What is Bulk Modulus?

Bulk Modulus is a measure of a substance's resistance to uniform compression, defined as the ratio of volumetric stress to volumetric strain.

What is the formula for Bulk Modulus?

The formula is $B = \frac{-\Delta P}{\Delta V/V}$, where $\Delta P$ is change in pressure, $\Delta V$ is change in volume, and $V$ is initial volume.

How do you convert cm³ to litres?

1 litre is equal to $1000$ cm³, so divide the volume in cm³ by $1000$ to get litres.

What is the SI unit of pressure?

The SI unit of pressure is Pascal (Pa), where $1 \text{ Pa} = 1 \text{ N/m}^2$.

What does GPa stand for?

GPa stands for Gigapascal, which is equal to $10^9$ Pascals.

Why is there a negative sign in the Bulk Modulus formula?

The negative sign indicates that an increase in pressure results in a decrease in volume.

What is the change in pressure in this problem?

The pressure changes from 1 atm to 5 atm, so $\Delta P = 4$ atm.

How do we handle the conversion of atm to Pa?

The problem specifies to take $1 \text{ atm} = 10^5 \text{ Pa}$, so $\Delta P = 4 \times 10^5 \text{ Pa}$.

Is the initial volume required in litres or m³?

The final answer is specifically requested in litres.

What is the value of 2 GPa in standard units?

$2 \text{ GPa} = 2 \times 10^9 \text{ Pa}$.

A sample of a liquid is kept at 1 atm. It is compressed to 5 atm which leads to a change of volume of 0.8 cm³. If the bulk modulus of the liquid is 2 GPa, the initial volume of the liquid was ___

A sample of a liquid is kept at 1 atm. It is compressed to 5 atm which leads to a change of volume of 0.8 cm³. If the bulk modulus of the liquid is 2 GPa, the initial volume of the liquid was ____ litre. (Take 1 atm = 10⁵ Pa) | JEE Main Physics

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PhysicsElasticity

QNumerical

A sample of a liquid is kept at $1$ atm. It is compressed to $5$ atm which leads to a change of volume of $0.8$ cm³. If the bulk modulus of the liquid is $2$ GPa, the initial volume of the liquid was ____ litre. (Take $1$ atm = $10^5$ Pa)

✅ Correct Answer

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Solution Steps

List the Given Data

Initial Pressure $P_1 = 1$ atm. Final Pressure $P_2 = 5$ atm.

Change in Pressure $\Delta P = P_2 – P_1 = 5 – 1 = 4$ atm.

Convert $\Delta P$ to Pascals: $\Delta P = 4 \times 10^5$ Pa.

Change in volume $\Delta V = 0.8$ cm³.

Bulk Modulus $B = 2$ GPa $= 2 \times 10^9$ Pa.

Apply Bulk Modulus Formula

The definition of Bulk Modulus is:

$$B = \frac{\Delta P}{(\Delta V / V)}$$

Where $V$ is the initial volume.

Rearrange to solve for Initial Volume $V$

$$V = \frac{B \times \Delta V}{\Delta P}$$

Substitute values to find $V$ in cm³

$$V = \frac{(2 \times 10^9) \times 0.8}{4 \times 10^5}$$

$$V = \frac{1.6 \times 10^9}{4 \times 10^5} = 0.4 \times 10^4 = 4000 \text{ cm}^3$$

Convert to Litres

Since $1000 \text{ cm}^3 = 1 \text{ litre}$:

$$V = \frac{4000}{1000} = 4 \text{ litres}$$

Final Answer: 4

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Theory

1. Definition of Bulk Modulus

Bulk Modulus ($B$ or $K$) characterizes a material’s response to uniform pressure. It is defined as the ratio of the infinitesimal pressure increase to the resulting relative decrease of the volume. Mathematically, $B = -V \frac{dP}{dV}$. For liquids, which are much less compressible than gases, the Bulk Modulus is very high, typically in the range of Gigapascals (GPa). A higher Bulk Modulus implies that the substance is more “stiff” or harder to compress.

2. Stress and Strain in Volume Deformation

In the context of Bulk Modulus, the ‘stress’ is the uniform pressure applied to all surfaces of the object (volumetric stress), and the ‘strain’ is the fractional change in volume ($\Delta V/V$), known as volumetric strain. Unlike Young’s Modulus which deals with linear deformation, or Shear Modulus which deals with shape deformation, Bulk Modulus deals specifically with changes in the size of the object while its shape remains constant.

3. Compressibility of Liquids

Compressibility ($k$) is the reciprocal of the Bulk Modulus ($k = 1/B$). While we often treat liquids as “incompressible” in basic fluid mechanics (Pascal’s Law), they do compress slightly under extreme pressures. This compressibility is vital in applications like hydraulic systems, deep-sea exploration, and underwater acoustics (sonar), where the speed of sound in a medium is directly related to its Bulk Modulus and density ($v = \sqrt{B/\rho}$).

4. Units and Conversions in Elasticity

JEE problems frequently test your ability to navigate unit systems. Pressure is often given in atmospheres (atm), where $1 \text{ atm} \approx 1.013 \times 10^5 \text{ Pa}$, though $10^5$ is often used for simplicity. Moduli are given in GPa ($10^9 \text{ Pa}$) or MPa ($10^6 \text{ Pa}$). Volume conversions are also critical: $1 \text{ m}^3 = 10^3 \text{ litres} = 10^6 \text{ cm}^3$. Ensuring all variables are in consistent SI units before calculation is the most important step in avoiding numerical errors.

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FAQs

What is the value of 1 GPa?

$1 \text{ GPa} = 10^9 \text{ Pascals} = 10^9 \text{ N/m}^2$.

Does Bulk Modulus apply to gases?

Yes, but for gases, it depends on whether the compression is isothermal ($B = P$) or adiabatic ($B = \gamma P$).

Why did we use $\Delta P = 4$ and not 5?

The formula requires the *change* in pressure ($\Delta P = P_{final} – P_{initial}$), which is $5 – 1 = 4$ atm.

What is the relationship between density and Bulk Modulus?

As pressure increases and volume decreases, density increases. The fractional change in density is equal to the volumetric strain: $\Delta \rho / \rho = \Delta P / B$.

What is the dimension of Bulk Modulus?

The dimensions are the same as pressure: $[M L^{-1} T^{-2}]$.

How many cm³ are in a litre?

There are exactly $1000 \text{ cm}^3$ (or mL) in $1$ litre.

What is the Bulk Modulus of an ideal rigid body?

An ideal rigid body cannot be compressed ($\Delta V = 0$), so its Bulk Modulus is infinite.

What if the liquid was water?

Water has a Bulk Modulus of about $2.2 \text{ GPa}$, which is very close to the $2 \text{ GPa}$ given in this problem.

Is Bulk Modulus always positive?

Yes, because an increase in pressure always causes a decrease in volume for stable matter.

Can I solve this in m³ directly?

Yes, $0.8 \text{ cm}^3 = 0.8 \times 10^{-6} \text{ m}^3$. Result would be $4 \times 10^{-3} \text{ m}^3$, which is 4 litres.

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