The correct answer is "  Photoelectric effect "

Einstein worked on all the given topics, i.e. Special theory of relativity, Bose – Einstein Statistics, Photoelectric effect, General relativity.

But  Einstein was conferred with the Nobel prize in physics for his work on the discovery of the law of the photoelectric effect.

i.e. \(\displaystyle h\nu=h\nu_o+\frac{1}{2}mv^2\)

The correct answer is " 9.1% "

Given, \(\displaystyle z=ab^2c^{-2}=\frac{ab^2}{c^2}\)

Now, % error in a, b and c are respectively, 2.1 %, 1.3% and 2.2%.

\(\therefore\) % error in z = \(\displaystyle \frac{\triangle z}{z}\times100=\frac{\triangle a}{a}\times100+2\times\frac{\triangle b}{b}\times100+2\times\frac{\triangle c}{c}\times100\)

\(\therefore\) \(\displaystyle \frac{\triangle z}{z}\times100=2.1+2\times1.3+2\times2.2=9.1\)%

The correct answer is "  a straight line passing through the origin. "

Given, \(a=g\;\;\;and\;\;\;u=0\)

From 1st equation of motion, \(v=u+at\)

\(\therefore\) \(v=0+10t=10t\)

\(\therefore\) \(v\propto t\)

So, this shows that, the nature of a graph drawn for a freely falling body with time on x –axis and speed on the y – axis will be a straight line passing through the origin.

The correct answer is " 3 m/s "

Plotting a schematic graph from given data, we get

Let, time of collision be 't'

Now, \(\displaystyle s=ut+\frac{1}{2}at^2\)

For particle A, \(\displaystyle 30=0+\frac{(0.4\times\cos60^{\circ})t^2}{2}\)............\(\displaystyle [\because acceleration \;perpendicular \;to\;\;\vec v \;is \;(a\cos\theta)]\)

\(\therefore\) \(\displaystyle 30=0.1t^2.............[\because \cos60^{\circ}=\frac{1}2{}]\)

\(\therefore\) \(t=\sqrt{300}s\)

For collision both A and B travel same horizontal distance

\(\therefore\) \(S_A=S_B\)

\(\therefore\) \(\displaystyle vt=0+\frac{1}{2}(a\sin60^{\circ})t^2\)

\(\therefore\) \(\displaystyle v=\frac{1}{2}(a\sin60^{\circ})t=\frac{1}{2}\times0.4\times \frac{\sqrt3}{2}\times\sqrt{300}=3\;m/s\)

The correct answer is " \(\displaystyle \frac{\pi}{2\sqrt2}\) "

Plotting diagram based on the data of the question,

We know that, \(\displaystyle v_{rms}= \frac{total \; distance}{total\;times\;taken}={\displaystyle\large\frac{\frac{2\pi R}{4}}{t}}\)

Again, \(\displaystyle v_{vv}=\frac{displacement }{t}=\frac{\sqrt2R}{t}\)

Now, \(\displaystyle \frac{v_{rms}}{v_{vv}}=\frac{\pi}{2\sqrt2}\)

The correct answer is " \(\displaystyle \frac{5\sqrt5}{2} N\) "

Given, m = 4 kg

u = \(3\hat{i}\;m/s\)

\(v= (8.0\hat{i}+10.0\hat{j})\;m/s\)

t = 8 sec

Now, from 1st equation of motion, \(\vec{v}=\vec{u}+\vec{a}t\)

\(\therefore\) \(8.0\hat{i}+10.0\hat{j}=3\hat{i}+a\times8\)

\(\therefore\) \(\displaystyle a = \frac{1}{8}\times(5\hat i+10\hat j)\;m/s^2\)

Now, \(\displaystyle \vec F=m\vec a=4\times \frac{1}{8}\times(5\hat i+10\hat j)= \frac{5}{2}\times(1\hat i+2\hat j)\)

\(\therefore\) \(\displaystyle | \vec F |=\frac{5}{2}\times\sqrt{1^2+2^2}=\frac{5\sqrt5}{2}\;N\)

The correct answer is " \(7\sqrt2 \;m\) "

Plotting diagram based on given data and mentioning Forces, we get 

From the diagram, we can find that

\(N = mg\cos\theta\)

and \(ma=mg\sin\theta-\mu mg \cos\theta\)

\(\therefore\) \(\displaystyle a=g\sin45^{\circ}-\mu g \cos45^{\circ}=10\times\frac{1}{\sqrt2}-0.3\times10\times \frac{1}{\sqrt2}\)

\(\therefore\) \(\displaystyle a=\frac{7}{\sqrt2} \;ms^{-2}\)

Now, \(\displaystyle s=ut+\frac{1}{2}at^2=0+\frac{1}{2}\times\frac{7}{\sqrt2}\times2^2=7\sqrt2\;m\)

The correct answer is "  Work done depends upon only on the initial and final positions. "

Conservative forces are those forces whose work dones are path independent, i.e. their work dones don't depend upon the path they take.

i.e. Work done depends upon only on the initial and final positions.

The correct answer is "  5 ln (20) km/s "

Given, \(\displaystyle \frac{dm}{dt}=100\;kg/s\)

\(u = 5km/s\)

\(\displaystyle m=\frac{1}{20}m_o\;\;\;\Rightarrow\frac{m_o}{m}=20\)

Now, gravitational and viscous forces are ignored. (Given)

So, \(\displaystyle v=u\times \left(\frac{m_0}{m}\right)=5\;in\;(20)\;km/s\)

The correct answer is " \(\displaystyle \frac{85}{6}\;m/s\) "

Presenting the given data and scenerio from the question, we get the following disgram,

The velocity of the second ball is given by the following relation as,

\(\displaystyle \vec {v_B}= \frac{m_A(1+e)}{m_A+m_B}\vec{u_A}+\frac{m_B-em_A}{m_A+m_B}\vec{u_B}\)

\(\therefore\) \(\displaystyle v_B=\frac{50 \left(1+\frac{2}{5} \right)}{50+10}\times10+\frac{10-50\times\frac{2}{5}}{50+10}\times(-15)\)

\(\therefore\) \(\displaystyle v_B=\frac{70}{6}+\frac{15}{6}\)

\(\therefore\) \(\displaystyle v_B=\frac{85}{6}\;m/s\)

The correct answer is " 56 J "

Total KE = \((KE)_{translational}+(KE)_{Rotational}\)

\(\therefore\) Total KE = \(\displaystyle \frac{1}{2}mv^2+\frac{1}{2}I_s\omega^2\)

\(\therefore\) Total KE = \(\displaystyle \frac{1}{2}mv^2+\frac{1}{2}\left(\frac{2}{5}mR^2 \right)\left(\frac{v}{R} \right)^2\)

\(\therefore\) Total KE = \(\displaystyle \frac{1}{2}5\times4^2+\frac{1}{2}\left(\frac{2}{5}\times5\times R^2 \right)\left(\frac{4^2}{R^2} \right)=56\;J\)

The correct answer is " \(\displaystyle \frac{1}{800\pi^2}\) "

Given, m = 0.3 kg, K= 50N/m

We know that, \(\displaystyle \omega =\sqrt{\frac{K}{m}} =\sqrt{\frac{50}{0.3}}=\sqrt{\frac{500}{3}}\;rad/s\)

Now, time t = \(\displaystyle 100\times T=100\times\frac{2\pi}{\omega}\)

Again, we know that amplitude, A = \(\displaystyle X_m e^{-\lambda t}\)

\(\therefore\) \(\displaystyle \frac{1}{e}X_m=X_m\frac{1}{e^{\lambda t}}\)

\(\therefore\) \(\displaystyle \frac{1}{e}=\frac{1}{e^{\lambda t}}\)

\(\therefore\) \(\lambda t=1\)

\(\therefore\) \(\displaystyle \lambda=\frac{1}{t}=\frac{\omega}{200\pi}\)

Again, \(\omega'=\sqrt{\omega^2-\lambda^2}\)

\(\therefore\) \(\displaystyle {\omega'}^2=\omega^2-\frac{\omega^2}{400\pi^2}=\omega^2 \left(1-\frac{1}{400\pi^2} \right)\)

\(\therefore\) \(\displaystyle \frac{\omega'^2}{\omega^2}=1-\frac{1}{400\pi^2}\)

\(\therefore\) \(\displaystyle 1-\frac{\omega'^2}{\omega^2}=\frac{1}{400\pi^2}=\frac{\omega^2-\omega'^2}{\omega^2}\)

\(\displaystyle \left(\frac{\omega^2-\omega'^2}{\omega^2} \right) \rightarrow 0, \;So, by\; binomial\;\left( \frac{\omega^2-\omega'^2}{\omega^2}\right) =2\left(\frac{\omega-\omega'}{\omega} \right) \)

\(\therefore\) \(\displa
Please choose your answer from the right side options

The correct answer is " 13.2 km "

We know that, \(\displaystyle v_{es}=\sqrt{\frac{2GM}{R}}\)

\(\therefore\) \(\displaystyle R=\frac{2GM}{v_{es}^2}=\frac{2\times6.6\times10^{-11}\times6.4\times10^{23}}{(8\times10^4)^2}=13.2\times10^3\;m\)

\(\therefore\) \(R=13.2\;km\)

The correct answer is " 0.02 mm "

We know that, \(\displaystyle \eta=\frac{F_L}{Ax}=\frac{mgL}{Ax}\)

\(\therefore\) \(\displaystyle 3\times10^{10}=\frac{(400\pi\times10)\times6\times10^{-2}}{\pi\times(2\times10^{-2})^2\times x}\)

\(\therefore\) \(\displaystyle x=\frac{4\pi\times10^3\times6\times10^{-2}}{4\pi\times10^{-4}\times 3\times10^{10}}\)

\(\therefore\) \(x = 2\times 10^{-5}=0.02\times10^{-3}=0.02\;mm\)

The correct answer is " 11 cm "

The expression for Reynold's number for the turbulent flow is given by,

\(\displaystyle R_e=\frac{\rho vD}{\eta}=\frac{3000\times10^{-3}}{10^3\times2\times10^{-3}}=1.5\)

Now the expression for velocity head is given by,

\(\displaystyle V= \frac{v^2}{2g}=\frac{(1.5)^2}{2\times10}=0.1125\;m \simeq0.11\;m\)

So, V = 11 cm

The correct answer is " \(15\times10^{-2}\;m/s\) "

We know that, terminal velocity, \(\displaystyle v=\frac{2r^2}{9\eta}g(\rho_c-\rho_o)\)

\(\therefore\)  \(\displaystyle v=\frac{2\times(3\times10^{-3})^2}{9\times1}\times10\times(9\times10^3-1.5\times10^3)\)

\(\therefore\) \(v=150\times10^{-3}=15\times10^{-2}\;m/s\)

The correct answer is " 3 K "

According to Wein's displacement law, \(\lambda T=b\)

\(\therefore\) \(10^{-3}\times T=3\times10^{-3}\)

\(\therefore\) T = 3 K

The correct answer is " \(\displaystyle \frac{\ln (5)}{\ln(5/2)}\) "

We know that, \(\displaystyle \ln \frac{T_1-T_o}{T(t)-T_o}=kt \)

For, \(T_1=0^{\circ}C\)

\(\therefore\) \(\displaystyle \ln \frac{-T_o}{T(t)-T_o}=kt \)

Now substituting mcT(t) for Q(t) in the above equation,

\(\displaystyle \ln \frac{-Q_o}{Q(t)-Q_o}=kt \)

According to question, \(\displaystyle \ln \frac{-Q_o}{\frac{60}{100}Q_o-Q_o}=kt_1 \)  and  \(\displaystyle \ln \frac{-Q_o}{\frac{80}{100}Q_o-Q_o}=kt_2 \)

So, \(\displaystyle\large \frac{t_2}{t_1}= \frac{\ln \frac{-Q_o}{\frac{80}{100}Q_o-Q_o}}{\ln \frac{-Q_o}{\frac{60}{100}Q_o-Q_o}}\)

\(\therefore\) \(\displaystyle\large \frac{t_2}{t_1}= \frac{\ln \frac{-Q_o}{-\frac{20}{100}Q_o}}{\ln \frac{-Q_o}{-\frac{40}{100}Q_o}}=\frac{\ln (5)}{\ln(5/2)}\)

The answer is " 2 "

we know that, efficiency of carnot engine is, \(\displaystyle \eta=1-\frac{T_2}{T_1}\)

For % error, \(\displaystyle \frac{\triangle \eta}{\eta}= \frac{\triangle T_1}{T_1}+ \frac{\triangle T_2}{T_2}\)

For change in \(T_1\) by 0.4%,

 \(\displaystyle \frac{\triangle \eta_1}{\eta}= \frac{0.4}{100}+ 0= \frac{0.4}{100}\)

For change in \(T_2\) by 0.2%,

 \(\displaystyle \frac{\triangle \eta_2}{\eta}= \frac{0.2}{100}+ 0= \frac{0.2}{100}\)

\(\therefore\) \(\displaystyle \frac{\triangle \eta_1}{\triangle \eta_2}=\frac{0.4}{0.2}=2\)

The correct answer is " 927 °C "

We know that, \(\displaystyle V_{rms}=\sqrt{\frac{3kT}{m}}\)

\(\therefore\) \(V \propto \sqrt T\)

According to question, \(\displaystyle \frac{V_1}{V_2}=\sqrt{\frac{T_1}{T_2}}\)

\(\therefore\) \(\displaystyle \frac{V_1}{2V_1}=\sqrt{\frac{300}{T_2}}\)...............[\(T_1=27^{\circ}C=27+273=300K\)]

\(\therefore\) \(\displaystyle T_2=300\times4=1200K\)

\(\therefore\) \(T_2=1200-273=927^{\circ}C\)

The correct answer is " 3.5 m/s "

Case 1: Source is moving toward a stationary observer

Here, \(\displaystyle n_1=\frac{v}{v-v_s}\times n_o\)

Case 1: Source is moving away from the same stationary observer

Here, \(\displaystyle n_2=\frac{v}{v+v_s}\times n_o\)

Now, beat = \(n_1-n_2\)

\(\therefore\) \(\displaystyle 5 = n_o\times v \left(\frac{1}{v-v_s}- \frac{1}{v+v_s} \right) =250\times 350\times \left(\frac{1}{350-v_s}- \frac{1}{350+v_s} \right)\)

\(\therefore\) \(\displaystyle 5 =250\times350\times \left(\frac{350+v_s-350+v_s}{(350-v_s)(350+v_s)} \right)\)

\(\therefore\)  \(\displaystyle 5 =250\times350\times \left(\frac{2v_s}{(350)^2-(v_s)^2}\right)\)

\(\therefore\) \({v_s}^2+35000v_s-122500=0\)

Now, \(\displaystyle v_s = {-35000 \pm \sqrt{(35000)^2-4\times1\times122500} \over 2\times1}\)

\(\therefore\) \(\displaystyle v_s = \frac{7}{2}\;m/s=3.5\;m/s\) 

The correct answer is " 816 Hz "

Here, Source (drone) is moving towards observer (drone operator)

So, \(\displaystyle n'=\frac{v}{v-v_s}\times n_o\)

\(\therefore\) \(\displaystyle n'=\frac{340}{340-15}\times 780=816 \;Hz\)

The correct answer is "  4 mm "

Consider the below diagram,

We know that, \(\displaystyle \frac{\mu_2}{v}+ \frac{\mu_1}{u}= \frac{\mu_2-\mu_1}{R}\)

\(\therefore\) \(\displaystyle \frac{1}{5}+\frac{1.6}{u}=\frac{1-1.6}{3}\)

\(\therefore\) \(u = -4\;mm\)

The correct answer is " 244 cm "

Formula for limit of resolution of a telescope is, \(\displaystyle \theta=\frac{1.22\lambda}{a}\)

\(\therefore\) \(\displaystyle a=\frac{1.22\lambda}{ \theta}=\frac{1.22\times500\times10^{-9}}{2.5\times10^{-7}}=2.44\;m\)

\(\therefore\) a = 244 cm

The correct answer is " 0.51 "

From the given data, the following diagram is presented.

The expression for the electric field is written as,

\(E=E_{Q_o}-E_{cavity}\)

\(\therefore\) \(\displaystyle E = \frac{Q_o}{4\pi \epsilon_o R^3} \left(\frac{R}{2} \right)-0\)

\(\therefore\) \(\displaystyle E = \frac{Q_o}{4\pi \epsilon_o R^2} \left(\frac{1}{2} \right)\)..........(i)

The expression for the charge after the creation of the cavity is,

\(\displaystyle \frac{Q}{Q_o}=\frac{\frac{4}{3}\pi \left(R^3-(R/4)^3 \right)}{\frac{4}{3}\pi R^3}\)

\(\therefore\) \(\displaystyle \frac{Q}{Q_o}=\frac{63}{64}\)

\(\therefore\) \(\displaystyle Q_o=\frac{64}{63}Q\).........(ii)

Putting value from eqn(ii) in eqn(i),

\(\displaystyle E = \frac{Q}{4\pi \epsilon_o R^2} \left(\frac{1}{2}\times\frac{64}{63} \right)=0.51 \frac{Q}{4\pi \epsilon_o R^2}\)

So, K = 0.51

The correct answer is " 25 pF "

The metal plate portion won't take part in parallel capacitor formation.

From the given data, the following is drawn,

Now, \(\displaystyle C_{eq}=\frac{C_1C_2}{C_1+C_2}\)

\(\therefore\) \(\displaystyle C_{eq}=\epsilon_oA {\left[ \large \frac{\frac{1}{d_1}\times \frac{1}{d_2}}{\frac{1}{d_1}+ \frac{1}{d_2}} \right]}=\frac{\epsilon_o A}{d_1+d_2}\)

\(\therefore\) \(\displaystyle C_{eq}=\frac{1}{4\pi\times9\times10^9}\times\frac{36\pi\times10^{-4}}{3\times10^{-3}+1\times10^{-3}}\)

\(\therefore\) \(\displaystyle C_{eq}=\frac{1}{4}\times10^{-10}=\frac{100}{4}\times10^{-12}\)

\(\therefore\) \(C_{eq}=25\;pF\)

The correct answer is " R "

We know that the equivalent resistance in

Series combination is, \(R_{eq}=R_1+R_2+...\)

Parallel combination is, \(\displaystyle \frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+...\)

Decoding and simplifying the given circuit, we get as follow results

So, The equivalent resistance is, R.

The correct answer is " \(\displaystyle V_2= -3V, I_3=1A\) "

Simplifying the given circuit a bit further,

no current flows through the battery with emf \(E_1\)

For the potential drop of 2 V takes place across the parallel branch AB, the expression is calculated as,

V = IR

\(\therefore\) \(2=I_32\)

\(\therefore\) \(I_3=1A\)

The voltage across the branch AB is calculated as,

\(V_{AB}=5-I_3R_2\)

\(\therefore 2=5-1\times R_2\)

\(\therefore\) \(R_2=3 \Omega\)

The voltage \(V_2\) across \(R_2\) is calculated as,

\(V_2=-I_3R_2=-3\times1=-3\;V\)

Now, \(\displaystyle V_2= -3V, I_3=1A\)

The correct answer is " 1.8 N "

We know that, \(F=ILB \sin \theta\)

\(\therefore\) \(F=6\times2\times30\times10^{-2}\times0.5\times\sin90^{\circ}\)..........[L = 2R]

\(\therefore\) F = 1.8 N

The correct answer is " \(\displaystyle \frac{\mu_o \sigma \omega R}{2}\) "

Consider the below diagram.

Let the radius of ring be r and thickness of the ring be dr.

The expression for the charge on the element is written as,

\(dq=\sigma (2\pi r)dr\)

The current associated with the rotating charge dq is calculated as,

\(\displaystyle di=\frac{dq}{T}=\frac{\omega dq}{2\pi}\)

\(\therefore \) \(di=\sigma \omega rdr\)

The expression for the magnetic field at the centre is calculated as,

\(\displaystyle dB=\frac{\mu_o di}{2r}=\frac{\mu_o\sigma\omega rdr}{2r}=\frac{\mu_o\sigma \omega dr}{2}\)

On integration,

\(\displaystyle B =\int dB=\frac{\mu_o\sigma\omega}{2}\int_0^R dr\)

\(\therefore \) \(\displaystyle B= \frac{\mu_o \sigma \omega R}{2}\)

The correct answer is " 45º "

We know that, \(B_V=B\cos\theta\)

\(\therefore\) \(5\times10^{-5}=\sqrt{10}\times10^{-5}\times\cos\theta\)

\(\therefore\) \(\displaystyle \cos\theta=\frac{\sqrt2}{\sqrt5} \approx \frac{1}{\sqrt2}\)

\(\therefore\) \(\theta = 45^{\circ}\)

The correct answer is " 6 H "

We know that, average emf induced in an inductor is, \(\displaystyle \varepsilon = L\frac{I_1-I_2}{\triangle t}\)

\(\therefore\) \(\displaystyle L = \frac{\varepsilon\triangle t}{I_1-I_2}=\frac{150\times0.2}{6-1}=6 H\)

The correct answer is " 3000 V "

We know that, \(\displaystyle \omega = 2\pi f=2\pi \times \frac{150}{\pi}=300\)

Current amplitude will be maximum, when resonance occurs.

During resonance, impedance z = R

\(\therefore\) \(\displaystyle I_{max}=\frac{V_{max}}{R}=\frac{200}{10}=20A\)

Now, Voltage difference across inductor is, \(V_L=I_{max}\omega L=20\times300\times0.5=3000V\)

The correct answer is " \(-3\times10^{-6}\cos(5\times10^{6}t)V\) "

Formula for EMF induced is, \(\displaystyle \varepsilon=-N\frac{d\phi}{dt}=-N\frac{d(BA)}{dt}=-NA\frac{dB}{dt}\)

\(\therefore\) \(\displaystyle \varepsilon = -300\times20\times10^{-4}\times \frac{d}{dt}[10^{-12}\sin(5\times10^6t)]\)

\(\therefore\) \(\displaystyle \varepsilon = -300\times20\times10^{-4}\times 10^{-12} \times \cos(5\times10^6t)\times \frac{d}{dt}(5\times10^6t)\)

\(\therefore\) \(\displaystyle \varepsilon = -300\times20\times10^{-4}\times 10^{-12} \times \cos(5\times10^6t)\times (5\times10^6)\)

\(\therefore\) \(\varepsilon = -3\times10^{-6}\cos(5\times10^{6}t)V\)

The correct answer is " 0.4 Å "

We know that, \(\displaystyle \lambda = \frac{h}{p}\;\;\;and\;\;\;p=\sqrt{2m(KE)}\)

\(\therefore\) \(\displaystyle \lambda = \frac{h}{\sqrt{2m(KE)}}\)

\(\therefore\) \(\displaystyle \lambda = \frac{6.4\times10^{-34}}{\sqrt{2\times8\times10^{-31}\times3\times10^3\times1.6\times10^{-19}}}\)

\(\therefore\)\(\lambda =\) 0.4 Å

The correct answer is " \(\displaystyle 15<\frac{\lambda_p}{\lambda_L}<16\) "

For longest wavelength in Paschen series, \(\displaystyle \frac{1}{\lambda_p}=R \left[\frac{1}{3^2}-\frac{1}{4^2} \right]=\frac{7}{9\times16}R\)..........(i)

For longest wavelength in Lyman series, \(\displaystyle \frac{1}{\lambda_L}=R \left[\frac{1}{1^2}-\frac{1}{2^2} \right]=\frac{3}{4}R\).....................(ii)

Now, \(\displaystyle \frac{eqn(ii)}{eqn(i)},\) \(\displaystyle \frac{\lambda_p}{\lambda_L}={\displaystyle \large\frac{(3/4)R}{(\frac{7}{9\times16})R}} = 15.42\)

So, \(\displaystyle 15<\frac{\lambda_p}{\lambda_L}<16\)

The correct answer is " 18 "

The effective half life when the radioactive sample has more than one radioactive material is given by,

\(\displaystyle \frac{1}{T_{eff \;T/2}}=\frac{1}{T_{1\frac{1}{2}}}+\frac{1}{T_{2\frac{1}{2}}}+\frac{1}{T_{3\frac{1}{2}}}+...\)

The effective half life of net material is calculated as,

\(\displaystyle {T_{eff \;T/2}}=\frac{0.7\times0.3}{0.7+0.3}=0.21\) hr

The relation between the average mean life and effective half life is given by,

\(\displaystyle T_{mean}=\frac{{T_{eff \;T/2}}}{0.693}=\frac{0.21}{0.693}=0.3 \;hr=18\;min\)

The correct answer is "  (A) is correct but (R) is incorrect. "

Both Si and GaAs are used in making solar cells.

The energy varies from 3 eV on the violet side to 1.5 eV on the red side, which implies that visible light can make solar energy.

The energy band gap of Si and GaAs are greater than photon energy of the radio waves, microwaves, and infrared waves.

This implies that the assertion is correct but the reason is incorrect.

The correct answer is " NAND gate "

Given,

Here, when both gates are closed(1), the output is '0'; otherwise '1'.

So, this represents the NAND gate.

The correct answer is " 45 m "

The maximum line of sight is calculated from the following expression as,

\(d_m=\sqrt{2Rh_T}+\sqrt{2Rh_R}\)

\(\therefore\) \(40\times10^3=\sqrt{2\times6.4\times10^6\times20}+\sqrt{2\times6.4\times10^6\times x}\)

\(\therefore\) \(\sqrt{2\times6.4\times10^6\times x}=24000\)

\(\therefore\) x = 45 m