Properties of Matter
An elastic spring under tension of $3 \mathrm{~N}$ has a length $a$. Its length is $b$ under tension $2 \mathrm{~N}$. For its length $(3 a-2 b)$, the value of tension will be _______ N.
Explanation:
To determine the tension for the elastic spring's length of $(3a - 2b)$, we can use Hooke's law which states that the force exerted by a spring is proportional to the extension or compression of the spring from its natural length. Mathematically, Hooke's law is given by:
$ F = k \cdot \Delta x $
where:
- $F$ is the force exerted by the spring.
- $k$ is the spring constant (a measure of the stiffness of the spring).
- $\Delta x$ is the displacement from the natural length.
Let's denote the natural (unstretched) length of the spring as $l_0$. Given that the spring's length is $a$ under a tension of $3 \mathrm{~N}$, and its length is $b$ under a tension of $2 \mathrm{~N}$, we can write the equations as:
$ 3 \mathrm{~N} = k \cdot (a - l_0) $
$ 2 \mathrm{~N} = k \cdot (b - l_0) $
Next, we need to find the length displacement $(3a - 2b)$ in terms of the natural length $l_0$. We express the displacements as:
$ x = (3a - 2b) - l_0 $
To solve for this, let's express $a - l_0$ and $b - l_0$ from the given conditions:
$ a - l_0 = \frac{3 \mathrm{~N}}{k} $
$ b - l_0 = \frac{2 \mathrm{~N}}{k} $
Substitute these into the displacement equation:
$ x = (3a - 2b) - l_0 = 3 \left( \frac{3 \mathrm{~N}}{k} + l_0 \right) - 2 \left( \frac{2 \mathrm{~N}}{k} + l_0 \right) - l_0 $
On simplifying, we get:
$ x = 3 \frac{3 \mathrm{~N}}{k} + 3 l_0 - 2 \frac{2 \mathrm{~N}}{k} - 2 l_0 - l_0 $
$ x = \frac{9 \mathrm{~N}}{k} + 3 l_0 - \frac{4 \mathrm{~N}}{k} - 3 l_0 $
$ x = \frac{5 \mathrm{~N}}{k} $
Finally, by Hooke's Law, the force corresponding to this displacement is:
$ F = k \cdot x = k \cdot \frac{5 \mathrm{~N}}{k} = 5 \mathrm{~N} $
Therefore, the tension required for the elastic spring's length $(3a - 2b)$ is 5 N.
[Area of cross section of wire $=0.005 \mathrm{~cm}^2, \mathrm{Y}=2 \times 10^{11} \mathrm{Nm}^{-2}$ and $\mathrm{g}=10 \mathrm{~ms}^{-2}$ ]
Explanation:
To solve this problem, we use the fact that the longitudinal strain in a wire is given by the formula:
$\text{Strain} = \frac{\Delta L}{L} = \frac{F}{AY}$where $\Delta L$ is the change in length, $L$ is the original length, $F$ is the force applied, $A$ is the area of cross-section of the wire, and $Y$ is the Young's modulus of the material of the wire.
The force applied by each load due to gravity is calculated using $F = mg$, where $m$ is the mass of the load and $g$ is the acceleration due to gravity ($10 \text{ m/s}^2$ in this case).
For the upper wire, the total force applied is the weight of both masses (2 kg and 1 kg):
$F_1 = (2 \text{ kg} + 1 \text{ kg}) \times 10 \text{ m/s}^2 = 30 \text{ N}$For the lower wire, the force applied is just the weight of the 1 kg mass:
$F_2 = 1 \text{ kg} \times 10 \text{ m/s}^2 = 10 \text{ N}$Since the area $A$ and Young's modulus $Y$ are the same for both wires, these values cancel out when we calculate the ratio of the strains. Therefore, the ratio of the strains is directly proportional to the ratio of the forces:
$\frac{\text{Strain of upper wire}}{\text{Strain of lower wire}} = \frac{F_1}{F_2} = \frac{30 \text{ N}}{10 \text{ N}} = 3$Hence, the ratio of longitudinal strain of the upper wire to that of the lower wire is 3.
(Take air density to be $1 \mathrm{~kg} \mathrm{~m}^{-3}$ and $\mathrm{g}=10 \mathrm{~ms}^{-2}$ )
Explanation:
To solve this problem, we need to employ Bernoulli's equation, which is applied to describe the behavior of fluid flow. For a fluid in steady flow, the principle states that the sum of the pressure potential energy density, kinetic energy density, and the gravitational potential energy density has the same value at all points along a streamline. Since the plane is in level flight, we can ignore changes in gravitational potential energy. Bernoulli's equation can be written as:
$ P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2, $
where,
- $P_1$ and $P_2$ are the static pressures on the lower and upper surfaces of the wing respectively,
- $v_1$ is the airspeed over the lower surface of the wing,
- $v_2$ is the airspeed over the upper surface of the wing,
- $\rho$ is the density of air,
- $P_2 > P_1$ because $v_2 > v_1$.
First, let's convert the airspeeds to $ \mathrm{m/s} $:
$ v_1 = 180 \frac{\mathrm{km}}{\mathrm{h}} \times \frac{1000 \mathrm{m}}{3600 \mathrm{s}} = 50 \mathrm{m/s}, $
$ v_2 = 252 \frac{\mathrm{km}}{\mathrm{h}} \times \frac{1000 \mathrm{m}}{3600 \mathrm{s}} = 70 \mathrm{m/s}. $
The pressure difference between the lower and upper wing surfaces can thus be calculated using Bernoulli's equation:
$ \Delta P = P_1 - P_2 = \frac{1}{2} \rho v_2^2 - \frac{1}{2} \rho v_1^2. $
Substitute the values (with $\rho = 1 \mathrm{~kg/m}^3$):
$ \Delta P = \frac{1}{2} (1 \mathrm{~kg/m}^3) (70 \mathrm{m/s})^2 - \frac{1}{2} (1 \mathrm{~kg/m}^3) (50 \mathrm{m/s})^2, $
$ \Delta P = \frac{1}{2} (4900 \mathrm{~kg/m \cdot s}^2) - \frac{1}{2} (2500 \mathrm{~kg/m \cdot s}^2), $
$ \Delta P = \frac{1}{2} (2400 \mathrm{~kg/m \cdot s}^2), $
$ \Delta P = 1200 \mathrm{~N/m}^2. $
The lift force generated by the pressure difference over one wing is $ \Delta P \times \text{wing area} $, and since there are two wings, we must double the lift force generated by one wing to find the total lift force sustaining the plane. The weight of the plane is effectively the lift force when in level flight at constant speed. So:
$ \text{Lift} = 2 \times \Delta P \times \text{wing area}, $
$ \text{Lift} = 2 \times 1200 \mathrm{~N/m}^2 \times 40 \mathrm{~m}^2, $
$ \text{Lift} = 2 \times 48000 \mathrm{N}, $
$ \text{Lift} = 96000 \mathrm{N}. $
To find the mass $m$ of the plane, we use Newton's second law, where lift force is equal to the weight ($ mg $, where $g$ is the acceleration due to gravity):
$ m g = \text{Lift}, $
$ m = \frac{\text{Lift}}{g}, $
$ m = \frac{96000 \mathrm{N}}{10 \mathrm{m/s}^2}, $
$ m = 9600 \mathrm{kg}. $
Therefore, the mass of the plane is 9600 kg.
Two blocks of mass $2 \mathrm{~kg}$ and $4 \mathrm{~kg}$ are connected by a metal wire going over a smooth pulley as shown in figure. The radius of wire is $4.0 \times 10^{-5} \mathrm{~m}$ and Young's modulus of the metal is $2.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$. The longitudinal strain developed in the wire is $\frac{1}{\alpha \pi}$. The value of $\alpha$ is _________. [Use $g=10 \mathrm{~m} / \mathrm{s}^2$]
Explanation:
$\begin{aligned} & \mathrm{T}=\left(\frac{2 \mathrm{~m}_1 \mathrm{~m}_2}{\mathrm{~m}_1+\mathrm{m}_2}\right) \mathrm{g}=\frac{80}{3} \mathrm{~N} \\ & \mathrm{~A}=\pi \mathrm{r}^2=16 \pi \times 10^{-10} \mathrm{~m}^2 \\ & \text { Strain }=\frac{\Delta \ell}{\ell}=\frac{\mathrm{F}}{\mathrm{AY}}=\frac{\mathrm{T}}{\mathrm{AY}} \\ & =\frac{80 / 3}{16 \pi \times 10^{-10} \times 2 \times 10^{11}}=\frac{1}{12 \pi} \\ & \alpha=12 \end{aligned}$
The depth below the surface of sea to which a rubber ball be taken so as to decrease its volume by $0.02 \%$ is _______ $m$.
(Take density of sea water $=10^3 \mathrm{kgm}^{-3}$, Bulk modulus of rubber $=9 \times 10^8 \mathrm{~Nm}^{-2}$, and $g=10 \mathrm{~ms}^{-2}$)
Explanation:
$\begin{aligned} & \beta=\frac{-\Delta \mathrm{P}}{\frac{\Delta \mathrm{V}}{\mathrm{V}}} \\ & \Delta \mathrm{P}=-\beta \frac{\Delta \mathrm{V}}{\mathrm{V}} \\ & \rho \mathrm{gh}=-\beta \frac{\Delta \mathrm{V}}{\mathrm{V}} \\ & 10^3 \times 10 \times \mathrm{h}=-9 \times 10^8 \times\left(-\frac{0.02}{100}\right) \\ & \Rightarrow \mathrm{h}=18 \mathrm{~m} \end{aligned}$
A big drop is formed by coalescing 1000 small identical drops of water. If $E_1$ be the total surface energy of 1000 small drops of water and $E_2$ be the surface energy of single big drop of water, then $E_1: E_2$ is $x: 1$ where $x=$ ________.
Explanation:
$\begin{aligned} & \rho\left({ }_3^4 \pi r^3\right) 1000={ }_3^4 \pi R^3 \rho \\ & R=10 r \\ & E_1=1000 \times 4 \pi r^2 \times S \\ & E_2=4 \pi(10 r)^2 S \\ & \frac{E_1}{E_2}=\frac{10}{1}, x=10 \end{aligned}$
Each of three blocks $\mathrm{P}, \mathrm{Q}$ and $\mathrm{R}$ shown in figure has a mass of $3 \mathrm{~kg}$. Each of the wires $\mathrm{A}$ and $\mathrm{B}$ has cross-sectional area $0.005 \mathrm{~cm}^2$ and Young's modulus $2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}$. Neglecting friction, the longitudinal strain on wire $B$ is ________ $\times 10^{-4}$. (Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$)
Explanation:
$\begin{aligned} & a=\frac{10}{3} \mathrm{~m} / \mathrm{s}^2 \\\\ & 30-T_1=3 \times a \\\\ & T_1=20 \mathrm{~N} \\\\ & \text { strain }=\frac{\text { stress }}{Y} \\\\ & =2 \times 10^{-4} \end{aligned}$
Two metallic wires $P$ and $Q$ have same volume and are made up of same material. If their area of cross sections are in the ratio $4: 1$ and force $F_1$ is applied to $P$, an extension of $\Delta l$ is produced. The force which is required to produce same extension in $Q$ is $\mathrm{F}_2$.
The value of $\frac{F_1}{F_2}$ is _________.
Explanation:
$\mathrm{Y}=\frac{\text { Stress }}{\text { Strain }}=\frac{\mathrm{F} / \mathrm{A}}{\Delta \ell / \ell}=\frac{\mathrm{F} \ell}{\mathrm{A} \Delta \ell}$
$\begin{aligned} & \Delta \ell=\frac{\mathrm{F} \ell}{\mathrm{AY}} \\ & \mathrm{V}=\mathrm{A} \ell \Rightarrow \ell=\frac{\mathrm{V}}{\mathrm{A}} \\ & \Delta \ell=\frac{\mathrm{FV}}{\mathrm{A}^2 \mathrm{Y}} \end{aligned}$
$Y ~\& V$ is same for both the wires
$\begin{aligned} & \Delta \ell \propto \frac{\mathrm{F}}{\mathrm{A}^2} \\ & \frac{\Delta \ell_1}{\Delta \ell_2}=\frac{\mathrm{F}_1}{\mathrm{~A}_1^2} \times \frac{\mathrm{A}_2^2}{\mathrm{~F}_2} \\ & \Delta \ell_1=\Delta \ell_2 \\ & \mathrm{~F}_1 \mathrm{~A}_2^2=\mathrm{F}_2 \mathrm{~A}_1^2 \\ & \frac{\mathrm{F}_1}{\mathrm{~F}_2}=\frac{\mathrm{A}_1^2}{\mathrm{~A}_2^2}=\left(\frac{4}{1}\right)^2=16 \end{aligned}$
In a test experiment on a model aeroplane in wind tunnel, the flow speeds on the upper and lower surfaces of the wings are $70 \mathrm{~ms}^{-1}$ and $65 \mathrm{~ms}^{-1}$ respectively. If the wing area is $2 \mathrm{~m}^2$, the lift of the wing is _________ $N$.
(Given density of air $=1.2 \mathrm{~kg} \mathrm{~m}^{-3}$)
Explanation:
$\begin{aligned} & \mathrm{F}=\frac{1}{2} \rho\left(\mathrm{v}_1^2-\mathrm{v}_2^2\right) \mathrm{A} \\ & \mathrm{F}=\frac{1}{2} \times 1.2 \times\left(70^2-65^2\right) \times 2 \\ & =810 \mathrm{~N} \end{aligned}$
The reading of pressure metre attached with a closed pipe is $4.5 \times 10^4 \mathrm{~N} / \mathrm{m}^2$. On opening the valve, water starts flowing and the reading of pressure metre falls to $2.0 \times 10^4 \mathrm{~N} / \mathrm{m}^2$. The velocity of water is found to be $\sqrt{V} \mathrm{~m} / \mathrm{s}$. The value of $V$ is _________.
Explanation:
$\begin{aligned} & \text { Change in pressure }=\frac{1}{2} \rho \mathrm{v}^2 \\ & 4.5 \times 10^4-2.0 \times 10^4=\frac{1}{2} \times 10^3 \times \mathrm{v}^2 \\ & 2.5 \times 10^4=\frac{1}{2} \times 10^3 \times \mathrm{v}^2 \\ & \mathrm{v}^2=50 \\ & \mathrm{v}=\sqrt{50} \\ & \text { Velocity of water }=\sqrt{\mathrm{V}}=\sqrt{50} \\ & =\mathrm{V}=50 \end{aligned}$
If average depth of an ocean is $4000 \mathrm{~m}$ and the bulk modulus of water is $2 \times 10^9 \mathrm{~Nm}^{-2}$, then fractional compression $\frac{\Delta V}{V}$ of water at the bottom of ocean is $\alpha \times 10^{-2}$. The value of $\alpha$ is _______ (Given, $\mathrm{g}=10 \mathrm{~ms}^{-2}, \rho=1000 \mathrm{~kg} \mathrm{~m}^{-3}$)
Explanation:
$\begin{aligned} & \mathrm{B}=-\frac{\Delta \mathrm{P}}{\left(\frac{\Delta \mathrm{V}}{\mathrm{V}}\right)} \\ & -\left(\frac{\Delta \mathrm{V}}{\mathrm{V}}\right)=\frac{\rho \mathrm{gh}}{\mathrm{B}}=\frac{1000 \times 10 \times 4000}{2 \times 10^9} \\ & =2 \times 10^{-2}[-\mathrm{ve} \text { sign represent compression }] \end{aligned}$
A table tennis ball has radius $(3 / 2) \times 10^{-2} \mathrm{~m}$ and mass $(22 / 7) \times 10^{-3} \mathrm{~kg}$. It is slowly pushed down into a swimming pool to a depth of $d=0.7 \mathrm{~m}$ below the water surface and then released from rest. It emerges from the water surface at speed $v$, without getting wet, and rises up to a height $H$. Which of the following option(s) is(are) correct?
[Given: $\pi=22 / 7, g=10 \mathrm{~m} \mathrm{~s}^{-2}$, density of water $=1 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}$, viscosity of water $=1 \times 10^{-3} \mathrm{~Pa}$-s.]
Explanation:
$\begin{aligned} & A v=\mathrm{av}_1 \\ & A\left(-\frac{d y}{d t}\right)=a \sqrt{2 g y} ; d t=\frac{A}{a \sqrt{2 g}} \cdot \frac{-d y}{\sqrt{y}} \\ & \int_\limits 0^{t_1} d t=\frac{A}{a \sqrt{2 g}} \int_\limits h^0-\frac{d y}{\sqrt{y}} \\ & t_1=\frac{A}{a \sqrt{2 g}} 2 \sqrt{h} ; t_1=\frac{A}{a} \sqrt{\frac{2 h}{g}} \end{aligned}$
$A v^{\prime}=a v_2$
$ \begin{aligned} & A\left(-\frac{d y}{d t}\right)=a \sqrt{2 g(H+y)} \\ & d t=-\frac{A}{a \sqrt{2 g}} \frac{d y}{\sqrt{H+y}} \\ & \int_\limits0^{t_2} d t=-\frac{A}{a \sqrt{2 g}} \int_\limits{\mathrm{H}}^0 \frac{d y}{\sqrt{H+y}} \\ & t_2=\frac{A}{a \sqrt{2 g}}(2)(\sqrt{H+h}-\sqrt{H}) \& H=\frac{16 h}{9} \\ & =\frac{A}{a} \sqrt{\frac{2 h}{g}}\left(\frac{5}{3}-\frac{4}{3}\right) \\ & t_2=\frac{A}{a} \sqrt{\frac{2 h}{g}}\left(\frac{1}{3}\right) \\ & \text { ratio } \frac{t_1}{t_2}=3 \end{aligned}$
Given below are two statements: one is labelled as Assertion $\mathbf{A}$ and the other is labelled as Reason $\mathbf{R}$
Assertion A : A spherical body of radius $(5 \pm 0.1) \mathrm{mm}$ having a particular density is falling through a liquid of constant density. The percentage error in the calculation of its terminal velocity is $4 \%$.
Reason R : The terminal velocity of the spherical body falling through the liquid is inversely proportional to its radius.
In the light of the above statements, choose the correct answer from the options given below
The figure shows a liquid of given density flowing steadily in horizontal tube of varying cross - section. Cross sectional areas at $\mathrm{A}$ is $1.5 \mathrm{~cm}^{2}$, and $\mathrm{B}$ is $25 \mathrm{~mm}^{2}$, if the speed of liquid at $\mathrm{B}$ is $60 \mathrm{~cm} / \mathrm{s}$ then $\left(\mathrm{P}_{\mathrm{A}}-\mathrm{P}_{\mathrm{B}}\right)$ is :
(Given $\mathrm{P}_{\mathrm{A}}$ and $\mathrm{P}_{\mathrm{B}}$ are liquid pressures at $\mathrm{A}$ and $\mathrm{B}$% points.
density $\rho=1000 \mathrm{~kg} \mathrm{~m}^{-3}$
$\mathrm{A}$ and $\mathrm{B}$ are on the axis of tube
Under isothermal condition, the pressure of a gas is given by $\mathrm{P}=a \mathrm{~V}^{-3}$, where $a$ is a constant and $\mathrm{V}$ is the volume of the gas. The bulk modulus at constant temperature is equal to
A body cools from $80^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$ in 5 minutes. The temperature of the surrounding is $20^{\circ} \mathrm{C}$. The time it takes to cool from $60^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$ is :
Eight equal drops of water are falling through air with a steady speed of $10 \mathrm{~cm} / \mathrm{s}$. If the drops coalesce, the new velocity is:-
Young's moduli of the material of wires A and B are in the ratio of $1: 4$, while its area of cross sections are in the ratio of $1: 3$. If the same amount of load is applied to both the wires, the amount of elongation produced in the wires $\mathrm{A}$ and $\mathrm{B}$ will be in the ratio of
[Assume length of wires A and B are same]
Given below are two statements:
Statement I : Pressure in a reservoir of water is same at all points at the same level of water.
Statement II : The pressure applied to enclosed water is transmitted in all directions equally.
In the light of the above statements, choose the correct answer from the options given below:
A hydraulic automobile lift is designed to lift vehicles of mass $5000 \mathrm{~kg}$. The area of cross section of the cylinder carrying the load is $250 \mathrm{~cm}^{2}$. The maximum pressure the smaller piston would have to bear is $\left[\right.$ Assume $\left.g=10 \mathrm{~m} / \mathrm{s}^{2}\right]$
An air bubble of volume $1 \mathrm{~cm}^{3}$ rises from the bottom of a lake $40 \mathrm{~m}$ deep to the surface at a temperature of $12^{\circ} \mathrm{C}$. The atmospheric pressure is $1 \times 10^{5} \mathrm{~Pa}$ the density of water is $1000 \mathrm{~kg} / \mathrm{m}^{3}$ and $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$. There is no difference of the temperature of water at the depth of $40 \mathrm{~m}$ and on the surface. The volume of air bubble when it reaches the surface will be:
An aluminium rod with Young's modulus $Y=7.0 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}$ undergoes elastic strain of $0.04 \%$. The energy per unit volume stored in the rod in SI unit is:
Given below are two statements: one is labelled as Assertion $\mathbf{A}$ and the other is labelled as Reason $\mathbf{R}$
Assertion A: When you squeeze one end of a tube to get toothpaste out from the other end, Pascal's principle is observed.
Reason R: A change in the pressure applied to an enclosed incompressible fluid is transmitted undiminished to every portion of the fluid and to the walls of its container.
In the light of the above statements, choose the most appropriate answer from the options given below
A small ball of mass $\mathrm{M}$ and density $\rho$ is dropped in a viscous liquid of density $\rho_{0}$. After some time, the ball falls with a constant velocity. What is the viscous force on the ball ?
The Young's modulus of a steel wire of length $6 \mathrm{~m}$ and cross-sectional area $3 \mathrm{~mm}^{2}$, is $2 \times 10^{11}~\mathrm{N} / \mathrm{m}^{2}$. The wire is suspended from its support on a given planet. A block of mass $4 \mathrm{~kg}$ is attached to the free end of the wire. The acceleration due to gravity on the planet is $\frac{1}{4}$ of its value on the earth. The elongation of wire is (Take $g$ on the earth $=10 \mathrm{~m} / \mathrm{s}^{2}$) :
A mercury drop of radius $10^{-3}~\mathrm{m}$ is broken into 125 equal size droplets. Surface tension of mercury is $0.45~\mathrm{Nm}^{-1}$. The gain in surface energy is :
If 1000 droplets of water of surface tension $0.07 \mathrm{~N} / \mathrm{m}$, having same radius $1 \mathrm{~mm}$ each, combine to from a single drop. In the process the released surface energy is :-
$\left( {\mathrm{Take}\,\pi = {{22} \over 7}} \right)$
The height of liquid column raised in a capillary tube of certain radius when dipped in liquid A vertically is, $5 \mathrm{~cm}$. If the tube is dipped in a similar manner in another liquid $\mathrm{B}$ of surface tension and density double the values of liquid $\mathrm{A}$, the height of liquid column raised in liquid $\mathrm{B}$ would be __________ m.
Choose the correct relationship between Poisson ratio $(\sigma)$, bulk modulus (K) and modulus of rigidity $(\eta)$ of a given solid object :
A fully loaded boeing aircraft has a mass of $5.4\times10^5$ kg. Its total wing area is 500 m$^2$. It is in level flight with a speed of 1080 km/h. If the density of air $\rho$ is 1.2 kg m$^{-3}$, the fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface in percentage will be. ($\mathrm{g=10~m/s^2}$)
Surface tension of a soap bubble is $2.0 \times 10^{-2} \mathrm{Nm}^{-1}$. Work done to increase the radius of soap bubble from $3.5 \mathrm{~cm}$ to $7 \mathrm{~cm}$ will be:
Take $\left[\pi=\frac{22}{7}\right]$
A bicycle tyre is filled with air having pressure of $270 ~\mathrm{kPa}$ at $27^{\circ} \mathrm{C}$. The approximate pressure of the air in the tyre when the temperature increases to $36^{\circ} \mathrm{C}$ is
Match List I with List II
| List I | List II | ||
|---|---|---|---|
| A. | Surface tension | I. | $\mathrm{kg~m^{-1}~s^{-1}}$ |
| B. | Pressure | II. | $\mathrm{kg~ms^{-1}}$ |
| C. | Viscosity | III. | $\mathrm{kg~m^{-1}~s^{-2}}$ |
| D. | Impulse | IV. | $\mathrm{kg~s^{-2}}$ |
Choose the correct answer from the options given below :
A bowl filled with very hot soup cools from 98$^\circ$C to 86$^\circ$C in 2 minutes when the room temperature is 22$^\circ$C. How long it will take to cool from 75$^\circ$C to 69$^\circ$C?
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R
Assertion A : Steel is used in the construction of buildings and bridges.
Reason R : Steel is more elastic and its elastic limit is high.
In the light of above statements, choose the most appropriate answer from the options given below
The frequency ($\nu$) of an oscillating liquid drop may depend upon radius ($r$) of the drop, density ($\rho$) of liquid and the surface tension (s) of the liquid as $\nu=r^a\rho^b s^c$. The values of a, b and c respectively are
A 100 m long wire having cross-sectional area $\mathrm{6.25\times10^{-4}~m^2}$ and Young's modulus is $\mathrm{10^{10}~Nm^{-2}}$ is subjected to a load of 250 N, then the elongation in the wire will be :
Explanation:
$\Delta P = 2 \frac{T}{R}$
where $\Delta P$ is the pressure difference, $T$ is the surface tension, and $R$ is the radius of the bubble.
Plugging in the given values:
$\Delta P = 2 \frac{0.075 \mathrm{~Nm}^{-1}}{1.0 \mathrm{~mm}} = 2 \frac{0.075 \mathrm{~Nm}^{-1}}{10^{-3} \mathrm{~m}} = 150 \mathrm{~Pa}$
Now, we need to account for the hydrostatic pressure due to the depth of the bubble below the free surface:
$P_{hydrostatic} = \rho g h$
where $\rho$ is the density of the liquid, $g$ is the acceleration due to gravity, and $h$ is the depth below the free surface.
Plugging in the given values:
$P_{hydrostatic} = 1000 \mathrm{~kg} \mathrm{~m}^{-3} \cdot 10 \mathrm{~ms}^{-2} \cdot 0.1 \mathrm{~m} = 1000 \mathrm{~Pa}$
So, the total pressure difference inside the bubble compared to atmospheric pressure is the sum of the pressure difference due to surface tension and hydrostatic pressure:
$\Delta P_{total} = \Delta P + P_{hydrostatic} = 150 \mathrm{~Pa} + 1000 \mathrm{~Pa} = 1150 \mathrm{~Pa}$
The elastic potential energy stored in a steel wire of length $20 \mathrm{~m}$ stretched through $2 \mathrm{~cm}$ is $80 \mathrm{~J}$. The cross sectional area of the wire is __________ $\mathrm{mm}^{2}$.
$\left(\right.$ Given, $\left.y=2.0 \times 10^{11} \mathrm{Nm}^{-2}\right)$
Explanation:
The stress can be given as $\text{stress} = Y \times \text{strain}$, where Y is the Young's modulus.
The energy stored in the wire can be written as:
$\text{Energy} = \frac{1}{2} \times \text{stress} \times \text{strain} \times \text{volume}$
Substituting the stress formula, we get:
$\text{Energy} = \frac{1}{2} \times Y \times \text{strain}^2 \times A \times L$
We are given that the energy stored is $80 \ \text{J}$, the original length of the wire is $20 \ \text{m}$, the elongation is $2 \ \text{cm}$, and the Young's modulus is $2.0 \times 10^{11} \ \text{Nm}^{-2}$. We need to find the cross-sectional area (A) of the wire.
$80 = \frac{1}{2} \times 2 \times 10^{11} \times \left(\frac{2 \times 10^{-2}}{20}\right)^2 \times A \times 20$
Now we can solve for A:
$A = \frac{80 \times 20^2}{(2.0 \times 10^{11}) \times (2 \times 10^{-2})^2} = 40 \times 10^{-6} \ \text{m}^2$
To convert the area to $\text{mm}^2$, we multiply by $10^6$:
$A = 40 \times 10^{-6} \times 10^6 = 40 \ \text{mm}^2$
So, the cross-sectional area of the wire is $40 \ \text{mm}^2$.
Glycerin of density $1.25 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$ is flowing through the conical section of pipe The area of cross-section of the pipe at its ends are $10 \mathrm{~cm}^{2}$ and $5 \mathrm{~cm}^{2}$ and pressure drop across its length is $3 ~\mathrm{Nm}^{-2}$. The rate of flow of glycerin through the pipe is $x \times 10^{-5} \mathrm{~m}^{3} \mathrm{~s}^{-1}$. The value of $x$ is ___________.
Explanation:
We can use the Bernoulli equation and continuity equation to solve this problem. The Bernoulli equation is given by:
$P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2$
The continuity equation is given by:
$A_1 v_1 = A_2 v_2$
From the given data, we have:
$P_1 - P_2 = 3 \mathrm{~Nm}^{-2}$ $A_1 = 10 \mathrm{~cm}^2 = 10 \times 10^{-4} \mathrm{~m}^2$ $A_2 = 5 \mathrm{~cm}^2 = 5 \times 10^{-4} \mathrm{~m}^2$ $\rho = 1.25 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}$
Rearrange the continuity equation to solve for $v_2$:
$v_2 = \frac{A_1}{A_2} v_1 = 2v_1$
Substitute $v_2$ and rearrange the Bernoulli equation:
$P_1 - P_2 = \frac{1}{2} \rho (v_2^2 - v_1^2)$
$3 = \frac{1}{2} \times 1.25 \times 10^3 (4v_1^2 - v_1^2)$
Now, solve for $v_1$:
$3 = \frac{1}{2} \times 1.25 \times 10^3 \times 3v_1^2$
$v_1^2 = \frac{3}{1.875 \times 10^3}$
$v_1 = \sqrt{\frac{3}{1.875 \times 10^3}}$
$v_1 \approx 0.0400 \mathrm{~m} \mathrm{~s}^{-1}$
Now, calculate the rate of flow of glycerin through the pipe (volume flow rate) using $v_1$ and $A_1$:
$Q = A_1 v_1$
$Q = 10 \times 10^{-4} \times 0.0400$
$Q = 4 \times 10^{-5} \mathrm{~m}^{3} \mathrm{~s}^{-1}$
So, the rate of flow of glycerin through the pipe is $4 \times 10^{-5} \mathrm{~m}^{3} \mathrm{~s}^{-1}$, and the value of $x$ is 4.
The surface tension of soap solution is $3.5 \times 10^{-2} \mathrm{~Nm}^{-1}$. The amount of work done required to increase the radius of soap bubble from $10 \mathrm{~cm}$ to $20 \mathrm{~cm}$ is _________ $\times ~10^{-4} \mathrm{~J}$.
$(\operatorname{take} \pi=22 / 7)$
Explanation:
$ W = T \Delta A $
where W is the work done, T is the surface tension, and ΔA is the change in surface area.
For a soap bubble, we need to consider both the inner and outer surfaces, so the surface area is doubled. The surface area of a sphere is:
$ A = 4\pi r^2 $
The initial surface area of the soap bubble is:
$ A_1 = 2 \cdot 4\pi (0.1\,\mathrm{m})^2 = 8\pi (0.1\,\mathrm{m})^2 $
The final surface area of the soap bubble is:
$ A_2 = 2 \cdot 4\pi (0.2\,\mathrm{m})^2 = 8\pi (0.2\,\mathrm{m})^2 $
The change in surface area is:
$ \Delta A = A_2 - A_1 = 8\pi(0.2\,\mathrm{m})^2 - 8\pi(0.1\,\mathrm{m})^2 $
Now, we can calculate the work done:
$ W = T \Delta A = (3.5 \times 10^{-2}\,\mathrm{Nm}^{-1})[8\pi(0.2\,\mathrm{m})^2 - 8\pi(0.1\,\mathrm{m})^2] $
Using the given value of π:
$ W = (3.5 \times 10^{-2}\,\mathrm{Nm}^{-1})[8(22/7)(0.2\,\mathrm{m})^2 - 8(22/7)(0.1\,\mathrm{m})^2] $
$ W = (3.5 \times 10^{-2}\,\mathrm{Nm}^{-1})[8(22/7)(0.04\,\mathrm{m^2}) - 8(22/7)(0.01\,\mathrm{m^2})] $
$ W = (3.5 \times 10^{-2}\,\mathrm{Nm}^{-1})[8(22/7)(0.03\,\mathrm{m^2})] $
$ \begin{aligned} & W=2 \times 1.32 \times 10^{-2} \\\\ &W =2 \times 132 \times 10^{-4} \mathrm{~J} \\\\ & W=264 \times 10^{-4} \mathrm{~J} \end{aligned} $
The work done to increase the radius of the soap bubble is 264 × 10⁻⁴ J.
A wire of density $8 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$ is stretched between two clamps $0.5 \mathrm{~m}$ apart. The extension developed in the wire is $3.2 \times 10^{-4} \mathrm{~m}$. If $Y=8 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}$, the fundamental frequency of vibration in the wire will be ___________ $\mathrm{Hz}$.
Explanation:
To determine the fundamental frequency of the vibrating wire, we need to first find the tension (T) in the wire and the wave velocity (V) in the wire.
- Tension in the wire (T): We used Young's modulus (Y) to relate the stress and strain in the wire. The formula for stress is:
$\text{stress} = Y \times \text{strain}$
Here, the strain is the extension ($\Delta L$) divided by the original length (L):
$\text{strain} = \frac{\Delta L}{L}$
Now, the tension (T) in the wire is the product of stress and cross-sectional area (A):
$T = \text{stress} \times A$
Combining the above equations, we get the expression for tension:
$T = \frac{Y \Delta L}{L} \times A$
- Wave velocity in the wire (V): The linear mass density ($\mu$) of the wire is given by:
$\mu = \frac{m}{L}$
We need to find the ratio $\frac{T}{\mu}$, which represents the square of the wave velocity. Using the expressions for tension and linear mass density, we get:
$\frac{T}{\mu} = \frac{Y \Delta L}{L} \times \frac{A}{m} = \frac{Y \Delta L}{L} \times \frac{1}{\rho}$
Here, $\rho$ is the density of the wire material. Plugging in the given values, we find the value of $\frac{T}{\mu}$, which is:
$\frac{T}{\mu} = 6.4 \times 10^3$
Now, we find the wave velocity (V) by taking the square root of $\frac{T}{\mu}$:
$V = \sqrt{T/\mu} = 80 \mathrm{~m/s}$
- Fundamental frequency (f): Finally, we find the fundamental frequency of the vibrating wire using the formula:
$f = \frac{V}{2L}$
Plugging in the values, we get the fundamental frequency (f) as:
$f = 80 \mathrm{~Hz}$
So, the fundamental frequency of vibration in the wire is 80 Hz.
The length of a wire becomes $l_{1}$ and $l_{2}$ when $100 \mathrm{~N}$ and $120 \mathrm{~N}$ tensions are applied respectively. If $10 ~l_{2}=11~ l_{1}$, the natural length of wire will be $\frac{1}{x} ~l_{1}$. Here the value of $x$ is _____________.
Explanation:
Given:
- When tension $T_1 = 100 \mathrm{~N}$, extension $= l_1 - l_0$.
- When tension $T_2 = 120 \mathrm{~N}$, extension $= l_2 - l_0$.
Now, let's write the equations using Hooke's Law:
$100 = k(l_1 - l_0)$
$120 = k(l_2 - l_0)$
Divide the first equation by the second equation:
$\frac{5}{6} = \frac{l_1 - l_0}{l_2 - l_0}$
Given the relationship between $l_1$ and $l_2$:
$10l_2 = 11l_1$
Now, let's solve for $l_0$:
$5l_2 - 5l_0 = 6l_1 - 6l_0$
$l_0 = 6l_1 - 5l_2$
Substitute the relationship between $l_1$ and $l_2$:
$l_0 = 6l_1 - 5\left(\frac{11l_1}{10}\right)$
$l_0 = 6l_1 - \frac{11l_1}{2}$
$l_0 = \frac{l_1}{2}$
Therefore, the natural length of the wire is $\frac{1}{x}l_1 = \frac{2}{1}l_1 = 2l_1$.
The value of $x$ is $2$.
Figure below shows a liquid being pushed out of the tube by a piston having area of cross section $2.0 \mathrm{~cm}^{2}$. The area of cross section at the outlet is $10 \mathrm{~mm}^{2}$. If the piston is pushed at a speed of $4 \mathrm{~cm} \mathrm{~s}^{-1}$, the speed of outgoing fluid is __________ $\mathrm{cm} \mathrm{s}^{-1}$
Explanation:
$ \begin{aligned} & \mathrm{A}_1 \mathrm{~V}_1=\mathrm{A}_2 \mathrm{~V}_2 \\\\ & \mathrm{~V}_2=\frac{2 \times 4}{10 \times 10^{-2}}=80 \mathrm{~cm} / \mathrm{s} \end{aligned} $
Two wires each of radius 0.2 cm and negligible mass, one made of steel and the other made of brass are loaded as shown in the figure. The elongation of the steel wire is __________ $\times$ 10$^{-6}$ m. [Young's modulus for steel = 2 $\times$ 10$^{11}$ Nm$^{-2}$ and g = 10 ms$^{-2}$ ]
Explanation:
$\begin{aligned} & r_1 =r_2=0.2 \mathrm{~cm}=2 \times 10^{-3} \mathrm{~m} \\\\ & y_1 \text { (steel) } =2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2} \\\\ & l_1 =1.6 \mathrm{~m}\end{aligned}$
At equilibrium,
$ \begin{aligned} &T_2 =1.14 \mathrm{~g}=11.4 \mathrm{~N} \\\\ &T_2+2 g =T_1 \\\\ &\therefore T_1 =11.4+20=31.4 \mathrm{~N} \end{aligned} $
$ \begin{aligned} & \Delta l_1=\frac{T_1 l_1}{Y_1 A_1} \\\\ = & \frac{31.4 \times 1.6}{2 \times 10^{11} \times \pi\left(2 \times 10^{-3}\right)^2} 1.1 \\\\ = & \frac{16}{8} \times 10^{-5} \\\\ = & 2 \times 10^{-5} \mathrm{~m}=20 \times 10^{-6} \mathrm{~m} \end{aligned} $