just suspended' => net force zero:

    qE + (4/3) pi r^3 sigma g = (4/3) pi r^3 rho g

    qE = (4/3) pi r^3 (rho - sigma) g …

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11. COMEDK 2023Set 2023-M1 markMCQ
    Q.A block of wood floats in water with $(4 / 5)$ th of its volume submerged. If the same block just floats in a liquid, the density of the liquid is (in $\mathrm{kgm}^{-3}$) (A) 1250 (B) 600 (C) 400 (D) 800

    ›Reveal solutionSolution

    The block density is $800\ \mathrm{kg/m^3}$; a liquid in which it just floats (fully submerged) must have the same density, $800\ \mathrm{kg/m^3}$.

    Floating in water: weight $=$ buoyant force of the submerged part,

    $\rho_{block}\,V\,g=\rho_w\left(\tfrac45 V\right)g\;\Rightarrow\;\rho_{block}=\tfrac45(1000)=800\ \mathrm{kg/m^3}.$ …

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12. MHT-CET 2023Set pcm-2023-05-11-M1 markMCQ
    Q.A body of density '$\rho$' is dropped from rest at a height '$h$' into a lake of density '$\sigma' (\sigma>\rho)$. The maximum depth to which the body sinks before returning to float on the surface is (neglect air dissipative forces) (A) $\frac{\mathrm{h} \rho}{(\sigma-\rho)}$ (B) $\frac{h \rho}{(\sigma+\rho)}$ (C) $\frac{\mathrm{h} \rho}{(\rho-\sigma)}$ (D) $\frac{2 \mathrm{~h} \rho}{(\sigma-\rho)}$

    ›Reveal solutionSolution

    Kinetic energy at surface dissipated by net upward force. …

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