k₁ and k₂ attached to a mass m as shown in figure. The results can be generalized to any number of springs in parallel.

Let the force F be applied towards right as shown in figure. In this case, both the springs elongate or compress by the same amount of displacement. Therefore, net force for the displacement of mass m is
F = - k_px  ..(1)
where kp is called effective spring constant.
Let the first spring be elongated by a displacement x due to force F₁ and second spring be elongated by the same displacement x due to force F₂' then the net force
F = - k₁x - k₂x ...(2)
Equating equations (2) and (1), we get
k_p= k₁ + k_{2     ...}(3)
Generalizing, for n springs connected in parallel
\({ k }_{ P }=\sum _{ i=1 }^{ n }{ { k }_{ i } } \) ....(4)
If all spring constants are identical i.e.,k₁ = k₂ = ... = kn = k then
k_p = n k    ...(5)
This implies that the effective spring constant increases by a factor n. Hence, for the springs in parallel connection, the effective spring constant is greater than individual spring constant.

Let a be the displacement amplitude and co be the angular frequency of S.H.M then maximum velocity, Vo = \(\omega \)A  \(\therefore \omega ={ V }_{ 0 }/A\)
Maximum acceleration, a_o= \({ \omega }^{ 2 }A={ \left( \frac { { V }_{ 0 } }{ A } \right) }^{ 2 }\)
A =\(\frac { { V }_{ 0 }^{ 2 } }{ A } \)
\(\therefore\) Displacement amplitude, A = \(\frac { { V }_{ 0 }^{ 2 } }{ { a }_{ 0 } } \)

The effective value of acceleration due to gravity g' = g cos \(\alpha\)
\(T=\quad 2\pi \sqrt { \frac { l }{ g' } } =2\pi \sqrt { \frac { 1 }{ cos\alpha } } \)

(i) The force experienced by a mass 'm' which is on the surface of the Earth is given by F= \(\frac { { GM }_{ 1 }m }{ { r }_{ E }^{ 2 } } \)
(ii) M_E-mass of the Earth, m - mass of the object, R_E- radius of the Earth.
Equating Newton's second law, F= mg, to equation we get,
mg= \(\frac { { GM }_{ 1 }m }{ { r }_{ E }^{ 2 } } \)
g=\(\frac { { GM }_{ E }m }{ { r }_{ E }^{ } } \)
(iii) Now the force experienced by some other object of mass M at a distance r from the center of the Earth is given by,
F= \(\frac { { GM }_{ E }m }{ { r }_{ }^{ 2 } } \)
Using the value of g in equation, the force F will be,
\(F=-gM\frac { { R }_{ E }^{ 2 } }{ { r }^{ 2 } } \)
From this it is clear that the force can be calculated simply by knowing the value of g. It is to. be. noted that in the above calculation G is not required.

Since, the system is going from A to B and that back to A, it is undergoing a cyclic change. Now in cyclic change, there is no change in internal energy (\(\triangle\)U = 0).From first law of thermodynamic 
\(\triangle\)Q = \(\triangle\)U =\(\triangle\)W
= \(\triangle\)W
The amount of heat absorbed by the system
= 10 cal- 30 J
=(10 x 4.2-30)J
= (42 - 30) J
\(\therefore\)  The net work done by the system = 12 J 

(i) N =  \(\mu \) N_A
Where N_A is Avogadro number (6.023\(\times\)1023mol^-1)
(ii) PV = N_A 11NAkT. Here N_A k = R called universal gas constant and its value is 8.314 j/mol.K
So  the ideal gas law can be written for u mole of gas as
PV = \(\mu \)RT
(iii) This.is called the equation of state for an ideal gas. It relates the pressure, volume and temperature of thermodynamic system at equilibrium. 

Accumulate to kinetic theory:
(i) The molecules of a gas are in a state of continuous random motion. 
(ii) They collide with one another and also with the walls of the vessel. Whenever a molecule collides with the wall. It returns with a changed momentum and an equal momentum is transfused to the wall (conservation of momentum).
Accumulate to Newton's law:
(i) The transfer of momentum to the wall is equal to the force excited on the wall.
(ii) The force excited per unit area of the wall is the pressure of the gas.
Hence a gas excites pres due to the continuous call of its molecules with the walls of the vessel.

It states that if the pressure remains const, then the v₀ of a g mass of a gas increases or decreases by \(\frac{1}{2723.15}\) of its volume at 0^oC for each 1°C rise or fall of temperature V_o - volume of the g_n mass
of a gas at O°C
Accumulate to charle's law, its volume at 1°C is 
\({ v }_{ 1 }={ v }_{ 0 }+\frac { { v }_{ 0 } }{ 273.15 } ={ v }_{ 0 }\left( 1+\frac { 1 }{ 273.15 } \right) \)
Volume of the gas at t^oC
\({ v }_{ 1 }={ v }_{ 0 }\left( 1+\frac { 1 }{ 273.15 } \right) \)
If T_o and T are temperature on kelvin scale corresponding to 0°C & t^oC, then
T_o= 27:3.15+0 = 273.15
T = 273.15 +t
\(\therefore { v }_{ t }={ v }_{ 0 }\left( \frac { T }{ { T }_{ 0 } } \right) or\frac { { v }_{ 1 } }{ T } =\frac { { v }_{ 0 } }{ { T }_{ 0 } } \)
\(\frac{v}{T}\) = constant i.e. v x T.
So, the law states that at constant pressure the volume of a g_n mass of a gas is directly proportional to its absolute temperature. The graph. between. V and T for a g_n mass of a gas at constant pressure is a S_t line.

(i) This is because a body with large reflectivity is poor absorber of heat, and poor aborbers of heat are emitters.
(ii) When we touch a brass tumber on a chilly day, heat flows from our body to the tumbler quickly (as thermal conductivity of brass is very high) and as a result, it appears colder.
(iii) On the other hand, as the wood a bad conductor, heat does not flow to the wooden tray from our body, on touching it.

(i) The oil used as a lubricant for heavy machinery parts should have a high viscous coefficient.
(ii) The highly viscous liquid is used to damp the motion of some instruments and is used as brake oil in hydraulic brakes.
(iii) Blood circulation through arteries and veins depends upon the viscosity of fluids.

(i) For the propagation of the waves, the medium must possess both inertia and elasticity, which decide the velocity of the wave in that medium.
(ii) In a given medium, the velocity of a wave is a constant whereas the constituent particles in that medium move with different velocities at different positions. Velocity is maximum at their mean position and zero at extreme positions.
(iii) Waves undergo reflections, refraction, interference, diffraction and polarization.

Let us first calculate the stiffness constant of the spring balance by using equation (10.29),
\(K=\frac { mg }{ l } =\frac { 25\times 11.5 }{ 0.25 } ={ 1150 }Nm^{ -1 }\)
The time period of oscillations is given by
\(T=2\pi \sqrt { \frac { M }{ k } } \)where M is the mass of the body.
Since, M is unknown, rearranging, we get 
\(M=\frac { k{ T }^{ 2 } }{ { 4\pi }^{ 2 } } =\frac { \left( 1150 \right) { \left( 0.5 \right) }^{ 2 } }{ { 4\pi }^{ 2 } } =7.3kg\)
The gravitational force acting on the body is W = Mg = 7.3\(\times\)11.5 = 83.95 N ≈ 84 N

The efficiency of heat engine is given by
\(\eta =1-\frac { { Q }_{ L } }{ { Q }_{ H } } \)
\(\eta =1-\frac { 300 }{ 500 } =1-\frac { 3 }{ 5 } \)
\(\eta \) = 1 – 0.6 = 0.4
The heat engine has 40% efficiency, implying that this heat engine converts only 40% of the input heat into work.

(a) We know that work done by the gas in an isothermal expansion
Since μ = 0.5
W = 0.5mol \(\times\)\(\frac { 8.31J }{ mol.K } \times 300 K\ In\ \left( \frac { 6L }{ 2L } \right) \)
W = 1.369 kJ
Note that W is positive since the work is done by the gas.
(b) From the First law of thermodynamics, in an isothermal process the heat supplied is spent to do work.Therefore, Q = W = 1.369 kJ. Thus Q is also positive which implies that heat flows into the system.
(c) For an isothermal process
P_iV_i = P_fV_f = μRT
\({ P }_{ f }=\frac { \mu RT }{ { V }_{ f } } =0.5mol\times \frac { 8.31J }{ mol.K } \times \frac { 300K }{ 6\times { 10 }^{ -3 }{ m }^{ 3 } } \)
= 207.75 k Pa

Let each side of the cube be l. The volume occupied by 3 cm depth of cube,
V  = (3cm) \(\times\) l² = 3l²cm
According to the principle of floatation, we have
V\(\rho\)g = mg \(\Rightarrow\) V\(\rho\) = m
\(\rho\) is density of water = 1000 kg m^-3
\(\Rightarrow\) (3l²\(\times\)10^-2m) \(\times\)(1000 kgm^-3) = 300\(\times\)10^-3 kg
\({ l }^{ 2 }=\frac { 300\times { 10 }^{ -3 } }{ 3\times { 10 }^{ -2 }\times 1000 } { m }^{ 2 }\Rightarrow { l }^{ 2 }=100\times { 10 }^{ -4 }{ m }^{ 2 }\)
l = 10\(\times\)10^-2 m = 10 cm
Therefore, volume of cube V = l³ = 1000 cm³

From the shadows it is revealed that Earth has spherical shape with bulging along equator and flat at poles.

The answer is wrong.
Actually, the seasons in the Earth arise due to the rotation of Earth around the sun with 23.5^o tilt.

30% of Power=\(\frac{W}{t}=\frac{mgh}{t}=\frac{V\rho gh}{t}\)
\(\frac{30}{100}\times P=\frac{V\rho gh}{t}\)
P=43.6 KW

(a) The potential energy U = mgh = 2 \(\times\) 10 \(\times\) 5 = 100 J
Here the positive sign implies that the energy is stored on the mass.
(b) This potential energy is transferred from external agency which applies the force on the mass.
(c) The external applied force \(\overrightarrow { { F }_{ a } } \) which takes the object to the height 5 m is \(\overrightarrow { { F }_{ a } } =-\overrightarrow { { F }_{ g } } \)
\(\overrightarrow { { F }_{ a } } =-(-mg\hat { j } )=mg\hat { j } \)
where, \(\hat { j } \) represents unit vector vertical upward direction.
(d) From the definition of potential energy, the object must be moved at constant velocity. So the net force acting on the object is zero.
\(\overrightarrow { { F }_{ g } } +\overrightarrow { { F }_{ a } } =0\)

When the block slides, the force acting on the block is kinetic friction which is equal to f_k= μ_smg.
From Newton's second law ma = -μ_smg
The negative sign implies that force acts on the opposite direction of motion.
The acceleration of the block while sliding a =-μ_kg
The negative sign implies that the acceleration is in opposite direction of the velocity. Note that the acceleration depends only on g and the coefficient of kinetic friction μ_k. We can apply the following kinematic equation
v=u+at
The final velocity is zero
0=u-u_kgt
t=\(\frac { u }{ { \mu }_{ k }g } \).

\(h=ut+\frac { 1 }{ 2 } { gt }^{ 2 }\)
find a by differentiating h twice w.r.t.
a = g
As F = ma so F = mg

If speed of an object increases with time, its acceleration is positive.
(Acceleration is in the direction of motion) and if speed of an object decreases with time its acceleration is negative (Acceleration is opposite to the direction of motion).

m = 66.0 gm.

Velocity of center of mass = \(\frac{31\hat{i}-34\hat{j}+15\hat{k}}{2} ms^{-1}\).

G = 6.67\(\times\)10^-11 N m²kg^-2
= 6.67\(\times\)10^-11 (kg m s^-2) (m² kg^-2)
=6.67\(\times\)10^-11 kg^-1 m³ s^-2
= 6.67\(\times\)10^-11 (1000 g)^-1(100 cm)³ (s^-2)
=6.67\(\times\)10^-11 x \(1\over 1000\) \(\times\)100 \(\times\)100\(\times\)100g^-1cm³ s^-2
=6.67\(\times\)10^-8 g^-1cm³s^-2.

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.

(i) As 360⁰ = 2\(\pi \)rad .
\(\therefore\) 1⁰= \(2\pi \over 360\) rad = 1.745\(\times\)10^-2 rad
(ii) 1⁰ = 60' = 1.745 \(\times\)10^-2 rad
\(\therefore\)1' = 2.908 X 10^{-4\(\approx\)}2.91\(\times\)10^-4 rad
(iii) 1'= 60" = 2.908\(\times\)10^-4rad
1" = 4.847\(\times\)10^-6 rad^\(\approx\)4.85\(\times\)10^-6 rad

Given:
t=3s
Distance s = Area of \(\triangle OAB\)

\(=\frac{1}{2}\times OA\times BA\)
\(=\frac{1}{2}\times 3\times 6=9m\)
If the speed varies with the time: Then \(v=\frac{ds}{dt}\Rightarrow ds=v\ dt\)
\(\Rightarrow \int { ds } =\int { v }\ dt\)
\(or\ s=\int { v }\ dt\)

\(a={F \over m}={20 \over 5}={4ms}^{-2}\)
v=u + a t = 0 + 4\(\times\)10 = 40 ms-1
\(AE={1\over2}\times5\times{(40)}^{2}=4000\) Joule

The force F has to act downward, because the body is to move up with an acceleration greater than 'g' acceleration due to gravity.
According to Newton's II law,
In a force acting an a downward} mg + F = ma
F = m(g+a)
= 20(9.8 + 4)
= 20(13.8)
= 276N

1. Greater the mass concentrated away from the axis, greater the moment of inertia.
2. Moment of inertia of a body about an axis of rotation resists a change in it. In rotational motion, moment of inertia increases with increase in torque to change it's rotation.

Newton's third law states that for every action there is an equal and opposite reaction.

Time becomes irrelevant. Because day and year based on spinning and revolution of the planet. So each planet has its own year length.
Eg: Uranus and Neptune move too slow.

(i) Displacement is the difference between the final and initial positions of the object in a given interval of time. It can also be defined as the shortest distance between these two positions of the object and its direction is from the initial to final position of the object, during the given interval of time. It is a vector quantity.
(ii) Distance is the actual path length travelled by an object in the given interval of time during the motion. It is a positive scalar quantity.

\(\Delta L=10\log { _{ 10 }\left[ \frac { { I }_{ 1 } }{ { I }_{ 0 } } \right] =50dB } \)
\(\log { _{ 10 }\left[ \frac { { I }_{ 1 } }{ { I }_{ 0 } } \right] =5dB } \)
\(\frac { { I }_{ 1 } }{ { I }_{ 0 } } ={ 10 }^{ 5 }\Rightarrow { I }_{ 1 }={ 10 }^{ 5 }{ I }_{ 0 }\times { 10 }^{ -12 }{ Wm }^{ -2 }\)
I₁ = 10^-7 Wm^-2
Since three musical instruments are played,
therefore, I_total = 3I₁ = 3\(\times\)10^-7 Wm^-2.  

Given y₁ = 5 sin(240\(\pi\)t) and y₂ = 4 sin(244\(\pi\)t)
Comparing with y = A sin(2\(\pi\) f₁t), we get
2\(\pi\)f₁ = 240π \(\Rightarrow\) f₁ = 120Hz
2\(\pi\)f₂ = 244π \(\Rightarrow\) f₂ = 122Hz
The number of beats produced is | f₁ − f₂| = |120 − 122| = |− 2|=2 beats per sec

Kinetic energy is KE\(=\frac { 1 }{ 2 } { mv }_{ x }^{ 2 }\)
Multiply numerator and denominator by m
\(KE=\frac { 1 }{ 2m } { m^{ 2 }v }_{ x }^{ 2 }=\frac { 1 }{ 2m } \left( { mv }_{ x } \right) ^{ 2 }=\frac { 1 }{ 2m } { P }_{ x }^{ 2 }\)
where, P_x is the linear momentum of the particle executing simple harmonic motion.
Total energy can be written as sum of kinetic energy and potential energy, therefore, from equation (10.73) and also from equation (10.75), we get
E = KE + U(x) \(=\frac { 1 }{ 2m } { P }_{ x }^{ 2 }+\frac { 1 }{ 2m } { m\omega ^{ 2 }v }_{ x }^{ 2 }\) = constant

From equation (7.34), we have 
h∝\(\frac { 1 }{ r } \Rightarrow hr=\)constant
Consider two capillary tubes with radius r₁ and r₂ which on placing in a liquid, capillary rises to height h₁ and h₂, respectively. Then,
h₁r₁ = h₂r₂ = constant
\(\Rightarrow { h }_{ 2 }=\frac { { h }_{ 1 }{ r }_{ 1 } }{ { r }_{ 2 } } =\frac { \left( 2\times { 10 }^{ -2 }m \right) }{ \frac { r }{ 3 } } \Rightarrow { h }_{ 2 }=6{ \times 10 }^{ 2 }m\)

The center of gravity of a body is the point at which the entire weight of the body acts irrespective of the position and orientation of the body.

(i) We know from postulates of kinetic theory that the molecules of a gas are in random motion and they collide with each other.
(ii) Between two successive collisions, a molecule moves along a straight path with uniform velocity.
(iii) This path is called mean free path. Consider a system of molecules each with diameter d. Let n be the number of molecules per unit volume.
(iv) Assume that only one molecule is in motion,and all others are at rest.

(v) If a molecule moves with average speed v in a time t, the distance travelled is vt.
(vi) In this time t, consider the molecule to move in an imaginary cylinder of volume nd2vr.
(vii) It collides with any molecule. whose center is within this cylinder. Therefore, the number of collisions is equal to the number of molecules in the volume of the imaginary cylinder.
(viii) It is equal to \(\pi\)d²vtn. The total path length divided by the number of collisions in time t is the mean free path.
Mean free pat, \(\lambda =\frac{distance \ travelled}{Number \ of \ collisions}\)
\(\lambda =\frac { vt }{ n{ \pi d }^{ 2 }vt } =\frac { 1 }{ n{ \pi d }^{ 2 } } \)   ...(1)
(ix) Though we have assumed that only one molecule is moving at a time and other molecules are at rest, in actual practice all the molecules are in random motion.
(x) So the average relative speed of one molecule with respect to other molecules has to be taken into account. After some detailed calculations (you will learn in higher classes) the correct expression for mean free path .
\(\therefore \ \lambda =\frac { 1 }{ \sqrt { 2 } n{ \pi d }^{ 2 } } \)   ...(2)
(xi) The equation (1) implies that the mean free path is inversely proportional to number density.
(xii) When the number density increases the molecular collisions increases and it decreases the distance travelled by the molecule before collisions:
Case1: Rearranging the equation (2) using 'm' (mass of the molecule)
\(\therefore \ \lambda =\frac { 1 }{ \sqrt { 2 } n{ \pi d }^{ 2} mm } \)
But mn = mass per unit volume = p (density of the gas)
\(\therefore \ \lambda =\frac { 1 }{ \sqrt { 2 } n{ \pi d }^{ 2 } p} \)
Also we know that PV = NkT
P =\(\frac{N}{V}\)KT= nKT
\(\therefore n =\frac{P}{KT}\)
Substituting n = \(\frac{P}{KT}\) in equation, we get
\(\lambda =\frac { 1 }{ \sqrt { 2 } n{ \pi d }^{ 2 }P } \)

Monoatomic molecule
A monoatomic molecule by virtue of its nature has only three translational degrees of freedom. Therefore f = 3
Example: Helium, Neon, Argon

Diatomic temperature
At Normal temperature.
A molecule of a diatomic gas consists of two atoms bound to each other by a force of attraction. Physically the molecule can be regarded as a system of two point masses fixed at the ends of a massless elastic spring.
The center of mass lies in the center of the diatomic molecule. so, the motion of the center of mass requires three translational degrees of freedom. In addition, the diatomic can rotate about three mutually perpendicular axes. But the moment of inertia about its own axis of rotation is negligible. Therefore, it has only two rotational degrees of freedom (one rotation is about Z axis and another rotation is about Y axis). Therefore totally there are five degrees of freedom. f = 5
At High Temperature
At a very high temperature such as 5000 K, the diatomic molecules possess additional two degrees of freedom due to vibrational motion [one due to kinetic energy of vibration and the other is due to potential energy]. So, totally are seven degrees of freedom. f = 7.
Examples: Hydrogen, Nitrogen, Oxygen
Triatomic molecules
There are two cases.
Linear triatomic molecule
In this type, two atoms lie on either side of the central atom.

Linear triatomic molecule has three translational degrees of freedom. It has two rotational degrees of freedom because it is similar to diatomic molecule except there is an addtional atom at the center. At normal temperature, linear triatomic molecule will have five degrees of freedom. At high temperature it has two additional vibrational degrees of freedom. So a linear triatomic molecule has seven degrees of freedom.
Example: Carbon dioxide
Non-Linear triatomic molecule
In this case, the three atoms lie at the vertices of a triangle.
In has three translational degrees of freedom and three rotational degrees of freedom about three mutually orthogonal axes. The total degrees of freedom f = 6

Example: Water, Sulphur dioxide

Force \(\overrightarrow { F } =mg\left( -\hat { j } \right) =-mg\hat { j } \)
Displacement vector \(d\vec{r}\) = dx\(\hat { i } \) + dy \(\hat { j } \)
(As the displacement is in two dimension; unit vectors \(\hat { j } \) and \(\hat { i } \)  are used)
(a) Since the motion is only vertical, horizontal displacement component dx s zero. Hence, work done by the force along path 1 (of distance h).
\({ W }_{ push\ \ 1 }=\int _{ A }^{ B }{ \overrightarrow { F } .d\overrightarrow { r } =\int _{ A }^{ B }{ (-mg\hat { j } ).(dy\hat{j})=-mg\int _{ 0 }^{ h }{ dy=-mgh } } } \)
Total work done for path 2 is
\({ W }_{ push\ \ 2 }=\int _{ A }^{ B }{ \overrightarrow { F } .d\overrightarrow { r } =\int _{ A }^{ C }{ \overrightarrow { F } .d\overrightarrow { r } +\int _{ C }^{ D }{ \overrightarrow { F } .d\overrightarrow { r } +\int _{ D }^{ B }{ \overrightarrow { F } .d\overrightarrow { r } } } } } \)
But \(\int _{ A }^{ C }{ \overrightarrow { F } .d\overrightarrow { r } =\int _{ A }^{ B }{ (-mg\hat { j } ).(dx\hat { i } )=0 } } \)
\(\int _{ A }^{ D }{ \overrightarrow { F } .d\overrightarrow { r } =\int _{ C }^{ D }{ (-mg\hat { j } ).(dy\hat { j } )=-mg\int _{ 0 }^{ h }{ dy } =-mgh } } \)
\(\int _{ D }^{ B }{ \overrightarrow { F } .d\overrightarrow { r } =\int _{ A }^{ B }{ (-mg\hat { j } ).(dx\hat { i } )=0 } } \)
Therefore, the total work done by the force along the path 2 is
\({ W }_{ push\ \ 2 }=\int _{ A }^{ B }{ \vec { F } } .d\overrightarrow { r } =-mgh\)
Note that the work done by the conservative force is independent of the path.