\(v_{\mathrm{A}}=500 \mathrm{~ms}^{-1} \quad \lambda_{A}=5 \mathrm{~m}\)
Frequency in medium B
\(\mathrm{f}_{\mathrm{B}}=\frac{v_{A}}{\lambda_{A}}=\frac{500}{5}=100 \mathrm{~Hz}\)
Wavelength in medium B
\(\lambda_{B}=\frac{v_{s}}{f_{s}}=\frac{600}{100}=6 \mathrm{~Hz}\)

\(\mathrm{F}_{\mathrm{d}} \propto \mathrm{v} \)
\(\mathrm{F}_{\mathrm{d}}=-\mathrm{bv} ; \quad \mathrm{F}_{\mathrm{d}}=\mathrm{kv} \)
\(\therefore \mathrm{k}=\frac{F_{d}}{v} \)
\(\text { Units of } k=\frac{k g m s^{-2}}{m s^{-1}}\)
\(\mathrm{k}=\mathrm{kg} \mathrm{s}^{-1}\)

\(\gamma=\frac{27}{17}\)
Number of moles of helium
\(\mathrm{n}=\frac{8}{4}=2\)
Number of moles of oxygen
\(n^{\prime}=\frac{16}{32}=\frac{1}{2}\)
For mono atomic Helium gas
\(\mathrm{f} =3 \)
\(\mathrm{C}_{\mathrm{V}} =\frac{f}{2} R \)
\(=\frac{3}{2} R \)
For diatomic oxygen gas
f = 5
\(\mathrm{C}_{\mathrm{V}} =\frac{f}{2} R \)
\(=\frac{5}{2} R \)
\(\mathrm{C}_{\mathrm{V}} \text { mixture } =\frac{n c_{v}+n^{\prime} C_{v}^{\prime}}{n+n^{\prime}} \)
\(=\frac{2 \times \frac{3}{2} R+\frac{1}{2} \times \frac{5}{2} R}{2+\frac{1}{2}} \)
\(C_{V} =\frac{3 R+\frac{5}{4} R}{\frac{5}{2}} \)
\(=\frac{17 R}{10} \)
\(\gamma =\frac{C_{p}}{C_{v}} \)
\(=1+\frac{R}{C_{V}} \)
\(=1+\frac{R}{\frac{17 R} {10}}\)
\(=1+\frac{10}{17} \)
\(=\frac{27}{17} \)

Moment of Inertia = MK²
K-Radius of gyration
During heating K would be increased.
Hence moment of inertia will increase

\(F=-\frac{d u}{d x} \quad F(x) =k x+a x^{3} \)
\(d u =-F d x \)
\(u(x) =-\int_{0}^{x}\left(-k x+a x^{3}\right) d x \)
\(=\int_{0}^{x} k x d x-a \int_{0}^{x} x^{3} d x \)
\(=\frac{k x^{2}}{2}-\frac{a x^{4}}{2} \)
\(U(x) =\frac{x^{2}}{2}\left(k-\frac{a x^{2}}{2}\right) \)
\(u(x)=0 \text { at } x=0 \text { and }\)
\(U(x) =0 ; k-\frac{a x^{2}}{2}=0 \)
\(=\frac{a}{2} x^{2}=-k \)
\(x^{2} =\frac{2 k}{a} \)
\(\therefore x =\sqrt{\frac{2 k}{a}} \)
\(\text { Clearly } u(x)=0 \text { at } x=0 \text { and }\)
\(x=\sqrt{\frac{2 k}{a}}\)
\(\text { For } x>\sqrt{\frac{2 k}{a}} U(x) \text { will be negative. } \)
\(\text { At } x=0 ; F=\frac{-d u}{d x}=0\)
(i.e.,) Slope of V - x graph is zero at x = 0
Hence the most appropriate answer is d.

It is a uniform circular motion. So the direction of velocity changes but not the magnitude. Therefore speed in considered constant. Again magnitude of acceleration does not change.

\(\text {Dimension of permittivity } \varepsilon=A^{2} C^{2} N^{-1} m^{-2}\)
\(\text {Dimension of Boltzmann constant } k_{B}=\left[\mathrm{ML}^{2} \mathrm{~T}^{-2} \mathrm{~K}^{-1}\right]\)
\(\text {Dimension of Absolute Temperature }=T\)
No. of charged particles per unit
Volume = No. of dimensions = a
Dimension of charge = AT
Dimension of length = L
\(\text { Dimension of }=\frac{\sqrt{\varepsilon k_{B} T}}{n q^{2}}\)
\(=\frac{\sqrt{\left[\mathrm{A}^{2} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}\right]}\left[\mathrm{ML}^{2} \mathrm{~T}^{-2} \mathrm{~K}^{-1}\right][\mathrm{T}]}{[\mathrm{AT}]^{2}}\)
\(=\sqrt{L^{2}}=L\)
\(\therefore \text { Dimension of length }=\mathrm{L}\)

In a closed room the sound is repeatedly reflected from the walls and it is even heard long after the sound source ceases to function. The residual sound remaining in an enclosure and the phenomenon of multiple reflections of sound is called reverberation.

\(v=\sqrt { \frac { y }{ \rho } } =\sqrt { \frac { 2\times { 10 }^{ 11 } }{ 7800 } } =\sqrt { 0.2564\times { 10 }^{ 8 } } \)
= 0.506\(\times\)10⁴ms^-1= 5\(\times\)10³ms^-1
Therefore, longitudinal waves travel faster in a solid than in a liquid or a gas. Now you may understand why a shepherd checks before crossing railway track by keeping his ears on the rails to safeguard his cattle.

For all the processes that occur in nature (irreversible process), the entropy always increases. For reversible process entropy will not change.

The average size of a class is 6m length, 5 m breadth and 4 m height. The volume of the room V = 6\(\times\)5 \(\times\)4 = 120m³. We can determine the number of mole. At room temperature 300K, the volume of a gas occupied by any gas is equal to 24.6L.
The number of mole \(\mu =\frac { 120{ m }^{ 2 } }{ 24.6\times { 10 }^{ -3 }{ m }^{ 3 } } \approx 4878mol\)
Air is the mixture of about 20% oxygen, 79% nitrogen and remaining one percent are argon, hydrogen, helium, and xenon. The molar mass of air is 29 g mol^-1.
So the total mass of air in the room m = 4878\(\times\)29 = 141.4kg

Stress: Stress is defined as the restoring force per unit area.
Stress, \(\sigma=\frac{Force}{Area}\)
Strain: Strain is defined as the rates of change in size to the original size if an object.
Strain, ε \(=\frac{\Delta l}{l}\)

We know that when the object falls towards the Earth, it experiences acceleration due to gravity g = 9.8 m s^-2 downward. We can choose the coordinate system as shown in the figure. The acceleration is along the negative y direction.
\(\vec { a } =g(-\hat { j } )-g\hat { j } \)

k=\(\frac { Heat }{ Temperature } =[k]=\frac { [{ ML }^{ 2 }{ T }^{ 2 }] }{ [K] } \)= [M¹L²T^-2K^-1]

(i) The energy possessed by the body due to gravitational force gives rise to gravitational potential energy.
(ii) The energy due to spring force and other similar forces give rise to elastic potential energy.
(iii) The energy due to electrostatic force on charges gives rise to electrostatic potential energy.

The weight of tree exerts a torque about the point where the cut is made. This causes rotation of the tree about the cut has to be made at say point A to weaken the tree trunk and to shift the centre of mass to the right and eventually move towards the ground on the right.

No, a single isolated force cannot exist in nature because it violates the Newton's third law.

One Newton is that force which produces an acceleration of 1 m/s^-2 in a body of mass 1 kilogram.
∴ 1 Newton = 1 kg m/s²

(i). It states that the rate of cooling of a body is directly proportional to the temperature differ between the body and the surroundings.
(ii) Consider a spherical calorimeter of mass m whose outer surface is blackened. It is filled with hot water of mass m, the calorimeter with thermometer is suspended from a stand.
(iii) The calorimeter & the hot water radiate heat energy to the surrounding. Using a stop clock, the temperature is noted for every 30 sec. interval of time tree the temperature falls by about 20°C. The readings are tabulated.
(iv) If the temperature falls from T₁ to T₂ in I see, the quantity of heat energy lost by radiation Q = (ms + m₁s₁)(T₁ - T₂), where 's' is the specific heat capacity of the material of the calorimeter & S₁ - specific heat capacity of water.
Rate of cooling =\(\frac{Heat energy lost}{timet_n}\)
\(\therefore \frac { Q }{ E } =\frac { \left( ms+{ m }_{ 1 }{ m }_{ 1 } \right) \left( { T }_{ 1 }+{ T }_{ 2 } \right) }{ t } \)
Room temperature - T_o
(v) Average excess temperature of the  colorimeter over that of the surroundings 
\(-\frac { { T }_{ 1 }-{ T }_{ 2 } }{ 2 } ={ T }_{ 0 }\)
(vi) Acceleration to Newton's law of cooling 
\(\frac { Q }{ T } =\left( \frac { { T }_{ 1 }+{ T }_{ 2 } }{ 2 } -{ T }_{ 0 } \right) \)
(vii) Assume the pressure of the gas remains constant during an infinitesimally small outward displacement dy then work done dW - F. dx = P.A. dx
dW = pdv
Total work done by the gas from volume
V₁ to v₂ is \(W=\int _{ { v }_{ 1 } }^{ { v }_{ 2 } }{ pdv } \)
(ix) But pv\(\gamma\) = constant (k)
\(W=\int _{ { v }_{ 1 } }^{ { v }_{ 2 } }{ k{ v }^{ \gamma } } dv=k{ \left[ \frac { { v }^{ \gamma -1 } }{ 1-\gamma } \right] }_{ { v }_{ 1 } }^{ { v }_{ 2 } }\quad \left[ \because p=\frac { k }{ { v }^{ \gamma } } \right] \)
\(\therefore W=\frac { k }{ 1-\gamma } \left[ { v }_{ 2 }^{ 1-\gamma }{ -v }_{ 1 }^{ 1-\gamma } \right] \)
\(W=\frac { 1 }{ 1-\gamma } \left[ { kv }_{ 2 }^{ 1-\gamma }{ -kv }_{ 1 }^{ 1-\gamma } \right] \)
\({ p }_{ 2 }{ v }_{ 2 }^{ \gamma }={ p }_{ 1 }{ v }_{ 2 }^{ \gamma }\)
(x) Subtract the value of k
\(\therefore W=\frac { 1 }{ 1-\gamma } \left[ { p }_{ 2 }{ v }_{ 2 }^{ \gamma },{ v }_{ 2 }^{ 1-\gamma }-{ p }_{ 1 }{ v }_{ 1 }^{ \gamma }{ v }_{ 1 }^{ 1-\gamma } \right] \)
\(W=\frac { 1 }{ 1-\gamma } \left[ { { p }_{ 2 }v }_{ 2 }^{ }{ -{ p }_{ 1 }v }_{ 1 }^{ } \right] \)
It T~ is the final temperature of the gas in adorable expansion, then
p₁v₁ = RT₁P₂v₂ = RT₂
\(\therefore W=\frac { 1 }{ 1-\gamma } \left[ { R }_{ 2 }^{ }{ -R }_{ 1 }^{ } \right] \)
This is the equation for the work done during adiabatic process.

According to Bernoulli's theorem, the sum of pressure energy, kinetic energy, and potentialenergy per unit mass of an incompressible, nonviscous fluid in a streamlined flow remains a constant. Mathematically,
\(\frac{P}{\rho}+\frac{1}{2}v^{2}+gh\) = constant
This is known as Bernoulli's equation.
Proof:

Let us consider a flow of liquid through a pipe AB as shown in Figure. Let V be the volume of the liquid when it enters A in a time t which is equal to the volume of the liquid leaving B in the same time. Let a_A, v_A and P_A be the area of cross section of the tube, velocity of the liquid and pressure exerted by the liquid at A respectively.
Let the force exerted by the liquid at A is
F_A= P_Aa_A
Distance travelled by the liquid in time t is
d = v_At
Therefore, the work done is
W = F_Ad = P_Aa_Av_A t
But a_Av_At = a_Ad = V, volume of the liquid entering at A.
Thus, the work done is the pressure energy (at A), W = F_Ad = P_AV
Pressure energy per unit volume at
A = \(\frac{Pressure \ energy}{volume}=\frac{P_{A} V}{V}=P_{A}\)
Pressure energy per unit mass at
A = \(\frac{Pressure \ energy}{volume}=\frac{P_{A} V}{m}=\frac{P_{A}}{\frac{m}{V}}=\frac{P_{A}}{\rho}\)
Since m is the mass of the liquid entering at A in a given time, therefore, pressure energy of the liquid at A is
\(E_{PA}=P_{A}V=P_{A}V\times (\frac{m}{m})=m \frac{P_{A}}{\rho}\)
Potential energy of the liquid at A, P_EA = mg h_A,
Due to the flow of liquid, the kinetic energy of the liquid at A,
\(KE_{A}=\frac{1}{2}m V_{A}^{2}\)
Therefore, the total energy due to the flow of liquid at A, E_A= EP_A+ KE_A + PE_A
\(E_{A}=m \frac{P_{A}}{\rho}+\frac{1}{2}m V^{2}_{A}+mg \ h_{A}\)
Similarly, let a_B, v_B, and P_B be the area of cross section of the tube, velocity of the liquid, and pressure exerted by the liquid at B. Calculating the total energy at EB, we get
\(EB=m \frac{P_{B}}{\rho}+\frac{1}{2}mv^{2}_{B}+mg h_{B}\)
From the law of conservation of energy,
EA = EB
\(m \frac{P_{A}}{\rho}+\frac{1}{2} mv^{2}_{A}+mgh_{A}=m\frac{P_{B}}{\rho}+\frac{1}{2}mv^{2}_{B}+mgh_{B}\)
\(\frac{P_{A}}{\rho}+\frac{1}{2}V^{2}_{A}+gh_{A}=\frac{P_{B}}{\rho}+\frac{1}{2}V^{2}_{B}+gh_{B}\) = constant
Thus, the above equation can be written as
\(\frac{P}{\rho g}+\frac{1}{2}\frac{v^{2}}{g}+h\) = constant
The above equation is the consequence of the conservation of energy which is true until there is no loss of energy due to friction. But in practice, some energy is lost due to friction. This arises due to the fact that in a fluid flow, the layers flowing with different velocities exert frictional forces on each other. This loss of energy is generally converted into heat energy. Therefore, Bernoulli's relation is strictly valid for fluids with zero viscosity or non-viscous liquids. Notice that when the liquid flows through a horizontal pipe, then \(\mathrm{h}=0 \Rightarrow \frac{P}{\rho g}+\frac{1}{2} \frac{v^{2}}{g}=\) constant.

The total energy of a satellite orbiting the Earth at a distance h from the surface of Earth is calculated as follows; The total energy of the satellite is the sum of its kinetic energy and the gravitational potential energy. The potential energy of the satellite is,
U = \(\frac { { GN }_{ s }{ M }_{ E } }{ ({ R }_{ E }+h) } \) ......(1)
Here M_s - mass of the satellite, M_E - mass of the Earth, R_E - radius of the Earth 
The Kinetic energy of the satellite is
K.E = \(\frac { 1 }{ 2 } \) M_sv²,  ......(2)
Here v is the orbital speed of the satellite and is equal to
v = \(\sqrt { \frac { { GM }_{ E } }{ (R_{ E }+h) } } \)   ......(3)
Substituting the value of v in (2) the kinetic energy of the satellite becomes,
K.E = \(\frac { 1 }{ 2 } \frac { { GM }_{ E }{ M }_{ s } }{ 2({ R }_{ E }+h) } \)
Therefore the total energy of the satellite is
E = \(\frac { 1 }{ 2 } \frac { { GM }_{ E }{ M }_{ s } }{ ({ R }_{ E }+h) } -\frac { { GM }_{ 5 }{ M }_{ E } }{ ({ R }_{ E }+h) } \)
E = \(\frac { { GM }_{ E }{ M }_{ s } }{ 2({ R }_{ E }+h) } \)   .........(4)
The negative sign in the total energy implies that the satellite is bound to the Earth and it cannot escape from the Earth.
As h approaches infinity the total energy tends to zero. Its physical meaning is that the satellite is completely free from the influence of Earth's gravity and is not bound to Earth at large distances.

The relative magnitude of forces should be indicated when the free body diagram for mass m is drawn.

Case (a): The forces F₁ and F₂ have equal length but opposite direction. So net force along y-direction is zero. Since the force is zero, acceleration is also zero along Y-direction (Newton's second law). Similarly in the x direction, F₃ and F₄ have equal length and opposite in direction. So 'net force is zero in the x direction. So there is no acceleration in x direction.
Case (b): The forces F₁ and F₂ are not equal in length and act opposite to each other. The figure (b) shows that there are unbalanced forces along the y-direction. So the particle has acceleration in the y-direction. The forces F₃ and F₄ are having equal length and act in opposite directions. So there is no net force along the x direction. So the particle has no acceleration in the x direction.
Case (c): The forces F₁ and F₂ are equal in magnitude and act opposite to each other. The net force is zero in y direction. F₃ So in y-direction there is no acceleration. F₄ The forces and F₃ are not equal in F₄ magnitude and is greater than. So there is a net acceleration in negative x direction.
Case (d): The forces F₁ and F₂ are equal in magnitude and act opposite to each other. The net force is zero in y direction. So there is no acceleration in y-direction. The forces F₃ and F₄ are not equal in magnitude. The force F₄ is greater than the force F₃. So there is a net acceleration in the positive x direction.

(i) Two vectors\(\vec A\) and \(\vec B\) are said to be equal when they have equal magnitude and same direction.
(ii) If two vectors \(\vec A\) and \(\vec B\) act in the same direction along the same line or on parallel lines, such that the angle between them is 0°. 
(iii) Two vectors are said to be anti-parallel when they are in opposite directions.
(iv) A vector divided by its magnitude is a unit vector. It has a magnitude equal to unit or one. \(\hat A=\frac{\vec A}{A}\).

Force, F = 1000 N
Displacement, s = 200 m
Angle e = 60°
Workdone =?
Workdone W = Fs cos ,
= 1000\(\times\)200\(\times\)cos 60°
= 1000\(\times\)200\(\times\)\(\frac{1}{2}\)
=1\(\times\)10⁵ J = 1\(\times\)10² kJ

Let the center of mass of two point masses m₁ and m₂, which are at positions x₁ and x₂ respectively on the X-axis. For this case, we can express the position of center of mass in the following three ways based on the choice of the coordinate system.
(i) When the masses are on positive X-axis: The origin is taken arbitrarily so that the masses m₁ and m₂ are at positions x₁ and x₂ on the positive X-axis as shown in Figure. The center of mass will also be on the positive X-axis at x_CM as given by the expression,
\({ x }_{ CM }=\frac { { m }_{ 1 }x_{ 1 }+{ m }_{ 2 }x_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \)
(ii) When the origin coincides with any one of the masses: The calculation could be minimised if the origin of the coordinate system is made to. coincide with any one of the masses as shown in Figure. When the origin coincides with the point mass m₁ its position x₁ is zero, (i.e. x₁ = 0). Then,
\({ x }_{ CM }=\frac { { m }_{ 1 }\left( 0 \right) +{ m }_{ 2 }x_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \)
The equation further simplifies as,
\({ x }_{ CM }=\frac { { m }_{ 2 }x_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \)
(iii) When the origin coincides with the center of mass itself:
If the origin of the coordinate system is made to coincide with the center of mass, then, x_CM =0 and the mass m₁ is found to be on the negative X-axis as shown in Figure. Hence, its position x₁ is negative, (i.e. -x₁).
\(0=\frac { { m }_{ 1 }\left( -{ x }_{ 1 } \right) +{ m }_{ 2 }{ x }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \)
0 = m₁(-x₁) + m₂x₂
m₁x₁ = m₂x₂
The expression given above is known as principle of moments.

For the particle to escape, K.E = P.E
\(\frac { 1 }{ 2 } { mV }_{ e }^{ 2 }=\frac { GMm }{ R+h } \)
But supphed K.E =\(=\frac { 1 }{ 2 } \times \frac { 1 }{ 2 } { mV }_{ e }^{ 2 }=\frac { GMm }{ 2R } \)
Suppose the particle rises to a height h, then
\(=\frac { 1 }{ 2 } \times \frac { 1 }{ 2 } { mV }_{ e }^{ 2 }=\frac { GMm }{ R+h } \)
\(\frac { GMm }{ R+h } =\frac { GMm }{ R+h } \)
h=R

(i) Every object emits heat radiations at all finite temperatures (except 0 K) as well as it absorbs radiations from the surroundings. For example, if you touch someone, they might feel your skin as either hot or cold.
(ii) A body at high temperature radiates more heat to the surroundings than it receives from it. Similarly, a body at a lower temperature receives more heat from the surroundings than it loses to it.
(iii) Prevost applied the idea of 'thermal equilibrium' to radiation. He suggested that all bodies radiate energy but hot bodies radiate more heat than the cooler bodies. At one point of time the rate of exchange of heat from both the bodies will become the same. Now the bodies are said to be in 'thermal equilibrium'.
(iv) Only at absolute zero temperature, a body will stop emitting.

 It states that the volume of a gn mass of a gas is inversely proportional to it pressure provided the temperature remains constant.
\(v\times \frac { 1 }{ P } (or)v=\frac { k }{ p } \) (or) pv = constant
Its value depends on
(i) mass of the gas
(ii) its temperature and
(iii) the units in which P and v are measured.
P₁ & V₁ - initial values of pressure and volume
P₂ & v₂ - Final values of pressure and volume
then accumulate the Boyle's law P₁ V₁ = P₂ v₂
Graph between P vs. V and P vs.\(\frac{1}{v}\) for a g_n mass a gas a constant temperature T are shown below.

Time period of a simple harmonic oscillator
\(T=2\pi \sqrt { \frac { m }{ k } } \)
m - mass k - force constant

a. Since k₁ and k₂ are parallel, k_u = k₁ + k₂ Similarly, k₃ and k₄ are parallel, therefore, k_d = k₃ + k₄ But k_u and k_d are in series,
therefore, \({ k }_{ eq }=\frac { { k }_{ u }{ k }_{ d } }{ { k }_{ u }+{ k }_{ d } } \)
If all the spring constants are equal then, k₁ = k₂ = k₃ = k₄ = k
Which means, k_u = 2k and k_d = 2k
Hence, \({ k }_{ eq }=\frac { { 4k }^{ 2 } }{ 4k } =k\)
b. Since k₁ and k₂ are parallel, k_A = k₁ + k₂ Similarly, k₄ and k₅ are parallel,
therefore, k_B = k₄ + k₅
But k_A, k₃, k_B, and k₆ are in series,
therefore, \(\frac { 1 }{ { k }_{ eq } } =\frac { 1 }{ { K }_{ A } } +\frac { 1 }{ { K }_{ 3 } } +\frac { 1 }{ { K }_{ B } } +\frac { 1 }{ { K }_{ 6 } } \)
If all the spring constants are equal
then, k₁ = k₂ = k₃ = k₄ = k₅ = k₆ = k
which means, k_A = 2k and k_B = 2k
\(\frac { 1 }{ { k }_{ eq } } =\frac { 1 }{ { 2K } } +\frac { 1 }{ { K } } +\frac { 1 }{ { 2K } } +\frac { 1 }{ { K } } =\frac { 3 }{ { K } } \)
\({ k }_{ eq }=\frac { k }{ 3 } \)

The slope in the position-time graph will give the speed of the particle.
In the graph (a) slope is zero. Graph (c) has higher slope than graphs (b) and (d). So we can arrange the speeds in ascending order as
v_a<v_b<v_d<v_c

m = 66.0 gm.

T ∝ m^a l^b g^c;
T = k.m^a l^b g^c
Here k is the dimensionless constant. Rewriting the above equation with dimensions
[T¹] = [M^a] [L^b] [LT^-2]^c
[M⁰L⁰T¹] = [M^aL^b+cT^-2c]
Comparing the powers of M, L and T on both sides, a = 0, b + c = 0, -2c = 1
Solving for a, b and c a = 0, b = 1/2, and c = -1/2
From the above equation T = k.m⁰ l^1/2 g^-1/2
T= k\(\left( \frac { 1 }{ g } \right) ^{ 1/2 }=k\sqrt { \frac { l }{ g } } \)
Experimentally k = 2π hence
T = \(2\pi \sqrt { l/g } \)

When the vehicle of mass m is moving with velocity v, the kinetic energy of the vehicle \(k=\left( {1\over 2} \right){mv}^{2}\) and if s is the stopping distance, then work done by friction.
W = \(fs\ \cos\theta=\mu mg\ s\ \cos 180°\)
So by work - Energy theorem, \(W=\triangle k = k_f-k_i\)
i.e, \(-\mu mgs=0-{1\over 2}{mv}^{2}\) or \(s={v^2\over 2\mu g}\)

10^-6m equivalent of 1 \(\mu\) m
1 metre is equivalent to \({{1}\over{{10}^{-6}}}={10}^{6}\mu\) m
In one metre 10⁶ microns are present