Q: For common metals such as aluminium and copper, which sequence correctly represents the relative magnitudes of the elastic moduli?

Shear modulus < Young’s modulus < bulk modulus.. For common metals the shear modulus is smallest, then Young's modulus, then the bulk modulus (e.g. copper: G~48, Y~117, K~140 GPa; aluminium: G~26, Y~70, K~76 GPa). So shear modulus < Young's modulus < bulk modulus.

Q: A steel sheet fails under a shear stress of 3.5 × 10⁸ N m⁻². The approximate force required to punch a circular hole of diameter 1 cm through a steel plate 0.3 cm thick is:

3.3 × 10⁴ N. The hole is punched by shearing the cylindrical surface of area pi*d*t. F = stress x area = 3.5e8 x pi x 0.01 x 0.003 ~ 3.3e4 N.

Q: A copper wire of fixed volume V is stretched out to a length l. If a constant force F is applied to this wire and it elongates by Δl, which of the following plots would be a straight line?

Δl against l². With fixed volume V, cross-section A = V/l, so Delta-l = F*l/(A*Y) = F*l^2/(V*Y). Hence Delta-l is proportional to l^2, giving a straight line for Delta-l vs l^2.

Q: A material has Poisson’s ratio equal to 0.5. When a tensile force is applied to a wire made of this material, its cross-sectional area decreases by 4%. The corresponding percentage increase in its length is:

4%. When Poisson's ratio is 0.5, it indicates that the material is incompressible, meaning that any decrease in cross-sectional area directly corresponds to an equal percentage increase in length. Therefore, a 4% decrease in area results in a 4% increase in length.

Q: A uniform wire of length L and weight W is fixed at its upper end to a point on the roof, and a load of weight W1 is hung from its lower end. If the wire has cross-sectional area s, what is the stress in the wire at a point located 3L/4 above the lower end?

(W1 + 3W/4)/s. The stress at a cross-section equals the total weight hanging below it divided by area. Below a point 3L/4 above the lower end there is length 3L/4 of wire (weight 3W/4) plus the load W1. So stress = (W1 + 3W/4)/s.

Q: Two wires A and B are made of the same material and have the same initial length. Their radii are in the ratio 1:2. If both wires are stretched by identical loads and the extension of wire A is 8 mm, what is the extension of wire B?

2 mm. Same material/length/load: e = FL/(A*Y) and A ~ r^2, so e ~ 1/r^2. B has twice the radius, hence 4x the area, so e_B = e_A/4 = 8/4 = 2 mm.

Q: Choose the statement(s) that are true: I. A liquid has zero shear modulus. II. The Young’s modulus of a substance falls as temperature increases. III. Poisson’s ratio has no unit.

I, II and III. I: a liquid cannot sustain shear, so its shear modulus is zero - true. II: Young's modulus decreases as temperature rises - true. III: Poisson's ratio is a ratio of strains, hence dimensionless - true. So I, II and III are all correct.

Q: A wire has a diameter of 12 mm and a length of 1 m. Its top end is fixed, and the free end is turned by 30°. What is the angle of shear produced in the wire?

0.18°. For torsion, arc length matching gives shear angle theta = r*phi/L. With r = 6 mm, L = 1000 mm, phi = 30 deg: theta = 6*30/1000 = 0.18 deg.

Question 1

Select the incorrect statement.

Accepted Answer

Gases are less easily compressed.. The incorrect statement is 'Gases are less easily compressed' - gases are in fact the most easily compressed state because of their large intermolecular spacing. The other statements (high bulk modulus of solids, near-incompressibility of solids, compressibility = 1/bulk modulus) are all true.

Question 2

A steel wire has length l and cross-sectional area A. When a certain load is applied, it elongates by 1 cm. If the same load is now applied to another steel wire made of the same material, having twice the length and half the cross-sectional area, what will be its extension?

Accepted Answer

4 cm. Extension dl = F l/(A Y). Doubling length and halving area gives dl' = F(2l)/((A/2)Y) = 4 * F l/(A Y) = 4 * 1 cm = 4 cm.

Question 3

For common metals such as aluminium and copper, which sequence correctly represents the relative magnitudes of the elastic moduli?

Accepted Answer

Shear modulus < Young’s modulus < bulk modulus.. For common metals the shear modulus is smallest, then Young's modulus, then the bulk modulus (e.g. copper: G~48, Y~117, K~140 GPa; aluminium: G~26, Y~70, K~76 GPa). So shear modulus < Young's modulus < bulk modulus.

Question 4

A steel sheet fails under a shear stress of 3.5 × 10⁸ N m⁻². The approximate force required to punch a circular hole of diameter 1 cm through a steel plate 0.3 cm thick is:

Accepted Answer

3.3 × 10⁴ N. The hole is punched by shearing the cylindrical surface of area pi*d*t. F = stress x area = 3.5e8 x pi x 0.01 x 0.003 ~ 3.3e4 N.

Question 5

A copper wire of fixed volume V is stretched out to a length l. If a constant force F is applied to this wire and it elongates by Δl, which of the following plots would be a straight line?

Accepted Answer

Δl against l². With fixed volume V, cross-section A = V/l, so Delta-l = F*l/(A*Y) = F*l^2/(V*Y). Hence Delta-l is proportional to l^2, giving a straight line for Delta-l vs l^2.

Question 6

A material has Poisson’s ratio equal to 0.5. When a tensile force is applied to a wire made of this material, its cross-sectional area decreases by 4%. The corresponding percentage increase in its length is:

Accepted Answer

4%. When Poisson's ratio is 0.5, it indicates that the material is incompressible, meaning that any decrease in cross-sectional area directly corresponds to an equal percentage increase in length. Therefore, a 4% decrease in area results in a 4% increase in length.

Question 7

A uniform wire of length L and weight W is fixed at its upper end to a point on the roof, and a load of weight W1 is hung from its lower end. If the wire has cross-sectional area s, what is the stress in the wire at a point located 3L/4 above the lower end?

Accepted Answer

(W1 + 3W/4)/s. The stress at a cross-section equals the total weight hanging below it divided by area. Below a point 3L/4 above the lower end there is length 3L/4 of wire (weight 3W/4) plus the load W1. So stress = (W1 + 3W/4)/s.

Question 8

Two wires A and B are made of the same material and have the same initial length. Their radii are in the ratio 1:2. If both wires are stretched by identical loads and the extension of wire A is 8 mm, what is the extension of wire B?

Accepted Answer

2 mm. Same material/length/load: e = FL/(A*Y) and A ~ r^2, so e ~ 1/r^2. B has twice the radius, hence 4x the area, so e_B = e_A/4 = 8/4 = 2 mm.

Question 9

Choose the statement(s) that are true:
I. A liquid has zero shear modulus.
II. The Young’s modulus of a substance falls as temperature increases.
III. Poisson’s ratio has no unit.

Accepted Answer

I, II and III. I: a liquid cannot sustain shear, so its shear modulus is zero - true. II: Young's modulus decreases as temperature rises - true. III: Poisson's ratio is a ratio of strains, hence dimensionless - true. So I, II and III are all correct.

Question 10

A wire has a diameter of 12 mm and a length of 1 m. Its top end is fixed, and the free end is turned by 30°. What is the angle of shear produced in the wire?

Accepted Answer

0.18°. For torsion, arc length matching gives shear angle theta = r*phi/L. With r = 6 mm, L = 1000 mm, phi = 30 deg: theta = 6*30/1000 = 0.18 deg.

Question 11

A steel rod used in construction has a radius of 10 mm and a length of 1.0 m. It is pulled by a force of 100 kN. Taking Young’s modulus of structural steel as 2 × 10¹¹ N m⁻², the approximate percentage strain in the rod is

Accepted Answer

0.16%. A = pi*r^2 = pi*(0.01)^2 = pi*10^-4 m^2. Strain = F/(A*Y) = 10^5/((pi*10^-4)*(2*10^11)) = 1/(2*pi*100) = 1.59*10^-3, i.e. about 0.16%.

Question 12

An elastic string has length a m when the longitudinal tension is 4 N and length b m when the longitudinal tension is 5 N. Its length, in metres, when the longitudinal tension is 9 N is

Accepted Answer

5b − 4a. Let L = L0 + kT. Then a = L0 + 4k and b = L0 + 5k, so k = b - a and L0 = 5a - 4b. At T = 9: L = L0 + 9k = (5a - 4b) + 9(b - a) = 5b - 4a.

Question 13

A uniform rope of length L and density ρ is suspended vertically from a fixed support. If the Young’s modulus of the rope material is Y, then the extension produced in the rope by its own weight is

Accepted Answer

(ρgL²)/(2Y). The correct option is derived from the relationship between the weight of the rope and its extension under its own weight, where the extension is proportional to the square of the length and inversely proportional to the Young's modulus. The factor of 1/2 arises from integrating the varying forces along the length of the rope.

Question 14

For a beam and a circular rod having the same cross-sectional area and subjected to the same load, what is the ratio of their depressions if the beam has a square cross-section?

Accepted Answer

3: π. For equal area A: I_square = a^4/12 = A^2/12, I_circle = pi*r^4/4 = A^2/(4*pi). Since delta ~ 1/I, delta_beam:delta_rod = I_circle:I_square = (1/(4*pi)):(1/12) = 3:pi.

Question 15

A cylindrical wire is stretched so that its length becomes twice the original value. Because of the resulting reduction in its diameter, what is the percentage increase in its resistance?

Accepted Answer

300%. When the length of the wire is doubled, its resistance increases due to the lengthening effect, while the diameter reduction leads to a decrease in cross-sectional area, further increasing resistance. The combined effect results in a total resistance increase of 300%.

Question 16

In the experiment used to find Young’s modulus of elasticity, what is observed as the applied load is gradually increased up to the breaking stress?

Accepted Answer

The wire’s cross-sectional area keeps reducing, the wire elongates, and eventually it snaps.. As the load increases, the wire undergoes plastic deformation, which causes it to elongate and reduce in cross-sectional area until it reaches its breaking point.

Question 17

A wire is stretched so that the stress in it is S and the Young’s modulus of its material is Y. What is the elastic energy stored per unit volume of the wire?

Accepted Answer

S²/2Y. The elastic energy stored per unit volume in a material under stress can be derived from the relationship between stress and strain, where the energy density is given by the formula U = (1/2) * stress * strain. Since strain can be expressed as stress divided by Young's modulus (strain = S/Y), substituting this into the energy density formula leads to U = (1/2) * S * (S/Y) = S²/(2Y).

Question 18

A wire elongates by mm when a load W is hanged from it. If the wire goes over a pulley and two weights W each are hung at the two ends, the elongation of the wire will be (in mm)

Accepted Answer

1. When two equal weights are hung from both ends of the wire, the tension in the wire effectively doubles compared to when a single weight is applied. Since elongation is directly proportional to the load, the wire will elongate by the same amount as it did with one weight, resulting in a total elongation of 1 mm.

Question 19

The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by 100°C is: [For steel Young’s modulus is 2 × 10¹¹ N m⁻² and coefficient of thermal expansion is 1.1 × 10⁻⁵ K⁻¹]

Accepted Answer

2.2 × 10⁸ Pa. The correct option is derived from the formula that relates thermal expansion and Young's modulus, which indicates that to maintain the original length of the wire despite temperature increase, a specific tensile stress must be applied. The calculated stress, based on the given values, results in 2.2 × 10⁸ Pa, making it the appropriate choice.

Question 20

A solid sphere of radius r is made of a soft material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless piston of area A floats on the surface of the liquid, covering entire cross-section of cylindrical container. When a mass m is placed on the surface of t

Accepted Answer

mg/3Ka. The correct option is derived from the relationship between pressure change and volume change in the sphere due to the applied force from the mass on the piston. The fractional decrement in the radius is proportional to the force applied (mg) divided by the bulk modulus (K) and the geometric factor related to the sphere's volume, which introduces the factor of 3 in the denominator.

Question 21

A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that his density remains same, the stress in the leg will change by a factor of

Accepted Answer

9. When the linear dimensions increase by a factor of 9, the volume increases by a factor of 9³ (729), and since density remains constant, the mass also increases by the same factor. Stress is defined as force per unit area, and since the area increases by a factor of 9² (81), the stress in the leg changes by a factor of 9.

Question 22

A cube at 0°C is subjected to an external pressure P, producing equal compression on all faces. If the cube has bulk modulus K and coefficient of linear expansion α, by how much must its temperature be increased to restore it to its original dimensions?

Accepted Answer

P/(3αK). Pressure causes fractional volume decrease dV/V = P/K. Restoring it by heating needs 3*alpha*dT = P/K, so dT = P/(3*alpha*K).

Question 23

A metal rod has initial length L at room temperature and a uniform cross-sectional area A. Its coefficient of linear expansion is α per °C. When the rod is heated through a temperature increase of ΔT K, equal compressive forces F are applied at both ends so that its length does not change

Accepted Answer

F/(A α ΔT). Preventing expansion means the elastic compressive strain equals the thermal strain alpha*dT. With stress = F/A, Young's modulus Y = stress/strain = (F/A)/(alpha*dT) = F/(A alpha dT).

Question 24

A cylindrical wire is stretched so that its length becomes twice the original length. If the resulting reduction in diameter is taken into account, by what percentage does its resistance change?

Accepted Answer

300%. With volume fixed, doubling length halves the area, so R = rho*L/A increases by factor (2)*(2) = 4, i.e. R -> 4R. The increase is 300%.

Question 25

A sonometer wire of length 1.5 m is made of steel. The tension in it produces an elastic strain of 1%. What is the fundamental frequency of steel if density and elasticity of steel are 7.7 × 10³ kg/m³ and 2.2 × 10¹¹ N/m² respectively ?

Accepted Answer

178.2 Hz. The fundamental frequency of a vibrating string is determined by its length, tension, and mass per unit length. Given the elastic properties and density of steel, the calculations yield a frequency of 178.2 Hz, which aligns with the physical principles governing the vibration of strings under tension.

Question 26

The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by 100°C is: (For steel Young's modules is 2 × 10¹¹ N m⁻² and coefficient of thermal expansion is 1.1 × 10⁻⁵ K⁻¹)

Accepted Answer

2.2 × 10⁸ Pa. The correct option is derived from the formula for thermal stress, which accounts for the change in temperature and the material's properties. By applying the Young's modulus and the coefficient of thermal expansion, we find that the pressure needed to maintain the wire's length at a higher temperature is 2.2 × 10⁸ Pa.

Question 27

A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that his density remains same, the stress in the leg will change by a factor of-

Accepted Answer

9. When the linear dimensions of an object increase by a factor of 9, its volume increases by a factor of 9³ (729), while the weight increases proportionally to the volume. Since stress is defined as force (weight) per unit area, and the area increases by a factor of 9² (81), the stress in the leg will increase by a factor of 9.

Question 28

A rod, of length L at room temperature and uniform area of cross section A, is made of a metal having coefficient of linear expansion α/°C. It is observed that an external compressive force F is applied on each of its ends, prevents any change in the length of the rod, when its temperature

Accepted Answer

F / A α ΔT. Thermal strain prevented = alpha*dT. Compressive stress = F/A. Young's modulus Y = stress/strain = (F/A)/(alpha*dT) = F/(A*alpha*dT). Answer: F/(A alpha dT).

Question 29

The elastic limit of brass is 379 MPa. What should be the minimum diameter of a brass rod if it is to support a 400 N load without exceeding its elastic limit?

Accepted Answer

1.16 mm. Stress = F/(pi d^2/4) <= 379e6 Pa. d^2 >= 4*400/(pi*379e6) = 1.34e-6, so d >= 1.16e-3 m = 1.16 mm (minimum diameter).

Question 30

A steel wire having a radius of 2.0 mm, carrying a load of 4 kg, is hanging from a ceiling. Given that g = 3.1 π m s⁻², what will be the tensile stress that would be developed in the wire ?

Accepted Answer

3.1 × 10⁶ N m⁻². F = mg = 4 * 3.1*pi N. Area = pi r^2 = pi*(2e-3)^2. Stress = F/A = (4*3.1*pi)/(pi*4e-6) = 3.1e6 N/m^2.

Question 31

A boy's catapult is made of rubber cord which is 42 cm long, with 6 mm diameter of cross-section and of negligible mass. The boy keeps a stone weighing 0.02 kg on it and stretches the cord by 20 cm by applying a constant force. When released, the stone flies off with a velocity of 20 m s⁻¹

Accepted Answer

10⁶ N m⁻². A = pi*(0.003)^2 = 2.83e-5 m^2. Y = m v^2 L /(A * dL^2) = (0.02*400*0.42)/(2.83e-5 * 0.04) ~ 3e6 N/m^2, i.e. order 10^6.

Question 32

A student determined Young's Modulus of elasticity using the formula Y = MgL³ / 4bd³δ. The value of g is taken to be 9.8 m/s², without any significant error, his observation are as following.

Accepted Answer

Thickness of bar (d): Least count of the Equipment used for measurement 0.01 mm; Observed value 0.4 cm. The thickness of the bar (d) is crucial for calculating Young's Modulus, as it directly affects the stiffness of the material. The observed value of 0.4 cm and a least count of 0.01 mm indicate a precise measurement, which is essential for accurate results in the formula.

Question 33

In order to determine the Young's Modulus of a wire of radius 0.2 cm (measured using a scale of least count = 0.001 cm) and length 1m (measured using a scale of least count = 1 mm), using a scale of least count = 1 mm, a weight of mass 1kg (measured using a scale of least count = 1g) was h

Accepted Answer

1.4%. The fractional error in the value of Young's Modulus is calculated by considering the uncertainties in the measurements of radius, length, mass, and elongation. Each measurement contributes to the overall uncertainty, and when combined, they yield a total fractional error of 1.4%, making option D the correct choice.

Question 34

The value of tension in a long thin metal wire has been changed from T1 to T2. The lengths of the metal wire at two different values of tension T1 and T2 are ℓ1 and ℓ2 respectively. The actual length of the metal wire is:

Accepted Answer

(T1ℓ2 − T2ℓ1)/(T1 − T2). With l = l0 + kT, l1 = l0 + kT1 and l2 = l0 + kT2. Eliminating k gives the natural length l0 = (T1 l2 - T2 l1)/(T1 - T2).

Question 35

If Y, K and η are the values of Young's modulus, bulk modulus and modulus of rigidity of any material respectively. Choose the correct relation for these parameters.

Accepted Answer

K = Yη/(9η - 3Y) N/m². This relation correctly expresses the bulk modulus in terms of Young's modulus and the modulus of rigidity, reflecting the interdependence of these elastic constants in material science.

Question 36

A uniform metallic wire is elongated by 0.04 m when subjected to a linear force F. The elongation, if its length and diameter are doubled and subjected to the same force will be ____ cm.

Accepted Answer

0.02. Elongation = FL/(A*Y). Doubling length multiplies by 2; doubling diameter quadruples area, dividing by 4. Net factor 2/4 = 1/2, so new elongation = 0.04/2 = 0.02.

Question 37

The normal density of a material is ρ and its bulk modulus of elasticity is K. The magnitude of increase in density of material, when a pressure P is applied uniformly on all sides, will be:

Accepted Answer

ρP/K. Bulk modulus K = -V dP/dV = rho dP/drho. For an applied pressure P this gives the density increase d(rho) = rho*P/K.

Question 38

The length of metallic wire is ℓ1 when tension in it is T1. It is ℓ2 when the tension is T2. The original length of the wire will be -

Accepted Answer

(T2ℓ1 - T1ℓ2)/(T2 - T1). With l = L0 + k T (k = L0/AY): l1-l2 = k(T1-T2). Solving, L0 = (T1*l2 - T2*l1)/(T1 - T2) = (T2*l1 - T1*l2)/(T2 - T1).

Question 39

A uniform heavy rod of weight 10 kg m s⁻², cross-sectional area 100 cm² and length 20 cm is hanging from a fixed support. Young modulus of the material of the rod is 2 × 10¹¹ N m⁻². Neglecting the lateral contraction, find the elongation of rod due to its own weight.

Accepted Answer

5 × 10⁻¹⁰ m. The elongation of the rod due to its own weight can be calculated using the formula for elongation under uniform load, which incorporates the weight of the rod, its length, cross-sectional area, and Young's modulus. The correct option reflects the calculated elongation based on these parameters, confirming that the rod stretches a small amount under its own weight.

Question 40

Two wires of same length and radius are joined end to end and loaded. The Young's moduli of the materials of the two wires are Y₁ and Y₂. The combination behaves as a single wire then Young's modulus is:

Accepted Answer

Y = 2Y₁Y₂ / (Y₁ + Y₂). Two equal wires in series each have stiffness Y_i*A/L; the combination (length 2L) has stiffness Y*A/(2L). Adding compliances: 2/Y = 1/Y1 + 1/Y2, giving Y = 2*Y1*Y2/(Y1 + Y2).

Question 41

A steel wire of length 3.2 m (Yₛ = 2.0 × 10¹¹ N m⁻²) and a copper wire of length 4.4 m (Y_c = 1.1 × 10¹¹ N m⁻²), both of radius 1.4 mm are connected end to end. When stretched by a load, the net elongation is found to be 1.4 mm. The load applied, in Newton, will be: (Given π = 22/7)

Accepted Answer

154. A = (22/7)(1.4e-3)^2 = 6.16e-6 m^2. L1/Y1 + L2/Y2 = 3.2/2.0e11 + 4.4/1.1e11 = 1.6e-11 + 4.0e-11 = 5.6e-11. F = elongation*A/sum = (1.4e-3)(6.16e-6)/(5.6e-11) = 154 N.

Question 42

If the length of a wire is made double and radius is halved of its respective values. Then, the Young’s modules of the material of the wire will:
(1) Remains same
(2) Become 8 times its initial value
(3) Become 1/4th of its initial value
(4) Become 4 times its initial value

Accepted Answer

Remains same. Young's modulus depends only on the material, not on the geometry of the specimen. Changing the wire's length or radius does not change its Young's modulus, so it remains the same.

Question 43

Match List I with List II.
List I: A. Torque, B. Stress, C. Latent Heat, D. Power
List II: I. Nms⁻¹, II. J kg⁻¹, III. Nm, IV. Nm⁻²
Choose the correct answer from the options given below:

Accepted Answer

A-III, B-IV, C-II, D-I. Torque has units N m (III), stress N m^-2 (IV), latent heat J kg^-1 (II), and power N m s^-1 (I). The correct matching is A-III, B-IV, C-II, D-I.

Question 44

A metal wire of length 0.5 m and cross-sectional area 10⁻⁴ m² has breaking stress 5 × 10⁸ N m⁻². A block of 10 kg is attached at one end of the string and is rotating in a horizontal circle. The maximum linear velocity of block will be ____ ms⁻¹.

Accepted Answer

50. The maximum linear velocity is determined by the breaking stress of the wire, which relates to the maximum tension the wire can withstand. By calculating the maximum force using the breaking stress and the cross-sectional area, and then applying the formula for centripetal force, we find that the maximum linear velocity of the block is 50 m/s.

Question 45

A wire of length L is hanging from a fixed support. The length changes to L1 and L2 when masses 1 kg and 2 kg are suspended respectively from its free end. Then the value of L is equal to -

Accepted Answer

2L1 - L2. The correct option, 2L1 - L2, is derived from the relationship between the elongation of the wire under different loads. When the first mass is added, the wire stretches to L1, and when the second mass is added, it stretches to L2. The difference in elongation allows us to express the original length L in terms of L1 and L2.

Question 46

The Young's modulus of a steel wire of length 6 m and cross-sectional area 3 mm², is 2 × 10¹¹ N/m². The wire is suspended from its support on a given planet. A block of mass 4 kg is attached to the free end of the wire. The acceleration due to gravity on the planet is 1/4 of its value on t

Accepted Answer

0.1 mm. The elongation of the wire can be calculated using the formula for elongation under a tensile load, which incorporates Young's modulus, the force applied (weight of the block), the original length, and the cross-sectional area. Given that the gravitational acceleration on the planet is 1/4 of that on Earth, the effective weight of the block is reduced, resulting in a smaller elongation, which correctly leads to the answer of 0.1 mm.

Question 47

Young’s moduli of the material of wires A and B are in the ratio of 1: 4, while its area of cross sections are in the ratio of 1: 3. If the same amount of load is applied to both the wires, the amount of elongation produced in the wires A and B will be in the ratio of - [Assume length of w

Accepted Answer

12: 1. The elongation of a wire is inversely proportional to its Young's modulus and directly proportional to its length and the applied load, while also being inversely proportional to its cross-sectional area. Given the ratios of Young's moduli (1:4) and cross-sectional areas (1:3), we can derive the elongation ratio as follows: elongation is proportional to the load divided by the product of Young's modulus and area. This results in a ratio of elongation for wires A and B of 12:1.

Question 48

The elastic potential energy stored in a steel wire of length 20 m stretched through 2 cm is 80 J. The cross sectional area of the wire is ______ mm². (Given, y = 2.0 × 10¹¹ N m⁻²)

Accepted Answer

40. The elastic potential energy stored in a stretched wire can be calculated using the formula U = (1/2) * (F² / y * A), where F is the force applied, y is Young's modulus, and A is the cross-sectional area. Given the values, the calculations confirm that the cross-sectional area of the wire is indeed 40 mm².

Question 49

A wire of length 'L' and radius 'r' is clamped rigidly at one end. When the other end of the wire is pulled by a force f, its length increases by 'ℓ'. Another wire of same material of length '2L' and radius '2r' is pulled by a force '2f'. Then the increase in its length will be:

Accepted Answer

ℓ. The increase in length of a wire under tension is proportional to the force applied and inversely proportional to its cross-sectional area and length. In this case, the second wire has double the length and quadruple the cross-sectional area, which balances the effect of doubling the force, resulting in the same increase in length as the first wire.

Question 50

Choose the correct relationship between Poisson ratio (σ), bulk modulus (K) and modulus of rigidity (η) of a given solid object

Accepted Answer

σ = (3K − 2η)/(6K + 2η). The correct option expresses the Poisson ratio in terms of the bulk modulus and modulus of rigidity, reflecting the relationship between these mechanical properties of materials, which is essential for understanding how materials deform under stress.

JEE Main Physics: Mechanical Properties of Solids questions with solutions

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