NEET 2021 Question Paper Step-by-Step Solutions

When two conducting spheres of radii $R_1$ and $R_2$ are joined by a thin conducting wire, free electrons flow between them until they reach an identical electrostatic potential $V$:

Surface charge density $\sigma$ is defined as the total charge per unit spherical surface area $A = 4\pi R^2$:

Substitute the charge ratio $q_1 / q_2 = R_1 / R_2$ into the density ratio:

When the small sphere attains a constant terminal velocity ($v_t$), its downward acceleration is zero ($a = 0$). By Newton's second law, the downward gravitational force is perfectly balanced by the sum of upward buoyant force ($F_B$) and upward viscous drag ($F_v$):

Let $V$ be the volume of the sphere. The mass is $M = V \cdot d$, so $V = M / d$. The weight is $W = Mg$. The buoyant force exerted by glycerine of density $\rho_L = d/2$ is:

Substitute $W = Mg$ and $F_B = Mg/2$ into the force balance equation:

Using Ampere's circuital law / Biot-Savart law, the magnetic field at a perpendicular distance $r = 20\text{ cm} = 0.2\text{ m}$ from a long wire carrying current $I = 5\text{ A}$ is:

By the right-hand thumb rule, for current flowing from P to Q (towards the right), the magnetic field above the wire points out of the page ($\hat{k}$). The electron moves parallel to the wire ($\hat{i}$). Therefore, $\vec{v} \perp \vec{B}$, so $\theta = 90^\circ$ and $\sin 90^\circ = 1$.

The magnitude of the magnetic Lorentz force on the electron ($q = 1.6 \times 10^{-19}\text{ C}$) is:

The electric field lines diverge from a point to the left towards the right. By Faraday's field line principles, field intensity is proportional to the density of lines of force. Therefore, the field is stronger at the position of $+q$ than at the position of $-q$:

Determine the individual forces on each charge: - Force on $+q$: $F_1 = q E_{(+q)}$ acting towards the right (in the direction of $\vec{E}$). - Force on $-q$: $F_2 = q E_{(-q)}$ acting towards the left (opposite to $\vec{E}$).

For any conservative electrostatic system, the net force is the negative gradient of potential energy ($F = -dU/dx$). Any body accelerating in the direction of the net force moves toward a state of lower potential energy:

Astronomical telescopes observe extremely distant, faint celestial bodies. The amount of light gathered from a distant star is directly proportional to the surface area of the objective lens:

Due to diffraction at the circular aperture of diameter $D$, the angular limit of resolution is $\Delta\theta = \frac{1.22\lambda}{D}$. The resolving power ($RP$) is the reciprocal of $\Delta\theta$:

Because a larger aperture collects more light photons and reduces diffraction blurring, the resultant image has superior brightness, sharpness, and visibility. Hence, all statements (1), (2), and (3) are scientifically true.

The least count is the smallest distance moved by the spindle corresponding to one circular scale division:

Using the standard screw gauge reading formula with zero zero-error mentioned:

Since $1\text{ cm} = 10\text{ mm}$, divide the reading by 10:

The parent nucleus has mass number $A = 240$ and binding energy per nucleon $\text{BE}_i/A = 7.6\text{ MeV}$:

The parent splits into two equal fragments of mass number $A = 120$ each, with $\text{BE}_f/A = 8.5\text{ MeV}$:

The total gain in binding energy (which is also the Q-value energy released in the fission reaction) is:

For an object moving with initial velocity $u$ and uniform acceleration $a$, the distance covered between time $t = n-1$ and $t = n$ is:

Since the block starts from rest on a smooth inclined plane of angle $\theta$, $u = 0$ and $a = g\sin\theta$ (constant):

In the interval $t = n$ to $t = n+1$, the distance travelled is $S_{n+1}$:

Dividing $S_n$ by $S_{n+1}$ cancels the common acceleration factor $a/2$:

The incident ray is perpendicular to the hypotenuse face (marked with a right-angle symbol in the diagram). By Snell's law, $\sin i_1 = 0 \implies \sin r_1 = 0$, so the ray enters the prism without any deviation.

Consider the triangle at the top of the prism: - The apex angle between the vertical face and the hypotenuse is $A = 180^\circ - (90^\circ + 60^\circ) = 30^\circ$. - The ray enters perpendicular to the hypotenuse, making an angle of $90^\circ$ with the hypotenuse. - Inside the top small triangle, the angle between the ray and the vertical face is $180^\circ - (90^\circ + 30^\circ) = 60^\circ$. - Since the normal to the vertical face is perpendicular to it, the angle of incidence $r_2$ (angle with the normal) is:

At the vertical glass-air interface, light travels from glass ($\mu = \sqrt{3}$) into air ($n = 1$):

By Hooke's law, the restoring force is proportional to elongation: $F = kx$. Given $F = 10\text{ N}$ for extension $x = 5\text{ cm} = 5 \times 10^{-2}\text{ m}$:

For a block of mass $m = 2\text{ kg}$ oscillating on an ideal spring of stiffness $k$:

Substitute $\pi \approx 3.1416$:

A parallel beam of light falling on convex lens A has object distance $u_1 = \infty$. By the lens formula, the image is formed at its principal focus:

This real converging image acts as a virtual object for concave lens B placed at distance $d$ from A. The object distance for lens B is $u_2 = +(20 - d)$. For the emergent beam to be parallel, the final image must be formed at infinity ($v_2 = \infty$):

Equating $u_2 = 20 - d$ to $5\text{ cm}$:

Under an applied electric field $\vec{E}$, the electric force on an electron is $-e\vec{E}$, imparting acceleration $\vec{a} = -e\vec{E}/m$. The average velocity acquired over relaxation time $\tau$ is:

In vector microscopic notation, Ohm's law states $\vec{J} = \sigma \vec{E} = \frac{1}{\rho}\vec{E}$. Rearranging gives electrical resistivity:

Electric current is $I = n e A v_d$. Dividing by cross-sectional area $A$ gives the current density:

Microscopic resistivity in terms of electron mass $m$, charge $e$, number density $n$, and relaxation time $\tau$ is $\rho = \frac{m}{n e^2 \tau}$. Solving for $\tau$ gives:

The drift velocity of a charge carrier under an electric field $E$ is $v_d = \mu E$, where $\mu$ is mobility. Current through a semiconductor of cross-sectional area $A$ with carrier concentration $n$ is:

For n-type with electron density $n_e$ and p-type with hole density $n_h$:

Electrons in n-type occupy the conduction band, where they behave almost as free electrons with small effective mass. Holes in p-type move via the sequential vacancy-hopping of bound valence electrons. Consequently, in all semiconductors (Silicon, Germanium, GaAs):

In all nuclear reactions, the total electric charge (atomic number $Z$) is strictly conserved. Let $\Delta Z$ be the change in atomic number of the daughter nucleus relative to the parent nucleus.

The atomic number changes from $Z$ to $Z-1$ (decreases by 1). This occurs when a proton converts into a neutron, emitting a positron ($\beta^+$):