Unit 7: Properties of Solids and Liquids — Online MCQ Test
PHYSICS · NEET (All) · NEET
Practice Unit 7: Properties of Solids and Liquids with a free chapter-wise online MCQ test.
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....
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Unit 7: Properties of Solids and Liquids — Important Questions & Answers
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).
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.
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.
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.
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.
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).
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}.
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}.
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
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}.
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.
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:
- A. MgL / (AL₁)
- B. MgL / (A(L₁ − L))
- C. MgL₁ / (AL)
- D. Mg(L₁ − L) / (AL)
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)).
Young's modulus Y = Stress/Strain = (F/A) / (ΔL/L) = (Mg/A) / ((L₁ − L)/L) = MgL / (A(L₁ − L)).