In longitudinal wave motion, the constituent of the medium oscillate or vibrate about their mean positions in a direction parallel to the direction of propagation (direction of energy transfer) of waves.
Example: Sound waves travelling in air.

\({ K }_{ 1 }=\frac { 2\pi }{ 1 } =6.28 \ rad \ { m }^{ -1 }\)
\({ K }_{ 2 }=\frac { 2\pi }{ 6 } =1.05 \ rad \ { m }^{ -1 }\)

Force constant of a spring is defined as the restoring force per unit length.

RMS velocity \(\mathrm{C} =\sqrt{\frac{3 R T}{M}} \)
\(\mathrm{T}=\mathrm{T}_{1} =273 \mathrm{~K}  \)
\(\therefore C =\sqrt{\frac{3 R \times 273}{M}}\)
When the RMS velocity is tripled
\(3 \mathrm{C}=\sqrt{\frac{3 R T}{M}}\)
Dividing equation is (2) by (1) we get
\(\frac{3 C}{C} =\sqrt{\frac{3 R T / M}{3 R \times 273 / M}}  \)
\(3 =\sqrt{\frac{T}{273}}  \)
\(\therefore 9 =\frac{T}{273}  \)
\(\therefore T =273 \times 9=2457 \mathrm{~K}\)
\(\therefore\) Temperature \(\mathrm{T}_{2}=2457 \mathrm{~K}, \mathrm{~T}_{1}=273 \mathrm{~K}\)

The minimum number of independent coordinates needed to specify the position and configuration of a thermodynamic system in space is called degree of freedom of the system.

Pressure arises due to momentum transfer to the wall of the container.

One calorie is defined as the amount of energy required to raise 1 gram of an object by 1^oc.

Two systems are said to be in thermal equilibrium with each other if they are at the same temperature, will not change with time.

(i) Acceleration to Boyle's law P \(\alpha\)\(\frac { 1 }{ V } \)
(ii) Acceleration to Charles' law V \(\alpha\) T. By combining these two equations we have PV = CT. Here C is a positive constant. 
(iii) So we can write the constant C as k times the number of particles N.
Here k is the Boltzmann constant (1.381\(\times\)10^-23 JK^-1) and it is found to be a universal constant. So the ideal gas law can be stated as follows PV = NKT

When oil comes out through a tin with one hole, the pressure inside the tin becomes less than the atmospheric pressure, soon the oil stops flowing out. When two holes are made in the tin, air keeps on entering the tin, through the other hole and maintains pressure inside.

Angle of contact is defined as the angle between the tangent to the liquid surface at the point of contact and the solid surface inside the liquid.

Pascal's law states that, if the pressure in a liquid is changed at a particular point, the change is transmitted to the entire liquid without being diminished in magnitude.

Hooke's law of elasticity states that within the elastic limit, the strain produced in a body is directly proportional to the stress applied.
strain ∝ stress

Here, the elastic modulus is Young modulus and due to stretching, stress is tensile stress and strain is tensile strain.
Within the elastic limit, stress is proportional to strain (obey Hooke’s law). Therefore, it shows a straight line behaviour. So, Young modulus can be computed by taking slope of these straight lines. Hence, calculating the slope for the straight line, we get
Slope of A > Slope of B > Slope of C
Which implies,
Young modulus of C < Young modulus of B < Young modulus of A
Notice that larger the slope, lesser the strain (fractional change in length). So, the material is much stiffer. Hence, the elasticity of wire A is greater than wire B which is greater than C. From this example, we have understood that Young’s modulus measures the resistance of solid to a change in its length.

The weight of an object is defined as the downward force whose magnitude W is equal to the upward force that must be applied to the object to hold it at rest or at constant velocity relative to the Earth.

Yes, the angular momentum of a planet is conserved.
During the orbitary motion of the planets around the Sun, the line of action of gravitational force passes through the axis, the external torque is zero. Hence the angular momentum is conserved.
Torque,  \( { \tau } =\frac { dL }{ dt }\)
If torque is zero then angular momentum (L) is constant i.e., it is conserved.

1. Law of orbits
Each planet moves around the Sun in an elliptical orbit with the Sun at one of the foci.
2. Law of area
The radial vector (line joining the Sun to a planet) sweeps equal areas in equal intervals of time.
3. Law of period
The square of the time period of revolution of a planet around the Sun in its elliptical orbit is directly proportional to the cube of the semi major axis of the ellipse. It can be written as :
\(T^{2} \propto a^{3}  \)
\(\frac{T^{2}}{a^{3}}=\text { constant. }\)

Mass m = 100 kg
Acceleration a = 50 cm s^-2 = 0.5 ms^-2
Using Newton's second law,
F= ma
F=100 kg\(\times\)0.5 ms^-2 = 50N

We can resolve all the vectors in x, y and z components with respect to Cartesian coordinate system.
Once we resolve the components we can separately equate the x components on both sides, y components on both sides, and z components on both the sides of the equation, we then get
\(F_{1x}+F_{2x}+F_{3x}=F_{4x}\)
\(F_{1y}+F_{2y}+F_{3y}=F_{4y}\)
\(F_{1z}+F_{2z}+F_{3z}=F_{4z}\)

Force \(\vec F=\vec i+2\vec j+3\vec k\)
Distance \(\vec S=4\vec i+6\vec j\)
Work done  \(\vec F.\vec S=(\vec i2\vec j+3\vec k).(4\vec i+6\vec j)=4+12+0=16J\)

three

four

Due to the added weight of rice sack, centre of gravity of the combined body weight and the carrying weight shifted to new position. Once he bends, the Centre of gravity realigns as with the body's axis making his body balanced.

No. 
If a car is moving at a constant speed, then net external force will be zero.
Because a = \(\frac{v-u}{t}\)
For constant speed v = u, then a = 0. (a-acceleration)
F = ma ∴ F = zero. i.e.. no external force.
W = FS = 0. So net work done is zero .

\(\text {Mass of a bullet } m=20 g=20 \times 10^{-3} \mathrm{~kg} =0.02 \mathrm{~kg}  \)
\(\text {Mass of a pendulum } M =5 \mathrm{~kg}  \)
\(\text {Height } \mathrm{h} =10 \mathrm{~cm}  \)
\(=10 \times 10^{-2}  \)
\(=0.1 \mathrm{~m}  \)

\(\text { K.E. of the block } =\text { P.E. of the block }  \)
\(\frac{1}{2} M v^{2} =M g h  \)
\(\therefore v^{2} =\sqrt{2 g h}  \)
\(=\sqrt{2 \times 9.8 \times 0.1}=\sqrt{1.96}=1.4 \mathrm{~m} / \mathrm{s}  \)
\(\therefore \text { Final speed } v =1.4 \mathrm{~m} / \mathrm{s}  \)
\(\text {Final speed } v =\frac{m_{1} u_{1}+m_{2} u_{2}}{\left(m_{1}+m_{2}\right)}  \)
\(v_{1} =\frac{0.02 u_{1}+5 \times 0}{(0.02+5)}=\frac{0.02}{5.02} u_{1}  \)
\(\text { But } v =1.4  \)
\(\therefore 1.4 =\frac{0.02}{5.02} u_{1}  \)
\(u_{1}=\frac{1.4 \times 5.02}{0.02}=\frac{7.028}{0.02}=351.4 \mathrm{~m} / \mathrm{s}\)
\(\therefore \text { Initial speed } =351.4 \mathrm{~m} / \mathrm{s}\)

Given:
Force mg = 30 N; height = 10 m
Work done to lift a load W = ?
W = F.S (or) mgh
= 30 \(\times\) 10
W = 300J

Force acting vertically to sling shot
Tension T in each string;
Now applying Lami's theorem, we get
\(\frac{2 T}{\sin \theta} =F  \)
\(\frac{2 T}{\sin (2 \times 30)} =50 N  \)
\(\frac{2 T}{\sin (60)} =50 N  \)
\(\frac{2 T}{\sqrt{3}} =50 N  \)
\(\frac{T}{2 T} =\frac{2}{\sqrt{3}} \times 50 N  \)
\(T=\frac{50 N}{\sqrt{3}}=28.268 N\)

Characteristics of elastic collision are
1. Total momentum remains conserved
2. Total kinetic energy remains conserved.
3. In elastic collision conservative forces are involved. Hence total kinetic energy is conserved.
4. In elastic collision, mechanical energy is not dissipated.
Characteristics of inelastic collision are
1. Total momentum is conserved.
2. Total kinetic energy is not conserved.
3. Forces involved are non-conservative forces
4. Mechanical energy is dissipated into heat, light, sound etc.

If an object is thrown vertically upward, the direction of motion is upward, but gravitational force is downward.

The surface area of parachute is very large. And when it descends downwards, the air provides resistance to it and so it descends slowly.

(i) According to Newton's third law when one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.
(ii) When you push a car from inside, the reaction force of your pushing is balanced out by your body moving backward and eventually the seat behind you pushes against to bring things to static equilibrium.

All units in atomic standards are more useful because they never change with time.

Least count of screw gauge
\(={{Pitch}\over{No. of\ divisions}}\)
\(={{1}\over{100}}=0.01\ mm\) (or) 0.001 cm
Least count of vernier calipers
\(=1MSD-1-1VSD=(1-19/20)MSD\)
\(={{1}\over{20}}\) = 0.05 cm.
So screw gauge is more precise than vernier.

Radius of the circle r = 3.12 m
Area of the circle A = \(\pi\) r²
= 3.14\(\times\)3.12\(\times\)3.12
= 30.566016 m²
According to the rule of significant 
A = 30.6 m² 

(i) A quantity which is described by both its magnitude and direction is called a vector quantity.
(ii) Geometrically, a vector is a directed line segment
Examples: Force, velocity displacement, acceleration, position vector, linear momentum and angular momentum.

When two or more springs are connected in parallel, we can replace (by removing) all these springs with an equivalent spring (effective spring) whose net effect is same as if all the springs are in parallel connection.
k_p = k₁ + k₂
If n springs are connected in parallel then k_p = nk.

(i) According to first law of thermodynamics, in an isothermal process the given heat is completely converted into work (Q = W). Is it a violation of the second law of thermodynamics.
(ii) No. For non-cyclic process like an isothermal expansion, the heat can be completely converted into work. But Second law of thermodynamics implies that 'In a cyclic process only a portion of the heat absorbed is converted into work'.

Quantity of heat \(\triangle\)Q = ms \(\triangle\)T
If internal energy is equivalent to heat transferred then internal energy.
U = \(\triangle\)Q = ms \(\triangle\)T
= C\(\triangle\)T

Thermal expansion is the tendency of matter to change in shape, area, and volume due to a change in temperature.

Angular momentum \(\mathrm{L} =5 t^{2} \hat{i}-6 t \hat{j}+3 \hat{k} \)
\(\text {Torque }  \propto \frac{d L}{d t}  \)
\(=\frac{d}{d t}\left[5 t^{2} \hat{i}-6 t \hat{j}+3 \hat{k}\right]=10 t \hat{i}-6 \hat{j}\)

The gravitational potential energy U(r) of a system of two masses m₁ and m₂ separated by a distance r as the amount of work done to bring the mass m₂ from infinity to a distance r assuming m₁ to be fixed in its position and is written as
\(\mathrm{U}(\mathrm{r})=-\frac{G m_{1} m_{2}}{r}\)

Angle \(\theta\) = 60 ^o
The distance between the tree and a point
x = 50m
Height of the tree (h)=?
For triangulation method tan\(\theta\) =\({h\over x}\)
h= x tan \(\theta\) = 50\(\times\)tan 60⁰ = 50\(\times\)1.732
h = 86.6m
The height of the tree is 86.6m.

Various types of potential energy are
(i) The energy possessed by the body due to gravitational force gives rise to gravitational potential energy.
The gravitational potential energy (U) at some height h is equal to the amount of work required to take the object from the ground to that height h.
U = mgh
(ii) The energy due to spring force and other similar forces give rise to elastic potential energy.
a) At the equilibrium position x = 0 potential energy is \(U=\frac{1}{2} k x^{2}\)
b) If the initial position is not zero and if the mass is changed from position x_i to x_f, if then elastic potential energy is \(U=\frac{1}{2} k\left(x_{f}^{2}-x_{i}^{2}\right)\)
(iii) The energy due to electrostatic force on charges gives rise to electrostatic potential energy.
Electrostatic potential energy is the work done to arrange two charges q₁ and q₂ at a separation \(r=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}\)

1. Equal mass balancing each side so reading shows Zero.
2. Mass of string acting downward direction in the inclined plane.
mg sin θ = T
T = 2kg x 9.8m/s² x sin(30^o)
T = 9.8N

360^o = 27 days
1 day = \(\frac{360^o}{27}=13^o.3'\)

(a) T - mg = ma
(b) \(T_{x}+F_{x}=A_{x}+B_{x} ; T_{y}+F_{y}=A_{y}+B_{y} ; T_{z}+F_{z}=A_{z}+B_{z}\)
(c) T + mg = ma
(d) \(T_{x}-F_{x}=A_{x}-B_{x} ; T_{y}-F_{y}=A_{y}-B_{y} ; T_{z}-F_{z}=A_{z}-B_{z}\)

\(\overrightarrow { { v }_{ 1 } } =5\hat { i } +6\hat { j } \)
\(\overrightarrow { { v }_{ 2 } } =2\hat { i } +3\hat { j } \)
\(\therefore \Delta \vec{r} =\vec{r}_{2}-\vec{r}_{1}=2 \hat{i}+3 \hat{j}-5 \hat{i}-6 \hat{j}  \)
\(=-3 \hat{i}-3 \hat{j}  \)
\(\Delta t =5 \mathrm{sec}  \)
\(\therefore \Delta v_{\text {avg }} s =\frac{\Delta \vec{r}}{\Delta t}  \)
\(=\frac{-3}{5}(\hat{i}+\hat{j})\)

(i) Systematic observation
(ii) Controlled experimentation
(iii) Reasoning (qualitative and quantitative)
(iv) Modelling (Mathematical)
(v) Prediction and verification (theories)

It 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 \(\overrightarrow { A } \) and \(\overrightarrow { B } \) are two vectors, then their vector product \(\vec{A} \times \vec{B}=\vec{C}=(A B \sin \theta) \hat{n}\).
The direction of the product vector \((\hat{n})\) is perpendicular to the plane containing the two vectors, in accordance with the right hand screw rule or right hand thumb rule.
1. It is not commutative, i.e., \(\vec{A} \times \vec{B} \neq \vec{B} \times \vec{A}. \ But \ \vec{A} \times \vec{B}=-[\vec{B} \times \vec{A}].\)
2. \((\vec{A} \times \vec{B})_{\max }=A B \hat{n}\), when \(\theta=90^{\circ}\) i.e., when \(\overrightarrow { A } \) and \(\overrightarrow { B } \) are orthogonal to each other.
3. \((\vec{A} \times \vec{B})_{\text {min }}=0\), when \(\theta=0^{\circ} \ or \ 180^{\circ}\) i.e., when the vectors are either parallel or antiparallel provided \(\overrightarrow { A } \) and \(\overrightarrow { B } \) are non-zero vectors.
4. The self - vector products of unit vectors are \(\hat{i} \times \hat{i}=\hat{j} \times \hat{j}=\hat{k} \times \hat{k}=0\)
5. In the case of orthogonal unit vectors, \(\hat{i} \times \hat{j}=\hat{k}, \hat{j} \times \hat{k}=i, \hat{k} \times \hat{i}=j\)
6. Torque \(\tau=\vec{r} \times \overrightarrow{\mathrm{F}}\), angular momentum \(\vec{L}=\vec{r} \times \vec{p}\ an \ \vec{V}=\vec{w} \times \vec{r}\) are examples of vector product.