Centrifugal force.

Two vectors having the same initial point are called co-initial vectors. Two vectors which either act along the same line or along parallel lines are called collinear vectors.

Velocity after 5 seconds
= v = u + at = 0 + 5 x 5 = 25 ms^-1
Distance travelled in 5 seconds
\(\mathrm{S}=u t+\frac{1}{2} a t^{2}\)
\(=0+\frac{1}{2} 5 \times 5^{2}=62.5 \mathrm{~m}\)

Power \(=\frac { \text{Work done }}{ \text{Time taken} } =\frac { mgh }{ t } \)
\(=\frac { 20\times 1000\times 9.8\times 180 }{ 60\times 60 }\)
=9800 W=9.8  kW

27657.1 mm²

The various forces acting on M and m are
T = Mg                ............(i)
T cos θ = mg          ..............(ii)
\(\mathrm{T} \sin \theta=\frac{m v^{2}}{r}=m r \omega^{2}\)            .............(iii)
where r is the radius of the circular path and ω is the angular velocity.
r = I sin θ
From (iii) T sin θ = mrω² = m(l sin θ)ω²,
But T = Mg
ஃ Mg sin θ = ml sin θ ω²
or  \(\omega^{2}=\frac{\mathrm{Mg}}{m l}\)
or  \(v=\frac{\omega}{2 \pi}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{Mg}}{m l}}\)
Thus, the frequency of rotation of m, so that M remains stationary is given by
\(v=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{Mg}}{m l}}\) 

The second stone will 'catch up' with the first stone when the distance covered by it in (t - n) second will equal the distance covered by the first stone in t second.
Now distance covered by the first stone in t second = '2 gt² and distance covered by the second stone in (t-n) second.
= u (t-n) + \(\frac { 1 }{ 2 } \)g (t-n)²
∴ \(\frac { 1 }{ 2 } \) gt² = u(t-n) + \(\frac { 1 }{ 2 } \)g (t-n)²
or \(\frac { 1 }{ 2 } \)g [t²- (t-n)²] = u(t-n)
or \(\frac { 1 }{ 2 } \)g [(2t-n)n] = u(t-n)
or gnt- \(\frac { 1 }{ 2 } \)or gn² = ut-un
or t(gn-u) = \(\left( \frac { 1 }{ 2 } gn-u \right) n\)
or \(t=\frac { n\left( \frac { 1 }{ 2 } gn-u \right) }{ (gn-u) } \)
The distance covered by the first stone in this time is
h   \(\frac { 1 }{ 2 } gt^{ 2 }=\frac { 1 }{ 2 } g\left[ \frac { n\left( \frac { 1 }{ 2 } gn-u \right) }{ (gn-u) } \right] ^{ 2 }\)
Thus the second stone will overtake the first at distance
\(\frac { 1 }{ 2 } g=\frac { 1 }{ 2 } g\left[ \frac { n\left( \frac { gn }{ 2 } -u \right) }{ (gn-u) } \right] ^{ 2 }\)below the top of the cliff.

The horizontally directed force 400 N and weight 20 kg of the block are resolved into two mutually perpendicular components, parallel and perpendicular to the plane as shown.
N =20 g cos 45° + 400 sin 45° = 421.4 N
The frictional force experienced by the block F = \(\mu\)N =\(\mu\) x 421.4 = 421.4 \(\mu\)N.

As the accelerated motion is taking place up the plane.
400 cos 45° - 20g sin 45° - f = 20a
\({400\over \sqrt{2}}-{20\times 9.8\over \sqrt{2}}-421.4\mu=20a=20\)
\(\mu=({400\over \sqrt{2}}-{196\over \sqrt{2}}-12)\times {1\over 421.4}\)
\(={282.8-138.6-12\over 421.4}=0.3137\)
The coefficient of sliding friction between the block and the incline = 0.3137

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.

We know that the usual dimensions of Y are
\(\frac { [ML{ T }^{ -2 }] }{ [{ { L }^{ 2 }] } } ,i.e.,[M{ L }^{ -1 }{ T }^{ -2 }]\)
To express these in terms of F, A and V, we must express, M, Land T in terms of these new 'fundamental' quantities.
\([V]=[L{ T }^{ -1 }],[A]=[L{ T }^{ -2 }],[F]=[ML{ T }^{ -2 }]\)
\(\\ M=F{ A }^{ -1 },T=V{ A }^{ -1 },L={ V }^{ 2 }{ A }^{ -1 }\)
\(\\ [Y]=[M{ L }^{ -1 }{ T }^{ -2 }]\)
\(=[F{ A }^{ -1 }][{ V }^{ 2 }{ A }^{ -1 }]{ [V{ A }^{ -1 }] }^{ -2 }\ \)
\(=F{ A }^{ 2 }{ V }^{ -4 }\)
Thus the 'new' dimensions of Young's modulus are \([F{ V }^{ -4 }{ A }^{ 2 }]\)

Mass of stone, m = 0.25 kg, Radius of the string, r = 1.5 m
Frequency, v = 40rev/min = \(\frac { 40 }{ 60 } \) rev/s = \(\frac { 2 }{ 3 } \) rev/s
Centripetal force required for circular motion is obtained from the tension in the string.
\(\therefore \) Tension in the string = Centripetal force
T = \(mr{ \omega }^{ 2 }\)
\(= mr\left( 2\pi n \right) ^{ 2 } \ \left[ \therefore \omega =2\pi v \right] \)
\(= mr4\pi ^{ 2 }{ v }^{ 2 }\)
\(T=0.25\times 1.5\times 4\times \left( \frac { 22 }{ 7 } \right) ^{ 2 }\times \left( \frac { 2 }{ 3 } \right) ^{ 2 }= 6.6N\)
Maximum tension which can be withstand by the string
\({ T }_{ max }=200\quad N=\frac { mv^{ 2 }max }{ r } \)
\(\\ { v }_{ max }=\sqrt { \frac { { T }_{ max }\times r }{ m } } =\sqrt { \frac { 200\times 1.5 }{ 0.25 } } =34.6{ m }/{ s }\)

The net force exerted on the stone by both of them will be
 \(F_{ R }=\sqrt { { F }^{ 2 }+{ F }^{ 2 }+2{ F }^{ 2 }\cos { { 60 }^{ 0 } } } \)
\( =\sqrt { 2{ F }^{ 2 }+2{ F }^{ 2 }\times 1/2 } \)
\(=\sqrt { 2{ F }^{ 2 }+{ F }^{ 2 } } =\sqrt { 3 } F\)

The correct option is A T=\(k \sqrt{ρr^3/S}\)
Dimensional formula of,
Time period, T
[T]=[M⁰L⁰T¹]
Surface tension, S
[S]=[ML⁰T⁻²]
Radius, r
[r]=[M⁰LT⁰]
Density, ρ
[ρ]=[M¹L⁻³T⁰]
Let us suppose the relation is
T=kρ^ar^bS^c
⇒[T]=[kρ^ar^bS^c]
⇒[M⁰L⁰T¹]=[M¹L⁻³T⁰]^a × [M⁰LT⁰]^b × [ML⁰T⁻²]^c
⇒[M⁰L⁰T¹]=[M^(a+c)L^(−3a+b)T^(−2c)]
On comparing, we get,
a+c=0(1)
−3a+b=0(2)
−2c=1(3)
On solving these equations, we get,
a=1/2,b=3/2 and c=−1/2
Thus,
T=kρ^ar^bS^c ⇒\(T=\frac { { pr }^{ 3 } }{ s } \)

(a) 5.3 m,
(b) 7.9 m and
(c) 69 m

v = \(\frac{dx}{dt}\)
Since v = \(\alpha\sqrt{x}\)
we have \( \frac{dx}{dt}=\alpha\sqrt{x} \) or \(\frac{dx}{\sqrt{x}}=\alpha dt\)
Intefrating from \( t=0(x=0) to \ t=t(x=x)\)
we have \( \overset { x }{ \underset { 0 }{ \int } } { x }^{ -1/2 }dx=\alpha \overset { t }{ \underset { 0 }{ \int } } dt\)
∴ \({ \left| \frac { { x }^{ 1/2 } }{ 1/2 } \right| }_{ 0 }^{ x }=\alpha t\)
or \( x = \frac{\alpha^2t^2}{4}\)
The time dependence of the velocity is obtained by differentiating both sides of this relation w.r.t. time t. Thus
\(v=\frac{dv}{dt}=\frac{\alpha.2t}{4}=\frac{\alpha^2}{2}t\)
The velocity x of the particle is thus increasing in direct proportion to time.
Similarly, the time dependence of acceleration is obtained by differentiating both sides of this relation w.r.t. 't', Thus
\( a=\frac{dv}{dt}=\frac{\alpha^2}{2}\)
The particle is thus moving with a constant acceleration.
To find the average velocity over the first s metre, we assume that the time taken to cover this distance is T. Using
\( x=\frac{\alpha^2t^2}{4}\)
we get s= \(\frac{\alpha^2T^2}{4} or \ T=\frac{2\sqrt{s}}{\alpha}\)
The average velocity v_av \((=s/T)\) is, therefore
\( v_{av}=(\frac{\alpha}{2}\sqrt{s}) \).

(i) Heat energy required for burning of casing of rocket comes from the rocket itself. As a result of work done against friction the kinetic energy of rocket continuously decreases and this work against friction reappears as heat energy.
(ii) The gravitational force is a conservative force, hence, work done by the gravitational force over one complete (closed) orbit of comet is zero. 
(iii) As an artificial satellite gradually loses its energy due to dissipation against atmospheric resistance, its potential energy decreases rapidly. As a result, kinetic energy of satellite slightly increases i.e. its speed increases progressively.
(iv) In figure, the man carries the mass of 15 kg on his hands and walks 2m. In this case, he is actually doing work against the friction force.  Friction force contribution by mass 
\(f=\mu N=\mu mg\times 15\times 9.8N\)
and work done against friction
 \({ W }_{ 1 }={ f }_{ S }=\mu \times 15\times 9.8\)= 294 \(\mu \)J
In figure (ii) the tension in string, T = mg =15 \(\times \) 9.8 N
Hence, force applied by man for pulling the rope F =T =15\(\times \) 9.8N
\(\therefore \)  Work done by man, Wz = Fs =15\(\times \) 9.8\(\times \) 2 = 294 J and additional work has to be done against friction also.
Thus, it is clear that W₂ > W₁.

(i) A fired cannon ball is a projectile and we want an equation that relates the launch angle \({ \theta }_{ 0 }\) to the ball horizontal displacement i. e. range as it moves from the cannon to the ship.
 \(\therefore \quad { \theta }_{ 0 }=\frac { 1 }{ 2 } \sin ^{ -1 }{ \left( \frac { gR }{ { u }_{ 0 }^{ 2 } } \right) } =\frac { 1 }{ 2 } \sin ^{ -1 }{ \left( \frac { 9.8\times 560 }{ { (82) }^{ 2 } } \right) } \)
= \(\frac { 1 }{ 2 } \sin ^{ -1 }{ (0.816) } =27°\)
If one angle is \(27°\), then other angle \(\left( 90°-{ \theta }_{ 0 } \right) \) is  = \(90°-27°\) = \(63°\)
(ii) Maximum range at \({ \theta }_{ 0 }\) = \(45°\)
\(\therefore \quad R=\frac { { u }^{ 2 } }{ g } \sin { 2{ \theta }_{ 0 } } =\frac { { (82) }^{ 2 } }{ 9.8 } \times \sin { 90° } \quad \)
= 686 m

(i) From the conservation of X and Y components of linear momentum
mu = mv₁ cos 30° + mv₂ cos 30°
\( \Rightarrow \ u=\left(v_{1}+v_{2}\right) \frac{\sqrt{3}}{2} \) 
and 0 = mv₁ sin 30 - mV₂ sin 30 \(\Rightarrow\) v₁ = v₂
\( v_{1}=\frac{u}{\sqrt{3}} \) 
(ii) Total K.E before collision is \( \mathrm{K}_{i}=\frac{1}{2} m u^{2} \) and after collision \( \mathrm{K}_{f}=\frac{1}{2} m v_{1}^{2}+\frac{1}{2} m v_{2}^{2} \) 
\( =\frac{1}{2} m\left(v_{1}^{2}+v_{2}^{2}\right)=\frac{1}{2} m\left(2 v_{1}^{2}\right)=\frac{m u^{2}}{3} \)
\( \frac{\mathrm{K}_{f}}{\mathrm{~K}_{i}}=\frac{2}{3} \) 
(iii) \( e=\frac{v_{2}-v_{1}}{u_{1}-u_{2}}=0 \) 
(iv) As the collision is inelastic, body loses some energy so that K.E of ball does not remain the same. However, total energy and total momentum of ball and earth system remain the same.

C)  Range will be maximum when θ = 45^∘and in this condition R = 4HÞ H = R/4(always) because R = 4 Hcot θ and θ = 45^∘ So maximum height is 25% of maximum range.It does not depends upon the velocity of projection.            

B)  If both assertion and reason are true but reason is not the correct explanation of the assertion.
If two protons are brought near one another, work has to be done against electrostatic force because same charge repel each other. This work done is stored as potential energy in the system.