JEE PYQs Integer Type

Hint

(d) To find the initial volume of the liquid, we use the definition of the Bulk Modulus, which relates the pressure applied to the fractional change in volume.
Step 1: Identify the Given Values and Convert Units
Initial pressure \(\left(P_1\right)=1 \mathrm{~atm}\)
Final pressure \(\left(P_2\right)=5 \mathrm{~atm}\)
Change in pressure \((\Delta P)=P_2-P_1=4 \mathrm{~atm}\)
Convert \(\Delta P\) to Pascals: \(4 \times 10^5 \mathrm{~Pa}\) (since \(1 \mathrm{~atm}=10^5 \mathrm{~Pa}\))
Change in volume \((\Delta V)=0.8 \mathrm{~cm}^3=0.8 \times 10^{-6} \mathrm{~m}^3\)
Bulk modulus \((B)=2 \mathrm{GPa}=2 \times 10^9 \mathrm{~Pa}\)
Step 2: Set up the Bulk Modulus Formula
The Bulk Modulus \((B)\) is given by:
\(
B=\frac{\Delta P}{\left(\frac{\Delta V}{V}\right)}
\)
Where \(V\) is the initial volume. Rearranging the formula to solve for \(V\) :
\(
V=\frac{B \cdot \Delta V}{\Delta P}
\)
Step 3: Calculate the Initial Volume
Substitute the values into the formula:
\(
\begin{gathered}
V=\frac{\left(2 \times 10^9 \mathrm{~Pa}\right) \times\left(0.8 \times 10^{-6} \mathrm{~m}^3\right)}{4 \times 10^5 \mathrm{~Pa}} \\
V=\frac{1.6 \times 10^3}{4 \times 10^5} \mathrm{~m}^3 \\
V=\frac{1600}{400,000} \mathrm{~m}^3 \\
V=0.004 \mathrm{~m}^3
\end{gathered}
\)
Step 4: Convert to Litres
Since \(1 \mathrm{~m}^3=1000\) litres:
\(
\begin{gathered}
V=0.004 \times 1000 \text { litres } \\
V=4 \text { litres }
\end{gathered}
\)
Final Answer: The initial volume of the liquid was \(\mathbf{4}\) litres.

Hint

(a)

Step 1: Identify the Shear Modulus Formula
The shear modulus \((\eta)\) of a material is defined as the ratio of shear stress \((\tau)\) to shear strain (\(\theta\)). For a force \(F\) applied parallel to a face of area \(A\), the relationship is given by:
\(
\eta=\frac{\tau}{\theta}=\frac{F}{A \theta}
\)
Step 2: Determine the Relevant Areas
The problem states the slabs have a square cross-section with sides \(l\) and thicknesses \(d_1, d_2\). Since the force is applied to the narrow faces, and the thickness \(d\) is less than \(l\) , the area \(A\) of the narrow face is:
\(
A=l \times d
\)
Step 3: Establish the Ratio of Shear Moduli
Using the formula from Step 1 for both materials:
\(
\begin{aligned}
& \eta_1=\frac{F}{l d_1 \theta_1} \\
& \eta_2=\frac{F}{l d_2 \theta_2}
\end{aligned}
\)
Taking the ratio of \(\eta_2\) to \(\eta_1\) :
\(
\frac{\eta_2}{\eta_1}=\frac{F /\left(l d_2 \theta_2\right)}{F /\left(l d_1 \theta_1\right)}=\frac{d_1 \theta_1}{d_2 \theta_2}
\)
Step 4: Calculate the Final Value
Substitute the given conditions \(d_2=2 d_1\) and \(\theta_2=2 \theta_1\) into the ratio:
\(
\frac{n_2}{n_1}=\frac{d_1 \cdot \theta_1}{\left(2 d_1\right) \cdot\left(2 \theta_1\right)}=\frac{1}{4}
\)
Given \(\eta_1=4 \times 10^9 \mathrm{~N} / \mathrm{m}^2\) :
\(
\eta_2=\frac{1}{4} \times 4 \times 10^9=1 \times 10^9 \mathrm{~N} / \mathrm{m}^2
\)
Comparing this with the form \(x \times 10^9 \mathrm{~N} / \mathrm{m}^2\), we find \(x=1\).

Hint

(a) To find the original (natural) length of the string, we use Hooke’s Law, which states that the extension of an elastic material is directly proportional to the tension applied to it.
Step 1: Set up the Linear Relationship
The stretched length (\(l\)) of a string is equal to its original length (\(L\)) plus the extension (\(\Delta L\)). According to Hooke’s Law:
\(
\Delta L=k \cdot T
\)
where \(k\) is a constant related to the material’s properties (\(L / A Y\)). Therefore:
\(
l=L+k T
\)
Step 2: Formulate Equations for Both Cases
We are given two states of tension and length:
For \(T_1=5 \mathrm{~N}\) and \(l_1=1.4 \mathrm{~m}\) :
\(
1.4=L+5 k \dots(1)
\)
For \(T_2=7 \mathrm{~N}\) and \(l_2=1.56 \mathrm{~m}\) :
\(
1.56=L+7 k \dots(2)
\)
Step 3: Solve for the Constant \(k\)
Subtract Equation 1 from Equation 2 to eliminate \(L\) :
\(
\begin{gathered}
(1.56-1.4)=(L+7 k)-(L+5 k) \\
0.16=2 k \\
k=\frac{0.16}{2}=0.08 \mathrm{~m} / \mathrm{N}
\end{gathered}
\)
Step 4: Calculate the Original Length \(L\)
Substitute the value of \(k\) back into Equation 1:
\(
\begin{gathered}
1.4=L+5(0.08) \\
1.4=L+0.40 \\
L=1.4-0.40 \\
L=1.0 \mathrm{~m}
\end{gathered}
\)
Final Answer: The original length of the string is \(\mathbf{1 . 0 ~ m}\).

Hint

(c) To find the elastic potential energy density of the wire, we need to determine the longitudinal strain using the Poisson’s ratio and then apply the energy density formula.
Step 1: Calculate the Longitudinal Strain
The Poisson’s ratio \((\sigma)\) is defined as the ratio of the transverse strain \((\beta)\) to the longitudinal strain \((\alpha)\) :
\(
\sigma=\frac{\text { Transverse Strain }}{\text { Longitudinal Strain }}=\frac{\beta}{\alpha}
\)
Rearranging to solve for the longitudinal strain \((\alpha)\) :
\(
\alpha=\frac{\beta}{\sigma}
\)
Given:
Transverse strain \((\beta)=10^{-3}\)
Poisson’s ratio \((\sigma)=0.2\)
\(
\alpha=\frac{10^{-3}}{0.2}=5 \times 10^{-3}
\)
Step 2: Identify the Energy Density Formula
The elastic potential energy density (\(u\)), which is the energy stored per unit volume, is given by the formula:
\(
u=\frac{1}{2} \times \text { Stress } \text { × } \text { Strain }
\)
Since Stress \(=Y \times\) Strain, we can write:
\(
u=\frac{1}{2} \cdot Y \cdot \alpha^2
\)
Step 3: Calculation
Substitute the given values into the formula:
Young’s Modulus \((Y)=2.0 \times 10^{11} \mathrm{Nm}^{-2}\)
Longitudinal Strain \((\alpha)=5 \times 10^{-3}\)
\(
\begin{gathered}
u=\frac{1}{2} \times\left(2.0 \times 10^{11}\right) \times\left(5 \times 10^{-3}\right)^2 \\
u=\left(1.0 \times 10^{11}\right) \times\left(25 \times 10^{-6}\right) \\
u=25 \times 10^5 \mathrm{~J} / \mathrm{m}^3
\end{gathered}
\)
Final Answer: The elastic potential energy density of the wire is \(25 \times 10^5\) (in SI units).

Hint

(d) To find the volume contraction of the copper cube, we use the definition of the Bulk Modulus, which relates the pressure applied to the fractional change in volume of the material.
Step 1: Identify the Given Values
Edge length of the cube \((L): 10 \mathrm{~cm}=0.1 \mathrm{~m}\)
Hydraulic pressure \((P): 7 \times 10^6 \mathrm{~Pa}\)
Bulk modulus (\(B\)): \(1.4 \times 10^{11} \mathrm{Nm}^{-2}\)
Step 2: Calculate the Initial Volume (\(V\))
The volume of a cube is given by \(V=L^3\) :
\(
V=(0.1 \mathrm{~m})^3=0.001 \mathrm{~m}^3=10^{-3} \mathrm{~m}^3
\)
Step 3: Apply the Bulk Modulus Formula
The Bulk Modulus (\(B\)) is defined as:
\(
B=\frac{P}{\left(\frac{\Delta V}{V}\right)}
\)
Rearranging the formula to solve for the volume contraction (\(\Delta V\)):
\(
\Delta V=\frac{P \cdot V}{B}
\)
Step 4: Calculate the Volume Contraction \(\mathrm{m}^3\)
Substitute the values into the equation:
\(
\begin{gathered}
\Delta V=\frac{\left(7 \times 10^6 \mathrm{~Pa}\right) \times\left(10^{-3} \mathrm{~m}^3\right)}{1.4 \times 10^{11} \mathrm{Nm}^{-2}} \\
\Delta V=\frac{7 \times 10^3}{1.4 \times 10^{11}} \\
\Delta V=5 \times 10^{-8} \mathrm{~m}^3
\end{gathered}
\)
Step 5: Convert the Result to \(\mathrm{mm}^3\)
Since \(1 \mathrm{~m}=10^3 \mathrm{~mm}\), then \(1 \mathrm{~m}^3=\left(10^3\right)^3 \mathrm{~mm}^3=10^9 \mathrm{~mm}^3\) :
\(
\begin{gathered}
\Delta V=\left(5 \times 10^{-8}\right) \times 10^9 \mathrm{~mm}^3 \\
\Delta V=5 \times 10^1 \mathrm{~mm}^3 \\
\Delta V=50 \mathrm{~mm}^3
\end{gathered}
\)
Final Answer: The volume contraction of the solid copper cube would be \(50 \mathrm{~mm}^3\).

Hint

(a) To find the value of \(c\) in the dimensional formula \(\left[M^a L^b T^c\right]\), we need to determine the dimensions of the quantity “modulus of elasticity per unit torque.”
Step 1: Dimensions of Modulus of Elasticity (\(E\))
The modulus of elasticity (whether Young’s, Bulk, or Shear modulus) is defined as the ratio of stress to strain.
\(
\text { Modulus of Elasticity }(E)=\frac{\text { Stress }}{\text { Strain }}
\)
Stress is force per unit area: \(\left[M L^{-1} T^{-2}\right]\).
Strain is a dimensionless quantity (\(\left[M^0 L^0 T^0\right]\)). Therefore, the dimensions of \(E\) are:
\(
[E]=\left[M^1 L^{-1} T^{-2}\right]
\)
Step 2: Dimensions of Torque (\(\tau\))
Torque is defined as the product of force and the perpendicular distance (lever arm).
\(
\operatorname{Torque}(\tau)=\text { Force × Distance }
\)
\(
\operatorname{Torque}(\tau)=\left[M L T^{-2}\right] \times[L]=\left[M^1 L^2 T^{-2}\right]
\)
So, the dimensions of \(\tau\) are:
\(
[\tau]=\left[M^1 L^2 T^{-2}\right]
\)
Step 3: Dimensions of the Measured Quantity
The problem asks for the dimensions of the modulus of elasticity per unit torque \(\left(\frac{E}{\tau}\right)\) :
\(
\begin{gathered}
\text { Dimension }=\frac{[E]}{[\tau]}=\frac{\left[M^1 L^{-1} T^{-2}\right]}{\left[M^1 L^2 T^{-2}\right]} \\
=\left[M^{1-1} L^{-1-2} T^{-2-(-2)}\right] \\
=\left[M^0 L^{-3} T^0\right]
\end{gathered}
\)
Step 4: Comparing with \(\left[M^a L^b T^c\right]\)
The calculated dimensions are \(\left[M^0 L^{-3} T^0\right]\). Comparing this with the given form \(\left[M^a L^b T^c\right]\) :
\(a=0\)
\(b=-3\)
\(c=0\)

Hint

(c) To find the increase in pressure required, we use the definition of the Bulk Modulus, which relates the pressure applied to the fractional change in volume.
Step 1: Identify the Bulk Modulus Formula
The Bulk Modulus \((B)\) is defined as the ratio of the change in pressure \((\Delta P)\) to the fractional change in volume (volumetric strain):
\(
B=\frac{\Delta P}{\left|\frac{\Delta V}{V}\right|}
\)
Rearranging the formula to solve for the change in pressure \((\Delta P)\) :
\(
\Delta P=B \times\left|\frac{\Delta V}{V}\right|
\)
Step 2: Given Values
Bulk Modulus (B): \(2.15 \times 10^9 \mathrm{Nm}^{-2}\)
Percentage decrease in volume: \(0.2 \%\)
Fractional change in volume \(\left(\frac{\Delta V}{V}\right): \frac{0.2}{100}=0.002=2 \times 10^{-3}\)
Step 3: Calculation
Substitute the values into the formula:
\(
\begin{gathered}
\Delta P=\left(2.15 \times 10^9 \mathrm{Nm}^{-2}\right) \times\left(2 \times 10^{-3}\right) \\
\Delta P=4.3 \times 10^6 \mathrm{Nm}^{-2}
\end{gathered}
\)
Step 4: Express in the Required Form
The question asks for the pressure in the form \(P \times 10^5 \mathrm{Nm}^{-2}\). We can rewrite our result as:
\(
\Delta P=43 \times 10^5 \mathrm{Nm}^{-2}
\)
Comparing this with \(P \times 10^5 \mathrm{Nm}^{-2}\), we get:
\(
P=43
\)

Hint

(b) To solve this, we must first determine the actual tension acting on the wire.
Step 1: Determine the Tension in the Wire
A common point of confusion in this type of problem is whether the forces “add up.”
When two people pull the ends of a wire with 200 N each in opposite directions, the wire is in equilibrium.
This is identical to one end being fixed to a wall and the other end being pulled with 200 N.
Therefore, the Tension (\(T\)) in the wire is 200 N , not 400 N .
Step 2: Identify the Given Values
Tension (\(F\)): 200 N
Original Length \((L): 2 \mathrm{~m}\)
Young’s Modulus \((Y): 1 \times 10^{11} \mathrm{Nm}^{-2}\)
Area \((A): 2 \mathrm{~cm}^2=2 \times\left(10^{-2} \mathrm{~m}\right)^2=2 \times 10^{-4} \mathrm{~m}^2\)
Step 3: Calculate the Extension (\(\boldsymbol{\Delta} \boldsymbol{L}\))
Using the formula for Young’s Modulus:
\(
Y=\frac{F \cdot L}{A \cdot \Delta L} \Longrightarrow \Delta L=\frac{F \cdot L}{A \cdot Y}
\)
Substitute the values:
\(
\begin{gathered}
\Delta L=\frac{200 \times 2}{\left(2 \times 10^{-4}\right) \times\left(1 \times 10^{11}\right)} \\
\Delta L=\frac{400}{2 \times 10^7} \\
\Delta L=200 \times 10^{-7} \mathrm{~m} \\
\Delta L=20 \times 10^{-6} \mathrm{~m}
\end{gathered}
\)
Step 4: Convert to Micrometers (\(\mu \mathrm{m}\))
Since \(1 \mu \mathrm{~m}=10^{-6} \mathrm{~m}\) :
\(
\Delta L=20 \mu \mathrm{~m}
\)
Final Answer: The wire will extend in length by \(20 \mu \mathrm{~m}\).

Hint

(a) 

To find the cross-sectional area \(A\), we need to analyze the tension in the wire created by the suspended mass and relate that tension to the elastic stretching of the wire.
Step 1: Analyze the Equilibrium at the Center
When the mass \(m\) is suspended at the midpoint, the wire sags. Let \(T\) be the tension in each half of the wire. The vertical components of the tension support the weight of the mass:
\(
2 T \sin \theta=m g
\)
Since \(\theta\) is very small (\(\theta=1 / 100 \mathrm{rad}\)), we can use the approximation \(\sin \theta \approx \theta\) :
\(
2 T \theta=m g \Longrightarrow T=\frac{m g}{2 \theta}
\)
Substituting the given values (\(m=2 \mathrm{~kg}, g=10 \mathrm{~m} / \mathrm{s}^2, \theta=0.01 \mathrm{rad}\)):
\(
T=\frac{2 \times 10}{2 \times 0.01}=\frac{20}{0.02}=1000 \mathrm{~N}
\)
Step 2: Determine the Longitudinal Strain
Let \(L\) be the original half-length of the wire \((1 \mathrm{~m})\) and \(L^{\prime}\) be the stretched half-length. From the geometry of the triangle formed by the sag:
\(
\cos \theta=\frac{L}{L^{\prime}} \Longrightarrow L^{\prime}=\frac{L}{\cos \theta}
\)
\(
\cos \theta=\frac{L}{L^{\prime}} \Longrightarrow L^{\prime}=\frac{L}{\cos \theta}
\)
The longitudinal strain \((\alpha)\) is:
\(
\alpha=\frac{L^{\prime}-L}{L}=\frac{\frac{L}{\cos \theta}-L}{L}=\frac{1}{\cos \theta}-1
\)
Using the small angle approximation \(\cos \theta \approx 1-\frac{\theta^2}{2}\) :
\(
\alpha \approx\left(1-\frac{\theta^2}{2}\right)^{-1}-1 \approx\left(1+\frac{\theta^2}{2}\right)-1=\frac{\theta^2}{2}
\)
Substituting \(\theta=0.01\) :
\(
\alpha=\frac{(0.01)^2}{2}=\frac{10^{-4}}{2}=5 \times 10^{-5}
\)
Step 3: Calculate the Area \(\boldsymbol{A}\)
Young’s Modulus (\(Y\)) is defined as:
\(
Y=\frac{\text { Stress }}{\text { Strain }}=\frac{T / A}{\alpha} \Longrightarrow A=\frac{T}{Y \alpha}
\)
Substitute the values \(\left(T=1000 \mathrm{~N}, Y=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2, \alpha=5 \times 10^{-5}\right)\) :
\(
\begin{gathered}
A=\frac{1000}{\left(2 \times 10^{11}\right) \times\left(5 \times 10^{-5}\right)} \\
A=\frac{10^3}{10 \times 10^6}=\frac{10^3}{10^7} \\
A=10^{-4} \mathrm{~m}^2
\end{gathered}
\)
Comparing this to the requested format \(A \times 10^{-4} \mathrm{~m}^2\). The value of \(A\) is 1.

Hint

(c) To find the maximum length of a wire that can hang under its own weight without breaking, we must ensure that the maximum stress (occurring at the top of the wire) does not exceed the breaking stress of the material.

Step 1: Identify the Stress at the Support
For a wire hanging vertically, the maximum tension \(T\) is at the point of suspension, as it must support the entire weight of the wire.
\(
\begin{gathered}
T=\text { Mass of the wire } \text { × } g_{\text {planet }} \\
T=(\text { Area } \text { × } \text { Length } \text { × } \text { Density }) \times g_{\text {planet }}
\end{gathered}
\)
The maximum stress \(\sigma\) is defined as tension per unit area:
\(
\begin{gathered}
\sigma=\frac{T}{A}=\frac{A \cdot L \cdot \rho \cdot g_{\text {planet }}}{A} \\
\sigma=L \cdot \rho \cdot g_{\text {planet }}
\end{gathered}
\)
Step 2: Set the Condition for Maximum Length
The wire will break if the stress exceeds the breaking stress \(\left(\sigma_b\right)\). Therefore, the maximum length \(L\) is achieved when the stress equals the breaking stress:
\(
\sigma_b=L \cdot \rho \cdot g_{\text {planet }}
\)
\(
L=\frac{\sigma_b}{\rho \cdot g_{\text {planet }}}
\)
Step 3: Given Values
Breaking Stress \(\left(\sigma_b\right): 1.2 \times 10^8 \mathrm{~N} / \mathrm{m}^2\)
Density ( \(\rho\) ): \(6 \times 10^4 \mathrm{~kg} / \mathrm{m}^3\)
Gravity on Earth (\(g\)): \(10 \mathrm{~m} / \mathrm{s}^2\)
Gravity on Planet \(\left(g_{\text {planet }}\right): \frac{1}{3} \times 10=\frac{10}{3} \mathrm{~m} / \mathrm{s}^2\)
Step 4: Calculation
Substitute the values into the formula:
\(
\begin{gathered}
L=\frac{1.2 \times 10^8}{\left(6 \times 10^4\right) \times\left(\frac{10}{3}\right)} \\
L=\frac{1.2 \times 10^8}{2 \times 10^5} \\
L=0.6 \times 10^3 \\
L=600 \mathrm{~m}
\end{gathered}
\)
Final Answer: The maximum length of the wire without breaking is 600 m.

Hint

(c) To find the tension for the given length, we can use Hooke’s Law, which describes the relationship between the force (tension) applied to a spring and its extension.
The Formula:
Hooke’s Law is expressed as:
\(
F=k\left(L-L_0\right)
\)
Where:
\(F\) is the tension (force).
\(k\) is the spring constant.
\(L\) is the current length.
\(L_0\) is the natural (unstretched) length of the spring.
Step 1: Set up the equations
From the problem, we have two distinct conditions:
Under tension 3 N , length is \(a\) :
\(
3=k\left(a-L_0\right)-(\text { Eq. 1 })
\)
Under tension \(\mathbf{2} \mathbf{N}\), length is \(b\) :
\(
2=k\left(b-L_0\right)
\)
Step 2: Solve for \(k\) and \(L_0\)
Divide (Eq. 1) by (Eq. 2) to eliminate \(k\) :
\(
\begin{gathered}
\frac{3}{2}=\frac{a-L_0}{b-L_0} \\
3\left(b-L_0\right)=2\left(a-L_0\right) \\
3 b-3 L_0=2 a-2 L_0
\end{gathered}
\)
Now, substitute \(L_0\) back into (Eq. 1) to find \(k\) :
\(
\begin{gathered}
3=k(a-(3 b-2 a)) \\
3=k(3 a-3 b)
\end{gathered}
\)
\(
k=\frac{1}{a-b}
\)
Step 3: Find the tension for length (\(3 a-2 b\))
Let the required tension be \(F^{\prime}\). Using the formula \(F^{\prime}=k\left(L^{\prime}-L_0\right)\), where \(L^{\prime}=3 a-2 b\) :
\(
F^{\prime}=k[(3 a-2 b)-(3 b-2 a)]
\)
Substitute the value of \(k\) :
\(
\begin{gathered}
F^{\prime}=\frac{1}{a-b}[3 a-2 b-3 b+2 a] \\
F^{\prime}=\frac{1}{a-b}[5 a-5 b] \\
F^{\prime}=\frac{5(a-b)}{a-b}
\end{gathered}
\)
\(
F^{\prime}=5
\)
The value of tension for the length \((3 a-2 b)\) will be 5 N.

Hint

(b)

To find the ratio of the longitudinal strain of the upper wire to that of the lower wire, we analyze the forces acting on each wire using the principles of elasticity.
Identify the Tension in each wire:
Lower Wire: This wire is attached to the 1 kg load. It only supports the weight of this load.
\(
T_1=M_2 \times g=1 \mathrm{~kg} \times 10 \mathrm{~m} / \mathrm{s}^2=10 \mathrm{~N}
\)
Upper Wire: This wire is attached to the ceiling and supports the 2 kg load, the lower wire, and the 1 kg load. Neglecting the mass of the wires, the total weight supported is the sum of both loads.
\(
T_2=\left(M_1+M_2\right) \times g=(2 \mathrm{~kg}+1 \mathrm{~kg}) \times 10 \mathrm{~m} / \mathrm{s}^2=30 \mathrm{~N}
\)
Longitudinal Strain Formula:
The longitudinal strain \((\epsilon)\) is defined as the change in length per unit original length. According to Young’s Modulus (\(Y\)):
\(
\begin{gathered}
Y=\frac{\text { Stress }}{\text { Strain }}=\frac{T / A}{\epsilon} \\
\epsilon=\frac{T}{A \cdot Y}
\end{gathered}
\)
Where:
\(T\) is the tension.
\(A\) is the area of cross-section \(\left(0.005 \mathrm{~cm}^2\right)\).
\(Y\) is the Young’s Modulus \(\left(2 \times 10^{11} \mathrm{Nm}^{-2}\right)\).
Calculate the Ratio:
Since both wires are “similar”, they have the same area of cross-section (\(\boldsymbol{A}\)) and the same Young’s Modulus (\(Y\)). Therefore, the strain is directly proportional to the tension:
\(
\epsilon \propto T
\)
The ratio of the strains is:
\(
\begin{gathered}
\frac{\epsilon_2}{\epsilon_1}=\frac{T_2}{T_1} \\
\frac{\epsilon_2}{\epsilon_2}=\frac{30 \mathrm{~N}}{10 \mathrm{~N}}=3
\end{gathered}
\)
Final Answer: The ratio of the longitudinal strain of the upper wire to that of the lower wire is 3.

Hint

(b) Step 1: Calculate the Tension in the Wire
In an Atwood machine system where two masses \(m_1=2 \mathrm{~kg}\) and \(m_2=4 \mathrm{~kg}\) are connected over a smooth pulley, the tension \(T\) in the wire is given by the formula:
\(
T=\frac{2 m_1 m_2}{m_1+m_2} g
\)
Substituting the given values \(\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)\) :
\(
T=\frac{2 \times 2 \times 4}{2+4} \times 10=\frac{16}{6} \times 10=\frac{80}{3} \mathrm{~N}
\)
Step 2: Calculate the Longitudinal Strain
Longitudinal strain \(\epsilon\) is defined as the ratio of stress to Young’s modulus \(Y\). The stress is the tension \(T\) divided by the cross-sectional area \(A=\pi r^2\).
\(
\epsilon=\frac{\text { Stress }}{Y}=\frac{T}{A Y}=\frac{T}{\pi r^2 Y}
\)
Given \(r=4.0 \times 10^{-5} \mathrm{~m}\) and \(Y=2.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^2\) :
\(
\begin{gathered}
\epsilon=\frac{80 / 3}{\pi\left(4.0 \times 10^{-5}\right)^2\left(2.0 \times 10^{11}\right)} \\
\epsilon=\frac{80}{3 \times \pi \times 16 \times 10^{-10} \times 2 \times 10^{11}} \\
\epsilon=\frac{80}{3 \times \pi \times 32 \times 10^1}=\frac{80}{960 \pi} \\
\epsilon=\frac{1}{12 \pi}
\end{gathered}
\)
Step 3: Determine the value of \(\alpha\)
Comparing the calculated strain to the given form \(\frac{1}{\alpha \pi}\) :
Step 3: Determine the value of \(\alpha\)
Comparing the calculated strain to the given form \(\frac{1}{\alpha \pi}\) :
\(
\begin{aligned}
\frac{1}{12 \pi} & =\frac{1}{\alpha \pi} \\
\alpha & =12
\end{aligned}
\)

Hint

(d) Step 1: Calculate the change in pressure using Bulk Modulus
The Bulk Modulus \(\boldsymbol{B}\) is defined as the ratio of the change in pressure \(\boldsymbol{\Delta} \boldsymbol{P}\) to the fractional change in volume \(\frac{\Delta V}{V}\). The formula is given by:
\(
B=\frac{\Delta P}{\frac{\Delta V}{V}}
\)
Given the volume decrease is \(0.02 \%\), the fractional change is:
\(
\frac{\Delta V}{V}=\frac{0.02}{100}=2 \times 10^{-4}
\)
Substituting the given values \(B=9 \times 10^8 \mathrm{Nm}^{-2}\) :
\(
\Delta P=B \times \frac{\Delta V}{V}=\left(9 \times 10^8\right) \times\left(2 \times 10^{-4}\right)=1.8 \times 10^5 \mathrm{~Pa}
\)
Step 2: Determine the depth from the hydrostatic pressure
The change in pressure at a depth \(h\) in a fluid of density \(\rho\) is given by the hydrostatic pressure formula:
\(
\Delta P=\rho g h
\)
We are given \(\rho=10^3 \mathrm{kgm}^{-3}\) and \(g=10 \mathrm{~ms}^{-2}\). Rearranging to solve for \(h\) :
\(
h=\frac{\Delta P}{\rho g}
\)
Substituting the calculated \(\Delta P\).
\(
h=\frac{1.8 \times 10^5}{10^3 \times 10}=\frac{1.8 \times 10^5}{10^4}=18 \mathrm{~m}
\)
The rubber ball must be taken to a depth of 18 m.

Hint

(b) 

\(
\begin{aligned}
& 3 g-T_1=3 a \\
& T_1-T_2=3 a \\
& T_2=3 a \\
& =3 g=9 a
\end{aligned}
\)
\(
a=g / 3 \Rightarrow T_2=3 g / 3=g=10 \mathrm{~N}
\)
\(
T_1=3 a + T_2=g+T_2=20 \mathrm{~N}
\)
Calculate the longitudinal strain on wire \(\boldsymbol{B}\):
The longitudinal strain is given by the formula Strain \(=\frac{\text { Stress }}{Y}=\frac{T_1}{A \cdot Y}\).
Given values:
\(A=0.005 \mathrm{~cm}^2=0.005 \times 10^{-4} \mathrm{~m}^2=5 \times 10^{-7} \mathrm{~m}^2\).
\(Y=2 \times 10^{11} \mathrm{Nm}^{-2}\).
Calculation:
\(
\begin{gathered}
\text { Strain }=\frac{20}{\left(5 \times 10^{-7}\right) \times\left(2 \times 10^{11}\right)} \\
\text { Strain }=\frac{20}{10 \times 10^4}=\frac{20}{10^5}=2 \times 10^{-4}
\end{gathered}
\)
The longitudinal strain on wire \(B\) is \(2 \times 10^{-4}\).

Hint

(b) To solve for the ratio of the forces \(\frac{F_1}{F_2}\), we need to use the relationship between Young’s Modulus, force, area, and length.
Step 1: The Core Formula
The Young’s Modulus (\(Y\)) is given by:
\(
Y=\frac{\text { Stress }}{\text { Strain }}=\frac{F / A}{\Delta l / l}=\frac{F l}{A \Delta l}
\)
Rearranging for Force \((F)\) :
\(
F=\frac{Y A \Delta l}{l}
\)
Step 2: Constraints from the Problem
Same Material: \(Y_P=Y_Q=Y\).
Same Volume \((V)\) : Since \(V=A \times l\), we can say \(l=\frac{V}{A}\).
Same Extension: \(\Delta l_P=\Delta l_Q=\Delta l\).
Area Ratio: \(\frac{A_p}{A_q}=\frac{4}{1}\).
Step 3: Expressing Force in terms of Area and Volume
Substitute \(l=\frac{V}{A}\) into the force equation:
\(
F=\frac{Y A \Delta l}{(V / A)}=\frac{Y A^2 \Delta l}{V}
\)
Since \(Y, \Delta l\), and \(V\) are constant for both wires in this specific scenario, we find that:
\(
F \propto A^2
\)
Step 4: Calculating the Ratio
Now, we can set up the ratio for \(F_1\) (force on \(P\)) and \(F_2\) (force on \(Q\)):
\(
\frac{F_1}{F_2}=\frac{A_P^2}{A_Q^2}=\left(\frac{A_P}{A_Q}\right)^2
\)
Given that \(\frac{A_p}{A_Q}=\frac{4}{1}\) :
\(
\frac{F_1}{F_2}=\left(\frac{4}{1}\right)^2=\frac{16}{1}
\)

Hint

(a) Step 1: Calculate the Pressure at the Bottom
The pressure change \(\Delta P\) at the bottom of the ocean, relative to the surface, is determined by the hydrostatic pressure formula \(\Delta P=\rho g h\).
\(
\Delta P=1000 \times 10 \times 4000=4 \times 10^7 \mathrm{Nm}^{-2}
\)
Step 2: Relate Pressure to Bulk Modulus
The Bulk Modulus (\(B\)) of a fluid relates the change in pressure to the fractional change in volume (compression). The formula for fractional compression is:
\(
\frac{\Delta V}{V}=\frac{\Delta P}{B}
\)
Step 3: Calculate the Fractional Compression
Substitute the known values into the equation to find the ratio:
\(
\frac{\Delta V}{V}=\frac{4 \times 10^7}{2 \times 10^9}=2 \times 10^{-2}
\)
Step 4: Determine the Value of \(\boldsymbol{\alpha}\)
The problem states that the fractional compression is expressed as \(\alpha \times 10^{-2}\). By comparing this to our calculated result:
\(
\begin{aligned}
\alpha \times 10^{-2} & =2 \times 10^{-2} \\
\alpha & =2
\end{aligned}
\)

Hint

(a) Step 1: Identify the formula for elastic potential energy
The elastic potential energy \(U\) stored in a stretched wire is related to its Young’s modulus \(Y\), cross-sectional area \(A\), original length \(L_{\text {, and extension }} \Delta L\) by the following expression:
\(
U=\frac{1}{2} \times \text { Stress } \text { × } \text { Strain } \text { × } \text { Volume }
\)
Since Stress \(=Y \times\) Strain and Strain \(=\frac{\Delta L}{L}\), we can rewrite the formula as:
\(
U=\frac{1}{2} \times Y \times\left(\frac{\Delta L}{L}\right)^2 \times(A \times L)
\)
Simplifying the expression for \(U\) :
\(
U=\frac{Y A(\Delta L)^2}{2 L}
\)
Step 2: Substitute the given values
We are given:
Length \(L=20 \mathrm{~m}\)
Extension \(\Delta L=2 \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m}\)
Energy \(U=80 \mathrm{~J}\)
Young’s modulus \(Y=2.0 \times 10^{11} \mathrm{Nm}^{-2}\)
Rearrange the formula to solve for the cross-sectional area \(\boldsymbol{A}\) :
\(
A=\frac{2 U L}{Y(\Delta L)^2}
\)
Substitute the numerical values:
\(
\begin{gathered}
A=\frac{2 \times 80 \times 20}{2.0 \times 10^{11} \times\left(2 \times 10^{-2}\right)^2} \\
A=\frac{3200}{2.0 \times 10^{11} \times 4 \times 10^{-4}} \\
A=\frac{3200}{8 \times 10^7} \\
A=40 \times 10^{-6} \mathrm{~m}^2
\end{gathered}
\)
Step 3: Convert units to \(\mathrm{mm}^2\)
To convert the area from square meters (\(\mathrm{m}^2\)) to square millimeters (\(\mathrm{mm}^2\)), we use the conversion factor \(1 \mathrm{~m}^2=10^6 \mathrm{~mm}^2\) :
\(
\begin{gathered}
A=40 \times 10^{-6} \times 10^6 \mathrm{~mm}^2 \\
A=40 \mathrm{~mm}^2
\end{gathered}
\)
The cross sectional area of the wire is \(40 \mathrm{~mm}^2\).

Hint

(d) Step 1: Relating Tension and Young’s Modulus
The tension \(T\) in the wire can be derived from the formula for Young’s Modulus \(Y\) :
\(
Y=\frac{\text { Stress }}{\text { Strain }}=\frac{T / A}{\Delta L / L}
\)
Rearranging for tension \(T\), where \(A\) is the cross-sectional area, \(L\) is the original length, and \(\boldsymbol{\Delta} \boldsymbol{L}\) is the extension:
\(
T=\frac{Y A \Delta L}{L}
\)
Step 2: Determining Wave Velocity
The velocity \(v\) of a transverse wave in a stretched wire depends on the tension \(T\) and the linear mass density \(\mu\) :
\(
v=\sqrt{\frac{T}{\mu}}
\)
Since \(\mu=\) density × area \(=\rho A\), we substitute the expression for \(T\) :
\(
v=\sqrt{\frac{Y A \Delta L / L}{\rho A}}=\sqrt{\frac{Y \Delta L}{\rho L}}
\)
Step 3: Calculating Fundamental Frequency
The fundamental frequency \(f\) for a wire fixed at both ends is given by:
\(
f=\frac{v}{2 L}=\frac{1}{2 L} \sqrt{\frac{Y \Delta L}{\rho L}}
\)
Substitute the given values: \(Y=8 \times 10^{10} \mathrm{~N} / \mathrm{m}^2, \rho=8 \times 10^3 \mathrm{~kg} / \mathrm{m}^3, L=0.5 \mathrm{~m}\), and \(\Delta L=3.2 \times 10^{-4} \mathrm{~m}:\)
\(
f=\frac{1}{2(0.5)} \sqrt{\frac{\left(8 \times 10^{10}\right)\left(3.2 \times 10^{-4}\right)}{\left(8 \times 10^3\right)(0.5)}}
\)
\(
f=1 \times \sqrt{\frac{2.56 \times 10^7}{4000}}=\sqrt{6400}=80 \mathrm{~Hz}
\)
The fundamental frequency of vibration is 80 Hz.

Hint

(b) To find the natural length of the wire, we can apply Hooke’s Law, which states that the extension of a wire is proportional to the tension applied.
Step 1: Set up the equations
Let \(L_0\) be the natural length of the wire and \(k\) be the force constant (stiffness) of the wire. The formula for length \(l\) under tension \(T\) is:
\(
T=k\left(l-L_0\right)
\)
We have two scenarios:
For tension 100 N , length is \(l_1\) :
\(
100=k\left(l_1-L_0\right) \dots(1)
\)
For tension \(\mathbf{1 2 0} \mathbf{N}\), length is \(\boldsymbol{l}_2\) :
\(
120=k\left(l_2-L_0\right) \dots(2)
\)
Step 2: Use the relationship between \(l_1\) and \(l_2\)
The problem states that \(10 l_2=11 l_1\), which means:
\(
l_2=\frac{11}{10} l_1
\)
Step 3: Solve for \(L_0\)
Divide (Eq. 1) by (Eq. 2) to eliminate \(k\) :
\(
\begin{gathered}
\frac{100}{120}=\frac{l_1-L_0}{l_2-L_0} \\
\frac{5}{6}=\frac{l_1-L_0}{l_2-L_0} \\
5\left(l_2-L_0\right)=6\left(l_1-L_0\right) \\
5 l_2-5 L_0=6 l_1-6 L_0 \\
L_0=6 l_1-5 l_2
\end{gathered}
\)
Now, substitute \(l_2=\frac{11}{10} l_1\) into the equation:
\(
\begin{gathered}
L_0=6 l_1-5\left(\frac{11}{10} l_1\right) \\
L_0=6 l_1-\frac{11}{2} l_1 \\
L_0=6 l_1-5.5 l_1 \\
L_0=0.5 l_1=\frac{1}{2} l_1
\end{gathered}
\)
Step 4: Determine the value of \(x\)
The problem gives the natural length as \(\frac{1}{x} l_1\). Comparing this to our result:
\(
\begin{aligned}
\frac{1}{x} l_1 & =\frac{1}{2} l_1 \\
x & =2
\end{aligned}
\)

Hint

(c)

Step 1: Tension in steel wire
\(
\begin{aligned}
& \mathrm{T}_2=2 \mathrm{~g}+\mathrm{T}_1 \\
& \mathrm{~T}_2=20+11.4 \\
& =31.4 \mathrm{~N}
\end{aligned}
\)
Step 2: Calculate Cross-Sectional Area
The area \(\boldsymbol{A}\) is given by:
\(
\begin{gathered}
A=\pi r^2=3.14 \times\left(0.2 \times 10^{-2}\right)^2=3.14 \times 0.04 \times 10^{-4} \mathrm{~m}^2 \\
A=0.1256 \times 10^{-4} \mathrm{~m}^2
\end{gathered}
\)
Step 3: Calculate Elongation (\(\boldsymbol{\Delta} \boldsymbol{L}\))
Using the formula for Young’s Modulus \(Y=\frac{T_2 \cdot L}{A \cdot \Delta L}\) :
\(
\Delta L=\frac{T_2 \cdot L}{A \cdot Y}
\)
Substitute the values:
\(
\Delta L=\frac{31.4 \times 1.6}{\left(3.14 \times 0.04 \times 10^{-4}\right) \times\left(2 \times 10^{11}\right)}
\)
Simplify the expression:
\(
\begin{gathered}
\Delta L=\frac{10 \times 1.6}{0.04 \times 10^{-4} \times 2 \times 10^{11}}=\frac{16}{0.08 \times 10^7}=\frac{16}{8 \times 10^5} \\
\Delta L=2 \times 10^{-5} \mathrm{~m}=20 \times 10^{-6} \mathrm{~m}
\end{gathered}
\)
The elongation of the steel wire is \(20 \times 10^{-6} \mathrm{~m}\).

Hint

(d) Step 1: Identify the formula for tensile stress
In an equilibrium state, the tensile stress \(\sigma\) in the wire is the ratio of the restoring force (equal to the weight of the block) to the cross-sectional area of the wire. The weight \(W\) is given by \(m g\) and the area \(A\) of a wire with diameter \(d\) is \(\frac{\pi d^2}{4}\). Thus:
\(
\sigma=\frac{F}{A}=\frac{m g}{\frac{\pi d^2}{4}}=\frac{4 m g}{\pi d^2}
\)
Step 2: Rearrange to solve for mass \(\boldsymbol{m}\)
To find the mass, we rearrange the equation to isolate \(m\) :
\(
m=\frac{\sigma \pi d^2}{4 g}
\)
Step 3: Substitute the given values and calculate
Given: \(\sigma=7 \times 10^5 \mathrm{Nm}^{-2}, d=14 \times 10^{-3} \mathrm{~m}, g=9.8 \mathrm{~ms}^{-2}\), and \(\pi=\frac{22}{7}\).
\(
\begin{gathered}
m=\frac{\left(7 \times 10^5\right) \times\left(\frac{22}{7}\right) \times\left(14 \times 10^{-3}\right)^2}{4 \times 9.8} \\
m=\frac{22 \times 10^5 \times 196 \times 10^{-6}}{39.2} \\
m=\frac{22 \times 196 \times 10^{-1}}{39.2}=\frac{4312 \times 0.1}{39.2} \\
m=\frac{431.2}{39.2} \\
m=11
\end{gathered}
\)

Hint

(c) Step 1: Calculate the Cross-sectional Area
The rod has a circular cross-section with a radius \(r=20 \mathrm{~mm}=2 \times 10^{-2} \mathrm{~m}\). The area \(\boldsymbol{A}\) is given by:
\(
A=\pi r^2=\pi\left(2 \times 10^{-2}\right)^2=4 \pi \times 10^{-4} \mathrm{~m}^2
\)
Given that \(F=62.8 \mathrm{kN} \approx 20 \pi \times 10^3 \mathrm{~N}\) (taking \(\pi \approx 3.14\)), we can use this approximation for cleaner calculation.
Step 2: Calculate the Longitudinal Strain
Young’s Modulus \(Y\) is defined as the ratio of longitudinal stress to longitudinal strain (\(\epsilon\)):
\(
Y=\frac{\text { Stress }}{\text { Strain }}=\frac{F / A}{\epsilon}
\)
Rearranging to solve for strain \(\epsilon\) :
\(
\epsilon=\frac{F}{A \cdot Y}
\)
Substitute the known values:
\(
\begin{gathered}
\epsilon=\frac{62.8 \times 10^3}{\left(4 \pi \times 10^{-4}\right) \cdot\left(2.0 \times 10^{11}\right)} \\
\epsilon=\frac{20 \pi \times 10^3}{8 \pi \times 10^7} \\
\epsilon=2.5 \times 10^{-4}
\end{gathered}
\)
To express this in the form \(x \times 10^{-5}\) :
\(
\epsilon=25 \times 10^{-5}
\)
The longitudinal strain is \(25 \times 10^{-5}\).

Hint

(a) To find the value of \(x\), we use the definition of the Bulk Modulus (\(B\)), which relates the pressure applied to a substance to the resulting fractional change in its volume.
Step 1: Identify the Formula
The Bulk Modulus (\(B\)) is given by:
\(
B=\frac{P}{\left(\frac{\Delta V}{V}\right)}
\)
Where:
\(P\) is the applied pressure.
\(\frac{\Delta V}{V}\) is the fractional compression (change in volume divided by original volume).
Step 2: Set up equations for both liquids
The problem states that the same pressure \(P\) is applied to both water and the liquid.
For Water: Fractional compression \(\left(\frac{\Delta V}{V}\right)_w=0.01 \%=\frac{0.01}{100}=10^{-4}\)
\(
B_w=\frac{P}{10^{-4}}
\)
For the Liquid: Fractional compression \(\left(\frac{\Delta V}{V}\right)_l=0.03 \%=\frac{0.03}{100}=3 \times 10^{-4}\)
\(
B_l=\frac{P}{3 \times 10^{-4}}
\)
Step 3: Calculate the ratio
The ratio of the Bulk Modulus of water to that of the liquid is:
\(
\begin{gathered}
\frac{B_w}{B_l}=\frac{P / 10^{-4}}{P /\left(3 \times 10^{-4}\right)} \\
\frac{B_w}{B_l}=\frac{3 \times 10^{-4}}{10^{-4}} \\
\frac{B_w}{B_l}=3
\end{gathered}
\)
Step 4: Determine the value of \(x\)
The problem states that the ratio is \(\frac{3}{x}\). Comparing our result to this:
\(
\frac{3}{x}=3
\)
\(
x=1
\)

Hint

(c) To find the value of mass \(M\) required to restore the rod to its original length after cooling, we can equate the thermal contraction to the mechanical elongation produced by the mass.
Step 1: Calculate the Thermal Contraction
When the rod is cooled, its length decreases. The thermal contraction (\(\Delta L_{\text {thermal }}\)) is given by:
\(
\Delta L_{\text {thermal }}=L \cdot \alpha \cdot \Delta T
\)
Where:
\(L=1 \mathrm{~m}\) (initial length)
\(\alpha=2 \times 10^{-5} \mathrm{~K}^{-1}\)
\(\Delta T=210^{\circ} \mathrm{C}-160^{\circ} \mathrm{C}=50 \mathrm{~K}\)
Step 2: Calculate the Mechanical Elongation
The mass \(M\) attached to the end of the rod creates a tension \(F=M g\). The elongation (\(\Delta L_{\text {mechanical }}\)) produced by this force is given by Hooke’s Law:
\(
\Delta L_{\text {mechanical }}=\frac{F \cdot L}{A \cdot Y}=\frac{M g \cdot L}{A \cdot Y}
\)
Where:
\(A=3 \times 10^{-6} \mathrm{~m}^2\)
\(Y=2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}\)
\(g=10 \mathrm{~m} \mathrm{~s}^{-2}\)
Step 3: Equate the two length changes
Since the rod’s length is restored to its original value of 1 m , the mechanical elongation must exactly compensate for the thermal contraction:
\(
\begin{gathered}
\Delta L_{\text {mechanical }}=\Delta L_{\text {thermal }} \\
\frac{M g \cdot L}{A \cdot Y}=L \cdot \alpha \cdot \Delta T
\end{gathered}
\)
Step 4: Solve for \(M\)
Rearranging the formula to solve for \(M\) (note that \(L\) cancels out):
\(
M=\frac{A \cdot Y \cdot \alpha \cdot \Delta T}{g}
\)
Substitute the given values:
\(
M=\frac{\left(3 \times 10^{-6}\right) \times\left(2 \times 10^{11}\right) \times\left(2 \times 10^{-5}\right) \times(50)}{10}
\)
\(
M=60 \mathrm{~kg}
\)

Hint

(d) Step 1: Analyze the Extension-Load Graph
The graph plots extension (\(l\)) against load (\(F\)). Since the curve makes an angle of \(45^{\circ}\) with the load axis, the slope of the graph is given by:
\(
\tan \left(45^{\circ}\right)=\frac{l}{F}=1
\)
From this relationship, we find that the ratio of load to extension is:
\(
\frac{F}{l}=1 \mathrm{~N} / \mathrm{m}
\)
Step 2: Determine Wire Dimensions
The given parameters of the wire are converted to SI units:
Length \((L): 62.8 \mathrm{~cm}=0.628 \mathrm{~m}\)
Diameter (d): \(4 \mathrm{~mm}=4 \times 10^{-3} \mathrm{~m}\)
Radius \((r): 2 \times 10^{-3} \mathrm{~m}\)
The cross-sectional area (\(\boldsymbol{A}\)) is:
\(
A=\pi r^2=\pi\left(2 \times 10^{-3}\right)^2=4 \pi \times 10^{-6} \mathrm{~m}^2
\)
Step 3: Calculate Young’s Modulus
The formula for Young’s modulus (1) is:
\(
Y=\frac{\text { Stress }}{\text { Strain }}=\frac{F / A}{l / L}=\left(\frac{F}{l}\right) \frac{L}{A}
\)
Substituting the known values:
\(
Y=(1) \frac{0.628}{4 \pi \times 10^{-6}}
\)
Using \(\pi \approx 3.14\), we note that \(0.628=0.2 \times 3.14=0.2 \pi\).
\(
Y=\frac{0.2 \pi}{4 \pi \times 10^{-6}}=\frac{0.2}{4} \times 10^6=0.05 \times 10^6=5 \times 10^4 \mathrm{Nm}^{-2}
\)
Comparing this result with the given form \(x \times 10^4\), we find \(x=5\).

Hint

(b) To find the maximum linear velocity, we need to consider the tension in the wire as it provides the necessary centripetal force for the block to rotate in a horizontal circle.
Step 1: Identify the Limiting Condition
The wire will break if the tension (\(T\)) exceeds the maximum force it can withstand. The maximum tension (\(T_{\text {max }}\)) is determined by the breaking stress:
\(
\begin{gathered}
\text { Breaking Stress }=\frac{T_{\max }}{\text { Area }} \\
T_{\max }=\text { Breaking Stress } \text { × } \text { Area }
\end{gathered}
\)
Given values:
Breaking Stress \(=5 \times 10^8 \mathrm{Nm}^{-2}\)
Area \((A)=10^{-4} \mathrm{~m}^2\)
\(
T_{\max }=\left(5 \times 10^8\right) \times\left(10^{-4}\right)=5 \times 10^4 \mathrm{~N}
\)
Step 2: Relate Tension to Centripetal Force
For a block rotating in a horizontal circle, the tension in the wire provides the centripetal force (\(F_c\)):
\(
T=F_c=\frac{m v^2}{R}
\)
To find the maximum linear velocity (\(v_{\text {max }}\)), we set the tension to its maximum value:
\(
T_{\max }=\frac{m v_{\max }^2}{L}
\)
(Note: In a horizontal circle with the wire fixed at one end, the radius \(R\) is equal to the length of the wire \(L\))
Step 3: Solve for Maximum Velocity (\(v_{\text {max }}\))
Rearranging the formula:
\(
\begin{gathered}
v_{\max }^2=\frac{T_{\max } \cdot L}{m} \\
v_{\max }=\sqrt{\frac{T_{\max } \cdot L}{m}}
\end{gathered}
\)
Substitute the values:
\(T_{\max }=5 \times 10^4 \mathrm{~N}\)
\(L=0.5 \mathrm{~m}\)
\(m=10 \mathrm{~kg}\)
\(
\begin{gathered}
v_{\max }=\sqrt{\frac{\left(5 \times 10^4\right) \times 0.5}{10}} \\
v_{\max }=\sqrt{\frac{25000}{10}} \\
v_{\max }=\sqrt{2500} \\
v_{\max }=50 \mathrm{~ms}^{-1}
\end{gathered}
\)
Final Answer: The maximum linear velocity of the block will be \(50 \mathrm{~ms}^{-1}\).

Hint

(b) Step 1: Calculate the Tension in the String
The speed of a transverse wave in a string is given by the formula \(v=\sqrt{\frac{T}{\mu}}\), where \(T\) is the tension and \(\mu\) is the linear mass density. Rearranging for tension:
\(
T=v^2 \mu=v^2\left(\frac{m}{L}\right)
\)
Given \(v=60 \mathrm{~m} / \mathrm{s}, m=10 \mathrm{~g}=0.01 \mathrm{~kg}\), and \(L=50 \mathrm{~cm}=0.5 \mathrm{~m}\) :
\(
T=(60)^2 \times \frac{0.01}{0.5}=3600 \times 0.02=72 \mathrm{~N}
\)
Step 2: Determine the Extension using Young’s
Modulus
Young’s modulus (\(Y\)) relates the stress and strain on the wire:
\(
Y=\frac{\text { Stress }}{\text { Strain }}=\frac{T / A}{\Delta L / L}
\)
Rearranging the formula to solve for the extension (\(\Delta L\)):
\(
\Delta L=\frac{T L}{A Y}
\)
Substitute the known values (\(A=2.0 \mathrm{~mm}^2=2.0 \times 10^{-6} \mathrm{~m}^2\) and \(Y=1.2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2\)):
\(
\begin{aligned}
& \Delta L=\frac{72 \times 0.5}{\left(2.0 \times 10^{-6}\right) \times\left(1.2 \times 10^{11}\right)} \\
& \Delta L=\frac{36}{2.4 \times 10^5}=15 \times 10^{-5} \mathrm{~m}
\end{aligned}
\)
Step 3: Find the Value of \(\boldsymbol{x}\)
The problem states the extension is in the form \(x \times 10^{-5} \mathrm{~m}\). Comparing this to our calculated result:
\(
15 \times 10^{-5} \mathrm{~m}=x \times 10^{-5} \mathrm{~m}
\)
Therefore, \(x=15\). 

Hint

(a)

To find the strain produced in the string when the body is at the bottom of its vertical circular path, we must first determine the tension acting on the string at that specific point.
Calculate Tension at the Bottom:
When a body moves in a vertical circle, the tension (\(T\)) at the bottom must overcome the weight of the body and provide the necessary centripetal force to keep it in orbit.
Step 1: Calculate the Tension at the Bottom
At the lowest point of a vertical circular path, the tension \(T\) in the string must overcome gravity and provide the necessary centripetal force. The equation of motion is:
\(
T-m g=\frac{m v^2}{R}
\)
Substituting the given values \(m=2 \mathrm{~kg}, g=10 \mathrm{~m} / \mathrm{s}^2, v=5 \mathrm{~m} / \mathrm{s}\), and \(R=0.5 \mathrm{~m}\) :
\(
\begin{aligned}
& T=m g+\frac{m v^2}{R}=(2 \times 10)+\frac{2 \times 5^2}{0.5} \\
& T=20+\frac{50}{0.5}=20+100=120 \mathrm{~N}
\end{aligned}
\)
Step 2: Calculate the Strain
Strain \((\epsilon)\) is defined as the ratio of stress to Young’s modulus \((Y)\). Stress is the tension divided by the cross-sectional area (\(\boldsymbol{A}\)).
\(
\text { Strain }=\frac{\text { Stress }}{Y}=\frac{T}{A \times Y}
\)
Given \(A=4 \mathrm{~mm}^2=4 \times 10^{-6} \mathrm{~m}^2\) and \(Y=10^{11} \mathrm{~N} / \mathrm{m}^2\) :
\(
\begin{aligned}
\text { Strain } & =\frac{120}{\left(4 \times 10^{-6}\right) \times 10^{11}} \\
\text { Strain } & =\frac{120}{4 \times 10^5}=30 \times 10^{-5}
\end{aligned}
\)
The strain produced is \(30 \times 10^{-5}\).

Hint

(b)

\(
\begin{aligned}
& Y=\frac{F \times l}{A \times \Delta l} \\
& =\frac{1}{A} \times \frac{\text { Wire length }}{\frac{\text { Extension }}{\text { load }}} \\
& Y=\frac{1}{A} \times\left(\frac{1}{0.25 \times 10^{-5}}\right) \\
& Y=10^{11} \times 2 \\
& \Rightarrow x=2
\end{aligned}
\)

Explanation: To solve for the Young’s modulus, we need to analyze the relationship between the slope of the extension-load curve and the length of the wire based on the properties of elasticity.
Step 1: Establish the Relationship
According to Hooke’s Law, the extension (\(\Delta L\)) of a wire is given by:
\(
\Delta L=\frac{F \cdot L}{A \cdot Y}
\)
Where:
\(F\) is the load (force).
\(L\) is the length of the wire.
\(A\) is the cross-sectional area.
\(Y\) is the Young’s modulus.
The slope (\(S\)) of the extension vs. load curve is defined as:
\(
S=\frac{\Delta L}{F}=\frac{L}{A \cdot Y}
\)
Step 2: Analyze the Given Graph
The problem describes a second graph where the slope \((S)\) is plotted against the length \((L)\). From the equation \(S=\left(\frac{1}{A \cdot Y}\right) L\), we can see that the slope of this new graph (let’s call it \(m\)) represents:
\(
m=\frac{S}{L}=\frac{1}{A \cdot Y}
\)
From the data provided in the JEE problem (where the graph typically shows a slope \(S= 2.5 \times 10^{-6} \mathrm{~m} / \mathrm{N}\) for a length \(L=1 \mathrm{~m}\)):
\(
\frac{S}{L}=2.5 \times 10^{-6} \mathrm{~N}^{-1}
\)
Step 3: Solve for Young’s Modulus (\(Y\))
Rearranging the equation to solve for \(Y\) :
\(
Y=\frac{1}{A \cdot(S / L)}
\)
Given values:
\(\operatorname{Area}(A)=2 \mathrm{~mm}^2=2 \times 10^{-6} \mathrm{~m}^2\)
\(\frac{S}{L}=2.5 \times 10^{-6} \mathrm{~N}^{-1}\)
\(
Y=\frac{1}{\left(2 \times 10^{-6}\right) \times\left(2.5 \times 10^{-6}\right)}
\)
\(
\begin{gathered}
Y=\frac{1}{5 \times 10^{-12}} \\
Y=0.2 \times 10^{12} \\
Y=2 \times 10^{11} \mathrm{Nm}^{-2}
\end{gathered}
\)
Step 4: Determine the Value of \(x\)
The problem states that \(Y=x \times 10^{11} \mathrm{Nm}^{-2}\). Comparing this to our result:
\(
x=2
\)

Hint

(c)

\(
\begin{aligned}
& Y=\frac{F l}{A \Delta x} \\
& \Delta x=\frac{F l}{Y A} \\
& =\frac{18 \times 10^4 \times 60 \times 10^{-2}}{25 \times 10^9 \times 60 \times 15 \times 10^{-4}} \\
& =48 \times 10^{-6} \mathrm{~m}
\end{aligned}
\)

Explanation:

Step 1: Identify Physical Parameters
From the problem description, we have a square aluminum slab with the following characteristics:
Shear Modulus (G): \(25 \times 10^9 \mathrm{Nm}^{-2}\)
Force (F): \(18.0 \times 10^4 \mathrm{~N}\)
Height of the slab (\(L\)): \(60 \mathrm{~cm}=0.6 \mathrm{~m}\)
Thickness (\(t\)): \(15 \mathrm{~cm}=0.15 \mathrm{~m}\)
Area of the narrow face (\(A\)): The force is applied to the narrow face (\(L \times t\)), so
\(
A=0.6 \mathrm{~m} \times 0.15 \mathrm{~m}=0.09 \mathrm{~m}^2
\)
Step 2: Apply the Shear Modulus Formula
The shear modulus is defined as the ratio of shear stress to shear strain:
\(
G=\frac{\text { Shear Stress }}{\text { Shear Strain }}=\frac{F / A}{\Delta x / L}
\)
Rearranging to solve for the displacement (\(\Delta x\)):
\(
\Delta x=\frac{F \cdot L}{A \cdot G}
\)
Step 3: Compute Numerical Values
Substitute the given values into the rearranged equation:
\(
\begin{gathered}
\Delta x=\frac{\left(18.0 \times 10^4\right) \cdot 0.6}{0.09 \cdot\left(25 \times 10^9\right)} \\
\Delta x=\frac{10.8 \times 10^4}{2.25 \times 10^9}=4.8 \times 10^{-5} \mathrm{~m}
\end{gathered}
\)
To convert the result to micrometers \((\mu \mathrm{m})\) :
\(
\Delta x=4.8 \times 10^{-5} \cdot 10^6 \mu \mathrm{~m}=48 \mu \mathrm{~m}
\)
The displacement of the upper edge is \(48 \mu \mathrm{~m}\).

Hint

(d) Step 1: Identify the formula for self-weight elongation
For a uniform rod of mass \(M\), length \(L\), and cross-sectional area \(A\), the tension \(T\) at a distance \(y\) from the free (bottom) end is given by \(T(y)=\frac{M g}{L} y\). The total elongation \(\Delta L\) is calculated by integrating the extension of an infinitesimal element \(d y\) :
\(
\Delta L=\int_0^L \frac{T(y) d y}{A Y}=\int_0^L \frac{M g y}{L A Y} d y=\frac{M g L}{2 A Y}
\)
Step 2: Substitute given values
Given the values \(M=20 \mathrm{~kg}, L=20 \mathrm{~m}, A=0.4 \mathrm{~m}^2, Y=2 \times 10^{11} \mathrm{Nm}^{-2}\), and \(g=10 \mathrm{~ms}^{-2}\), we substitute them into the formula:
\(
\begin{gathered}
\Delta L=\frac{20 \times 10 \times 20}{2 \times 0.4 \times 2 \times 10^{11}} \\
\Delta L=\frac{4000}{1.6 \times 10^{11}} \\
\Delta L=2500 \times 10^{-11}=25 \times 10^{-9} \mathrm{~m}
\end{gathered}
\)
Comparing the result \(\Delta L=25 \times 10^{-9} \mathrm{~m}\) with the given form \(x \times 10^{-9} \mathrm{~m}\), we find the value of \(x\) is 25.

Hint

(b)

\(
\begin{aligned}
& \frac{F / A}{l / L}=Y, A=\pi D^2 \\
& \frac{\Delta Y}{Y}=\frac{\Delta F}{F}+\frac{2 \Delta D}{D}+\frac{\Delta l}{e}+\frac{\Delta L}{L} \\
& =2 \times \frac{0.01}{0.4}+\frac{0.02}{0.4} \\
& =\frac{0.04}{0.4}=\frac{1}{10} \\
& Y=\frac{F l}{A \Delta l} \\
& =\frac{10 \times 1}{\pi(0.1 \mathrm{~mm})^2 \times 0.4 \mathrm{~mm}} \\
& =1.988 \times 10^{11} \\
& \approx 2 \times 10^{11} \\
& \frac{\Delta Y}{Y}=\frac{1}{10} \\
& \Delta Y=\frac{Y}{10}=2 \times 10^{10}
\end{aligned}
\)
The problem asked for the value of \(x\) in \(x \times 10^{10} \mathrm{Nm}^{-2}\). Comparing we get, \(x=2\).

Hint

(d)

\(
\begin{aligned}
& \frac{F / A}{\Delta L / L}=Y \\
& \Rightarrow \Delta L=\frac{F L}{A Y} \\
& \frac{\Delta L_2}{\Delta L_1}=\left(\frac{F_2}{F_1}\right) \times\left(\frac{L_2}{L_1}\right) \times\left(\frac{A_1}{A_2}\right) \\
& =4 \times 4 \times \frac{1}{16}=1 \\
& \Delta L_2=\Delta L_1=5 \mathrm{~cm}
\end{aligned}
\)

Explanation:

Step 1: Relate Extension to Physical Properties
The extension \(\Delta L\) of a wire under an applied force \(F\) is determined by Young’s Modulus \(Y\), which is defined as the ratio of stress to strain:
\(
Y=\frac{F / A}{\Delta L / L}=\frac{F L}{A \Delta L}
\)
Rearranging this formula to solve for the extension \(\Delta L\), where \(A=\pi r^2\) :
\(
\Delta L=\frac{F L}{\pi r^2 Y}
\)
Step 2: Substitute Values for the Second Wire
Given the first wire has an extension \(\Delta L_1=5 \mathrm{~cm}\) with parameters \(F, L\), and \(r\). For the second wire, the parameters are changed to \(F_2=4 F, L_2=4 L\), and \(r_2=4 r\). Since the material is the same, \(Y\) remains constant. Substituting these into the extension formula:
\(
\begin{gathered}
\Delta L_2=\frac{(4 F)(4 L)}{\pi(4 r)^2 Y} \\
\Delta L_2=\frac{16 F L}{\pi\left(16 r^2\right) Y} \\
\Delta L_2=\frac{16}{16} \cdot \frac{F L}{\pi r^2 Y} \\
\Delta L_2=\Delta L_1=5 ~cm
\end{gathered}
\)
The extension of the second wire is the same as the first, which is 5 cm.

Hint

(b) Step 1: Determine Young’s Modulus (\(Y\))
From the provided stress-strain graph given, the material follows a linear elastic relationship where the slope represents Young’s Modulus. Taking a point from the graph (e.g., stress \(\sigma=20 \mathrm{~N} / \mathrm{m}^2\) at strain \(\epsilon=1 \times 10^{-10}\)):
\(
Y=\frac{\text { Stress }}{\text { Strain }}=\frac{20}{1 \times 10^{-10}}=20 \times 10^{10} \mathrm{~N} / \mathrm{m}^2
\)
Step 2: Calculate Energy Density
The energy density \((u)\) stored in a linear elastic material is given by the formula:
\(
\begin{aligned}
&u=\frac{1}{2} \times \text { Stress } \text { × } \text { Strain }\\
&\text { Since Stress }=Y \times \text { Strain, we can rewrite it as: }\\
&u=\frac{1}{2} \times Y \times(\text { Strain })^2
\end{aligned}
\)
Substituting the given strain \(\epsilon=5 \times 10^{-4}\) :
\(
\begin{gathered}
u=\frac{1}{2} \times\left(20 \times 10^{10}\right) \times\left(5 \times 10^{-4}\right)^2 \\
u=10 \times 10^{10} \times 25 \times 10^{-8} \\
u=250 \times 10^2=25000 \mathrm{~J} / \mathrm{m}^3
\end{gathered}
\)
Step 3: Convert to \(\mathbf{k J} / \mathbf{m}^3\)
Since \(1 \mathrm{~kJ}=1000 \mathrm{~J}\) :
\(
\text { Energy density }=\frac{25000}{1000}=25 \mathrm{~kJ} / \mathrm{m}^3
\)

Hint

(d) Step 1: Establish the Relationship for Elongation
The elongation \(\Delta L\) of a wire under a load \(M\) is governed by Young’s modulus \(Y\). The formula for elongation is:
\(
\Delta L=\frac{F L}{A Y}=\frac{M g L}{A Y}
\)
Since the wire’s material (\(Y\)), dimensions (\(L, A\)), and the suspended mass ( \(M\) ) remain constant, the elongation is directly proportional to the acceleration due to gravity (\(g\)):
\(
\Delta L \propto g
\)
Step 2: Set up the Ratio
We can compare the elongation on Earth (\(\Delta L_e\)) and the planet (\(\Delta L_p\)) using the ratio of their respective gravitational accelerations (\(g_e\) and \(g_p\)):
\(
\frac{\Delta L_p}{\Delta L_e}=\frac{g_p}{g_e}
\)
Step 3: Calculate the Gravity on the Planet
Substitute the given values into the ratio:
\(
\begin{gathered}
\Delta L_e=10^{-4} \mathrm{~m} \\
\Delta L_p=6 \times 10^{-5} \mathrm{~m} \\
g_e=10 \mathrm{~ms}^{-2}
\end{gathered}
\)
Rearranging for \(g_p\) :
\(
\begin{gathered}
g_p=g_e \times \frac{\Delta L_p}{\Delta L_e} \\
g_p=10 \times \frac{6 \times 10^{-5}}{10^{-4}} \\
g_p=10 \times 0.6=6 \mathrm{~ms}^{-2}
\end{gathered}
\)
The acceleration due to gravity on the planet is \(6 \mathrm{~ms}^{-2}\).

Hint

(b) To find the tension produced in the rod, we need to calculate the thermal stress that develops when a material is heated but prevented from expanding.
Step 1: Identify the Thermal Stress Formula
When a rod of length \(L\) is heated by a temperature difference \(\Delta T\) and is fixed between rigid supports, it tries to expand by \(\Delta L=L \alpha \Delta T\). Since it is prevented from expanding, a compressive thermal stress is produced.
The thermal stress \((\sigma)\) is given by:
\(
\sigma=Y \alpha \Delta T
\)
The tension (or thermal force \(F\)) produced is:
\(
F=\text { Stress } \text { × } \text { Area }=Y A \alpha \Delta T
\)
Step 2: Substitute the Given Values
Young’s Modulus \((Y)=2.0 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}\)
Coefficient of linear expansion \((\alpha)=10^{-5}{ }^{\circ} \mathrm{C}^{-1}\)
\(\operatorname{Area}(A)=10 \mathrm{~cm}^2=10 \times 10^{-4} \mathrm{~m}^2=10^{-3} \mathrm{~m}^2\)
Change in temperature \((\Delta T)=400^{\circ} \mathrm{C}-0^{\circ} \mathrm{C}=400 \mathrm{~K}\)
Note: The length \(L\) cancels out in the stress calculation and does not affect the final force when the rod is completely restricted.
Step 3: Perform the Calculation
\(
\begin{gathered}
F=\left(2.0 \times 10^{11}\right) \times\left(10^{-3}\right) \times\left(10^{-5}\right) \times(400) \\
F=2.0 \times 400 \times 10^{11-3-5} \\
F=800 \times 10^3 \\
F=8 \times 10^5 \mathrm{~N}
\end{gathered}
\)
Step 4: Determine the Value of \(x\)
The problem states the tension is \(x \times 10^5 \mathrm{~N}\). Comparing this to our result:
\(
x=8
\)

Hint

(c) To find the depth at which the rubber ball undergoes a specific volume change, we combine the definition of Bulk Modulus with the formula for Hydrostatic Pressure.
Step 1: Identify the Formula for Bulk Modulus
The Bulk Modulus \((B)\) is defined as the ratio of the change in pressure \((\Delta P)\) to the fractional change in volume \(\left(\frac{\Delta V}{V}\right)\) :
\(
B=\frac{\Delta P}{\left(\frac{\Delta V}{V}\right)}
\)
From this, we can solve for the required increase in pressure:
\(
\Delta P=B \times\left(\frac{\Delta V}{V}\right)
\)
Step 2: Calculate the Required Pressure (\(\triangle P\))
Given values:
Bulk Modulus of rubber \((B)=9.8 \times 10^8 \mathrm{Nm}^{-2}\)
Percentage decrease in volume \(=0.5 \%\)
Fractional decrease \(\left(\frac{\Delta V}{V}\right)=\frac{0.5}{100}=5 \times 10^{-3}\)
\(
\begin{gathered}
\Delta P=\left(9.8 \times 10^8\right) \times\left(5 \times 10^{-3}\right) \\
\Delta P=49 \times 10^5 \mathrm{Nm}^{-2}
\end{gathered}
\)
Step 3: Relate Pressure to Depth (\(h\))
The gauge pressure at a depth \(h\) in a fluid is given by:
\(
\Delta P=\rho g h
\)
Where:
\(\rho\) (density of sea water) \(=10^3 \mathrm{kgm}^{-3}\)
\(g\) (acceleration due to gravity) \(=9.8 \mathrm{~m} / \mathrm{s}^2\)
\(h\) is the depth in meters.
Step 4: Solve for Depth (\(h\))
Equating the two expressions for \(\Delta P\) :
\(
\begin{gathered}
\rho g h=49 \times 10^5 \\
\left(10^3\right) \times(9.8) \times h=49 \times 10^5 \\
9.8 \times 10^3 \times h=49 \times 10^5
\end{gathered}
\)
\(
h=500 \mathrm{~m}
\)
Final Answer: When a rubber ball is taken to a depth of 500 m in deep sea, its volume decreases by \(0.5 \%\).

Hint

(c)

\(
\begin{aligned}
& B . S_1=\frac{T_{1 \max }}{8 \times 10^{-7}} \Rightarrow T_{1 \max }=8 \times 1.25 \times 100=1000 \mathrm{~N} \\
& B . S_2=\frac{T_{2 \max }}{4 \times 10^{-7}} \Rightarrow T_{2 \max }=4 \times 1.25 \times 100=500 \mathrm{~N} \\
& m=\frac{500-100}{10}=40 \mathrm{~kg}
\end{aligned}
\)

Explanation: To solve this, we must identify which wire is most restricted by its breaking stress when a mass \(M\) is added to the pan. In the typical setup for this problem, \(W_1\) is the upper wire (supporting the 20 kg mass, \(W_2\), the 10 kg mass, and the pan) and \(W_2\) is the lower wire (supporting the 10 kg mass and the pan).
Step 1: Calculate the Breaking Force for each wire
The breaking force \(\left(F_b\right)\) is the maximum tension a wire can withstand:
\(
F_b=\text { Breaking Stress } \text { × } \text { Area }
\)
For \(W_1\) :
\(
F_{b 1}=\left(1.25 \times 10^9\right) \times\left(8 \times 10^{-7}\right)=1000 \mathrm{~N}
\)
For \(W_2\) :
\(
F_{b 2}=\left(1.25 \times 10^9\right) \times\left(4 \times 10^{-7}\right)=500 \mathrm{~N}
\)
Step 2: Determine Tension in terms of added mass \(M\)
Let \(M\) be the mass placed in the pan.
Tension in \(W_2\left(T_2\right)\) : Supports the 10 kg mass and the pan mass \(M\).
\(
T_2=(10+M) g=(10+M) \times 10
\)
Tension in \(W_1\left(T_1\right)\) : Supports the 20 kg mass, the 10 kg mass, and the pan mass \(M\).
\(
T_1=(20+10+M) g=(30+M) \times 10
\)
Step 3: Find the limiting mass \(M\)
We check the breaking condition for both wires:
Condition for \(W_2\) :
\(
\begin{gathered}
T_2 \leq F_{b 2} \\
(10+M) \times 10 \leq 500 \\
10+M \leq 50 \Longrightarrow M \leq 40 \mathrm{~kg}
\end{gathered}
\)
Condition for \(W_1\) :
\(
\begin{gathered}
T_1 \leq F_{b 1} \\
(30+M) \times 10 \leq 1000 \\
30+M \leq 100 \Longrightarrow M \leq 70 \mathrm{~kg}
\end{gathered}
\)
Step 4: Conclusion
To ensure neither wire breaks, we must take the smaller of the two values. If we exceed 40 kg , wire \(W_2\) will snap even though \(W_1\) is still safe.
Final Answer: The maximum mass that can be placed in the pan is 40 kg.

Hint

(b) To find the velocity of the projected stone, we apply the principle of Conservation of Energy. The elastic potential energy stored in the stretched rubber of the catapult is converted into the kinetic energy of the stone upon release.
Step 1: Calculate the Elastic Potential Energy (\(U\))
The energy stored in a stretched wire (or rubber) is given by:
\(
U=\frac{1}{2} \times \text { Stress } \text { × } \text { Strain } \text { × } \text { Volume }
\)
Alternatively, using Young’s modulus (\(Y\)):
\(
U=\frac{1}{2} \times Y \times(\text { Strain })^2 \times \text { Volume }
\)
Given Values:
\(Y=0.5 \times 10^9 \mathrm{~N} / \mathrm{m}^2\)
Length \((L)=0.1 \mathrm{~m}\)
\(\operatorname{Area}(A)=10^{-6} \mathrm{~m}^2\)
Extension \((\Delta L)=0.04 \mathrm{~m}\)
Strain \((\epsilon)=\frac{\Delta L}{L}=\frac{0.04}{0.1}=0.4\)
Volume \((V)=A \times L=10^{-6} \times 0.1=10^{-7} \mathrm{~m}^3\)
Calculating \(U\) :
\(
\begin{gathered}
U=\frac{1}{2} \times\left(0.5 \times 10^9\right) \times(0.4)^2 \times\left(10^{-7}\right) \\
U=\frac{1}{2} \times 0.5 \times 0.16 \times 10^2 \\
U=0.25 \times 16=4 \mathrm{~J}
\end{gathered}
\)
Step 2: Relate Potential Energy to Kinetic Energy
When the stone is released, all stored energy \(U\) is converted into kinetic energy (\(K\)):
\(
U=\frac{1}{2} m v^2
\)
Given Mass:
\(m=20 \mathrm{~g}=0.02 \mathrm{~kg}\)
Solving for Velocity (\(v\)):
\(
\begin{aligned}
&\begin{gathered}
\begin{array}{c}
4=\frac{1}{2} \times 0.02 \times v^2 \\
4=0.01 \times v^2
\end{array} \\
v^2=\frac{4}{0.01}=400 \\
v=\sqrt{400}=20 \mathrm{~m} / \mathrm{s}
\end{gathered}\\
&\text { Final Answer: The velocity of the projected stone is } 20 \mathrm{~m} / \mathrm{s} \text {. }
\end{aligned}
\)

Hint

(d) To find the ratio of the lengths of wires \(A\) and \(B\), we use the expression for Young’s Modulus (\(Y\)), which relates force, length, area, and extension.
Step 1: Identify the Formula
The extension (\(\Delta L\)) of a wire is given by:
\(
\Delta L=\frac{F \cdot L}{A \cdot Y}=\frac{F \cdot L}{\left(\pi r^2\right) \cdot Y}
\)
Rearranging this to solve for the length \((L)\) :
\(
L=\frac{Y \cdot \pi r^2 \cdot \Delta L}{F}
\)
Step 2: Establish the Ratio
Since both wires are made of the same material (\(Y\) is constant) and are subjected to the same force (\(F=2 \mathrm{~N}\)), we can say:
\(
L \propto r^2 \cdot \Delta L
\)
Therefore, the ratio of the length of wire \(A\) to wire \(B(a: b)\) is:
\(
\frac{a}{b}=\frac{L_A}{L_B}=\frac{r_A^2 \cdot \Delta L_A}{r_B^2 \cdot \Delta L_B}
\)
Step 3: Substitute the Given Values
Extension of wire \(A\left(\Delta L_A\right)=2 \mathrm{~mm}\)
Extension of wire \(B\left(\Delta L_B\right)=4 \mathrm{~mm}\)
Radius of wire \(B\left(r_B\right)=4 \times\) Radius of wire \(A\left(r_A\right) \Longrightarrow \frac{r_A}{r_B}=\frac{1}{4}\)
Substitute these into the ratio:
\(
\begin{gathered}
\frac{a}{b}=\left(\frac{r_A}{r_B}\right)^2 \times\left(\frac{\Delta L_A}{\Delta L_B}\right) \\
\frac{a}{b}=\left(\frac{1}{4}\right)^2 \times\left(\frac{2}{4}\right) \\
\frac{a}{b}=\frac{1}{16} \times \frac{1}{2} \\
\frac{a}{b}=\frac{1}{32}
\end{gathered}
\)
Step 4: Determine the Value of \(x\)
The problem states that the ratio \(\frac{a}{b}\) can be expressed as \(\frac{1}{x}\).
\(
\begin{aligned}
& \frac{1}{32}=\frac{1}{x} \\
& x=32
\end{aligned}
\)

Hint

(b)

Step 1: Identify the Relationship for Elongation
The elongation \(\Delta L\) of a wire under a force \(F\) is determined by Young’s Modulus \(Y\), which is given by the formula:
\(
Y=\frac{\text { Stress }}{\text { Strain }}=\frac{F / A}{\Delta L / L}=\frac{F L}{A \Delta L}
\)
Rearranging for elongation \(\Delta L\), where the cross-sectional area \(A=\frac{\pi D^2}{4}\) :
\(
\Delta L=\frac{4 F L}{\pi D^2 Y}
\)
For a wire of the same material, \(Y\) and \(F\) remain constant, so the relationship is:
\(
\Delta L \propto \frac{L}{D^2}
\)
Step 2: Compare Initial and Final States
Let the initial length be \(L_1\) and diameter be \(D_1\). The initial elongation is \(\Delta L_1=0.04 \mathrm{~m}\). The new dimensions are \(L_2=2 L_1\) and \(D_2=2 D_1\). The ratio of the elongations is:
\(
\frac{\Delta L_2}{\Delta L_1}=\frac{L_2}{L_1} \times\left(\frac{D_1}{D_2}\right)^2
\)
Substituting the given changes:
\(
\frac{\Delta L_2}{\Delta L_1}=\frac{2 L_1}{L_1} \times\left(\frac{D_1}{2 D_1}\right)^2=2 \times \frac{1}{4}=\frac{1}{2}
\)
Step 3: Calculate the Final Value
Calculate \(\Delta L_2\) and convert it to centimeters:
\(
\Delta L_2=\frac{1}{2} \times 0.04 \mathrm{~m}=0.02 \mathrm{~m} = 2 \mathrm{~cm}
\)

Hint

(d) Step 1: Convert Units and Relate Stress to Tension
First, convert the density \(\rho\) from \(\mathrm{kg} \mathrm{cm}^{-3}\) to SI units \(\left(\mathrm{kg} \mathrm{m}^{-3}\right)\) :
\(
\rho=9 \times 10^{-3} \mathrm{~kg} \mathrm{~cm}^{-3}=9 \times 10^{-3} \times 10^6 \mathrm{~kg} \mathrm{~m}^{-3}=9000 \mathrm{~kg} \mathrm{~m}^{-3}
\)
Young’s modulus \(Y\) is defined as the ratio of stress to strain. If \(T\) is the tension and \(A\) is the cross-sectional area:
\(
Y=\frac{\text { Stress }}{\text { Strain }}=\frac{T / A}{\text { Strain }} \Longrightarrow \frac{T}{A}=Y \times \text { Strain }
\)
Step 2: Determine the Wave Speed
The speed of a transverse wave in a stretched wire is given by \(v=\sqrt{\frac{T}{\mu}}\), where \(\mu\) is the linear mass density \((\mu=\rho A)\). Substituting the expression for tension:
\(
v=\sqrt{\frac{T}{\rho A}}=\sqrt{\frac{Y \times \operatorname{Strain}}{\rho}}
\)
Substitute the given values:
\(
v=\sqrt{\frac{\left(9 \times 10^{10}\right) \times\left(4.9 \times 10^{-4}\right)}{9000}}=\sqrt{\frac{4.41 \times 10^7}{9000}}=\sqrt{4900}=70 \mathrm{~m} / \mathrm{s}
\)
Step 3: Calculate the Fundamental Frequency
The lowest (fundamental) frequency \(f\) for a wire of length \(L\) fixed at both ends is:
\(
f=\frac{v}{2 L}
\)
Given \(L=1 \mathrm{~m}\) :
\(
f=\frac{70}{2 \times 1}=35 \mathrm{~Hz}
\)
The lowest frequency of the transverse vibrations is \(\mathbf{3 5 ~ H z}\).