Properties of Matter

A. Young's Modulus $=\frac{\text { Stress }}{\text { Strain }}=\frac{F L}{A \Delta L}$

B. Compressibility $=\frac{1}{\text { Bulk modulus }}=-\frac{1}{\Delta P}\left(\frac{\Delta V}{V}\right)$

C. Bulk Modulus $=-P\left(\frac{V}{\Delta V}\right)$

D. Poisson's Ratio $=\frac{\text { Lateral strain }}{\text { Longitudinal strain }}=\frac{\Delta d}{\Delta L}\left(\frac{L}{d}\right)$

$P=P_0+\rho g h$

$ \begin{aligned} & \Rightarrow 100 \times 10^5=10^5+10^3 \times 10 \times h \\ & \Rightarrow 10^7=10^5+10^4 h \\ & \Rightarrow 10^3=10+h \\ & \Rightarrow h=1000-10=990 \mathrm{~m} \end{aligned} $

$\begin{aligned} & R O C=\text { Radius of curvature at point } A \\ & \text { Curvature }=\frac{1}{R O C}=\frac{\left|\frac{d^2 y}{d x^2}\right|}{\left(1+\left(\frac{d y}{d x}\right)^2\right)^{\frac{3}{2}}}=\frac{\left|\frac{d^2 y}{d x^2}\right|}{(1+0)^{\frac{3}{2}}}=\frac{d^2 y}{d x^2} \quad\left[\because \frac{d y}{d x}=\tan \theta=0\right] \\ & \Delta P=S \times \text { curvature } \\ & \Rightarrow \rho g y=S \frac{d^2 y}{d x^2} \\ & \therefore \frac{d^2 y}{d x^2}=\frac{\rho g y}{S} \end{aligned}$

Alternate Solution :

For the given element, (consider length d in Z direction) Net force in upward direction $=$ Weight $(\mathrm{S} \sin (\theta+\mathrm{d} \theta)-\mathrm{S} \sin \theta) \mathrm{d}=\mathrm{mg}$

Angle is small $\therefore \sin \theta \approx \theta$

$ \begin{aligned} & \Rightarrow \frac{d \theta}{y d x}=\frac{\rho g}{S} \\ & \tan \theta=\frac{d y}{d x} \Rightarrow \text { Differentiating wrt } x \\ & \sec ^2 \theta \frac{d \theta}{d x}=\frac{d^2 y}{d x^2} \end{aligned} $

Put $d \theta$ from (2) in (1) $d$ take $\cos \theta \approx 1$, we get

$ \frac{d^2 y}{d x^2}=\frac{\rho g y}{S} $

According to Bernoulli's principle,

Kinetic energy per unit volume + Potential energy per unit volume + Pressure $=$ Constant

$\frac{1}{2} \rho V^2+\rho g h+P=\text { constant }$

Apply Bernoulli's principle at point $X$ and $Y$,

$\begin{aligned} & P+K_1+\rho g(0)=P+K_2+\rho g(\mathrm{~h}) \\ & K_1=K_2+\rho g h \\ & K_1>K_2 \end{aligned}$

First, we need to find the maximum force that can be applied to the steel wire within its elastic limit. This force can be calculated using the given area under stress and the stress limit provided by the elastic limit of steel.

Let's assume the area of cross-section of the wire is $A$. The force exerted can be given by:

$ F = \sigma \times A $

where:

$ \sigma = 8 \times 10^8 \mathrm{~N/m}^2 $ (Elastic limit of steel)

To calculate the elongation ($ \Delta L $) under this force, we use Hooke's Law, which relates force, elongation, cross-sectional area, original length, and Young's modulus as follows:

$ F = \frac{Y \times A \times \Delta L}{L} $

where:

$ Y = 2 \times 10^{11} \mathrm{~N/m}^2 $ (Young's modulus of steel),

$ L = 1 \mathrm{~m} $ (original length of the wire).

Substituting for $ F $ from the earlier expression and rearranging the formula, we get:

$ \sigma \times A = \frac{Y \times A \times \Delta L}{L} $

$ \sigma = \frac{Y \times \Delta L}{L} $

$ \Delta L = \frac{\sigma \times L}{Y} $

$ \Delta L = \frac{8 \times 10^8 \times 1}{2 \times 10^{11}} \mathrm{~meters} $

$ \Delta L = 0.004 \mathrm{~meters} $

$ \Delta L = 4 \mathrm{~mm} $

Thus, the maximum elongation of the wire within the elastic limit is $4 \mathrm{~mm}$.

The answer is: Option A - 4 mm.

$ \begin{aligned} & \text { Excess force }=T \times 2 \pi R \\ & =\frac{7}{100} \times 2 \times 3.14 \times \frac{4.5}{100} \\ & =197.82 \times 10^{-4} \\ & =19.8 \times 10^{-3} \mathrm{~N} \\ & =19.8 \mathrm{~mN} \end{aligned} $

Given the properties and conditions of the metallic bar, we are required to calculate the compressive force developed due to heating. Key inputs include the Young's modulus (E), coefficient of linear thermal expansion (α), change in temperature ($\Delta T$), and the original dimensions of the bar.

First, compute the linear expansion of the bar if it were free to expand. The change in length ($\Delta L$) due to thermal expansion can be computed through the formula:

$ \Delta L = \alpha L_0 \Delta T $

Given:

• $ \alpha = 10^{-5} { }^{\circ} C^{-1} $
• $ L_0 = 1 \text{ m} $
• $ \Delta T = 100^{\circ} C $

Thus:

$ \Delta L = 10^{-5} \times 1 \times 100 = 0.001 \text{ m} $

This is the change in length that the bar would undergo if not constrained.

However, in this scenario, the bar is constrained and does not actually expand. This constraint induces a compressive stress (constrained thermal stress) in the bar, which can be calculated using the formula relating stress, Young's modulus, and strain:

$ \sigma = E \epsilon $

Where the strain ($\epsilon$) under constrained conditions due to thermal expansion is:

$ \epsilon = \frac{\Delta L}{L_0} = \frac{0.001}{1} = 0.001 $

Therefore:

$ \sigma = 0.5 \times 10^{11} \times 0.001 = 5 \times 10^7 \mathrm{ N/m}^2 $

This stress is the force per unit area. To find the compressive force, we need to multiply this stress by the cross-sectional area of the bar:

$ F = \sigma A = 5 \times 10^7 \mathrm{ N/m}^2 \times 10^{-3} \mathrm{ m}^2 = 5 \times 10^4 \mathrm{ N} $

Thus, the compressive force developed in the bar is $50 \times 10^3 \mathrm{~N}$.

Hence, the correct answer is Option B:

$50 \times 10^3 \mathrm{~N}$

$ \begin{aligned} & \frac{\text { E.P.E }}{\text { Volume }}=\frac{1}{2}(\text { stress)(strain) } \\ & =\frac{1}{2}(Y)(\text { strain })^2 \\ & =\frac{1}{2}(\mathrm{Y})\left(\frac{\Delta \mathrm{L}}{\mathrm{L}}\right)^2 \\ & =\frac{1}{2}\left(2 \times 10^{11}\right)\left(\frac{1 \times 10^{-3}}{100 \times 10^{-2}}\right)^2 \\ & =10^5 \end{aligned} $

Pressure is a scalar quantity.

$ \begin{aligned} & F=6 \pi \mathrm{r} \eta v \\ & =(6)(3.14)\left(\frac{1 \times 10^{-3}}{2}\right)\left(0.8 \times 10^{-1}\right)(2) \\ & =1.5 \times 10^{-3} \mathrm{~N} \end{aligned} $

$E=2 T\left(4 \pi \mathrm{R}^{2}\right)$

$=2(0.03)(4)(3.14)\left(2 \times 10^{-2}\right)^{2}$

$=3.01 \times 10^{-4} \mathrm{~J}$

When a wire is suspended from the ceiling and stretched by a weight W attached at its free end, the force acting on the wire is equal to the weight W. This force causes longitudinal stress in the wire.

Longitudinal stress is defined as the force acting on an object divided by its cross-sectional area. In this case, the force acting on the wire is W, and the cross-sectional area of the wire is A.

Thus, the longitudinal stress at any point of the wire with cross-sectional area A is given by :

$\text{Longitudinal Stress} = \frac{\text{Force}}{\text{Area}} = \frac{W}{A}$

In hydrostatic condition,

$P = {P_0} + \rho gh$

Here, P : absolute pressure at depth h, P₀ is atmospheric pressure

Since, height of liquid in both vessel is same therefore pressure on the base of both vessel will be same.

F_viscous = $-$6$\pi$$\eta$av

F_viscous = 6 $\times$ 3.14 $\times$ 0.9 $\times$ 5 $\times$ 10^$-$3 $\times$ 10 $\times$ 10^$-$2

= 84.78 $\times$ 10^$-$4

= 8.478 $\times$ 10^$-$3 N

Initial speed of ball is zero and it finally attains terminal speed

Excess pressure inside the bubble $ = \Delta P - {{4T} \over R}$

${P_{in}} = {P_{out}} + {{4T} \over R}$

as 'R' increases 'P' decreases

It is true that stretching of spring is determined by shear modulus of the spring as when coil spring is stretched neither its length nor its volume changes, there is only change in its shape.

Tensile strength of steel is more than that of copper.

Hence Assertion is true and Reason is false.

Given, Height of small orifice from ground $(H–h) = 0.25$ m

Total height of water tank,

$(H)=1 \mathrm{~m}$

$\therefore$ Range of water stream,

$\begin{aligned} R & =2 \sqrt{h(H-h)}=2 \sqrt{(H-0.25)(0.25)} \\\\ & =2 \sqrt{(1-0.25) 0.25}=2 \sqrt{0.75 \times 0.25} \\\\ & =0.866 \mathrm{~m}=86.6 \mathrm{~cm} \end{aligned}$

Given :

Diameter at the first end, $ d_1 = 5 \, \text{cm} $

Velocity at the first end, $ v_1 = 4 \, \text{m/s} $

Diameter at the second end, $ d_2 = 2 \, \text{cm} $

$ \Delta P = ? $

Step 1 : Calculate the radii

$ r_1 = \frac{d_1}{2} = 2.5 \, \text{cm} = 2.5 \times 10^{-2} \, \text{m} $

$ r_2 = \frac{d_2}{2} = 1 \, \text{cm} = 10^{-2} \, \text{m} $

Step 2 : Apply the principle of continuity

$ A_1 v_1 = A_2 v_2 $

$ \pi r_1^2 v_1 = \pi r_2^2 v_2 $

$ \left(2.5 \times 10^{-2}\right)^2 \times 4 = \left(10^{-2}\right)^2 \cdot v_2 $

$ v_2 = 25 \, \text{m/s} $

Step 3 : Use Bernoulli's theorem

$ p_1 + \frac{1}{2} \rho v_1^2 = p_2 + \frac{1}{2} \rho v_2^2 $

$ p_1 - p_2 = \frac{1}{2} \rho \left(v_2^2 - v_1^2\right) $

Given $ \rho = 10^3 \, \text{kg/m}^3 $:

$ p_1 - p_2 = \frac{1}{2} \times 10^3 \left(25^2 - 4^2\right) $

$ = 304500 \, \text{Pa} $

Given, surface tension of water, $T=0.1 \mathrm{~N} / \mathrm{m}$

Radius of small drops, $r=1 \mathrm{~mm}=10^{-3} \mathrm{~m}$

If $R$ be the radius of bigger drops, then volume remains conserved.

$\begin{aligned} \text{i.e.,} \quad V_1 & =V_2 \\ \frac{4}{3} \pi R^3 & =2 \times \frac{4}{3} \pi r^3 \\ R^3 & =2 r^3=2 \cdot\left(10^{-3}\right)^3 \\ \therefore \quad R & =2^{1 / 3} \times 10^{-3} \mathrm{~m} \end{aligned}$

$\therefore$ Change in energy during isothermal combination,

$\begin{aligned} \Delta E & =T \cdot \Delta A \\ & =T\left[A_2-A_1\right] \\ & =0.1\left[2 \times 4 \pi r^2-4 \pi R^2\right] \\ & \begin{aligned} & =0.1 \times 4 \pi\left[2 r^2-R^2\right] \\ & =0.4 \pi\left[2 \times 10^{-6}-2^{2 / 3} \times 10^{-6}\right] \\ & =0.4 \pi \times 10^{-6} \times 0.41 \\ & =0.52 \times 10^{-6} \mathrm{~J}=0.52 \mu \mathrm{J} \simeq 0.5 \mu \mathrm{J} \end{aligned} \end{aligned}$

Surface tension is responsible for the shape of tiny water drops when falling freely in the absence of other forces like gravity, shape of water drops is spherical because droplets of water tend to be pulled into a spherical shape by the cohesive forces of the surface layer.

In the presence of gravity, the shape of heavy water drops is not spherical when falling freely. It is oval shaped surface tension of methyl alcohol is lower than surface tension of water.

Hence, water has not minimum surface tension among all liquids.

Therefore, both Assertion and Reason are false.

Sometimes, insects can walk on the surface of water due to surface tension, when legs of insects are not being wet. In this situation, the gravitational force on insect is balanced by force due to surface tension. Hence, both assertion and reason are true and reason is correct explanation of assertion.

As excess pressure,

$p=\frac{2 s}{R} \Rightarrow p \propto \frac{1}{R}$

$\therefore$ Excess pressure inside smaller drop is large due to which smaller force resist deforming forces better.