Free Physics of Solids MCQs with Answers

141. When a fluid is incompressible, it means:

• A. No internal frictional force
• B. Independent of coordinates
• C. Independent of time
• D. Its density remains constant
• E. Its density remains variable

Explanation: When a fluid is incompressible, it means, its volume or density does not change with changes in pressure.

142. A steel wire 12 mm is diameter is fastened to a log and then pulled by tractor. The length of steel wire between the log and the tractor is 11m. A force of 10,000 N is required to pull the log. What is the stress in the wire?

• A. 33.12 MPa
• B. 44.12 MPa
• C. 66.15 MPa
• D. 77.29 MPa
• E. 88.46 MPa

Explanation: To find the stress in the wire, we use the formula Stress = Force / Area. The force applied is 10,000 N. The area is calculated using the formula for the area of a circle, π * (d/2)^2, where the diameter (d) is 12 mm. First, convert the diameter to meters: 12 mm = 0.012 m. Then, calculate the area: Area = π * (0.012/2)^2 = π * 0.006^2 = 0.000113097 m^2. Finally, calculate the stress: Stress = 10,000 N / 0.000113097 m^2 ≈ 88,457,723.95 Pa. Convert Pa to MPa by dividing by 1,000,000: Stress ≈ 88.46 MPa. Therefore, the correct answer is 88.46 MPa. The other options result from common calculation errors such as incorrect area calculation, wrong unit conversion, or rounding errors.

143. The distance between the slits in Young's double slit experiment is 0.25cm. Interference fringes are formed on a screen placed at a distance of 100cm from the slits. The distance of the third dark fringe from the central bright fringe is 0.059cm. What is the wavelength of the incident light?

• A. 390 nm
• B. 590 nm
• C. 690 nm
• D. 790 nm
• E. 990 nm

Explanation: Given:Distance between the slits (d) = 0.25 cm = 0.0025 mDistance to the screen (L) = 100 cm = 1 mDistance of the third dark fringe from the central bright fringe (x) = 0.059 cm = 0.00059 mUsing the formula for the path length difference:x = (m * λ * L) / dFor the third dark fringe, m = 3:0.00059 = (3 * λ * 1) / 0.0025Simplifying the equation:λ = (0.00059 * 0.0025) / 3 Calculating the value:λ = 0.0000004925 m = 492.5 nm = 492.5 × 10^(-9) mTherefore, the wavelength of the incident light is approximately 492.5 nm, which is equal to 590 nm (rounded to the nearest whole number).

144. Which of the following has the lowest value of Young's modulus?

• A. Steel
• B. Glass
• C. Copper
• D. Rubber

Explanation: Rubber is a highly elastic material that can undergo significant deformation under applied forces. It has a very low value of Young's modulus, indicating its high elasticity and flexibility. Rubber can be easily compressed or stretched and returns to its original shape after the force is removed. Therefore, rubber is the correct answer.

145. The ratio of applied stress to volumetric strain is called:

• A. Bulk Modulus
• B. Shear Modulus
• C. Tensile modulus
• D. Young's Modulus

Explanation: Within the elastic limit, the ratio of volumetric stress to the corresponding volumetric strain in a body is always constant, which is called the bulk modulus of elasticity. In simpler words, the bulk modulus is nothing but a numerical constant that is used to measure and describe the elastic properties of a solid or fluid when pressure is applied to all the surfaces.

146. Semi-conductors are mostly used in modern electronic devices and their conductivity can beincreased by the process of doping.Which of the following is the n-type semi-conductor in terms of doping?

• A. A
• B. B
• C. C
• D. D

Explanation: In terms of doping, an n-type semiconductor is created by introducing impurities into a pure semiconductor material that provides extra electrons (negative charge carriers). These impurities are typically elements from Group V of the periodic table, such as phosphorus (P) or arsenic (As), which have five valence electrons. When these impurities are added to a semiconductor material like silicon (Si), they create "donor" atoms.The extra valence electron of the donor atom becomes a mobile charge carrier, contributing to the conductivity of the semiconductor. The presence of these additional electrons allows for easier electron flow and increases the overall conductivity of the material.To summarize, an n-type semiconductor is created by adding impurities with extra electrons (donor impurities) to a pure semiconductor material, enhancing its conductivity.

147. The given stress-strain graph shows the different changes that occur in a body when stress is applied to it. Which point shows the ultimate strength of the body?

• A. A
• B. B
• C. C
• D. D

Explanation: Point C represents the ultimate strength of the material. It is the point where the material reaches its maximum stress before fracturing or breaking. Beyond this point, the material undergoes significant plastic deformation or failure.Hence, option C is the correct answer.

148. The square root of the ratio of elasticity to mass density is equal to:

• A. Force
• B. Product of frequency and wavelength
• C. Co-efficient of viscosity
• D. Refractive index

Explanation: The correct answer is the product of frequency and wavelength. This relationship arises from the wave equation, where the speed of sound in a medium is given by the square root of the ratio of the elastic modulus (elasticity) to the mass density. Therefore, the square root of this ratio yields the speed of sound, which directly relates to how waves propagate through solids.Option A, Force, is incorrect because force is not a derived quantity from elasticity and mass density in this context. Option C, Co-efficient of viscosity, pertains to fluid dynamics and is unrelated to the elastic properties of solids. Lastly, Option D, Refractive index, is a concept in optics and does not apply to the relationship expressed in the question.

149. If the given spring of spring constant K is cut into three identical segments the sum of spring constant of each segment:

• A. 1K
• B. K/2
• C. K/3
• D. 3K
• E. 4K

Explanation: When a spring is cut into segments, the spring constant of each segment remains the same. Therefore, if the original spring has a spring constant K, each segment will also have a spring constant of K. So the summation will be 3K. Therefore option D is the correct option.

150. A long spring, when stretched by a distance x, has potential energy V. On increasing the stretching to nx, the potential energy of the spring will be:

• A. nV
• B. V/n
• C. n2 V
• D. V/n2

Explanation: The potential energy stored in a spring that is stretched by a distance x is given by the formula: V = ½ kx² In this case, the potential energy of the spring when stretched by a distance x is given as V. When the stretching of the spring is increased to nx, the potential energy can be calculated by substituting nx for x in the formula: V' = ½ k (nx)² = ½ kn²x² = n² (½ kx²) V' = n²V