Organised by MechTRIX-7.0.

#### Mathematics Section

1. (Sequence,Series and Mathematical Induction)

Let $a$, $b$, $c$ be positive real numbers such that $abc = 1$. Prove that $a^2 + b^2 + c^2 \geq a + b + c$. Also, state the condition for equality.($10$ points)

Solution:

a,b,c are real numbers

$(a-1)^2 + (b-1)^2 + (c-1)^2 \geq 0$

or, $a^2 -2a +1 + b^2 - 2b +1 + c^2 - 2c +1 \geq 0$

or, $a^2 + b^2 + c^2 \geq 2(a+b+c) -3$

or, $a^2 + b^2 + c^2 \geq (a+b+c) + (a+b+c) -3$...(i)

From AM-GM inequality,

$\dfrac{a+ b+c}{3} \geq \sqrt[3]{abc}$

or, $a+b+c \geq 3 \sqrt[3]{abc}$

but, $abc =1$

so,

$a+b+c \geq 3$

We can write equation (i) as;

or, $a^2 + b^2 + c^2 \geq (a+b+c) + 3 -3$

or, $a^2 + b^2 + c^2 \geq a+b+c$

Condition for equality is, $a = b = c = 1$

2.(Complex Numbers)

Find all pairs ($a$,$b$) of real numbers such that $(a + bi)^5 = b + ai$.($10$ points)

Solution:

Let,

$a + ib = r(\cos \theta + i \sin \theta)$

$\mid (a+ib)^5 \mid = \mid b+ ia \mid$

or, $r^5 = r$

or, $r(r^4-1) =0$

either $r =0$

which means $a=b=0$

(0,0) is one pair.

or,$r =1$

$(a+ib)^5 = (\cos \theta + i \sin \theta)^5 = \cos 5\theta + i \sin 5\theta$

$(b+ia) = \sin \theta + i \cos \theta = \cos (90^{\circ} - \theta) + i \sin (90^{\circ} - \theta)$

so,

$5\theta = 90^{\circ} - \theta + 360^{\circ} k$

or, $6\theta = 90^{\circ} + 360^{\circ} k$

or, $\theta = 15^{\circ} + 60^{\circ} k$

where, $k$ is integer, so

$\theta = 15^{\circ}, 75^{\circ}, 135^{\circ}, 195^{\circ}, 255^{\circ}, 315^{\circ}$

when r=1,

(a,b) = $(\pm \dfrac{1}{\sqrt{2}} \mp \dfrac{1}{\sqrt{2}}), (\pm \dfrac{\sqrt{6} \pm \sqrt{2}}{4}, \pm \dfrac{\sqrt{6} \mp \sqrt{2}}{4})$

when $r=0$,

$(a,b) = (0,0)$

These are the pairs.

3. (Calculus)

Let $f$ be continuous one to one function defined on an interval, and suppose that $f$ is differentiable at $f^{-1}(b)$, with the derivative $f'(f^{-1}(b)) \neq 0$. Then, prove that $f^{-1}$ is differentiable at $b$, and

$(f^{-1})'(b) = \dfrac{1}{f'(f^{-1}(b))}$

Solution:

Let $b = f(a)$,

Then,

$\lim_{h \to 0} \dfrac{f^{-1}(b+h) - f^{-1}(b)}{h}$

= $\lim_{h \to 0} \dfrac{f^{-1}(b+h) - a}{h}$

For every h there is unique k such that

$f(a+k) = b+h$

= $\lim_{k \to 0} \dfrac{f^{-1}(f(a+k)) - a}{f(a+k) - b}$

= $\lim_{k \to 0} \dfrac{a+ k - a}{f(a+k) - f(a)}$

= $\lim_{k \to 0} \dfrac{k}{f(a+k) - f(a)}$

= $\dfrac{1}{\lim_{k \to 0} \dfrac{f(a+k) - f(a)}{k}}$

= $\dfrac{1}{f'(a)}$

but, $a = f^{-1}(b)$

= $\dfrac{1}{f'(f^{-1} (b))}$

#### Physics section

1. A projectile is thrown from a point P. It moves in such a way that its distance from P is always increasing. Find the maximum angle above the horizontal with which the projectile could have been thrown. You can ignore air resistance. Take $g$ = 10 $ms^{-2}$.($10$ points)

Solution:

Let projectile be projected from point P making angle $\theta$ with the horizontal with initial speed of $u$ m/s.

At any instant, its horizontal distance is $x(t)$ and vertical distance is $y(t)$.

The distance from point P is $r(t) = \sqrt{x(t)^2 + y(t)^2}$

$x(t) = u \cos \theta \times t$

$y(t) = u \sin \theta \times t - \dfrac{1}{2} g t^2$

Then,

$r(t) = \sqrt{(u \cos \theta \times t)^2 + (u \sin \theta \times t - \dfrac{1}{2} gt^2)^2}$

= $\sqrt{u^2 \cos^2 \theta \times t^2 + u^2 \sin^2 \theta \times t^2 - u \sin \theta \times t \times gt^2 + \dfrac{1}{4}g^2 \times t^4}$

= $\sqrt{u^2 t^2 - u g \sin\theta t^3 + \dfrac{1}{4} g^2 t^4}$

= $\dfrac{1}{2} \sqrt{4u^2t^2 - 4 ug \sin \theta t^3 + g^2t^4}$

$\dfrac{d r(t)}{dt} = \dfrac{1}{2(\sqrt{4u^2t^2 - 4 ug \sin \theta t^3 + g^2t^4})} 8u^2t - 12 ug \sin \theta t^2 + 4g^2 t^3$

$\dfrac{d r(t)}{dt} = \dfrac{2t}{(\sqrt{4u^2t^2 - 4 ug \sin \theta t^3 + g^2t^4})} 2u^2 - 3 ug \sin \theta t + g^2 t^2$

The distance is always increasing,

$\dfrac{d r(t)}{dt} > 0$

or, $\dfrac{2t}{(\sqrt{4u^2t^2 - 4 ug \sin \theta t^3 + g^2t^4})} 2u^2 - 3 ug \sin \theta t + g^2 t^2 > 0$

or, $2u^2 - 3ug \sin \theta t + g^2t^2 > 0$

$a = g^2 > 0$

so, for quadratic equation to be positive,

$(-3 u \sin\theta)^2 - 4 \times g^2 \times 2 u^2 < 0$

or, $9u^2 \sin^2 \theta - 8 u^2g^2 < 0$

or, $\sin^2 \theta < \dfrac{8}{9}$

or, $\sin \theta < \sqrt{\dfrac{8}{9}}$

or, $\theta < \sin^{-1}(\sqrt{\dfrac{8}{9}})$

or, $\theta < 70.52^{\circ}$

So, the maximum angle is $70.52^{\circ}$

2. Consider a hot-air balloon with fixed volume $V_B = 1.1 m^3$. The mass of the balloon envelope, whose volume is to be neglected in comparison to $V_B$, is $m_H$ = 0.187 kg.

The balloon shall be started, where the external air temperature is $\theta_1 = 20^{\circ}$ C and the normal external air pressure is $P_0 = 1.013 \times 10^5$ Pa. Under these conditions the density of air is $\rho_1$ = $1.2 kgm^{-3}$

a) What temperature $\theta_2$ must the warmed air inside the balloon have to make the balloon just float?

b) First the balloon is held fast to the ground and the internal air is heated to a steady-state temperature of $\theta_3$ = $110^{\circ}$C. The balloon is fastened with a rope. Calculate the force on the rope. ($10$ points)

solution:

Mass of envelope($m_H$) = 0.187 kg

Initial Density of air ($\rho_1$) = $1.2 kg m^{-3}$

Initial temperature($\theta_1$) = $20^{\circ}$

Volume of air($V_B$) = $1.1 m^3$

a. For floating condition,

The total mass of the balloon, consisting of the mass of the envelope $m_H$ and the mass

of the air quantity of temperature $\theta_2$ must equal the mass of the displaced air quantity

with temperature $\theta_1$ = $20^{\circ}$ C.

$V_B \rho_2 + m_H = V_B ⋅ \rho_1$

or, $\rho_2 = \rho_1 - \dfrac{m_H}{V_B}$

or, $\rho_2 = 1.2 - \dfrac{0.187}{1.1}$

or, $\rho_2 = 1.03 kg m^{-3}$

We know,

$\dfrac{\theta_1}{\theta_2} = \dfrac{\rho_2}{\rho_1}$

or, $\dfrac{20+273}{\theta_2} = \dfrac{1.03}{1.2}$

or, $\theta_2 = 341.45$ K

or, $\theta_2 = 68.35^{\circ}$ C.

b.

The force $F_B$ acting on the rope is the difference between the buoyant force $F_A$ and the weight force $FG$:

It follows with $\rho_3 ⋅ \theta_3 = \rho_1 ⋅ \theta_1$

or, $\rho_3 \times 383 = 1.2 \times 293$

or, $\rho_3 = 0.918$

or,

$F_B = V_B \rho_1 ⋅g - (V_B ⋅\rho_3 + m_H) ⋅ g$ ... (3)

or, $F_B = 1.1 \times 1.2 \times 10 - (1.1 \times 0.918 + 0.187)\times 10$

or, $F_B = 13.2 - 11.968$

or, $F_B = 1.232$ N.

3. A dust particle of mass $m = 5.0 \times 10^{-9}$ kg and charge is $q_0= 2.0$nC starts from rest at point a and moves in a straight line to point b. what is its speed $v$ at point b ? ($10$ points)

Solution:

Mass of dust particle(m) = $5.0 \times 10^{-9}$ kg

Charge on dust particle($q_0$) = $2.0$nC

Potential at a $V_a$ = $\dfrac{q_1}{4\pi \epsilon_0 r_1} + \dfrac{q_2}{4 \pi \epsilon_0 r_2}$

= $\dfrac{3 \times 10^{-9}}{4\pi \epsilon_0 1 \times 10^{-2}} + \dfrac{- 3 \times 10^{-9}}{4\pi \epsilon_0 2 \times 10^{-2}}$

= $(9 \times 10^9 )(\dfrac{3 \times 10^{-9}}{ 1 \times 10^{-2}} + \dfrac{- 3 \times 10^{-9}}{2 \times 10^{-2}})$

= $1350$ V

Potential at b $V_b$ = $\dfrac{3 \times 10^{-9}}{4\pi \epsilon_0 2 \times 10^{-2}} + \dfrac{- 3 \times 10^{-9}}{4\pi \epsilon_0 1 \times 10^{-2}}$

= $(9 \times 10^9 )(\dfrac{3 \times 10^{-9}}{ 2 \times 10^{-2}} + \dfrac{- 3 \times 10^{-9}}{1 \times 10^{-2}})$

= $-1350$ V

Using Conservation of energy at a and b

$E_a = E_b$

or, $q_0 V_a + 0 = q_0 V_b + \dfrac{1}{2}m v^2$

or, $q_0(V_a - V_b) = \dfrac{1}{2} \times 5.0 \times 10^{-9} v^2$

or, $2 \times 10^{-9}(1350+1350) = \dfrac{1}{2} \times 5.0 \times 10^{-9} v^2$

or, $v = 46$ m/s.