Free Physics of Solids MCQs with Answers

161. A body floats in a liquid contained in a beaker. The whole system, as shown, falls freely under gravity. The upthrust on the body due to the liquid is:

• A. Zero
• B. Equal to the weight of the liquid displaced
• C. Equal to the weight of the body in air
• D. Equal to the weight of the immersed position of the body

Explanation: This is the correct answer according to Archimedes' principle. The upthrust on a body submerged in a liquid is equal to the weight of the liquid displaced by the body. This is a fundamental principle in fluid mechanics.

162. What is the Young's modulus of elasticity for a perfectly rigid body?

• A. Infinity
• B. Zero
• C. 1
• D. -1

Explanation: Young's modulus measures the stiffness or rigidity of a material, but it doesn't apply to perfectly rigid bodies, and its value does not extend to infinity. Since strain is zero therefore Y is infinite.

163. Two stretched membranes of area 2 cm2 and 3 cm2 are placed in a liquid at the same depth. The ratio of pressures on them is:

• A. 1 : 1
• B. 2 : 3
• C. 3 : 2
• D. 22 : 32

Explanation: This is the correct answer. According to Pascal's law, when two stretched membranes are placed at the same depth in a liquid, the pressure on them is determined by the depth and the density of the liquid, but it's not influenced by the areas of the membranes. Therefore, the ratio of the pressures on the two membranes is 1:1, meaning they experience the same pressure. Pressure is independent of area of cross section.

164. Water enters through end A with speed v1 and leaves through end B with speed v2 of a cylindrical tube AB. The tube is always completely filled with water. In case I the tube is horizontal and in case II it is vertical with end A upwards and in case III it is vertical with end B upwards. We have v1=v2 for:

• A. Case I
• B. Case II
• C. Case III
• D. Each case

Explanation: The correct option is D. each case. Take any orientation of cylindrical tube [uniform cross section] Now, in any orientation the continuity equation is valid. Therefore, A1v1=A2v2 But A1=A2 ⇒v1=v2 Hence option (D) is correct. This happens in accordance with the equation of continuity and this equation was derived on the principle of conservation of mass and it is true in every case, whether the tube remains horizontal or vertical.

165. Two solid spherical balls of radius r1 & r2 (r2 < r1) and density σ are tied up with a string and released in a viscous liquid of lesser density ρ and coefficient of viscosity η, with the string just taut as shown. The terminal velocity of the spheres is:

• A. Option A: vt = (2/9) * (σ - ρ) * g * r1² / η
• B. Option B: vt = (2/9) * (σ - ρ) * g * r2² / η
• C. Option C: vt = (2/9) * (σ - ρ) * g * (r1² + r2²) / (2 * η)
• D. Option D: vt = (4/9) * (σ - ρ) * g * r1 * r2 / η

Explanation: The correct answer is Option C. Since the two spheres are tied together and moving through a viscous fluid, their terminal velocity depends on the combined effect of their radii. The terminal velocity formula for a sphere moving through a viscous fluid is modified to account for both spheres. At terminal velocity net force is zero.6πη(r₁+r₂)V┬+⅔π(r₁³+r₂³)ρg = ⅔π(r₁³+r₂³)σgOption A and Option B are incorrect as they only consider one sphere each, ignoring the connection between them. Option D incorrectly combines the radii, which does not reflect the physics of the situation.

166. Two elastic rods are joined between fixed supports as shown in the figure. The condition for no change in the lengths of individual rods with the increase of temperature is: (α1,α2 = linear expansion coefficient, A1, A2 = area of rods, Y1,Y2 = Young's modulus)

• A. A1/A2 = (α1 * Y1) / (α2 * Y2)
• B. A1/A2 = (L1 * α1 * Y1) / (L2 * α2 * Y2)
• C. A1/A2 = (L2 * α2 * Y2) / (L1 * α1 * Y1)
• D. A1/A2 = (α2 * Y2) / (α1 * Y1)

Explanation: The correct answer is d) A1/A2 = (α2 * Y2) / (α1 * Y1)For no change in length, the compressive force in both rods should be equal:A1 * Y1 * α1 * ΔT = A2 * Y2 * α2 * ΔT=> A1/A2 = (α2 * Y2) / (α1 * Y1)

167. A wire of length '2m' is clamped horizontally between two fixed supports. A mass m = 5 kg is hung from the middle of the wire. The vertical depression in the wire in equilibrium is: (Young's modulus of wire = 2.4 x 10^9 N/m2, cross-sectional area = 1 cm2)

• A. 4.68 cm
• B. 1.52 cm
• C. 1.12 cm
• D. 0.58 cm

Explanation: The vertical depression in the wire can be calculated using the formula for the depression of a wire with a central load: δ = (W * L3) / (48 * Y * I), where W is the weight of the mass, L is the length of the wire, Y is Young's modulus, and I is the moment of inertia (I = A2 / 4π for a circular cross-section). Using the given parameters, the correct depression is found to be 4.68 cm.equation 2T sin θ = mg⇒ 2(YA/a) x sin θ. sin θ = mg⇒ 2YA/a x. x²/a² = mg⇒ x = [a³mg / 2YA](1/3) = [1m × 5kg × 10m/s² / 2 × (2.4 × 10⁹ N/m²) × 10⁻⁴ m²](1/3) = 4.68 cmThe other options are incorrect as they do not align with this calculation using the provided values.

168. A gas undergoes a process in which the pressure and volume are related by VPn = constant. The bulk modulus of the gas is :

• A. nP
• B. P1/n
• C. P/n
• D. Pn

Explanation: The bulk modulus (B) of a gas is defined as the negative of the volume change ratio to the pressure change, given by B = -V(dp/dV). For processes where VPn = constant, we differentiate this equation to find that dp/dV = -nP/V. Substituting this into the formula for bulk modulus, we find B = P/n. The other options do not fit this derivation:VPⁿ = (V + ΔV)(P + ΔP)ⁿVPⁿ = VPⁿ (1 + ΔV/V)(1 + nΔP/P)∴ ΔV/V = -n ΔP/PK = - ΔP / (ΔV/V) = P/nOption A (nP): This form is unrelated to the bulk modulus expression.Option B (P1/n): This does not match the derived formula for the bulk modulus.Option D (Pn): This option confuses the multiplication of P and n with their division.

169. The adjacent graph shows the extension (∆l) of a wire of length λ m suspended from the top of a roof at one end and with a load W connected to the other end. If the cross-sectional area of the wire is 10-6 m2, calculate the Young's modulus of the material of the wire:

• A. 2 x 10^11 λ N/m2
• B. 2 x 10^-11 λ N/m2
• C. 3 x 10^12 λ N/m2
• D. 2 x 10^13 λ N/m2

Explanation: To calculate Young's modulus (Y), use the formula Y = Stress/Strain. Here, Stress = Force (F) / Area (A) and Strain = Extension (Δl) / Original Length (λ). From the graph, you get the values needed for these parameters:Stress = F/A = W / 10-6 m2 and Strain = Δl / λ.Insert these into the Young's modulus formula:Y = (W / 10-6) / (Δl / λ) = (W * λ) / (Δl * 10-6).Using the data, calculate to get Y = 2 × 1011 λ N/m2.Option A is correct because it accurately applies the formula and concepts. Options B, C, and D are incorrect due to miscalculations or misinterpretations of the data or formula.

170. Two wires of equal length and cross-section are suspended as shown. Their Young's moduli are Y1 and Y2, respectively. The equivalent Young's modulus will be:

• A. Y1 + Y2
• B. (Y1 + Y2) / 2
• C. (Y1 * Y2) / (Y1 + Y2)
• D. √(Y1 * Y2)

Explanation: The correct answer is Option B: (Y1 + Y2) / 2. Option A is incorrect because it implies a linear addition of moduli, which is applicable in parallel arrangements, not series. Option C is incorrect because it reflects a division of stress in a different context. Option D shows the geometric mean.