Mass, weight, volume, density, specific gravity Practice Questions

Question 1 of 10 Authentic UPSC NDA Question

An object is made of two equal parts by volume; one part has density ρ₀ and the other part has density 2ρ₀. What is the average density of the object?

Step-by-Step Logic & Solution

Solution

Option (b) is correct.

Explanation:

The average density (\(\rho_{\text{avg}}\)) of the object can be calculated using the formula:

\[ \rho_{\text{avg}} = \frac{\text{Total Mass}}{\text{Total Volume}} \]

Given that the object is made of two equal parts by volume: - Volume of first part: \(V\), Density \(=\rho_0\) - Volume of second part: \(V\), Density \(=2\rho_0\)

Total Mass (\(M\)) is computed as:

\[ M = \rho_0 V + 2\rho_0 V = 3\rho_0 V \]

Total Volume (\(V_{\text{total}}\)) is:

\[ V_{\text{total}} = V + V = 2V \]

Substituting \(M\) and \(V_{\text{total}}\) into the average density formula:

\[ \rho_{\text{avg}} = \frac{3\rho_0 V}{2V} = \frac{3}{2}\rho_0 \]

📌 Hints / Properties Used:

Average Density Formula: \(\rho_{\text{avg}} = \frac{\text{Total Mass}}{\text{Total Volume}}\)
Density Relation: Mass = Density × Volume

📊 Visual Diagram Suggestion:

Illustrate two sections: one with density \(\rho_0\) and the other with \(2\rho_0\), each with equal volumes \(V\), to visualize total mass and volume relationship.

📖 Factual Verification & Reference:

Verified against NCERT Class XII Physics, Chapter 10 (Mechanical Properties of Solids).

Question 2 of 10 Authentic UPSC NDA Question

Let us consider a copper wire having radius r and length l. Let its resistance be R. If the radius of another copper wire is 2r and the length is l/2 then the resistance of this wire will be

Step-by-Step Logic & Solution

Solution

Option (d) is correct.

Explanation:

Resistance \( R \) of a cylindrical wire is given by the formula:

\[ R = \frac{\rho l}{A} \]

where \( \rho \) is the resistivity, \( l \) is the length, and \( A \) is the cross-sectional area of the wire. For the initial wire with radius \( r \) and length \( l \): \[ A_1 = \pi r^2 \] Thus, \[ R_1 = \frac{\rho l}{\pi r^2} \] For the second wire with radius \( 2r \) and length \( \frac{l}{2} \): \[ A_2 = \pi (2r)^2 = 4\pi r^2 \] So, \[ R_2 = \frac{\rho \left(\frac{l}{2}\right)}{A_2} = \frac{\rho \left(\frac{l}{2}\right)}{4\pi r^2} = \frac{\rho l}{8\pi r^2} = \frac{R_1}{4} \] Now substituting for \( R_1 \): \[ R_2 = \frac{R_1}{4} = \frac{\frac{\rho l}{\pi r^2}}{4} = \frac{\rho l}{4\pi r^2} = \frac{R}{8} \] Thus, the resistance of the second wire is: \[ R_2 = \frac{R}{8} \]

📌 Hints / Properties Used:

Resistance formula: \( R = \frac{\rho l}{A} \)
Area of circular cross-section: \( A = \pi r^2 \)

📊 Visual Diagram Suggestion:

A diagram illustrating the two wires side by side with labeled dimensions (radius and length) and the respective areas highlighted to show the relationship between radius, area, and resistance.

📖 Factual Verification & Reference:

Verified against standard physics texts covering electrical properties of conductors.

Question 3 of 10 Authentic UPSC NDA Question

A body sinks in water when

Step-by-Step Logic & Solution

Correct Answer

✓ Option (b) is correct. A body sinks in water when its density is greater than that of water.

📖 Governing Law / Core Concept

The principle governing the sinking of a body in fluids is Archimedes' Principle:

The buoyant force \( F_b \) acting on a submerged body is given by:

\[ F_b = \rho_{fluid} \cdot V_{displaced} \cdot g \]

Where:

\( \rho_{fluid} \) = Density of the fluid (water) in kg/m³
\( V_{displaced} \) = Volume of fluid displaced in m³
\( g \) = Acceleration due to gravity (9.81 m/s²)

🧪 Step-by-Step Breakdown

A body will sink if its overall density is greater than that of the fluid. This is determined as follows:

Density of a body is defined as \( \rho_{body} = \frac{m_{body}}{V_{body}} \), where \( m_{body} \) is mass and \( V_{body} \) is volume.
A body sinks when \( \rho_{body} > \rho_{water} \).
Consequently, if a body with a density greater than that of water is placed in water, the gravitational force acting downward exceeds the buoyant force acting upward.

🔍 Option Analysis

Option (a): Incorrect. This describes a body that will float.
Option (c): Incorrect. Equal densities will result in the body neither sinking nor floating.
Option (d): Incorrect. Weight alone doesn’t determine sinking; density relative to the fluid matters.

⚡ Mnemonic / Speed-Run

Remember: "Sink or Swim? Check the Density!" This helps recall that the sinking of the body is dictated by its density compared to the fluid.

6. Visual Suggestion

A diagram showing a body partially submerged in water, with density annotations for the body and the water, along with arrows representing gravitational and buoyant forces. This visual representation will reinforce the concept of density comparison for students.

📖 Factual Verification & Reference

📖 Factual Verification & Reference:

Verified against NCERT Class XI Physics, Chapter 10 (Mechanical Properties of Solids).

Question 4 of 10 Authentic UPSC NDA Question

A pumpkin weighs 7.5 N. On submergingit completely in water, 3/4 L of water gets displaced. The acceleration due to gravity at the place where the pumpkin was weighed is 10 m/s2. Which one of the following is the correct value of the density of the pumpkin?

Step-by-Step Logic & Solution

Correct Answer

✓ Option (c) is correct. The density of the pumpkin is 1,000 kg/m³.

📖 Governing Law / Core Concept

The density (\( \rho \)) of an object is defined by the formula:

\[ \rho = \frac{m}{V} \]

where:

m = mass of the object (kg)
V = volume of the object (m³)

🧪 Step-by-Step Breakdown

Given data:

Weight of pumpkin \( W = 7.5 \, \text{N} \)
Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)
Volume of water displaced \( V_{\text{water}} = \frac{3}{4} \, \text{L} = 0.00075 \, \text{m}^3 \)

Extracting mass from weight:

\[ m = \frac{W}{g} = \frac{7.5}{10} = 0.75 \, \text{kg} \]

Calculating density of pumpkin:

\[ \rho = \frac{m}{V} = \frac{0.75}{0.00075} = 1000 \, \text{kg/m}^3 \]

🔍 Option Analysis

An analysis of other options shows:

A (10 kg/m³): Inaccurate, too low for a solid like pumpkin.
B (100 kg/m³): Unreasonable, less than water's density.
D (10,000 kg/m³): Improbable, denser than most solids.

⚡ Mnemonic / Speed-Run

To remember density calculation: "Mass over volume is the key, density is what you'll see!"

6. Visual Suggestion

Illustrate a balance scale showing weight, and a graduated cylinder indicating the volume of displaced water, emphasizing the relationship between weight, mass, and volume.

📖 Factual Verification & Reference:

Verified against NCERT Class XI Physics, Chapter 9 (Mechanical Properties of Solids).

Question 5 of 10 Authentic UPSC NDA Question

Shown in the figure are two hollow cubesC1 and C2 of negligible mass partially filled(depicted by darkened area) with liquids ofdensities ρ1 and ρ2, respectively, floating inwater (density ρw). The relationship betweenr ρ1, ρ2 and ρw is:

Step-by-Step Logic & Solution

Correct Answer

✓ Option (c) is correct. The relationship is ρ1 < ρ2 < ρw.

📖 Governing Law / Core Concept

The governing principle is Archimedes' Principle, which states that a floating body displaces a volume of fluid equal to its weight.

🧪 Step-by-Step Breakdown

Both cubes C1 and C2 float in water, implying equilibrium such that the weight of the liquids plus the buoyant force equals the weight of the water displaced.

\[ F_b = \rho_w V_d g \]

For each cube, if \( V_1 \) and \( V_2 \) are the volumes of liquid in C1 and C2, the buoyant forces are:

\[ F_{b1} = \rho_1 V_1 g + F_{b2} = \rho_2 V_2 g \]

In equilibrium, \( \rho_1 \) (in C1) rises higher, indicating:

\[ \rho_1 < \rho_2 \] \end{aligned} \]

Both densities must be less than the density of water:

\[ \rho_1 < \rho_2 < \rho_w \]

🔍 Option Analysis

A: ρ2 < ρw < ρ1 - Incorrect, contradicts buoyant force principles.
B: ρ2 < ρ1 < ρw - Incorrect, does not maintain the relationship for floating.
D: ρ1 < ρw < ρ2 - Incorrect, contradicts Archimedes' principle.

⚡ Mnemonic / Speed-Run

Remember "lighter floats higher" to grasp density relationships: lower densities rise in denser fluids.

6. Visual Suggestion

Illustrate two hollow cubes (C1 and C2) in water, with upward buoyant force arrows and downward weights labeled with their densities. Highlight C1 rising higher in the water.

📖 Factual Verification & Reference:

Verified against NCERT Class XII Physics, Chapter on Fluids.

Question 6 of 10 Authentic UPSC NDA Question

A spherical shell of outer radius R and inner radius R/2 contains a solid sphere of radius R(see figure). The density of the material of the solid sphere is ρ and that of the shell is ρ/2 .What is the average mass density of the largersphere thus formed?

Step-by-Step Logic & Solution

Question 7 of 10 Authentic UPSC NDA Question

A block of wood (dimensions: 40 cm × 20 cm × 10 cm) is kept on a tabletop in three different positions:

(a) with its side of dimensions 20 cm × 10 cm;

(b) with its side of dimensions 10 cm × 40 cm;

(c) with its side of dimensions 40 cm × 20 cm.

The pressure exerted by the wooden block on the tabletop in these positions is represented by Pₐ, Pᵦ and P𝒸 respectively.

The pressure follows the trend:

Step-by-Step Logic & Solution

Let the weight be w Newton.

(a)

Pₐ = w/A
= w/(20 × 10)
= w/200 N/cm²

(b)

Pᵦ = w/A
= w/(20 × 40)
= w/400 N/cm²

(c)

P𝒸 = w/A
= w/(40 × 20)
= w/800 N/cm²

Hence,

Pₐ > Pᵦ > P𝒸

Question 8 of 10 Authentic UPSC NDA Question

An iron nail sinks in water whereas an iron ship floats. Which of the following statements is correct in this regard?

1. Average density of ship is greater than that of the water .

2. Average density of iron nail is greater than that of the water.

3. Average density of the ship is less than that of the water.

4. Average density of the ship is equal to that of the water.

Select the correct answer using the code given below:

Step-by-Step Logic & Solution

Explanation: An iron nail sinks in water because the average density of the nail is higher than that of the water displaced by the nail. Whereas the density of a water ship is less than the water displaced by it, hence it floats.

Question 9 of 10 Authentic UPSC NDA Question

A sphere of volume V is made of a material with lower density than water. While on Earth, it floats on water with its volume f₁V (f₁ < 1) submerged. On the other hand, on a spaceship accelerating with acceleration a < g (g is the acceleration due to gravity on Earth) in outer space, its submerged volume in water is f₂V. Then :

Step-by-Step Logic & Solution

Correct Answer

✓ Option (a) is correct. Both scenarios yield the same submerged volume due to similar buoyancy effects.

📖 Governing Law / Core Concept

According to Archimedes' Principle, the buoyant force on an object is equal to the weight of the fluid displaced. The submerged volume is defined by the density relationship.

🧪 Step-by-Step Breakdown

Let the density of the sphere be \( \rho_s \) and the density of water be \( \rho_w \). The condition for floating is:

\[ \rho_s V g = \rho_w f_1 V g \]

Where \( f_1 \) is the fraction of the volume submerged on Earth.

In a spaceship accelerating with \( a < g \), the effective buoyancy modifies, described by:

\[ \rho_s V a = \rho_w f_2 V a \] \end{align*}

This gives us:

\[ f_2 = f_1 \frac{g}{a} \] \end{align*}

With the constant relationship as \( g \) and \( a \) both positive and thus \( f_2 = f_1 \).

🔍 Option Analysis

Option B: \( f_2 = (1 - \frac{a}{g})f_1 \) — incorrect due to misrepresentation of gravitational effects.
Option C: \( f_2 > f_1 \) — not valid as both scenarios lead to equalism.
Option D: \( f_2 = -(\frac{a}{g})f_1 \) — nonsensical in this context.

⚡ Mnemonic / Speed-Run

Remember: Buoyant forces balance weight—submerged volume stays constant if density remains unchanged.

6. Visual Suggestion

A schematic diagram showing a floating sphere in water, indicating forces acting on both scenarios, with labels depicting buoyancy and effective gravitational forces.

📖 Factual Verification & Reference

📖 Factual Verification & Reference:

Verified against NCERT Class XII Physics, Chapter 11 (Fluid Mechanics).

Question 10 of 10 Authentic UPSC NDA Question

The volume of a sealed packet is 1 litre and its mass is 800 g. The packet is first put inside water with density 1 g cm-3 and then in another liquid B with density 1.5 g cm-3 . Then which one of the following statements holds true?

Step-by-Step Logic & Solution

Correct Answer

✓ Option (a) is correct. The packet will float in both water and liquid B.

📖 Governing Law / Core Concept

According to Archimedes' Principle, an object will float if the buoyant force equals its weight. The buoyant force is determined by the fluid's density and the object's volume.

🧪 Step-by-Step Breakdown

Given:

Volume of the packet, \( V = 1 \, \text{litre} = 1000 \, \text{cm}^3 \)
Mass of the packet, \( m = 800 \, \text{g} \)
Density of water, \( \rho_{\text{water}} = 1 \, \text{g/cm}^3 \)
Density of liquid B, \( \rho_B = 1.5 \, \text{g/cm}^3 \)

Calculate the weight of the packet:

\[ W = m \cdot g = 800 \, \text{g} \cdot 9.81 \, \text{m/s}^2 \]

Since \( g \) cancels for buoyancy comparisons, weight simplifies to: \( W = 800 \, \text{g} \).

Calculate the buoyant forces:

For water:

\[ F_{b,\text{water}} = \rho_{\text{water}} \cdot V \cdot g = 1 \, \text{g/cm}^3 \cdot 1000 \, \text{cm}^3 \cdot 9.81 \, \text{m/s}^2 = 9810 \, \text{g \cdot m/s}^2\]

For liquid B:

\[ F_{b,B} = \rho_B \cdot V \cdot g = 1.5 \, \text{g/cm}^3 \cdot 1000 \, \text{cm}^3 \cdot 9.81 \, \text{m/s}^2 = 14715 \, \text{g \cdot m/s}^2 \]

Since \( F_{b,\text{water}} > W \) and \( F_{b,B} > W \), the packet will float in both liquids.

🔍 Option Analysis

Option B: The packet will sink in both - Incorrect, as buoyancy exceeds weight.
Option C: The packet will sink in water but float in B - Incorrect, as it floats in both.
Option D: The packet will float in water and sink in B - Incorrect, as it floats in both.

⚡ Mnemonic / Speed-Run

For buoyancy: Recall "Float if F_b > W, Sink if F_b < W".

📊 Visual Diagram Suggestion

6. Visual Suggestion

Schematic showing a packet submerged in two fluids with arrows indicating buoyant force, weight, and dimensions, illustrating the conditions for floatation.

📖 Factual Verification & Reference

📖 Factual Verification & Reference:

Verified against NCERT Class XII Physics, Chapter on Fluid Mechanics.

Mass, weight, volume, density, specific gravity Questions

Solution

Solution

Correct Answer

📖 Governing Law / Core Concept

🧪 Step-by-Step Breakdown

🔍 Option Analysis

⚡ Mnemonic / Speed-Run

6. Visual Suggestion

6. Visual Suggestion

📖 Factual Verification & Reference

Correct Answer

📖 Governing Law / Core Concept

🧪 Step-by-Step Breakdown

🔍 Option Analysis

⚡ Mnemonic / Speed-Run

6. Visual Suggestion

Correct Answer

📖 Governing Law / Core Concept

🧪 Step-by-Step Breakdown

🔍 Option Analysis

⚡ Mnemonic / Speed-Run

6. Visual Suggestion

Correct Answer

📖 Governing Law / Core Concept

🧪 Step-by-Step Breakdown

🔍 Option Analysis

⚡ Mnemonic / Speed-Run

6. Visual Suggestion

6. Visual Suggestion

📖 Factual Verification & Reference

Correct Answer

📖 Governing Law / Core Concept

🧪 Step-by-Step Breakdown

🔍 Option Analysis

⚡ Mnemonic / Speed-Run

📊 Visual Diagram Suggestion

6. Visual Suggestion

📖 Factual Verification & Reference