Q.The two thigh bones (femurs), each of cross-sectional area $10\ \text{cm}^{2}$ support the upper part of a human body of mass $40\ \text{kg}$. Estimate the average pressure sustained by the femurs.

Showing the 12 most recent of 14 on this concept.

1. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.The pressure applied on a cube from all sides is 'P'. In order to keep its volume constant, the temperature of cube is to be raised by (The Bulk modulus and the coefficient of volume expansion of the material of the cube are $\beta$ and $\alpha$ respectively) (A) $\frac{P}{\alpha\beta}$ (B) $\frac{P\alpha}{\beta}$ (C) $\frac{P\beta}{\alpha}$ (D) $\frac{\alpha\beta}{P}$

   ›Reveal solutionSolution

   To keep the cube's volume fixed while pressure squeezes it, we need thermal expansion to exactly offset the pressure-induced shrinkage; the required temperature rise is $\Delta T = \dfrac{P}{\alpha\beta}$.

   Concept and Intuition

   Bulk modulus $\beta$ tells us how much fractional volume change a given pressure causes; the coefficient of volume expansion $\alpha$ tells us how much fractional volume change a given temperature rise causes. If both effects act on the cube simultaneously and we want zero NET change in volume, the two fractional changes must be equal and opposite.

   Step-by-Step Solution

   1. Bulk modulus is defined as $\beta = -\dfrac{\Delta P}{\Delta V/V}$, so the fractional volume decrease due to pressure $P$ is $\dfrac{\Delta V}{V}\bigg|_{pressure} = \dfrac{P}{\beta}$ (magnitude).
   2. Coefficient of volume expansion: $\alpha = \dfrac{1}{V}\dfrac{\Delta V}{\Delta T}$, so the fractional volume increase due to a temperature rise $\Delta T$ is $\dfrac{\Delta V}{V}\bigg|_{thermal} = \alpha \Delta T$. …

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2. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.When a sphere is taken to the bottom of a sea of depth 1 km, it contracts in volume by 0.01%, then the bulk modulus of the material of the sphere is (Acceleration due to gravity $=10\ \mathrm{m\,s^{-2}}$) (A) $10\times10^{6}\ \mathrm{N\,m^{-2}}$ (B) $1.2\times10^{10}\ \mathrm{N\,m^{-2}}$ (C) $10\times10^{10}\ \mathrm{N\,m^{-2}}$ (D) $10\times10^{11}\ \mathrm{N\,m^{-2}}$

   ›Reveal solutionSolution

   Using $\Delta P=\rho g h$ for the extra pressure at 1 km depth and $K=\Delta P/(\Delta V/V)$ with the given 0.01% volume contraction gives $K=10\times10^{10}\ \mathrm{N\,m^{-2}}$.

   Concept and Intuition

   Bulk modulus measures a material's resistance to uniform compression: $K = \dfrac{\Delta P}{-\Delta V/V}$ (the minus sign just reflects that volume shrinks under increased pressure; we work with magnitudes). Taking an object deeper into a sea increases the pressure on it by the hydrostatic pressure of the water column above, $\Delta P=\rho g h$. The tiny fractional volume change tells us how stiff the sphere's material is.

   Step-by-Step Solution

   1. Extra pressure at depth $h=1\,\mathrm{km}=1000$ m: $\Delta P = \rho g h = 1000\times10\times1000 = 10^{7}\ \mathrm{N\,m^{-2}}$.
   2. Fractional volume contraction: $\dfrac{\Delta V}{V} = 0.01\% = 1\times10^{-4}$. …

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3. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.If the pressure on a body is increased from 200 kPa to 250 kPa, the volume of the body decreases by 0.25%. The compressibility of the material of the body is (in $\mathrm{m^2\,N^{-1}}$) (A) $2 \times 10^{7}$ (B) $2 \times 10^{-7}$ (C) $5 \times 10^{8}$ (D) $5 \times 10^{-8}$

   ›Reveal solutionSolution

   Tests the definition of compressibility (reciprocal of bulk modulus) computed from a fractional volume change under a pressure change. Answer: $5\times10^{-8}\ \text{m}^2\text{N}^{-1}$.

   Concept and Intuition

   Bulk modulus $B=-\dfrac{\Delta P}{\Delta V/V}$ measures a material's resistance to uniform compression; compressibility $k=1/B$ measures how readily it compresses. A larger $k$ means the material shrinks more for a given pressure increase.

   Step-by-Step Solution

   1. Change in pressure: $\Delta P = 250\ \text{kPa}-200\ \text{kPa}=50\ \text{kPa}=5\times10^{4}\ \text{Pa}$.
   2. Fractional volume decrease: $\dfrac{\Delta V}{V}=0.25\%=2.5\times10^{-3}$.
   3. Compressibility: $k=\dfrac{\Delta V/V}{\Delta P}=\dfrac{2.5\times10^{-3}}{5\times10^{4}}$. …

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4. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.Depth of a river is 100 m. Magnitude of compressibility of the water is $0.5 \times 10^{-9}\ \mathrm{N^{-1}\,m^2}$. The fractional compression in water at the bottom of the river is (Acceleration due to gravity $= 10\ \mathrm{ms^{-2}}$) (A) $0.9 \times 10^3$ (B) $0.5 \times 10^{-3}$ (C) $2 \times 10^{-3}$ (D) $1.3 \times 10^{-2}$

   ›Reveal solutionSolution

   Fractional volume compression equals hydrostatic pressure times compressibility; computing $P=\rho gh$ and multiplying gives $0.5\times10^{-3}$.

   Concept and Intuition

   Bulk modulus $K$ is defined by $P = -K\dfrac{\Delta V}{V}$, so compressibility ($1/K$) directly converts an applied pressure into the fractional volume change: $\dfrac{\Delta V}{V} = P \times \text{compressibility}$. The pressure doing the compressing here is simply the hydrostatic pressure at that depth.

   Step-by-Step Solution

   1. Hydrostatic pressure at depth $h=100\ \text{m}$: $P = \rho g h = 1000 \times 10 \times 100 = 10^6\ \text{Pa}$.
   2. Fractional compression: $\dfrac{\Delta V}{V} = P \times \text{compressibility} = 10^6 \times 0.5\times10^{-9}$. …

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5. AP EAPCET 2023Set ap-2023-05-22-AN1 markMCQ
   Q.The bulk modulus (B) and compressibility (k) are related by (A) $k = \sqrt{\dfrac{1}{B}}$ (B) $k = B^2$ (C) $k = \dfrac{1}{B}$ (D) $k = \dfrac{1}{B^2}$

   ›Reveal solutionSolution

   Compressibility is, by definition, the reciprocal of bulk modulus.

   Concept and Intuition

   Bulk modulus $B=-\dfrac{\Delta P}{\Delta V/V}$ measures how much a material resists uniform compression — a large $B$ means the material is hard to compress. Compressibility $k$ is defined as exactly how easy it is to compress, i.e. the inverse of $B$: $k=1/B$.

   Step-by-Step Solution

   1. Definition of bulk modulus: $B=\dfrac{-\Delta P}{\Delta V/V}$.
   2. Definition of compressibility: $k=\dfrac{-\Delta V/V}{\Delta P}$. …

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6. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.A spherical ball of volume 2000 cm$^3$ is subjected to a hydraulic pressure of 15 atm. If the change in volume is $5 \times 10^{-2}$ cm$^3$, the bulk modulus of the material of the spherical ball is (1 atm = $10^5$ Nm$^{-2}$) (A) $6 \times 10^{10}$ Nm$^{-2}$ (B) $2 \times 10^{10}$ Nm$^{-2}$ (C) $5 \times 10^{10}$ Nm$^{-2}$ (D) $15 \times 10^{10}$ Nm$^{-2}$

   ›Reveal solutionSolution

   Applying the bulk modulus formula directly with the given pressure change and fractional volume change gives $6\times10^{10}\,\text{Nm}^{-2}$.

   Concept and Intuition

   Bulk modulus measures a material's resistance to uniform compression: $B = \dfrac{\Delta P}{\Delta V/V}$ (the volumetric strain here is small enough that we drop the conventional negative sign, since we're asked for magnitude). Since $V$ and $\Delta V$ are given in the same units (cm$^3$), their ratio is dimensionless and unit conversion is needed only for $\Delta P$.

   Step-by-Step Solution

   1. Given: $V=2000\,\text{cm}^3$, $\Delta P = 15\,\text{atm} = 15\times10^5\,\text{Nm}^{-2}$ (using $1\,\text{atm}=10^5\,\text{Nm}^{-2}$), $\Delta V = 5\times10^{-2}\,\text{cm}^3 = 0.05\,\text{cm}^3$.
   2. Bulk modulus: $B = \dfrac{\Delta P}{\Delta V/V} = \dfrac{\Delta P \cdot V}{\Delta V}$. …

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7. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
   Q.The maximum possible height of a mountain on earth is approximately (elastic limit of mountain rock $= 30\times10^{7}\ \mathrm{Nm}^{-2}$, average density of mountain rock $= 3\times10^{3}\ \mathrm{kgm}^{-3}$, $g = 10\ \mathrm{ms}^{-2}$) (A) 9 km (B) 10 km (C) 12 km (D) 8.8 km

   ›Reveal solutionSolution

   Equating the base pressure $\rho gh$ of the mountain's own weight to the rock's elastic limit gives the maximum possible mountain height, $10$ km.

   Concept and Intuition

   A mountain's base experiences a compressive stress from the weight of the rock above it, exactly like a column under its own weight: $P=\rho gh$ (density times $g$ times height, analogous to fluid pressure at depth $h$). If this stress exceeds the rock's elastic limit, the rock would deform/flow, so the theoretical maximum height is when this stress just equals the elastic limit.

   Step-by-Step Solution

   1. Stress at the base due to self-weight: $P = \rho g h$.
   2. Set $P =$ elastic limit $= 30\times10^7\text{ Nm}^{-2}$.
   3. $h = \dfrac{30\times10^7}{\rho g} = \dfrac{30\times10^7}{(3\times10^3)(10)}$. …

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8. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
   Q.Isothermal bulk modulus of a gas at a pressure P is ($\gamma$ - ratio of specific heat capacities of the gas) (A) $\gamma$ (B) $\gamma P$ (C) P (D) $\dfrac{\gamma}{P}$

   ›Reveal solutionSolution

   Differentiating the isothermal gas law $PV=\text{const}$ shows the isothermal bulk modulus equals the pressure itself, $B_{iso}=P$ (while the adiabatic bulk modulus is $\gamma P$).

   Concept and Intuition

   Bulk modulus measures a substance's resistance to uniform compression: $B=-V\dfrac{dP}{dV}$. For an ideal gas undergoing an isothermal change, Boyle's law $PV=\text{constant}$ applies. Differentiating this relation directly gives the isothermal bulk modulus in terms of pressure alone — no $\gamma$ appears, since $\gamma$ only enters the adiabatic case ($PV^\gamma=\text{const}$), where $B_{adiabatic}=\gamma P$.

   Step-by-Step Solution

   1. Isothermal condition: $PV=k$ (constant).
   2. Differentiate: $P\,dV+V\,dP=0\Rightarrow \dfrac{dP}{dV}=-\dfrac{P}{V}$.
   3. Bulk modulus definition: $B=-V\dfrac{dP}{dV}=-V\left(-\dfrac{P}{V}\right)=P$. …

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9. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.The fractional change in the volume of a glass slab when subjected to hydraulic pressure of 14 atm is (Bulk modulus of glass $= 40 \times 10^{9}\ \text{Nm}^{-2}$) (A) $1.44 \times 10^{-5}$ (B) $3.54 \times 10^{-5}$ (C) $2.74 \times 10^{-5}$ (D) $3.14 \times 10^{-5}$

   ›Reveal solutionSolution

   Using $\Delta V/V = P/B$ with $P$ converted from atmospheres to pascals gives a fractional volume change of $3.54\times10^{-5}$.

   Concept and Intuition

   Bulk modulus $B$ relates the applied (hydraulic) pressure to the resulting fractional change in volume: $B = \dfrac{P}{\Delta V/V}$, so $\Delta V/V = P/B$. A large $B$ (like glass) means a small fractional volume change even under significant pressure.

   Step-by-Step Solution

   1. Convert pressure: $P = 14\ \text{atm} = 14\times1.013\times10^5\ \text{Pa} \approx 1.418\times10^6$ Pa.
   2. Given $B = 40\times10^9$ Pa.
   3. Fractional volume change: $\dfrac{\Delta V}{V} = \dfrac{P}{B} = \dfrac{1.418\times10^6}{40\times10^9}$.
   4. Compute: $\dfrac{1.418\times10^6}{4\times10^{10}} \approx 3.545\times10^{-5}$.

   Common Mistakes …

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10. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
    Q.The pressure required to decrease the volume of 4000 cc water by 0.05% is (Bulk modulus of water = $2.2\times10^{9}\ \mathrm{Nm^{-2}}$) (A) $11\times10^{6}\ \mathrm{Nm^{-2}}$ (B) $5\times10^{5}\ \mathrm{Nm^{-2}}$ (C) $2.2\times10^{6}\ \mathrm{Nm^{-2}}$ (D) $1.1\times10^{6}\ \mathrm{Nm^{-2}}$

    ›Reveal solutionSolution

    This is a direct application of the bulk modulus definition to find the pressure needed for a given fractional volume change. The required pressure is $1.1\times10^{6}\ \mathrm{Nm^{-2}}$.

    Concept and Intuition

    Bulk modulus measures a substance's resistance to uniform compression: $B=\dfrac{\Delta P}{\Delta V/V}$ (magnitude form). A larger $B$ means more pressure is needed to achieve the same fractional volume decrease. Note that the actual volume (4000 cc here) is a distractor — only the fractional change $\Delta V/V$ matters, so the 4000 cc value never needs to be used numerically.

    Step-by-Step Solution

    1. Fractional volume change: $\dfrac{\Delta V}{V}=0.05\%=\dfrac{0.05}{100}=5\times10^{-4}$.
    2. Bulk modulus relation: $\Delta P = B\times\dfrac{\Delta V}{V}$. …

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11. AP EAPCET 2022Set ap-2022-07-12-AN1 markMCQ
    Q.A metal cube has an edge length of 90 cm. If $2\times10^9$ Nm$^{-1}$ of pressure (or stress) is required to reduce the edge length to 89.5 cm, then the bulk modulus of the metal is (A) $1\times10^{11}$ Nm$^{-2}$ (B) $2.5\times10^{10}$ Nm$^{-2}$ (C) $9\times10^{11}$ Nm$^{-2}$ (D) $1.2\times10^{11}$ Nm$^{-2}$

    ›Reveal solutionSolution

    Bulk modulus is stress divided by volumetric (fractional volume) strain; for a cube, fractional volume change is three times the fractional length change for small deformations.

    Concept and Intuition

    For a cube of edge $L$, volume $V=L^3$. Differentiating, $\dfrac{dV}{V} = 3\dfrac{dL}{L}$ for small changes — this is why linear strain must be tripled to get volumetric strain for a uniform (isotropic) compression.

    Step-by-Step Solution

    1. Linear strain: $\dfrac{\Delta L}{L} = \dfrac{90-89.5}{90} = \dfrac{0.5}{90} = \dfrac{1}{180}$.
    2. Volumetric strain: $\dfrac{\Delta V}{V} = 3\times\dfrac{1}{180} = \dfrac{1}{60}$. …

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12. AP EAPCET 2022Set ap-2022-07-12-FN1 markMCQ
    Q.A metal cube has an edge length 0.1 m. The bulk modulus of the metal is $1.4\times10^{11}\ \mathrm{N\,m^{-2}}$. When subjected to hydraulic pressure, the volume contraction of the cube is $5\times10^{-2}\ \mathrm{cm^3}$. The applied pressure is (A) $2\times10^6$ Pa (B) $1.4\times10^7$ Pa (C) $7\times10^6$ Pa (D) $5\times10^5$ Pa

    ›Reveal solutionSolution

    Bulk modulus directly relates applied pressure to fractional volume change: $B=\Delta P/(\Delta V/V)$. Careful unit conversion of the small volume change is the key step. Answer: $7\times10^6\ \mathrm{Pa}$.

    Concept and Intuition

    Bulk modulus $B$ measures a material's resistance to uniform compression: the more pressure needed to produce a given fractional volume change, the stiffer (higher $B$) the material. Rearranging $B=\Delta P/(\Delta V/V)$ gives the pressure directly from the given volume contraction.

    Step-by-Step Solution

    1. Original volume: $V = (0.1\,\mathrm{m})^3 = 1\times10^{-3}\ \mathrm{m^3}$.
    2. Volume contraction: $\Delta V = 5\times10^{-2}\ \mathrm{cm^3}$. Since $1\ \mathrm{cm^3}=10^{-6}\ \mathrm{m^3}$, $\Delta V = 5\times10^{-2}\times10^{-6}=5\times10^{-8}\ \mathrm{m^3}$.
    3. Fractional volume change: $\dfrac{\Delta V}{V} = \dfrac{5\times10^{-8}}{1\times10^{-3}} = 5\times10^{-5}$. …

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