Q.A $50\ \text{kg}$ girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter $1.0\ \text{cm}$. What is the pressure exerted by the heel on the horizontal floor?

Showing the 12 most recent of 37 on this concept.

1. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.A water drop when pressed between two parallel glass plates spreads in to a circle of diameter 10 cm. If the surface tension of water is $70 \times 10^{-3} \ \mathrm{Nm}^{-1}$ and the force required to separate the two glass plates is 2.2 N, then the volume of the drop is nearly (A) $5 \times 10^{-4} \ \mathrm{m}^3$ (B) $3.93 \times 10^{-6} \ \mathrm{m}^3$ (C) $62.8 \times 10^{-6} \ \mathrm{m}^3$ (D) $2 \times 10^{-6} \ \mathrm{m}^3$

   ›Reveal solutionSolution

   The plates are held together by the reduced pressure inside the curved water film, $\Delta P = 2T/d$. Setting $F = \Delta P\cdot\pi r^2$ gives the film thickness $d$, and then $V = \pi r^2 d \approx 3.93\times10^{-6}\ \mathrm{m^3}$, option (B).

   Concept & intuition

   The squeezed drop forms a thin circular film of radius $r$ and thickness $d$. Its edge is a concave (cylindrical) meniscus of radius of curvature $d/2$, so the pressure inside the film is lower than atmospheric by $\Delta P = 2T/d$. Atmospheric pressure therefore pushes the plates together; the force needed to separate them is this pressure drop acting over the film area $\pi r^2$.

   Watch out

   The attractive force is not surface tension pulling along the rim's circumference — that would give a value far too small. It arises from the pressure jump across the curved liquid surface (Young–Laplace), which acts over the whole contact area.

   1. Pressure difference across the meniscus For the cylindrical edge, one radius of curvature is $d/2$ and the other is effectively infinite:

   $\Delta P = T\left(\frac{1}{d/2}\right) = \frac{2T}{d}.$

   2. Force on the plates $F = \Delta P \cdot \pi r^2 = \frac{2T}{d}\,\pi r^2,$ …

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2. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.The percentage error in the measurement of length when a metal scale calibrated at $30\,^\circ\mathrm{C}$ is used at $-10\,^\circ\mathrm{C}$ is (Coefficient of linear expansion of the metal $= 12 \times 10^{-6}\,^\circ\mathrm{C}^{-1}$) (A) $0.024$ (B) $0.036$ (C) $0.048$ (D) $0.012$

   ›Reveal solutionSolution

   The percentage error arises because the scale contracts when cooled, making each marked length shorter than its true value. The error equals the thermal strain: $\alpha \Delta T = 12\times10^{-6} \times 40 = 0.00048$, which as a percentage is $0.048\%$. The correct option is (C).

   Concept & Intuition

   A metal scale is calibrated at a reference temperature (here $30^\circ$C). At that temperature, the markings (say 1 cm apart) are exactly correct. When the scale is used at a different temperature, the metal expands or contracts. If the scale is cooled, it shrinks: the physical distance between two markings becomes less than the number printed. So when you measure an object, you read a larger number than the true length — the error is positive. The fractional error in any length measurement is simply the fractional change in the scale’s length, i.e., $\alpha \Delta T$. Multiply by 100 to get the percentage.

   ---

   Step-by-step solution

   1. Identify the temperature change The scale was calibrated at $T_0 = 30^\circ$C and used at $T = -10^\circ$C. The change in temperature is

   $\Delta T = T - T_0 = -10 - 30 = -40^\circ\text{C}.$

   The negative sign indicates cooling (contraction).

   2. Recall the linear expansion formula For a small temperature change, the change in length of a solid is

   $\Delta L = \alpha L_0 \Delta T,$

   where $\alpha = 12 \times 10^{-6}\,^\circ\mathrm{C}^{-1}$ is the coefficient of linear expansion, and $L_0$ is the original length at $T_0$.

   3. Find the fractional change in the scale’s length The fractional change is

   $\frac{\Delta L}{L_0} = \alpha \Delta T = (12 \times 10^{-6}) \times (-40) = -4.8 \times 10^{-4}.$

   The negative sign means the scale contracts: each interval (e.g., 1 cm mark) is now physically shorter by $0.048\%$ of its original length.

   4. Interpret the percentage error in measurement …

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3. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.If the apparent weight of a cube of mass 400 g immersed in water is 3.36 N, then the density of the material of the cube (in kg m$^{-3}$) is (Acceleration due to gravity = 10 ms$^{-2}$) (A) 8350 (B) 9250 (C) 7500 (D) 6250

   ›Reveal solutionSolution

   The apparent weight is the actual weight minus the buoyant force. Using the given apparent weight and the density of water, we find the cube's volume and then its density. The density of the material is 6250 kg/m³, which corresponds to option (D).

   The key here is understanding what "apparent weight" means. When an object is immersed in a fluid, the fluid exerts an upward buoyant force on it. This buoyant force reduces the object's effective weight — what a scale would read if you weighed it underwater. That reading is the apparent weight.

   The buoyant force is given by Archimedes' principle: it equals the weight of the fluid displaced by the object. For a fully immersed object, the displaced volume equals the object's own volume. So if we know the apparent weight and the actual weight, we can find the buoyant force, then the volume, and finally the density.

   Let's work through it step by step.

   1. Find the actual weight of the cube.

      Mass $m = 400\ \text{g} = 0.4\ \text{kg}$.

      Actual weight $W = mg = 0.4 \times 10 = 4\ \text{N}$.

   2. Find the buoyant force.

      Apparent weight $W_{\text{app}} = W - F_b$, where $F_b$ is the buoyant force.

      So $F_b = W - W_{\text{app}} = 4 - 3.36 = 0.64\ \text{N}$.

   3. Relate buoyant force to volume.

      $F_b = \rho_{\text{water}} \cdot V \cdot g$, where $\rho_{\text{water}} = 1000\ \text{kg/m}^3$ and $g = 10\ \text{m/s}^2$.

      So $0.64 = 1000 \times V \times 10$ …

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4. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.If the fractional compression of water at the bottom of an ocean is $1.5 \times 10^{-2}$, then the depth of the ocean is (Bulk modulus of water $= 2.2 \times 10^9 \ \mathrm{Nm}^{-2}$ and acceleration due to gravity $= 10 \ \mathrm{ms}^{-2}$) (A) 3.3 km (B) 1.7 km (C) 1.1 km (D) 2.4 km

   ›Reveal solutionSolution

   The fractional compression is related to the pressure increase via the bulk modulus; equating that pressure to the hydrostatic pressure $P = \rho g h$ gives the depth. The ocean depth is 3.3 km (option A).

   The key idea here is that the bulk modulus $B$ of a fluid tells you how much pressure is needed to produce a given fractional change in volume. For water at the bottom of the ocean, the pressure comes from the weight of the water column above — that’s hydrostatic pressure. So we link the two: the pressure that causes the compression is exactly $P = \rho g h$, where $h$ is the depth we want.

   The fractional compression is given as $\frac{\Delta V}{V} = 1.5 \times 10^{-2}$. The bulk modulus is defined as

   $B = -\frac{\Delta P}{\Delta V / V}.$

   The negative sign means that an increase in pressure ($+\Delta P$) causes a decrease in volume ($-\Delta V$), so the ratio $\Delta V / V$ is negative. But here we are given the magnitude of the fractional compression, so we take $|\Delta V / V| = 1.5 \times 10^{-2}$ and write

   $\Delta P = B \times \left|\frac{\Delta V}{V}\right|.$

   Now let’s work through it step by step.

   1. Find the pressure increase at the bottom. Using the bulk modulus formula:

   $\Delta P = B \times \frac{\Delta V}{V} = (2.2 \times 10^9) \times (1.5 \times 10^{-2}).$

   Compute:

   $\Delta P = 2.2 \times 1.5 \times 10^{7} = 3.3 \times 10^{7} \ \mathrm{N/m^2}.$

   2. Relate this pressure to the depth of the ocean. Hydrostatic pressure at depth $h$ in a fluid of density $\rho$ is $P = \rho g h$. For water, $\rho = 1000 \ \mathrm{kg/m^3}$ (standard value, though not explicitly given — it’s assumed). With $g = 10 \ \mathrm{m/s^2}$, we have:

   $\rho g h = 3.3 \times 10^{7}.$

   So …

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5. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.If a brass sphere of radius 36 cm is submerged in a lake at a depth where the pressure is $10^7$ Pa, then the change in the radius of the sphere is (Bulk modulus of brass = 60 GPa) (A) $4 \times 10^{-2}$ cm (B) $2 \times 10^{-3}$ cm (C) $4 \times 10^{-3}$ cm (D) $2 \times 10^{-2}$ cm

   ›Reveal solutionSolution

   Volumetric strain under pressure is $\Delta V/V = P/K$, and for a sphere $\Delta V/V = 3\,\Delta r/r$. This gives $\Delta r = 2 \times 10^{-3}\ \mathrm{cm}$ — option (B).

   ---

   Concept

   Immersion pressure $P$ compresses the sphere. The bulk modulus relates pressure to volumetric strain, $K = \dfrac{P}{\Delta V/V}$. For a sphere $V \propto r^3$, so $\dfrac{\Delta V}{V} = 3\dfrac{\Delta r}{r}$.

   Solution

   Volumetric strain:

   $\frac{\Delta V}{V} = \frac{P}{K} = \frac{10^7}{60 \times 10^9} = \frac{1}{6000}.$

   Radial strain: …

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6. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.If a radioactive substance decays 10% in every 16 hours, then the percentage of the radioactive substance that remains after 2 days is (A) 82.2 (B) 18.8 (C) 27.1 (D) 72.9

   ›Reveal solutionSolution

   The substance decays by 10% every 16 hours, meaning 90% remains after each 16-hour period. Two days = 48 hours = three 16-hour periods, so the remaining fraction is $0.9^3 = 0.729$, or 72.9%. The correct option is (D).

   The key concept here is exponential decay in discrete steps. When a substance loses a fixed percentage of its current amount each time interval, the amount remaining after $n$ intervals is the initial amount multiplied by $(1 - \text{decay rate})^n$. This is not a linear subtraction — you don’t just subtract 10% three times from 100%, because each 10% loss is taken from a smaller amount.

   Let’s work through it step by step.

   1. Understand the decay rate and time interval.

      The substance decays 10% every 16 hours. That means after 16 hours, 90% of what was there at the start of that period remains. So the “retention factor” per 16-hour period is $0.9$.

   2. Convert the total time into the number of decay periods.

      Two days = 48 hours. How many 16-hour periods fit into 48 hours?

   $\frac{48}{16} = 3$

   So there are three full decay periods.

   3. Apply the decay repeatedly.

      After the first 16 hours: remaining fraction = $0.9$.

      After the second 16 hours: remaining fraction = $0.9 \times 0.9 = 0.9^2$.

      After the third 16 hours: remaining fraction = $0.9^3$.

   4. Calculate the numerical value.

      $0.9^3 = 0.9 \times 0.9 \times 0.9 = 0.81 \times 0.9 = 0.729$ …

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7. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
   Q.If the relative error in the determination of area of a body is $0.08$, then the percentage error in the determination of its volume is (A) $6$ (B) $12$ (C) $16$ (D) $4$

   ›Reveal solutionSolution

   The problem asks for the percentage error in volume given the relative error in area. Since area is proportional to the square of a linear dimension ($L^2$) and volume is proportional to the cube of the same linear dimension ($L^3$), we use error propagation rules. The relative error in volume is found to be $0.12$, which corresponds to a percentage error of $\boxed{12}$.

   When we measure physical quantities, there's always some uncertainty or error. Error analysis helps us understand how these uncertainties propagate when we perform calculations involving the measured quantities.

   The core idea here is how errors propagate when a quantity is raised to a power.

   • Relative Error: For a quantity $X$, if $\Delta X$ is the absolute error in its measurement, then the relative error is defined as $\frac{\Delta X}{X}$. It's a dimensionless quantity.
   • Percentage Error: This is simply the relative error expressed as a percentage: $\frac{\Delta X}{X} \times 100\%$.

   If a physical quantity $Y$ depends on another quantity $X$ such that $Y = c X^n$, where $c$ is a constant and $n$ is a power, then the relative error in $Y$ is related to the relative error in $X$ by:

   $\frac{\Delta Y}{Y} = n \frac{\Delta X}{X}$

   This formula is derived by taking the natural logarithm of both sides and then differentiating.

   For any given body, its area ($A$) is proportional to the square of its characteristic linear dimension ($L$), and its volume ($V$) is proportional to the cube of the same characteristic linear dimension. For example, for a sphere, $A = 4\pi R^2$ and $V = \frac{4}{3}\pi R^3$. For a cube, $A = 6s^2$ and $V = s^3$. In both cases, $A \propto L^2$ and $V \propto L^3$, where $L$ is the radius $R$ or side $s$. The constants of proportionality ($4\pi$, $\frac{4}{3}\pi$, $6$, $1$) do not affect the relative errors.

   Let's apply this concept to solve the problem.

   1. Relate Area to a Linear Dimension: Let $L$ be a characteristic linear dimension of the body (e.g., its length, radius, or side). The area $A$ of the body will be proportional to the square of this linear dimension:

   $A \propto L^2$

   We can write this as $A = k L^2$, where $k$ is a dimensionless constant that depends on the shape of the body.
   

   2. Apply Error Propagation for Area:

   Using the error propagation formula for powers, the relative error in area is related to the relative error in the linear dimension:

   $\frac{\Delta A}{A} = 2 \frac{\Delta L}{L}$

   We are given that the relative error in the determination of the area is $0.08$.
   

   $0.08 = 2 \frac{\Delta L}{L}$

   3. Calculate Relative Error in Linear Dimension: From the above equation, we can find the relative error in the linear dimension $L$: …

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8. TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
   Q.If the maximum and minimum amplitudes of an amplitude modulated wave are 9 V and 3 V respectively, then the modulation index is (A) $\dfrac{1}{2}$ (B) $3$ (C) $2$ (D) $\dfrac{1}{3}$

   ›Reveal solutionSolution

   The modulation index is found from the ratio of the difference to the sum of the maximum and minimum amplitudes. For $A_{\text{max}} = 9\ \text{V}$ and $A_{\text{min}} = 3\ \text{V}$, the modulation index is $\frac{1}{2}$.

   In amplitude modulation, the carrier wave’s amplitude is varied in proportion to the message signal. The modulation index $m$ tells us how deeply the carrier is modulated — it’s the ratio of the message amplitude to the carrier amplitude. When you look at the modulated waveform on an oscilloscope, the envelope shows a maximum amplitude $A_{\text{max}}$ and a minimum amplitude $A_{\text{min}}$. These are directly related to the carrier amplitude $A_c$ and the message amplitude $A_m$:

   • The peak of the envelope is $A_{\text{max}} = A_c + A_m$
   • The trough of the envelope is $A_{\text{min}} = A_c - A_m$

   From these two equations, you can solve for $A_c$ and $A_m$, and then $m = A_m / A_c$. But there’s a cleaner way: add and subtract the two equations to get $A_c$ and $A_m$ directly, then form the ratio.

   1. Write the given values:

      $A_{\text{max}} = 9\ \text{V}$, $A_{\text{min}} = 3\ \text{V}$.

   2. The carrier amplitude is the average of the max and min:

   $A_c = \frac{A_{\text{max}} + A_{\text{min}}}{2} = \frac{9 + 3}{2} = 6\ \text{V}$

   3. The message amplitude is half the difference:

   $A_m = \frac{A_{\text{max}} - A_{\text{min}}}{2} = \frac{9 - 3}{2} = 3\ \text{V}$

   4. The modulation index is: $m = \frac{A_m}{A_c} = \frac{3}{6} = \frac{1}{2}$ …

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9. TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
   Q.A wooden cube is floating in a bucket of water with $\dfrac{3}{4}$ of its volume immersed. If this bucket with the wooden block is now placed in a lift moving down with an acceleration of $\dfrac{g}{2}$, the fraction of volume of wooden cube immersed in water is (A) $\dfrac{3}{4}$ (B) $\dfrac{3}{8}$ (C) $\dfrac{3}{2}$ (D) $\dfrac{1}{2}$

   ›Reveal solutionSolution

   The fraction of the cube’s volume immersed depends only on the relative densities of the cube and water, not on the effective gravity. Since the lift’s acceleration changes both the weight of the cube and the buoyant force equally, the immersed fraction remains $\frac{3}{4}$.

   The key idea here is that buoyancy in a fluid arises from the pressure difference, which itself depends on the effective gravity acting on the fluid. When the entire system (bucket + water + cube) accelerates together, the effective gravity changes for both the cube and the water in exactly the same way. So the ratio of densities — which determines the immersed fraction — stays unchanged.

   Let’s work through it carefully.

   1. Understand the floating condition in a stationary lift

      When the lift is at rest (or moving with constant velocity), the cube is in equilibrium. The weight of the cube is balanced by the buoyant force.

      Let the density of the cube be $\rho_c$, the density of water be $\rho_w$, the volume of the cube be $V$, and the immersed volume be $V_{\text{im}}$.

      Weight: $W = \rho_c V g$

      Buoyant force: $F_b = \rho_w V_{\text{im}} g$

      At equilibrium: $\rho_c V g = \rho_w V_{\text{im}} g$

      Cancelling $g$: $\rho_c V = \rho_w V_{\text{im}}$

      So the fraction immersed is $\frac{V_{\text{im}}}{V} = \frac{\rho_c}{\rho_w}$

      Given that this fraction is $\frac{3}{4}$, we have $\frac{\rho_c}{\rho_w} = \frac{3}{4}$.

   2. What changes when the lift accelerates downward?

      The lift moves down with acceleration $a = \frac{g}{2}$. In the non-inertial frame of the lift, every object experiences a pseudo-force upward equal to $m a$. So the effective gravity in this frame is $g_{\text{eff}} = g - a = g - \frac{g}{2} = \frac{g}{2}$.

      Both the cube and the water inside the bucket experience this same effective gravity.

   3. Re-write the equilibrium condition in the accelerating lift

      In the lift’s frame, the cube is again in equilibrium (it floats stationary relative to the water). …

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10. TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
    Q.If the length of a cylinder made with a material of Poisson’s ratio 0.4 is increased by 5%, then the decrease in its diameter is (A) $0.5\%$ (B) $2.0\%$ (C) $1.0\%$ (D) $1.5\%$

    ›Reveal solutionSolution

    When a material is stretched longitudinally, its lateral dimensions (like diameter) tend to contract. Poisson's ratio quantifies this phenomenon. Given a Poisson's ratio of 0.4 and a 5% increase in length, the diameter will decrease by 2.0%.

    In materials science, when a material is subjected to a tensile force along one axis, it elongates in that direction. Simultaneously, it tends to contract in the perpendicular directions. This phenomenon is described by Poisson's ratio.

    Concept and Intuition

    Imagine pulling on a rubber band. As it gets longer, you'll notice it also gets thinner. This thinning is the lateral contraction. Poisson's ratio ($\nu$) is a fundamental material property that quantifies this relationship. It's defined as the negative ratio of the lateral strain to the longitudinal strain.

    • Longitudinal strain ($\epsilon_L$) is the fractional change in length along the direction of the applied force. If the original length is $L$ and the change in length is $\Delta L$, then $\epsilon_L = \frac{\Delta L}{L}$.
    • Lateral strain ($\epsilon_D$) is the fractional change in the dimension perpendicular to the applied force (e.g., diameter or width). If the original diameter is $D$ and the change in diameter is $\Delta D$, then $\epsilon_D = \frac{\Delta D}{D}$.

    Poisson's ratio is given by:

    $\nu = -\frac{\text{lateral strain}}{\text{longitudinal strain}} = -\frac{\Delta D / D}{\Delta L / L}$

    The negative sign is crucial because when a material is stretched ($\Delta L > 0$), its diameter typically decreases ($\Delta D < 0$). This makes the ratio $\frac{\Delta D / D}{\Delta L / L}$ negative, so the negative sign in the formula ensures that Poisson's ratio $\nu$ is a positive value for most common materials.

    Let's apply this concept to the given problem.

    1. Identify the given values:

       • Poisson's ratio, $\nu = 0.4$.
       • Percentage increase in length = 5%. This means the longitudinal strain, $\frac{\Delta L}{L}$, is $0.05$.

    2. Relate longitudinal and lateral strain using Poisson's ratio:

       We use the formula for Poisson's ratio:

    $\nu = -\frac{\Delta D / D}{\Delta L / L}$

    We want to find the percentage decrease in diameter, which is $100 \times \left|\frac{\Delta D}{D}\right|$. Let's rearrange the formula to solve for the lateral strain $\frac{\Delta D}{D}$:
    

    $\frac{\Delta D}{D} = -\nu \times \frac{\Delta L}{L}$

    3. Substitute the known values: …

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11. TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
    Q.A 31.4 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter of 2 cm. The pressure exerted by the heel on the horizontal floor is (Acceleration due to gravity $= 10\, \mathrm{ms}^{-2}$) (A) $10^6\, \mathrm{Pa}$ (B) $2.5 \times 10^5\, \mathrm{Pa}$ (C) $10^5\, \mathrm{Pa}$ (D) $2.5 \times 10^6\, \mathrm{Pa}$

    ›Reveal solutionSolution

    Pressure is force per unit area. The girl's weight acts over the tiny circular area of the heel, producing a pressure of $10^6 \, \mathrm{Pa}$.

    Pressure measures how concentrated a force is. When the same force acts over a smaller area, the pressure increases dramatically — which is exactly why high heels can damage soft floors while flat shoes do not. Here, the entire weight of the girl is supported by a single small circular heel.

    The pressure is defined as:

    $P = \frac{F}{A}$

    where $F$ is the perpendicular force and $A$ is the area over which it acts.

    Step-by-step solution

    1. Find the force exerted by the heel

       The heel must support the girl's entire weight. The gravitational force is:

    $F = mg = 31.4 \times 10 = 314 \, \mathrm{N}$

    2. Calculate the area of the circular heel

       The heel has diameter $d = 2 \, \mathrm{cm} = 0.02 \, \mathrm{m}$, so radius $r = 0.01 \, \mathrm{m}$.

       The area of a circle is:

    $A = \pi r^2 = \pi \times (0.01)^2 = \pi \times 10^{-4} \, \mathrm{m}^2$

    3. Compute the pressure

       Substituting into the pressure formula:

    $P = \frac{314}{\pi \times 10^{-4}} = \frac{314 \times 10^4}{\pi}$

    Using $\pi \approx 3.14$: …

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12. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
    Q.A wooden cube of side 10 cm floats at the interface between water and oil with its lower surface 3 cm below the interface. If the density of oil is $0.9~\mathrm{g~cm^{-3}}$, the mass of the wooden cube is (A) $940~\mathrm{g}$ (B) $900~\mathrm{g}$ (C) $1000~\mathrm{g}$ (D) $930~\mathrm{g}$

    ›Reveal solutionSolution

    The cube floats with part in oil and part in water; the buoyant force equals its weight. Using the submerged volumes in each fluid and their densities gives the cube’s mass as 930 g, so option (D) is correct.

    Concept & Intuition

    When an object floats at the interface of two immiscible fluids, it experiences an upward buoyant force from both fluids. The total buoyant force is the sum of the weights of the displaced volumes of each fluid. For equilibrium, this total must equal the weight of the object. Here, the cube’s lower surface is 3 cm below the oil–water interface, meaning 3 cm of its height is in water and the remaining 7 cm is in oil. We can directly compute the mass from the displaced masses.

    Step-by-step solution

    1. Determine the submerged volumes The cube has side length $10~\mathrm{cm}$.
       • Height in water: $h_w = 3~\mathrm{cm}$ → volume in water:

    $V_w = 10 \times 10 \times 3 = 300~\mathrm{cm^3}$

    • Height in oil: $h_o = 10 - 3 = 7~\mathrm{cm}$ → volume in oil:

    $V_o = 10 \times 10 \times 7 = 700~\mathrm{cm^3}$

    2. Write the equilibrium condition Buoyant force = weight of the cube:

    $\rho_w g V_w + \rho_o g V_o = m g$

    Cancel $g$:

    $m = \rho_w V_w + \rho_o V_o$

    3. Insert densities Density of water: $\rho_w = 1.0~\mathrm{g\,cm^{-3}}$ …

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