Let OAB be the inclined plane making an angle \(\alpha\) to the horizontal and let the projectile hit it at a point A where OA = r (Fig.).
At the instant the projectile hits the inclined plane, its horizontal and vertical displacements are ON (= r cos \(\alpha\)) and NA (= r sin \(\alpha\)) respectively.
Now the time taken to cover a horizontal distance r cos \(\alpha\) is clearly r cos \(\alpha\)/ v cos \(\theta\). The vertical distance moved in this time being r sin \(\alpha\), we have

\(r \ sin \alpha=(v \ sin \theta)({r \ cos \alpha \over v \ cos \theta})-{1\over2}g({r \ cos \alpha\over v cos \theta})^2\)
or \(r(sin \alpha - tan \theta cos \alpha)=-{1\over2}g{\cos^2\alpha\over v^2 \cos^2 \theta }r^2\)
\(\therefore r={2v^2cos^2\theta\over g \ cos^2 \alpha}[tan \theta cos \alpha-sin \alpha]\)
\(={2v^2cos^2\theta\over g \ cos^2 \alpha}[{sin \theta cos \alpha-cos \theta sin \alpha\over cos \theta}]\)
\(={2v^2cos^2\theta\over g \ cos^2 \alpha} sin (\theta - \alpha)\)
Thus the range of the projectile on the inclined plane is
\(={2v^2cos^2\theta\over g \ cos^2 \alpha} sin (\theta - \alpha)\)
The time at which the projectile hits the inclined plane being (r cos \(\alpha\) / v cos \(\theta\)), we have
\(t={r \ cos \alpha \over v \ cos \theta}={2v^2cos\theta\over g \ cos^2\alpha} sin (\theta - \alpha){cos \alpha\over v cos \theta}\)
or \(t={2 \ v \ sin (\theta - \alpha)\over g \ cos \alpha}\)
A confirmation of these results is obtained by putting \(\alpha\) = 0 for which case we get
\(r={2 \ v^2 cos \theta \ sin \theta \over g}={v^2sin 2\theta\over g}and \ t={2 \ v \ sin \theta \over g}\)
the usual expressions for the range and time of flight of a projectile.

Total momentum imparted to B
2 x 3.63 x 3.05 kg ms^-1
Velocity of B \(=\frac{2\times3.63\times3.05}{22.7}ms^{-1}=0.975\ ms^{-1}\)
Velocity of A when the cat jumps away from A
\(=\frac{3.63\times3.05}{22.7}ms^{-1}=0.4877\ ms^{-1}\)
When the cat comes back to A
Velocity of A \(=\frac{22.7\times0.4877+3.63\times3.05}{22.7+3.63}\ ms^{-1}\)
= 0.841 ms^-1

Let the particle executingSHMstarts oscillating from its mean position. Then displacement equation is
x = A sin ωt
∴ Particle velocity, v = Aພcosωt
∴ Instantaneous K.E., K = \(\frac { 1 }{ 2 } { mv }^{ 2 }=\frac { 1 }{ 2 } { mA }^{ 2 }{ \omega }^{ 2 }{ cos }^{ 2 }\omega t\)
∴ Average value of K.E. over one complete cycle
\({ K }_{ av }=\frac { 1 }{ T } \int _{ o }^{ T }{ \frac { 1 }{ 2 } { mA }^{ 2 }{ \omega }^{ 2 }{ cos }^{ 2 }\omega tdt=\frac { { mA }^{ 2 }{ \omega }^{ 2 } }{ 2T } = } \int _{ 0 }^{ T }{ { { cos }^{ 2 }\omega tdt } } \)
\(=\frac { { mA }^{ 2 }{ \omega }^{ 2 } }{ 2T } \int _{ 0 }^{ T }{ \frac { (1+cos2\omega t) }{ 2 } dt } \)
\(=\frac { { mA }^{ 2 }{ \omega }^{ 2 } }{ 4T } { \left[ t+\frac { sin2\omega t }{ 2\omega } \right] }_{ 0 }^{ T }\)
\(=\frac { { mA }^{ 2 }{ \omega }^{ 2 } }{ 4T } \left[ (T-0)+\left( \frac { sin2\omega t-sin0 }{ 2\omega } \right) \right] \)
\(=\frac { 1 }{ 4 } { mA }^{ 2 }{ \omega }^{ 2 }...(i)\)
Again instantaneous P.E., U = \(\frac { 1 }{ 2 } { kx }^{ 2 }=\frac { 1 }{ 2 } { m\omega }^{ 2 }{ x }^{ 2 }=\frac { 1 }{ 2 } { m\omega }^{ 2 }{ A }^{ 2 }{ sin }^{ 2 }\omega t\)
∴ Average value of P.E. over one complete cycle
\(U_{ av }=\frac { 1 }{ T } \int _{ o }^{ T }{ \frac { 1 }{ 2 } { m\omega }^{ 2 }{ A }^{ 2 }{ sin }^{ 2 }\omega t=\frac { { mA }^{ 2 }{ \omega }^{ 2 } }{ 2T } = } \int _{ 0 }^{ T }{ { { sin }^{ 2 }\omega tdt } } \)
\(=\frac { { m\omega }^{ 2 }{ A }^{ 2 } }{ 2T } \int _{ 0 }^{ T }{ \frac { (1-cos2\omega t) }{ 2 } dt } \)
\(=\frac { { m\omega }^{ 2 }{ A }^{ 2 } }{ 4T } { \left[ t-\frac { sin2\omega t }{ 2\omega } \right] }_{ 0 }^{ T }\)
\(=\frac { { m\omega }^{ 2 }{ A }^{ 2 } }{ 4T } \left[ (T-0)-\left( \frac { sin2\omega t-sin0 }{ 2\omega } \right) \right] \)
\(=\frac { 1 }{ 4 } { m\omega }^{ 2 }A^{ 2 }...(ii)\)
Simple comparison of (i) and (ii), shows that
\(K_{ av }=U_{ av }=\frac { 1 }{ 4 } { m\omega }^{ 2 }A^{ 2 }\) 

We know that C_p - C_v = R
and \({C_P\over C_v}=γ\)
From eqn. (ii) C_p = γC_v and sustituting this value in (i),
We have \(⋎C_v-C_v=R⇒C_v={R\over (⋎-1)}\)
\(C_v=γ.C_v={γR\over(γ-1)}\)

Here volume of room V = 25.0 m³ , temperature, T = 27⁰C = 300 K and
Pressure, P = 1 atm  = 1.01 x 10⁵ pa
According to gas equation,
PV = \(\mu \)RT = \(\mu \)N_A.K_B T
Hence, total number of air molecules in the volume of given gas,
N=\(\mu \). N_A = \(\frac { PV }{ { k }_{ B }T } \)
\(\therefore N=\frac { 1.01\times { 10 }^{ 5 }\times 25.0 }{ (1.38\times { 10 }^{ -23 })\times 300 } =6.1\times { 10 }^{ 26 }\) 

For first stone
x (0) 200 m, v (0) = 15 ms^-1, a = -10 ms^-2
x₁ (t) X(0) + v (0) t + \(\frac{1}{2}\) a t²
x₁ (t) 200 + 15t - 5t²
When the first stone hits the ground, x₁ (t) = a
∴ 5t² + 15t + 200 a
On simplification, t = 8 s
For second stone, X(0) = 200 m, v (0) = 30 ms^-1, a = -10 ms^-2
 x₂ (t) 200 + 30t - 5t²
When this stone hits the ground, x₂(t) = a
∴  -5t² + 30t + 200 = 0
Relative position of second stone w.r.t. first is given by x₂ (t) - x₁(t) = 15t
Since there is a linear relationship between x₂(t) - X₁ (t) and i, therefore the graph is a straight line.
For maximum separation, t = 8 s
So maximum separation is 120 m
After 8 second, only the second stone would be in motion. So, the graph is in accordance with the quadratic equation.

As, \(v=\frac { 1 }{ 2\pi } \sqrt { \frac { k }{ m } } \)
\(k=4\pi ^{ 2 }m{ v }^{ 2 }\)
For maximum displacement \({ y }_{ max }=A\) 
Maximum restoring force,
F = - kA =- mg
or \(A=\frac { mg }{ k } =\frac { mg }{ 4{ \pi }^{ 2 }{ mv }^{ 2 } } =\frac { g }{ 4\pi ^{ 2 }{ v }^{ 2 } }\)
\( \\ =\frac { 9.8 }{ 4\times { (3.14) }^{ 2 }\times { (0.50) }^{ 2 } } =0.99m\)

Here, \(\mu =5.0,\quad T={ 7 }^{ \circ }C=273+7\quad =280K\)
Number of atoms =\(\mu { N }_{ A }=5.0\times 6.02\times { 10 }^{ 23 }\approx 30\times { 10 }^{ 23 }\)

Given, T₁ = 27 °C
L_T1 = 5.231 m
L_T2 = 5.243 m
So, L_T2 =L_T1 [1+αl (T₂ –T₁ )]
5.243 m = 5.231 m [1 + 1.20×10^–5 K^–1 (T2 –27 °C)] or T² = 218 °C

As the triple point of water on Celsius is 0.01⁰C (and not 0⁰ C) and on kelvin scale 273.16 and the size of degree on the two scale is same, so 
t_c - 0.01 = T - 273.16
t_c = T - 273.15

Let y be the depth of cube in mercury. The depth of cube in water will be (0.06 - y) m. The buoyant force on cube due to mercury can be find out by using
Archimedes' principle.
B₁ = (0.06)² x y x (13.6 x 10³)x  9.8 N
Similarly, due to water the buoyant force on cube is
B₂ = (0.06)² x (0.06 - y) x 10³ x 9.8 N
When cube is in equilibrium, the sum of both the buoyant forces will be equal to the total weight of iron
cube. B₁ + B₂ = weight of iron cube
or  = (0.06)² x 10³ x 9.8 [13.6 Y + (0.06 - y)]
= (0.06)³ x 7.7 x 10³ x 9.8
 \(y=\frac { 0.396 }{ 12.6 } =0.032m\)
Fraction of cube inside mercury = \(\frac { 0.032 }{ 0.06 } =0.533=\) 53%

\(1.76\times 10^{ -2 }{ N }/{ m }\)

Given, mass(m) = 14 .5kg
Length of wire (l ) = 1 m
Angular frequency (v) = 2 revls
Angular velocity (\(\omega \)) = 2\(\pi \)v
= 2\(\pi \)\(\times \)2 rad/s = 4\(\pi \) rad/s
 
Area of cross-section of wire (A) = 0.065 cm²
= 6.5 \(\times \)10^-6 m²
Young's modulus for steel (Y) = 2 \(\times \) 10¹¹ N/m².
At lowest point of the vertical circle, T - mg = ml\({ \omega }^{ 2 }\)
or  T= mg +  m\({ \omega }^{ 2 }\)
= (14.5\(\times \)9.8)+14.5\(\times \)1\(\times \)\(​​{ (4\pi ) }^{ 2 }\)
= 14.5(9.8+16\({ \pi }^{ 2 }\))
= 14.5(9.8\(\times \)16\(\times \)9.87)          [\(\because \) \({ \pi }^{ 2 }\)=9.87]
= 14.5\(\times \) 167.72N=2431.94 N
Young's modulus (Y) =\(\frac { Stress }{ Strain } =\frac { (T/A) }{ \Delta l/l } =\frac { Tl }{ A.\Delta l } \)
\(\therefore \Delta l=\frac { T.l }{ A.Y } =\frac { 2431.94\times 1 }{ 6.5\times { 10 }^{ -6 }\times 2\times { 10 }^{ 11 } } \)
= 1.87\(\times \)10^-3 m =1.87 mm

(i) Here, M = 125 kg,
v₁ = 4 m/s,
m = 85 kg
Applying, conservation of linear momentum, we get
 p_initial = p_final  i.e., Mv₁ = ( M + m ) v_f
The value of final speed
v_f = \(\frac { M{ v }_{ i } }{ M+m } \) = \(\frac { 125\times 4 }{ (125+85) } \) = 2.38 m / s
(ii) Average force exerted to the quarterback's femur
So,   F_av x \(\Delta \) t = M ( v_f - v_i )
i.e.  Fav = \(\frac { M({ v }_{ f }-{ v }_{ i }) }{ \Delta t } \)  = \(\frac { 125(4-2.38) }{ 0.1 } \) 
= \(\frac { 125\times 1.62 }{ 0.1 } \) = 2.03 x 10³ N
Shearing stress = \(\frac { F }{ A } \) = \(\frac { 2.03\times { 10 }^{ 3 } }{ 5\times { 10 }^{ -4 } } \) 
= 4.06 x 10⁶ Pa

\(g=\frac { GM }{ { R }^{ 2 } } \ or\ \ M=\frac { gR^{ 2 } }{ G } \)
This relation is true not only to the earth but for any heavenly body which is assumed to be spherical
\(g=1.67ms^{ -2 },\ R=1.74\times 10^{ 6 }m\)
\(\\ G=6.67\times 10^{ 11 }Nm^{ -2 }kg^{ -2 }\)
\(\\ \therefore Mass\ of\ the\ moon,\ M=\frac { 1.67\times (1.74\times 10^{ 6 })^{ 2 } }{ 6.67\times 10^{ -11 } } kg\)
\(\\ =7.58\times 10^{ 22 }kg\)

To solve this question, we use the Kepler’s third law. This states that the square of the period of the revolution of all the planets about the sun is directly proportional to the cube of the mean distance between the planets and the sun. This can be mathematically given as
T2 ∝ R3
⇒T2=kR3 where  T is the period of the revolution of a planet around the sun,  R is the mean radius of the planet to the sun.
Hence, we can write by comparison between two planets, that
\(\frac{T_1^2}{T_2^2}=\frac{R_1^3}{R_2^3}\)
We can use any planet with a known distance and period as the second planet. We choose earth.
\(\frac{T_N^2}{T_E^2}=\frac{R_N^3}{R_E^3}\) where the subscript N and E stands for Neptune and Earth respectively.
For earth, the period is 1 year. Hence, write that
\( \frac{165^2}{1^2}=\frac{R_N^3}{R_E^3} \)
\( \Rightarrow R_N^3=165^2 R_E^3\)
Hence, by finding the cube root of both sides, we have
\(R_N=\sqrt[3]{165^2} R_E\)
⇒RN=30RE
Radius of the earth is about 1.50×10¹¹m
Hence, RN=30(1.50×1011)=4.5×10¹²m

Work done by torque in turing rotor by angle d \(\theta \) is
= \(\tau \)  d \(\theta \)
So, power delivered by engine
 P = \(\frac { Work\ done }{ Time\ taken } =\tau \frac { d\theta }{ dt } \)
[dt= time for turing by angle dθ]     
or   P = \(\tau \omega \)
So, power required = 180 x 200 = 36000 W
= 36 k W     [ 1 k W = 1000 W]

In this question we are given a solid uniform cone of radius R and height h, we need to find the center of mass from the vertex. We need to use the below diagram to understand the problem.
Let us consider the radius of the base of the cone to be R and height is h.
We will consider a small circular cross section of radius ‘z’ and thickness ‘dr’ from a distance ‘r’ from the vertex of the solid cone.
Consider two triangles ABC and AOD. These triangles are similar triangles, so we can write,
AB/AO=BC/OD
This can be expressed as,
r/h=z/R…… (1)
The volume of the small volume element considered is, dV=πz²dr,
which can be written in terms of r from equation (1). So we get,
\(d V=\frac{\pi R^2 r^2}{h^2} d r\) ...(2)
Center of mass for continuous mass distribution is given by the formula,
\(C . M=\frac{1}{M} \int r \rho \frac{\pi R^2 r^2}{h^2} d r \)
\( \Longrightarrow C . M=\frac{\pi R^2 \rho}{M h^2} \int r^3 d r\)
The limits of integration are from 0 to h, so
\(C . M=\frac{\pi R^2 \rho}{M h^2} \int_0^h r^3 d r\)
On integrating and applying the limits we get,
\(C . M=\frac{\pi R^2 \rho h^2}{4 M}\)
Substituting M as \(M=\rho V=\frac{\rho}{3 \pi R^2 h}\), we can write
\(\mathrm{C} \cdot \mathrm{M}=\frac{3 h}{4}\)
At height \(\frac { h }{ 4 } \) from the point, O. O is at the centre of the base of the cone

Here, m₁ = 4 kg, m₂ = 5 kg
Force applied on block P = 12 N
This force must be equal to the kinetic .friction applied on P by Q.
\(12={ F }_{ k }={ \mu }_{ k }R={ \mu }_{ k }{ m }_{ 1 }g \)
\(\\ 12={ \mu }_{ k }\times 4g\ or\ { \mu }_{ k }={ \frac { 12 }{ 4g } }=\frac { 3 }{ g }\)
acceleration of P=\(\frac { 12 }{ 4 } =3m/s\)
 
When force F is applied on Q to create common motion in P & Q, the forces created (friction) is shown in the diagram above. F, force will be created on P and F₂ force will be created on Q.
Considering forces on Q, we have
 \(F-{ F }_{ 2 }(net\quad force)={ m }_{ 2 }\times a={ m }_{ 2 }\times 3\)
Since,
 \({ F }_{ 1 }={ F }_{ 2 }={ \mu }_{ 1 }{ m }_{ 1 }g\) 
\(\\ F-{ \mu }_{ k }{ m }_{ 2 }g={ m }_{ 2 }\times 3\)
\(\\ F={ \mu }_{ k }{ m }_{ 2 }g+{ m }_{ 2 }\times 3=\frac { 3 }{ g } \times 4\times g+5\times 3=12+15=27N\)
As this force moves both the blocks together on a smooth table, so the acceleration produced is
\(a=\frac { F }{ { m }_{ 1 }+{ m }_{ 2 } } =\frac { 27 }{ 4+5 } =3m/s\) 

Given, mass of the particle (m) = 500 g = 0.5 kg
x-t graph of the particle is a straight line.
Hence, particle is moving with a uniform velocity along x-axis, i. e. its acceleration along x-axis is zero and hence, force acting along x-axis is zero.
y- t graph of particle is a parabola. Therefore, particle is in accelerated motion along y-axis.
At     t = 0, u_y= 0
Along y-axis, at t = 2s, y =4m
Using equation of motion,
 \(y={ u }_{ y }t+\frac { 1 }{ 2 } { a }_{ y }{ t }^{ 2 }\) 
\(\\ 4=0\times 2+\frac { 1 }{ 2 } \times { a }_{ y }\times { (2) }^{ 2 }\)
\(\\ { a }_{ y }=2m/{ s }^{ 2 }\)
Force acting y-axis (f_y) = ma_y
 \(0.5\times 2=1.0N\)   (along y-axis)

Here, v = 27 kmh^-1 = 27\(\times \)(1000 m) \(\times \)(60\(\times \)60 s)^-1 = 7.5 ms^-1, r = 80 m
Centripetal acceleration, a_c=\(\frac { { v }^{ 2 } }{ r } =\frac { { (7.5) }^{ 2 } }{ 80 } \) = 0.7 ms^-2
Let the cyclist applies the brakes at the point P of the circular turn, then tangential acceleration a_T will act opposite to velocity.
Acceleration along the tangent, a_T = 0.5 ms^-2
Angle between both the accelerations is \(90°\)
Therefore, the magnitude of resultant acceleration \(a=\sqrt { { a }_{ C }^{ 2 }+{ a }_{ T }^{ 2 } } =\sqrt { { (0.7) }^{ 2 }+{ (0.5) }^{ 2 } } \)
   
Let the resultant acceleration make an angle \(\beta \) with the tangent i. e. the direction of net acceleration of the cyclist then, \(\tan { \beta } =\frac { { a }_{ C } }{ { a }_{ T } } =\frac { 0.7 }{ 0.5 } \) = 1.4 or \(\beta =54°28\prime \)  

The number of strands of hair on our head
If we assume a uniform distribution of strands of hair on head then, the number of strands of hair
= Area of the head/Area of cross-section of hair
The thickness of a strand of hair is measured by an appropriate instrument, if it is obtained
d = 5 x 10^-5 = 5 x 10^-3 cm
Then, area of cross-section of human hair
 \(=\pi \left( \frac { d }{ 2 } \right) ^{ 2 }=\frac { \pi d^{ 2 } }{ 4 } \)
\(\\ =\frac { 3.14\times (5\times 10^{ -3 })^{ 2 } }{ 4 } =\frac { 3.14\times 25 }{ 4 } \times 10^{ -6 }cm^{ 2 }\)
Average radius of human head (r)=8 cm
 \(\because \) Area of human head \(=\pi r^{ 2 }=3.14\times (8)^{ 2 }\)
3.14 x 64 cm²
\(\therefore \)The number of strands of hair \(=\frac { 3.14\times 64 }{ 3.14\times \frac { 25 }{ 4 } \times 10^{ -6 } } \)
 \(\approx 10\times 10^{ 6 }=10^{ 7 }\)

(i) Mass vaccination is good as, it is used to make the socity free from the diseases like small pox.
(ii) Television for eradication of illiteracy and for mass communication of news and ideas is good as, it is a medium which is easily within the reach of common man and also they are very habitual to it.
(iii) Prenatal sex determination is bad because people are misusing it. Some of the people after determination of sex of child, think to abort. They do it especially with girl child.
(iv) Computer for increase in work efficiency is good as using the computer, a man can do much more work with greater efficiency and accuracy.
(v) Putting artificial satellite into orbits around the Earth is good for development as these satellites serve many purpose like remote sensing, weather foresting.
(vi) Development of nuclear weapons is bad as they can be used in mass destruction.
(vii) Development of new and powerful tool of chemical and biological warfare are bad, as they can also be used for mass destruction.
(viii) Purification of water for drinking purpose is good as we can save ourself from the diseases which we can have due to drinking of the water.
(ix) Plastic surgery is good as with the help of it a man or women can remove the skin defects occurring due to accident or some other reasons. It has some bad effects too but they are not very considerable.
(x) Cloning is good as far as animals are concerned. With the help of it, we can develop some species of animals which can be used to serve some specific purpose. But it is not good for human beings.

We know, when the lift accelerates upward, the reaction R] on the base is given by
R₁ = Mg + Ma = M (g + a)
⇒ R₁ = 60 (9.8 + 2) = 60 x 11.8 = 708 N.
When the lift accelerates downward with acceleration a, the reaction R₂ on the base given by
R₂ = Mg - Ma = M (g - a)
⇒ R₂ = 60 (9.8 - 2) = 60 x 7.8 = 468 N
Therefore, R₁ > R₂.

Yes, e.g., circular motion is periodic but not oscillatory.

No, since torque cannot be applied.

This is because sound is transmitted both through air and medium.

As the energy is recoverable therefore the force is conservative force.

Thermal conductivity, electrical conductivity and compressibility.

Parsec is bigger unit than light year (1 parsec = 3.26 light year).

68.25 K

Yes, it is possible to describe the difference of length to remain constant. So, the change in length of each rod must br equal at all temperature. Let L_b and L_s be the length of the brass and the steel rod \(\alpha _{ b }\)and  \(\alpha _{ s }\)be the coefficients of linear expansion of the two metals. Let there is change in temperature be \(\triangle T\)
Then \(\alpha _{ b }L_{ b }\triangle T=\alpha _{ s }L_{ s }\triangle T\)
or \(\quad \alpha _{ b }L_{ b }=\alpha _{ s }L_{ s }\quad \quad \Rightarrow \quad L_{ b }/L_{ s }=\alpha _{ s }/\alpha _{ b }\)
Hence, the length of the rods must be in the inverse ratio of the coefficient of linear expansion of their materials.

Given, depth of ocean (h) = 3km = 3000m
Density of water\((\rho )\) = \({ 10 }^{ 3 }kg/m^{ 3 }\)
Pressure exerted by water column
\(p=h\rho g=3000\times 10^{ 3 }9.8\)
\(\\=29.4\times 10^{ 6 }Pa=2.94\times 10^{ 7 }Pa\)
Maxim stress which can be withstand by the vertical off-shore structure = \({ 10 }^{ 9 }Pa\)
As, \({ 10 }^{ 9 }Pa>2.9\times { 10 }^{ 7 }Pa\)
Therefore, the vertical structure is suitable for putting up in top of an oil well in the ocean.

The safety of a person depends upon the momentum with which the person hits the planet. Since, the mass of the person is constant, therefore the maximum velocity v is the limiting factor.

We infer that we should avoid junk food and do physical work instead of depending on machines.

Here, actual path length travelled, s = 23 km; Displacement = 10 km;
Time taken, t  = 28 min = \(\frac{28}{60} h\)
(a) Average speed of taxi = \(\frac{\text { actual path length }}{\text { time taken }}=\frac{23}{\frac{28}{60}} k \frac{m}{h}=49.3 \mathrm{~km} / \mathrm{h}\)
(b) Magnitude of average velocity = \(=\frac{\text { displacement }}{\text { time taken }}=\frac{10}{\frac{28}{60}} \mathrm{~km} / \mathrm{h}=21.4 \mathrm{~km} / \mathrm{h}\)
The average speed is not equal to the magnitude of average velocity. The two are equal for the motion of taxi along a straight path in one direction.

143.4

Given  v = krt
Since velocity changes with time, the motion in circular path involves tangential acceleration.
So it is a non-uniform circular motion.
\(a_r={v^2\over r}=k^2 \ rt^2\)
or  \(a_k={dv\over dt}=kr\)
Net acceleration = an =\(a_n=\sqrt{a^2_r+a^2_t}\)
\(=\sqrt{(k^2rt^2)^2+(kr)^2}=kr\sqrt{1+k^2t^4}\)

The entire structure of the kinetic theory of gases is based on the following assumptions which were first stated by Classius.
1. A gas consists of a very large number of molecules (of the order of Avogadro's number, 10²³), which are perfect elastic spheres. They are identical in all respects for a given gas and are different for different gases.
2. The molecules of a gas are in a state of incessant random motion. They move in all directions with different speeds, (of the order of 500 m/s) and obey Newton's laws of motion.
3. The size of the gas molecules is very small as compared to the distance between them. If typical size of a molecule is 2 \(\dot { A } \) .average distance between the molecules is \(\ge \)20 A. Hence volume occupied by the molecules is negligible in comparison to the volume of the gas.
4. The molecules do not exert any force of attraction or repulsion on each other, except during collision.
5. The collisions of the molecules with themselves and with the walls of the vessel are perfectly elastic. As such the momentum and the kinetic energy of the molecules are conserved during collisions, though their velocities change.
6. There is no concentration of the molecules at any point inside the container i.e., molecular density is uniform throughout the gas.
7. A molecule moves along a straight line between two successive collisions and the average straight distance covered between two successive collisions is called the mean free path of the molecules
8. The collisions are almost instantaneous, i.e., the time of collision of two molecules is negligible as compared to time interval between two successive collisions.
Numerical:
We know that
\(P=\frac { 1 }{ 3 } \rho { v }^{ 2 }\)
\(\\ \therefore { v }_{ rms }=\sqrt { ({ v }^{ 2 }) } =\left( \frac { 3p }{ \rho } \right) \)
Given that P = 1,0 x 10⁵  newton / meter² and
\(\rho =1.98\ kg/meter^{ 3 }\)
\(\\ \therefore \ { v }_{ rms }=\sqrt { \left[ \left\{ \frac { 3\times (1.0\times { 10 }^{ 5 } }{ 1.98 } \right\} \right] } =389\ meter/sec\)
From kinetic theory of gases, the root mean square speed is directly proportional to the square root of absolute temperature
\({ v }_{ rms }\propto \sqrt { T } \)
\(\\ \therefore \ \frac { { (v }_{ rms })_{ 30 } }{ { (v }_{ rms })_{ 0 } } =\sqrt { \left[ \left( \frac { 273+30 }{ 273+0 } \right) \right] } =\sqrt { \left[ \left( \frac { 303 }{ 273 } \right) \right] } =1.053\)
\(\\ or\ { (v }_{ rms })_{ 30 }={ (v }_{ rms })_{ 0 }\times 1.053\)
\(\\ =389\times 1.053=410\ meter/sec\)

The linear speed at 3 seconds is
v = 2 x 3 = 6 m/s
The radial acceleration at 3 seconds
\(={v^2\over r}={6\times6\over 0.2}=180 m/s^2\)
The tangential acceleration is given by
\({dv\over dt}=2, Since \ v=2t\)
\(\therefore\) tangential acceleration is 2 m/s².
Total acceleration \(=\sqrt{a^2_r+a_t^2}=\sqrt{180^2+2^2}=\sqrt{32400+4}\)
\(=\sqrt{32404} ms^{-2}\)

Here, Young's modulus, \(\Upsilon \) = 2 x 10¹¹Nm^-2
Area of cross-section, A = 5 x 10^-5 m²
Let the initial length of wire be L. Then, increase in length of wire,
ΔL= L
Now, \(\Upsilon \)=\(\frac { F\times L }{ A\times \triangle L } \)
∴ F = \(\frac { \Upsilon \times A\times \triangle L }{ L } \)
⇒ F = \(\frac { 2\times 10^{ 11 }\times 5\times 10^{ -5 }\times L }{ L } \)
or F = 10⁷ N.

\(\theta =2000"\)
\(\\ =\frac { 2000 }{ 3600 } \times \frac { \pi }{ 180 } rad\)
\(d=1.496\times { 10 }^{ 11 }\)

From the figure,
\(\theta =\frac { D }{ d } \)
\(\\ D=\theta d\)
= 9.7 x 10^-3 x 1.496 x 10¹¹
= 1.45 x 10⁹ m

Total energy at B = kinetic energy + potential energy
= 0 + mgh
= 2 \(\times\) 9.8 \(\times\)10
= 196 J
As it descends half the height, it loses potential energy which is given by
\(=mg\frac { h }{ 2 } \)
\(=\frac { 1 }{ 2 } mgh=98\ J\)
\(\therefore\) its potential energy at C = (196 - 98) = 98 J
The loss of potential energy = gain in kinetic energy
= 196 - 98
= 98 J
But \(K.E.=\frac { 1 }{ 2 } { mv }^{ 2 }\)
\(\therefore \quad \frac { 1 }{ 2 } \times 2\times { v }^{ 2 }=98\)
\(\Rightarrow \ { v }^{ 2 }=98\ or\ v=7\sqrt { 2 } m/s\)

An extremely thin layer of liquid can be considered as the collection of large number of hemispherical drops. In case of a spherical drop, the excess of pressure = \(\frac{2T}{r}\) But in case of thin layer of liquid, which is a combination of hemispherical drops, the excess pressure is \(P=\frac{T}{r},\) where \(r=\frac{d}{2}\)
Therefore,
\(P=\frac{T}{\frac{d}{2}}=\frac{2T}{d}\)
Force due to surface tension pushing the two plates together is
\(F=P\times A=\frac{2TA}{d}\) 

Many phenomena at microscopic level are not explained by classical theory. So quantum theory is needed, e.g., photoelectric effect, interaction among elementary particles.

Given, h = 300m, g = 9.8m/s^2 , v = 340ms^-1
t₁ = time taken by stone to strike the water surface 
 \(t_{ 1 }=\sqrt { \frac { 2h }{ g } } =\sqrt { \frac { 300 }{ 49 } } =7.82s\left( as\quad h=0+\frac { 1 }{ 2 } gt^{ 2 }_{ 1 } \right) \)
t₂ = time taken by the splash's sound to reach top of the tower
\(t_{ 2 }=\frac { h }{ v } =\frac { 300 }{ 340 } =0.882\quad \left[ v=\frac { h }{ t_{ 2 } } \right] \)
Total time, t = time to hear splash of sound 
= t₁ + t₂ =7.82 + 0.882
= 8.702

(i)\(\therefore \) process is adiabatic
\(\therefore \)    \(\Delta \)Q = 0
Work done on the gas,\(\Delta \)W = -500J
According to the first law of thermodynamics.
\(\Delta \)Q = \(\Delta \)U +\(\Delta \)W \(\Rightarrow \) \(\Delta \)U = -\(\Delta \)W = 500J
(ii) This is because heat absorbed by a substance (coolant) is directly proportional to the specific heat of the substance.
(iii) This is because in a harbour town, the relative humidity is more than in a desert town. Hence, the climate of a harbour town is without extremes of hot and cold.

As the brass scale of a barometer gives correct reading at T₁ = 0⁰ C , hence at temperature T₂ = 27⁰ C., the scale will expand and will not give correct reading
In such, true value
= Observed scale reading x\((1+\alpha \triangle T)\)
True pressure = 75.00 cm x [1 + 2.0 x 10^-5 x (27 - 0)]
= 75 x (1 + 2.0 x 10^-5 x 27]
= 75.00(1 + 54 x 10^-5) cm = 75.04 cm

(i) According to law of conservation of angular momentum.L = mvr = constant, therefore the comet moves faster when it is close to the sun and moves slower when it is father away from the sun. Therefore, the speed of the comet does not remain constant
(ii) As the linear speed varies, the angular speed also varies.Therefore, angular speed of the comet does not remain constant
(iii) As no external torque is acting on the comet, therefore, according to law of conservation of angular momentum, the angular momentum of the comet remain constant
(iv) Kinetic energy of the comet = \(\frac { 1 }{ 2 } { mv }^{ 2 }\)
As the linear of the comet changes, its kinetic energy also changes.Therefore, its KE does not remain constant
(v) Potential energy of the comet changes as its kinetic energy changes.
(vi) Only angular momentum and total energy of a comet remain constant throughout its orbit.

Let t be the time taken by the ball to reach the highest point where height from ground be s.
Taking vertical upward motion of the ball, we have u =- 29.4ms^-, a = 9.8ms^-, v = 0, s = S,t = ?
As, v²-u² = 2 as
0 -(-29.4)² = 2 x 9.8 x S
or \(S=\frac { -{ (29.4) }^{ 2 } }{ 2\times 9.8 } =-44.1m\)
Here, negative sign shows that the distance is covered in upward direction.
As, v = u + at
\(\therefore \quad 0=-29.4+9.8\times t\quad or\quad t=\frac { 29.4 }{ 9.8 } =3s\)
It means time of ascent = 3s
When an object moves under the effect of gravity alone, the time of ascent is always equal to the time of descent.
Therefore, total time after which the ball returns to the player's hand = 3 + 3 = 6s.