The linear acceleration along the inclined plane can be computed by
\(a=\frac{g sin\theta}{1+\frac{K^{2}}{R^{2}}}\)
For a thin uniform circular ring, axis passing through its center is I = MR².
\(\therefore K^{2}=R^{2} \Rightarrow \frac{K^{2}}{R^{2}}=1\)
And the angle of inclination, \(\theta=45 ^{0}\)
\(\Rightarrow (sin 45^{0}=\frac{1}{\sqrt{2}})\)
Hence, \(a=\frac{g+\frac{1}{\sqrt{2}}}{1+1}\)
\(a=\frac{g}{2\sqrt{2}}\)ms^-2.

For both persons, the Kinematic equations are the same, with u=0,a_c =g and a_moon  =\(g\over 6\)
a_e = g and a_m =\(g\over 6\)
For a person on earth, V_earth = \(\sqrt{2gh}=\sqrt{2\times 10\times 100}\)
Hence, V_earth =\(\sqrt{2000} ms^{-1}\) gives the velocity at the ground, on earth.
Similarly, for a person on the moon,
 V_earth =\(\sqrt{2gh\over6}={\sqrt{2000}\over \sqrt{6}}ms^{-1}\)
The person on earth reaches ground with greater velocity than the person on the moon.

(i) Parallel axis theorem states that the moment of inertia of a body about any axis is equal to the sum of its moment of inertia about a parallel axis through its center of mass and the product of the mass of the body and the square of the perpendicular distance between the two axes.
(ii) If I_C is the moment of inertia of the body of mass M about an axis passing through the center of mass, then the moment of inertia I about a parallel axis at a distance d from it is given by the relation,
I = I_C + Md²
(iii) Let us consider a rigid body as shown in Figure. Its moment of inertia about an axis AB passing through the center of mass is I_C DE is another axis parallel to AB at a perpendicular distance d from AB. The moment of inertia of the body about DE is I. We attempt to get an expression for I in terms of I_C For this, let us consider a point mass m on the body at position x from its center of mass.

(iv) The moment of inertia of the point mass about the axis DE is, m(x + d)². The moment of inertia I of the whole body about DE is the summation of the above expression.
\(I=\sum { m\left( x+d \right) ^{ 2 } } \)
This equation could further be written as,
\(I=\sum { m\left( { x }^{ 2 }+{ d }^{ 2 }+2xd \right) } \)
\(I=\sum { \left( { mx }^{ 2 }+m{ d }^{ 2 }+2dmx \right) } \)
\(I=\sum { { mx }^{ 2 }+\sum { m{ d }^{ 2 } } +2d\sum { mx } } \)
(v) Here, \(\sum { mx^{ 2 } } \) is the moment of inertia of the body about the center of mass. Hence, 
I_C = \(\sum { mx= } 0\) because, x can take positive and negative values with respect to the axis AB. The summation \(\left( \sum { mx } \right) \) will be zero.
Thus, I = I_c + \(\sum { md^{ 2 } } \) = I_C + \(\left( \sum { m } \right) d^{ 2 }\)
(vi) Here, \(\sum { m } \) is the entire mass M of the object \(\left( \sum { m=M } \right) \)
I = I_C + Md²
Hence the parallel axis theorem is proved.

Work-Kinetic Energy Theorem
Work and energy are equivalents. This is true in the case of kinetic energy also. To prove this, let us consider a body of mass m at rest on a frictionless horizontal surface.
The work (W) done by the constant force (F) for a displacement (s) in the same direction is,
W = Fs
The constant force is given by the equation,
F = ma
The third equation of motion can be written as,
\(v^{2} =u^{2}+2 a s  \)
\(a =\frac{v^{2}-u^{2}}{2 s}\)
Substituting for a in equation (2),
\(F=m\left(\frac{v^{2}-u^{2}}{2 s}\right)\)
Substituting equation (2), (1)
\(w=m\left(\frac{v^{2}}{2 s} s\right)-m\left(\frac{u^{2}}{2 s} s\right)  \)
\(w=\frac{1}{2} m v^{2}-\frac{1}{2} m u^{2}\)
The expression for kinetic energy:
The term \(\left(\frac{1}{2} m v^{2}\right)\) in the above equation is the kinetic energy of the body of mass (m) moving with velocity(v).
\(K E=\frac{1}{2} m v^{2}\)
Kinetic energy of the body is always positive. From equations (4) and (5)
\(\Delta K E =\frac{1}{2} m v^{2}-\frac{1}{2} m u^{2}  \)
\(\text {Thus, } W =\Delta K E\)
The expression on the right hand side (RHS) of equation (6) is the change in kinetic energy (\(\Delta\)KE) of the body.
This implies that the work done by the force on the body changes the kinetic energy of the body, This is called work-kinetic energy theorem.
The work-kinetic energy theorem implies the following.
1. If the work done by the force on the body is positive then its kinetic energy increases.
2. If the work done by the force on the body is negative then its kinetic energy decreases.
3. If there is no work done by the force on the body then there is no change in its kinetic energy, which means that the body has moved at constant speed provided its mass remains constant.

(i) One of the important applications is suitcases with rolling on coasters. Rolling wheels makes it easier than carrying luggage.
(ii) When an object moves on a surface, essentially it is sliding on it. But wheels move on the -surface through rolling motion.
(iii) In rolling motion when a wheel moves on a surface, the point of contact with surface is always at rest.
(iv) Since the point of contact is at rest, there is no relative motion between the wheel and surface. Hence the frictional force is very less. At the same time if an object moves without a wheel, there is a relative motion between the object and the surface.
(v) As a result frictional force is larger. This makes it difficult to move the object.
(vi) Ideally in pure rolling, motion of the point of contact with the surface should be at rest, but in practice it is not so.
(vii) Due to the elastic nature of the surface at the point of contact there will be some deformation on the object at this point on the wheel or surface.
(viii) Due to this deformation, there will be minimal friction between wheel and surface. It is called 'rolling friction'. In fact, 'rolling friction' is much smaller than kinetic friction.

Scalar product:
Scalar product or dot product of two vectors in defined as the product of the magnitudes of both the vectors and the cosine of the angle between them.
If \(\vec{A}\) and \(\vec{B}\) are two vector having an angle \(\theta\) between them, then \(\vec{A} \vec{B}=A B \cos \theta\) where A and B are magnitudes of \(\vec{A}\) and \(\vec{B}\) .
Example: work, energy and electric flux.
Properties:
1. \(\vec{A}\).\(\vec{B}\)  is always a scalar. It is positive if \(\theta\)<90 and it is negative if \(90^{\circ}<\theta<180^{\circ}\)
2. When the vectors are parallel, \(\theta=0^{\circ} \ and \ \cos 0^{\circ}=1 \therefore(\vec{A} \cdot \vec{B})_{\text {mat }}=A B\).
3. When the vectors are anti-parallel, \(\theta=180^{\circ}\ and \ \cos 180^{\circ}=-1 \therefore(\vec{A} \cdot \vec{B})_{\min }=-A B\)
4. When the vectors are perpendicular to each other, \(\theta=90^{\circ}\ and \ \cos 90^{\circ}=0\therefore \vec{A} \vec{B}=0\).
5. Scalar product is commutative i.e, \(\vec{A} \cdot \vec{B}=\vec{B} \cdot \vec{A}\)
6. It obeys distributive law i.e., \(\vec{A} \cdot(\vec{B}+\vec{C})=\vec{A} \cdot \vec{B}+\vec{A} \cdot \vec{C}\)
7. Self dot product is given by \(\vec{A} \cdot \vec{A}=A A \cos \theta=A^{2}, \ here\ \theta=0^{\circ}\). The magnitude of the vector \(\vec{A}\ is \ (\vec{A})=A=\sqrt{\vec{A} \cdot \vec{A}}\)
8. In the case of orthogonal unit vectors \(\vec{i}, \vec{j} \ and \ \vec{k}\)
\(\hat{i} \hat{j}=\hat{j} \hat{j}=\hat{k} \cdot \hat{k}=1 \text { and } \)
\(\vec{i} \cdot \vec{j}=\hat{j} \hat{k}=\hat{k} \hat{i}=0\)
9. The angle between the vectors \(\theta=\cos ^{-1}\left[\frac{\vec{A} \cdot \vec{B}}{A B}\right]\)
10. In terms of components,
\(\vec{A} \cdot \vec{B} =\left(A_{x} \hat{i}+\mathrm{A}_{y} \hat{j}+A_{z} \hat{k}\right)\left(B_{x} \hat{i}+\mathrm{B}_{y} \hat{j}+B_{z} \hat{k}\right) \)
\(=A_{x} B_{x}+A_{y} B_{y}+A_{i} B_{z} \text {, with all other terms zero. }\)
The magnitude of A is given by \(|\vec{A}|=A=\sqrt{A_{x}^{2}+A_{y}^{2}+A_{2}^{2}}\) and \(|\vec{B}|=B=\sqrt{B_{x}^{2}+B_{y}^{2}+B_{2}^{2}}\) Vector product:
The vector product or cross product of two vectors is defined as another vector having a magnitude equal to the product of the magnitudes of two vectors and the sine of the angle between them.
If \(\vec{A}\) and \(\vec{B}\) are two vectors, then \(\vec{A} \times \vec{B}=\vec{C}=(A B \sin \theta) \hat{n}\).
The direction \(\hat{n}\ of \ \vec{A} \times \vec{B}\) is perpendicular to the plane containing the vectors \(\vec{A}\) and \(\vec{B}\) and is determined by the right hand screw rule or right hand thumb rule.
Example: Torque \(\tau=\vec{r} \times \vec{F}\) and Angular momentum \(\vec{L}=\vec{r} \times \vec{p}\)
Properties:
1. The resultant of the vector product is always another vector whose direction is perpendicular to the plane containing these two vectors \(\vec{A}\) and \(\vec{B}\) even though the vectors \(\vec{A}\) and \(\vec{B}\) may or may not be mutually orthogonal.
2. It is not commutative. \(\vec{A} \times \vec{B} \neq \vec{B} \times \vec{A}\). But \(\vec{A} \times \vec{B}=-[\vec{B} \times \vec{A}]\)
3. When the vectors \(\vec{A}\) and \(\vec{B}\) are orthogonal to each other the vector product will have maximum magnitude as \(\theta=90^{\circ}\  and \ \sin \theta=1\).
\((\vec{A} \times \vec{B})_{\max }=A B \hat{n}\)
4. The vector product of two non-zero vectors will be minimum when (sin \(\theta\))=0, i.e., \(\theta=0^{\circ} \ or \ 180^{\circ}(\vec{A} \times \vec{B})_{\min }=0\).
It means that the vector product of two non-zero vectors vanishes if the vectors are parallel or anti parallel.
5. The self-cross product is a null vector. \(\vec{A} \times \vec{A}=A A \sin 0^{\circ} \hat{n}=\overrightarrow{0}\)
6. The self-vector products of unit vectors are then zero \(\hat{i} \times \hat{i}=\hat{j} \times \hat{j}=\hat{k} \times \hat{k}=0\).
7. In the case of orthogonal unit vectors, \(\vec{i}, \vec{j} \ and \ \hat{k}\)
\(\hat{i} \times \hat{j}=\hat{k}, \hat{j} \times \hat{k}=\hat{i} \ and \ \hat{k} \times \hat{i}=\hat{j}\) and
\(\hat{j} \times \hat{i}=-\hat{k}, \hat{k} \times \hat{j}=-\hat{i} \ and \ \hat{i} \times \hat{k}=-\hat{j}\)
8. In terms of components,
\(\vec{A} \times \vec{B}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ A_{x} & A_{y} & A_{z} \\ B_{x} & B_{y} & B_{z} \end{array}\right|=\begin{array}{r} +\hat{i}\left(A_{y} B_{z}-A_{z} B_{y}\right) \\ +\hat{j}\left(A_{z} B_{x}-A_{x} B_{z}\right) \\ +\hat{k}\left(A_{0} B_{y}-A_{z} B_{x}\right) \end{array}\)
9. If two vectors \(\vec{A}\) and \(\vec{B}\) form adjacent sides of a parallelogram, then magnitude \((\vec{A} \times \vec{B})\) is equal to the area of the parallelogram.
10. If two vectors \(\vec{A}\) and \(\vec{B}\) are represented by the two sides of a triangle taken in order, then the area of the triangle is equal to \(\frac{1}{2}|\vec{A} \times \vec{B}|\)

i) Systematic errors
Systematic errors are reproducible inaccuracies that are consistently in the same direction. These occur often due to a problem that persists throughout the experiment. Systematic errors can be classified as follows
1) Instrumental errors
When an instrument is not calibrated properly at the time of manufacture, instrumental errors may arise. If a measurement is made with a meter scale whose end is worn out, the result obtained will have errors. These errors can be corrected by choosing the instrument carefully.
2) Imperfections in experimental technique or procedure
These errors arise due to the limitations in the experimental arrangement. As an example, while performing experiments with a calorimeter, if there is no proper insulation, there will be radiation losses. This results in errors and to overcome these, necessary correction has to be applied.
3) Personal errors
These errors are due to individuals performing the experiment, may be due to incorrect initial setting up of the experiment or carelessness of the individual making the observation due to improper precautions.
4) Errors due to external causes
The change in the external conditions during an experiment can cause error in measurement. For example, changes in temperature, humidity or pressure during measurements may affect the result of the measurement.
5) Least count error
Least count is the smallest value that can be measured by the measuring instrument, and the error due to this measurement is least count error. The instrument's resolution hence is the cause of this error. Least count error can be reduced by using a high precision instrument for the measurement.
ii) Random errors
Random errors may arise due to random and unpredictable variations in experimental conditions like pressure, temperature, voltage supply etc. Errors may also be due to personal errors by the observer who performs the experiment. Random errors are sometimes called "chance error". When different landings are obtained by a person every time he occurs. For example, consider the case of the thickness of a wire measured using a screw gauge. The readings taken may be different for different trials. In this case, a large number of measurements are made and then the arithmetic mean is taken.
If n number of trial readings are taken in an experiment, and the readings are a₁, a₂, a₃,...............a_n. The arithmetic mean is
\( a_{m}=\frac{a_{1}+a_{2}+a_{3}+\ldots \ldots \ldots \ldots \ldots \ldots a_{n}}{n} \text { or } \)
\(a_{m}=\frac{1}{n} \sum_{i=1}^{t=n} a_{i}\)
Usually this arithmetic mean is taken as the best possible true value of the quantity.
Certain procedures to be followed to minimize experimental errors, along with examples are shown in Table.
Type of error Example How to minimize it Random error Suppose you measure the mass of a ring three times using the same balance and get slightly different values. 15.46 g, 15.42 g, 15.44 g Take more data. Random errors can be evaluated through statistical analysis and can be reduced by averaging over a large number of observations. Systematic error Suppose the cloth tape measure that you use to measure the length of an object has been stretched out from years of use. (As a result all of the length measurements are not correct). Systematic errors are difficult to detect and cannot be analysed statistically, because all of the data is in the same direction. (Either too high or too low) iii) Gross Error
The error caused due to the sheer carelessness of an observer is called gross error.
For example
(i) Reading an instrument without setting it properly.
(ii) Taking observations in a wrong manner without bothering about the sources of errors and the precautions.
(iii) Recording wrong observations.
(iv) Using wrong values of the observations in calculations.
These errors can be minimized only when an observer is careful and mentally alert.

(i) Torque about point D is zero. (as F passes through D).
Torque about point C is zero. (as F passes through C).
Torque about point O is \((\overrightarrow {OC})\times \overrightarrow{F}\) and direction is along GH.
Torque about point B is \((\overrightarrow {BD})\times \overrightarrow{F}\) and direction is along GH.
(The perpendicular distance of \(\overrightarrow {BD}\) with respect to \(\overrightarrow {F}\) is \(\overrightarrow {OC}\).
(ii) Torque about axis CD is zero (as F is parallel to CD).
Torque about axis OC is zero (as F intersects OC).
Torque about axis AB is zero (as F is parallel to AB).
Torque about axis GH is  \((\overrightarrow {OC})\times \overrightarrow{F}\) and direction is along GH.

A collection of forces is said to be concurrent, if the lines of forces act at a common point. Concurrent forces need not be in the same plane. If they are in the same plane, they are concurrent as well as coplanar forces.

A body, under the action of concurrent forces, is said to be in equilibrium, when there is no change in the state of rest or of uniform motion along a straight line.
The necessary condition for the equilibrium of a body under the action of concurrent forces is that the vector sum of all the forces acting on the body must be zero.
Lami's Theorem: For concurrent forces \(\frac { { F_{ 1 } } }{ sin\ \alpha } =\frac { { F_{ 2 } } }{ sin \ \beta } =\frac { { F_{ 3 } } }{ sin\ \gamma } \). If a system of three concurrent and coplanar forces is in equilibrium, then Lami's theorem states that the magnitude of each force of the system is proportional to sine of the angle between the other two forces. The constant of proportionality is same for all three forces.

In cgs system 76 cm of mercury pressure = 76\(\times\)13.6\(\times\)980 dyne cm^-2
The dimensional formula of pressure P is [ML^-1T^-2]; so P₁ [M₁^a L₁^b T₁^c ]= P₂[M₂^a L₂^bT₂^c]
We have P₂ = P₁ \([{M_1\over M_2}]^a[{L_1\over 2}]^b[{T_1\over T_2} ]^c\)
M₁ = 1g, M₂ =1kg
L₁ = 1 cm, L₂ = 1m
T₁ = 1 s, T₂ = 1s
So, a = 1, b = -1, and c =-2
Then, P₁ = 76\(\times\)13.6\(\times\)980
\([{1g\over 1kg}]^1[{1cm\over 1m}]^{-1}[{1s\over 1s}]^{-1}=76\times13.6\times980[{10^{-3}kg\over 1kg}]^1[{10^{-2}m\over1m}]^{-1}[{1s\over 1s}]^{-2}\)
= 76\(\times\)13.6\(\times\)980\(\times\)[10^-3]\(\times\)10²
P₂ = 1.01\(\times\)10⁵ Nm^-2