NEET-UG 2021

E = energy = [ML²T^$-$2]

G = Gravitational constant = [M^$-$1L³T^$-$2]

So, $\frac{E}{G}=\frac{[E]}{[G]}=\frac{M{L}^2{T}^{-2}}{M^{-1}{L}^3{T}^{-2}}=[M^2{L}^{-1}{T}^0]$

L.C. = $\frac{Pitch}{CSD}$

$=\frac{1mm}{100}$ = 0.01 m = 0.001 cm

Radius = M.S. + CSR(L.C)

= 0 + 52 (0.001)

= 0.052 cm

E $\propto$ F^a A^b T^c

[M¹L²T^$-$2] $\propto$ [M¹L¹T^$-$2]^a [LT^$-$2]^b [T]^c

[ML²T^–2] $\propto$ [M^a L^{a + b} T^{–2a – 2b + c}]

a = 1

a + b = 2 $\Rightarrow$ b = 1

$-$2a $-$ 2b + c = $-$ 2

$\Rightarrow$ c = 2

a = 1, b = 1, c = 2

E $\propto$ [F] [A] [T²]

u = 0

a = 5

t = 4

velocity of car at t = 4 sec is

V = u + at

V = 0 + 5 $\times$ 4

V = 20 m/s

For ball :

At t = 4 s, A ball is dropped out of a window so velocity of ball at this instant is 20 ms–1 along horizontal.

After 2 seconds of motion :

Horizontal velocity of ball , V_x = 20 m/sec

V_y = u + ut

= 10 $\times$ 2

Vertical velocity of ball, V_y = 20 m/sec

So magnitude of velocity of ball

V = $\sqrt{V_x^2+V_y^2}$ = 20$\sqrt{2}$

and Acceleration of ball at t = 6 s is g = 10 m/sec² As ball is under free fall.

Take the surface of the Earth as the reference level for potential energy.

Initially, the particle is released from height $S$, so

• Initial potential energy $=mgS$

• Initial kinetic energy $=0$

So, total mechanical energy is

$$E=mgS$$

Now suppose at some instant the particle is at height $h$ above the surface.

Then,

• Potential energy at that instant $=mgh$

• Kinetic energy at that instant is given to be three times the potential energy:

$$K=3U=3mgh$$

Using conservation of mechanical energy:

$$K+U=mgS$$

Substitute $K=3mgh$ and $U=mgh$:

$$3mgh+mgh=mgS$$

$$4mgh=mgS$$

$$h=\frac{S}{4}$$

So the height from the surface is

$$\frac{S}{4}$$

Now find the speed at that instant.

Since

$$K=\frac{1}{2}m{v}^2$$

and also

$$K=3mgh$$

So,

$$\frac{1}{2}m{v}^2=3mgh$$

Cancel $m$:

$$\frac{1}{2}{v}^2=3gh$$

$$v^2=6gh$$

Now put $h=\frac{S}{4}$:

$$v^2=6g\cdot \frac{S}{4}=\frac{3gS}{2}$$

Therefore,

$$v=\sqrt{\frac{3gS}{2}}$$

Hence, the correct answer is:

$$\boxed{\frac{S}{4},\sqrt{\frac{3gS}{2}}}$$

So, Option A is correct.

E = mgh

P_input = $\frac{mgh}{t}$

$=\frac{15\times 10\times 60}{1}=9000=9$ kW

10% loss = 0.9 $\times$ 10³

P_output = 9 $\times$ 10³ $-$ 0.9 $\times$ 10³ = 8.1 kW

To calculate the impulse imparted to the ball, we need to look at the change in momentum during the collision with the ground. The impulse $I$ can be determined by the following relationship:

$$I=\Delta p$$

Where:

• $\Delta p$ is the change in momentum.
• Momentum $p$ is defined as the product of the mass $m$ of an object and its velocity $v$.

Since the ball rebounds to the same height, its speed just before it hits the ground and just after it leaves the ground will be the same (ignoring air resistance), although the direction of the velocity will change.

Let's calculate the speed of the ball just before the impact. The ball drops from a height $h$ with initial velocity $u=0$ m/s. Using the kinematic equation for constant acceleration under gravity $g$, we have:

$$v^2=u^2+2gh$$

Plugging in the values for $u=0$, $g=10$ m/s², and $h=10$ m, we get:

$$v^2=0^2+2\cdot 10\cdot 10$$

$$v^2=200$$

$$v=\sqrt{200}$$

$v=10\sqrt{2}$ m/s

Since the ball bounces back to the same height, its speed on the way up just after the collision will be $10\sqrt{2}$ m/s but in the opposite direction.

The change in velocity $\Delta v$ is:

$$\Delta v=v_{\text{final}}-v_{\text{initial}}$$

$$\Delta v=10\sqrt{2}-(-10\sqrt{2})$$

$\Delta v=20\sqrt{2}$ m/s

Given that the mass $m$ is 0.15 kg, we can now calculate $\Delta p$:

$$\Delta p=m\Delta v$$

$$\Delta p=0.15\cdot 20\sqrt{2}$$

$$\Delta p=3\sqrt{2}$$

$$\Delta p\approx 3\cdot 1.414$$

$\Delta p\approx 4.242$ kg m/s

The magnitude of the impulse imparted to the ball is therefore approximately 4.242 kg m/s, which most closely matches:

Option C: 4.2 kg m/s.

Given that,

Mass of Ring = M; Radius of Ring = R

Now 90° arc is removed from circular ring, then

Mass removed = $\frac{M}{4}$

$\therefore$ Mass of remaining portion = $\frac{3M}{4}$



I = MR²

I' = $\frac{3M}{4}$R²

I' = $\frac{3}{4}$I

$\therefore$ K = $\frac{3}{4}$

By balancing torque

2g $\times$ 20 = 0.5g $\times$ 60 + mg $\times$ 120

$\Rightarrow$ m = $\frac{0.5}{6}$ kg = $\frac{1}{12}$ kg

Let F_v be the viscous force and F_B be the Bouyant force acting on the ball.



Then, when body moves with constant velocity

Mg = F_B + F_v        [a = 0]

F_v = Mg – F_B

= $dVg-\frac{d}{2}Vg$  (Mg = dVg)   V = volume of ball.

= $\frac{d}{2}Vg$

$\Rightarrow$ F_v = $\frac{Mg}{2}$

(A) $V_{rms}=\sqrt{\frac{3RT}{M}}$

(B) $P=$ $\frac{1}{3}nm{\overline{v}}^2$

(C) $E=\frac{3}{2}kT$

(D) $E_{Total}=n\frac{f}{2}RT=\frac{5}{2}RT$

Displacement equation of SHM of frequency 'n'

x = A sin ($\omega$t) = A sin (2$\pi$nt)

Now,

Potential energy

$U=\frac{1}{2}k{x}^2=\frac{1}{2}K{A}^2{\sin }^2(2\pi nt)$

$=\frac{1}{2}k{A}^2\left[\frac{1-\cos (2\pi (2n)t)}{2}\right]$

So frequency of potential energy = 2n

F = Kx

10 = K $\times$ 0.05

K = $\frac{1000}{5}=200$

$T=2\pi \sqrt{\frac{m}{K}}=2\pi \sqrt{\frac{2}{200}}$

$=\frac{2\pi }{10}=\frac{6.28}{10}$

= .628 s

Polar molecules have centres of positive and negative charges separated by some distance, so they have permanent dipole moment.

$\left|{\overrightarrow{E}}_1\right|>\left|{\overrightarrow{E}}_2\right|$

As field lines are closed at charge +q.

So, net force on the dipole acts towards right side. A system always moves to decrease it's potential energy.

Let charge and radius of smaller drop is q and r respectively

For smaller drop, V = $\frac{kq}{r}=220V$

Let R be radius of bigger drop,

As volume remains the same

$\left(\frac{4}{3}\pi r^3\right)\times 27=\frac{4}{3}\pi R^3$

$\Rightarrow$ R = 3r

Now, using charge conservation,

Q = 27q

V_{big drop} = $\frac{kQ}{r}=\frac{k\left(27q\right)}{r}=9\left(\frac{kq}{r}\right)$

= 9 $\times$ 220 = 1980 V

$\varphi$ = constant

E_unknown = $\varphi$ I_b

$\Rightarrow$ E_unknown $\propto$ I_b

$\frac{E_1}{E_2}=\frac{I_1}{I_2}\Rightarrow \frac{1.5}{2.5}=\frac{36}{x}$

$x=\frac{36\times 5}{3}=60$ cm

$\frac{R}{4}$ = .25

R = 1

Now these four resistances are arranged in series

R_S = R + R + R + R = 4R

= 4 $\times$ 1 = 4$\Omega$

Current density, J = $\frac{I}{A}$ = nev_d

Drift velocity, V_d = $\frac{eE}{m}$$\tau$ ;

Electrical resistivity, $\rho$ = $\frac{m}{n{e}^2\tau }$ or $\rho$ = $\frac{E}{J}$

Relaxation period, $\tau =\frac{m}{n{e}^2\rho }$

A $\to$ R

B $\to$ S

D $\to$ Q

C $\to$ P

Energy = Energy density $\times$ volume

= $\frac{1}{2}{\epsilon }_0{E}^{2}Ad$

C_AB = 2C

From Ampere's circuital law

$B_{in}=\frac{{\mu }_{0}Ir}{2\pi R^2}$

$B_0=\frac{{\mu }_{0}I}{2\pi r}$

$B=\frac{{\mu }_{0}I}{2\pi R}$

F = BVq sin$\theta$

$\theta$ = 90${}^{\circ }$

F = BVq

$F=\frac{{\mu }_{0}I}{2\pi R}\times V\times e=\frac{2\times {10}^{-7}\times 5}{20\times {10}^{-2}}\times 10^5\times 1.6\times 10^{-19}$

F = 8 $\times$ 10^$-$20 N

Given q = 1 and $\overrightarrow{v}=2\hat{i}+4\hat{j}+6\hat{k}$ and $\overrightarrow{F}=4\hat{i}-20\hat{j}+12\hat{k}$

Also given, $\overrightarrow{F}=q\left(\overrightarrow{v}\times \overrightarrow{B}\right)$

$=q\overrightarrow{v}\times \left(B\hat{i}+B\hat{j}+B_0\hat{k}\right)$

$\Rightarrow$ $\left(4\hat{i}-20\hat{j}+12\hat{k}\right)=-1\times \left[\left(2\hat{i}+4\hat{j}+6\hat{k}\right)\times \left(B\hat{i}+B\hat{j}+B_0\hat{k}\right)\right]$

Thus, calculating values of RHS,

$\overrightarrow{v}\times \overrightarrow{B}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 4& 6\\ B& B& B_0\end{array}\right|$

= $\hat{i}$(4B₀ - 6B) - $\hat{j}$(2B₀ - 6B) + $\hat{k}$(2B - 4B)

Comparing L.H.S and R.H.S,

4B₀ – 6B = 4 .....(1)

–(2B₀ – 6B)= –20 .....(2)

2B – 4B = 12 $\Rightarrow$ B = –6 .....(3)

From (2) and (3) B = –6 and B₀ = –8

$\therefore$ $\overrightarrow{B}$ = $-6\hat{i}-6\hat{j}-8\hat{k}$

Two concentric coils are of radius R₁ and R₂ as shown


Let current in outer loop be i

Magnetic field at center = B = $\frac{{\mu }_{0}i}{2R_1}$

Magnetic flux through inner coil = B $\times$ $\pi R_2^2$

$\Rightarrow$ $\varphi$ = $\frac{{\mu }_{0}i}{2R_1}\times \pi R_2^2$

As per definition, $\varphi$ = Mi

$\Rightarrow$ M = $\left(\frac{{\mu }_0\pi }{2}\right)\frac{R_2^2}{R_1}$

$\therefore$ M $\propto$ $\frac{R_2^2}{R_1}$

I₀ = 10$\sqrt{2}$ A

I_RMS = $\frac{I}{\sqrt{2}}$ = 10A

$V_{RMS}=\sqrt{V_R^2+{(V_L-V_C)}^2}$

$=\sqrt{{(40)}^2+{(40-10)}^2}$

= 50 V

Impedance, $Z=\frac{V_{RMS}}{I_{RM}}=\frac{50V}{10V}=5\Omega$

Given, V = V₀sin$\omega$t

We know, $q=CV$

Now displacement current I_d is given by,

I_d = $\frac{dq}{dt}=\frac{CdV}{dt}$

$I_d=C(V_0\omega \cos \omega t)$

$=V_0\omega C\cos \omega t$

Power loss = 0

$\eta$ = 100%

P_in = P_o/p

V_PI_p = V_SI_s

220 $\times$ I_p = 44

I_p = $\frac{44}{220}=\frac{1}{5}$A = .2A

$Q=\frac{\omega }{\Delta \omega }=\frac{\omega L}{R}\Rightarrow \Delta \omega =R/L=\frac{50}{4}=8$ rad/sec

${\omega }_0=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{5\times 80\times {10}^{-6}}}=50$ rad/sec

${\omega }_{\min }={\omega }_0-\frac{\Delta \omega }{2}=46$ rad/sec

${\omega }_{\max }={\omega }_0-\frac{\Delta \omega }{2}=54$ rad/sec

Wave in x direction

C = $\overrightarrow{E}$ $\times$ $\overrightarrow{B}$

$\left(-\hat{j}+\hat{k}\right)\times \left(-\hat{j}-\hat{k}\right)$

$=\hat{i}+\hat{i}$

$=2\hat{i}$

from Snell's law

$\sqrt{3}$ sin 30^o = 1 $\times$ sin r

$\Rightarrow$ r = 60^o

Parallel beam of light after refraction from convex lens converge at the focus of convex lens. In question it is given light after refraction pass through concave lens becomes parallel. Therefore light refracted from convex lens virtually meet at focus of concave lens.

According to above ray diagram, d = f_A $-$ f_B

= 20 $-$ 5

= 15 cm

With larger aperture of objective lens, the light gathering power in telescope is high.

Also, the resolving power or the ability to observe two objects distinctly also depends on the diameter of the objective. Thus objective of large diameter is preferred.

Also, with large diameters fainter objects can be observed. Hence it also contributes to the better quality and visibility of images.

Hence, all options are correct.

First for image formation from lens

u = $-$60 cm

f = +30 cm

$\Rightarrow v=\frac{uf}{u+f}=\frac{-60\times 30}{-60+30}=60$ cm

This real image formed by lens acts as virtual object for mirror.



Real image from plane mirror is formed 20 cm in front of mirror, hence at 20cm distance from lens. Now for second refraction from lens,

u = $-$20 cm

f = +30 cm

$v=\frac{uf}{u+f}=\frac{-20\times 30}{-20+30}=-60$ cm

So, final virtual image is 60 cm from lens, or 20 cm behind mirror.

Given binding energy per nucleon of X, Y & Z are

7.6MeV, 8.5MeV & 8.5MeV respectively,

Gain in binding energy is :

Q = Binding Energy of products $-$ Binding energy of reactants

= (120 $\times$ 8.5 $\times$ 2) $-$ (240 $\times$ 7.6) MeV

= 216 MeV

$$\beta$$⁺ decreases atomic number by 1.

$\alpha$ decreases atomic number by 2.

$\beta$^$$-$$ increases atomic number by 1.

$\frac{hc}{\lambda }=k_{\max }+\varphi$ [given $\varphi$ is negligible]

So, $\frac{hc}{\lambda }=K_{\max }$

${\lambda }_d=\frac{h}{\sqrt{2m{K}_{\max }}}\Rightarrow K_{\max }=\frac{h^2}{2m{\lambda }_d^2}$

$\left(\frac{hc}{\lambda }\right)=\frac{h^2}{2m{\lambda }_d^2}\Rightarrow \lambda =\left(\frac{2mc}{h}\right){\lambda }_d^2$

In reverse biased, after breakdown, voltage across the zener diode becomes constant. Therefore zener diode is connected in reverse biased when used as voltage regulator.

for Ge Potential barrier V₀ = 0.3 V

Si Potential barrier V₀ = 0.7 V

In N type semiconductor majority charge carriers are e^$$-$$ and P type semiconductor majority charge carriers are holes.

I = neAV_d = neA ($\mu$E)

As $\mu$_e > $\mu$_h $\Rightarrow$ I_e > I_h

Based on above calculation, possible empirical formula is CH₃.

Energy of electromagnetic radiation (E)

$=\frac{hc}{\lambda }=h\gamma$

So, $\frac{c}{\lambda }=\gamma$

$\Rightarrow \lambda =\frac{c}{\gamma }$

$\lambda =\frac{3\times {10}^8}{1368\times {10}^3}=219.3$ m

$\bullet$ Isoelectronic species have some number of electrons.

According to Boyle's law, at constant temperature, pressure of a gas of fixed amount varies inversely with volume.

p $\propto$ $\frac{1}{V}$ $\Rightarrow$ p = $\frac{k}{V}$ $\Rightarrow$ pV = k

where k is proportionality constant and equal to nRT.


As temperature increases, gas expands i.e. volume of gas increases.

So, the product pV increases and the p vs V curve of Boyle’s law shifts upwards.

$n_{O_2}=\frac{4}{32}=\frac{1}{8}$

$n_{H_2}=\frac{2}{2}=1$

$n_t=\frac{1}{8}+1=\frac{9}{8}$

$P_{t}V=n_{t}RT$

$P_t=\frac{\frac{9}{8}\times 0.082\times 273}{1}=25.18$ atm

Dimethylammonium acetate is a salt of weak acid and weak base whose pH can be calculated as

pH = 7 + $\frac{1}{2}$(pK_a $-$ pK_b)

pK_a of acetic acid = 4.77

pK_b of dimethyl amine = 3.27

pH = 7 + $\frac{1}{2}$(4.77 $-$ 3.27)

= 7.75

$\bullet$ Osmotic pressure ($\pi$) = iCRT where C is molar concentration of the solution

$\bullet$ With increase in molar concentration of solution osmotic pressure increases.

$\bullet$ Since, weight of all solutes and its solution volume are equal, so higher will be the molar mass of solute, smaller will be molar concentration and smaller will be the osmotic pressure.

$\bullet$ Order of molar mass of solute decreases as Sucrose > Glucose > Urea

$\bullet$ So, correct order of osmotic pressure of solution is P₃ > P₁ > P₂

Given : $n_{C_6{H}_6}:n_{C_8{H}_{18}}=3:2$

So, ${\chi }_{C_6{H}_6}=\frac{3}{5},{\chi }_{C_8{H}_{18}}=\frac{2}{5}$

Total vapour pressure of solution,

$p_s=p_{C_6{H}_6}^o{\chi }_{C_6{H}_6}+p_{C_8{H}_{18}}^o{\chi }_{C_8{H}_{18}}$

$=280\times \frac{3}{5}+420\times \frac{2}{5}$

$=168+168$

$=336$ mm of Hg