By using the triangular law of addition \(\hat i+\hat j\) as shown in the following figure,

The definition of unit vector is \(\hat A.\hat A=1\)
But here,\((i+j).(i+j)=\hat i.\hat i+\hat i.\hat j+\hat j.\hat i+\hat j.\hat j=1+0+0+1=2\)
So, \(\hat i+\hat j\) is not a unit vector.
To make any vector to a unit vector, must divide the vector by its magnitude, \(\hat A=\frac{\vec A}{|\vec A|}\).
The magnitude of the vector \(\hat i+\hat j=\sqrt 2.\)
Hence, thee unit vector is \(\frac{\hat i+\hat j}{\sqrt 2}.\)

The following figure shows the forces acting on the horizontal and vertical direction.
When the car takes turn with the speed v, the centripetal force is exerted by horizontal component of normal force and static frictional force. It is given by

N sinθ+f cosθ=\(\frac { mv^{ 2 } }{ r } \)        ......(1)
In the vertical direction, there is no acceleration. It implies that the vertical component of normal force is balanced by downward gravitational force and downward vertical component of frictional force. This can be expressed as
N cosθ=mg+f sinθ
or N cosθ-f sinθ=mg
Diving the equation (1) by equation (2), we get
\(\frac { Nsin\theta +fcos\theta }{ Ncos\theta -fsin\theta } =\frac { { v }^{ 2 } }{ rg } \)
To calculate the maximum speed for the safe turn, we can use the maximum static friction is given by. By substituting this relation in equation (3), we get
\(\frac { Nsin\theta +{ \mu }_{ s }Ncos\theta }{ Ncos\theta -{ \mu }_{ s }Nsin\theta } =\frac { { v }_{ max }^{ 2 } }{ rg } \)
By taking outside the bracket in L.H. S of equation
\(\frac { Ncos\left\{ \left( \frac { Nsin\theta }{ Ncos\theta } \right) +{ \mu }_{ s } \right\} }{ Ncos\theta \left( 1-{ \mu }_{ s }\frac { Nsin\theta }{ Ncos\theta } \right) } =\frac { { v }_{ max }^{ 2 } }{ rg } \)
\(\frac { (tan\theta +{ \mu }_{ s }) }{ 1-{ \mu }_{ s }tan\theta } =\frac { { v }_{ max }^{ 2 } }{ rg } \)
The Maximum speed for safe turn is given. by
vmax=\(\sqrt { rg\frac { (tan\theta +{ \mu }_{ s }) }{ (1-{ \mu }_{ s }tan\theta ) } } \)
Suppose we neglect the effect of friction (μ_s = 0), then safe speed
vsafe=\(\sqrt { rgtan\theta } \)
Note that the maximum speed with which the car takes safe turn is increased by friction (equation (4)). Suppose the car turns with speed v < _-safe then the stati~ friction acts up in the slope to prevent from inward skidding.
If the car turns with the speed little greater than, then the static friction acts down the slope to prevent outward skidding. But if the car turns with the speed greater than then static friction cannot prevent from outward skidding.

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.

Consider an object moving along a circle of radius r. In a time ∆t, the object travels in an arc distance ∆s as shown in figure. The corresponding angle subtended is ∆\(\theta\)
The ∆s can be written in terms of  ∆\(\theta\)

  ∆s =  r∆\(\theta\)   ...............(i)
In a time ∆t, we have
    \(\frac { \Delta s }{ \Delta t } =t\frac { \Delta \theta }{ \Delta t } \)  .......(ii)
In the limit Δt⟶0 the above equation becomes
       \(\frac { ds }{ dt } =r\omega \)   ........(iii)
Here\(\frac { ds }{ dt } \) is linear speed (v) which is tangential to the circle and \(\omega \) is angular speed. So equation (iii) becomes
v_r = rω  ........(iv)
which gives the relation between linear speed and angular speed
Eq (iv) is true only for circular motion. In general the relation between linear and angular velocity is given by
\(\vec { v } =\vec { \omega } \times \vec { r } \)...........(v)
For circular motion eq. (v) reduces to eq. (iv) since \(\vec { \omega } \) and \(\vec { r } \) are perpendicular to each other Differentiating the eq. (iv) with respect to time, we get (since r is constant)
\(\frac { dv }{ dt } =\frac { rd\omega }{ dt } =r\alpha \)
Here\(\frac { dv }{ dt } \) is the tangential acceleration and is denoted as a_t = \(\frac { d\omega }{ dt } \) is the angular acceleration Then eq.(v) becomes
a_t = r∝   ......(vii)

Consider a projectile, say a ball, thrown horizontally with an initial velocity \(\vec{u}\) from the top of a tower of height h.
As the ball moves, it covers a horizontal distance due to its uniform horizontal velocity u, and a vertical downward distance because of constant acceleration due to gravity g. Thus, under the combined effect the ball moves along the path OPA. The motion is in a 2-dimensional plane. Let the ball take-time t to reach the ground at point A, Then the horizontal distance travelled by the ball is x(t) = x, and the vertical distance travelled is y(t) =y.

We can apply the kinematic equations along the x direction and y direction separately. Since this is two-dimensional motion, the velocity will have both horizontal component U_x and vertical component u_y.
Motion along horizontal direction: The particle has zero acceleration along x direction. So, the initial velocity U_x remains constant throughout the motion.
The distance traveled by the projectile at a time t is given by the equation x = u_x  t+\(1\over 2\) at ².
Since a = 0 along x direction, we have
x = u_xt ................(i)
Motion along downward direction: Here u _y = 0 (initial velocity has no downward component), a = g (we choose the +ve y-axis in downward direction), and distance y at time t.
\(\therefore\) y=\(u_xt+{1\over2}at^2,\) we get
\(y={1\over 2}at^2\) ..............(ii)
Substituting the value of t from equation (i) in equation (ii) we have
\(y={1\over 2}g{x^2\over u^2_x}=({g\over 2u^2_x})x^2\)
y = Kx²
where K =  \(g\over 2u^2_x\) is constant.
Equation (iii) is the equation of a parabola. Thus, the path followed by the projectile is a  parabola.
1. Time of Flight: The time taken for the projectile to complete its trajectory or time taken by the projectile to hit the ground is called time of flight. Consider the example of a tower and projectile. Let h be the height .of a tower. Let T be the time taken by the projectile to hit the ground, after being thrown horizontally from the tower.
We know that S_y = u_yt +\(1\over2\) at² for vertical motion. Here S_y = h, t = T, u_y = 0 (i.e., no initial vertical velocity). Then
\(h={1\over2}gt^2 \ or \ T=\sqrt{2h\over g}\)
Thus, the time of flight for projectile motion depends on the height of the tower, but is independent of the horizontal. velocity of projection. If one ball falls vertically and another ball is projected horizontally with some velocity, both the balls will reach the bottom at the same time.

2. Horizontal range: The horizontal distance covered by the projectile from the foot of the tower to the point where the projectile hits the ground is called horizontal range.
For horizontal motion, we have
\(s_x=u_xt+{1\over2}at^2\)
Here, S_x = R (range), U_x = u, a = 0 (no horizontal acceleration) T is time of flight. Then horizontal range = uT.
Since the time of flight T =\(\sqrt{2h\over g}\) ,we substitute this and we get the horizontal range of the particle as R = u\(\sqrt{2h\over g}\)
The above equation implies that the range R is directly proportional to the initial velocity u and inversely proportional to acceleration due to gravity g.

3. Resultant Velocity (Velocity of projectile at any time): At any instant t, the projectile has velocity components along both x-axis and y-axis. The resultant of these two components gives the velocity of the projectile at that instant t
The velocity component at any t along horizontal (x-axis) is v_x= u_x+a_xt.
Since, U_x = u, a_x = 0 , we get
v_x= u
The component of velocity along vertical direction (y-axis) isv_y = u_y + a_y t
Since, u_y = 0, a_y = g, we get
v_y = gt
Hence the velocity of the particle at any instant is
\(\vec{v}=u\hat{i}+gt\hat{j}\)
The speed of the particle at any instant t is given by
\(v=\sqrt{v^2_x+v^2_y}\)
\(\therefore v=\sqrt{u^2+g^2t^2}\)
4. Speed of the projectile when it hits the ground: When the projectile hits the ground after initially thrown horizontally from the top of tower of height h, the time of flight is
\(t=\sqrt{2h\over g}\)
The horizontal component velocity of the projectile remains the same i.e V_x = u.  
The vertical component velocity of the projectile at time T is
\(v=gT=g\sqrt{2h\over g}=\sqrt{2gh}\)
The speed of the particle when it reaches the ground is
\(v=\sqrt{u^2+2gh}\)

When two objects A and B are moving with different velocities, then the velocity of one object A with respect to another object B is called relative velocity of object A with respect to B.
Case 1
Consider two objects A and B moving with uniform velocities V_A and V_B as shown, along straight tracks in the same direction \(\vec { { V }_{ A } } \) ,\(\vec { { V }_{ B } } \) with respect to ground.
The relative velocity of object A with respect to object B is \(\vec { { V }_{ AB } } =\vec { { V }_{ A } } -\vec { { V }_{ B } } \)
The relative velocity of object B with respect to object A is \(\vec { { V }_{ BS } } =\vec { { V }_{ B } } -\vec { { V }_{ A } } \)
Thus, if two objects are moving in the same direction, the magnitude of relative velocity of one object with respect to another is equal to the difference in magnitude of two velocities.
Case 2
Consider two objects A and B moving with uniform velocities V_A and V_B along the same straight tracks but opposite in direction.

The relative velocity of an object A with respect to object B is
.  \(\vec { { V }_{ AB } } =\vec { { V }_{ A } } -(-\vec { { V }_{ B } } )=\vec { { V }_{ A } } +\vec { { V }_{ B } } \)
The relative velocity of an object B with respect to object A is
   \(\vec { { V }_{ AB } } =-\vec { { V }_{ A } } -\vec { { V }_{ A } } =-(\vec { { V }_{ A } } +\vec { { V }_{ B } } )\)
Thus, if two objects are moving in opposite directions, the magnitude of relative velocity of one object with respect to other is equal to the sum of magnitude of their velocities.
Case 3
Consider the velocities \(\vec { { V }_{ B } } \) and \(\vec { { V }_{ A } } \) at an angle θ between their directions.
The relative velocity of A with respect to B, \(\vec { { V }_{ AB } } =\vec { { V }_{ A } } -\vec { { V }_{ B } } \)
Then, the magnitude and direction of is given by \({ V }_{ AB }=\sqrt { { v } _{ A }^{ 2 }+ { v } _{ B }^{ 2 }-2{ v }_{ A }{ v }_{ B }cos\theta } \)
and tan\(\beta =\frac { { v }_{ B } \ sin\theta }{ { v }_{ A }-{ v }_{ B } \ cos\theta } \)  (Here β is angle between\(\vec { { V }_{ AB } } \) and \(\vec { { V }_{ B } } \))
(i) When \(\theta \) = 0^o, the bodies move along parallel straight lines in the same direction,
We have v_AB=(v_A - v_B) in the direction of \(\vec { { V }_{ A } } \) .Obviously \({ V }_{ BA }=({ V }_{ B }+{ V }_{ A })\) in the direction of \(\vec { { V }_{B } } \)
(ii) When\(\theta \) =180^o, the bodies move along parallel straight lines in opposite directions,
We have \({ V }_{ AB }=({ V }_{ A}+{ V }_{B })\)in the direction of \(\vec { { V }_{ A } } \) 
Similarly, \({ V }_{ BA }=({ V }_{ B}-{ V }_{A })\)in the direction of \(\vec { { V }_{ B } } \)
(iii) If the two bodies are moving at right angles to each other, then \(\theta \)= 90^o.The magnitude of the relative velocity of A with respect to B = \({ V }_{ AB }=\sqrt { { v }_{ A }^{ 2 }+{ v }_{ B }^{ 2 } } \)
(iv) Consider a person moving horizontally with velocity \(\vec { { V }_{ M } } \). Let rain fall vertically with velocity \(\vec { { V }_{ R } } \). An umbrella is held to avoid the rain. Then the relative velocity of the rain with respect to the person is,

which has magnitude 
 \(\vec { { V }_{ RM } } =\vec { { V }_{ R } } -\vec { { V }_{ M } } \)
which has magnitude
\(\mathrm{v}_{\mathrm{RM}}=\sqrt{v_{R}^{2}+v_{M}^{2}}\)
and direction \(\theta ={ tan }^{ -1 }\left( \frac { { V }_{ M } }{ { V }_{ R } } \right) \)with the vertical.
In order to save himself from the rain, he should hold an umbrella at an angle θ with the vertical.

(a) Linear motion: An object is said to be in linear motion if it moves in a straight line.
Examples:
1. An athlete running on a straight track
2. AA particle falling vertically downwards to the Earth
(b) Circular motion: Circular motion is defined as a motion described by an object traversing a circular path.
Examples:
1. The whirling motion of a stone attached to a string.
2. The motion of a satellite around the Earth.
These two circular motions are shown in figure.

(c) Rotational motion: If any object moves in a rotational motion about an axis, the motion is called 'rotation'. During rotation every point in the object transverses a circular path about an axis, (except the points located on the axis).
Examples:
1. Rotation of a disc about an axis through its center
2. Spinning of the Earth about its own axis.
These two rotational motions are shown in figure.

(d) Vibratory motion: If an object or particle executes a to-and-fro motion about a fixed point, it is said to be in vibratory motion. This is sometimes also called oscillatory motion.
Examples:
1.Vibration of a string on a guitar
2. Movement of a swing
These motions are shown in figure

Other types of motion like elliptical motion and helical motion are also possible.

The principle of homogeneity of dimensions states that the dimensions of all the terms in a physical expression should be the same. For example, in the physical expression v²= u² + 2as, the dimensions of v², u² and 2 as are the same and equal to [L²T^-2].
This method is used to
(i) Convert a physical quantity from one system of units to another.
(ii) Check the dimensional correctness of a given physical equation.
(iii) Establish relations among various physical quantities.
(i) To convert a physical quantity from one system of units to another: This is based on the fact that the product of the numerical values (n) and its corresponding unit (u) is a constant. i.e, n₁[u₁] = constant (or) n, n₁[u₁ ] = n₂[u₂].
Consider a physical quantity which has dimension 'a' in mass, 'b' in length and 'c' in time.
If the fundamental units in one system are M₁, L₁ and T₁ and the other system are M₂, L_2, and T₂ respectively, then we can write, n₁ [M₁^a L₁^b T₁^c] = n₂ [ M ₂^a  L₂^b T₂^c]
We have thus converted the numerical value of physical quantity from one system of units into the other system.
Example: Convert 76 cm of mercury pressure into Nm^-2 using the method of dimensions.
Solution: 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]
\(P_{1}\left[M_{1}^{a} L_{1}^{b} T_{1}^{c}\right]=P_{2}\left[M_{2}^{a} L_{2}^{b} T_{2}^{c}\right]\)
We have
\(P_{2} =\left[\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}\right]^{a}\left[\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}\right]^{b}\left[\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right]^{c} \)
\(M_{1} =1 \mathrm{~g}, \mathrm{M}_{2}=1 \mathrm{~kg}\)
\(L_{1}=1 \mathrm{~cm}, \mathrm{~L}_{2}=1 \mathrm{~m} \)
\(T_{1}=1 \mathrm{~s}, T_{2}=1 \mathrm{~s}\)
So a=1, b=1 and c=-2
Then
\(P_{2} =76 \times 13.6 \times 980\left[\frac{\mathrm{g}}{1 \mathrm{~kg}}\right]^{1}\left[\frac{\mathrm{cm}}{1 \mathrm{~m}}\right]^{-1}\left[\frac{1 \mathrm{~s}}{1 \mathrm{~s}}\right]^{-2} \)
\(=76 \times 13.6 \times 980\left[\frac{10^{-3} \mathrm{~kg}}{1 \mathrm{~kg}}\right]^{1}\left[\frac{10^{-2} \mathrm{~m}}{1 \mathrm{~m}}\right]^{-1}\left[\frac{1 \mathrm{~s}}{1 \mathrm{~s}}\right]^{-2} \)
\(=76 \times 13.6 \times 980 \times\left[10^{-3}\right] \times 10^{2} \)
\(P_{2} =1.01 \times 10^{5} \mathrm{Nm}^{-2}\)
(ii) To check the dimensional correctness of a given physical equation:
Example: The equation \(1\over 2\) mv² = mgh can be checked by using this method as follows.
Solution: Dimensional formula for
\(\boxed{{1\over 2}mv^2=[M][LT^{-1}]^2=[ML^2T^{-2}]}\)
Dimensional formula for
\(\boxed {mgh=[M][LT^{-2}][L]=[ML^{2}T^{-2}] \\ [ML^{2}T^{2}]=[ML^{2}T^{-2}]}\)
Both sides are dimensionally the same, hence the equations\(1\over 2\) mv² = mgh is dimensionally correct.
(iii) To establish the relation among various physical quantities:
 If the physical quantity Q depends upon the quantities Q₁, Q₂ and Q₃ ie. Q is proportional to Q₁, Q₂ and Q₃.
Then,
\(Q \alpha Q_{1}^{a} Q_{2}^{b} Q_{3}^{c}  \)
\(Q=k Q_{1}^{a} Q_{2}^{b} Q_{3}^{c}\)
where k is a dimensionless constant. When the dimensional formula of Q₁, Q₂ and Q₃ are substituted, then according to the principle of homogeneity, the powers of M, L, T are made equal on both sides of the equation. From this, we get the values of a, b, c.
Example:
Obtain an expression for the time period T of a simple pendulum. The time period T depend upon (i) mass 'm' of the bob (ii) length 'l' of the pendulum and (iii) acceleration due to gravity g at the place where the pendulum is suspended. (Constant k=2π ) i.e
Solution:
\(\boxed{T \alpha m^a l^b g^c \\ T=k.m^al^bg^c}\)
Here k is the dimensionless constant. Rewriting the above equation with dimensions.
\(\boxed{[T^1]=[M^a][L^b][LT^{-2}]^c\\ [M^oL^oT^1]=[M^aL^{b+c}T^{-2c}]}\)
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. m^o l^1/2 g-^1/2
T=\(k{1\over g}^{1\over 2}=k\sqrt{1\over g}\)
Experimentally k = 2\(\pi\) , hence \(T=2\pi \sqrt{l\over g}\)

n\(\propto\) I^aT^bm^c, [I] = [M^oL¹T^o]
[T] = [M¹L¹T^-2] (force)
[M] = [M¹L^-1T^o]
[M^o L^oT^-1] = [M^oL¹T^o]^a [M¹L¹T^-2]^b [M^oL^-1T^o]^C
b + c = 0
a + b - c = 0
-2b = -1 \(\Rightarrow\) b = \(1\over2\)
c =\(-{1\over2}a=1\)
n\(\propto\) \({1\over l}{\sqrt{T\over m}}\)

A^o= 10^-10 m
Atomic volume of 1 mole of hydrogen
= Avagadro's number\(\times\)volume of hydrogen molecule
= 6.023\(\times\)10²³ \(\times\)\(4\over 3\) \(\times\) \(\pi\) x (10^-10 m)³
= 25.2\(\times\)10^-7 m³
Molar volume = 22.4 L = 22.4\(\times\)10^-3 m³
\(Molar \ volume \over Atomic\ volume\)= 0.89x10⁴\(\approx \)104
This ratio is large because actual size of gas molecule is negligible in comparison to the inter molecular separation.

The equivalent resistance of a parallel combination
\(R'=\frac{R_1R_2}{R_1+R_2}=\frac{150\times220}{150+220}=\frac{33000}{370}=89.1\ Ohm\)
We know that, \(\frac{1}{R'}=\frac{1}{R_1}+\frac{1}{R_2}\)
\(\frac { \triangle { R }^{ ' } }{ \left( { R }^{ ' } \right) ^{ 2 } } =\frac { \triangle { R }_{ 1 } }{ { R }_{ 1 }^{ 2 } } +\frac { \triangle { R }_{ 2 } }{ { R }_{ 2 }^{ 2 } } \)
\(\triangle { R }^{ ' }=\left( { R }^{ ' } \right) ^{ 2 }\frac { \triangle { R }_{ 1 } }{ { R }_{ 1 }^{ 2 } } +\left( { R }^{ ' } \right) ^{ 2 }\frac { \triangle { R }_{ 2 } }{ { R }_{ 2 }^{ 2 } } =\left( \frac { { R }^{ ' } }{ { R }_{ 1 } } \right) ^{ 2 }\triangle { R }_{ 1 }+\left( \frac { { R }^{ ' } }{ { R }_{ 2 } } \right) ^{ 2 }\triangle { R }_{ 2 }\)
Substituting the value,
\(\triangle { R }^{ ' }=\left[ \frac { 89.1 }{ 150 } \right] ^{ 2 }\times 2+\left[ \frac { 89.1 }{ 220 } \right] ^{ 2 }\times 6=0.070+0.098=0.168\)
R' = 89.1 ± 0.168 Ohm.

Let  T = Kr^aM^bG^c  ..... (1)
Where  K = a dimensionless constant
dimensions of the various quantities are [T] = T, [r] = L, [M] = M
[G] \(={Fr^2\over m_1m_2}={MLT^{-2}.L\over MM}=M^{-1}L^3T^{-2}\)
Substituting these dimensions in equation (1) we get,
[T] = [L]^a [M]^b [M^-1 L³ T^-2]^c
M^OL^oT^I=M^b-c=M^b-cL^a+3cT^-2c
Equating the dimensions of M, L and T, we get b - c = 0, a + 3c = 0, - 2c = 1
on solving a=\(3\over2\) b=\(-{1\over2}\) c=\(-{1\over2}\) T=Kr^3/2M^-1/2 G^-l/2 or T²=\({K^2R^3\over MG} \Rightarrow \therefore T^2 \propto r^3\)

(i) Given \(\lambda ={h\over mv}\)
As wavelength is a distance, 
LHS = \(\therefore [\lambda]=L\)
Also, RHS =\([{h\over mv}] ={Planck's \ constant \over mass \times velocity}={ML^2T^{-1}\over MLT^{-1}}=L\)
\(\therefore\) Dimensions of LHS = Dimensions of RHS.
Hence the given equation is dimensionally consistent.
(ii) Given  v = \(\sqrt{2GM\over R}\)
LHS = [V] = LT^-1
\(RHS=[{2GM\over R}]^{1\over2}=[{M^{1\over2}L^3T^{-2}.M \over L}]^{1\over 2}=[L^2T^{-2}]{1\over2}=LT^{-1}\)
\(\therefore\) Dimensions of LHS = Dimensions of RHS.
Hence the given equation is dimensionally correct.

As F \(=G{m_1m_2\over r^2}\)
\(G={F.r^2\over m_1m_2}\)
\([G]={{MLT}^{-2}.L^2\over MM}={M}^{-1}L^3{T}^{-2}\)
\(\therefore\) a = -1, b ~ 3, c = -2

CGS units	SI units
n1 = 6.67\(\times\) 10-8	n2 =?,
m1=1g	m2 = 1 kg=1000 g
L1 = 1cm	L2 = 1 cm = 100cm
T1=1s	T2=1s

\(\therefore\) \(n_2=n_1{\left[ {M_1 \over M_2} \right]}^{a}{\left[ {L_1 \over L_2} \right]}^{b}{\left[ {T_1 \over T_2} \right]}^{c}\)
\(=6.67\times{10}^{-8}{\left[ {{1\over 1000}} \right]}^{-1}{\left[ {{1\over 100}} \right]}^{3}\left[ {1\over 1} \right]^{-2}=6.67\times{10}^{}-11\)
Hence in SI units,   G = 6.67\(\times\)10^-11Nm² kg^-2

In the figure, Velocity of rain, \(\vec V_R=\vec OA=3.5ms^{-1}\)(vertically downwards)
Velocity of wind, \(\vec V_w=\vec OB=12 ms^{-1}\)east to west

The magnitude of the resultant velocity is
\(V=\sqrt{V_R^2+V_W^2}=\sqrt{35^2+12^2}=37\ ms^{-1}\)
Let the resultant velocity \(\vec v=(\vec {OC})\) make an angle θ with the vertical, then
\(tan\theta=\frac{AC}{OA}=\frac{V_W}{V_R}=\frac{12}{35}=0.343\)
\(\therefore\theta=tan^{-1}(0.343)\simeq19^0\)
Hence the boy should hold umbrella bending it towards east making an angle of about 19⁰ with the vertical.

Given:
F = 50N
m₁ = 5 kg
m₂ = 10 kg
m₃ = 15 kg
All the blocks move with common acceleration a under the force F = 50N
∴ F = (m₁ + m₂ + m₃)a
or a = \(\frac { F }{ { m }_{ 1 }+{ m }_{ 2 }+{ m }_{ 3 } } =\frac { 50 }{ 5+10+15 } =\frac { 5 }{ 3 } \) ms^-2
To determine T₁ Refer to the free-body diagram for m₁ shown in the diagram. Clearly, the tension T₁ produces acceleration a in mass m₁.
∴ T₁ = m₁a = \(5\times \frac { 5 }{ 3 } =\frac { 25 }{ 3 } \) = 8.33 N

To determine T₂ Refer to the free-body diagram for m₃ shown in the diagram (b). Force F acts towards right and tension T₂ acts towards left.
∴ F - T₂ = m₃ a (or) 50 - T₂ = 15 x \(\frac { 5 }{ 3 } \) or T₂ =25N

Given:
u = 100 ms^-1, g = -10 ms^-2
At highest point,v = 0
As v = u + gt
∴ 0 = 100 - 10\(\times\)t
∴ Time taken to reach highest point,
\(t=\frac{100}{10}\) =10 s
The ball will return to the ground at t = 20 s
Velocities of the ball at different instants at time will be as follows:
At t = 0, v = 100 -10\(\times\)0 = 100 ms^-1
At t = 5 s, v = 100 -10\(\times\)5 = 50 ms^-1
At t = 10 s, v = 100 - 10\(\times\)10 = 0
At t = 15 s,v = 100 -10\(\times\)15 = - 50 ms^-1
At t = 20 s, v = 100 - 10\(\times\)20 = - 100 ms^-1
The velocity-time graph will be as shown in the figure.

(i) Maximum height attained by the ball
= Area of \(\Delta\)OAB
=\(\frac{1}{2}\)\(\times\)10s\(\times\)100ms^-1= 500m
(ii) Height attained after 15 s
= Area of \(\Delta\)AOB + Area of \(\Delta\)BCD
= 500 + \(\frac{1}{2}\)(15-10) \(\times\)(-50) = 500 -125 = 375m

(i) Formula: From O to P \(Average\ velocity=\frac{Displacement}{Time\ interval}\)
\(=\frac{+360m}{18s}=+20\ ms^-1\)
\(Average\ speed=\frac{Path\ length}{Time\ interval}=\frac{360m}{18s}=20ms^{-1}\)
(ii) From O to P and back to Q.
\(Average\ velocity=\frac{Displacement}{Time\ interval}\)
\(=\frac{OQ}{18+6}=\frac{+240m}{24s}=10ms^{-1}\)
\(Average\ speed=\frac{Path\ length}{Time\ interval}\)
\(=\frac{OP+OQ}{18+6}=\frac{(360+120)}{24s}=20ms^{-1}\)

(i) Error in the difference of two quantities:
Let \(\triangle\)A and \(\triangle\)B be the absolute errors in the two quantities, A and B, respectively. Then,
Measured value of A = A \(\pm \triangle\)A
Measured value of B = B \(\pm \triangle\)B
Consider the difference, Z =A - B
The error \(\triangle\)Z in Z is then given by
\(Z\pm\triangle Z=(A\pm\triangle A)-(B\pm\triangle B)\)
\(=(A-B)\pm(\triangle A+\triangle B)\)
\(=Z\pm(\triangle A + \triangle B)\)
(or) \(\triangle Z=\triangle A+\triangle B\)
(ii) Error in the division or quotient of two quantities: Let \(\triangle\)A and \(\triangle\)B be the absolute errors in the two quantities A and B respectively.
Consider the quotient, Z = \(A\over B\)
The error \(\triangle\)Z in Z is given by \(Z\pm\triangle Z={A\pm\triangle A\over B\pm\triangle B}={A{(1\pm{\triangle A \over A})\over B{(\pm {\triangle B\over B})}}}={A\over B}(1\pm{\triangle A\over A})(1\pm{\triangle B\over B})^{-1}\)
\(or Z\pm \triangle Z=Z(1\pm {\triangle A\over A})(1\mp{\triangle B\over B}) \)        [using (1 +x)ⁿ=1+ nx, when x «1]   
Dividing both sides by Z, we get, \(1\pm{\triangle Z\over Z}=(1\pm{\triangle A\over A})(1\mp{\triangle B\over B})=1\pm {\triangle A\over A}\mp {\triangle B\over B}\pm{\triangle A\over A}{\triangle B\over B}\)
As the terms \(\triangle\)A / A and \(\triangle\)B/B are small, their product term can be neglected,
The maximum fractional error in Z is given by \({\triangle Z \over Z}=({\triangle A \over A}+{\triangle B\over B})\)
(iii) Error in the power of a quantity: Consider the n^th power of A, Z = Aⁿ The error\(\triangle\) Zin Z is given by
Z\(\pm \triangle\)Z=(A\(\pm \triangle\)A)ⁿ=Aⁿ =\(=(1\pm{\triangle A\over A})^n=Z(1\pm n{\triangle A\over A})\)
We get [(1+x)ⁿ + nx, when x« 1] neglecting remaining terms, Dividing both sides by Z
\(1\pm{\triangle Z \over Z}=1\pm n{\triangle A \over A}or{\triangle Z \over Z}=n{\triangle A \over A}\)

When a vehicle travels in a curved path, there must be a centripetal force acting on it. This centripetal force is provided by the frictional force between tyre and surface of the road. Consider a vehicle of mass 'm' moving at a speed 'v' in the circular track of radius 'r'. There are three forces acting on the vehicle when it moves as shown in the figure.
(i) Gravitational force (mg) acting downwards
(ii) Normal force (mg) acting upwards
(iii) Frictional force (F_s) acting horizontally inwards along the road
Suppose the road is horizontal then the normal force and gravitational force are exactly equal and opposite. The centripetal force is provided by the force of static friction Fs between the tyre and surface of the road which acts towards the center of the circular track,
\(\frac { m{ v }^{ 2 } }{ r } ={ F }_{ s }\)
As we have already seen in the previous section, the static friction can increase from zero to a maximum value
F_s ≤ μ_smg
There are two conditions possible namely without skidding and skidding.
For without skidding \(\frac { m{ v }^{ 2 } }{ r } \le { \mu }_{ s }mg\), or \({ \mu }_{ s }\ge \frac { { v }^{ 2 } }{ rg } \) or \(\sqrt { { \mu }_{ s }rg } \ge v\)
The static friction would be able to provide necessary-centripetal force to bend the car on the road.

Consider a projectile, say a ball, thrown horizontally with an initial velocity \(\vec u\) from the top of a tower of height h in the figure (graphical representation of horizontal projection).
Motion along horizontal direction: The particle has zero acceleration along x direction. So, the initial velocity u, remains constant throughout the motion. The distance traveled by the projectile at a time t is given by the equation \(x={ u }_{ x }t+\frac { 1 }{ 2 } a{ t }^{ 2 }\). Since a = 0 along r direction, we have
x = u_xt.................(1)

Motion along downward direction: Here u_y = 0 (initial velocity has no downward component), a = g (we choose the positive y-axis in downward direction), and distance y at time t.
∴ From the equation,\(y={ u }_{ y }t+\frac { 1 }{ 2 } a{ t }^{ 2 }\)
\(y=\frac{1}{2}gt^2\)..............(2)
Substituting the value oft from equation (1) in equation (2) we have
\(y=\frac { 1 }{ 2 } g\frac { { x }^{ 2 } }{ { u }_{ x }^{ 2 } } =\left( \frac { g }{ { 2u }_{ x }^{ 2 } } \right) { x }^{ 2 }\)
y = Kx²................(3)
where \(K=\frac { g }{ { 2u }_{ x }^{ 2 } } \) is constant
Equation (3) is the equation of a parabola. Thus, the path followed by the projectile is a parabola.
(i) Time of Flight: The time taken for the projectile to complete its trajectory or time taken by the projectile to hit the ground is called time of flight.
Consider the example of a tower and projectile. Let h be the height of a tower. Let T be the time taken by the projectile to hit the ground, after being thrown horizontally from the tower.
We know that \(s_y={ u }_{ y }t+\frac { 1 }{ 2 } a{ t }^{ 2 }\) for vertical motion. Here S_y = h, t = T, u_y = 0 (i.e., no initial vertical velocity).
Then \(h=\frac { 1 }{ 2 } g{ T }^{ 2 }\left( or \right) T=\sqrt { \frac { 2h }{ g } } \)
(ii) Horizontal range: The horizontal distance covered by the projectile from the foot of the tower to the point where the projectile hits the ground is called horizontal range. For horizontal motion, we have
\(s_x={ u }_{ x }t+\frac { 1 }{ 2 } a{ t }^{ 2 }\)
Here, s = R (range), u_x= u, a = 0 (no horizontal acceleration) T is time of flight. Then horizontal range = uT.
Since the time of flight \(T=\sqrt\frac{2h}{2},\) we substitute this and we get the horizontal range of the particle as \(R=u\sqrt\frac{2h}{2}.\)

(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. \(\hat A=\frac{\vec A}{A}\).

Consider two blocks of masses m₁ and m₂ (m₁  > m₂) kept in contact with each other on a smooth, horizontal frictionless surface as shown in the figure.

By the application of a horizontal force F, both the blocks are set into motion with acceleration 'a' simultaneously in the direction ofthe force F.
To find the acceleration \(\vec { a } \), Newton's second law has to be applied to the system (combined mass m = m₁ + m₂)
\(\vec { F } =m\vec { a } \)
If we choose the motion of the two masses along the positive x direction
\(F\hat { i } =ma\hat { i } \)
By comparing components on both sides of the above equation
F = ma where m = m₁ + m₂
The acceleration of the system is given by
∴ a = \(\frac { F }{ { m }_{ 1 }+{ m }_{ 2 } } \) ..............(1)
The force exerted by the block m₁ on m₂ due to its motion is called force of contact (\(\vec { f_{ 21 } } \)). According to Newton's third law, the block m₂ will exert an equivalent opposite reaction force (\(\vec { f_{ 12 } } \)) on block m₁..Figure shows the free body diagram of block m₁.
Figure shows the free body diagram of block m₁
∴ \(F\hat { i } ={ \vec { f } }_{ 12 }\hat { i } ={ m }_{ 1 }a\hat { i } \)
By comparing the components on both sides. of the above equation, we get
F-f₁₂ = m₁a
f₁₂ = F-m₁a      .....(2)
Substituting the value of accelera~ion from equation (1) in (2) we get
f₁₂= \(F-{ m }_{ 1 }\left( \frac { F }{ { m }_{ 1 }+{ m }_{ 2 } } \right) \)
f₁₂=\(f\left[ 1-\frac { { m }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } \right] \)
f₁₂=\(\frac { F{ m }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \)    ....(3)
Equation (3) shows that the magnitude of contact force depends on mass m: which provides the reaction force. Note that this force is acting along the negative x direction.
In vector notation, the reaction force on mass m₁ is given by \(\vec { { fi }_{ 12 } } =\frac { F{ m }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \)
For mass m₂ there is only one force acting on it in the x direction and it is denoted by f₂₁ This force is exerted by mass m₁. The free body diagram for mass m₂ is shown in the figure.

Applying Newton's second law for mass m₂
\(f_{ 21 }\hat { i } ={ m }_{ 2 }a\hat { i } \)
By comparing the components on, both sides of the above equation
f₂₁ = m₂a     .....(4)
Substituting for acceleration from equation (1) in equation (4), we get \({ f }_{ 21 }=\frac { F{ m }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \)
In this case the magnitude of the contact force is \({ f }_{ 21 }=\frac { F{ m }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \).
Free body diagram of block of mass m₂

When an object of mass m slides on a frictionless surface inclined at an angle e as shown in the figure, the forces acting on it decides the
(a) acceleration of the object 
(b) speed of the object when it reaches the bottom.
The force acting on the object is
(i) Downward gravitational force (mg)
(ii) Normal force perpendicular to inclined surface (N)

To draw the free body diagram, the block is assumed to be a point mass [in figure (a)]. Since the motion is on the inclined surface, we have to choose the coordinate system parallel to the inclined surface as shown
in the figure (b).
The gravitational force mg is resolved in to parallel component mg sine along the inclined plane and perpendicular component mg coss perpendicular to the inclined surface [figure (b)].
Note that the angle made by the gravitational force (mg) with the perpendicular to the surface is equal to the angle of inclination e' as shown in figure (c).
There is no motion (acceleration) along the y axis. Applying Newton's second law in the y direction
\(-mg\ cos\theta \hat { j } +N\hat { j } \) =0 (No acceleration)
By comparing the components on both sides, N-mg cos ፀ=0
N=mg cosፀ
The magnitude ~fnormal force (N) exerted by the surface is equivalent to mg cosፀ, Tlie object slides (with an acceleration) along the x direction. Applying Newton's second law in the x direction.
\(mg\ sin\theta\ \hat { i } =ma\hat { i } \)
By comparing the components on both sides, we can equate
mg sinፀ = ma
The acceleration of the sliding object is
a = g sinፀ

Note that the acceleration depends on the angle of inclination ፀ.

The uncertainty in a measurement is called an error. The three possible errors are 
(i) Systematic error
(ii) Random error and
(iii )Gross error
(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,
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.
Imperfections in experimental technique or procedure: These errors arise due to the limitations iri 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.
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. 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.
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 readings are obtained by a person every time he repeats the experiment, personal error 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 
\(a_m={a_1+a_2+a_3+...a_n\over n}(or)a_m={1\over n}\sum _{ i=1 }^{ i=n }{ { a }_{ i } } \)
Usually this arithmetic mean is taken as the best way to minimize the error.
(iii) Gross Error: The error caused clue to the shear carelessness of an observer is called gross error.
for example
(a) Reading an instrument without setting it properly.
(b) Taking observations in a wrong manner without bothering about the sources of errors and the precautions.
(c) Recording wrong observations.
(d) Using wrong values of the observations in calculations.
These errors can. be minimized only when an observer is careful and mentally alert.

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. 15.46g,15.42g, 15.44g.	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)

C is the centre of the Earth A and B are two diametrically opposite places on the surface of the Earth. From A and B, the parallaxes \(\theta _1\) and \(\theta _2\) respectively of Moon M with respect to some distant star are determined with the help of an astronomical telescope.
Thus, the total parallax of the Moon subtended on Earth \(\angle\)AMB = \(\theta _1\)+ \(\theta _2\) =\(\theta \)

If \(\theta\) is measured in radians, then\(\theta ={AB\over MC};AM\approx MC\)
\(\theta ={AB\over MC} or M.C={AB\over\theta}\)
Knowing the values of AB and \(\theta\) , we can calculate the distance MC of Moon from the Earth. 

Consider the horse as the 'system', then there are three forces acting on the horse
(i) Downward gravitational force (m_hg)
(ii) Force exerted by the road (F_r)
(iii) Backward force exerted by the cart (F_c)
It is shown in the following figure.
F_r - Force exerted by the road on the horse
F_c - force exerted by the cart on the horse
F_r^丄 - Perpendicular component of F_r = N
F_r^||-Parallel component of F, which is reason for forward movement.

The force exerted by .the road can be resolved into parallel and perpendicular components, The perpendicular component balances the downward gravitational force. There is parallel component along the forward direction. It is greater than the backward force (F_c). So there is net force along the forward direction which causes the forward movement of the horse.
If we take the cart as the system, then there are three forces acting on the cart.
(i) Downward gravitational force (m_cg)
(ii) Force exerted by the road (F_r')
(iii) Force exerted by the horse (F_h)

It is shown in the figure
The force exerted by the road (\(\vec { { F }_{ r } } \)) can be resolved into parallel and perpendicular components. The perpendicular component cancels the downward gravity (m_cg)
Parallel component acts backwards and the force exerted by the horse (\(\vec { { F }_{ h } } \)) acts forward. Force (\(\vec { { F }_{ h } } \)) is greater than the parallel component acting in the opposite direction. So there is an overall unbalanced force in the forward direction which causes the cart to accelerate forward.
If we take the cart + horse as a system, then there are two forces acting on the system.
(i) Downward gravitational force (m_h + m_c)g
(ii) The force exerted by the road (F_r) on the system.
It is shown in the following figure.

(iii) In this case the force exerted by the road (F_r) on the system (cart + horse) is resolved in to parallel and perpendicular components. The perpendicular component is the normal force which cancels the downward gravitational force (m_h +m_c)g. The parallel component of the force is not balanced, hence the system (cart + horse) accelerates and moves forward due to this force.

(i) Choose a suitable inertial coordinate system to analyse the problem. For most of the cases we can take Earth as an inertial coordinate system.
(ii) Identify the system to which Newton's laws need to be applied. The system can be a single object or more than one object.
(iii) Draw the free body diagram.
(iv) Once the forces acting on the system are identified, and the free body diagram is drawn, apply Newton's second law. In the left hand side of the equation, write the forces acting on the .system in vector notation and equate it to the right hand side of equation which is the product of mass .and acceleration. Here, acceleration should also be in vector notation.
(v) If acceleration is given, the force can be calculated. If the force is given, acceleration can be calculated.
By following the above steps: We fix the inertial coordinate system on the. ground as shown in the figure.

The forces acting on the mango are
(i) Gravitational force exerted by the Earth on the mango acting downward along negative y-axis.
(ii) Tension (in the cord attached to the mango) acts upward along positive y-axis.
The free body diagram for the mango is shown in the figure.
\(\vec { { F }_{ g } } =mg(-\hat { j } )=-mg\hat { j } \)
Here, mg is the magnitude of the gravitational force and \((-\hat { j } )\) represents the unit vector in negative y direction.
\(\vec { T } =T\hat { j } \)



Here T is the magnitude of the tension force and \((-\hat { j } )\) represents the unit vector in positive y direction.
\({ \vec { F } }_{ net }={ \vec { F } }_{ s }+{ \vec { T } }_{ g }=-mg\hat { j } +T\hat { j } =(T-mg)\hat { j } \)
From Newton's second law \({ \vec { F } }_{ net }=m\vec { a } \)
Since the mango is at rest with respect to us (inertial coordinate system) the acceleration is zero (\(\vec { a } =0\))
So \({ \vec { F } }_{ net }=m\vec { a } =0\)
\((T-mg)\hat { j } =0\)
By comparing the components on both sides of the above equation, we get T - mg = 0
So the tension force acting on the mango is given by T - mg
Mass of the mango m = 400g and g = 9.8 ms^-2 Tension acting on the mango is T = 0.4\(\times\)9.8 = 3.92 N.

In the Cartesian coordinate system any vector \(\vec A\) can be resolved into three components along x, y and z directions. This is shown in the figure (b) (Components of a vector in 2 dimensions and 3 dimensions).

Consider a 3-dimensional coordinate system. With respect to this a vector can be written in component form as
\(\vec { A } ={ A }_{ x }\hat { i } +{ A }_{ y }\hat { j } +A_{ z }\hat { k } \)
\(A=\sqrt { { A }_{ x }^{ 2 }+{ A }_{ y }^{ 2 }+{ A }_{ z }^{ 2 } } \)

(i) If the object is at rest and no external force is applied on the object, the static friction acting on the object is zero (f_s = 0).
(ii) If the object is at rest, and there is an external force applied parallel to the surface, then the force of static friction acting on the object is exactly equal to the external force applied on the object (f_s = F_ext). But still the static friction Is is less than μ_sN.
(iii) When object begins to slide, the static friction (f_s) acting on the object attains maximum.

If a force (F) acts on the object in a very short interval of time (M), from Newton's second law in magnitude form
Fdt = dp
Integrating over time from an initial time t_i to a final time t_f, we get
\(\int _{ i }^{ f }{ dp } =\int _{ { t }_{ i } }^{ { t }_{ f } }{ Fdt } \)
pf-pi = \(\int _{ { t }_{ i } }^{ { t }_{ f } }{ Fdt } \)
pi = initial momentum of the object at time t_i
P_t = final momentum of the object at time t_f.
pf - pi = Δp change in momentum of the object during the time interval
tf - ti = Δt
The integral \(\int _{ { t }_{ i } }^{ { t }_{ f } }{ Fdt } \)=J is called the impulse and it is equal to change in momentum of the object.
If the force is constant over the time interval, then
\(\int _{ { t }_{ i } }^{ { t }_{ f } }{ F } dt=\int _{ i }^{ f }{ dp } \) = F(t_f - t_i) = FΔt
FΔt = Δp

Galileo's experiment with. the second plane (a) at same inclination angle Cisthe first (b) with increased smoothness (c) with reduced angle of inclination (d) with zero angle of inclination
When a ball rolls from the top of an inclined plane to its bottom, after reaching the ground it moves some distance and continues to move on to another inclined plane of same angle of inclination as shown in the Figure (a). By increasing the smoothness of both the inclined planes, the ball reach almost the same height (h) from where it was released (L1) in the second plane (L2) [figure (b)]. The motion of the ball is then observed by varying the angle of inclination of the second plane keeping the same smoothness. If the angle of inclination is reduced, the ball travels longer distance in the second plane to reach the same height [figure (c)). When the angle of inclination is made zero, the ball moves forever in the horizontal direction [figure (d)]. If the Aristotelian idea were true, the ball would not have moved in the second plane even if its smoothness is made maximum since no force acted on it in the horizontal direction.

Given position of the particle is
\(r=2.00t\ \hat{t}-1.00t^2\hat{j}+3.00\hat{k}\)
Velocity: The rate of change of acceleration is called velocity
velocity v = \({{dr}\over{dt}}={d\over dt}(2.00t \ \hat{t}-1.00t^2\hat{j}+3.00\hat{k})\)
\(=\left[ 2.00\hat{i}-2.00t\ \hat{j}+0 \right]{ms}^{-1}\)
\(v=\left[ 2.00 \hat{i}-2.00t\ \hat {j}\right]{ms}^{-1}\)
Acceleration: The rate of change of position of the particle is called acceleration.
Acceleration \(a={dv \over dt}={d \over dt}(2.00\hat{i}-2.00t\hat{j})\)
\(=0-2.00\hat{j}\)
\(=-2.00\hat{j}\ {ms}^{-1}\)
At time_t t = 2s:
velocity \(=\left[ 2.00\hat{i}-2.00\hat{j} \right]{ms}^{-1}.\)
\(=\left[ 2.00\hat{i}-2.00\times2\hat{j} \right]\)
\(=\left[ 2.00\hat{i}-4.00\hat{j} \right]\)
Magnitude and direction of the particles.
\(v=\sqrt{{(2)}^{2}+{(-4)}^{2}}=\sqrt{4+16}\)
\(=\sqrt{20}=4.47{ms}^{-1}\)
If \(\theta\) is the angle which v makes with x-axis, then.
\(\tan\ \theta={{v_y}\over{v_x}}={-4 \over 2}=-2=\tan\ 63.5°\)
\(\theta=63.5°\) below the x-axis.

Time taken by person to travel 4 m
\({ t }_{ 1 }=\frac { \left( \frac { d }{ 2 } \right) }{ { v }_{ 1 } } =\frac { d }{ 2{ v }_{ 1 } } \)
\(=\frac { 4 }{ 2\times 1 } =\frac { 4 }{ 2 } =2\quad sec\)
Time taken by person to travel 3 m,
\({ t }_{ 2 }=\frac { \left( \frac { d }{ 2 } \right) }{ { v }_{ 2 } } =\frac { d }{ 2{ v }_{ 2 } } \)
\(=\frac { 4 }{ 2\times 1 } =\frac { 4 }{ 2 } =2\quad sec\)
Time taken by person to travel 3 m,
\({ t }_{ 2 }=\frac { \left( \frac { d }{ 2 } \right) }{ { v }_{ 2 } } =\frac { d }{ 2{ v }_{ 2 } } \)
\(=\frac { 3 }{ 2\times 0.7 } =\frac { 3 }{ 1.4 } =2.14\)
\(\because\) Total time = t₁ + t₂
= 2 + 2.14
= 4.14 sec.
Mean (or) average velocity,
\(\frac { d }{ { t }_{ 1 }+{ t }_{ 2 } } =\frac { 2{ v }_{ 1 }{ v }_{ 2 } }{ ({ v }_{ 1 }+{ v }_{ 2 }) } \)
\(=\frac { 2\times (1\times 0.7) }{ (1+0.7) } =\frac { 1.4 }{ 1.7 } \)
= 0.82 ms^-1

\(\overrightarrow { A } =2\hat { i } -3\hat { j } -4\hat { k } \)
\(\overrightarrow { B } =6\hat { i } -4\hat { j } -4\hat { k } \)
Sum of the two vectors \(\overrightarrow { A } \ and\ \overrightarrow { B } \) gives,
\(\overrightarrow { A } +\overrightarrow { B } =(2\hat { i } -3\hat { j } -4\hat { k } )+(6\hat { i } -4\hat { j } -4\hat { k } )\)
\(=2(8\hat { i } -\hat { j } -0)\)
\(=8\hat { i } -\hat { j } \)
Required vector,
\(=\hat { i } -(8\hat { i } -\hat { j } )\)
\(=\hat { i } -8\hat { i } +\hat { j } \)
\(=-7\hat { i } +\hat { j } \)

Speed u = 120 km/h = 120\(\times\)\(\frac { 5 }{ 18 } \)
u = 33 m/s
Final velocity v =0, distances = 200 m
We know,
v²=u² +2as
0²=(33)²+2a \(\times\) (200)
a = (-33)²/(2\(\times\)200)
\(a=\frac { 1089 }{ 400 } =2.722\quad { ms }^{ -2 }\)
As
v = u + at
0 = 33+(2.72)t
\(t=\frac { 33 }{ 2.72 } \)
t=12.13s

(i) In the case of whirling motion of a stone tied to a string, the centripetal force on the particle is provided by the tensional force on the string. In circular motion in an amusement park, the centripetal force is provided by.the tension in the iron ropes.
(ii) In motion of satellites around the Earth, the centripetal force is given by Earth's gravitational force on the satellites. Newton's second law for satellite motion is F = Earth's gravitational force = \(\frac { { mv }^{ 2 } }{ r } \)
Where r- a distance of the planet from the center of the Earth

(iii) When a car is moving on a circular track the centripetal force is given by the frictional force between the road and the tires Newton second law for this case is 

Frictional force = \(\frac { { mv }^{ 2 } }{ r } \)
m-mass of the car
v-speed of the car
r-radius of curvature of track.
Even when the car moves on a curved track, the car experiences the centripetal force which is provided by a frictional force between the surface and the tire of the car.
(iv) When the planets orbit around the Sun, they experience the centripetal force towards the center of the Sun. Here the gravitational force of the Sun acts as a centripetal force on the planets.
Newton's second law for this motion Gravitational force of Sun on the planet = \(\frac { { mv }^{ 2 } }{ r } \)

(i)	Least count error	Associated with the poor resolution of the instrument
(ii)	Instrumental errors	Associated with the faulty calibration or change in conditions
(iii)	Random errors	Getting difficult results for the same measurement done repeatedly
(iv)	Personal errors	Associated with the individual performing the experiments ie. Improper precautions, incorrect initial set up of experiment
(v)	Systematic errors	Which tends to be in the same direction

Static Friction Kinetic Friction It opposes the starting of motion It opposes the relative motion of the object with respect to the surface Independent of surface of contact Independent of surface of contact u_s depends on the nature of materials in mutual contact U_k depends on nature of materials and temperature of the surface Depends on the magnitude of applied force  Independent of magnitude of applied force It can take values from zero to It can never be zero and always equals to whatever be the speed (true <10 m/s) \({ f }_{ s }^{ max }>{ f }_{ k }\) It is less than maximal value of static friction   \({ u }_{ s }>{ u }_{ k }\) Coefficient of kinetic friction is less than coefficient of static friction

(i) The centripetal acceleration is given by \(a=\frac { { v }^{ 2 } }{ r } \) This expression explicitly depends on Moon's speed which is non trivial. We can work with the formula
ω²R_m = a_m 
(ii) a_m is centripetal acceleration of the Moon due to Earth's gravity.
ω is angular velocity
(iii) R_m the distance between Earth and the Moon, which is 60 times the radius of the Earth.
R_m = 60R = 60 \(\times\)6.4 \(\times\)10⁶ = 384 \(\times\)10⁶ m
(iv) As we know the angular velocity ω = \(\frac { 2\pi }{ T } \)and
T = 27.3 days = 27.3 \(\times\) 24 \(\times\) 60 \(\times\) 60 second
=  2.358 \(\times\) 10⁶ sec
(v) By substituting these values in the formula for acceleration 
a_m = \(\frac { (4\pi ^{ 2 })(384\times { 10 }^{ 6 }) }{ (2.358\times { 10 }^{ 6 }) } \)  = 0.00272 ms^-2
= 2.72 x 10^-3ms^-2
The centripetal acceleration of Moon towards the Earth is 0.00272 m s^-2

Consider an inclined plane on which an object is placed as shown in the figure, Let the angle which this plane makes with the horizontal be ፀ. For small angle of ፀ, the object may not slide down. As ፀ is increased, for a particular value of ፀ, the object begins to slide down. This value is called angle of repose. Hence, the angle of repose is the angle of the inclined plane with the horizontal such that an object placed on it begins to slide.

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.

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

Radius (r) of moon orbit from the centre of earth
\(r =(384,000 \mathrm{~km}) \times \frac{1000 \mathrm{~m}}{1 \mathrm{~km}} \)
= 384,000,000 m
Time Period (T) = (27 days ) \(\times \frac{24 \text { hours }}{1 \text { day }} \times \frac{60 \mathrm{~min}}{1 \mathrm{hr}} \times \frac{60 \mathrm{sec}}{1 \mathrm{~min}}\)
= 2,332,800 sec
\(velocity (v)=\frac{\text { circumference }}{\text { Time period }}=\frac{2 \pi \mathrm{r}}{\mathrm{T}}\)
\(=\frac{2 \pi \times 384,000,000 m}{2,332,800 \mathrm{~s}}=329 \pi \mathrm{m} / \mathrm{s}\)
Centripetal acceleration \(\left(a_{\perp}\right)=\frac{v^{2}}{r}=\frac{(329 \pi \mathrm{m} / \mathrm{s})^{2}}{384,000,000 \mathrm{~m}}\)
\(a_{\perp}=2.78 \times 10^{-3} \mathrm{~m} / \mathrm{s}^{2}\)
Tension due to Centripetal force
\(F_{c} =T=m a_{\perp}=7.34 \times 10^{22} \mathrm{~kg} \times 2.78 \times 10^{-3} \mathrm{~m} / \mathrm{s}^{2} \)
\(\mathrm{~T} =2.04052 \times 10^{20} N\)

For the system to be in equilibrium for block of mass M1 frictionless Tension in the string T
T = μ_s (M₃ + M₁)g
For the equilibrium of block of mass M₂
T = M₂g .....(2)
(1) = (2)
μ_s (M₃ + M₁)g = M₂g
M₃ =\(\frac { { m }_{ 2 } }{ { \mu }_{ 2 } } -{ m }_{ 1 }\) 

Max.height = \(\frac { { u }^{ 2 }{ sin }^{ 2 }\theta }{ 2g } \)
\(Range \ R=\frac { { u }^{ 2 }{ sin 2 }\theta }{ 2g } \)
Here, R = 4 x Max.height
\(\therefore \frac { { u }^{ 2 }{ sin 2 }\theta }{ g } =\frac { {4\times u }^{ 2 }{ sin }^{ 2 }\theta }{ 2g } \)
\(sin2\theta =2{ sin }^{ 2 }\theta \)
\(2sin\theta cos\theta =2{ sin }^{ 2 }\theta \)
\(\therefore \ cos\theta = sin \theta\)
\(\therefore \theta =45°\)

 \(\vec { { r }_{ 1 } } =3\hat { i } +4\hat { j } \)
\(\vec { { r }_{ 2 } } =\hat { i } +2\hat { j } \)    
\(\Delta \overrightarrow { r } =\overrightarrow { { r }_{ 2 } } -\overrightarrow { { r }_{ 1 } } \)

\(=(\hat { i } +2\hat { j } )-(3\hat { i } +4\hat { j } )\)
\(\Delta \overrightarrow { r } =-2\hat { i } +\hat { j } \)

Length of a simple pendulum \(l=20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m}\)
\(\text {Accuracy }=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}\)
Time for one oscillation \(=\frac{40}{50}=0.88\)
\(T =2 \pi \sqrt{\frac{l}{g}} ; T^{2}=4 \pi^{2}\left(\frac{l}{g}\right) \)
\(g =\frac{4 \pi^{2} l}{T^{2}}\)
\(\frac{\Delta g}{g} \times 100 ==\frac{\Delta l}{l} \times 100+2 \frac{\Delta T}{T} \times 100  \)
\(l =20 \mathrm{~cm} \Delta l=2 \mathrm{~mm}=0.2 \mathrm{~cm} \Delta T=1 \mathrm{~s}  \)
\(\frac{\Delta g}{g} =\frac{\Delta l}{l}+2 \frac{\Delta T}{T}  \)
\(=\frac{0.2}{20}+2 \times \frac{1}{40} \)
= 0.01 + 0.05 = 0.06
\(\therefore\) Percentage accuracy in the determination of g=0.06 x 100=6 %

(a) Unit: An arbitrarily chosen standard of measurement of a quantity, which is accepted internationally is called unit of the quantity.

For all parts of this problem, it will be convenient to use different coordinate systems for the two different blocks. For block 1, take the positive x -direction to be up the incline, parallel to the plane, and the positive y -direction to be perpendicular to the plane, directed with a positive upward component. Take the positive direction of the position of block 2 to be downward.
The normal component N of the contact force between block 1 and the ramp will be
N=m₁g cosθ
The net x-component of the force on block 1 is then
F_1x =T-f_friction - m₁g sinθ
where T is the tension in the string.
For the just-slipping condition, the frictional force has magnitude
f_friction = μ_sN = μ_sg cosθ
The tension in the string is the gravitational force of the suspended mass,
T=m₂g
For the just-slipping condition, the net force on block 1 must be zero. Equations (2), (3) and . (4) gives
0 = m₂g-μsm₁g cosθ-m₁gsinθ
m₂=m₁(μ_s cosθ+sinθ)

As shown in the figure, there are four forces acting on the mass m₁
(a) Downward gravitational force along the negative y-axis (m₁g)
(b) Upward normal force along the positive y-axis (N)
(c) Tension force due to mass m₂ along the positive x axis
(d) Frictional force along the negative x-axis
Since the mass m₁ has no vertical motion, m₁g = N

Free body diagram for mass m₁

To determine whether the mass m₁ moves on the surface, calculate the maximum static friction exerted by the table on the mass m₁ If the tension on the mass m₁ is equal to or greater than this maximum static friction, the object will move.
f_s^max = μ_sN = μ_sm₁g
f_s^max = 0.9\(\times\)7\(\times\)9.8 =61.74 N
The tension T = m₂g= 5\(\times\)9.8 = 49 N
T < f_s^max
The tension acting on the mass mi is less than the maximum static friction. So the mass m₁ will not move.
To move the mass m₁, T > f_s^max where T = m₂g
m₂ = \(\frac { { \mu }_{ s }{ m }_{ 1 }g }{ g } \)=μ_sm₁
m₂ = 0.9\(\times\)7 = 6.3 kg
If the mass m₂ is 6.3 kg then the mass m₁ will begin to slide. Note that if there is no friction on the surface, the mass m₁ will move for m₂ even for just 1 kg.
The values of coefficient of static friction for pairs of materials are presented in Table 3.1. Note that the ice and ice pair have very low coefficient of static friction. This means a block of ice can move easily over another block of ice.

Let us take two points given by
x₁= 2 and x₂ = 3, then y₁ = 4 and y₂ = 9
Here \(\Delta x\) = 1 and \(\Delta y\) = 5.
Then \(\frac{\Delta y}{\Delta x}=\frac{9-4}{3-2}=5\)
If we take x₁= 2 and x₂ = 2.5, then y₁= 4 and y₂ = (2.5)² = 6.25
Here \(\Delta x=0.5=\frac{1}{2}\) and \(\Delta y\) = 2.25
Then \(\frac{\Delta y}{\Delta x}=\frac{6.25-4}{0.5}=4.5\)
If we take x₁= 2 and x₂ = 2.25, then y₁= 4 and y₂ = 5.0625.
Here \(\Delta x=0.25=\frac{1}{4},\Delta y=1.0625\)
\(\frac{\Delta y}{\Delta x}=\frac{5.0625-4}{0.25}=\frac{(5.0625-4)}{\frac{1}{4}}\) = 4(5.0625 - 4) = 4.25
If we take x₁= 2 and x₂ = 2.1, then y₁= 4 and y₂ = 4.41.
Here \(\Delta x=0.1=\frac{1}{10}\)
and \(\frac{\Delta y}{\Delta x} = \frac{(4.41-4)}{\frac{1}{10}}\) = 10(4.41 - 4) = 4.1
These results are tabulated as shown below:

x1	x2	\(\Delta x\)	y1	y2	\(\frac{\Delta y}{\Delta x}\)
2	2.25	0.25	4	50625	4.25
2	2.1	0.1	4	4.41	4.1
2	2.01	0.01	4	4.0401	4.01
2	2.001	0.001	4	4.004001	4.001
2	2.0001	0.0001	4	4.00040001	4.0001

From the above table, the following inferences can be made.
(i) As \(\Delta x\) tends to zero, \(\frac{\Delta y}{\Delta x}\) approaches the limit given by the number 4.
(ii) At a point x = 2, the derivative \(\frac{dy}{dx}=4\)
(iii) It should also be mentioned here that \(\Delta x\rightarrow 0\) does not mean that \(\Delta x =0\)
This is because, if we substitute \(\Delta x=0,\frac{\Delta y}{\Delta x}\) because indeterminate.
In general, we can obtain the derivative of the function y = x², as follows:
\(\frac{\Delta y}{\Delta x}=\frac{(x+\Delta x)^2-x^2}{\Delta x}=\frac{x^2+2x\Delta x+\Delta x^2-x^2}{\Delta x}=\frac{2x\Delta x+\Delta x^2}{\Delta x}=2x+\Delta x\)
\(\frac { dy }{ dx } =\lim _{ \Delta x\rightarrow 0 }{ 2x+\Delta x=2x } \)

(a) Earth alone as a system: Earth orbits the Sun due to gravitational attraction of the Sun. If we consider Earth as a system, then Sun's gravitational force is an external force. If we take the Moon into account, it also exerts an external force on Earth.

(b) (Earth + Sun) as a system: In this case, there are two internal forces which form an action and reaction pair-the gravitational force exerted by the Sun on Earth and gravitational force exerted by the Earth on the Sun.

(c) Our body as a system: While walking, we exert a force on the Earth and Earth exerts an equal and opposite force on our body. If our body alone is considered as a system, then the force exerted by the earth on our body is external.

(d) (Our body + Earth) as a system: In this case, there are two internal forces present in the system. One is the force exerted by our body on the Earth and the other is the equal and opposite force exerted by the Earth on our body.

Let us assume that the Energy E depends on mass m and velocity of light c.
\(E\alpha m^ac^b\)
\(E=km^ac^b\) where K a constant
Dimensions of E = [ML²T^-2]
Dimensions of m = [M]
Dimensions ofc = [LT^-1]
Substituting the values in the above equation
[ML²T^-2] = K[M]^a [LT^-1]^b
By equating the dimensions
a = 1
b = 2
-b = -2
E = k.mc²
The value of constant k = 1
E = mc². This is Einstein's mass energy relation.

t₁ = 2.63s, t₂ = 2.56s,
t₃ = 2.42s, t₄ = 2.71s, t₅ = 2.80s
(i) T_m =\({t_1+t_2+t_3+t_4+t_5\over 5}={2.63+2.56+2.42+2.71+2.80\over 5}\)
T_m = 2.62s (Rounded off to 2^nd decimal place)
(ii) Absolute error
\(\triangle\)T = Tm - t
\(\triangle\)T_I = 2.62 - 2.63 = +0.01s
\(\triangle\)T₂ = 2.62 - 2.56 = +0.06s
\(\triangle\)T₃ = 2.62 - 2.42 = +0.20s
\(\triangle\)T₄ = 2.62 - 2.71 = +0.09s
\(\triangle\)T₅ = 2.62 - 2.80 = +0.18s
(iii) Mean absolute error = \({\sum |\triangle T_1|\over n}\)
\(\triangle\)T_m = \({0.01+0.06+0.20+0.09+0.18\over 5}\)
\(\triangle\)T_m​​​​=\({0.54\over 5}=0.108s=0.11s\) (Rounded off to 2^nd decimal place)
(iv) Relative error: S_T =\({\triangle T_m\over T_m}={0.11\over 2.62}=0.0419\)
S_T = 0.04
(v) Percentage error in T = 0.04\(\times\)100% = 4%
(vi) Time period of simple pendulum = T = (2.62 ± 4%)s

u = 20 ms^-1.
\(\theta\)  = 30°
Distance of goal Post = 40 m
Range  = \(\frac { { u }^{ 2 }sin2\theta }{ g } \)
\(=\frac {400\times \sqrt{3}}{9.8 \times2}=35.35m\)
The ball will not reach the goal post.

Frequency of a vibrating body \(\gamma \alpha \frac{1}{l} \sqrt{\frac{F}{M}}\)
\( a\text { Dimension of frequency } =\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{-1}  \)
\(\text {Dimension of length } =\mathrm{L}=\mathrm{M}^{0} \mathrm{LT}^{0}  \)
\(\text {Dimension of Force } =\mathrm{MLT}^{-2}  \)
\(\text {Dimension of Mass } =\mathrm{M}^{1} \mathrm{~L}^{0} \mathrm{~T}^{0}  \)
\(\text {Frequency } \gamma =x  \)
\(\gamma =\mathrm{K}\left[\mathrm{F}^{x}\right][\mathrm{M}]^{y}[\mathrm{~L}]^{z}\)
Using dimensions we get
\(\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{-1}=\left[\mathrm{MLT}^{-2}\right]^{x}[\mathrm{M}]^{y}[\mathrm{~L}]^{z}  \)
\(\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{-1}=\mathrm{M}^{x+y} \mathrm{~L}^{x+z} \mathrm{~T}^{-2 x}\)
Comparing the powers we get
x + y = 0  z = -x = \(\frac{1}{2} \)
x + z = 0  y = -x = \(\frac{-1}{2} \)
-2 x = -1 
\(\therefore x =\frac{-1}{-2}=\frac{1}{2}  \)
\(\therefore \gamma =1 \times[\mathrm{F}]^{1 / 2}[\mathrm{M}]^{-1 / 2}[\mathrm{~L}]^{-1 / 2}  \)
\(\gamma =\frac{1}{l} \sqrt{\frac{F}{m}}\)

The speed of sound in water v = 1460 ms^-1
Time taken t = 80 s
\(=\frac{2 d}{t} \)
2d = u x t 
= 1460 x 80 
2d =11680 m
\(\therefore \text { Distance of enemy submarine } d =\frac{11680}{2} \)
= 58.40 km