Q: A metal cube of side $ 10 \mathrm{cm} $ is subjected to a shearing stress of $ 10^{4} N m^{-2} $. The modulus of rigidity if the top of the cube is displaced by $ 0.05 \mathrm{cm} $ with respect to its bottom is

$ 2 \times 10^{6} \mathrm{Nm}^{-2} $. Shear strain is the relative displacement divided by the height of the cube: $0.05\text{ cm}/10\text{ cm} = 0.005$. The modulus of rigidity is $\tau/\gamma = 10^4/0.005 = 2\times10^6\,\text{N m}^{-2}$.

Question 1

Assertion (A): Ductile metals are used to prepare thin wires. Reason (R) : In the stress-strain curve of ductile metals, the length between the points representing elastic limit and breaking point is very small.

Accepted Answer

(A) is true but (R) is false.. Ductile metals are used for thin wires because they can be drawn out extensively without breaking. The reason is false: ductile metals typically have a large plastic region between the elastic limit and breaking point, not a very small one.

Question 2

To determine the Young modulus of a wire, several measurements are taken. In which row can the measurement not be taken directly with the stated apparatus?

Accepted Answer

measurement: area of cross-section of wire apparatus : micrometer screw gauge. A micrometer screw gauge measures the wire’s diameter, not its cross-sectional area directly. The area must be calculated from the diameter using the formula for a circle.

Question 3

The property to restore the natural shape or to oppose the deformation is called:

Accepted Answer

elasticity. Elasticity is the ability of a material to regain its original shape after the deforming force is removed. Plasticity and ductility describe permanent deformation behaviors, so they do not fit.

Question 4

Assertion For small deformations, the stress and strain are proportional to each other Reason A class of solids called elastomers does not obey Hooke's law.

Accepted Answer

Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion. The assertion is true because, for sufficiently small deformations, many solids show a linear stress-strain relationship. The reason is also true since elastomers often have nonlinear behavior, but that does not explain why small deformations are proportional; it only gives an exception to Hooke’s law.

Question 5

A metal cube of side $ 10 \mathrm{cm} $ is subjected to a shearing stress of $ 10^{4} N m^{-2} $. The modulus of rigidity if the top of the cube is displaced by $ 0.05 \mathrm{cm} $ with respect to its bottom is

Accepted Answer

$ 2 \times 10^{6} \mathrm{Nm}^{-2} $. Shear strain is the relative displacement divided by the height of the cube: $0.05\text{ cm}/10\text{ cm} = 0.005$. The modulus of rigidity is $\tau/\gamma = 10^4/0.005 = 2\times10^6\,\text{N m}^{-2}$.

Question 6

A metal wire upon excess stress moves to a region of permanent set. Before yielding to the fracture stress, it undergoes an extension equal to twice its length in its plastic region. The nature of the metal is

Accepted Answer

Ductile. A ductile metal undergoes substantial plastic deformation after yielding, so it can extend a lot before fracture. Brittle materials fracture with little plastic strain, while perfectly elastic or perfectly plastic are idealized behaviors that do not match this description.

Question 7

The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied?

Accepted Answer

Length = 100 cm, diameter = 1 mm. The extension of a wire under tension is proportional to its length and inversely proportional to the square of its diameter. Wire A has the smallest diameter and a significant length, leading to the largest extension.

Question 8

When a block of mass M is suspended by a long wire of length L, the length of the wire becomes (L + l). The elastic potential energy stored in the extended wire is:

Accepted Answer

1/2 Mgl. The elastic potential energy stored in the wire is given by the work done in stretching it. Since the force increases linearly with extension, the average force is (Mg)/2, and the work done is (1/2) × Mg × l.

Question 9

When an elastic material with Young's modulus Y is subjected to stretching stress S, elastic energy stored per unit volume of the material is

Accepted Answer

S² / 2Y. The elastic energy stored per unit volume in a stretched material is given by (stress²) / (2 × Young's modulus). Substituting S for stress and Y for Young's modulus, the formula becomes S² / 2Y.

Question 10

The bulk modulus of a spherical object is 'B'. If it is subjected to uniform pressure 'p', the fractional decrease in radius is

Accepted Answer

p / 3B. The bulk modulus (B) is defined as the ratio of pressure (p) to the volumetric strain. For a sphere, the fractional decrease in radius is one-third of the volumetric strain, leading to the result p / 3B.

Question 11

Two wires A and B are of the same material. Their lengths are in the ratio 1 : 2 and the diameter are in the ratio 2 : 1. If they are pulled by the same force, then increase in length will be in the ratio

Accepted Answer

8 : 1. The increase in length is proportional to the applied force and inversely proportional to the cross-sectional area and directly proportional to the length. Since the lengths are in the ratio 1:2 and diameters in the ratio 2:1, the cross-sectional areas are in the ratio 4:1. Thus, the increase in length will be in the ratio (1/4) × 2 = 1:8.

Question 12

A wire of length L₁, 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:

Accepted Answer

MgL / A(L₁ − L). Young's modulus (Y) is defined as stress divided by strain. Stress is the force per unit area (Mg/A), and strain is the relative change in length ((L₁ − L)/L). Substituting these into the formula Y = (stress)/(strain), we get Y = (Mg/A) / ((L₁ − L)/L), which simplifies to Y = MgL / A(L₁ − L).

Question 13

Two wires are made of the same material and have the same volume. The first wire has cross-sectional area A and the second wire has cross-sectional area 3A. If the length of the first wire is increased by ΔL on applying a force F, how much force is needed to stretch the second wire by the

Accepted Answer

9 F. The force required to stretch a wire is inversely proportional to its cross-sectional area for the same material and elongation. Since the second wire has three times the cross-sectional area of the first wire, the force required will be 9 times greater to achieve the same elongation.

Question 14

The Young's modulus of steel is twice that of brass. Two wires of same length and of same area of cross section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weights added to the steel and brass

Accepted Answer

1 : 2. The extension in a wire is inversely proportional to Young's modulus for the same length, area, and applied force. Since the Young's modulus of steel is twice that of brass, the force (weight) on the steel wire must be half that on the brass wire to achieve the same extension. Thus, the ratio is 1:2.

Question 15

The stress-strain curves are drawn for two different materials X and Y. It is observed that the ultimate strength point and the fracture point are close to each other for material X but are far apart for material Y. We can say that materials X and Y are likely to be (respectively):

Accepted Answer

Brittle and ductile. Material X, with ultimate strength and fracture points close together, is brittle as it breaks without significant deformation. Material Y, with these points far apart, is ductile as it undergoes significant deformation before breaking.

Question 16

Copper of fixed volume ‘V’, is drawn into wire of length ‘l’. When this wire is subjected to a constant force ‘F’, the extension produced in the wire is ‘Δl’. Which of the following graphs is a straight line?

Accepted Answer

Δl versus 1 / l². The extension Δl in the wire is inversely proportional to the square of its length (l²) for a constant volume and force, as per the relation Δl ∝ 1/l². Thus, the graph of Δl versus 1/l² will be a straight line.

Question 17

Diamond is very hard because

Accepted Answer

it has large cohesive energy. Diamond is very hard due to its large cohesive energy, which arises from the strong covalent bonds between carbon atoms in its tetrahedral structure.

Question 18

Which one of the following is the weakest kind of bonding in solids

Accepted Answer

Van der Waal’s. Van der Waal's forces are the weakest type of bonding in solids as they arise from temporary dipole interactions and are much weaker compared to ionic, metallic, or covalent bonds.

Question 19

The elongation of a material is directly proportional to its length and inversely proportional to its cross-sectional area, assuming which factors remain unchanged?

Accepted Answer

The force applied and Young's modulus are constant.. The extension (ΔL) of a material is directly proportional to its original length (L) and inversely proportional to its cross-sectional area (A), provided the force (F) and Young's modulus (Y) are constant. This follows from the formula ΔL = (F × L) / (A × Y).

Question 20

Given the formula Y = F / (A × Δl / l), rearrange to express Δl in terms of other variables.

Accepted Answer

As a result, plotting Δl against l² will yield a linear graph.. The equation Δl = Fl² / VY shows that Δl is directly proportional to l². Hence, a graph of Δl versus l² will be a straight line.

NEET Physics: Mechanical Properties of Solids questions with solutions

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