(a) laws of limiting friction.
1.The magnitude of the force of limiting friction (F) between any two bodies in constant is directly proportional to the normal reaction (R) between them, i.e., F ∝ R.
2. The direction of limiting friction is always opposite to the direction of motion.
3. The force of limiting friction is independent of the apparent area of contact; so long as normal reaction between the two bodies in contact remains the same.
4. The limiting friction depends on the nature of material of the surfaces in contact (i.e. force of adhesion) and their roughness or smoothness.
(b) Suppose m is the mass of a body that has to be moved up a rough plane AB, inclined to the horizontal at an angle e. The various forces involved are:

(i) Weight (mg) of the body, acting vertically downwards.
(ii) Normal reaction (R), acting perpendicular to the inclined plane AB.
(iii) Force of friction (F), acting down the inclined plane AB, as the body moves up the plane. In equilibrium,
R = mg cos θ              ..............(i)
and P = mg sin θ + F         ............(ii)
Where P is the force required to be applied in moving the body up the rough inclined plane AB. Under the action of this force, suppose the body moves through a distance S up the plane.
Work done = Force x Distance
W = P x S
W = (mg sin θ + F)S           ............(iii)
W = (mg sin θ + μ)S
Where μ is the coefficient of friction between the two surfaces in contact.
W = (mg sin θ + μ mg cos θ)S       [Using (i)]
W = mg(sin θ + μ cos θ)S

When a body is projected in the air in any direction, then the body is called a projectile.
(a) Suppose a body is projected with velocity u at an angle θ with the horizontal, P(x, y) is any point on its trajectory at time t.
Horizontal component of velocity is unaffected by gravity, but the vertical component (u sin θ) changes due to gravity.
ஃ x = (u cos θ)t.
\(y=(u \sin \theta) t-\frac{1}{2} g t^{2}\)
\(=u \sin \theta \times \frac{x}{u \cos \theta}-\frac{1}{2} g\left(\frac{x}{u \cos \theta}\right)^{2}\)
\(y=x \tan \theta-\frac{g x^{2}}{2 u^{2} \cos ^{2} \theta}\)                 ............(i)
(b) The greatest vertical distance attained by the projectile above the horizontal plane from the point of projection is called maximum height.
Maximum height, LN = H
(i) At maximum height
v = 0
\(\therefore \quad v^{2}-u_{y}^{2}=-2 g H, \text { where }\)
u_y = u sin θ 
or (u sinθ )² = 2 gH
\(\text { or } \quad H=\frac{u^{2} \sin ^{2} \theta}{2 g}\)
(ii) At maximum height
v = 0
ஃ 0 = u sin θ - gt
\(\text { or } \quad t=\frac{u \sin \theta}{g}\)
But time of flight
\(\mathrm{T}=2 t=\frac{2 u \sin \theta}{g}\)
(iii) When the body returns to the same horizontal level y = 0
\(\therefore \quad 0=x \tan \theta-\frac{g x^{2}}{2 u^{2} \cos ^{2} \theta} \quad[\text { From }(i)]\)
\(\text { or } x \tan \theta=\frac{g x^{2}}{2 u^{2} \cos ^{2} \theta}\)
\(\text { or } \quad x=\frac{2 u^{2} \sin \theta \cos \theta}{g}=\frac{u^{2} \sin 2 \theta}{g}\)
But coordinates of M are (R, 0). Putting x = R,
we have
\(\mathrm{R}=\frac{u^{2} \sin 2 \theta}{g}\)
(c) θ = 45°
(d) When an object is projected with velocity u making an angle θ with horizontal direction
\(R_{1}=\frac{u^{2} \sin 2 \theta}{g}\)             ...........(i)
When an object is projected with u making an angle (90^o - θ)
\(\mathrm{R}_{2}=\frac{u^{2} \sin 2\left(90^{\circ}-\theta\right)}{g}\)
\(=\frac{u^{2}}{g} \sin \left(180^{\circ}-2 \theta\right)\)
\(=\frac{u^{2}}{g} \sin 2 \theta\)                       ............(ii)
from (i) and (ii) R₁ = R₂
ஃ The horizontal range IS same for two complementary angles.

Let OX and OY be two perpendicular axes and YO = 490 m. A body projected horizontally from O with velocity u (= 98 ms^-1 ) meets the ground at A following a parabolic path shown in figure.

(i) Let T be the time of flight of the projectile i.e. time taken by projectile to go from O to A
Taking vertical downward motion (i.e. motion along OY axis) of projectile from 0 to A, we have
Y₀ =0, y = 490m,u_y =0, a_y = 9.8m/s ² , t = T
As, \(y=y_{0}+u_{y} t+\frac{1}{2} a_{y} t^{2}\)
\(\therefore \quad 490=0+0 \times T+\frac{1}{2} \times 9.8 \times T^{2}=4.9 T^{2}\)
or \(T=\sqrt{\frac{490}{4.9}}=10 \mathrm{~s}\) 
(ii) Taking horizontal motion (i. e. motion along OX axis) of projectile from O to A, we have
X₀ = 0, x = R (say), U_x = 98 m/s, t = T = 10 s , a_x = 0
As, \(x=x_{0}+u_{x} t+\frac{1}{2} a_{x} t^{2}\)
\(\therefore \quad R=0+98 \times 10+\frac{1}{2} \times 0 \times 10^{2}=980 \mathrm{~m}\)
(iii) Let v _x, v_y be the horizontal and vertical component velocity of the projectile at A.
Using the relation,
V_x = u_x + a_x t = 98 + 0 x 10 = 98m/s
Represented by AB
Using the relation
V_y = u_y + a_yt = 0 + 9.8 x 10 = 98m/s
Represented by AC.
\(\therefore\)Resultant velocity
\(v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{98^{2}+98^{2}}=98 \sqrt{2} \mathrm{~m} / \mathrm{s}\)
If \(\beta\) is the angle which v makes with the horizontal direction, then tan \(\beta=\frac{v_{y}}{v_{x}}=\frac{98}{98}=1 \text { or } \beta=45^{\circ}\) with the horizontal.

(i) Velocity attained by a particle after time t:
Let dt: be the change in velocity of the particle in time dt. Therefore, the acceleration of the particle is given by
\(a=\frac{dv}{dt} \ or \ dv=a \ dt\)
By integrating both sides, we get
\(\int{dv}= \int{a dt}\)
or \(\int{dv}= a \int{ dt}\)
or    v = at + k   --- (i)
where k is constant of integration.
when t = 0, v = u
Putting these values in equation (i), we get
k = u
Now putting the value of k in equation (i), we get
v = u + at
(ii) Displacement of the particle after time t:
Let dx be the displacement of the particle in time dt. Therefore, the velocity of the particle is given by
\(v=\frac{dx}{dt}\ or \ dx=vdt\)
Since v  =u + at
∴ dx=(u+at)dt
Integrating both sides, we get
\(\int{dx}=\int{(u+at)}dt\)
or  \(\int{dx}=\int{u}dt+\int{at \ dt}\) 
\(x= u\int{dt}+a \int{t\ dt}\)     [∵ u and a are constants]
or \(x=ut+a \frac{t^{2}}{2}+k\)
where k is constant of proportionality
where t = 0, x = x₀
∴ from equation (ii), we get
\(x=x_{0}+ut+\frac{1}{2}at^{2}\)
or \(x-x_{0}=ut+\frac{1}{2}at^{2}\)
since  x-x₀= S, displacement of the particle in the time interval t.
S = \(ut+\frac{1}{2}at^{2}\)
(iii) Velocity attained by a particle after travelling a distance S:
We know, \(v=\frac{dx}{dt}\)
Multiplying and dividing R.H.S. by dv, we get
\(v=\frac{dx}{dt}.\frac{dv}{dv}=\frac{dx}{dv}.\frac{dv}{dt}\)
As \(\frac{dv}{dt}=a (acceleration)\)
∴ v = a\(\frac { dx }{ dv } \) or v dv=a dx
Integrating both sides, we get \(\int { v\ dv=\int { a\ dx=a\int { dx } } } \)
or = \(\frac { v^{ 2 } }{ 2 } \)ax+k
when x = 0,v = u
Then,from eqn.(i),k = \(\frac { u^{ 2 } }{ 2 } \)
Putting the value of k in eqn. (i), we get
\(\frac { v^{ 2 } }{ 2 } \)-ax+\(\frac { u^{ 2 } }{ 2 } \)
or \(\frac { v^{ 2 } }{ 2 } -\frac { u^{ 2 } }{ 2 } =ax\)
or v²-u²= 2ax
x = s, then
v²-u² = 2 aS.

(a) We know that \({ \mu }_{ k }=\frac { frictional\ force }{ normal\ reaction } \)
\(\therefore\) frictional force = \({ \mu }_{ k }\times normal\ reaction\)
= 0.1 \(\times\) 2 kg wt = 0.1 \(\times\) 2 \(\times\) 9.8 N = 1.96 N
net effective force = (7-1.96)N = 5.04 N
acceleration = \(\frac { 5.04 }{ 2 } { ms }^{ -2 }=2.52\ { ms }^{ -2 }\)
distance, \(s=\frac { 1 }{ 2 } \times 2.52\times 10\times 10=126\ m\)
work done by applied force = 7 \(\times\) 126 J = 882 J
(b) Work done by friction = 1.96 \(\times\) 126 = -246.96 J
(c) Work done by net force = 5.04 \(\times\) 126 = 635.04 J
(d) Change in the kinetic energy of the body
= work done by the net force in 10 seconds
= 635.04 J (This is in accordance with work-energy theorem).

We are given mass of neutron star,
i.e  \(M=2.5\times 2\times 10^{ 30 }kg\)
\(=5\times 10^{ 30 }kg\)
\(\\ Radius \ of \ star, \ R=12\ km=1.2\times 10^{ 4 }m\)
\(\\ Frequency\ of\ rotation,\ v=1.5\)

\(2.7972\times { 10 }^{ -7 }N\)

(i) Acceleration due to gravity at altitude h from the earth's surface is given by
\({ g }^{ 1 }=\frac { g }{ \left( 1+\frac { h }{ { R }_{ e } } \right) ^{ 2 } } \)
where, \({ R }_{ e }\)is the radius of the earth.
Therefore, acceleration due to gravity decrease with increasing altitude.
(ii) Acceleration due to gravity at depth d from the earth's surface is given by
\({ g }^{ ' }=g\left( 1-\frac { d }{ { R }_{ e } } \right) \)
Therefore, acceleration due to gravity decreases with increasing depth.
(iii) Acceleration due to gravity is independent of the mass of the body.
(iv) The formula, 
\(-GMm\left( \frac { 1 }{ { r }_{ 2 } } -\frac { 1 }{ { r }_{ 1 } } \right) \)is more accurate than the formula \(mg\left( { r }_{ 2 }-{ r }_{ 1 } \right) \)for the difference of potential energy between two points \({ r }_{ 2 }\ and{ r }_{ 1 }\)distance away from the centre of the earth.

The velocity of the rain and the wind are represented by the vectors v_r and v_w in Fig. and are in the direction specified by the problem. Using the rule of vector addition, we see that the resultant of v_r and v_w is R as shown in the figure. The magnitude of R is
R = \(\sqrt { v_r^{ 2 }+{ v_w }^{ 2 } } =\sqrt { (35)^{ 2 }+{ (12) }^{ 2 } } \)  ms^-1 = 37 ms^-1
The direction θ that R makes with the vertical is given by
\(\tan { \theta } =\frac { v_w }{ v_r } = \frac { 12 }{ 35 } \)  = 0.343 or
 \(\theta =\tan ^{ -1 }{ \left( 0.343 \right) } =19°\) 
There, the boy should hold his umbrella in the vertical plane at an angle \(19°\) with vertical towards East.

Let the frequency of vibration of the string be given by 
\(\nu =Kl^{ a }m^{ b }T^{ c }\) ......(i)
where K = a dimensionless constant
Dimensions of the various quantities are
\(\nu =\left[ T^{ -1 } \right] ,l=\left[ L \right] ,T=\left[ T \right] \)
Force =\(\left[ MLT^{ -2 } \right] \)
and \(m=\frac { mass }{ length } =\left[ ML^{ -1 } \right] \)
Substituting these dimensions in equation(i), we get
\(\left[ T^{ -1 } \right] ={ \left[ L \right] }^{ a }\left[ ML^{ -1 } \right] ^{ b }\left[ MLT^{ -2 } \right] ^{ c }\)
or \(\left[ M^{ 0 }L^{ 0 }T^{ -1 } \right] =\left[ { M }^{ b+c }{ L }^{ a-b+c }{ T }^{ -2c } \right] \)
Equating the dimensions of M,L and T, we get
b + c = 0, a - b + c=0 and - 2c = - 1
on solving,\(a=-1,b=-\frac { 1 }{ 2 } and\quad c=\frac { 1 }{ 2 } \)
\(\therefore \quad \left( \nu \right) =Kl^{ -1 }m^{ { -1 }/{ 2 } }T^{ { 1 }/{ 2 } }\quad or\quad \left( \nu \right) =\frac { K }{ l } \sqrt { \frac { T }{ m } } \) 

Jet planes use atmospheric oxygen for burning fuel but rockets carry their own fuel and oxygen and do not depend on atmospheric oxygen.

When the external force is removed after the push, the mass gets detached when spring obtain its natural length and say, mass m rises h height from the pushed position.
      
Loss in potential energy of spring = Gain in gravitational potential energy
\(\frac { 1 }{ 2 } k[{ x }^{ 2 }]\ =mgh\)
\(\frac { 1 }{ 2 } \times 200[{ 0.1 }^{ 2 }]\quad =0.5\times 10\times h\)
1 = 5 h \(\Rightarrow \)  h = 0.2 m

\(R=41.3\ and \ \theta =24,1^{ o }\)

-8 m/s²

\(Impulse=force\times time=\left[ MLT^{ -2 } \right] \times \left[ T \right] =\left[ MLT^{ -1 } \right] \)

(i) Law of orbit: The planets including earth, go around the sun in elliptical orbits.
(ii) Law of area: The line joining the sun and the planet sweeps equal area in equal intervals of time.
(iii) Law of periods: The square of time period of revolution is directly proportional to the cube of semi-major axis of the elliptical orbit.

Angular momentum, L = r x p = r x mv = m (r x v)
\(\text { Now, } \mathbf{r} \times \mathbf{v}=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 0 & -2 \\ -5 & 0 & 5 \end{array}\right|\)
\(=\hat{\mathrm{i}}[(0 \times 5)-(0 \times-2)]-\hat{\mathrm{j}}[(2 \times 5)-(-2 \times-5)]\) \(+\hat{\mathbf{k}}[2 \times 0-(0 \times-5)]\)
= 0
So, angular momentum of particle is zero
\(\text { and } \tau=\text { torque }=\mathbf{r} \times \mathbf{F}=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 0 & -2 \\ 0 & 4 & 0 \end{array}\right|=8 \hat{\mathbf{i}}+8 \hat{\mathbf{k}} \mathrm{N}-\mathrm{m}\)

20m/s ²

Here \(\frac { 1 }{ 2 } { kx }^{ 2 }=\frac { 1 }{ 2 } { mv }^{ 2 }\)
or \(v=\sqrt { \frac { k }{ m } { x }^{ 2 } } =\sqrt { \frac { 100 }{ 100 } \times \frac { 25\times { 10 }^{ -4 } }{ 10^{ -3 } } } { ms }^{ -1 }\)
\(=\sqrt { 25 } \) ms^-1
Height = 2m = \(\sqrt { \frac { 2h }{ g } } =\sqrt { 0.4 } \)
The horizontal length covered = \(\sqrt { 0.4 } \times \sqrt { 25 } m\) 
= 1 m.

The weight will become zero under the following circumstances :
(i) during free fall
(ii) at the centre of the earth
(iii) in an artificial satellite
(iv) at a point where gravitational pull of earth is equal to the gravitational pull of the Moon.

 Time taken by man to go from his home to market
 \({ t }_{ 1 }=\frac { Distance }{ Speed } =\frac { 2.5 }{ 5 } h\)
Time taken by man to go from mae=rket to his home
\({ t }_{ 2 }=\frac { 2.5 }{ 5 } =\frac { 1 }{ 3 } h\)
Total time taken,t₁+ t₂ \(=\frac { 1 }{ 2 } +\frac { 1 }{ 3 } =\frac { 5 }{ 6 } h=50min\)
Distance moved in 30 in (from home to market)
= 25 km
Distance moved in 10 min (from market to home) with speed
7.5 km/h \(=7.5\times \frac { 10 }{ 60 } =1.25km\)
So, displacement = 2.5-1.25 = 1.25km
Distance travelled = 2.5+1.25 = 3.75 km
(a) Average velocity \(=\frac { 1.25 }{ (40/60) } =1.875km/h\)
(b) Average speed \(=\frac { 3.75 }{ (40/60) } =5.625km/h\)

(i) W = Fs cos \(\theta\) = 100 x 4 cos 0° = 400 J
(ii) Resistive force opposes the applied force. Box moves at 180^o to the resistive force.
Therefore W = Fs .cos 180 = -400 J. Since motion is along horizontal and gravity is along the vertical, therefore workdone by gravity is
W = Fs cos 90 = zero.
(iii) Force applied = tension in coupling = 2000 N
As P = Fv cos\(\theta\) = 2000 x 20 cos 0 ^o = 40000W = 40 kW
(iv) Work done = Power x time = 40 kW x 60 s = 2400 kJ

A) If both assertion and reason are true and the reason is the correct explanation of the assertion. 
If the soldiers while crossing a suspended bridge march in steps, the frequency of marching steps of soldiers may match with the natural frequency of oscillations of the suspended bridge. In that situation resonance will take place, then the amplitude of oscillation of the suspended bridge will increase enormously, which may cause the collapsing of the bridge. To avoid situations the soldiers are advised to go out steps on suspended bridge.

C) If assertion is true but reason is false.      
In circular motion the frictional force acting towards the centre of the horizontal circular path provides the centripetal force and avoid overturning of vehicle. Due to the change in direction of motion, velocity changes in circular motion.    

E) If assertion is false but reason is true.
Work done and power developed is zero in uniform circular motion only.

A) If both assertion and reason are true and the reason is the correct explanation of the assertion.