Exemplar MCQs

Hint

As the liquid is ideal, hence it does not have frictional force among its layers, thus the tangential forces are zero as there is no stress developed. Hence modulus of rigidity will be zero.

Hint

Breaking stress is a constant for a given material and it does not depend upon the dimension (length or thickness) of wire.

Hint

We know that, with increase in temperature, length of a wire changes as \(\mathrm{L}_{\mathrm{t}}=\mathrm{L}_0(1+\alpha \Delta \mathrm{T})\) where, \(\Delta \mathrm{T}\) is change in the temperature, \(L\) is original length, \(\alpha\) is coefficient of linear expansion and \(L_t\) is length at temperature T.
Now, we can write, \(\Delta \mathrm{L}=\mathrm{L}_{\mathrm{t}}-\mathrm{L}_{\circ}=\mathrm{L}_{\circ} \alpha \Delta \mathrm{T}\) Young’s modulus
\(
(\mathrm{Y})=\frac{\text { Stress }}{\text { Strain }}=\frac{\mathrm{FL}_0}{\mathrm{~A} \times \Delta \mathrm{L}}=\frac{\mathrm{FL}_0}{\mathrm{AL}_0 \alpha \Delta \mathrm{T}} \propto \frac{1}{\Delta \mathrm{T}}
\)
As, \(Y \propto \frac{1}{\Delta \mathrm{T}}\)
When temperature \((\Delta \mathrm{T})\) increases, \(Y\) decreases.

Hint

When a spring is stretched by a load its shape (shear) and length change. So strain produced is shearing and longitudinal strain.

Hint

(b) As the bar is supported symmetrically by the three wires, therefore extension in each wire is same. Let \(T\) be the tension in each wire and diameter of the wire is \(D\), then Young’s modulus is
\(
\begin{aligned}
Y & =\frac{\text { Stress }}{\text { Strain }}=\frac{F / A}{\Delta L / L}=\frac{F}{A} \times \frac{L}{\Delta L} \\
& =\frac{F}{\pi(D / 2)^2} \times \frac{L}{\Delta L}=\frac{4 F L}{\pi D^2 \Delta L} \\
\Rightarrow \quad D^2 & =\frac{4 F L}{\pi \Delta L Y} \Rightarrow D=\sqrt{\frac{4 F L}{\pi \Delta L Y}}
\end{aligned}
\)
As \(F\) and \(\frac{L}{\Delta L}\) are constants.
Hence, \(D \propto \sqrt{\frac{1}{Y}}\)
or \(D=\frac{K}{\sqrt{Y}}(K\) is the proportionality constant \()\)
Now, we can find ratio as \(\frac{D_{\text {copper }}}{D_{\text {iron }}}=\sqrt{\frac{Y_{\text {iron }}}{Y_{\text {copper }}}}\)

Hint

\(
\begin{aligned}
& \Delta L=(A O+B O-A B) \\
& \Delta L=2[A O-L] \\
& \left.\Delta L=2\left(L^2+x^2\right)^{\frac{1}{2}}-1\right)
\end{aligned}
\)
\(
\Delta L=2 L\left[1+\frac{x^2}{2 L^2}-1\right]=\frac{x^2}{L}
\)
Strain \(=\frac{\Delta L}{2 L}=\frac{x^2}{2 L^2}\). Hence option a is correct.

Hint

Let \(M\) be the mass of rectangular frame and \(\theta\) be the angle which the tension \(T\) in the string makes with the horizontal as shown in the figure.
\(
\therefore \quad 2 T \sin \theta=M g
\)
or \(\quad T=\frac{M g}{2 \sin \theta}\)
or \(T \propto \frac{1}{\sin \theta}\)
\(T\) is least if \(\sin \theta\) has maximum value i.e., \(\sin \theta=1=\sin 90^{\circ}\).

Hint

The steel rod will elongate, without any perceptible change in shape, but the rubber rod will elongate with the shape of the bottom edge tapered to a tip at the centre.

Hint

From graphs, it is clear that the ultimate strength of the material (ii) is greater than that of material (i). Therefore, the elastic behaviour of material (ii) is over a larger region of strain as compared to material (i). If the fracture point of a material is closer to the ultimate strength point, then the material is a brittle material. Therefore, material (ii) is more brittle than material (i).

Hint

\(
\text { Stress } \sigma=\frac{\text { Force }}{\text { Area }}=\frac{F}{A}
\)

\(
\text { Tension }=\text { Applied force }=F
\)

Hint

\(
\begin{aligned}
& T_b x=T_a(l-x) \\
& \frac{T_b}{T_a}=\left(\frac{l}{x}-1\right)
\end{aligned}
\)
For wire A, stress
\(
=\frac{T_a}{T_b}
\)
For wire \(B\), stress
\(
\frac{T_b}{A_b}=\frac{T_b}{2 A_a}
\)
Since stress on steel \(=\) stress on \(\mathrm{Al}\)
\(
\begin{aligned}
& \frac{T_a}{A_a}=\frac{T_b}{2 A_a} \rightarrow T_a=\frac{T_b}{2} \\
& \frac{T_b}{T_a}=2 \\
& \frac{l}{x}-1=\frac{2}{1}
\end{aligned}
\)
\(x=\frac{l}{3}\) from point \(b\)
so, distance from \(\mathrm{A}=l-x=l-\frac{l}{3}=\frac{2 l}{3}\)
hence \(\mathrm{m}\) is closer to B than A. so, option b is correct.
the rod remains in a horizontal and balanced. so, the strain will also be equal.
Strain (A) = Strain (B)
\(
\frac{S_a}{Y_a}=\frac{S_b}{Y_b}
\)
\(
\begin{aligned}
& Y \text { steel } / Y A l=\left(\frac{T_a}{T_b}\right)\left(\frac{A_b}{A_a}\right)=\left(\frac{x}{l}-x\right)\left(\frac{2 A a}{A a}\right) \\
& \frac{200 \times 10^9}{70 \times 10^9}=\frac{2 x}{l-x} \\
& 14 x=20 l-20 x \\
& 34 x=20 l  \\
& x=\frac{20 l}{34}=\frac{10 l}{17} \\
& (l-x)=\frac{7}{17}
\end{aligned}
\)
So, for an equal amount of strain, mass \(\mathrm{m}\) will be closer to \(\mathrm{A}\)
Hence option d is correct.

Hint

The answer is the option (a) and (d).
A property of an ideal liquid is that it is not compressible. (B)Bulk modulus \(=\frac{-p(V)}{\Delta V}\) \(\Delta V=0\) and \(\mathrm{B}=\) infinite for an ideal liquid.
No tangential forces act on this liquid, the shearing strain \(\Delta \theta=0\) and \(F=0\)
\(
n=\frac{\frac{F}{A}}{\Delta \theta}=\frac{0}{0}=\text { indeterminate }
\)
hence \(a\) and \(d\) are true.

Hint

The answer is the option (a) and (d).
As we know stress \(=\)
\(
\frac{F}{A}
\)
The areas of cross-sections of both wires are the same, and the force with which they get stretched is also the same. Hence the total stress for both is the same. Hence option a.
Strain \(=\) Stress \(/ Y \quad\)
As stresses are same on both,
\(
\text { Strain } \propto \frac{1}{Y} \text { (steel) and Strain } \propto \frac{1}{Y} \text { (Al) }
\)
Strain (steel) \(/\) Strain \(({Al})={Y}({Al}) / {Y}(\) steel \()\)
Now, \(Y(\mathrm{Al})<Y\) (Steel). Hence, strain(steel) \(<\) Strain (Al)
Hence, option d is correct.

Hint

\(Y=\) stress \(/\) strain and \(Y \propto\) stress
Y (steel)/ \(Y(\) rubber) = Stress(steel)/Stress(rubber)
We know that \(Y\) steel \(>Y\) rubber
Y steel \(/\) Y rubber \(>1\)
Hence, stress (steel)/stress (rubber) \(>1\)
So, stress (steel) > stress (rubber)

Hint

Stress = Magnitude of internal reaction force / Area of cross-section
Hence, stress is not a scalar quantity not a vector quantity, it is a tensor quantity.

Hint

Work Done in stretching a Wire or Spring:
In stretching a wire work is done against internal restoring forces. This work is stored in the wire as elastic potential energy or strain energy.
If a force \(F\) acts along the length \(L\) of the wire of cross-section \(A\) and stretches it by \(x\), then
Work done in stretching a wire is given by \(W=1 / 2 (\mathrm{F} \times \Delta \mathrm{l})\)
As springs of steel and copper are equally stretched. Therefore, for same force \((F)\),
\(
\begin{aligned}
W & \propto \Delta l \dots(i)\\
\text { Young’s modulus } & (Y)=\frac{F}{A} \times \frac{l}{\Delta l} \\
\text { or } & \Delta l=\frac{F}{A} \times \frac{l}{Y}
\end{aligned}
\)
As both springs are identical,
\(
\therefore \quad \Delta l \propto \frac{1}{Y} \dots(ii)
\)
From Eqs. (i) and (ii), we get \(W \propto \frac{1}{Y}\)
\(
\begin{array}{ll}
\therefore & \frac{W_{\text {steel }}}{W_{\text {copper }}}=\frac{Y_{\text {copper }}}{Y_{\text {steel }}}<1\left(\text { As } Y_{\text {steel }}>Y_{\text {copper }}\right) \\
\text { or } & W_{\text {steel }}<W_{\text {copper }}
\end{array}
\)
Therefore, more work will be done for stretching copper spring.

Hint

According to Hooke’s law
\(
(Y)=\frac{\text { stress }}{\text { longitudinal strain }}=\frac{F}{A} \times \frac{l}{\Delta l}
\)
For a perfectly rigid body, change in length \(\Delta l=0\), therefore longitudinal strain is zero.
\(
\therefore \quad Y=\frac{F}{A} \times \frac{l}{0}=\infty
\)
Hence, Young’s modulus for a perfectly rigid body is infinite \((\infty)\).

Hint

Bulk modulus is given by
\(
(B)=\frac{\text { Stress }}{\text { volume strain }}=\frac{p}{\Delta V / V}=\frac{p V}{\Delta V}
\)
For perfectly rigid body, change in volume \(\mathrm{AV}=0\), therefore volumetric strain is zero.
\(
\therefore \quad B=\frac{p V}{0}=\infty
\)
Hence, bulk modulus for a perfectly rigid body is infinity

Hint

We have to apply Hooke’s law to compare the extension in each wire.

\(
\begin{aligned}
& \frac{l}{\Delta L_2}=\left(\frac{\frac{F_2 L_2}{A_2 Y_2}}{\frac{F_1 L_1}{A_1 Y_1}}\right)=\frac{2 f 2 L}{f L} \times \pi r^2 \times \frac{Y}{4 \pi r^2 \times Y} \\
& \frac{\Delta L_2}{l}=\frac{4}{4}=1
\end{aligned}
\)
Hence, \(\Delta L_2=l\), which means the change of length in the second wire is the same.

Hint

\(
\begin{aligned}
& L_t=L(1+\alpha \Delta t) \\
& \Delta L=1 \times 10^{-5} \times 200 \\
& Y=\frac{F L}{A \Delta L}\left(L=1 \mathrm{~m}, A=0.0001 \mathrm{~m}^2\right) \\
& Y=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2  \\
& F=2 \times 10^{11} \times 10^{-5} \times 200 \times \frac{0.0001}{1}=4 \times 10^4 \mathrm{~N}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \frac{\Delta V}{V}=0.1 \%=\frac{0.1}{100}=10^{-3} \\
& \rho=10^3 \mathrm{~kg} \mathrm{~m}^{-3}, h=?
\end{aligned}
\)
Let the rubber ball be taken up to depth \(h\).
\(\therefore\) Change in pressure \((p)=h \rho g\)
We know, \(B=\frac{\Delta P}{(\Delta V / V)} \Rightarrow \Delta P=B \times \frac{\Delta V}{V}\)
\(
\Rightarrow \quad \Delta P=9.8 \times 10^8 \times 10^{-3}=9.8 \times 10^5 \mathrm{Nm}^{-2}
\)
Also, \(\Delta P=\rho g h\)
\(
h=\frac{\Delta P}{\rho g}=\frac{9.8 \times 10^5}{10^3 \times 9.8} \Rightarrow h=10^2 \mathrm{~m}=100 \mathrm{~m}
\)

Hint

According to the problem,
Length of steel cable I \(=9.1 \mathrm{~m}\)
Radius \(r=5 \mathrm{~mm}=5 \times 10 \sim^3 \mathrm{~m}\)
Tension in the cable \(\mathrm{F}=800 \mathrm{~N}\)
Young’s modulus for steel \(Y=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2\)
Change in length \(\Delta l=\) ?
Young’s modulus if given by
\(
\begin{aligned}
& Y=\frac{F / A}{\Delta L / L} \Rightarrow \Delta L=\frac{F L}{Y\left(\pi r^2\right)} \\
& \text { CBSELabs.com } \\
& \Delta L=\frac{800 \times 9.1}{\left(2 \times 10^{11}\right)\left(3.14 \times 25 \times 10^{-6}\right)}=4.64 \times 10^{-4} \mathrm{~m} \\
& \Rightarrow \quad \Delta L \approx 5 \times 10^{-4} \mathrm{~m} \approx 0.5 \times 10^{-3} \mathrm{~m}
\end{aligned}
\)

Hint

Since, ivory ball is more elastic than wet-clay ball, therefore, it tends to regain its original shape quickly. Due to this reason, more energy and momentum is transferred to the ivory ball in comparison to the wet-clay ball and hence, ivory ball will rise higher after striking the floor even though both are dropped from the same height.

Hint

According to the problem force \(F\) is applied along horizontally, so we resolve it in two perpendicular components—one is parallel to the inclined plane and other one is perpendicular to the inclined plane as shown in the diagram. Now, we can easily calculate the tensile and shearing stress. Here,
\(
F_{\perp}=F \sin \theta, F_{\|}=F \cos \theta
\)
Let the cross-sectional area of the bar be \(A\). Consider the equilibrium of the plane \(a a^{\prime}\). Here, \(F_{\perp}\) produces tensile stress and \(F_{\|}\) produces shear stress, on the plane \(a a^{\prime}\).
Let the area of the face \(a a^{\prime}\) be \(A\), then
\(
\therefore \quad \sin \theta=\frac{A}{A^{\prime}} \Rightarrow A^{\prime}=\frac{A}{\sin \theta}
\)
Tensile stress on the plane \(a a^{\prime}\)
\(
a a^{\prime}=\frac{F_{\perp}}{A^{\prime}}=\frac{F \sin \theta}{A / \sin \theta}=\frac{F}{A} \sin ^2 \theta
\)
Shearing stress on the plane \(a a^{\prime}\),
\(
\begin{aligned}
\text { Shearing stress } & =\frac{\text { Parallel force }}{\text { Area }} \\
& =\frac{F_{\|}}{A^{\prime}}=\frac{F \cos \theta}{A / \sin \theta}=\frac{F \sin \theta \cos \theta}{A}=\frac{F(2 \sin \theta \cos \theta)}{2 A} \\
& =\frac{F \sin 2 \theta}{2 A}
\end{aligned}
\)
(a) For tensile stress to be maximum,
\(
\sin ^2 \theta=1 \Rightarrow \sin \theta=1 \Rightarrow \theta=\frac{\pi}{2} \text { or } \theta=90^{\circ}
\)
(b) For shearing stress to be maximum,
\(
\sin 2 \theta=1 \Rightarrow 2 \theta=\frac{\pi}{2} \Rightarrow \theta=\frac{\pi}{4} \text { or } \theta=45^{\circ}
\)
important point: Here we are not applying the direct formula for stress. Because to analyze different types of stresses, we have to divide the force into components. In normal stress, the force is applied normally to the surface. But in shear stress deforming force is applied tangentially to one of the faces

Hint

\(
\text { Consider the diagram when a small element of length } d x \text { is considered at } x \text { from the load }(x=0) \text {. }
\)
Let \(T(x)\) and \(T(x+d x)\) are tensions on the two cross-sections a distance \(d x\) apart, then \(-T(x+d x)+T(x)=\) \(d m g=\mu d x g\) (where \(\mu\) is the mass/length) \(\ldots \ldots .(\because d m=\mu d x)\)
\(
\begin{aligned}
& d T=\mu g d x \ldots . .[\because d T=T(x+d x)-T(x)] \\
& \Rightarrow T(x)=\mu g x+C \ldots .(O n \text { integrating }) \\
& \text { At } x=0, T(0)=M g \\
& \Rightarrow C=m g
\end{aligned}
\)
\(
\therefore T(x)=\mu g x+M g
\)
Let the length \(d x\) at \(x\) increase by \(d r\), then
Young’s modulus \(Y=\frac{\text { Stress }}{\text { Strain }}\)
\(
\begin{aligned}
& \frac{\frac{T(x)}{A}}{\frac{d r}{d x}}=Y \\
& \Rightarrow \frac{d r}{d x}=\frac{1}{Y A} T(x) \\
& \Rightarrow r=\frac{1}{Y A} \int_0^L(\mu g x+M g) d x \\
& =\frac{1}{Y A}\left[\frac{\mu g x^2}{2}+M g x\right]_0^L \\
& =\frac{1}{Y A}\left[\frac{m g L^2}{2}+M g L\right] \\
& A=\pi \times\left(10^{-3}\right)^2 m^2 \\
& Y=200 \times 10^9 N^{-2} \\
\end{aligned}
\)
\(
\begin{aligned}
& m=\pi \times\left(10^{-3}\right)^2 \times 10 \times 7860 \mathrm{~kg} \\
& \therefore r=\frac{1}{2 \times 10^{11} \times \pi \times 10^{-6}} \left[\frac{\pi \times 786 \times 10^{-3} \times 10 \times 10}{2}+25 \times 10 \times 10\right] \\
& =\left[196.5 \times 10^{-6}+3.98 \times 10^{-3}\right] \\
& =4 \times 10^{-3} \mathrm{~m}
\end{aligned}
\)

Hint

Clearly, tension will be maximum at \(x=L\)
\(
\therefore \mathrm{T}=\mu \mathrm{gL}+\mathrm{Mg}=(\mathrm{m}+\mathrm{M}) \mathrm{g} \quad \ldots \ldots[\because \mathrm{m}=\mu \mathrm{L}]
\)
The yield force \(=(\) Yield strength \(Y)\) area \(=250 \times 10^6 \times \pi \times\left(10^{-3}\right)^2=25 \times \pi \mathrm{N}\)
At the yield point, \(T=\) Yield force
\(
\begin{aligned}
& \Rightarrow(\mathrm{m}+\mathrm{M}) \mathrm{g}=250 \times \pi \\
& \mathrm{m}=\pi \times\left(10^{-3}\right)^2 \times 10 \times 7860<<\mathrm{M} \\
& \therefore \mathrm{Mg}=250 \times \pi
\end{aligned}
\)
Hence, \(M=\frac{250 \times \pi}{10}=25 \times \pi=75 \mathrm{~kg}\).

Hint

Centripetal force acting on this element,
\(
\mathrm{dF}=\mathrm{dm} \cdot \mathrm{x} \omega^2
\)
\(
\mathrm{dF}=\left(\frac{\mathrm{M}}{2l}\right) \mathrm{dx} \cdot \mathrm{x} \omega^2
\)
Here, \(\mathrm{dF}\) is provided by tension in element \(\mathrm{dx}\) of the rod due to elasticity.
Let tension in the rod be \(\mathrm{F}\) at a distance \(\mathrm{x}\) from the axis of rotation.
\(F\) is due to the centripetal force acting on all the elements from \(x\) to l , i.e
\(
F=\frac{M \omega^2}{2l} \int_x^1 x d x=\frac{M \omega^2}{4l}\left(l^2-x^2\right)
\)
If \(\mathrm{d}(\mathrm{r})\) is the extension in the element of length \(\mathrm{dx}\) at position \(\mathrm{x}\), then,
\(
\mathrm{d}(\mathrm{r})=\frac{\mathrm{Fdx}}{\mathrm{YA}}\left[\because \mathrm{y}=\frac{\mathrm{F} / \mathrm{A}}{\mathrm{d}(\mathrm{r}) / \mathrm{dx}}\right]
\)
Hence, extension in the half of the rod (from axis to point A) is given by
\(
\begin{aligned}
& \Delta \mathrm{r}=\int_0^1 \mathrm{~d}(\mathrm{r})=\int_0^1 \frac{\mathrm{F} d \mathrm{x}}{\mathrm{YA}} \\
& =\frac{\mathrm{M} \omega^2}{4 \mathrm{YAl}}\left[l^2(\mathrm{x})-\frac{\mathrm{x}^3}{3}\right]_0^l=\frac{\mathrm{M} \omega^2}{4 \mathrm{YAl}}\left[l^3-\frac{l^3}{3}\right]=\frac{\mathrm{M} \omega^2 l^2}{6 \mathrm{YA}}
\end{aligned}
\)
Hence, total extension in entire rod of length 21.
\(
2 \Delta=\frac{M \omega^2 l^2}{3 \mathrm{YA}}
\)
\(\therefore\) Total change in length \(=\frac{2}{3 \mathrm{YA}} \mu \omega^2 l^2\)
First we have to consider a small element on the rod of mass \(\mathrm{dm}\) to find tension in the rod at this element and then calculate for the whole rod. Let us consider an element of width \(\mathrm{dx}\) at a distance \(\mathrm{x}\) from the given axis of rotation as shown in the diagram.

Hint

As the temperature of the rods increases length of each side will change, hence the angle corresponding to any vertex also changes as shown in the diagram.
Before heating, \(A B=B C=C A=1\)
After heating temperature of each rod is changed by \(\triangle T\) and sides of \(\triangle A B C\) is changed.
Let \(A B=I_1, B C=I_3, C A=I_2\), using cosine formula,
Let \(l_1=\mathrm{AB}, l_2=\mathrm{AC}, l_3=\mathrm{BC}\)
\(\cos \theta=\frac{l_3{ }^2+l_1^2-l_2^2}{2 l_3 l_1}\)
Or, \(2 l_3 l_1 \cos \theta=l_3^2+l_1^2-l_2^2\)
Differenciating
\(2\left(l_3 d l_1+l_1 d l_3\right) \cos \theta-2 l_1 l_3 \sin \theta d \theta\)
\(2 l_3 d l_3+2 l_1 d l_3+2 l_1 \alpha_1-2 l_2 \alpha_2\)
Now,
\(
\begin{aligned}
& d l_1=l_1 \alpha_1 \Delta t \\
& d l_2=l_2 \alpha_1 \Delta t \\
& d l_3=l_3 \alpha_2 \Delta t
\end{aligned}
\)
and \(l_1=l_2=l_3=l\)
\(
\begin{aligned}
& \left(l^2 \alpha_1 \Delta t+l^2 \alpha_1 \Delta t\right) \cos \theta+l^2 \sin \theta d \theta=l^2 \alpha_1 \Delta t+l^2 \alpha_1 \Delta t-l^2 \alpha_2 \Delta t \\
& \sin \theta d \theta=2 \alpha_1 \Delta t(1-\cos \theta)-\alpha_2 \Delta t
\end{aligned}
\)
Putting \(\theta=60^{\circ}\)
\(
d \theta \frac{\sqrt{3}}{2}=2 \alpha_1 \Delta t \times(1 / 2)-\alpha_2 \Delta t
\)
\(
=\left(\alpha_1-\alpha_2\right) \Delta t
\)
\(
d \theta=\frac{2\left(\alpha_1-\alpha_2\right) \Delta t}{\sqrt{3}}
\)

Hint

According to the problem, the elastic torque or the bending torque is given and we have to find the torque caused by the weight due to bending.
The diagram of the given situation is as shown.
The bending torque on the trunk of radius \(r\) of tree \(=Y \pi R^4 / 4 R\) where \(\mathrm{R}\) the radius of curvature of the bent surface.
\(
\tau=W d=\frac{Y \pi r^4}{4 R}
\)
Here deforming torque is equal to elastic torque (restoring torque) caused by bending of tree about its central axis when the tree is about to buckle.
Let \(h\) be the height of tree. If \(R \gg>\), then the centre of gravity of tree is at a height \(h / 2\) from the ground.
Refer figure, in \(\triangle A B C\)
\(
R^2=(R-d)^2+(h / 2)^2=R^2-2 R d+d^2+h^2 / 4
\)
Since, \(d<<R\), therefore the term \(d^2\) being very very small can be neglected. \(\therefore R^2=R^2-2 R d+h^2 / d\)
or \(\quad d=\frac{h^2}{8 R} \dots(i)\)
If \(\omega_0\) is the weight/volume, then
\(
\begin{aligned}
& \qquad \frac{Y \pi r^4}{4 R}=W_0\left(\pi r^2 h\right) \frac{h^2}{8 R} \quad[\because \text { Torque is caused by the weight] } \\
& \Rightarrow \quad h \simeq\left(\frac{2 Y}{W_0}\right)^{1 / 3} r^{2 / 3} \\
& \text { Hence, critical height }=h=\left(\frac{2 Y}{W_0}\right)^{1 / 3} r^{2 / 3}
\end{aligned}
\)

Hint

In this problem, the given string is elastic. Consider the diagram the stone is dropped from point P.
When the stone is dropped, then it covers distance Y before coming to rest, for the first instant.

\(
Y=L+(Y-L)
\)
First it covers the distance \(L\) equal to length of string distance in free fall and a further distance \((Y-L)\) due to extension in the string. So it covers a total distance \(Y\) until it instantaneously comes to rest at Q.
We have to find \(y\), so by applying energy conservation principle,
Loss in potential energy of stone = Gain in elastic potential energy in string
\(
\begin{aligned}
& \text { a) } m g y=\frac{1}{2} K(y-L)^2 \\
& m g y=\frac{1}{2} K\left(y^2+L^2-2 y L\right) \\
& 2 m g y=K y^2-2 K y L+K L^2 \\
& K y^2-2(K L+m g) y+K L^2=0
\end{aligned}
\)
Eliminating the negative sign, we are only taking positive sign only,
\(
y=\frac{(k L+m g)+\sqrt{2 m g k L+m^2 g^2}}{k}
\)

Hint

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 \(m g-k x=0\), where \(r\) is the extension from \(L\).
So, \(\mathrm{F}=0\)
Here the force of the string is balances by the force of gravity which makes, \(\mathrm{mg}=\mathrm{Kx}\)
If we assume \(v\) to be equal to maximum velocity then by the law of conservation of energy,
Kinetic energy of stone + Potential energy gained by the string = potential energy lost from \(P\) to \(Q^{\prime}\) \(\frac{1}{2} m v 2+\frac{1}{2} K x^2=m g(L+x) \quad\)
\(m v^2+K x^2=2 m g(L+x)\)
\(x=\frac{m g}{K}\)
\(m v^2=2 m g L+\frac{2 m^2 g^2}{K}-\frac{m^2 g^2}{K}\)
\(v=\left[2 g L+\frac{m g^2}{K}\right]^{\frac{1}{2}}\)

Hint

c. When the stone is at the lowest position i.e., at instantaneous distance \(\mathrm{Y}\) from \(\mathrm{P}\), then the equation of motion of the stone is \(\frac{m d^2 y}{d t^2}=m g-k(y-L)\)
\(
\Rightarrow \frac{d^2 y}{d t^2}+\frac{k}{m}(y-L)-g=0
\)
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 the above equation will be of the type \(z=A \cos (\omega t+\phi)\), where \(\omega=\sqrt{\frac{k}{m}}\) \(y=\left(L+\frac{m g}{k}\right)+A^{\prime} \cos (\omega t+\pi)\)
Thus, the stone will perform SHM with angular frequency \(\omega=\sqrt{\frac{k}{m}}\) about a point \(y_0=L+\frac{m g}{k}\).

Hint

We know, Stress versus strain curve for a metal wire (shown in the figure below).

For small deformations as seen in the region OA \(\frac{\text { stress }}{\text { strain }}=\tan \theta_1=\) constant When deformation is increased, \(\frac{\text { stress }}{\text { strain }}=\tan \theta_2\) (at point \(B\) ) From the curve, it is evident that \(\theta_2<\theta_1 \Rightarrow \tan \theta_2<\tan \theta_1\) Therefore, the ratio will decrease when deformation increases.

Hint

Stress \(=M g / A\), {Where \(A\) is the cross-sectional area of the wire \(\}\) Strain \(=l / L\).
Since the elastic potential energy
\(U=1 / 2(\text { Stress }) \times (\) Strain \() \times (\) Volume \()\)
\(\rightarrow U=1 / 2 \times (\mathrm{Mg} / \mathrm{A}) \times (\mathrm{l} / \mathrm{L}) \times (\mathrm{AL})=1 / 2 \mathrm{Mgl}\)
The loss in gravitational potential energy will be Mgl only when the load is suddenly released from the initial position i.e. when elongation is zero. The elastic potential energy stored in the wire \(=1 / 2 \mathrm{Mgl}\) when the elongation is l, the rest of the gravitational potential energy \(\mathrm{Mgl}-1 / 2 \mathrm{Mgl}=1 / 2 \mathrm{Mgl}\) is converted to the kinetic energy. At the time elongation is \(l\), the mass \(M\) will have a certain speed say v. So \(1 / 2 Mv^2\) \(=1 / 2 \mathrm{Mgl}\)
\(\rightarrow \mathrm{V}^2=\mathrm{gl}\)
\(\rightarrow \mathrm{V}=\sqrt{ }(g l)\)
This point as a mean position the mass will be in a simple harmonic motion vertically. Slowly it will come to rest at the mean position by dissipating this energy in the form of heat.

Hint

Let \(\mathrm{my}\) length \(=\mathrm{L}\)
Let the increase in length \(=l\)
Strain
\(
\Rightarrow {l}=\frac{L}{100}
\)
So, the increase in length will be \(\frac{L}{100}\).
Yes, the yield point is much less than the \(1 \%\) strain because the human body consists of joints and not one uniform solid structure.

Hint

In a compressed spring the molecules are under stress due to the elastic potential energy stored in it. When the acid dissolves the spring molecules the P.E. stored in it is converted to K.E. and the new molecule it forms with acid ion has this extra K.E. Due to this the temperature of the acid increases slightly. It means that the elastic potential energy is converted to heat energy.

Hint

Let the Young’s modulus of the material of the wire be \(Y\).
Force \(=\) Weight \(=\mathrm{W}(\) given \()\)
Let cross-sectional area = A
\(\mathrm{x}=1 \mathrm{~mm}=\) Elongation in the first case
Length \(=\mathrm{L}\)
\(
Y=\frac{\frac{W}{A}}{\frac{x}{L}}=\frac{W L}{A x}
\)
Let \(y\) be the elongation on one side of the wire when put in a pulley.
When put in a pulley, the length of the wire on each side \(=\frac{L}{2}\)
\(
\begin{aligned}
& \frac{\frac{W}{A}}{\frac{y}{\frac{L}{2}}}=Y \\
& \Rightarrow \frac{\frac{W}{A}}{\frac{y}{\frac{L}{2}}}=\frac{W L}{\mathrm{Ax}} \\
& \Rightarrow y=\frac{x}{2}
\end{aligned}
\)
Total elongation in the wire \(=2 \mathrm{y}=2\left(\frac{x}{2}\right)=x=1 \mathrm{~mm}\)

Hint

As the rod is of uniform mass distribution and stretched by its own weight, the topmost part of the rod experiences maximum stress due to the weight of the entire rod. This stress leads to lateral strain and the rod becomes thinner. Moving down along the length of the rod, the stress decreases because the lower parts bear lesser weight of the rod. With reduced stress, the lateral strain also reduces. Hence, the diameter of the rod gradually increases from top to bottom.

Hint

\(\frac{\Delta l}{\mathrm{I}}\)
cross sectional areaLength \(=A\), Length =\(l\)
Volume of the wire \(\mathrm{V}={Al}\)
Assuming no lateral strain when longitudinal strain occurs:
Increase in volume: \(\Delta \mathrm{V}=\mathrm{A} \Delta{l}\)
\(\Rightarrow \frac{\Delta V}{V}=\frac{A \Delta{l}}{{Al}}=\frac{\Delta{l}}{{l}}\)
So, \(\frac{\Delta V}{V}\) is directly proportional to \(\frac{\Delta l}{{l}}\).

Hint

Let the initial length of the metal wire is \(\mathrm{L}\).
The strain at tension \(T_1\) is \(\triangle \mathrm{L}_1=\mathrm{L}_1-\mathrm{L}\)
The strain at tension \(T_2\) is \(\triangle \mathrm{L}_2=\mathrm{L}_2-\mathrm{L}\)
Suppose, the Youngs modulus of the wire is \(\mathrm{Y}\),
\(
\frac{\frac{T_1}{A}}{\frac{\Delta L_1}{L}}=\frac{\frac{T_2}{A}}{\frac{\Delta L_2}{L}}
\)
Where \(\mathrm{A}\) is a cross-section of the wire assumed to be the same at all the situations.
\(
\begin{aligned}
& \Rightarrow \frac{T_1}{A} \times \frac{L}{\Delta L_1}=\frac{T_2}{A} \times \frac{L}{\Delta L_2} \\
& \Rightarrow \frac{T_1}{\left(L_1-\mathrm{L}\right)}=\frac{T_2}{\left(L_2-L\right)} \\
& T_1\left(L_2-L\right)=T_2\left(L_1-L\right) ; L=\frac{T_2 L_1-T_1 L_2}{T_2-T_1}
\end{aligned}
\)

Hint

Let us consider a vertical circular motion of speed \(v\)

Let \(A, B, C\) and \(D\) be the points on a vertical circle as shown in the figure. By applying \(\sum F=\) ma along radial direction at \(A\), we get \(\mathrm{T}-\mathrm{mg}=\frac{\mathrm{mv}^2}{\mathrm{r}}
\Rightarrow \mathrm{T}=\frac{\mathrm{mv}^2}{\mathrm{r}}+\mathrm{mg} \dots(1)\).
at \(B\), we get
\(
\mathrm{T}=\frac{m v^2}{\mathrm{r}} \dots(2)
\)
at \(\mathrm{C}\), we get
\(
\mathrm{T}+\mathrm{mg} \cos 30^{\circ}=\frac{m v^2}{r} \Rightarrow \mathrm{T}=\frac{m v^2}{r}-m g \cos 30^{\circ} \dots(3)
\)
at D, we get
\(
\mathrm{T}+\mathrm{mg}=\frac{\mathrm{mv}^2}{\mathrm{r}} \Rightarrow \mathrm{T}=\frac{\mathrm{mv}^2}{\mathrm{r}}-\mathrm{mg} \dots(4)
\)
From (1), (2), (3) and (4), we can see that the value of tension in the string is maximum at point \(A\) i.e at lowest point.
\(\therefore\) wire has more chance to break at this point.
Hence, option (b) is the correct answer.

Hint

When a mass is hanged by a wire the gravitational potential energy decreased is \(\mathrm{mgl}\) (where l is the elongation) of the wire. But the elastic potential energy increased in the wire \(=1 / 2 \mathrm{mgl}\). Hence the reduced gravitational potential energy will never completely appear as increased kinetic energy or increased elastic potential energy or as heat.

Hint

All options are correct.
(a) When a weight is loaded on a wire, the length of the wire increases. The relationship between weight and length is linear.
(b) When a weight is loaded, it produces stress on the wire. The relationship between stress and increase in length is also linear.
(c) When stress is applied, strain develops. Therefore, both are linearly related.
(d) Since the value of \(Y\) for the wire is unknown, \(X\) may also be the increase in its length. Nevertheless, they still show the same linear relationship.