For the initial equi-convex lens in air, focal length $f = 5 \text{ cm}$.
Using Lens Maker's formula: $\frac{1}{f} = (\mu_g - 1)\left(\frac{1}{R} - \frac{1}{-R}\right) \Rightarrow \frac{1}{5} = (1.5 - 1)\left(\frac{2}{R}\right) = 0.5 \times \frac{2}{R} = \frac{1}{R} \Rightarrow R = 5 \text{ cm}$.
When cut perpendicular to its principal axis, it forms two plano-convex lenses. For one part, $R_1 = 5 \text{ cm}$ and $R_2 = \infty$.
When dipped in water ($\mu_w = 4/3$), the new focal length $f_w$ is:
$\frac{1}{f_w} = \left(\frac{\mu_g}{\mu_w} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \left(\frac{1.5}{4/3} - 1\right)\left(\frac{1}{5} - 0\right)$
$\frac{1}{f_w} = \left(\frac{4.5}{4} - 1\right)\left(\frac{1}{5}\right) = \left(\frac{9/8} - 1\right)\left(\frac{1}{5}\right) = \left(\frac{1}{8}\right)\left(\frac{1}{5}\right) = \frac{1}{40}$.
Therefore, $f_w = 40 \text{ cm}$.
In a single slit diffraction pattern, the angular position of the first minimum is $\theta = \frac{\lambda_1}{a}$.
The angular position of the first secondary maximum is $\theta = \frac{3\lambda_2}{2a}$.
Since these positions coincide, we equate them:
$\frac{\lambda_1}{a} = \frac{3\lambda_2}{2a} \Rightarrow \lambda_1 = \frac{3}{2}\lambda_2 \Rightarrow \lambda_2 = \frac{2}{3}\lambda_1$.
Given $\lambda_1 = 6000 \mathring{A}$, $\lambda_2 = \frac{2}{3}(6000) = 4000 \mathring{A}$.
The power incident on the surface is $P = \text{Intensity} \times \text{Area} = I \times (\pi r^2)$.
$P = 1.4 \times 10^3 \text{ W/m}^2 \times \pi(2)^2 = 1.4 \times 10^3 \times 4\pi = 5.6\pi \times 10^3 \text{ W}$.
For a perfectly absorbing surface, the radiation force is $F = \frac{P}{c}$.
$F = \frac{5.6\pi \times 10^3}{3 \times 10^8} \approx \frac{17.59 \times 10^3}{3 \times 10^8} \approx 5.86 \times 10^{-5} \text{ N}$.
Using $\pi \approx 3.14$, the value closely aligns with $5.88 \times 10^{-5} \text{ N}$.
De-Broglie wavelength is given by $\lambda = \frac{h}{p}$. Since both have the same $\lambda = 1 \mathring{A}$, their momenta are equal ($p_e = p_p$).
Kinetic energy in terms of momentum is $K = \frac{p^2}{2m}$.
For the same momentum, Kinetic Energy is inversely proportional to mass ($K \propto \frac{1}{m}$).
$\frac{K_e}{K_p} = \frac{m_p}{m_e} = \frac{2000 m_e}{m_e} = \frac{2000}{1}$.
The ratio is $2000 : 1$.
Option (1) states that there are no free electrons at room temperature. This is incorrect. At room temperature (approx 300K), thermal energy is sufficient to break some covalent bonds, creating electron-hole pairs, which means some free electrons are available for conduction. The other statements correctly describe semiconductor behavior.
Initial radius $R = 1 \text{ cm} = 10^{-2} \text{ m}$. Number of droplets $n = 10^6$.
By conservation of volume: $\frac{4}{3}\pi R^3 = n \cdot \frac{4}{3}\pi r^3 \Rightarrow R^3 = 10^6 r^3 \Rightarrow r = \frac{R}{100} = 10^{-4} \text{ m}$.
Increase in surface area $\Delta A = n(4\pi r^2) - 4\pi R^2 = 4\pi (10^6 \times 10^{-8} - 10^{-4}) = 4\pi (10^{-2} - 10^{-4}) = 4\pi(0.0099) \approx 0.1244 \text{ m}^2$.
Energy expended $E = T \times \Delta A = 35 \times 10^{-3} \times 0.1244 \approx 4.354 \times 10^{-3} \text{ J}$.
Let the fundamental frequencies of the open pipes be $f_A = \frac{v}{2L_A}$ and $f_B = \frac{v}{2L_B}$. Given beats $|f_A - f_B| = 2$.
When pipe A is closed, its fundamental frequency halves: $f_A' = \frac{v}{4L_A} = \frac{f_A}{2}$.
Given the new beat frequency is still 2: $|f_A/2 - f_B| = 2$.
Case solving for $f_A > f_B$: $f_A - f_B = 2$ and $f_B - f_A/2 = 2$.
Adding both equations: $f_A - f_A/2 = 4 \Rightarrow f_A/2 = 4 \Rightarrow f_A = 8 \text{ Hz}$. Then $f_B = 6 \text{ Hz}$.
Ratio of lengths $\frac{L_A}{L_B} = \frac{f_B}{f_A} = \frac{6}{8} = \frac{3}{4}$.
Angular momentum about the origin is $L = mvr_{\perp}$, where $r_{\perp}$ is the constant perpendicular distance from the origin to the line of motion (P-Q).
Since the particle has a constant positive acceleration along the path from P to Q, its velocity is continuously increasing ($v_B > v_A$).
Because $m$ and $r_{\perp}$ are constant, $L \propto v$. Therefore, $L_B > L_A$, which means $L_A < L_B$.
The universal law of gravitation $F = \frac{GM_1M_2}{r^2}$ states that the gravitational force between two masses depends only on their masses and the distance between them. It is completely independent of the nature of the intervening medium. Therefore, the force remains F.
The potential at any point $x$ on the axis is $V(x) = \frac{k(+q)}{|x - (-a)|} + \frac{k(-q)}{|x - a|} = \frac{kq}{|x+a|} - \frac{kq}{|x-a|}$.
At origin $x=0$, $V = \frac{kq}{a} - \frac{kq}{a} = 0$.
As $x$ approaches $-a$, the first term dominates positively ($V \to +\infty$).
As $x$ approaches $+a$, the second term dominates negatively ($V \to -\infty$).
Graph (2) correctly depicts this asymptotic behavior and zero crossing at the origin.
Total EMF = $3 \times 2\text{V} = 6\text{V}$.
(1) Key K is open: The $6\Omega$ resistor is disconnected. The circuit has $1\Omega$ and $3\Omega$ in series. Total resistance $R = 1 + 3 = 4\Omega$. Current $I = \frac{6\text{V}}{4\Omega} = 1.5 \text{ A}$.
(2) Key K is closed: The $6\Omega$ and $3\Omega$ resistors are in parallel. $R_p = \frac{6 \times 3}{6+3} = 2\Omega$. Total circuit resistance $R_{eq} = 1\Omega + 2\Omega = 3\Omega$. Main current $I_{main} = \frac{6\text{V}}{3\Omega} = 2\text{A}$.
The ammeter reads the current through the $3\Omega$ branch: $I_A = I_{main} \times \frac{6}{6+3} = 2 \times \frac{6}{9} = \frac{12}{9} = \frac{4}{3} \text{ A}$.
Initial kinetic energy $K_0 = mgh$ (since all energy becomes potential at max height).
Let the required height be $y$. The kinetic energy at this height is $K = K_0 - U = mgh - mgy$.
Given $K = 75\% \text{ of } K_0 = 0.75 mgh$.
$0.75 mgh = mgh - mgy \Rightarrow mgy = 0.25 mgh \Rightarrow y = \frac{h}{4}$.
To find mutual inductance, assume a current $I_2$ flows in the larger outer coil ($r_2$).
The magnetic field produced by the outer coil at its center is $B_2 = \frac{\mu_0 I_2}{2r_2}$.
Since the inner coil ($r_1$) is very small ($r_1 \ll r_2$), this field is approximately uniform over its area $A_1 = \pi r_1^2$.
Magnetic flux linked with the inner coil is $\phi_1 = B_2 A_1 = \left(\frac{\mu_0 I_2}{2r_2}\right) (\pi r_1^2)$.
Mutual inductance $M = \frac{\phi_1}{I_2} = \frac{\mu_0 \pi r_1^2}{2r_2}$.
Speed of sound in a gas is $v = \sqrt{\frac{\gamma RT}{M}}$.
Root mean square speed of gas molecules is $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Dividing the two expressions: $\frac{v}{v_{rms}} = \frac{\sqrt{\gamma RT / M}}{\sqrt{3RT / M}} = \sqrt{\frac{\gamma}{3}}$.
Linear mass density of the string $\mu = \frac{m}{L} = \frac{0.035 \text{ kg}}{5.5 \text{ m}} = \frac{35}{5500} \text{ kg/m}$.
Speed of transverse wave $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{77}{35/5500}} = \sqrt{\frac{77 \times 5500}{35}} = \sqrt{11 \times \frac{5500}{5}} = \sqrt{11 \times 1100} = \sqrt{12100}$.
$v = 110 \text{ m/s}$.
The time period of revolution in Bohr's model is $T = \frac{2\pi r}{v}$.
We know that orbit radius $r \propto \frac{n^2}{Z}$ and orbital velocity $v \propto \frac{Z}{n}$.
Substituting these proportionalities: $T \propto \frac{n^2 / Z}{Z / n} = \frac{n^3}{Z^2}$.
The radius of the helical path depends on the perpendicular component of velocity: $R = \frac{m v_{\perp}}{qB} = \frac{mv\sin\theta}{qB}$.
$v_{\perp} = 4 \times 10^5 \sin 30^{\circ} = 2 \times 10^5 \text{ m/s}$.
$R = \frac{1.6 \times 10^{-27} \times 2 \times 10^5}{1.6 \times 10^{-19} \times 0.3} = \frac{2 \times 10^{-22}}{0.3 \times 10^{-19}} = \frac{2}{0.3} \times 10^{-3} \text{ m} = \frac{20}{3} \times 10^{-3} \text{ m}$.
$R \approx 6.66 \times 10^{-3} \text{ m} = 0.66 \text{ cm}$.
The weight of the particle acts vertically downwards. For the particle to be balanced, the electric force must act vertically upwards.
Since the particle is positively charged, it experiences an electric force in the same direction as the electric field.
Therefore, the electric field must point vertically upward.
The circuit diagram represents a Wheatstone bridge configuration. Since all four arms have identical resistance ($2200 \Omega$), the bridge is balanced.
The central branch (containing the $2200 \Omega$ diagonal resistor) will have no current flowing through it.
The equivalent resistance of the bridge is $R_{eq} = \frac{(2200+2200) \times (2200+2200)}{(2200+2200) + (2200+2200)} = \frac{4400 \times 4400}{8800} = 2200 \Omega$.
The main current drawn from the source is $I = \frac{V}{R_{eq}} = \frac{440 \text{ V}}{2200 \Omega} = \frac{1}{5} \text{ A}$.
The ammeter is in series with the main circuit, so it reads $1/5 \text{ A}$.
The relationship between charge, capacitance, and potential difference is $Q = CV$.
For a given capacitor, its capacitance $C$ is a constant determined by its geometry and dielectric medium.
Therefore, $V = \frac{Q}{C}$. If the charge $Q$ is increased, the potential difference $V$ between the plates must increase proportionally.
Initial velocity $u = 36 \text{ km/h} = 36 \times \frac{5}{18} = 10 \text{ m/s}$. Final velocity $v = 0$. Time $t = 4.0 \text{ s}$.
Retardation $a = \frac{v - u}{t} = \frac{0 - 10}{4} = -2.5 \text{ m/s}^2$.
Total mass $M = m_{vehicle} + m_{driver} = 400 \text{ kg} + 65 \text{ kg} = 465 \text{ kg}$.
Average retarding force $F = Ma = 465 \times 2.5 = 1162.5 \text{ N} = 1.1625 \times 10^3 \text{ N}$.
Rounding to the closest given option, it is $1.2 \times 10^3 \text{ N}$.
The area under the acceleration-time graph gives the change in velocity.
From $t=0$ to $t=2\text{s}$, acceleration $a = -10 \text{ m/s}^2$. Velocity at $t=2$ is $v_2 = v_0 + at = 0 + (-10 \times 2) = -20 \text{ m/s}$. The speed (magnitude of velocity) is $20 \text{ m/s}$.
From $t=2$ to $t=4\text{s}$, acceleration $a = +10 \text{ m/s}^2$. Velocity at $t=4$ is $v_4 = v_2 + a(t) = -20 + (10 \times 2) = 0 \text{ m/s}$.
The maximum speed achieved during this motion is $20 \text{ m/s}$, which occurs at $t=2 \text{ s}$.
Given $F \propto s^{-1/3}$. Since $F = ma = mv\frac{dv}{ds}$, we have $v\frac{dv}{ds} \propto s^{-1/3}$.
Integrating both sides: $\int v\,dv \propto \int s^{-1/3}ds \Rightarrow \frac{v^2}{2} \propto \frac{s^{2/3}}{2/3} \Rightarrow v^2 \propto s^{2/3} \Rightarrow v \propto s^{1/3}$.
Power delivered is $P = F \times v$.
Substituting the proportionalities: $P \propto (s^{-1/3}) \times (s^{1/3}) = s^0$.
Thus, power is constant and independent of displacement.
The velocity of the ball just before hitting the water is $v = \sqrt{2gh}$.
Since it doesn't change speed upon entering, this must equal its terminal velocity in water: $v_t = \frac{2 r^2 (\rho - \sigma) g}{9 \eta}$.
Given $r = 3 \times 10^{-4} \text{ m}$, $\rho = 10^4 \text{ kg/m}^3$, $\sigma = 10^3 \text{ kg/m}^3$ (water), $\eta = 9.8 \times 10^{-6}$.
$v_t = \frac{2 \times (9 \times 10^{-8}) \times (10000 - 1000) \times 9.8}{9 \times 9.8 \times 10^{-6}} = \frac{18 \times 10^{-8} \times 9000 \times 9.8}{9 \times 9.8 \times 10^{-6}} = \frac{162 \times 10^{-5}}{9 \times 10^{-6}} = 180 \text{ m/s}$.
Equating: $180 = \sqrt{2 \times 9.8 \times h} \Rightarrow 32400 = 19.6 \times h \Rightarrow h = \frac{32400}{19.6} \approx 1653 \text{ m} = 1.65 \times 10^3 \text{ m}$.
The P-V graph is a straight line from $(V_0, 2P_0)$ to $(2V_0, P_0)$. The equation of this line is $P = -\frac{P_0}{V_0}V + 3P_0$.
From ideal gas law, $T \propto PV = V\left(-\frac{P_0}{V_0}V + 3P_0\right) = 3P_0V - \frac{P_0}{V_0}V^2$.
This is a downward-opening parabola for $T$ as a function of $V$. The maximum temperature occurs at $V = 1.5V_0$.
Therefore, as the volume increases from $V_0$ to $2V_0$, the temperature first increases (is heated) until $V=1.5V_0$, and then decreases (is cooled) until $V=2V_0$.
Mechanical volume strain due to uniform pressure is $\left(\frac{\Delta V}{V}\right)_p = \frac{P}{K}$.
Thermal volume strain due to temperature increase $\Delta T$ is $\left(\frac{\Delta V}{V}\right)_T = \gamma \Delta T = 3\alpha \Delta T$.
To restore original volume, the thermal expansion must equal the mechanical compression:
$3\alpha \Delta T = \frac{P}{K} \Rightarrow \Delta T = \frac{P}{3K\alpha}$.
If no heat flows through AB, the nodes A and B must be at the same temperature. Thus, $T_B = T_A = 20^{\circ}\text{C}$.
Heat flows from D ($90^{\circ}\text{C}$) to B ($20^{\circ}\text{C}$) and then entirely to C ($0^{\circ}\text{C}$). The heat current is constant: $H_{DB} = H_{BC}$.
Using $H = \frac{KA\Delta T}{L}$:
$\frac{KA(90 - 20)}{L_{BD}} = \frac{KA(20 - 0)}{L_{BC}} \Rightarrow \frac{70}{L_{BD}} = \frac{20}{L_{BC}}$.
Ratio $\frac{L_{BD}}{L_{BC}} = \frac{70}{20} = \frac{7}{2}$.
First wave: $y_1 = 10 \sin(3\pi t + \pi/4)$. Its amplitude is $A_1 = 10$.
Second wave: $y_2 = 5 [\sin(3\pi t) + \sqrt{3}\cos(3\pi t)]$. Using trigonometric addition, $y_2 = 5 \times 2 \left[ \frac{1}{2}\sin(3\pi t) + \frac{\sqrt{3}}{2}\cos(3\pi t) \right] = 10 \sin(3\pi t + \pi/3)$.
Its amplitude is $A_2 = 10$.
The ratio of their amplitudes is $10 : 10 = 1 : 1$.
The binding energy of the first electron in He is given as 24.6 eV.
After removing the first electron, we are left with $He^+$ ion, which is a hydrogen-like species with atomic number $Z=2$.
The energy required to remove the remaining electron from its ground state is $E_2 = 13.6 \times Z^2 = 13.6 \times 2^2 = 13.6 \times 4 = 54.4 \text{ eV}$.
Total energy required to remove both electrons $= 24.6 \text{ eV} + 54.4 \text{ eV} = 79.0 \text{ eV}$.
When reflected and refracted rays are at $90^{\circ}$ to each other, Brewster's law is satisfied. Thus, the angle of incidence $i$ is Brewster's angle $i_p$.
According to Brewster's law: $\tan i = \frac{\mu_{refracted}}{\mu_{incident}}$. Since light goes from denser to rarer, $\tan i = \frac{\mu_r}{\mu_d}$.
The critical angle $C$ is defined by $\sin C = \frac{\mu_r}{\mu_d}$.
Therefore, $\sin C = \tan i \Rightarrow C = \sin^{-1}(\tan i)$. Since angle of incidence $i$ equals angle of reflection $r$, $C = \sin^{-1}(\tan r)$ is also valid. Both (1) and (2) are correct.
Given $\gamma = 1.5 = \frac{3}{2}$. We know $\gamma = \frac{C_p}{C_v}$.
So, $\frac{C_p}{C_v} = 1.5 \Rightarrow C_p = 1.5 C_v$.
Using Mayer's relation: $C_p - C_v = R \Rightarrow 1.5 C_v - C_v = R \Rightarrow 0.5 C_v = R \Rightarrow C_v = 2R$.
Then, $C_p = 1.5 \times (2R) = 3R$.
Number of turns per unit length $n = 15 \text{ turns/cm} = 1500 \text{ turns/m}$. Loop Area $A = 2 \times 10^{-4} \text{ m}^2$.
Magnetic field of solenoid $B = \mu_0 n I$. The flux linked with the loop is $\phi = BA = \mu_0 n I A$.
Induced emf $e = \frac{d\phi}{dt} = \mu_0 n A \frac{dI}{dt}$.
$e = \mu_0 \times 1500 \times 2 \times 10^{-4} \times \left(\frac{4 - 2}{0.1}\right) = \mu_0 \times 0.3 \times 20 = 6\mu_0 \text{ V}$.
The electric potential due to a dipole at a point defined by position vector $\vec{r}$ relative to the dipole center is given by:
$V = \frac{1}{4\pi\epsilon_0} \frac{p \cos\theta}{r^2}$.
Since $\vec{p} \cdot \vec{r} = p r \cos\theta$, we can rewrite it as $V = K \frac{\vec{p} \cdot \vec{r}}{r^3}$.
The voltage drop across resistor R is $V_R$. Since 5A flows through it, $V_R = 5R$.
The right parallel branch has $15 \Omega$ and $26 \Omega$ in series, so its resistance is $41 \Omega$. It is in parallel with R, so voltage across it is also $5R$.
Current through the right branch is $I_{right} = \frac{5R}{41}$.
Total current from the source $I_{total} = 5 + \frac{5R}{41}$.
Applying KVL: $50 = 1.5 \times I_{total} + 5R = 1.5\left(5 + \frac{5R}{41}\right) + 5R = 7.5 + \frac{7.5R}{41} + 5R$.
$42.5 = 5R + \frac{7.5R}{41}$. Multiplying by 41: $1742.5 = 205R + 7.5R = 212.5R \Rightarrow R = 8.2 \Omega$.
The cross product $\vec{A} \times \vec{B}$ results in a vector that is perpendicular to both $\vec{A}$ and $\vec{B}$.
Therefore, the dot product of $\vec{A}$ with this perpendicular vector $(\vec{A} \times \vec{B})$ must be zero.
$\vec{A} \cdot (\vec{A} \times \vec{B}) = 0$. (Note: Based on the English text structure, the "+ B" in Hindi was an OCR anomaly).
By Ohm's Law, Resistance $R = \frac{V}{I}$.
Voltage $V = \frac{\text{Work}}{\text{Charge}} = \frac{W}{I \cdot t}$. Its dimensions are $\frac{[M L^2 T^{-2}]}{[A T]} = [M L^2 T^{-3} A^{-1}]$.
So, $R = \frac{[M L^2 T^{-3} A^{-1}]}{[A]} = [M L^2 T^{-3} A^{-2}]$.
Using $I$ instead of $A$ for current dimension gives $[M L^2 T^{-3} I^{-2}]$.
A spring balance measures the tension within the spring. When it is pulled by 10 N from both ends in opposite directions, the entire spring experiences a uniform tension of 10 N (in static equilibrium).
Therefore, the reading on the spring balance will be 10 N.
A fan revolves around a fixed central axis. The axis itself does not undergo any translational motion through space. Therefore, the motion of the fan blades is purely rotational.
Acceleration due to gravity at height $h$ is $g_h = g\left(\frac{R}{R+h}\right)^2$. With $h = 1600 \text{ km}$ and $R = 6400 \text{ km}$, $g_h = g\left(\frac{6400}{8000}\right)^2 = g\left(\frac{4}{5}\right)^2 = \frac{16}{25}g$.
We need depth $d$ where $g_d = \frac{1}{2}g_h = \frac{1}{2}\left(\frac{16}{25}g\right) = \frac{8}{25}g$.
Formula for gravity at depth is $g_d = g\left(1 - \frac{d}{R}\right)$.
$\frac{8}{25}g = g\left(1 - \frac{d}{R}\right) \Rightarrow 1 - \frac{d}{R} = \frac{8}{25} \Rightarrow \frac{d}{R} = \frac{17}{25}$.
$d = \frac{17}{25} \times 6400 = 17 \times 256 = 4352 \text{ km}$.
The energy of an electron in the $n^{\text{th}}$ orbit of hydrogen is $E_n = -\frac{13.6}{n^2} \text{ eV}$.
Given $E_n = -3.4 \text{ eV} \Rightarrow -\frac{13.6}{n^2} = -3.4 \Rightarrow n^2 = 4 \Rightarrow n = 2$.
Angular momentum $L = \frac{nh}{2\pi} = \frac{2 \times 6.63 \times 10^{-34}}{2\pi} = \frac{6.63 \times 10^{-34}}{3.14} \approx 2.11 \times 10^{-34} \text{ J-s}$.
By Einstein's photoelectric equation, max kinetic energy is $K = h\nu - W$.
When frequency is doubled ($2\nu$), the velocity doubles, meaning kinetic energy increases by a factor of 4: $K' = 4K$.
$4K = h(2\nu) - W \Rightarrow 4(h\nu - W) = 2h\nu - W \Rightarrow 4h\nu - 4W = 2h\nu - W$.
$2h\nu = 3W \Rightarrow W = \frac{2h\nu}{3}$.
The lens forms a real image on a screen, which is characteristic of a convex lens.
Object distance $u = -25 \text{ cm}$. Image distance $v = +75 \text{ cm}$.
Using the thin lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{75} - \frac{1}{-25} = \frac{1}{75} + \frac{3}{75} = \frac{4}{75}$.
$f = \frac{75}{4} = +18.75 \text{ cm}$.
Orbital velocity is $v = \sqrt{\frac{GM}{r}}$.
The frequency of revolution is $f = \frac{1}{T} = \frac{v}{2\pi r}$.
Substituting $v$: $f = \frac{\sqrt{GM/r}}{2\pi r} = \frac{\sqrt{GM}}{2\pi r^{3/2}}$.
Thus, the frequency varies inversely with the square root of the cube of the radius: $f \propto \frac{1}{r^{3/2}} = \frac{1}{\sqrt{r^3}}$.
The electric field due to a single infinite sheet of charge density $\sigma$ is $E = \frac{\sigma}{2\epsilon_0}$.
For two intersecting sheets at right angles, they create two mutually perpendicular uniform electric fields, $E_1 = \frac{\sigma}{2\epsilon_0}$ and $E_2 = \frac{\sigma}{2\epsilon_0}$.
The resultant magnitude is the vector sum: $E_{net} = \sqrt{E_1^2 + E_2^2} = \sqrt{\left(\frac{\sigma}{2\epsilon_0}\right)^2 + \left(\frac{\sigma}{2\epsilon_0}\right)^2} = \sqrt{2} \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\sqrt{2}\epsilon_0}$.
Maximum height $H = \frac{u^2 \sin^2\theta}{2g}$.
Time of flight $T = \frac{2u\sin\theta}{g} \Rightarrow T^2 = \frac{4u^2\sin^2\theta}{g^2}$.
Ratio $\frac{H}{T^2} = \frac{\frac{u^2\sin^2\theta}{2g}}{\frac{4u^2\sin^2\theta}{g^2}} = \frac{g}{8}$.
Substituting $g = 10 \text{ m/s}^2$: $\frac{H}{T^2} = \frac{10}{8} = \frac{5}{4}$. The ratio is $5 : 4$.