Grade 11 : Physics - Solids (Set 4)

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Grade 11 : Physics - Solids (Set 4)— Questions & Detailed Solutions

Material Young's Modulus ($Y$) Yield Strength ($σ_y$)
Steel $2.0 \times 10^{11}\text{ N/m}^2$ $2.5 \times 10^8\text{ N/m}^2$
Brass $1.0 \times 10^{11}\text{ N/m}^2$ $1.5 \times 10^8\text{ N/m}^2$
Wire A (Steel) and Wire B (Brass) have equal original length $L = 2m$ and cross-sectional area $A = 1\text{ mm}^2$. Under a tensile load of 100 N, what is the ratio of elongation of Wire B to Wire A ($\Delta L_B / \Delta L_A$)?

Material	Young's Modulus ($Y$)	Yield Strength ($σ_y$)
Steel	$2.0 \times 10^{11}\text{ N/m}^2$	$2.5 \times 10^8\text{ N/m}^2$
Brass	$1.0 \times 10^{11}\text{ N/m}^2$	$1.5 \times 10^8\text{ N/m}^2$

(A)

0.50

(B)

1.00

(C)

2.00

(D)

4.00

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✓ Correct Option: (C)

Explanation:Elongation is given by $\Delta L = \frac{F L}{A Y}$. Since $F$, $L$, and $A$ are identical for both wires, elongation is inversely proportional to Young's modulus:$$\frac{\Delta L_B}{\Delta L_A} = \frac{Y_{\text{Steel}}}{Y_{\text{Brass}}} = \frac{2.0 \times 10^{11}}{1.0 \times 10^{11}} = 2.00$$Thus, Wire B stretches twice as much as Wire A.

A rigid horizontal bar of length $L = 1.2m$ is supported horizontally at its ends by two vertical wires: Wire 1 (Steel, $A_1 = 1\text{ mm}^2$) at $x=0$ and Wire 2 (Aluminum, $A_2 = 2\text{ mm}^2$) at $x=1.2m$.
At what distance $x$ from Wire 1 must a load $W$ be hung so that both wires experience equal tensile stress?

(A)

0.80 m

(B)

0.60 m

(C)

0.40 m

(D)

0.90 m

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✓ Correct Option: (A)

Explanation:For equal stress: $\frac{F_1}{A_1} = \frac{F_2}{A_2} \implies \frac{F_1}{1} = \frac{F_2}{2} \implies F_2 = 2 F_1$.
Total load $W = F_1 + F_2 = 3 F_1$.
Taking torque about Wire 1 ($x=0$):$$W \cdot x = F_2 \cdot L \implies 3 F_1 \cdot x = (2 F_1) \times 1.2 \implies x = \frac{2.4}{3} = 0.80\text{ m}$$

A composite cylinder consists of a solid steel core ($r_1 = 10\text{ mm}$) surrounded by a hollow brass sleeve (inner radius $r_1 = 10\text{ mm}$, outer radius $r_2 = 20\text{ mm}$) bonded to rigid end plates. Length $L = 1m$. $Y_{\text{steel}} = 2 \times 10^{11}\text{ N/m}^2$, $Y_{\text{brass}} = 1 \times 10^{11}\text{ N/m}^2$.
An axial force $F = 6\pi \times 10^4\text{ N}$ is applied to the end plates. Calculate the total elongation $\Delta L$ of the composite cylinder.

(A)

0.6 mm

(B)

0.8 mm

(C)

1.5 mm

(D)

1.2 mm

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✓ Correct Option: (D)

Explanation:Areas: $A_s = \pi r_1^2 = 10^{-4}\pim^2$, $A_b = \pi (r_2^2 - r_1^2) = 3 \times 10^{-4}\pim^2$.
Both materials stretch by the same $\Delta L$:$$F = F_s + F_b = \frac{\Delta L}{L} (Y_s A_s + Y_b A_b)$$$$Y_s A_s + Y_b A_b = (2 \times 10^{11})(10^{-4}\pi) + (1 \times 10^{11})(3 \times 10^{-4}\pi) = 5\pi \times 10^7\text{ N}$$$$\Delta L = \frac{6\pi \times 10^4}{5\pi \times 10^7} = 1.2 \times 10^{-3}\text{ m} = 1.2\text{ mm}$$

Material Young's Modulus $Y$ ($\times 10^{10}\text{ N/m}^2$)
P 6.0
Q 12.0
R 20.0
S 8.0
Four wires of identical dimensions are each subjected to a strain of $\epsilon = 0.001$. Based on the table above, which material stores the maximum elastic energy density?

Material	Young's Modulus $Y$ ($\times 10^{10}\text{ N/m}^2$)
P	6.0
Q	12.0
R	20.0
S	8.0

(A)

Material P

(B)

Material Q

(C)

Material R

(D)

Material S

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✓ Correct Option: (C)

Explanation:Elastic potential energy density is given by $u = \frac{1}{2} Y \epsilon^2$.
Since the strain $\epsilon = 0.001$ is constant for all materials, $u$ is directly proportional to Young's Modulus $Y$.
Material R has the highest Young's Modulus ($20.0 \times 10^{10}\text{ N/m}^2$), so it stores the maximum elastic energy density.

A composite rod made of a Steel segment ($L_1 = 0.5m, Y_1 = 2 \times 10^{11}\text{ N/m}^2, \alpha_1 = 12 \times 10^{-6}\text{ K}^{-1}$) and a Brass segment ($L_2 = 0.5m, Y_2 = 1 \times 10^{11}\text{ N/m}^2, \alpha_2 = 24 \times 10^{-6}\text{ K}^{-1}$) with cross-section $A = 2 \times 10^{-4}m^2$ is clamped tightly between two rigid unyielding walls.
If the temperature of the assembly is raised by $\Delta T = 50\text{ K}$, calculate the thermal compressive force developed in the rod.

(A)

12 kN

(B)

18 kN

(C)

24 kN

(D)

36 kN

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✓ Correct Option: (C)

Explanation:Free thermal expansion:$$\Delta L_{\text{th}} = \Delta T (L_1 \alpha_1 + L_2 \alpha_2) = 50 (0.5 \times 12 \times 10^{-6} + 0.5 \times 24 \times 10^{-6}) = 9 \times 10^{-4}\text{ m}$$Compressive contraction due to reaction force $F$:$$\Delta L_{\text{mech}} = \frac{F}{A} \left(\frac{L_1}{Y_1} + \frac{L_2}{Y_2}\right) = \frac{F}{2 \times 10^{-4}} (0.25 \times 10^{-11} + 0.50 \times 10^{-11}) = F \times 3.75 \times 10^{-8}$$Equating $\Delta L_{\text{mech}} = \Delta L_{\text{th}}$:$$F = \frac{9 \times 10^{-4}}{3.75 \times 10^{-8}} = 24,000\text{ N} = 24\text{ kN}$$

A vertical steel wire of original length $L = 2m$ and cross-sectional area $A = 1\text{ mm}^2$ supports a suspended mass $M = 10kg$ hanging in equilibrium ($g = 10m/s^2, Y = 2 \times 10^{11}\text{ N/m}^2$).
Calculate the total elastic potential energy stored in the stretched wire.

(A)

0.025 J

(B)

0.050 J

(C)

0.100 J

(D)

0.500 J

View Correct Answer & Detailed Solution

✓ Correct Option: (B)

Explanation:Tension force $F = M g = 100\text{ N}$.
Elongation:$$\Delta L = \frac{F L}{A Y} = \frac{100 \times 2}{(10^{-6}) \times (2 \times 10^{11})} = 10^{-3}\text{ m} = 1\text{ mm}$$Elastic potential energy:$$U = \frac{1}{2} F \Delta L = \frac{1}{2} \times 100 \times 10^{-3} = 0.050\text{ J}$$

A solid rubber cube of side length $L = 10cm$ has its lower face fixed to a floor. A tangential force $F = 500\text{ N}$ is applied uniformly to its upper face. The shear modulus of rubber is $\eta = 2 \times 10^6\text{ N/m}^2$.
What is the lateral displacement $\Delta x$ of the upper face relative to the fixed bottom face?

(A)

2.50 mm

(B)

1.25 mm

(C)

5.00 mm

(D)

7.50 mm

View Correct Answer & Detailed Solution

✓ Correct Option: (A)

Explanation:Area of face $A = L^2 = (0.1)^2 = 0.01m^2$.
Shear stress $\tau = \frac{F}{A} = \frac{500}{0.01} = 5 \times 10^4\text{ N/m}^2$.
Shear strain $\theta = \frac{\tau}{\eta} = \frac{5 \times 10^4}{2 \times 10^6} = 0.025\text{ rad}$.
Lateral displacement:$$\Delta x = L \cdot \theta = 0.1 \times 0.025 = 0.0025\text{ m} = 2.50\text{ mm}$$

A solid sphere of volume $V_0 = 0.5m^3$ is submerged in an oceanic chamber where the pressure is increased by $\Delta P = 4.0 \times 10^7\text{ N/m}^2$. The bulk modulus of the material is $B = 1.6 \times 10^{11}\text{ N/m}^2$.
Find the magnitude of the volumetric compression (change in volume $\Delta V$) of the sphere.

(A)

62.5 cm³

(B)

125 cm³

(C)

250 cm³

(D)

500 cm³

View Correct Answer & Detailed Solution

✓ Correct Option: (B)

Explanation:Bulk modulus is defined as $B = \frac{\Delta P}{\Delta V / V_0}$.
Solving for $\Delta V$:$$\Delta V = \frac{\Delta P \cdot V_0}{B} = \frac{(4.0 \times 10^7) \times 0.5}{1.6 \times 10^{11}} = 1.25 \times 10^{-4}\text{ m}^3$$Converting to $cm^3$: $1.25 \times 10^{-4} \times 10^6 = 125cm^3$.

Cycle Stage Work per unit volume ($u$)
Loading Work Density ($W_{\text{load}}$) $8.5 \times 10^4\text{ J/m}^3$
Unloading Work Density ($W_{\text{unload}}$) $3.5 \times 10^4\text{ J/m}^3$
A polymer sample of volume $V = 1 \times 10^{-5}m^3$ undergoes one complete elastic hysteresis cycle. How much thermal energy is dissipated in the sample during this cycle?

Cycle Stage	Work per unit volume ($u$)
Loading Work Density ($W_{\text{load}}$)	$8.5 \times 10^4\text{ J/m}^3$
Unloading Work Density ($W_{\text{unload}}$)	$3.5 \times 10^4\text{ J/m}^3$

(A)

0.25 J

(B)

0.85 J

(C)

1.20 J

(D)

0.50 J

View Correct Answer & Detailed Solution

✓ Correct Option: (D)

Explanation:Energy dissipated per unit volume is the area of the hysteresis loop:$$\Delta u = W_{\text{load}} - W_{\text{unload}} = 8.5 \times 10^4 - 3.5 \times 10^4 = 5.0 \times 10^4\text{ J/m}^3$$Total dissipated energy:$$E = \Delta u \times V = (5.0 \times 10^4\text{ J/m}^3) \times (1 \times 10^{-5}\text{ m}^3) = 0.50\text{ J}$$

Q10

Two identical copper wires $W_1$ and $W_2$ ($L = 1.5m, A = 0.5\text{ mm}^2$ each) are connected in series end-to-end to form a single continuous vertical suspension carrying a mass $m = 5kg$ ($g = 10m/s^2, Y = 1.2 \times 10^{11}\text{ N/m}^2$).
What is the total stretch of the combined two-wire assembly?

(A)

1.25 mm

(B)

2.50 mm

(C)

3.75 mm

(D)

5.00 mm

View Correct Answer & Detailed Solution

✓ Correct Option: (B)

Explanation:Total original length $L_{\text{total}} = 2 \times 1.5 = 3.0m$.
Stretching force $F = m g = 50\text{ N}$.
Total elongation:$$\Delta L_{\text{total}} = \frac{F L_{\text{total}}}{A Y} = \frac{50 \times 3.0}{(0.5 \times 10^{-6}) \times (1.2 \times 10^{11})} = \frac{150}{6 \times 10^4} = 2.50 \times 10^{-3}\text{ m} = 2.50\text{ mm}$$

Q11

Parameter Value
Original Diameter ($D$) 2.00 mm
Original Length ($L$) 1.00 m
Longitudinal Strain ($\epsilon_z$) 0.002
Poisson's Ratio ($\nu$) 0.30
Under axial tension, what is the magnitude of the decrease in the wire's radius ($\Delta r$)?

Parameter	Value
Original Diameter ($D$)	2.00 mm
Original Length ($L$)	1.00 m
Longitudinal Strain ($\epsilon_z$)	0.002
Poisson's Ratio ($\nu$)	0.30

(A)

0.30 \mu m

(B)

0.60 \mu m

(C)

1.20 \mu m

(D)

2.40 \mu m

View Correct Answer & Detailed Solution

✓ Correct Option: (B)

Explanation:Original radius $r = D / 2 = 1.00\text{ mm} = 10^{-3}m$.
Lateral strain:$$\epsilon_{\text{lat}} = \nu \cdot \epsilon_z = 0.30 \times 0.002 = 0.0006$$Change in radius:$$\Delta r = r \cdot \epsilon_{\text{lat}} = (10^{-3}\text{ m}) \times 0.0006 = 6.0 \times 10^{-7}\text{ m} = 0.60\text{ \mu m}$$

Q12

A uniform heavy cable of density $\rho = 8000\text{ kg/m}^3$ and breaking stress $\sigma_{\text{break}} = 4.0 \times 10^8\text{ N/m}^2$ hangs vertically under its own weight ($g = 10m/s^2$).
What is the maximum length $L_{\text{max}}$ of cable that can hang freely without snapping under its own weight?

(A)

5.0 km

(B)

2.5 km

(C)

8.0 km

(D)

10.0 km

View Correct Answer & Detailed Solution

✓ Correct Option: (A)

Explanation:Maximum stress occurs at the top support:$$\sigma_{\text{max}} = \frac{W}{A} = \frac{\rho A L g}{A} = \rho g L$$Setting $\sigma_{\text{max}} = \sigma_{\text{break}}$:$$8000 \times 10 \times L_{\text{max}} = 4.0 \times 10^8 \implies L_{\text{max}} = \frac{4.0 \times 10^8}{8 \times 10^4} = 5000\text{ m} = 5.0\text{ km}$$

Q13

A helical spring of stiffness $k_s = 4000\text{ N/m}$ is connected in series with a steel wire of stiffness $k_w = \frac{A Y}{L} = 200,000\text{ N/m}$. A tensile force $F$ stretches the combined spring-wire assembly.
What is the ratio of work done in stretching the spring to the work done in stretching the steel wire ($W_{\text{spring}} / W_{\text{wire}}$)?

(A)

(B)

(C)

(D)

100

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✓ Correct Option: (C)

Explanation:For elements in series, the stretching force $F$ is identical in both.
Work done on an elastic element of stiffness $k$ under force $F$ is $W = \frac{F^2}{2 k}$.
Therefore:$$\frac{W_{\text{spring}}}{W_{\text{wire}}} = \frac{k_w}{k_s} = \frac{200,000}{4000} = 50$$

Q14

Searle's Apparatus Variable Value
Wire Length ($L$) 2.0 m
Wire Diameter ($d$) 1.0 mm
Applied Load Mass ($m$) 4.0 kg
Observed Elongation ($\Delta L$) 0.50 mm
Using the experimental data in the table above ($g = 9.8m/s^2$), determine Young's Modulus ($Y$) of the wire material.

Searle's Apparatus Variable	Value
Wire Length ($L$)	2.0 m
Wire Diameter ($d$)	1.0 mm
Applied Load Mass ($m$)	4.0 kg
Observed Elongation ($\Delta L$)	0.50 mm

(A)

$1.0 \times 10^{11}\text{ N/m}^2$

(B)

$1.5 \times 10^{11}\text{ N/m}^2$

(C)

$2.0 \times 10^{11}\text{ N/m}^2$

(D)

$2.5 \times 10^{11}\text{ N/m}^2$

View Correct Answer & Detailed Solution

✓ Correct Option: (C)

Explanation:Cross-sectional area $A = \frac{\pi d^2}{4} = \frac{\pi (10^{-3})^2}{4} \approx 0.7854 \times 10^{-6}m^2$.
Force $F = m g = 4.0 \times 9.8 = 39.2\text{ N}$.
Young's Modulus:$$Y = \frac{F L}{A \Delta L} = \frac{39.2 \times 2.0}{(0.7854 \times 10^{-6}) \times (0.50 \times 10^{-3})} \approx 2.0 \times 10^{11}\text{ N/m}^2$$

Q15

A uniform rod of length $L = 1m$, density $\rho = 8000\text{ kg/m}^3$, and $Y = 2 \times 10^{11}\text{ N/m}^2$ rotates in a horizontal plane about a vertical axis passing through one end with angular velocity $\omega = 100\text{ rad/s}$.
Calculate the total elongation $\Delta L$ of the rod caused by centrifugal stress.

(A)

0.067 mm

(B)

0.133 mm

(C)

0.267 mm

(D)

0.400 mm

View Correct Answer & Detailed Solution

✓ Correct Option: (B)

Explanation:Tension at distance $x$ from axis: $T(x) = \frac{1}{2} \rho A \omega^2 (L^2 - x^2)$.
Integrating strain over length:$$\Delta L = \int_0^L \frac{T(x)}{A Y} dx = \frac{\rho \omega^2 L^3}{3 Y}$$$$\Delta L = \frac{8000 \times (100)^2 \times 1^3}{3 \times (2 \times 10^{11})} = \frac{8 \times 10^7}{6 \times 10^{11}} = 1.33 \times 10^{-4}\text{ m} = 0.133\text{ mm}$$

Q16

A composite wire consists of a Steel segment ($L_s = 2m, r_s = 1\text{ mm}, Y_s = 2 \times 10^{11}\text{ N/m}^2$) connected in series to a Brass segment ($L_b = 1m, r_b = 2\text{ mm}, Y_b = 1 \times 10^{11}\text{ N/m}^2$) supporting a hanging mass.
What is the ratio of elongation in the steel segment to that in the brass segment ($\Delta L_s / \Delta L_b$)?

(A)

(B)

(C)

(D)

View Correct Answer & Detailed Solution

✓ Correct Option: (D)

Explanation:Since force $F$ is identical:$$\Delta L = \frac{F L}{\pi r^2 Y} \implies \frac{\Delta L_s}{\Delta L_b} = \left(\frac{L_s}{L_b}\right) \left(\frac{r_b}{r_s}\right)^2 \left(\frac{Y_b}{Y_s}\right)$$$$\frac{\Delta L_s}{\Delta L_b} = (2) \times (2)^2 \times \left(\frac{1 \times 10^{11}}{2 \times 10^{11}}\right) = 2 \times 4 \times 0.5 = 4$$

Q17

A solid metallic sphere is subjected to a uniform hydrostatic pressure $P = 1.5 \times 10^8\text{ N/m}^2$. The Bulk Modulus of the material is $B = 1.0 \times 10^{11}\text{ N/m}^2$.
Calculate the percentage decrease in the radius of the sphere.

(A)

0.02%

(B)

0.05%

(C)

0.15%

(D)

0.45%

View Correct Answer & Detailed Solution

✓ Correct Option: (B)

Explanation:Fractional volume strain:$$\frac{\Delta V}{V} = \frac{P}{B} = \frac{1.5 \times 10^8}{1.0 \times 10^{11}} = 1.5 \times 10^{-3} = 0.15\%$$For a sphere $V = \frac{4}{3}\pi R^3$, so radial strain is $\frac{\Delta R}{R} = \frac{1}{3} \frac{\Delta V}{V}$.
Percentage change in radius $= \frac{1}{3} \times 0.15% = 0.05%$.

Q18

Characteristic Point Stress ($\text{N/m}^2$) Strain
Proportional Limit $2.0 \times 10^8$ 0.0010
Yield Point $2.4 \times 10^8$ 0.0015
Ultimate Tensile Strength $4.0 \times 10^8$ 0.1500
From the stress-strain data above, determine Young's Modulus ($Y$) of the material in its elastic regime.

Characteristic Point	Stress ($\text{N/m}^2$)	Strain
Proportional Limit	$2.0 \times 10^8$	0.0010
Yield Point	$2.4 \times 10^8$	0.0015
Ultimate Tensile Strength	$4.0 \times 10^8$	0.1500

(A)

$2.0 \times 10^{11}\text{ N/m}^2$

(B)

$1.6 \times 10^{11}\text{ N/m}^2$

(C)

$2.4 \times 10^{11}\text{ N/m}^2$

(D)

$4.0 \times 10^{11}\text{ N/m}^2$

View Correct Answer & Detailed Solution

✓ Correct Option: (A)

Explanation:Hooke's law applies up to the Proportional Limit where stress is linearly proportional to strain:$$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{2.0 \times 10^8\text{ N/m}^2}{0.0010} = 2.0 \times 10^{11}\text{ N/m}^2$$

Q19

Wire X has length $L$ and radius $r$. Wire Y is made of the same material but has length $2L$ and radius $2r$. Both wires are stretched by the same tensile load $F$.
What is the ratio of elastic potential energy stored in Wire X to Wire Y ($U_X / U_Y$)?

(A)

0.5

(B)

1.0

(C)

2.0

(D)

4.0

View Correct Answer & Detailed Solution

✓ Correct Option: (C)

Explanation:Stored energy $U = \frac{F^2 L}{2 A Y} = \frac{F^2 L}{2 \pi r^2 Y} \propto \frac{L}{r^2}$.
For Wire X: $U_X \propto \frac{L}{r^2}$.
For Wire Y: $U_Y \propto \frac{2L}{(2r)^2} = \frac{2L}{4r^2} = \frac{1}{2} \frac{L}{r^2}$.
Ratio: $U_X / U_Y = 1 / 0.5 = 2.0$.

Q20

A steel wire ($L = 1m, A = 1\text{ mm}^2, \mu = 0.01\text{ kg/m}, Y = 2 \times 10^{11}\text{ N/m}^2, \alpha = 10^{-5}\text{ K}^{-1}$) is fixed rigidly between two walls at temperature $T_1$. It is then cooled by $\Delta T = 50\text{ K}$.
Calculate the fundamental frequency ($f_1$) of transverse vibrations setup on this cooled wire.

(A)

25 Hz

(B)

50 Hz

(C)

100 Hz

(D)

200 Hz

View Correct Answer & Detailed Solution

✓ Correct Option: (B)

Explanation:Thermal stress $\sigma = Y \alpha \Delta T = (2 \times 10^{11}) \times 10^{-5} \times 50 = 10^8\text{ N/m}^2$.
Thermal tension $T = \sigma A = 10^8 \times 10^{-6} = 100\text{ N}$.
Wave velocity $v = \sqrt{T / \mu} = \sqrt{100 / 0.01} = 100m/s$.
Fundamental frequency:$$f_1 = \frac{v}{2 L} = \frac{100}{2 \times 1} = 50\text{ Hz}$$

Grade 11 : Physics - Solids (Set 4)

Grade 11 : Physics - Solids (Set 4)— Questions & Detailed Solutions

Material Young's Modulus $Y$ ($\times 10^{10}\text{ N/m}^2$)
P 6.0
Q 12.0
R 20.0
S 8.0
Four wires of identical dimensions are each subjected to a strain of $\epsilon = 0.001$. Based on the table above, which material stores the maximum elastic energy density?

A vertical steel wire of original length $L = 2m$ and cross-sectional area $A = 1\text{ mm}^2$ supports a suspended mass $M = 10kg$ hanging in equilibrium ($g = 10m/s^2, Y = 2 \times 10^{11}\text{ N/m}^2$).
Calculate the total elastic potential energy stored in the stretched wire.

Parameter Value
Original Diameter ($D$) 2.00 mm
Original Length ($L$) 1.00 m
Longitudinal Strain ($\epsilon_z$) 0.002
Poisson's Ratio ($\nu$) 0.30
Under axial tension, what is the magnitude of the decrease in the wire's radius ($\Delta r$)?

Searle's Apparatus Variable Value
Wire Length ($L$) 2.0 m
Wire Diameter ($d$) 1.0 mm
Applied Load Mass ($m$) 4.0 kg
Observed Elongation ($\Delta L$) 0.50 mm
Using the experimental data in the table above ($g = 9.8m/s^2$), determine Young's Modulus ($Y$) of the wire material.

A solid metallic sphere is subjected to a uniform hydrostatic pressure $P = 1.5 \times 10^8\text{ N/m}^2$. The Bulk Modulus of the material is $B = 1.0 \times 10^{11}\text{ N/m}^2$.
Calculate the percentage decrease in the radius of the sphere.

Characteristic Point Stress ($\text{N/m}^2$) Strain
Proportional Limit $2.0 \times 10^8$ 0.0010
Yield Point $2.4 \times 10^8$ 0.0015
Ultimate Tensile Strength $4.0 \times 10^8$ 0.1500
From the stress-strain data above, determine Young's Modulus ($Y$) of the material in its elastic regime.

Wire X has length $L$ and radius $r$. Wire Y is made of the same material but has length $2L$ and radius $2r$. Both wires are stretched by the same tensile load $F$.
What is the ratio of elastic potential energy stored in Wire X to Wire Y ($U_X / U_Y$)?

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Grade 11 : Physics - Solids (Set 4)

Grade 11 : Physics - Solids (Set 4)— Questions & Detailed Solutions

MaterialYoung's Modulus $Y$ ($\times 10^{10}\text{ N/m}^2$)P6.0Q12.0R20.0S8.0Four wires of identical dimensions are each subjected to a strain of $\epsilon = 0.001$. Based on the table above, which material stores the maximum elastic energy density?

A vertical steel wire of original length $L = 2m$ and cross-sectional area $A = 1\text{ mm}^2$ supports a suspended mass $M = 10kg$ hanging in equilibrium ($g = 10m/s^2, Y = 2 \times 10^{11}\text{ N/m}^2$).Calculate the total elastic potential energy stored in the stretched wire.

ParameterValueOriginal Diameter ($D$)2.00 mmOriginal Length ($L$)1.00 mLongitudinal Strain ($\epsilon_z$)0.002Poisson's Ratio ($\nu$)0.30Under axial tension, what is the magnitude of the decrease in the wire's radius ($\Delta r$)?

Searle's Apparatus VariableValueWire Length ($L$)2.0 mWire Diameter ($d$)1.0 mmApplied Load Mass ($m$)4.0 kgObserved Elongation ($\Delta L$)0.50 mmUsing the experimental data in the table above ($g = 9.8m/s^2$), determine Young's Modulus ($Y$) of the wire material.

A solid metallic sphere is subjected to a uniform hydrostatic pressure $P = 1.5 \times 10^8\text{ N/m}^2$. The Bulk Modulus of the material is $B = 1.0 \times 10^{11}\text{ N/m}^2$.Calculate the percentage decrease in the radius of the sphere.

Characteristic PointStress ($\text{N/m}^2$)StrainProportional Limit$2.0 \times 10^8$0.0010Yield Point$2.4 \times 10^8$0.0015Ultimate Tensile Strength$4.0 \times 10^8$0.1500From the stress-strain data above, determine Young's Modulus ($Y$) of the material in its elastic regime.

Wire X has length $L$ and radius $r$. Wire Y is made of the same material but has length $2L$ and radius $2r$. Both wires are stretched by the same tensile load $F$.What is the ratio of elastic potential energy stored in Wire X to Wire Y ($U_X / U_Y$)?

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Material Young's Modulus $Y$ ($\times 10^{10}\text{ N/m}^2$)
P 6.0
Q 12.0
R 20.0
S 8.0
Four wires of identical dimensions are each subjected to a strain of $\epsilon = 0.001$. Based on the table above, which material stores the maximum elastic energy density?

A vertical steel wire of original length $L = 2m$ and cross-sectional area $A = 1\text{ mm}^2$ supports a suspended mass $M = 10kg$ hanging in equilibrium ($g = 10m/s^2, Y = 2 \times 10^{11}\text{ N/m}^2$).
Calculate the total elastic potential energy stored in the stretched wire.

Parameter Value
Original Diameter ($D$) 2.00 mm
Original Length ($L$) 1.00 m
Longitudinal Strain ($\epsilon_z$) 0.002
Poisson's Ratio ($\nu$) 0.30
Under axial tension, what is the magnitude of the decrease in the wire's radius ($\Delta r$)?

Searle's Apparatus Variable Value
Wire Length ($L$) 2.0 m
Wire Diameter ($d$) 1.0 mm
Applied Load Mass ($m$) 4.0 kg
Observed Elongation ($\Delta L$) 0.50 mm
Using the experimental data in the table above ($g = 9.8m/s^2$), determine Young's Modulus ($Y$) of the wire material.

A solid metallic sphere is subjected to a uniform hydrostatic pressure $P = 1.5 \times 10^8\text{ N/m}^2$. The Bulk Modulus of the material is $B = 1.0 \times 10^{11}\text{ N/m}^2$.
Calculate the percentage decrease in the radius of the sphere.

Characteristic Point Stress ($\text{N/m}^2$) Strain
Proportional Limit $2.0 \times 10^8$ 0.0010
Yield Point $2.4 \times 10^8$ 0.0015
Ultimate Tensile Strength $4.0 \times 10^8$ 0.1500
From the stress-strain data above, determine Young's Modulus ($Y$) of the material in its elastic regime.

Wire X has length $L$ and radius $r$. Wire Y is made of the same material but has length $2L$ and radius $2r$. Both wires are stretched by the same tensile load $F$.
What is the ratio of elastic potential energy stored in Wire X to Wire Y ($U_X / U_Y$)?