Hc Verma I for Class 12 Science Physics Chapter 14 - Some Mechanical Properties Of Matter

Answer:

The ratio of stress to strain will decrease. 

Beyond the elastic limit, the body loses its ability to restore completely when subjected to stress. Thus, there occurs more strain for a given stress. At some points, however, the body undergoes strain without the application of stress. So, the ratio of stress to strain decreases. 

Answer:

Let the CSA of the wire be A.

$$\begin{gathered} \mathrm{Stress} = \frac{\mathrm{Force} }{\mathrm{Area}}=\frac{Mg}{A} \\ \mathrm{Strain} = \frac{l}{L} \\ \mathrm{Volume} = AL \\ \mathrm{We} \mathrm{need} \mathrm{to} \mathrm{calculate} \mathrm{the} \mathrm{elastic} \mathrm{potential} \mathrm{energy} \mathrm{stored} \mathrm{in} \mathrm{the} \mathrm{wire} \mathrm{which} \mathrm{is} \mathrm{given} \mathrm{to} \mathrm{be} \mathrm{equal} \mathrm{to} \frac{1}{2}\times \mathrm{Stress}\times \mathrm{Strain}\times \mathrm{Volume}. \\ \mathrm{Elastic} \mathrm{potential} \mathrm{energy} = \frac{1}{2}\times \mathrm{Stress}\times \mathrm{Strain}\times \mathrm{Volume} \\ =\frac{1}{2}\times \frac{Mg}{A}\times \frac{l}{L}\times AL \\ =\frac{1}{2}Mgl \end{gathered}$$

The other $\frac{1}{2}Mgl$ is converted into kinetic energy of the mass. 

When the mass leaves its initial point on the spring, it acquires a velocity as it moves down. The velocity reaches its maximum at the end point. The spring oscillates. Finally, when the kinetic energy is dissipated into heat, the spring comes to rest. 

Answer:

The elephant has a greater weight than a mouse, but the material that makes their bones is the same. This means that in order to sustain an elephant's weight, one's bones need to suffer less stress. Stress = Force/area. A greater cross-sectional area reduces stress on the bones. This is why an elephant's bones are thicker.     

Answer:

Let my length = L
Let the increase in length = l
Strain
 $$\begin{gathered} =\frac{l}{L}=\frac{1}{100} \\ \Rightarrow l=\frac{L}{100} \\ \mathrm{So}, \mathrm{the} \mathrm{increase} \mathrm{in} \mathrm{length} \mathrm{will} \mathrm{be} \frac{L}{100}. \end{gathered}$$

Yes, the yield point is much less than the 1% strain because the human body consists of joints and not one uniform solid structure.

Answer:

The energy of vibration dissipates as heat from the shock absorber.  

Answer:

When a compressed spring dissolves in an acid, the acid molecules leave the sold lattice of the spring faster than the uncompressed spring. This in turn increases the kinetic energy of the solution. As a result, the temperature of the acid also increases. However, this temperature increase will be very small because the mechanical energy content in the spring is lesser than its chemical energy content.    

Answer:

It floats because of the surface tension of water. The surface of water behaves like a stretched membrane. When a blade is placed on the water surface, it's unable to pierce the stretched membrane of water due to its low weight and remains floating. 

However, if the blade is placed below the surface of water, it no longer experiences the surface tension and sinks to the bottom as the density of the blade is greater than that of water. 

Answer:

A liquid wets a surface when the angle of contact of the liquid with the surface is small or zero. Due to its fibrous nature, cloth produces capillary action when in contact with water. This makes clothes have very small contact angles with water. When wax is rubbed over cloth, the water does not wet the cloth because wax has a high contact angle with water.  

Answer:

No, the water will neither rise nor fall in the silver capillary.

$$\begin{gathered} \mathrm{According} \mathrm{to} \mathrm{Jurin}\text{'}\mathrm{s} \mathrm{law}, \mathrm{the} \mathrm{level} \mathrm{of} \mathrm{water} \mathrm{inside} \mathrm{a} \mathrm{capillary} \mathrm{tube} \mathrm{is} \mathrm{given} \mathrm{by} \\ \mathrm{h}=\frac{2\mathrm{Tcos\theta }}{\mathrm{r\rho g}} \\ \mathrm{Here}, \mathrm{\theta } = {90}^0 \\ \Rightarrow \mathrm{h} =\frac{2\mathrm{Tcos}{90}^0}{\mathrm{r\rho g}} \\ \Rightarrow \mathrm{h}=0 \end{gathered}$$

Thus, the water level neither rises nor falls.

Answer:

No, we cannot conclude the surface tension to be zero solely by the fact that the liquid neither rises nor falls in a capillary.

The height of the liquid inside a capillary tube is given by $h=\frac{2T\cos \theta }{r\rho g}$. From the equation, we see that the height (h) of the liquid may also be zero if the contact angle $\theta$ between the liquid and the capillary tube is ${90}^0 \mathrm{or} {270}^0$. 

Answer:

When water is poured in a glass, it reaches the brim and rises further. The edge of the glass lies below the water level. In this case, the force of attraction due to molecules of the glass surface is not perpendicular to the solid. Here, the contact angle can be greater than the standard contact angle for a pair of substances.  

Answer:

As the angle of contact is 0, there is no force between the surface of the tube and the liquid. The diameter of the liquid surface is pulled on both sides by equal and opposite forces of surface tension. This results in no net force remaining on the surface of the liquid. Hence, the liquid stays in equilibrium. 

Answer:

No, it does not violate the principle of conservation of energy.
 
There is a force of attraction between glass and water, which is why the liquid rises in the tube. However, when water and glass are not in contact, there exists a potential energy in the system. When they are brought into contact, this potential energy is first converted into kinetic energy, which lets the liquid rush upwards in the tube, and then into gravitational potential energy. Therefore,  energy is not created in the process.

Answer:

A mosquito thrown into water has its wings wetted. Now, wet wing surfaces tend to stick together because of the surface tension of water. This does not let the mosquito fly. 

Answer:

The forces act tangentially to the bubble surface on both sides of a given line but they have one component normal to the bubble surface. This component balances the force due to excess pressure inside the bubble. 

In the figure, let us consider a small length AB on the surface of the spherical bubble. Let the surface forces act tangentially along A and B. On producing the forces backwards, they meet at a point O. By the parallelogram law of forces, we see that the resultant force acts opposite to the normal. This balances the internal forces due to excess pressure. 

Answer:

No. The average intermolecular distances do not increase with an increase in the surface area.

A soap bubble's layer consists of several thousand layers of molecules. An increase in the surface area causes the surface energy to also increase. This in turn allows more and more molecules from the inner liquid layers of the bubble to attain potential energy, enabling them to enter the outer surface of the bubble. Hence, the surface area increases.

Answer:

No. For a liquid at rest, no viscous forces exist.

Viscous forces oppose relative motion between the layers of a liquid. These layers do not exist in a liquid that is at rest. Therefore, it is obvious that viscous forces are non-existent in a static liquid. 

Answer:

The motion of any liquid is dependent upon the amount of stress acting on it. The motion of one layer of liquid is resisted by the other due to the property of viscosity. A river bed remains in a static state. Therefore, any immediate layer of liquid in contact with the river bed will also remain static due to the frictional force. However, the next layer of liquid above this static layer will have a greater velocity due to lesser resistance offered by the static layer. Moving upwards, subsequent layers provide lesser and lesser resistance to the movement of the layers above it. Finally, the topmost layer acquires the maximum velocity. Therefore, for a river, the surface waters flow the fastest. 

Answer:

Castor oil will come to rest more quickly because it has a greater coefficient of viscosity than water.

Castor oil has a higher viscosity than water. It will therefore, lose kinetic energy and come to rest faster than water. 

Answer:

Correct option: (d) 2000 N

$$\begin{gathered} {\mathrm{F}}_1=500\mathrm{N} \\ \mathrm{Let} \mathrm{the} \mathrm{required} \mathrm{breaking} \mathrm{force} \mathrm{on} \mathrm{the} 2 \mathrm{cm} \mathrm{wire} \mathrm{be} \mathrm{F}. \\ \mathrm{Breaking} \mathrm{stress} \mathrm{in} 1 \mathrm{cm} \mathrm{wire} = \frac{{\mathrm{F}}_1}{{\mathrm{A}}_1}=\frac{500}{\mathrm{\pi }{\left({\displaystyle \frac{0.01}{2}}\right)}^2} \\ \mathrm{Breaking} \mathrm{stress} \mathrm{in} 2 \mathrm{cm} \mathrm{wire} = \frac{{\mathrm{F}}_2}{{\mathrm{A}}_2}=\frac{{\mathrm{F}}_2}{\mathrm{\pi }{\left({\displaystyle \frac{0.02}{2}}\right)}^2} \\ \mathrm{The} \mathrm{breaking} \mathrm{stress} \mathrm{is} \mathrm{the} \mathrm{same} \mathrm{for} \mathrm{a} \mathrm{material}. \\ \Rightarrow \frac{500}{\mathrm{\pi }{\left({\displaystyle \frac{0.01}{2}}\right)}^2}=\frac{{\mathrm{F}}_2}{\mathrm{\pi }{\left({\displaystyle \frac{0.02}{2}}\right)}^2} \\ =>{\mathrm{F}}_2=2000\mathrm{N} \end{gathered}$$

Answer:

Correct option: (a)

Breaking stress depends upon the intermolecular/ inter-atomic forces of attraction within materials. In other words, it depends upon the material of the wire. 

Answer:

Correct option: (b) 20 kg

As the wire is cut into two equal parts, both have equal cross-sectional areas. Therefore, a weight of 20 kg exerts a force of 20g on both the pieces. Breaking stress depends upon the material of the wire. Since 20g of force is exerted on wires with equal cross-sectional areas, both the wires can sustain a weight of 20 kg. 

Answer:

Correct option: (a)  1/8

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{Young}\text{'}\mathrm{s} \mathrm{modulus} \mathrm{of} \mathrm{the} \mathrm{wire}\text{'}\mathrm{s} \mathrm{material} \mathrm{be} \mathrm{Y}. \\ \mathrm{Here}: \\ \mathrm{Force} = \mathrm{F} \\ {\mathrm{A}}_1={\mathrm{\pi r}}^2 \\ {\mathrm{L}}_1=\mathrm{l} \\ {\mathrm{A}}_2=\mathrm{\pi }{\left(\frac{\mathrm{r}}{2}\right)}^2=\frac{{\mathrm{\pi r}}^2}{4} \\ {\mathrm{L}}_2=2\mathrm{l} \\ \mathrm{Let} \mathrm{the} \mathrm{elongation} \mathrm{in} \mathrm{A} \mathrm{be} \mathrm{x} \mathrm{and} \mathrm{that} \mathrm{in} \mathrm{B} \mathrm{be} \mathrm{y}. \\ \mathrm{Since} \mathrm{the} \mathrm{Young}\text{'}\mathrm{s} \mathrm{modulus} \mathrm{for} \mathrm{both} \mathrm{the} \mathrm{wires} \mathrm{is} \mathrm{the} \mathrm{same}: \\ \mathrm{Y} =\frac{ {\displaystyle \raisebox{1ex}{$\mathrm{F}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{A}}_1$}\right.}}{{\displaystyle \raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$l$}\right.}}=\frac{ {\displaystyle \raisebox{1ex}{$\mathrm{F}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{A}}_2$}\right.}}{{\displaystyle \raisebox{1ex}{$\mathrm{y}$}\!\left/ \!\raisebox{-1ex}{$2\mathrm{l}$}\right.}} \\ \Rightarrow \frac{\mathrm{x}}{\mathrm{y}}=\frac{{\mathrm{A}}_2}{2{\mathrm{A}}_1} \\ \Rightarrow \frac{\mathrm{x}}{\mathrm{y}}=\frac{1}{8} \end{gathered}$$

Answer:

Correct option: (b) 1.0 mm

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{Young}\text{'}\mathrm{s} \mathrm{modulus} \mathrm{of} \mathrm{the} \mathrm{material} \mathrm{of} \mathrm{the} \mathrm{wire} \mathrm{be} \mathrm{Y}. \\ \mathrm{Force} = \mathrm{Weight} = \mathrm{W} \left(\mathrm{given}\right) \\ \mathrm{Let} C.S.A.\hspace{0.17em}= \mathrm{A} \\ \mathrm{x} = 1 \mathrm{mm} = \mathrm{Elongation} \mathrm{in} \mathrm{the} \mathrm{first} \mathrm{case} \\ \mathrm{Length} = \mathrm{L} \\ \mathrm{Y} = \frac{{\displaystyle \raisebox{1ex}{$\mathrm{W}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{A}$}\right.}}{{\displaystyle \raisebox{1ex}{$\mathrm{x}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right.}}=\frac{\mathrm{WL}}{\mathrm{Ax}} \\ \mathrm{Let} \mathrm{y} \mathrm{be} \mathrm{the} \mathrm{elongation} \mathrm{on} \mathrm{one} \mathrm{side} \mathrm{of} \mathrm{the} \mathrm{wire} \mathrm{when} \mathrm{put} \mathrm{in} \mathrm{a} \mathrm{pulley}. \\ \mathrm{When} \mathrm{put} \mathrm{in} \mathrm{a} \mathrm{pulley}, \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{wire} \mathrm{on} \mathrm{each} \mathrm{side} = \frac{\mathrm{L}}{2} \\ \frac{{\displaystyle \raisebox{1ex}{$\mathrm{W}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{A}$}\right.}}{{\displaystyle \raisebox{1ex}{$\mathrm{y}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \frac{\mathrm{L}}{2}}$}\right.}} = \mathrm{Y} \\ \Rightarrow \frac{{\displaystyle \raisebox{1ex}{$\mathrm{W}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{A}$}\right.}}{{\displaystyle \raisebox{1ex}{$\mathrm{y}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \frac{\mathrm{L}}{2}}$}\right.}}=\frac{\mathrm{WL}}{\mathrm{Ax}} \\ \Rightarrow \mathrm{y} = \frac{\mathrm{x}}{2} \\ \mathrm{Total} \mathrm{elongation} \mathrm{in} \mathrm{the} \mathrm{wire} = 2\mathrm{y} =2\left(\frac{\mathrm{x}}{2}\right)= \mathrm{x} = 1\mathrm{mm} \end{gathered}$$

Answer:

Correct option: (a) s​mallest at the top and gradually increases down the rod

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. 

Answer:

Correct option: (a)  ∆ll

$$\begin{gathered} C.S.A.\hspace{0.17em}= \mathrm{A} \\ \mathrm{Length} = \mathrm{l} \\ \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{wire} \mathrm{V} = \mathrm{Al} \\ \mathrm{Assuming} \mathrm{no} \mathrm{lateral} \mathrm{strain} \mathrm{when} \mathrm{longitudinal} \mathrm{strain} \mathrm{occurs}: \\ \mathrm{Increase} \mathrm{in} \mathrm{volume}: ∆\mathrm{V}= \mathrm{A}∆\mathrm{l} \\ \Rightarrow \frac{ ∆\mathrm{V}}{ \mathrm{V}}=\frac{\mathrm{A}∆\mathrm{l}}{\mathrm{Al}}=\frac{∆\mathrm{l}}{\mathrm{l}} \\ \mathrm{So}, \frac{ ∆\mathrm{V}}{ \mathrm{V}} \mathrm{is} \mathrm{directly} \mathrm{proportional} \mathrm{to} \frac{∆\mathrm{l}}{\mathrm{l}}. \end{gathered}$$

Answer:

Correct option: (c)

)$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{Young}\text{'}\mathrm{s} \mathrm{modulus} \mathrm{be} \mathrm{Y}. \\ C.S.A.\hspace{0.17em}= \mathrm{A} \\ \mathrm{Actual} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{wire} = \mathrm{L} \\ \mathrm{For} \mathrm{tension} {\mathrm{T}}_1: \\ \mathrm{Y}=\frac{{\displaystyle \raisebox{1ex}{${\mathrm{T}}_1$}\!\left/ \!\raisebox{-1ex}{$\mathrm{A}$}\right.}}{{\displaystyle \raisebox{1ex}{$\left(\mathrm{L}-{\mathrm{l}}_1\right)$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right.}}...\left(1\right) \\ \mathrm{For} \mathrm{tension} {\mathrm{T}}_2: \\ \mathrm{Y}=\frac{{\displaystyle \raisebox{1ex}{${\mathrm{T}}_2$}\!\left/ \!\raisebox{-1ex}{$\mathrm{A}$}\right.}}{{\displaystyle \raisebox{1ex}{$\left(\mathrm{L}-{\mathrm{l}}_2\right)$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right.}}...\left(2\right) \\ \mathrm{From} \left(1\right) \mathrm{and} \left(2\right): \\ \frac{{\displaystyle \raisebox{1ex}{${\mathrm{T}}_1$}\!\left/ \!\raisebox{-1ex}{$\mathrm{A}$}\right.}}{{\displaystyle \raisebox{1ex}{$\left(\mathrm{L}-{\mathrm{l}}_1\right)$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right.}}=\frac{{\displaystyle \raisebox{1ex}{${\mathrm{T}}_2$}\!\left/ \!\raisebox{-1ex}{$\mathrm{A}$}\right.}}{{\displaystyle \raisebox{1ex}{$\left(\mathrm{L}-{\mathrm{l}}_2\right)$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right.}} \\ \Rightarrow \frac{{\mathrm{T}}_1}{\left(\mathrm{L}-{\mathrm{l}}_1\right)}=\frac{{\mathrm{T}}_2}{\left(\mathrm{L}-{\mathrm{l}}_2\right)} \\ \Rightarrow \mathrm{L}=\frac{{\mathrm{T}}_2{\mathrm{l}}_1-{\mathrm{T}}_1{\mathrm{l}}_2}{{\mathrm{T}}_2-{\mathrm{T}}_1} \end{gathered}$$

Answer:

Correct option: (b)

If the velocity of the mass is a maximum at the bottom, then the string experiences tension due to both the weight of the mass and the high centrifugal force. Both these factors weigh the mass downwards. The tension is therefore, maximum at the lowest point, causing the string to most likely break at the bottom. 

Answer:

Correct option: (d)

None of these is the correct option. The decreased gravitational potential energy transforms partly as elastic energy, partly as kinetic energy and also in the form of dissipated heat energy. 

Answer:

The correct option is (c).

The surface  of a liquid refers to the layer of molecules that have higher potential energy than the bulk of the liquid. This layer is typically 10 to 15 times the diameter of the molecule. Now, the size of an average molecule is around 1 nm = ${10}^{-9}$m, so a diameter of 10 to 15 times would be of order ​$10\times {10}^{-9}={10}^{-8}$ m.

Answer:

Correct option: (b)
As the ice cube melts completely, the water thus formed will have minimum surface area due to its surface tension. Any state of matter that has a minimum surface area to its volume takes the shape of a sphere. Therefore, as the ice melts, it will take the shape of a sphere. 

Answer:

Correct option: (a) 

As the water droplets merge to form a single droplet, the surface area decreases. With this decrease in surface area, the surface energy of the resulting drop also decreases. Therefore, extra energy must be liberated from the drop in accordance with the conservation of energy. 

Answer:

Correct option: (c)
​Dimension of modulus of elasticity: $\frac{{\displaystyle \raisebox{1ex}{$F$}\!\left/ \!\raisebox{-1ex}{$A$}\right.}}{{\displaystyle \raisebox{1ex}{$∆l$}\!\left/ \!\raisebox{-1ex}{$l$}\right.}}=\frac{\left[ML{T}^{-2}\right]}{L^2}=\left[M{L}^{-1}{T}^{-2}\right]$
Dimension of moment of force: ​$FL=\left[ML{T}^{-2}\right]\left[L\right]=\left[M{L}^2{T}^{-2}\right]$
Dimension of surface tension: ​$\frac{F}{{\displaystyle L}}=\frac{\left[ML{T}^{-2}\right]}{L}=\left[M{T}^{-2}\right]$
Dimension of coefficient of viscosity: ​$\frac{FL}{{\displaystyle Av}}=\frac{\left[ML{T}^{-2}\right]\left[L\right]}{\left[L^2\right]\left[L{T}^{-1}\right]}=\left[M{L}^{-1}{T}^{-1}\right]$

Answer:

Correct option: (d)

$$\begin{gathered} \mathrm{No}. \mathrm{of} \mathrm{surfaces} \mathrm{of} \mathrm{a} \mathrm{soap} \mathrm{bubble} = 2 \\ \mathrm{Increase} \mathrm{in} \mathrm{surface} \mathrm{area} = 4\mathrm{\pi }(2\mathrm{r}{)}^2-4\mathrm{\pi }(\mathrm{r}{)}^2=12{\mathrm{\pi r}}^2 \\ \mathrm{Total} \mathrm{increase} \mathrm{in} \mathrm{surface} \mathrm{area} = 2\times 12{\mathrm{\pi r}}^2=24{\mathrm{\pi r}}^2 \\ \mathrm{Work} \mathrm{done} = \mathrm{change} \mathrm{in} \mathrm{surface} \mathrm{energy} \\ =\mathrm{S}\times 24{\mathrm{\pi r}}^2 = 24{\mathrm{\pi r}}^2\mathrm{S} \end{gathered}$$

Answer:

Correct option: (a)

 Excess pressure inside a bubble is given by: $P=\frac{4T}{r}$.
When air is pushed into the bubble, it grows in size. Therefore, its radius increases. An increase in size causes the pressure inside the soap bubble to decrease as pressure is inversely proportional to the radius. 

Answer:

Correct option: (c)

The smaller bubble has a greater inner pressure than the bigger bubble. Air moves from a region of high pressure to a region of low pressure. Therefore, air moves from the smaller to the bigger bubble. 

Answer:

Correct option: (c)



$$\begin{gathered} \mathrm{Here}: \\ \mathrm{Radius} \mathrm{of} \mathrm{the} \mathrm{tube} = \mathrm{r} \\ \mathrm{Net} \mathrm{upward} \mathrm{force} \mathrm{due} \mathrm{to} \mathrm{surface} \mathrm{tension} = \mathrm{Scos\theta }\times 2\mathrm{\pi r} \\ \mathrm{Upward} \mathrm{pressure} = \frac{\mathrm{Scos\theta }\times 2\mathrm{\pi r}}{{\mathrm{\pi r}}^2}=\frac{2\mathrm{Scos\theta }}{\mathrm{r}} \\ \mathrm{Net} \mathrm{downward} \mathrm{pressure} \mathrm{due} \mathrm{to} \mathrm{atmosphere} = {\mathrm{P}}_{\mathrm{o}} \\ \Rightarrow \mathrm{Net} \mathrm{pressure} \mathrm{at} \mathrm{A} = {\mathrm{P}}_{\mathrm{o}} -\frac{2\mathrm{Scos\theta }}{\mathrm{r}} \\ \mathrm{Since} \mathrm{\theta } \mathrm{is} \mathrm{small}, \\ \mathrm{cos\theta }\approx 1. \\ \Rightarrow \mathrm{Net} \mathrm{pressure} = {\mathrm{P}}_{\mathrm{o}} -\frac{2\mathrm{S}}{\mathrm{r}} \end{gathered}$$

Answer:

Correct option: (d)
Let the excess pressure inside the second bubble be P.
​∴ Excess pressure inside the first bubble = 2P
Let the radius of the second bubble be R.
Let the radius of the first bubble be x.

$$\begin{gathered} \mathrm{Excess} \mathrm{pressure} \mathrm{inside} \mathrm{the} 2\mathrm{nd} \mathrm{soap} \mathrm{bubble}: \\ \mathrm{P} = \frac{4\mathrm{S}}{\mathrm{R}}...\left(1\right) \\ \mathrm{Excess} \mathrm{pressure} \mathrm{inside} \mathrm{the} 1\mathrm{st} \mathrm{soap} \mathrm{bubble}: \\ 2\mathrm{P} = \frac{4\mathrm{S}}{\mathrm{x}} \\ \mathrm{From} \left(1\right), \mathrm{we} \mathrm{get}: \\ 2\left(\frac{4\mathrm{S}}{\mathrm{R}}\right)=\frac{4\mathrm{S}}{\mathrm{x}} \\ \Rightarrow \mathrm{x} = \frac{\mathrm{R}}{2} \\ \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{first} \mathrm{bubble} = \frac{4}{3}{\mathrm{\pi x}}^3 \\ \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{second} \mathrm{bubble} = \frac{4}{3}{\mathrm{\pi R}}^3 \\ \Rightarrow \frac{4}{3}{\mathrm{\pi x}}^3=\mathrm{n}\frac{4}{3}{\mathrm{\pi R}}^3 \\ \Rightarrow {\mathrm{x}}^3={\mathrm{nR}}^3 \\ \Rightarrow {\left(\frac{\mathrm{R}}{2}\right)}^3={\mathrm{nR}}^3 \\ \Rightarrow \mathrm{n}=\frac{1}{8}=0.125 \end{gathered}$$

Answer:

Correct option: (c)

$$\begin{gathered} \mathrm{The} \mathrm{relationship} \mathrm{between} \mathrm{height} \mathrm{h} \mathrm{and} \mathrm{radius} \mathrm{r} \mathrm{is} \mathrm{given} \mathrm{by}: \\ \mathrm{h}=\frac{2\mathrm{Scos\theta }}{\mathrm{r\rho g}} \\ \mathrm{If} \mathrm{S}, \mathrm{\theta }, \mathrm{\rho } \mathrm{and} \mathrm{g} \mathrm{are} \mathrm{considered} \mathrm{constant}, \mathrm{we} \mathrm{have}: \\ \mathrm{h}\propto \frac{1}{\mathrm{r}} \end{gathered}$$

This equation has the characteristic of a rectangular hyperbola. Therefore, curve (c) is a rectangular hyperbola. 

Answer:

Correct option: (b)


$$\begin{gathered} \mathrm{Given}: \\ \mathrm{l} = 10 \mathrm{cm} \\ \mathrm{\alpha }={45}^0 \\ \mathrm{Rise} \mathrm{in} \mathrm{water} \mathrm{level} \mathrm{after} \mathrm{the} \mathrm{tube} \mathrm{is} \mathrm{tilted} = h \\ \Rightarrow \mathrm{l} =\mathit{ }h\cos {45}^0 \\ \Rightarrow h = \frac{\mathrm{l}}{\cos {45}^0}=\frac{10}{\left({\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}\right)}=10\sqrt{2 }\mathrm{cm} \end{gathered}$$

Answer:

Correct option: (d)

$$\begin{gathered} \mathrm{Height} \mathrm{of} \mathrm{water} \mathrm{column} \mathrm{in} \mathrm{capillary} \mathrm{tube} \mathrm{is} \mathrm{given} \mathrm{by}: \\ \mathrm{h}= \frac{2\mathrm{Tcos\theta } }{\mathrm{r\rho g}} \\ \mathrm{A} \mathrm{free} \mathrm{falling} \mathrm{elevator} \mathrm{experiences} \mathrm{zero} \mathrm{gravity}. \\ \Rightarrow \mathrm{h} = \frac{2\mathrm{Tcos\theta } }{\mathrm{r\rho }0}=\infty \\ \mathrm{But}, \mathrm{h} = 20 \mathrm{cm} \left(given\right) \\ \mathrm{Therefore}, \mathrm{the} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{water} \mathrm{column} \mathrm{will} \mathrm{remain} \mathrm{at} \mathrm{a} \mathrm{maximum} \mathrm{of} 20 \mathrm{cm}. \end{gathered}$$

Answer:

Correct option: (d)

Viscosity is one property of fluids. Fluids include both liquids and gases.

Answer:

Correct option: (a)

The force of viscosity arises from molecular interaction between different layers of fluids that are in motion. Molecular forces are electromagnetic in nature. Therefore, viscosity must also be electromagnetic.

Answer:

Correct option: (c)
The viscous force acts tangentially between two parallel layers of a liquid. In terms of force on a material, it is analogous to a shearing force.   

Answer:

Correct option: (c)
Air has viscosity. During rainfall, the raindrops acquire acceleration due to gravity. However, the increase in velocity is hindered by the viscous force acting upwards. A gradual balance between the two opposing forces causes the raindrops to attain a terminal velocity, thus, falling with a uniform velocity.   

Answer:

Correct option: (b)
 
The density of wood is less than that of water.When a piece of wood is immersed deep inside a long column of water and released, it experiences a buoyant force that gives it an upward acceleration. The velocity of wood increases as its motion is accelerated by the buoyant force. However, the viscous drag force acts simultaneously to oppose its upward motion. As a result, the initial acceleration decreases and the wood rises with a decreasing upward acceleration.

Answer:

Correct option: (d)

In vacuum, no viscous force exists. The sphere therefore, will have constant acceleration because of gravity. An accelerated motion implies that it won't have uniform velocity throughout its motion. In other words, there will be no terminal velocity. 

Answer:

Correct option: (b)
Initially, when the ball starts moving, its velocity is small. Gradually, the velocity of the ball increases due to acceleration caused by gravity. However, as the velocity increases, the viscous force acting on the ball also increases. This force tends to decelerate the ball. Therefore, after reaching a certain maximum velocity, the ball slows down. 

Answer:

Correct option: (a), (b), (c), (d)

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.  

Answer:

Correct option: (c) & (d)

(c) The surface molecules acquire air and liquid molecules in their sphere of influence. 
(d) The surface molecules have different magnitudes of forces pulling them from the top and the bulk. So, they are affected by a net finite force in one direction. 

Answer:

Correct option: (a), (b), (d)

$$\begin{gathered} \mathrm{Height} \mathrm{of} \mathrm{the} \mathrm{liquid} \mathrm{in} \mathrm{the} \mathrm{capillary} \mathrm{tube} \mathrm{is} \mathrm{given} \mathrm{by}: \\ \mathrm{h} = \frac{2\mathrm{Scos\theta }}{\mathrm{r\rho g}} \\ \mathrm{h} = \mathrm{Height} \\ \mathrm{S} = \mathrm{Surface} \mathrm{tension} \\ \mathrm{r} =\mathrm{Inner} \mathrm{radius} \mathrm{of} \mathrm{the} \mathrm{tube} \\ \mathrm{\rho }= \mathrm{Density} \mathrm{of} \mathrm{the} \mathrm{liquid} \\ \mathrm{g} = \mathrm{Acceleration} \mathrm{due} \mathrm{to} \mathrm{gravity} \\ a) \mathrm{\theta } \mathrm{and} \mathrm{\rho } \mathrm{depend} \mathrm{upon} \mathrm{the} \mathrm{material} \mathrm{of} \mathrm{the} \mathrm{capillary} \mathrm{tube} \mathrm{and} \mathrm{the} \mathrm{liquid}. \\ \mathrm{b}) \mathrm{h} \mathrm{is} \mathrm{dependent} \mathrm{on} \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{tube}. \mathrm{If} \mathrm{the} \mathrm{length} \mathrm{is} \mathrm{insufficient}, \mathrm{then} \mathrm{h} \mathrm{will} \mathrm{be} \mathrm{low}. \\ \mathrm{d}) \mathrm{r} \mathrm{is} \mathrm{the} \mathrm{inner} \mathrm{radius} \mathrm{of} \mathrm{the} \mathrm{tube}. \end{gathered}$$

Answer:

Correct option: (a), (b)

The angle of contact between a solid and a liquid depends upon the molecular forces of both the substances. Therefore,  it depends upon the material of the solid and the liquid. 

Answer:

Correct option: (c)


Let the height of the liquid-filled column be L.
Let the radius be denoted by R.

$$\begin{gathered} \mathrm{Total} \mathrm{perimeter} \mathrm{of} \mathrm{the} \mathrm{curved} \mathrm{part}=\mathrm{semi}-\mathrm{circumference} \mathrm{of} \mathrm{upper} \mathrm{area}= \mathrm{\pi r} \\ \mathrm{Total} \mathrm{surface} \mathrm{tension} \mathrm{force} =\pi RS \\ \mathrm{Total} \mathrm{perimeter} \mathrm{of} \mathrm{the} \mathrm{flat} \mathrm{part} =2R \\ \mathrm{Total} \mathrm{surface} \mathrm{tension} \mathrm{force} = 2RS \\ \mathrm{Ratio} \mathrm{of} \mathrm{curved} \mathrm{surface} \mathrm{force} \mathrm{to} \mathrm{flat} \mathrm{surface} \mathrm{force} = \frac{\mathit{\pi }\mathit{R}\mathit{S}}{\mathit{2}\mathit{R}\mathit{S}}=\frac{\mathrm{\pi }}{2} \end{gathered}$$

Answer:

Correct option: (c), (d)

If the liquid level does not rise, it may be assumed that the surface tension is zero or the contact angle is 90°, or both. However, we cannot tell for sure whether the surface tension of the liquid is zero or the contact angle is 0°.

Answer:

Correct option: (b), (c), (d)

(b) There is no gravitational force acting downwards. However, when the starting velocity is 20 m/s, the viscous force, which is directly proportional to velocity, becomes maximum and tends to accelerate the ball upwards.

 $$\begin{gathered} \mathrm{When} \mathrm{the} \mathrm{ball} \mathrm{fall}s \mathrm{under} \mathrm{gravity}, \\ \mathrm{neglecting} \mathrm{the} \mathrm{density} \mathrm{of} \mathrm{air}: \\ \mathrm{Mass} \mathrm{of} \mathrm{the} \mathrm{sphere} =\mathit{ }m \\ \mathrm{Radius} =\mathit{ }r \\ \mathrm{Viscous} \mathrm{drag} \mathrm{coeff}.= \eta \\ \mathrm{Terminal} \mathrm{velocity} \mathrm{is} \mathrm{given} \mathrm{by}: \\ mg = 6\pi \eta r{v}_{\mathit{T}}\mathit{ } \\ \Rightarrow \frac{6\mathrm{\pi }\eta r{v}_{\mathit{T}}}{m}=g...\left(1\right) \\ \mathrm{Now}, \mathrm{at} \mathrm{terminal} \mathrm{velocity}, \mathrm{the} \mathrm{acceleration} \mathrm{of} \mathrm{the} \mathrm{ball} \mathrm{due} \mathrm{to} \mathrm{the} \mathrm{viscous} \mathrm{force} \mathrm{is} \mathrm{given} \mathrm{by}: \\ a\mathit{ }\mathit{=}\mathit{ }\frac{\mathit{6}\mathit{\pi }\mathit{\eta }\mathit{r}{\mathit{v}}_{\mathit{T}}}{\mathit{m}} \\ \mathrm{Comparing} \mathrm{equations} \left(1\right) \mathrm{and} \left(2\right), \mathrm{we} \mathrm{find} \mathrm{that}: \\ a\mathit{ }\mathit{=}\mathit{ }g \end{gathered}$$

Thus, we see that the initial acceleration of the ball will be 9.8 ms${}^{-2}$.

(c) The velocity of the ball will decrease with time because of the upward viscous drag. As the force of viscosity is directly proportional to the velocity of the ball, the acceleration due to the viscous force will also decrease.

(d) When all the kinetic energy of the ball is radiated as heat due to the viscous force, the ball comes to rest. 

Answer:

Given:
Mass of the load (m) = 10 kg
Length of wire (L) = 3 m
Area of cross-section of the wire (A) = 4 mm² = 4.0 × 10⁻⁶ m²
Young's modulus of the metal Y = 2.0 × 10¹¹ N m⁻²

(a) Stress = F/A  

     F = mg
        = $10\times 10$ = 100 N    (g = 10 m/s²)
     $$\begin{gathered} \therefore \frac{F}{A}=\frac{100}{4\times {10}^{-6}} \\ =2.5\times {10}^7 \mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$

(b) Strain = $\frac{\Delta L}{L}$
​
Or, $\mathrm{Strain} = \frac{\mathrm{Stress}}{Y}$

$$\begin{gathered} \mathrm{Strain} =\frac{2.5\times {10}^7}{2\times {10}^{11}} \\ =1.25\times {10}^{-4} \mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$

(c) Let the elongation in the wire be $∆L$.
$$\begin{gathered} \mathrm{Strain}=\frac{\Delta L}{L} \\ \Rightarrow \mathrm{\Delta }L=\left(\mathrm{Strain}\right)\times L \\ =1.25\times {10}^{-4}\times 3 \\ =3.75\times {10}^{-4} \mathrm{m} \end{gathered}$$

Answer:

Given:
Radius of cylinder (r) = 2 cm = $2\times {10}^{-2} \mathrm{m}$
Length of cylinder (L) = 2 m
Mass of the load = 100 kg
Young's modulus of the metal =  $2\times {10}^{11} \mathrm{N}/{\mathrm{m}}^2$

(a) Stress(ρ) is given by: $\frac{F}{A}$
      Here, F is the force given by mg = $100\times 10 =1000 \mathrm{N}$ ( Taking g = 10 m/s²)
                 A is the area of cross-section = πr²  = 4π$\times {10}^{-4} {\mathrm{m}}^2$
  
$$\begin{gathered} \Rightarrow S\mathrm{tress} \rho =\frac{mg}{A} \\ =\frac{\left(100\times 10\right)}{\left(4\mathrm{\pi }\times {10}^{-4}\right)} \\ =7.96\times {10}^5 \mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$

(b) Strain is given by:

$$\begin{gathered} \mathrm{Strain}=\frac{\rho }{Y}=\frac{\left(7.96\times {10}^5\right)}{\left(2\times {10}^{11}\right)} \\ =4\times {10}^{-6} \end{gathered}$$

(c) Compression of the cylinder:
                           ΔL = strain × L
                                = 4 × 10⁻⁶ × 2 = 8 × 10⁻⁶ m

Answer:

Given:
$$\begin{gathered} \mathrm{Elastic} \mathrm{limit} \mathrm{of} \mathrm{steel}\frac{ F}{A}=8\times {10}^5 \mathrm{N}/{\mathrm{m}}^2 \\ \mathrm{Young}\text{'}\mathrm{s} \mathrm{modulus} \mathrm{of} \mathrm{steel} Y=2\times {10}^{11} \mathrm{N}/{\mathrm{m}}^2 \\ \mathrm{Length} \mathrm{of} \mathrm{steel} \mathrm{wire} L=\frac{1}{2}\mathrm{m}=0.5 \mathrm{m} \end{gathered}$$

The elastic limit of steel indicates the maximum pressure that steel can bear.
Let the maximum elongation of steel wire be $∆L$.
$$\begin{gathered} Y=\frac{F}{A} \frac{L}{∆L} \\ \Rightarrow ∆L=\frac{FL}{AY} \\ \Rightarrow ∆L=\frac{8\times {10}^5\times \left(0.5\right)}{2\times {10}^{11}} \\ =2\times {10}^{-3} \mathrm{m}=2 \mathrm{mm} \end{gathered}$$

Hence, the required elongation of steel wire is 2 mm.

Answer:

Given:
Young's modulus of steel = 2 × 10¹¹ N m⁻²
Young's modulus of copper = 1.3 × 10 11 N m⁻²
Both wires are of equal length and equal cross-sectional area. Also, equal tension is applied on them.
 As per the question:
$$\begin{gathered} L_{\mathrm{steel}}=L_{\mathrm{Cu}} \\ {A}_{\mathrm{steel}}=A_{\mathrm{Cu}} \\ {F}_{\mathrm{Cu}}=F_{\mathrm{Steel}} \end{gathered}$$

Here: L_steel and L_Cu denote the lengths of steel and copper wires, respectively.
           A_steel and A_Cu denote the cross-sectional areas of steel and copper wires, respectively.
           F_steel and F_Cu  denote the tension of steel and cooper wires, respectively.

$\left(\mathrm{a}\right)\frac{\mathrm{Stress} \mathrm{of} \mathrm{Cu}}{\mathrm{Stress} \mathrm{of} \mathrm{Steel}}=\frac{F_{\mathrm{Cu}}}{A_{\mathrm{Cu}}}\frac{A_{\mathrm{Steel}}}{F_{\mathrm{Steel}}}=1$

(b)
$$\begin{gathered} \frac{\mathrm{Strain} \mathrm{of} \mathrm{Cu}}{\mathrm{Strain} \mathrm{of} \mathrm{steel}} = \frac{{\displaystyle \frac{∆L_{\mathrm{Steel}}}{L_{\mathrm{Steel}}}}}{{\displaystyle \frac{∆L_{\mathrm{cu}}}{L_{\mathrm{cu}}}}}=\frac{F_{\mathrm{Steel}} L_{\mathrm{Steel}} A_{cu} Y_{cu}}{A_{\mathrm{Steel}} Y_{\mathrm{Steel}} F_{cu} L_{cu}} \\ \left(\mathrm{Using} \frac{∆L}{L}=\frac{F}{AY}\right) \\ \Rightarrow \frac{\mathrm{Strain} \mathrm{of} \mathrm{Cu}}{\mathrm{Strain} \mathrm{of} \mathrm{steel}}=\frac{{\mathrm{Y}}_{\mathrm{cu}}}{{\mathrm{Y}}_{\mathrm{Steel}}}=\frac{1.3\times {10}^{11}}{2\times {10}^{11}} \\ \Rightarrow \frac{\mathrm{Strain} \mathrm{of} \mathrm{C}\mathrm{u}}{\mathrm{Strain} \mathrm{of} \mathrm{steel}}=\frac{13}{20} \\ \Rightarrow \frac{\mathrm{Strain} \mathrm{of} \mathrm{steel}}{\mathrm{Strain} \mathrm{of} \mathrm{Cu}}=\frac{20}{13} \end{gathered}$$

Hence, the required ratio is 20 : 13.

Answer:

Given that both wires are of equal length and equal cross-sectional area,
the block applies equal tension on both of them.

∴ $$\begin{gathered} L_{\mathrm{steel}}=L_{\mathrm{Cu}} \\ {A}_{\mathrm{steel}}=A_{\mathrm{Cu}} \\ {F}_{\mathrm{Cu}}=F_{\mathrm{Steel}} \end{gathered}$$
$$\begin{gathered} \frac{\mathrm{Strain} \mathrm{of} Cu}{\mathrm{Strain} \mathrm{of} \mathrm{steel}} = \frac{{\displaystyle \frac{∆L_{\mathrm{Steel}}}{L_{\mathrm{Steel}}}}}{{\displaystyle \frac{∆L_{\mathrm{cu}}}{L_{\mathrm{cu}}}}}=\frac{F_{\mathrm{Steel}} L_{\mathrm{Steel}} A_{cu} Y_{cu}}{A_{\mathrm{Steel}} Y_{\mathrm{Steel}} F_{cu} L_{cu}} \\ \left(\mathrm{Using} \frac{∆L}{L}=\frac{F}{AY}\right) \\ \Rightarrow \frac{\mathrm{Strain} \mathrm{of} Cu}{\mathrm{Strain} \mathrm{of} \mathrm{steel}}=\frac{{\mathrm{Y}}_{\mathrm{cu}}}{{\mathrm{Y}}_{\mathrm{Steel}}}=\frac{1.3\times {10}^{11}}{2\times {10}^{11}} \\ \Rightarrow \frac{\mathrm{Strain} \mathrm{of} \mathrm{steel} }{\mathrm{Strain} \mathrm{of} \mathrm{C}\mathrm{u}}=\frac{20}{13}=1.54 \end{gathered}$$

Hence, the required ratio of the longitudinal strains is 20 : 13.

Answer:

(a) Given:
$$\begin{gathered} \mathrm{Breaking} \mathrm{stress} \mathrm{of} \mathrm{wire} =8\times {10}^8 \mathrm{N}/{\mathrm{m}}^2 \\ \mathrm{Area} \mathrm{of} \mathrm{cross}-\mathrm{section} \mathrm{of} \mathrm{upper} \mathrm{wire} \left(A_{\mathrm{u}}\right)=0.006 {\mathrm{cm}}^2=6\times {10}^{-7} \mathrm{m} \\ \mathrm{Area} \mathrm{of} \mathrm{cross}-\mathrm{section} \mathrm{of} \mathrm{lower} \mathrm{wire} \left(A_{\mathrm{l}}\right)=0.003 {\mathrm{cm}}^2=3\times {10}^{-7} \mathrm{m} \\ {m}_1=10 \mathrm{kg}, m_2=20 \mathrm{kg} \end{gathered}$$

Tension in lower wire $T_l=m_{1}g+w$
Here: g is the acceleration due to gravity
           w is the load
∴ Stress in lower wire$=\frac{{\mathrm{T}}_l}{{\mathrm{A}}_l}=\frac{m_{1}g+w}{{\mathrm{A}}_l}$
$$\begin{gathered} \Rightarrow \frac{m_{1}g+w}{A_l}=8\times {10}^8 \\ \Rightarrow w=\left[\left(8\times {10}^8\right)\times \left(3\times {10}^{-7}\right)\right]-100 \\ \Rightarrow w=140 \mathrm{N} \mathrm{or} 14 \mathrm{kg} \end{gathered}$$

Now, tension in upper wire $T_2=m_{1}g+m_{2}g+w$

∴ Stress in upper wire$=\frac{{\mathrm{T}}_u}{{\mathrm{A}}_u}=\frac{m_{\mathit{2}}g+m_{1}g+w}{{\mathrm{A}}_u}$
$$\begin{gathered} \Rightarrow \frac{m_2 g+m_{1}g+w}{{\mathrm{A}}_u}=8\times {10}^8 \\ \Rightarrow w=180 \mathrm{N} \mathrm{or} 18 \mathrm{kg} \end{gathered}$$

For the same breaking stress, the maximum load that can be put is 140 N or 14 kg. The lower wire will break first if the load is increased.

(b) $\mathrm{If} m_1=10 \mathrm{kg} \mathrm{and} m_2=36 \mathrm{kg}$:
Tension in lower wire $T_l=m_{1}g+w$
Here: g is the acceleration due to gravity
           w is the load
∴ Stress in lower wire:

$$\begin{gathered} \Rightarrow \frac{{\mathrm{T}}_l}{{\mathrm{A}}_l}=\frac{m_{1}g+w}{{\mathrm{A}}_l}=8\times {10}^5 \\ \Rightarrow w=140 \mathrm{N} \end{gathered}$$

Now, tension in upper wire $T_2=m_{1}g+m_{2}g+w$

∴ Stress in upper wire:

$$\begin{gathered} \Rightarrow \frac{{\mathrm{T}}_u}{{\mathrm{A}}_u}=\frac{m_{2}g+m_{1}g+w}{{\mathrm{A}}_u}=8\times {10}^5 \\ \Rightarrow w=20 \mathrm{N} \end{gathered}$$

For the same breaking stress, the maximum load that can be put is 20 N or 2 kg. The upper wire will break first if the load is increased.

Answer:

Given:
Force (F) applied by two persons on the rope = 100 N
$$\begin{gathered} \mathrm{Original} \mathrm{length} \mathrm{of} \mathrm{rope} L=2 \mathrm{m} \\ \mathrm{Extension} \mathrm{in} \mathrm{the} \mathrm{rope} ∆L=0.01 \mathrm{m} \\ \mathrm{Area} \mathrm{of} \mathrm{cross}-section \mathrm{of} \mathrm{the} \mathrm{rope} A=2\times {10}^{-4} \\ \mathrm{We} \mathrm{know} \mathrm{that}: \\ \mathrm{Young}\text{'}\mathrm{s} \mathrm{modulus}\mathit{ }Y=\frac{F}{A}\times \frac{L}{∆L} \\ =\frac{100}{2\times {10}^{-4}}\times \frac{2}{0.01} \\ \Rightarrow Y=1\times {10}^8 \mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$

Hence, the required Young's modulus  for the rope is $1\times {10}^8 \mathrm{N}/{\mathrm{m}}^2$.

Answer:

Given:
Cross-sectional area of steel rod A = 4 cm² = 4 × 10⁻⁴ m²
Length of steel rod L = 2 m
Compression during night hours ΔL = 0.1 cm = 10⁻³ m
Young modulus of steel Y = 1.9 × 10¹¹ N m⁻²

Let the tension developed at night be F.
$$\begin{gathered} Y=\frac{F}{A}\times \frac{L}{∆L} \\ \Rightarrow F=\frac{YA∆L}{L} \\ =\frac{1.9\times {10}^{11}\times 4\times {10}^{-4}\times {10}^{-3}}{2} \\ =3.8\times {10}^4 \mathrm{N} \end{gathered}$$

∴ Required tension developed in steel rod during night hours = 3.8 × 10⁴ N.

Answer:

Given:
Force (F) = m₂g/2
Area of cross-section of the string = A
 Young's modulus = Y
Let a be the acceleration produced in block m₂ in the downward direction and T be the tension in the string.
From the free body diagram:
$$\begin{gathered} m_{2}g-T=m_{2}a ...\left(\mathrm{i}\right) \\ T-F=m_{1}a ...\left(\mathrm{ii}\right) \end{gathered}$$

From equations (i) and (ii), we get:
$$\begin{gathered} a=\frac{m_{2}g-F}{m_1+m_2} \\ \mathrm{Applying} F=\frac{m_{2}g }{2} \\ \Rightarrow a=\frac{m_{2}g}{2\left(m_1+m_2\right)} \end{gathered}$$

Again, T = F + m₁a

On applying the values of F and a, we get:

$\Rightarrow T=\frac{m_{2}g}{2}+m_1\frac{m_{2}g}{2\left(m_1+m_2\right)}$

We know that:

$$\begin{gathered} Y=\frac{FL}{A∆L} \\ \Rightarrow \mathrm{Strain}=\frac{∆L}{L}=\frac{F}{AY} \\ \Rightarrow \mathrm{Strain}=\frac{\left(m_2^2+2m_1{m}_2\right)g}{2\left(m_1+m_2\right) AY} \\ =\frac{m_{2}g\left(2m_1+m_2\right)}{2AY \left(m_1+m_2\right)} \end{gathered}$$

∴ Required strain developed in the string $=\frac{m_{2}g\left(2m_1+m_2\right)}{2AY \left(m_1+m_2\right)}$.

Answer:

Given:
Mass of sphere (m) = 20 kg
Length of metal wire (L) = 4 m
Diameter of wire (d = 2r) = 1 mm
⇒ r = 5 × 10⁻⁴ m
Young's modulus of the metal wire = 2.0 × 10¹¹ N m⁻²
Tension in the wire in equilibrium = T
T= mg
When it is moved at an angle θ and released, let the tension at the lowest point be T'​.
$\Rightarrow T\text{'}=mg+\frac{m{v}^2}{r}$
The change in tension is due to the centrifugal force.
∴​ $∆T=T\text{'}-T$
$∆T=\frac{m{v}^2}{r} ...\left(\mathrm{i}\right)$

Now, using work energy principle:
$$\begin{gathered} \frac{1}{2}m{v}^2-0=mgr\left(1-\cos \theta \right) \\ \Rightarrow v^2=2gr\left(1-\cos \theta \right) ...\left(2\right) \end{gathered}$$

Applying the value of v² in (i):

$$\begin{gathered} ∆T=\frac{m \left[2gr\left(1-cos\theta \right)\right]}{r} \\ =2mg\left(1-cos\theta \right) \end{gathered}$$

$$\begin{gathered} \mathrm{Now}, F=∆T \\ \mathrm{Also}, F=\frac{YA∆L}{L} \\ \Rightarrow \frac{YA∆L}{L}=2mg\left(1-\cos \theta \right) \\ \Rightarrow \cos \theta =1-\frac{YA∆L}{L\left(2mg\right)} \\ \Rightarrow cos\theta =1-\left[\frac{2\times {10}^{11}\times 4\times 3.14\times {\left(5\right)}^2\times {10}^{-8}\times 2\times {10}^{-3}}{4\times 2\times 20\times 10}\right] \\ \Rightarrow \cos \theta =0.80 \\ \mathrm{Or}, \theta =36.4° \end{gathered}$$

Hence, the required maximum value of θ is 35.4˚.

Answer:

Given:
Original length of steel wire (L) = 1 m
Area of cross-section (A) = 4.00 mm² = 4 × 10⁻² cm²
Load = 2.16 kg
Young's modulus of steel (Y) = 2 × 10¹¹ N/m²
Acceleration due to gravity (g) = 10 m s⁻²

Let T be the tension in the string after the load is suspended and θ be the angle made by the string with the vertical, as shown in the figure:


$\cos \theta =\frac{x}{\sqrt{x^2+l^2}}=\frac{x}{l}{\left\{1+\frac{x^2}{l^2}\right\}}^{-1/2}$
Expanding the above equation using the binomial theorem:
$$\begin{gathered} \cos \theta =\frac{x}{l}\left\{1-\frac{1}{2}\frac{x^2}{l^2}\right\} \left(\mathrm{neglecting} \mathrm{the} \mathrm{higher} \mathrm{order} \mathrm{terms}\right) \\ \mathrm{Since} x<<l, \frac{x^2}{l^2} \mathrm{can} \mathrm{be} \mathrm{neglected}. \\ \Rightarrow \cos \theta =\frac{x}{l} \end{gathered}$$
Increase in length:
ΔL = (AC + CB) − AB
AC = (l² + x²)^1/2
ΔL = 2 (l² + x²)^1/2 − 2l

We know that:
$$\begin{gathered} Y=\frac{F}{A}\frac{L}{∆L} \\ \Rightarrow 2\times {10}^{12}=\frac{T\times 100}{\left(4\times {10}^{-2}\right)\times \left[2{\left({50}^2+x^2\right)}^{1/2}-100\right]} \end{gathered}$$

From the free body diagram:
$$\begin{gathered} 2T\cos \theta =mg \\ 2T\left(\frac{x}{50}\right)=2.16\times {10}^3\times 980 \\ \Rightarrow \frac{2\times \left(2\times {10}^{12}\right)\times \left(4\times {10}^{-2}\right)\times \left[2\left({50}^2+x^2 {\displaystyle \frac{1}{2}}\right)-100\right]x}{100\times 50}=\left(2.16\right)\times {10}^3\times 980 \end{gathered}$$

On solving the above equation, we get x = 1.5 cm.
  Hence, the required vertical depression is 1.5 cm.

Answer:

Given:
Cross-sectional area of copper wire A = 0.01 cm² = 10⁻⁶ m²
Applied tension T = 20 N
Young modulus of copper Y = 1.1 × 10¹¹ N/m²
Poisson ratio σ = 0.32

We know that:
$Y=\frac{FL}{A∆L}$

$$\begin{gathered} \Rightarrow \frac{∆L}{L}=\frac{F}{AY} \\ =\frac{20}{{10}^{-6}\times 1.1\times {10}^{11}}=18.18\times {10}^{-5} \end{gathered}$$

$$\begin{gathered} \mathrm{Poisson}\text{'}\mathrm{s} \mathrm{ratio},\sigma =\frac{{\displaystyle \frac{∆d}{d}}}{{\displaystyle \frac{∆L}{L}}}=0.32 \\ \mathrm{Where} d \mathrm{is} \mathrm{the} \mathrm{transverse} \mathrm{length} \\ \mathrm{So}, \frac{∆d}{d}=\left(0.32\right)\times \frac{∆L}{L} \\ =0.32\times \left(18.18\right)\times {10}^{-5}=5.81\times {10}^{-5} \\ \mathrm{Again}, \frac{∆A}{A}=\frac{2∆r}{r}=\frac{2∆d}{d} \\ \Rightarrow ∆A=\frac{2∆d}{d}A \\ \Rightarrow ∆A=2\times \left(5.8\times {10}^{-5}\right)\times \left(0.01\right) \\ =1.164\times {10}^{-6} {\mathrm{cm}}^2 \end{gathered}$$

Hence, the required decrease in the cross -sectional area is $1.164\times {10}^{-6} {\mathrm{cm}}^2$.

Answer:

Given:
Bulk modulus of water (B) = $2.1\times {10}^9 {\mathrm{Nm}}^{-2}$
In order to decrease the volume (V) of a water sample by 0.01%, let the increase in pressure be P.
 
$$\begin{gathered} \frac{V\times 0.01}{100}=∆V \\ \Rightarrow \frac{∆V}{V}={10}^{-4} \\ \mathrm{From} B=\frac{PV}{∆V}, \mathrm{we} \mathrm{have}: \\ \Rightarrow P=B\left(\frac{∆V}{V}\right) \\ =2.1\times {10}^9\times {10}^{-4} \\ =2.1\times {10}^5 \mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$

Hence, the required increase in pressure is $2.1\times {10}^5 {\mathrm{Nm}}^{-2}$.

Answer:

Given:
$\mathrm{Bulk} \mathrm{modulus} \mathrm{of} water B=2\times {10}^9 \mathrm{N}/{\mathrm{m}}^2$
Depth (d) = 400 m
Density of water at the surface (ρ₀) = 1030 kg/m³
We know that:
$$\begin{gathered} \mathrm{Density} \mathrm{at} \mathrm{surface} {\rho }_0=\frac{m}{{\mathrm{V}}_0} \\ \mathrm{Density} \mathrm{at} \mathrm{depth} {\rho }_{\mathrm{d}}=\frac{m}{{\mathrm{V}}_{\mathrm{d}}} \\ \Rightarrow \frac{{\rho }_{\mathrm{d}}}{{\rho }_0}=\frac{V_0}{V_d} ...\left(\mathrm{i}\right) \end{gathered}$$

Here: ρ_d = density of water at a depth
            m =  mass
            V₀ = volume at the surface
            V_d = volume at a depth

$$\begin{gathered} \mathrm{Pressure} \mathrm{at} \mathrm{a} \mathrm{depth} d={\mathrm{\rho }}_{0}gd \\ \mathrm{Acceleration} \mathrm{due} \mathrm{to} \mathrm{gravity} g = 10 {\mathrm{ms}}^2 \\ \mathrm{Volume} \mathrm{strain} = \frac{V_0-V_d}{V_0} \\ B=\frac{\mathrm{Pressure}}{\mathrm{Volume} \mathrm{strain}} \\ \Rightarrow B=\frac{{\mathrm{\rho }}_{0}gd}{\left(\frac{V_0-V_d}{V_0}\right)} \\ \Rightarrow 1-\frac{V_d}{V_0}=\frac{{\mathrm{\rho }}_{0}gd}{B} \\ \Rightarrow \frac{V_d}{V_0}=\left(1-\frac{p_0 gd}{B}\right) ...\left(\mathrm{ii}\right) \end{gathered}$$
Using equations (i) and (ii), we get:

$$\begin{gathered} \frac{{\rho }_{\mathrm{d}}}{{\rho }_0}=\frac{1}{\left(1-{\displaystyle \frac{{\rho }_{0}gd}{B}}\right)} \\ \Rightarrow {\rho }_d=\frac{1}{\left(1- {\displaystyle \frac{{\rho }_{0}gh}{B}}\right)}{\rho }_0 \\ \Rightarrow {\rho }_d=\frac{1030 }{\left(1-{\displaystyle \frac{1030\times 10\times 400}{2\times {10}^9}}\right)} \approx 1032 \mathrm{kg}/{\mathrm{m}}^3 \\ \mathrm{Change} \mathrm{in} \mathrm{density} ={\rho }_{\mathrm{d}}-{\rho }_0 \\ =1032-1030=2 \mathrm{kg}/{\mathrm{m}}^3 \end{gathered}$$

Hence, the required density at a depth of 400 m below the surface is 2 kg/m³.​

Answer:

Given:
Face area of steel plate A = 4 cm² = 4 × 10⁻⁴ m²
Thickness of steel plate d = 0.5 cm = 0.5 × 10⁻² m
Applied force on the upper surface F   = 10 N
Rigidity modulus of steel = 8.4 × 10¹⁰ N m⁻²
Let θ be the angular displacement. 
Rigidity modulus $m=\frac{F}{A\theta }$
$$\begin{gathered} \Rightarrow m=\left(\frac{10}{4\times {10}^{-4}\theta }\right) \\ \Rightarrow \theta =\frac{10}{4\times {10}^{-4}\times 8.4\times {10}^{10}} \\ =0.297\times {10}^{-6} \end{gathered}$$

∴ Lateral displacement of the upper surface with respect to the lower surface = θ × d
⇒ (0.297) × 10⁻⁶ × (0.5) × 10⁻²
⇒ 1.5 × 10⁻⁹ m

Hence, the required lateral displacement of the steel plate is 1.5 × 10⁻⁹ m.

Answer:

Given:
Length of thread l = 5 cm = 5 × 10⁻² m
Surface tension of water T = 0.76 N/m
We know that:
F = T × l = 0.76 × 5 × 10⁻²
   = 3.8 × 10⁻³ N
Therefore, the water surface on one side of the thread pulls it with a force of 3.8 × 10⁻³ N.

Answer:

Given:
$$\begin{gathered} \mathrm{Radius} \mathrm{of} \mathrm{mercury} \mathrm{drop} r=2 \mathrm{mm}=2\times {10}^{-3} \mathrm{m} \\ \mathrm{Radius} \mathrm{of} \mathrm{soap} \mathrm{bubble} r=4 \mathrm{mm}=4\times {10}^{-3} \mathrm{m} \\ \mathrm{Radius} \mathrm{of} \mathrm{air} \mathrm{bubble} r=4 \mathrm{mm}=4\times {10}^{-3} \mathrm{m} \\ \mathrm{Surface} \mathrm{tension} \mathrm{of} \mathrm{mercury} T_{\mathrm{Hg}}=0.465 \mathrm{N}/\mathrm{m} \\ \mathrm{Surface} \mathrm{tension} \mathrm{of} \mathrm{soap} \mathrm{solution} T_s=0.03 \mathrm{N}/\mathrm{m} \\ \mathrm{Surface} \mathrm{tension} \mathrm{of} \mathrm{water} T_a=0.076 \mathrm{N}/\mathrm{m} \end{gathered}$$

(a) Excess pressure inside mercury drop:
$$\begin{gathered} P=\frac{2T_{Hg}}{r} \\ =\frac{0.465\times 2}{2\times {10}^{-3}}=465 \mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$

(b) Excess pressure inside the soap bubble:
  $$\begin{gathered} P=\frac{4T_s}{r} \\ =\frac{4\times 0.03}{4\times {10}^{-3}}=30 \mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$

(c) Excess pressure inside the air bubble:
$$\begin{gathered} P=\frac{2T_a}{r} \\ =\frac{2\times 0.076}{4\times {10}^{-3}}=38 \mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$

Answer:

Given:
Surface area of mercury drop, A = 1 mm² = 10⁻⁶ m²
Radius of mercury drop, r = 4 mm = 4 × 10⁻³ m
Atmospheric pressure, P₀ = 1.0 × 10⁵ P_a
Surface tension of mercury, T = 0.465 N/m

(a) Force exerted by air on the surface area:
 F = P₀A
⇒ F= 1.0 × 10⁵ × 10⁻⁶ = 0.1 N

(b) Force exerted by mercury below the surface area:
$$\begin{gathered} \mathrm{Pressure} P\text{'} = P_0+\frac{2T}{r} \\ F=P\text{'}A=\left(P_0+\frac{2T}{r}\right)A \\ =\left(0.1+\frac{2\times 0.465}{4\times {10}^{-3}}\right)\times {10}^{-6} \\ =0.1+0.00023=0.10023 \mathrm{N} \end{gathered}$$

(c) Force exerted by mercury surface in contact with it:
$$\begin{gathered} P=\frac{2T}{r} \\ F=PA=\frac{2T}{r}A \\ =\frac{2\times 0.465}{4\times {10}^{-3}}\times {10}^{-6}=0.00023 \mathrm{N} \end{gathered}$$

Answer:

Given:
Surface tension of water T = 7.5 × 10⁻² N/m
Taking cos θ = 1:
Radius of capillary A (r_A) = 0.5 mm = 0.5 × 10⁻³ m

Height of water level in capillary A:
$$\begin{gathered} h_{\mathrm{A}}=\frac{2T \cos \theta }{r_{\mathrm{A}}\rho g} \\ =\frac{2\times 7.5\times {10}^{-2}}{0.5\times {10}^{-3}\times 1000\times 10} \\ =3\times {10}^{-2} \mathrm{m}=3 \mathrm{cm} \end{gathered}$$

Radius of capillary B (r_B) = 1 mm = 1 × 10⁻³ m

Height of water level in capillary B:
$$\begin{gathered} h_{\mathrm{B}}=\frac{2T\cos \theta }{r_{\mathrm{B}}\rho g} \\ =\frac{2\times 7.5\times {10}^{-2}}{1\times {10}^{-3}\times {10}^3\times 10} \\ =15\times {10}^{-3} \mathrm{m}=1.5 \mathrm{cm} \end{gathered}$$

Radius of capillary C (r_C) = 1.5 mm = 1.5 × 10⁻³ m
 Height of water level in capillary C:
$$\begin{gathered} h_{\mathrm{C}}=\frac{2T \cos \mathrm{\theta }}{r_{\mathrm{C}}\rho g} \\ =\frac{2\times 7.5\times {10}^{-2}}{1.5\times {10}^{-3}\times {10}^3\times 10} \\ =\frac{15}{1.5}\times {10}^{-3} \mathrm{m}=1 \mathrm{cm} \end{gathered}$$

Answer:

Let T be the surface tension, r be the inner radius of the capillary tube and ρ be the density of the liquid.

For cos θ = 1, height (h) of the liquid level is given as:

$h=\frac{2T\cos \theta }{r\mathrm{\rho }g}$

Now, for mercury:

$h_{\mathrm{Hg}}=\frac{2T_{\mathrm{Hg}} }{r{\mathrm{\rho }}_{\mathrm{Hg}} g}$     ...(i)

For water:

$h_w=\frac{2T_w}{r{\mathrm{\rho }}_{w}g}$    ...(ii)

Dividing (ii) by (i), we get:

$$\begin{gathered} \frac{h_w}{h_{\mathrm{Hg}}}=\frac{T_w}{T_{\mathrm{Hg}}}\times \frac{{\rho }_{\mathrm{Hg}}}{{\rho }_w} \\ =\left(\frac{0.075}{0.465}\right)\times \left(13.6\right) \\ =2.19 \end{gathered}$$

Height of the water level:

h_w = 2 × 2.19 = 4.38 cm

Hence, the required rise in the water level in the capillary tube is 4.38 cm.

Answer:

Given:
Radius of tube r = 1.0 mm
Atmospheric pressure = 76 cm of Hg
Contact angle of mercury with glass θ = 135°
Surface tension of mercury T = 0.465 N/m
Density of mercury = 13600 kg m⁻³

Let h be the rise in level in the barometer.

$$\begin{gathered} h=\frac{2T\cos \theta }{r\rho g} \\ =\frac{2\times 465\times \left(1/\sqrt{2}\right)}{{10}^{-3}\times 13600\times 10}=0.0048 \mathrm{m} \\ =0.48 \mathrm{cm} \end{gathered}$$

∴ Net rise in level in the barometer tube = H − h
                                                            = 76 − 0.48
                                                            = 75.52 cm

Answer:

Given:
Radius of capillary tube r = 0.5 mm = 5 × 10⁻⁴ m
Depth (where pressure is to be found) h = 5.0 cm = 5 × 10⁻² m
Surface tension of water T = 0.075 N/m
Excess pressure at 5 cm before the surface:
P = ρhg = 1000 × (5 × 10⁻²) × 9.8 = 490 N/m²
Excess pressure at the surface is given by:
$$\begin{gathered} P_0=\frac{2T}{r}=\frac{2\times \left(0.75\right)}{\left(5\times {10}^{-4}\right)} \\ =300 \mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$
Difference in pressure: P₀ − P $=490-300=190 \mathrm{N}/{\mathrm{m}}^2$

Hence, the required difference in pressure is 190 N/m².

Answer:

Given:
Radius of cylindrical vessel, r  = 6.0 cm = 0.06 m
Surface tension of water, T = 0.075 J/m²
 Area, A = πr² = π × (0.06)²
Surface energy = T × A
                             = (0.075) × (3.14) × (0.06)²
                             = 8.5 × 10⁻⁴ J
Therefore, the surface energy of water kept in a cylindrical vessel is 8.5 × 10⁻⁴ J.

Answer:

Given:
Initial radius of mercury drop R = 2 mm = 2 × 10⁻³ m
Surface tension of mercury T = 0.465 J/m²
Let the radius of a small drop of mercury be r.
As one big drop is split into 8 identical droplets:
 volume of initial drop = 8 × (volume of a small drop)
$\left(\frac{4}{3}\right)\pi R^3=\left(\frac{4}{3}\right)\mathrm{\pi }{r}^3\times 8$

Taking cube root on both sides of the above equation:

$r=\frac{R}{2}=10$^-3 m 
Surface energy = T × surface area

∴ Increase in surface energy = TA' − TA
                                              = (8 × 4πr² − 4πR²) T
                                             $$\begin{gathered} =4\pi T\left[8\times \left(\frac{R^2}{4}\right)-R^2\right] \\ =4\pi T{R}^2 \end{gathered}$$
                                              = 4 × (3.14) × (0.465) × (4 × 10⁻⁶)
                                              = 23.36 × 10⁻⁶
                                              = 23.4 μJ

Hence, the required increase in the surface energy of the mercury droplets is 23.4 μJ.

Answer:

Given:
Radius of capillary tube r = 1 mm = 10⁻³ m

(a) Let T be the surface tension and ρ be the density of the liquid.

Then, for cos θ = 1, height (h) of liquid level:

$h=\frac{2T}{r\rho g}$    ...(i), 
where g is the acceleration due to gravity

1. $$\begin{gathered} \Rightarrow h=\frac{2\times \left(0.076\right)}{{10}^{-3}\times 10\times 100} \\ =1.52 \mathrm{cm} \\ =1.52\times {10}^{-2} \mathrm{m} \\ =1.52 \mathrm{cm} \end{gathered}$$