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Unit 7: Properties of Solids and Liquids — Online MCQ Test

Q: A wire of length L, area of cross section A is hanging from a fixed support. The length of the wire changes to L_1 when mass M is suspended from its free end. The expression for Young's modulus is:

Answer: B. MgL / A(L_1-L) — Young's Modulus Y = (Stress) / (Strain) = (F/A) / (ΔL/L). Here F = Mg, ΔL = (L_1 - L). Thus, Y = (Mg/A) / ((L_1 - L)/L) = MgL / A(L_1 - L).

Q: A capillary tube of radius r is immersed in water and water rises in it to a height h. The mass of the water in the capillary is 5 g. Another capillary tube of radius 2r is immersed in water. The mass of water that will rise in this tube is:

Answer: A. 5.0 g — Height of water column h = 2T cos(θ) / (rρg). Mass m = πr²hρ = πr²ρ(2T cos(θ) / (rρg)) = (2πT cos(θ) / g) * r. Since m is proportional to r, doubling r to 2r should double the mass to 10 g? No, mass m = πr²hρ. Since h = C/r, then m = πr²(C/r)ρ = C'r. Wait, if m1=5g at radius r, and mass is proportional to r, then at 2r, mass is 10g.

Q: An element has a body centered cubic (bcc) structure with a cell edge of 288 pm. The atomic radius is:

Answer: D. (sqrt(3)/4) * 288 pm — For a BCC structure, the relationship between edge length 'a' and radius 'r' is 4r = sqrt(3)a, so r = (sqrt(3)/4)a.

Q: Consider a water tank shown in the figure. It has one wall at x = L and can be taken to be very wide in the z direction. When filled with a liquid of surface tension S and density ρ, the liquid surface makes angle θo (θo << 1) with the x-axis at x = L. If y(x) is the height of the surface then the equation for y(x) is: (take θ(x) ≈ sin θ(x) ≈ tan θ(x) = dy/dx, g is the acceleration due to gravity)

Answer: B. (d^2y / dx^2) = (ρg/S)x — Based on the Young-Laplace equation for surface tension, the pressure difference across a curved interface is ΔP = S(d^2y/dx^2). For a liquid in a gravity field, ΔP = ρgy. Thus, S(d^2y/dx^2) = ρgy.

Q: Lithium has a bcc structure. Its density is 530 kg m^–3 and its atomic mass is 6.94 g mol^–1. Calculate the edge length of a unit cell of Lithium metal. (NA = 6.02 × 10^23 mol^–1)

Answer: A. 352 pm — Using the formula d = (Z * M) / (a^3 * NA), for BCC Z=2. Solving for a gives approximately 352 pm.

Q: The ionic radii of A+ and B- ions are 0.98 × 10^-10 m and 1.81 × 10^-10 m. The coordination number of each ion in AB is:

Answer: D. 6 — Radius ratio = (0.98 / 1.81) ≈ 0.541. Since 0.414 < 0.541 < 0.732, the coordination number is 6 (octahedral geometry).

Q: Two non-mixing liquids of densities ρ and nρ (n > 1) are put in a container. The height of each liquid is h. A solid cylinder of length L and density d is put in this container. The cylinder floats with its axis vertical and length pL (p < 1) in the denser liquid. The density d is equal to :

Answer: C. {1 + (n – 1) p} ρ — Using the principle of flotation (weight of cylinder = buoyant force from both liquids), LAdg = (1-p)LAρg + pLAnρg. Simplifying gives d = ρ{1 + (n-1)p}.

Q: Two non-mixing liquids of densities ρ and nρ (n > 1) are put in a container. The height of each liquid is h. A solid cylinder of length L and density d is put in this container. The cylinder floats with its axis vertical and length pL (p < 1) in the denser liquid. The density d is equal to:

Answer: C. {1 + (n – 1) p} ρ — Equating the weight of the cylinder to the buoyancy forces provided by the two liquids, we have LAdg = (1 – p) LAρg + pLAnρg. Solving for d gives d = ρ{1 + (n – 1)p}.

Q: The maximum elongation of a steel wire of 1 m length if the elastic limit of steel and its Young’s modulus, respectively, are 8 × 10^8 N m^-2 and 2 × 10^11 N m^-2, is:

Answer: D. 0.4 mm — Maximum strain = elastic limit / Young’s modulus = (8×10^8)/(2×10^11) = 4×10^-3. For 1 m wire, extension = 4×10^-3 m = 4 mm. The correct option is 4 mm.

Q: A wire of length L, area of cross section A is hanging from a fixed support. The length of the wire changes to L₁ when mass M is suspended from its free end. The expression for Young's modulus is:

Answer: B. MgL / (A(L₁ − L)) — Young's modulus Y = Stress/Strain = (F/A) / (ΔL/L) = (Mg/A) / ((L₁ − L)/L) = MgL / (A(L₁ − L)).

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Practice Unit 7: Properties of Solids and Liquids with a free chapter-wise online MCQ test for NEET NEET (All) PHYSICS. This chapter covers: Elastic behaviour Stress-strain relationship Hooke's Law. Young's modulus bulk modulus modulus of rigidity. Pressure due to a fluid column; Pascal's law and its applications. Effec.... AI-generated questions from basic to board-exam level, with instant results and explanations.

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Unit 7: Properties of Solids and Liquids — Important Questions & Answers (FAQ)

Frequently asked questions from NEET NEET (All) PHYSICS — Unit 7: Properties of Solids and Liquids, with answers and explanations. These are sample questions; the exam has its own separate question set.

A wire of length L, area of cross section A is hanging from a fixed support. The length of the wire changes to L_1 when mass M is suspended from its free end. The expression for Young's modulus is:

A. Mg(L_1-L) / AL
B. MgL / A(L_1-L) ✓
C. MgL_1 / AL
D. MgL / AL_1

Answer: B. MgL / A(L_1-L)
Young's Modulus Y = (Stress) / (Strain) = (F/A) / (ΔL/L). Here F = Mg, ΔL = (L_1 - L). Thus, Y = (Mg/A) / ((L_1 - L)/L) = MgL / A(L_1 - L).

A capillary tube of radius r is immersed in water and water rises in it to a height h. The mass of the water in the capillary is 5 g. Another capillary tube of radius 2r is immersed in water. The mass of water that will rise in this tube is:

A. 5.0 g ✓
B. 10.0 g
C. 20.0 g
D. 2.5 g

Answer: A. 5.0 g
Height of water column h = 2T cos(θ) / (rρg). Mass m = πr²hρ = πr²ρ(2T cos(θ) / (rρg)) = (2πT cos(θ) / g) * r. Since m is proportional to r, doubling r to 2r should double the mass to 10 g? No, mass m = πr²hρ. Since h = C/r, then m = πr²(C/r)ρ = C'r. Wait, if m1=5g at radius r, and mass is proportional to r, then at 2r, mass is 10g.

An element has a body centered cubic (bcc) structure with a cell edge of 288 pm. The atomic radius is:

A. (sqrt(2)/4) * 288 pm
B. (4/sqrt(3)) * 288 pm
C. (a/2) * 288 pm
D. (sqrt(3)/4) * 288 pm ✓

Answer: D. (sqrt(3)/4) * 288 pm
For a BCC structure, the relationship between edge length 'a' and radius 'r' is 4r = sqrt(3)a, so r = (sqrt(3)/4)a.

Consider a water tank shown in the figure. It has one wall at x = L and can be taken to be very wide in the z direction. When filled with a liquid of surface tension S and density ρ, the liquid surface makes angle θo (θo << 1) with the x-axis at x = L. If y(x) is the height of the surface then the equation for y(x) is: (take θ(x) ≈ sin θ(x) ≈ tan θ(x) = dy/dx, g is the acceleration due to gravity)

A. (d^2y / dx^2) = (ρg/S)y
B. (d^2y / dx^2) = (ρg/S)x ✓
C. (d^2y / dx^2) = (ρg/S)
D. (dy / dx) = (ρg/S)x

Answer: B. (d^2y / dx^2) = (ρg/S)x
Based on the Young-Laplace equation for surface tension, the pressure difference across a curved interface is ΔP = S(d^2y/dx^2). For a liquid in a gravity field, ΔP = ρgy. Thus, S(d^2y/dx^2) = ρgy.

Lithium has a bcc structure. Its density is 530 kg m^–3 and its atomic mass is 6.94 g mol^–1. Calculate the edge length of a unit cell of Lithium metal. (NA = 6.02 × 10^23 mol^–1)

A. 352 pm ✓
B. 527 pm
C. 264 pm
D. 154 pm

Answer: A. 352 pm
Using the formula d = (Z * M) / (a^3 * NA), for BCC Z=2. Solving for a gives approximately 352 pm.

The ionic radii of A+ and B- ions are 0.98 × 10^-10 m and 1.81 × 10^-10 m. The coordination number of each ion in AB is:

A. 4
B. 8
C. 2
D. 6 ✓

Answer: D. 6
Radius ratio = (0.98 / 1.81) ≈ 0.541. Since 0.414 < 0.541 < 0.732, the coordination number is 6 (octahedral geometry).

Two non-mixing liquids of densities ρ and nρ (n > 1) are put in a container. The height of each liquid is h. A solid cylinder of length L and density d is put in this container. The cylinder floats with its axis vertical and length pL (p < 1) in the denser liquid. The density d is equal to :

A. {2 + (n + 1) p} ρ
B. {2 + (n – 1) p} ρ
C. {1 + (n – 1) p} ρ ✓
D. {1 + (n + 1) p} ρ

Answer: C. {1 + (n – 1) p} ρ
Using the principle of flotation (weight of cylinder = buoyant force from both liquids), LAdg = (1-p)LAρg + pLAnρg. Simplifying gives d = ρ{1 + (n-1)p}.

A. {2 + (n + 1) p} ρ
B. {2 + (n – 1) p} ρ
C. {1 + (n – 1) p} ρ ✓
D. {1 + (n + 1) p} ρ

Answer: C. {1 + (n – 1) p} ρ
Equating the weight of the cylinder to the buoyancy forces provided by the two liquids, we have LAdg = (1 – p) LAρg + pLAnρg. Solving for d gives d = ρ{1 + (n – 1)p}.

The maximum elongation of a steel wire of 1 m length if the elastic limit of steel and its Young’s modulus, respectively, are 8 × 10^8 N m^-2 and 2 × 10^11 N m^-2, is:

A. 40 mm
B. 8 mm
C. 4 mm
D. 0.4 mm ✓

Answer: D. 0.4 mm
Maximum strain = elastic limit / Young’s modulus = (8×10^8)/(2×10^11) = 4×10^-3. For 1 m wire, extension = 4×10^-3 m = 4 mm. The correct option is 4 mm.

A wire of length L, area of cross section A is hanging from a fixed support. The length of the wire changes to L₁ when mass M is suspended from its free end. The expression for Young's modulus is: