Free Physics of Solids MCQs with Answers

111. If S is strain and Y is Young's modulus of elasticity of wirematerial, then energy stored in the wire per unit volume is:

• A. s²/2y
• B. 2y/s²
• C. s/2y
• D. 2s²y

Explanation: The energy stored per unit volume in a wire under elasticdeformation is given by the formula U = (1/2) ×Y ×S2, whereU is the energy per unit volume, Y is Young's modulus, and Sis the strain. Rearranging this formula gives U = S2 / (2Y).The correct answer is thus s² / 2y.

112. Breaking strength of a wire is 105 N if its diameter is halved,what is the strength?

• A. 4 x 10^5
• B. 2 x 10^5
• C. 1 x 10^5
• D. None of these

Explanation: The breaking strength of a wire is directly proportional to itscross-sectional area. When the diameter of the wire ishalved, the cross-sectional area is reduced by a factor offour (since the area is proportional to the square of thediameter). Therefore, the breaking strength also reduces bya factor of four, resulting in a new breaking strength of 1×105N.

113. 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: (α₁α₂ = linear expansion coefficient, A₁ A₂ = area of rods, Y₁Y₂ = Young's modulus)

• A. A
• B. B
• C. C
• D. D

Explanation: The correct condition for no change in the lengths of theindividual rods when the temperature increases is given bythe equation A1Y1α1 = A2Y2α2. This equation ensures thatthe forces due to thermal expansion in both rods arebalanced. If this condition is met, the rods will notexperience any net change in length, as the expansion ofone rod will be perfectly countered by the expansion of theother.Other options fail to correctly balance these forces becausethey either swap the Young's modulus values, ignore thecross-sectional area, or omit the Young's modulusaltogether, leading to an incorrect assessment of the forces involved.

114. The adjacent graph shows the extension (Δl) of a wire of length 1 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. A. 2 x 10^11 Pa
• B. B. 1 x 10^11 Pa
• C. C. 3 x 10^11 Pa
• D. D. 5 x 10^10 Pa

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 getthe 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 × 1011λ N/m2 Option A is correct because it accurately applies the formulaand concepts. Options B, C, and D are incorrect due tomiscalculations or misinterpretations of the data or formula.

115. 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 would be:

• A. Y₁ + Y₂
• B. Y₁ + Y₂ / 2
• C. Y₁Y₂ / (Y₁ + Y₂)
• D. under root Y₁Y₂

Explanation: The correct answer is Option B: 2(Y1Y2)/(Y1 + Y2). In ascenario where two wires are of equal length and cross-section, the effective or equivalent Young's modulus whenthey are arranged in series is calculated using the formulafor the harmonic mean of the two moduli. This accounts forthe distribution of stress and strain across both wires.Option A is incorrect because it implies a linear addition ofmoduli, which is applicable in parallel arrangements, notseries. Option C is incorrect because it reflects a division ofstress in a different context. Option D suggests averagingthe moduli, which does not apply here.

116. In the case of linear deformation, the ratio of tensile stress to tensile strain is called:

• A. Energy stored in a stretched wire
• B. Young's double slit phenomenon
• C. Bulk Modulus
• D. Young's Modulus

Explanation: The Young's modulus (E) is a property of the material that tells us how easily it can stretch and deform and is defined as the ratio of tensile stress (σ) to tensile strain (ε). Where stress is the amount of force applied per unit area (σ = F/A) and strain is extension per unit length (ε = dl/l).

117. Electric conduction is high in:

• A. Solid nuclei
• B. Sugar solution
• C. Solid graphite
• D. None

Explanation: The conduction of electricity depends on the movement of electrons and other charged particles (ions). Metals have free electrons and thus can conduct electricity in a solid state but as the temperature increases, the collision between the electrons also increases, and thus the resistance to the flow of current increases. Liquids can conduct electricity if they are ionic (liquid mercury is metal and hence the conduction is by the same mechanism as for metals). Salt solutions, molten salts, etc, conduct electricity. Gases are insulators and don't conduct electricity unless the ionization potential is exceeded. Lightning is an example of the breakdown of the insulation of air.

118. If a substance can undergo plastic deformation, until it breaks, it is:

• A. Ductile substance
• B. Brittle substance
• C. Crystalline substance
• D. Polymeric substance

Explanation: Substances that elongate considerably and undergo plastic deformation before they break are known as ductile substances (A).

119. If stress is applied to a body then the ratio of change in volume to original volume will be:

• A. Polymeric strain
• B. Volumetric strain
• C. Parallel strain
• D. Tensile strain

Explanation: The ratio of the change of volume of the body to the original volume is known as volumetric strain. The strain produced by shear stress is known as shear strain. The stress may be normal stress or shear stress.

120. To convert the Si crystal into p-type semi-conductor, which group element will be doped:

• A. Trivalent Element
• B. Second Group Element
• C. Fourth Group Element
• D. Pentavalent Element

Explanation: Explanation:To convert a silicon (Si) crystals into a p-type semi-conductor, it needs to be doped with a trivalent element, which means an element that has three velence electrons in its outermost shell. This is because the Trivalent impurity atom will create a hole in the valence band, which behaves likes a positively charged particle. Second group elements have two valence electrons, so they cannot be used to create holes in the valence band. They are used to create n-type semiconductors by introducing an extra electron to the conduction band. Fourth group elements have four valence electrons, so they cannot be used to create holes in the valence band. This option is also incorrect. Pentavalent elements have five valence electrons, which means they can be used to create n-type semiconductors by providing an extra electron to the conduction band.