NEET 2022 Physics Question Paper with Answer and Solution

(B) Statement $I$ is incorrect because the law of radioactive decay states that the rate of decay $(dN/dt)$ is directly proportional to the number of nuclei present $(N)$, i.e., $dN/dt = -\lambda N$.
Statement $II$ is incorrect because the definition provided in the statement describes the mean life (average life), not the half-life. The half-life $(T_{1/2})$ is defined as the time required for the number of radioactive nuclei to reduce to half of its initial value.
Therefore, both statements are incorrect.

(A) The power consumed by the lamp is given by $P = V_{rms} \times I_{rms}$.
Given $P = 100\,W$ and $V_{rms} = 200\,V$.
Therefore,the root mean square current $I_{rms} = \frac{P}{V_{rms}} = \frac{100}{200} = 0.5\,A$.
The peak current $I_{peak}$ is related to the $RMS$ current by the formula $I_{peak} = I_{rms} \times \sqrt{2}$.
Substituting the value of $I_{rms}$,we get $I_{peak} = 0.5 \times \sqrt{2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \approx 0.707\,A$.

(A) In Bohr's model,the radius of the $n^{th}$ stationary orbit is given by the formula $R_n = a_0 \frac{n^2}{Z}$,where $a_0$ is the Bohr radius and $Z$ is the atomic number.
Since $a_0$ and $Z$ are constant for a given atom,the radius is proportional to the square of the principal quantum number: $R_n \propto n^2$.
For the second orbit $(n_1 = 2)$,the radius is $R_1 \propto 2^2 = 4$.
For the fourth orbit $(n_2 = 4)$,the radius is $R_2 \propto 4^2 = 16$.
Therefore,the ratio is $\frac{R_1}{R_2} = \frac{n_1^2}{n_2^2} = \frac{2^2}{4^2} = \frac{4}{16} = \frac{1}{4} = 0.25$.

(D) The magnetic flux is given by $\phi = 2t^3 + 4t^2 + 2t + 5$.
According to Faraday's law of electromagnetic induction,the induced $emf$ $(e)$ is given by the magnitude of the rate of change of magnetic flux:
$e = |\frac{d\phi}{dt}|$
$e = |\frac{d}{dt}(2t^3 + 4t^2 + 2t + 5)|$
$e = |6t^2 + 8t + 2|$
Now,substitute $t = 5 \; s$ into the expression for $e$:
$e = 6(5)^2 + 8(5) + 2$
$e = 6(25) + 40 + 2$
$e = 150 + 40 + 2 = 192 \; V$.
Thus,the magnitude of the induced $emf$ at $t = 5 \; s$ is $192 \; V$.

(D) Given:
Focal length of objective,$f_0 = 20\,m = 2000\,cm$.
Focal length of eyepiece,$f_e = 2\,cm$.
For normal adjustment:
$1$. The distance between the objective and the eyepiece is $L = f_0 + f_e = 2000\,cm + 2\,cm = 2002\,cm = 20.02\,m$. Thus,statement $(a)$ is correct.
$2$. The magnifying power (magnification) is $M = \frac{f_0}{f_e} = \frac{2000}{2} = 1000$. Thus,statement $(b)$ is correct.
$3$. In an astronomical telescope,the final image formed is inverted and magnified relative to the object. Thus,statement $(c)$ is incorrect.
$4$. The aperture of the eyepiece is kept smaller than that of the objective to ensure that all light rays collected by the objective enter the eye. Thus,statement $(d)$ is correct.
Therefore,the correct statements are $(a), (b),$ and $(d)$.

(A) The work done in bringing a charge $q_0$ from infinity to a point is given by $W = q_0 V$,where $V$ is the electric potential at that point due to the existing charges.
In a regular hexagon of side $d$,the distance of each corner from the centre is $d$.
The electric potential $V$ at the centre due to the six charges is the sum of the potentials due to each individual charge:
$V = \frac{1}{4 \pi \varepsilon_0} \left( \frac{+q}{d} + \frac{-q}{d} + \frac{+q}{d} + \frac{-q}{d} + \frac{+q}{d} + \frac{-q}{d} \right)$
$V = \frac{1}{4 \pi \varepsilon_0 d} (q - q + q - q + q - q)$
$V = 0$
Therefore,the work done $W = q_0 \times 0 = 0$.

(C) Given collector current $I_C = 24\,mA$.
The current gain $\alpha$ is the ratio of collector current to emitter current,which is given as $80\% = 0.8$.
We know that $I_C = \alpha I_E$,so $I_E = \frac{I_C}{\alpha}$.
Substituting the values: $I_E = \frac{24\,mA}{0.8} = 30\,mA$.
Using Kirchhoff's current law for a transistor,$I_E = I_B + I_C$,so $I_B = I_E - I_C$.
$I_B = 30\,mA - 24\,mA = 6\,mA$.
In an $n-p-n$ transistor,the emitter current enters the transistor,while the collector and base currents leave the transistor. However,the question asks for the base current magnitude and direction relative to the base terminal. Since $I_E = I_B + I_C$,the base current $I_B$ flows into the base region to compensate for the recombination of charge carriers. Thus,the current is $6\,mA$ and entering the base.

(C) The formula for the force per unit length between two parallel current-carrying conductors is given by:
$\frac{F}{\ell} = \frac{\mu_0 i_1 i_2}{2 \pi r}$
Given:
$i_1 = 5 \text{ A}$
$i_2 = 10 \text{ A}$
$r = 10 \text{ cm} = 0.1 \text{ m}$
$\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$
Substituting the values:
$\frac{F}{\ell} = \frac{4\pi \times 10^{-7} \times 5 \times 10}{2\pi \times 0.1}$
$\frac{F}{\ell} = \frac{2 \times 10^{-7} \times 50}{0.1} = \frac{100 \times 10^{-7}}{0.1} = 1000 \times 10^{-7} = 10^{-4} \text{ N/m}$
Since the currents are in the same direction,the force between the conductors is attractive.

(A) The magnetic field on the axis of a circular loop at a distance $x$ from the center is given by $B = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.
Given: $R = 100\,cm = 1\,m$,$x = 1\,m$,$I = \sqrt{2}\,A$,and $\mu_0 = 4\pi \times 10^{-7}\,T\cdot m/A$.
Substituting the values:
$B = \frac{4\pi \times 10^{-7} \times \sqrt{2} \times (1)^2}{2(1^2 + 1^2)^{3/2}}$
$B = \frac{4\pi \times 10^{-7} \times \sqrt{2}}{2(2)^{3/2}}$
Since $(2)^{3/2} = 2\sqrt{2}$,we have:
$B = \frac{4\pi \times 10^{-7} \times \sqrt{2}}{2 \times 2\sqrt{2}}$
$B = \frac{4\pi \times 10^{-7}}{4} = \pi \times 10^{-7}\,T$
$B = 3.14 \times 10^{-7}\,T$.

(A) The sliding contact $C$ divides the wire $AB$ into two parts: $AC$ and $CB$. Since $C$ is at one-fourth of the length,the resistance of $AC$ is $R_{AC} = R_0 / 4$ and the resistance of $CB$ is $R_{CB} = 3 R_0 / 4$.
The resistor $R$ is connected in parallel with the part $AC$. The equivalent resistance of the parallel combination of $R$ and $R_{AC}$ is $R_p = \frac{R \cdot (R_0 / 4)}{R + (R_0 / 4)} = \frac{R R_0}{4 R + R_0}$.
Now,the circuit consists of $R_p$ and $R_{CB}$ in series across the voltage source $V_0$. The potential drop $V$ across the resistor $R$ (which is the same as the potential drop across $R_p$) is given by the voltage divider rule:
$V = V_0 \cdot \frac{R_p}{R_p + R_{CB}}$
Substituting the values:
$V = V_0 \cdot \frac{\frac{R R_0}{4 R + R_0}}{\frac{R R_0}{4 R + R_0} + \frac{3 R_0}{4}}$
Multiplying the numerator and denominator by $4(4 R + R_0)$:
$V = V_0 \cdot \frac{4 R R_0}{4 R R_0 + 3 R_0(4 R + R_0)} = V_0 \cdot \frac{4 R R_0}{4 R R_0 + 12 R R_0 + 3 R_0^2} = V_0 \cdot \frac{4 R R_0}{16 R R_0 + 3 R_0^2}$
Dividing by $R_0$:
$V = \frac{4 V_0 R}{16 R + 3 R_0}$

(A) The ratio of electrostatic force $F_e$ to gravitational force $F_G$ is given by:
$\frac{F_e}{F_G} = \frac{\frac{K q_1 q_2}{r^2}}{\frac{G m_1 m_2}{r^2}} = \frac{K}{G} \cdot \frac{q_1 q_2}{m_1 m_2}$
Given $\frac{F_e}{F_G} = 2.4 \times 10^{39}$,$q_1 = q_2 = 1.6 \times 10^{-19} \; C$,$m_1 = 9.11 \times 10^{-31} \; kg$,and $m_2 = 1.67 \times 10^{-27} \; kg$.
Substituting the values:
$2.4 \times 10^{39} = \frac{K}{G} \cdot \frac{(1.6 \times 10^{-19})^2}{(9.11 \times 10^{-31}) \times (1.67 \times 10^{-27})}$
$\frac{K}{G} = \frac{2.4 \times 10^{39} \times (9.11 \times 1.67) \times 10^{-58}}{(1.6)^2 \times 10^{-38}}$
$\frac{K}{G} = \frac{2.4 \times 15.2137 \times 10^{-19}}{2.56 \times 10^{-38}} = \frac{36.51288}{2.56} \times 10^{19} \approx 14.26 \times 10^{19} = 1.426 \times 10^{20}$
Thus,the ratio is approximately $10^{20}$.