Answer (b) No frictional (viscous) force exists in case of ideal fluid, hence, tangential forces are zero so there is no stress developed.

• (a) Infinity: If the modulus of rigidity were infinity, it would imply that the liquid is perfectly rigid and cannot be deformed by any amount of shear stress, which contradicts the nature of liquids that can flow and deform under shear stress.

• (c) Unity: A modulus of rigidity of unity would imply that the liquid has a specific, non-zero resistance to shear stress, which is not the case for ideal liquids that have no resistance to shear stress.

• (d) Some finite small non-zero constant value: This would imply that the liquid has some measurable resistance to shear stress, which contradicts the definition of an ideal liquid that has no viscosity and thus no resistance to shear stress.

Answer (d) We know that, Breaking stress $=\frac{\text{ Breaking force }}{\text{ Area of cross-section }}$

When length of the wire changes, area of cross-section remains same.

Hence, breaking force will be same when length changes.

• Option (a) is incorrect because the breaking force is determined by the breaking stress and the area of cross-section, which remains unchanged when the length is halved. Therefore, the breaking force does not double.

• Option (b) is incorrect because halving the length of the wire does not affect the breaking stress or the area of cross-section. Thus, the breaking force does not become half.

• Option (c) is incorrect because the breaking force is not dependent on the length of the wire but on the breaking stress and the area of cross-section. Therefore, the breaking force does not become four times.

Answer (d) We know that with increase in temperature length of a wire changes as

$$L_t=L_o(1+\alpha \mathrm{\Delta }T)$$

where $\mathrm{\Delta }T$ is change in the temperature, $L_0$ is original length, $\alpha$ is coefficient of linear expansion and $L_t$ is Length at temperature $T$.

Now we can write

$$\mathrm{\Delta }L=L_t-L_0=L_0\alpha \mathrm{\Delta }T$$

where $\alpha$ is coefficient of linear expansion.

Young’s modulus $(Y)=\frac{\text{ Stress }}{\text{ Strain }}=\frac{F{L}_0}{A\times \mathrm{\Delta }L}=\frac{F{L}_0}{A{L}_0\alpha \mathrm{\Delta }T}\propto \frac{1}{\mathrm{\Delta }T}$

As,

$$Y\propto \frac{1}{\mathrm{\Delta }T}$$

When temperature increases $\mathrm{\Delta }T$ increases, hence, $Y$ decreases.

• (a) will also double: This option is incorrect because Young’s modulus of elasticity is inversely proportional to the change in temperature. As temperature increases, Young’s modulus decreases, not increases.

• (b) will become four times: This option is incorrect because Young’s modulus does not increase with temperature. Instead, it decreases as temperature increases. The relationship is inverse, not direct or exponential.

• (c) will remain same: This option is incorrect because Young’s modulus is affected by temperature changes. As temperature increases, the modulus decreases, so it does not remain the same.

Answer (c) Consider the diagram where a spring is stretched by applying a load to its free end. Clearly the length and shape of the spring changes.

The change in length corresponds to longitudinal strain and change in shape corresponds to shearing strain.

• (a) Volumetric: Volumetric strain refers to a change in the volume of a material. When a spring is stretched, its volume does not change significantly; rather, it is the length and shape that are primarily affected.

• (b) Shear: While shear strain is indeed produced in the spring, it is not the only type of strain. The spring also experiences longitudinal strain due to the change in length. Therefore, this option is incomplete.

• (d) Longitudinal: Longitudinal strain is produced in the spring due to the change in length. However, this option is incomplete because it does not account for the shear strain that also occurs due to the change in shape of the spring.

Answer (b) We know that Young’s modulus

$$\begin{array}{rl}& Y=\frac{\text{ Stress }}{\text{ Strain }}=\frac{F/A}{\mathrm{\Delta }L/L}=F/A\times \frac{L}{\mathrm{\Delta }L}\\ & =\frac{F}{\pi (D/2{)}^2}\times \frac{L}{\mathrm{\Delta }L}=\frac{4FL}{\pi D^2\mathrm{\Delta }L}\\ & \Rightarrow D^2=\frac{4FL}{\pi \mathrm{\Delta }LY}\Rightarrow D=\sqrt{\frac{4FL}{\pi \mathrm{\Delta }LY}}\\ & \text{ As }F\text{ and }\frac{L}{\mathrm{\Delta }L}\text{ are constants. }\end{array}$$

Hence,

$$D\propto \sqrt{\frac{1}{Y}}$$

Now, we can find ratio as $\frac{D_{\text{copper }}}{D_{\text{iron }}}=\sqrt{\frac{Y_{\text{iron }}}{Y_{\text{copper }}}}$

• Option (a) $Y_{\text{copper }}/Y_{\text{iron }}$: This option is incorrect because it suggests that the ratio of the diameters is directly proportional to the ratio of the Young’s moduli. However, the correct relationship is that the diameter is inversely proportional to the square root of the Young’s modulus. Therefore, this option does not correctly represent the inverse square root relationship.

• Option (c) $\frac{Y^2\text{ iron }}{Y^2\text{ copper }}$: This option is incorrect because it suggests that the ratio of the diameters is proportional to the square of the ratio of the Young’s moduli. The correct relationship involves the square root of the ratio of the Young’s moduli, not the square of the ratio. Hence, this option is not valid.

• Option (d) $\frac{Y_{\text{iron }}}{Y_{\text{copper }}}$: This option is incorrect because it implies a direct proportionality between the diameters and the Young’s moduli. The correct relationship is that the diameter is inversely proportional to the square root of the Young’s modulus, not directly proportional to the Young’s modulus itself. Therefore, this option does not accurately reflect the correct relationship.

Thinking Process

We will assume the vertical displacement $x$ to be very small compared to $L$. . Change in the length will be calculated by difference of final total length and initial length $2L$.

Answer (a) Consider the diagram below

Hence, change in length

$$\begin{array}{ccc}\mathrm{\Delta }L& =BO+OC-(BD+DC)& \\ & =2BO-2BD\\ & =2[BO-BD]\\ & =2[(x^2+L^2{)}^{1/2}-L]\\ & =2L[(1+\frac{x^2}{L^2}{)}^{1/2}-1]\\ & \approx 2L[1+\frac{1}{2}\frac{x^2}{L^2}-1]=\frac{x^2}{L}& \\ \therefore \text{ Strain }& =\frac{\mathrm{\Delta }L}{2L}=\frac{x^2/L}{2L}=\frac{x^2}{2L^2}& [\because x«L]\end{array}$$

• Option (b) $\frac{x}{L}$: This option is incorrect because it suggests a linear relationship between the strain and the horizontal displacement ( x ). However, the strain in the wire is derived from the change in length due to the vertical displacement caused by the mass ( m ), which involves a quadratic relationship with ( x ). The correct expression for strain involves ( x^2 ), not ( x ).

• Option (c) $x^2/L$: This option is incorrect because it does not account for the factor of 2 in the denominator that arises from the derivation of the strain. The correct strain formula is derived by dividing the change in length by the original length ( 2L ), leading to an additional factor of 2 in the denominator.

• Option (d) $x^2/2L$: This option is incorrect because it suggests that the strain is directly proportional to ( x^2 ) divided by ( 2L ). However, the correct strain formula involves dividing the change in length by the original length ( 2L ), which results in an additional factor of ( L ) in the denominator, making the correct expression ( \frac{x^2}{2L^2} ).

Answer (c) Consider the FBD diagram of the rectangular frame

Balancing vertical forces $2T\sin \theta -mg=0$

$2T\sin \theta =mg$

Total horizontal force $=T\cos \theta -T\cos \theta =0$

Now from Eq. (i), $T=\frac{mg}{2\sin \theta }$

As $mg$ is constant

$$\begin{array}{rl}\Rightarrow T\propto \frac{1}{\sin \theta }& \Rightarrow T_{max}=\frac{mg}{2\sin {\theta }_{\text{min }}}\\ \sin {\theta }_{\text{min }}=0& \Rightarrow {\theta }_{\text{min }}=0\end{array}$$

No option matches with $\theta =0^{\circ }$

$$\begin{array}{rl}& T_{min}=\frac{mg}{2\sin {\theta }_{max}}\\ & \sin {\theta }_{max}=1\Rightarrow \theta ={90}^{\circ }\\ & (\text{ since, }\sin {\theta }_{max}=1)\end{array}$$

Matches with option (b)

Hence, tension is least for the case (b).

Note We should be careful when measuring the angle, it must be in the direction as given in the diagram.

• Option (a): This option is incorrect because the tension in the strings depends on the angle (\theta) between the string and the vertical. As (\theta) changes, the value of (\sin \theta) changes, which in turn affects the tension (T). Therefore, the tension cannot be the same in all cases.

• Option (b): This option is incorrect because, as derived from the equation (T = \frac{mg}{2 \sin \theta}), the tension is inversely proportional to (\sin \theta). The tension is least when (\sin \theta) is maximum, which occurs at (\theta = 90^\circ). In case (b), (\theta) is not (90^\circ), so the tension is not the least in this case.

• Option (d): This option is incorrect because, similar to option (b), the tension is least when (\sin \theta) is maximum. In case (c), (\theta) is (90^\circ), which makes (\sin \theta = 1), resulting in the least tension. Therefore, the tension is not least in case (d).

Answer (d) Consider the diagram.

A mass $M$ is attached at the centre. As the mass is attached to both the rods, both rod will be elongated, but due to different elastic properties of material rubber changes shape also.

• Option (a) is incorrect: While both rods will indeed elongate, the rubber rod will also change shape due to its higher elasticity compared to steel. The rubber rod will not maintain its original cylindrical shape as it stretches.

• Option (b) is incorrect: The steel rod will elongate without a significant change in shape due to its high rigidity. The rubber rod, however, will not only elongate but also change shape, contradicting the statement that it will only elongate.

• Option (c) is incorrect: The rubber rod will elongate and change shape, but the shape of the bottom edge will not change to an ellipse. Instead, it will taper to a tip at the center due to the distribution of the mass and the elastic properties of rubber.

Thinking Process

We have to compare ultimate tensile strength for each material. Material having more ultimate tensile strength will be elastic over larger region.

Answer (c, $d)$

It is clear from the two graphs, the ultimate tensile strength for material (ii) is greater, hence material(ii) is elastic over larger region as compared to material (i). For material(ii) fracture point is nearer, hence it is more brittle.

• (a) Material (ii) is more elastic than material (i) and hence material (ii) is more brittle:

  • Incorrect because elasticity is not directly related to brittleness. Elasticity refers to the ability of a material to return to its original shape after deformation, while brittleness refers to the tendency of a material to fracture without significant deformation. The graph does not provide information that directly correlates elasticity with brittleness.

• (b) Material (i) and (ii) have the same elasticity and the same brittleness:

  • Incorrect because the graphs show different stress-strain behaviors for the two materials. Material (ii) has a higher ultimate tensile strength and a different fracture point compared to material (i), indicating that they do not have the same elasticity or brittleness.

Answer $(a,d)$

As shown in the diagram

Clearly, forces at each cross-section is $F$.

Now applying formula,

$$\begin{array}{rl}\text{ Stress }& =\frac{\text{ Tension }}{\text{ Area }}=\frac{F}{A}\\ \text{ Tension }& =\text{ Applied force }=F\end{array}$$

• (b) Tensile stress at any cross-section is zero: This option is incorrect because tensile stress is defined as the force per unit area. Since there is a force ( F ) acting on the wire and the wire has a cross-sectional area ( A ), the tensile stress cannot be zero. It is given by ( \frac{F}{A} ).

• (c) Tensile stress at any cross-section ( A ) of the wire is ( 2F / A ): This option is incorrect because the tensile stress is calculated as the force divided by the cross-sectional area. The force acting on the wire is ( F ), not ( 2F ). Therefore, the correct tensile stress is ( \frac{F}{A} ), not ( \frac{2F}{A} ).

Answer $(b,d)$

Let the mass is placed at $x$ from the end $B$.

Let $T_A$ and $T_B$ be the tensions in wire $A$ and wire $B$ respectively.

For the rotational equilibrium of the system,

$$\begin{array}{rl}& \sum \tau =0\\ & \Rightarrow T_{B}x-T_A(l-x)=0\\ & \Rightarrow \frac{T_B}{T_A}=\frac{l-x}{x}\\ & \text{ Stress in wire }A=S_A=\frac{T_A}{a_A}\\ & \text{ Stress in wire }B=S_B=\frac{T_B}{a_B}\end{array}$$

where $a_A$ and $a_B$ are cross-sectional areas of wire $A$ and $B$ respectively.

By question $a_B=2a_A$

Now, for equal stress

$$\begin{array}{rl}& \Rightarrow \frac{T_A}{a_A}=\frac{T_B}{a_B}\Rightarrow \frac{T_B}{T_A}=\frac{a_B}{a_A}=2\\ & \Rightarrow \frac{l-x}{x}=2\Rightarrow \frac{l}{x}-1=2\\ & \Rightarrow x=\frac{l}{3}\Rightarrow l-x=l-l/3=\frac{2l}{3}\end{array}$$

Hence, mass $m$ should be placed closer to $B$.

For equal strain, $(strain{)}_A$= $(strain{)}_B$

$\Rightarrow \frac{(Y_A)}{S_A}=\frac{Y_B}{S_B}\text{ (where }{Y}_A\text{ and }{Y}_B\text{ are Young modulii) }$

$\Rightarrow \frac{Y_{\text{steel }}}{T_A/a_A}=\frac{Y_{Al}}{T_B/a_B}$

$\Rightarrow \frac{Y_{\text{steel }}}{Y_{\text{Al }}}=\frac{T_A}{T_B}\times \frac{a_B}{a_A}=(\frac{x}{l-x})(\frac{2a_A}{a_A})$

$\Rightarrow \frac{200\times {10}^9}{70\times {10}^9}=\frac{2x}{l-x}\Rightarrow \frac{20}{7}=\frac{2x}{l-x}$

$\Rightarrow \frac{10}{7}=\frac{x}{l-x}\Rightarrow 10l-10x=7x$

$\Rightarrow 17x=10l\Rightarrow x=\frac{10l}{17}$

$l-x=l-\frac{10l}{17}=\frac{7l}{17}$

Hence, mass $m$ should be placed closer to wire $A$.

• Option (a) is incorrect: The mass should not be suspended close to wire A to have equal stresses in both wires. The correct placement for equal stress is closer to wire B, as derived in the solution. This is because the cross-sectional area of wire B is larger, and to balance the stress, the tension in wire B must be higher, which is achieved by placing the mass closer to wire B.

• Option (c) is incorrect: Suspending the mass at the middle of the wires will not result in equal stresses in both wires. The stress in each wire depends on the tension, which is influenced by the position of the mass. Placing the mass in the middle would not account for the differences in cross-sectional areas and Young’s moduli of the two wires, leading to unequal stresses.

• Option (d) is correct: This option is actually correct, as derived in the solution. The mass should be suspended closer to wire A to have equal strain in both wires. This is because the Young’s modulus and cross-sectional area differences require the mass to be positioned such that the strains (which are stress divided by Young’s modulus) are equal.

Answer $(a,d)$

As an ideal liquid is not compressible.

Hence, change in volume, $\mathrm{\Delta }V=0$

$$\begin{array}{rl}& \text{ Bulk modulus }B=\frac{\text{ Stress }}{\text{ Volume strain }}=\frac{F/A}{\mathrm{\Delta }V/V}=\frac{F}{A}\times \frac{V}{\mathrm{\Delta }V}=\mathrm{\infty }\\ & \text{ Compressibility, }K=\frac{1}{B}=\frac{1}{\mathrm{\infty }}=0\end{array}$$

As there is no tangential (viscous) force exists in case of an ideal fluid, hence, shear modulus $=0$.

• Option (b) the bulk modulus is zero: This is incorrect because for an ideal liquid, the bulk modulus is actually infinite. An ideal liquid is incompressible, meaning any applied pressure does not change its volume. Therefore, the bulk modulus, which measures resistance to compression, is infinitely large.

• Option (c) the shear modulus is infinite: This is incorrect because for an ideal liquid, the shear modulus is zero. An ideal liquid cannot resist shear stress; it flows under any applied tangential force, indicating that it has no rigidity against shear deformation.

Answer $(a,d)$

Consider the diagram where a deforming force $F$ is applied to the combination.

where $F$ is tension in each wire and $A$ is cross-section area of each wires.

As $F$ and $A$ are same for both the wire, hence, stress will be same for both the wire.

As,

$(Strain{)}_{steel}$ = $\frac{Stress}{Y_{steel}}$,

$(Strain{)}_{copper}$= $\frac{Stress}{Y_{copper}}$

$Y_{\text{steel }}\ne Y_{\text{copper }}$

Hence, the two wires will have differents strain.

• Option (b) different stress: This option is incorrect because the stress in both wires is the same. Stress is defined as the force per unit area, and since the force $F$ and the cross-sectional area $A$ are the same for both wires, the stress must also be the same.

• Option (c) the same strain: This option is incorrect because the strain in the two wires is different. Strain is defined as the ratio of the change in length to the original length, and it depends on the material’s Young’s modulus ($Y$). Since the Young’s modulus of steel and copper are different ($Y_{\text{steel}}\ne$ $Y_{\text{copper}}$), the strain in each wire will be different even though the stress is the same.

Answer Young’s modulus $(Y)=\frac{\text{ Stress }}{\text{ Longitudinal strain }}$

For same longitudinal strain,

But

$\therefore$

$Y\propto \text{ stress }$

$\frac{Y_{\text{steel }}}{Y_{\text{rubber }}}=\frac{(\text{ stress }{)}_{\text{steel }}}{(\text{ stress }{)}_{\text{rubber }}}$

Therefore, from Eq. (i),

$\frac{(\text{ stress }{)}_{\text{steel }}}{(\text{ stress }{)}_{\text{rubber }}}>1$ $\Rightarrow (\text{ stress }{)}_{\text{steel }}>{\text{ (stress) }}_{\text{rubber }}$

Answer Stress $=\frac{\text{ Magnitude of internal reaction force }}{\text{ Area of cross }-\text{ section }}$

Therefore, stress is a scalar quantity not a vector quantity.

Answer Work done in stretching a wire is given by $W=\frac{1}{2}F\times \mathrm{\Delta }l$

[where $F$ is applicd force and $\mathrm{\Delta }l$ is extension in the wire] As springs of steel and copper are equally stretched. Therefore, for same force $(F)$,

$$W\propto \mathrm{\Delta }l$$

$$\text{ Young’s modulus }(Y)=\frac{F}{A}\times \frac{l}{\mathrm{\Delta }l}\Rightarrow \mathrm{\Delta }l=\frac{F}{A}\times \frac{l}{Y}$$

As both springs are identical, $\mathrm{\Delta }l\propto \frac{1}{Y}$

From Eqs. (i) and (ii), we get $W\propto \frac{1}{Y}$

$$\begin{array}{ccc}\therefore & \frac{W_{\text{steel }}}{W_{\text{copper }}}=\frac{Y_{\text{copper }}}{Y_{\text{steel }}}<1& \text{ (As, }.Y_{\text{steel }}>Y_{\text{copper }})\\ \Rightarrow & W_{\text{steel }}<W_{\text{copper }}\end{array}$$

Therefore, more work will be done for stretching copper spring.

Answer Young’s modulus $(Y)=\frac{F}{A}\times \frac{l}{\mathrm{\Delta }l}$

For a perfectly rigid body, change in length $\mathrm{\Delta }l=0$

$$Y=\frac{F}{A}\times \frac{l}{0}=\mathrm{\infty }$$

Therefore, Young’s modulus for a perfectly rigid body is infinite( $(\mathrm{\infty })$.

Answer Bulk modulus $(K)=\frac{p}{\mathrm{\Delta }V/V}=\frac{pV}{\mathrm{\Delta }V}$

For perfectly rigid body, change in volume $\mathrm{\Delta }V=0$

$\therefore K=\frac{pV}{0}=\mathrm{\infty }$

Therefore, bulk modulus for a perfectly rigid body is infinity $(\mathrm{\infty })$.

Thinking Process

In this problem, we have to apply Hooke’s law and then elongation in each wire will be compared.

Answer The situation is shown in the diagram.

Now, $$ Young’s modulus $(Y)=\frac{f}{A}\times \frac{L}{l}$

For first wire,

$Y=\frac{f}{\pi r^2}\times \frac{L}{l}$

For second wire,

$Y=\frac{2f}{\pi (2r{)}^2}\times \frac{2L}{l^{\prime }}$

$=\frac{f}{\pi r^2}\times \frac{L}{l^{\prime }}$

From Eqs. (i) and (ii),

$$\frac{f}{\pi r^2}\times \frac{L}{l}=\frac{f}{\pi r^2}\times \frac{L}{l^{\prime }}$$ $$l=l^{\prime }$$

( $\because$ Both wires are of same material, hence, Young’s modulus will be same.)

Thinking Process

Due to increase in temperature of the rod, length, increases. The equation of thermal expansion for linear expansion will be applied.

Answer Given, Young’s modulus of steel $Y=2.0\times {10}^{11}N/m^2$

Coefficient of thermal expansion $\alpha ={10}^{-5}{}^{\circ }{C}^{-1}$

Length $l=1m$

Area of cross-section $A=1c{m}^2=1\times {10}^{-4}{m}^2$

Increase in temperature $\mathrm{\Delta }t={200}^{\circ }C-0^{\circ }C={200}^{\circ }C$

Tension produced in steel rod $(F)=YA\alpha \mathrm{\Delta }t$

$$\begin{array}{rl}& =2.0\times {10}^{11}\times 1\times {10}^{-4}\times {10}^{-5}\times 200\\ & =4\times {10}^{4}N\end{array}$$

Answer Given, Bulk modulus of rubber $(K)=9.8\times {10}^{8}N/m^2$

Density of sea water $(\rho )={10}^{3}kg/m^3$

Percentage decrease in volume, $(\frac{\mathrm{\Delta }V}{V}\times 100)=0.1\Rightarrow \frac{\mathrm{\Delta }V}{V}=\frac{0.1}{100}$

$\Rightarrow \frac{\mathrm{\Delta }V}{V}=\frac{1}{1000}$

Let the rubber ball be taken upto depth $h$.

$\therefore$ Change in pressure $(p)=h\rho g$

$\therefore$ Bulk modulus $(K)=|\frac{p}{\mathrm{\Delta }V/V}|=\frac{h\rho g}{(\mathrm{\Delta }V/V)}$

$\Rightarrow h=\frac{K\times (\mathrm{\Delta }V/V)}{\rho g}=\frac{9.8\times {10}^8\times \frac{1}{1000}}{{10}^3\times 9.8}=100m$

Answer

$$\begin{array}{rl}\text{ Length of steel cable }l& =9.1m\\ \text{ Radius }r& =5mm=5\times {10}^{-3}m\\ \text{ Tension in the cable }F& =800N\\ \text{ Young’s modulus for steel }Y& =2\times {10}^{11}N/m^2\\ \text{ Change in length }\mathrm{\Delta }l& =?\\ \text{ Young’s modulus }(Y)& =\frac{F}{A}\times \frac{l}{\mathrm{\Delta }l}\Rightarrow \mathrm{\Delta }l=\frac{F}{\pi r^2}\times \frac{l}{Y}\\ & =\frac{800}{3.14\times (5\times {10}^{-3}{)}^2}\times \frac{9.1}{2\times {10}^{11}}\\ & =4.64\times {10}^{-4}m\end{array}$$

Thinking Process

To solve this question, we have to resolve the force along and perpendicular to the inclined plane. Now, we can easily calculate the tensile and shearing stress.

Answer Consider the adjacent diagram.

Let the cross-sectional area of the bar be $A$. Consider the equilibrium of the plane aa’. A force $F$ must be acting on this plane making an angle $\frac{\pi }{2}-\theta$ with the normal $ON$. Resolving $F$ into components, along the plane $(FP)$ and normal to the plane.

$$\begin{array}{rl}& F_P=F\cos \theta \\ & F_N=F\sin \theta \end{array}$$

Let the area of the face aa’ be $A^{\prime }$, then

$$\begin{array}{rl}\frac{A}{A^{\prime }}& =\sin \theta \\ \therefore \text{ The tensile stress }& =\frac{A}{\sin \theta }\\ & =\frac{\text{ Normal force }}{A/\sin \theta }=\frac{F\sin \theta }{A^{\prime }}\\ \text{ Shearing stress }& =\frac{\text{ Parallel force }}{\text{ Area }}\\ & =\frac{F\cos \theta }{A/\sin \theta }=\frac{F}{A}\sin \theta \cdot \cos \theta \\ & =\frac{F}{2A}(2\sin \theta \cdot \cos \theta )=\frac{F}{2A}\sin 2\theta \end{array}$$

(a) For tensile stress to be maximum, ${\sin }^2\theta =1$

$$\begin{array}{rlr}\Rightarrow & \sin \theta & =1\\ \Rightarrow & \theta & =\frac{\pi }{2}\end{array}$$

(b) For shearing stress to be maximum,

$$\begin{array}{cccc}\Rightarrow & & \sin 2\theta & =1\\ \Rightarrow & 2\theta & =\frac{\pi }{2}\\ \Rightarrow & & \theta & =\frac{\pi }{4}\end{array}$$

Note We must not apply the formula for stress directly, forces must be resolved.

Thinking Process

When the wire is hanging different part of the wire will be acted upon by different forces. To find the tension due to whole wire, consider a small element and find the tension there and then integrate for the whole wire.

Answer Consider the diagram when a small element of length $dx$ is considered at $x$ from the load $(x=0)$.

(a) Let $T(x)$ and $T(x+dx)$ are tensions on the two cross-sections a distance $dx$ apart, then-$-T(x+dx)+T(x)=dmg=\mu dxg$ (where $\mu$ is the mass/length).

$(\because dm=\mu dx)$

$$\begin{array}{cccc}\Rightarrow & & T(x)& =\mu gx+C\\ \text{ At }x=0,T(0)=Mg\Rightarrow & C& =Mg\\ \therefore & & T(x)& =\mu gx+Mg\end{array}$$

Let the length $dx$ at $x$ increase by $dr$, then

$$\begin{array}{rl}\text{ Young’s modulus }Y& =\frac{\text{ Stress }}{\text{ Strain }}\\ \frac{T(x)/A}{dr/dx}& =Y\\ \Rightarrow \frac{dr}{dx}& =\frac{1}{YA}T(x)\\ r& =\frac{1}{YA}{\int }_0^L(\mu gx+Mg)dx\\ & =\frac{1}{YA}[\frac{\mu g{x}^2}{2}+Mgx{]}_0^L\\ & =\frac{1}{YA}[\frac{mg{L}^2}{2}+MgL]\end{array}$$

$$\begin{array}{c}[\because dT=T(x+dx)-T(x)]\\ \text{ (on integrating) }\end{array}$$

( $m$ is the mass of the wire)

$$A=\pi \times ({10}^{-3}{)}^2{m}^2$$

$$\begin{array}{rl}& Y=200\times {10}^{9}N{m}^{-2}\\ & m=\pi \times ({10}^{-3}{)}^2\times 10\times 7860kg\end{array}$$

$$\begin{array}{rl}\therefore r& =\frac{1}{2\times {10}^{11}\times \pi \times {10}^{-6}}[\frac{\pi \times 786\times {10}^{-3}\times 10\times 10}{2}+25\times 10\times 10]\\ & =[196.5\times {10}^{-6}+3.98\times {10}^{-3}]\approx 4\times {10}^{-3}m\end{array}$$

(b) Clearly tension will be maximum at $x=L$

$$\therefore T=\mu gL+Mg=(m+M)g$$

$$\text{The yield force =(Yield strength Y) area}=250\times {10}^6\times \pi \times ({10}^{-3}{)}^2=250\times \pi N$$
$$\text{At yield point}[\because m=\mu L]$$
$$\Rightarrow (m+M)g=250\times \pi$$
$$\Rightarrow m=\pi \times ({10}^{-3}{)}^2\times 10\times 7860\ll M$$
$$\therefore Mg\approx 250\times \pi$$
$$\text{Hence},M=\frac{250\times \pi }{10}=25\times \pi \approx 75kg$$.

Thinking Process

To solve this question, we have to consider a small element, find tension in the rod at this element and then calculate for the whole rod.

Answer Consider an element of width $dr$ at $r$ as shown in the diagram. Let $T(r)$ and $T(r+dr)$ be the tensions at $r$ and $r+dr$ respectively.

Net centrifugal force on the element $={\omega }^{2}rdm$

(where $\omega$ is angular velocity of the rod)

$$\begin{array}{rl}& ={\omega }^{2}r\mu dr\\ & \Rightarrow T(r)-T(r+dr)=\mu {\omega }^{2}rdr\\ & \Rightarrow -dT=\mu {\omega }^{2}rdr\\ & [\because \text{ Tension and centrifugal forces are opposite] }\\ & \therefore -{\int }_{T=0}^{T}dT={\int }_{r=l}^{r=r}\mu {\omega }^{2}rdr[\because T=0\text{ at }r=l]\\ & \Rightarrow T(r)=\frac{\mu {\omega }^2}{2}(l^2-r^2)\end{array}$$

Let the increase in length of the element $dr$ be $\mathrm{\Delta }r$

$$\begin{array}{rl}& \text{ So, }\text{ Young’s modulus }Y=\frac{\text{ Stress }}{\text{ Strain }}=\frac{T(r)/A}{\frac{\mathrm{\Delta }r}{dr}}\\ & \therefore \frac{\mathrm{\Delta }r}{dr}=\frac{T(r)}{A}=\frac{\mu {\omega }^2}{2YA}(l^2-r^2)\\ & \therefore \mathrm{\Delta }r=\frac{1}{YA}\frac{\mu {\omega }^2}{2}(l^2-r^2)dr\\ & \therefore \mathrm{\Delta }=\text{ change in length in right part }=\frac{1}{YA}\frac{\mu {\omega }^2}{2}{\int }_0^l(l^2-r^2)dr\\ & =(\frac{1}{YA})\frac{\mu {\omega }^2}{2}[l^3-\frac{l^3}{3}]=\frac{1}{3YA}\mu {\omega }^2{L}^2\\ & \therefore \text{ Total change in length }=2\mathrm{\Delta }=\frac{2}{3YA}\mu {\omega }^2{l}^2\end{array}$$

Thinking Process

Due to increase in temperature length of each side will change, hence, the angle corresponding to any vertex also changes.

Answer Consider the diagram shown

Let

$$\begin{array}{rl}{l}_1& =AB,l_2=AC,l_3=BC\\ \cos \theta & =\frac{l_3^2+l_1^2-l_2^2}{2l_3{l}_1}\end{array}$$

$$\Rightarrow 2l_3{l}_1\cos \theta =l_3^2+l_1^2-l_2^2$$

Differentiating $2(l_{3}d{l}_1+l_{1}d{l}_3)\cos \theta -2l_1{l}_3\sin \theta d\theta$

(assume $\mathrm{\angle }ABC=\theta$ )

Now,

and

$$=2l_{3}d{l}_1+2l_{1}d{l}_1-2l_{2}d{l}_2$$

$d{l}_1=l_1{\alpha }_1\mathrm{\Delta }t$ (where, $\mathrm{\Delta }t=$ change in temperature)

$d{l}_2=l_2{\alpha }_1\mathrm{\Delta }t\Rightarrow d{l}_3=l_3{\alpha }_2\mathrm{\Delta }t$

$l_1=l_2=l_3=l$

$(l^2{\alpha }_1\mathrm{\Delta }t+l^2{\alpha }_1\mathrm{\Delta }t)\cos \theta +l^2\sin \theta d\theta =l^2{\alpha }_1\mathrm{\Delta }t+l^2{\alpha }_1\mathrm{\Delta }t-l^2{\alpha }_2\mathrm{\Delta }t$

Putting

$$\sin \theta d\theta =2{\alpha }_1\mathrm{\Delta }t(1-\cos \theta )-{\alpha }_2\mathrm{\Delta }t$$

$$\theta ={60}^{\circ }$$

(for equilateral triangle)

$$d\theta \times \sin {60}^{\circ }=2{\alpha }_1\mathrm{\Delta }t(1-\cos {60}^{\circ })-{\alpha }_2\mathrm{\Delta }t$$

$$=2{\alpha }_1\mathrm{\Delta }t\times \frac{1}{2}-{\alpha }_2\mathrm{\Delta }t=({\alpha }_1-{\alpha }_2)\mathrm{\Delta }t$$

$\Rightarrow d\theta =$ change in the angle $\mathrm{\angle }ABC$

$$=\frac{({\alpha }_1-{\alpha }_2)\mathrm{\Delta }T}{\sin {60}^{\circ }}=\frac{2({\alpha }_1-{\alpha }_2)\mathrm{\Delta }T}{\sqrt{3}}(\because \mathrm{\Delta }t=\mathrm{\Delta }T\text{ given })$$

Thinking Process

In this question, the elastic torque is given we have to find the torque caused by the weight due to bending and equate with the given value.

Answer Consider the diagram according to the question, the bending torque on the trunk of radius $r$ of the tree $=\frac{Y\pi r^4}{4R}$

where $R$ is the radius of curvature of the bent surface.

When the tree is about to buckle $Wd=\frac{Y\pi r^4}{4R}$

If $R\gg h$, then the centre of gravity is at a height $l\approx \frac{1}{2}h$ from the ground.

From $\mathrm{△}ABC{R}^2\simeq (R-d{)}^2+(\frac{1}{2}h{)}^2$

if $d\ll R,R^2\simeq R^2-2Rd+\frac{1}{4}{h}^2$

$\therefore d=\frac{h^2}{8R}$

If ${\omega }_0$ is the weight/volume

$$\begin{array}{rl}& \frac{Y\pi r^4}{4R}={\omega }_0(\pi r^{2}h)\frac{h^2}{8R}[\because \text{ Torque is caused by the weight }]\\ & \Rightarrow h\simeq (\frac{2Y}{{\omega }_0}{)}^{1/3}{r}^{2/3}\\ & \text{ Hence, critical height }=h=(\frac{2Y}{{\omega }_0}{)}^{1/3}{r}^{2/3}.\end{array}$$

Thinking Process

In this problem ,the given string is elastic and there is no dissipation of energy we can apply conservation of mechanical energy.

Answer Consider the diagram the stone is dropped from point $p$.

(a) Till the stone drops through a length $L$ it will be in free fall. After that the elasticity of the string will force it to a SHM. Let the stone come to rest instantaneously at $y$.

The loss in PE of the stone is the PE stored in the stretched string.

$$mgy=\frac{1}{2}k(y-L{)}^2$$

$$\begin{array}{rl}\Rightarrow mgy& =\frac{1}{2}k{y}^2-kyL+\frac{1}{2}k{L}^2\Rightarrow \frac{1}{2}k{y}^2-(kL+mg)y+\frac{1}{2}k{L}^2=0\\ y& =\frac{(kL+mg)\pm \sqrt{(kL+mg{)}^2-k^2{L}^2}}{k}=\frac{(kL+mg)\pm \sqrt{2mgkL+m^2{g}^2}}{k}\end{array}$$

Retain the positive sign.

$$\therefore y=\frac{(kL+mg)+\sqrt{2mgkL+m^2{g}^2}}{k}$$

(b) In SHM, the maximum velocity is attained when the body passes, through the “equilibrium, position” i.e., when the instantaneous acceleration is zero. That is $mg-kx=0$, where $x$ is the extension from $L$.

$\Rightarrow$

$$mg=kx$$

Let the velocity be $v$. Then,

Now,

$$\begin{array}{rl}& \frac{1}{2}m{v}^2+\frac{1}{2}k{x}^2=mg(L+x)\text{ (from conservation of energy) }\\ & \frac{1}{2}m{v}^2=mg(L+x)-\frac{1}{2}k{x}^2\end{array}$$

$$\begin{array}{rl}\therefore \frac{1}{2}m{v}^2& =mg(L+\frac{mg}{k})-\frac{1}{2}k\frac{m^2{g}^2}{k^2}=mgL+\frac{m^2{g}^2}{k}-\frac{1}{2}\frac{m^2{g}^2}{k}\\ \frac{1}{2}m{v}^2& =mgL+\frac{1}{2}\frac{m^2{g}^2}{k}\\ \therefore v^2& =2gL+m{g}^2/k\\ v& =(2gL+m{g}^2/k{)}^{1/2}\end{array}$$

(c) When stone is at the lowest position i.e., at instantaneous distance $Y$ from $P$, then equation of motion of the stone is

$$\frac{m{d}^{2}y}{d{t}^2}=mg-k(y-L)\Rightarrow \frac{d^{2}y}{d{t}^2}+\frac{k}{m}(y-L)-g=0$$

Make a transformation of variables, $z=\frac{k}{m}(y-L)-g$

$$\therefore \frac{d^{2}z}{d{t}^2}+\frac{k}{m}z=0$$

It is a differential equation of second order which represents SHM.

Comparing with equation $\frac{d^{2}z}{d{t}^2}+{\omega }^{2}z=0$

Angular frequency of harmonic motion $\omega =\sqrt{\frac{k}{m}}$

The solution of above equation will be of the type $z=A\cos (\omega t+\varphi )$; where $\omega =\sqrt{\frac{k}{m}}$

$$y=(L+\frac{mg}{k})+A^{\prime }\cos (\omega t+\varphi )$$

Thus, the stone will perform SHM with angular frequency $\omega =\sqrt{k/m}$ about a point $y_0=L+\frac{mg}{k}$.