datavorous/entrance-exam-dataset huggingface dataset:
A virus with ssDNA is discovered. Its DNA is amplified by running 10 cycles of PCR. How many dsDNA strands will be formed at the end?

Options:
A 10
B 512
C 1024
D 102

The correct answer is:
512

Solution:
Polymerase chain reaction (PCR) is a process to amplify nucleic acid sequences. In this process, the DNA is replicated multiples times by adding synthesized primer which attach to the denatured DNA strand and are extended using a thermostable polymerase enzyme -Taq polymerase obtained from a thermophilic bacterium, Thermus aquaticus. DNA is denatured by heating and breaking the H-bonds.
In each cycle of PCR the dsDNA is doubled, so after n cycles the copies of DNA will be 2n .
Now the given sample of viral DNA is single stranded, therefore, in the first cycle it will form a dsDNA. And in the consecutive cycles the dsDNA will double in copies. Therefore, after 10 cycles the number of copies will be 29=512 .

Topic: Biology, Biotechnology - Principles and Processes, NEET
A voltage of peak value 283 V and varying frequency is applied to a series LCR combination in which R=3 Î©, L=25 mH and C=400 Î¼F. The frequency f (in Hz) of the source at which maximum power is dissipated in the above, is

Options:
A 48.4
B 52.7
C 47.4
D 50.3

The correct answer is:
50.3

Solution:
A series resonance circuit admits maximum current, as
P= i rms 2 âzâcosâ 0 o (Z = Impedance)
So, power dissipated is maximum at resonance.
So, frequency of the source at which maximum power is dissipated in the circuit is
f=12ÏLC
=12Ã3.1425Ã10-3Ã400Ã10-6
=12Ã3.1410-5=50.3 Hz

Topic: Physics, Alternating Current, NEET
A voltmeter having a resistance of 998 Î© is connected to a cell of emf 2 V and internal resistance 2 Î©. The error in the measurement of emf will be

Options:
A 4Ã10-1 V
B 2Ã10-3 V
C 4Ã10-3 V
D 2Ã10-1 V

The correct answer is:
4Ã10-3 V

Solution:
I=2998+2=21000=2Ã10-3A
=2 mA
Internal resistance is 2 Î©
Hence, the measure of emf will be
E=2-IÃ2
=2-2 mAÃ2
Therefore, error in measured value =4Ã10-3 V

Topic: Physics, Magnetic Effects of Current, NEET
A volume of a healthy human blood is

Options:
A 5 - 6 Litres
B 8 - 9 Litres
C 2 - 3 Litres
D 10 - 12 Litres

The correct answer is:
5 - 6 Litres

Solution:
Through various techniques such as injecting a harmless opaque substance of specific quantity and noting its degree of dilution, the blood volume in adult human body is measured to be near about 5 litres.

Topic: Biology, Body Fluids and Circulation, NEET
A wasteful, light-induced respiratory process releasing CO2 is called as

Options:
A Richmond Lang effect
B Photorespiration
C PCR
D Pasteur's effect

The correct answer is:
Photorespiration

Solution:
Photorespiration is a wasteful process which occurs in C-3 plants at high temperature and high light intensity. It occurs by glycolate cycle (which occurs in chloroplast, peroxisomes and mitochondria) in which synthesis of glucose is decreased and energy is lost in the form of heat.

Topic: Biology, Photosynthesis in Higher Plants, NEET
A water drop is divided into eight equal droplets. The pressure difference between the inner and outer side of the big drop will be

Options:
A Same as for smaller droplet
B 12 of that for smaller droplet
C 14 of that for smaller droplet
D twice that for smaller droplet

The correct answer is:
12 of that for smaller droplet

Solution:
When break into eight drop radius of small drop is r=Rn13⇒r=R813=R2
For large drop ΔP1=2TR
For small drop ΔP2=2Tr=2TR2=22TR

Topic: Physics, Mechanical Properties of Fluids, NEET
A water pump, rated 400 W, has an efficiency of 75%. If it is employed to raise water through a height of 40 m, the volume of water drawn in 10 min is

Options:
A 0.45 m3
B 0.75 m3
C 0.62 m3
D 0.02 m3

The correct answer is:
0.45 m3

Solution:
Output power =η× input power
=0.75×400=300 W
Now, from P=dwdt=d(mgh)dt
dmdt=Pgh=30010×40=34 kg
=34×60 kg  min-1
=45 kg  min-1
Hence, water drawn in 10 min
=45×10=450  kg  m-3
Since, density of water =1000 kg  m-3
Therefore, from m=ρV
V=mρ=4501000=0.45 m3

Topic: Physics, Work Power Energy, NEET
A water sample has Mn concentration as 0.5ppm. Is it safe to drink?

Options:
A Safe
B Not safe
C Cannot say
D None of the above

The correct answer is:
Not safe

Solution:
Correct Option is : 2
Not safe

Topic: Chemistry, Environmental Chemistry, NEET
A water sample has ppm level concentration of following anions, F-=10;SO42-=100;NO3-=50 . The anion/anions that make/makes the water sample unsuitable for drinking is/are

Options:
A only NO3-
B only F-
C only SO42-
D both SO42- and NO3-

The correct answer is:
only F-

Solution:
Above 500 ppm of SO42- ions in drinking water, can cause laxative effect otherwise lesser ppm value is permissible for drinking.
Maximum limit of NO3- ions in drinking water is 50 ppm, above this limit it can cause the disease like methamoglobinemia.
More than 1 ppm F- ions in drinking water are not fit for drinking, it can cause decay of bones and teeth.

Topic: Chemistry, Environmental Chemistry, NEET
A wave has velocity u in medium P and velocity 2u in medium Q. If the wave is refracted from medium P to medium Q at an angle of incidence 30°, then the angle of refraction will be

Options:
A 30°
B 45°
C 60°
D 90°

The correct answer is:
90°

Solution:
From Snell's law,
μ P  sin 30 o = μ Q sinr
sinr= μ P μ Q sin 30 o
as μ∝ 1 v
sinr= 2u u 1 2 =1
r= 90 o

Topic: Physics, Ray Optics, NEET
A wave in a string has an amplitude of $2 \mathrm{~cm}$. The wave travels in the +ve direction of $x$ axis with a speed of $128 \mathrm{~ms}^{-1}$ and it is noted that 5 complete waves fit in $4 \mathrm{~m}$ length of the string. The equation describing the wave is

Options:
A $y=(0.02) \mathrm{m} \sin (7.85 x+1005 \mathrm{t})$
B $\mathrm{y}=(0.02) \mathrm{m} \sin (15.7 \mathrm{x}-2010 \mathrm{t})$
C $y=(0.02) \mathrm{m} \sin (15.7 \mathrm{x}+2010 \mathrm{t})$
D $y=(0.02) m \sin (7.85 x-1005 \mathrm{t})$

The correct answer is:
$y=(0.02) m \sin (7.85 x-1005 \mathrm{t})$

Solution:
Key Idea Find the parameters and put in the general wave equation.
Here,
$$
\begin{aligned}
A & =2 \mathrm{~cm} \\
\text { direction } & =+ \text { ve direction } \\
\mathrm{v} & =128 \mathrm{~ms}^{-1}
\end{aligned}
$$
and
$$
5 \lambda=4
$$
Now,
$$
\mathrm{k}=\frac{2 \pi}{\lambda}=\frac{2 \pi \times 5}{4}=7.85
$$
and
$$
\mathrm{v}=\frac{\omega}{\mathrm{k}}=128 \mathrm{~ms}^{-1}
$$
$$
\begin{aligned}
\Rightarrow \quad \omega & =\mathrm{v} \times \mathrm{k}=128 \times 7.85 \\
& =1005
\end{aligned}
$$
$$
\begin{array}{lll}
\text { As, } & y & =\mathrm{A} \sin (\mathrm{kx}-\omega \mathrm{t}) \\
\therefore & & \mathrm{y}=2 \sin (7.85 \mathrm{x}-1005 \mathrm{t}) \\
& & =(0.02) \mathrm{m} \sin (7.85 \mathrm{x}-1005 \mathrm{t})
\end{array}
$$

Topic: Physics, Waves and Sound, NEET, NEET 2009 (Screening)
A wave is represented by the equation given by y=7sin7Ït-0.04Ïx+Ï3 where x is in metres and t is in seconds. The speed of the wave is

Options:
A 175 m/s
B 49π m/s
C 49/π m s
D 0.28π m/s

The correct answer is:
175 m/s

Solution:
speed of wave =coefficient of tcoefficient of x

âv= 7Ï0.04Ï= 175 m/s.

Topic: Physics, Waves and Sound, NEET
A wave is represented by y=Asin2⁡kx-ωt+ϕ. The amplitude and wavelength of wave is given by

Options:
A 2A, 2Ïk
B A, 2Ïk
C A2, 2Ïk
D A2, πk

The correct answer is:
A2, πk

Solution:
y=Asin2⁡kx-ωt+ϕ can be rewritten as
y=A2-A2cos⁡2kx-2ωt+2ϕ
Therefore the amplitude of the wave is A2 and wavelength of wave is πk

Topic: Physics, Waves and Sound, NEET
A wave of frequency 500 Hz has a velocity of 350 m s-1. The distance between two nearest points on the wave which are 60Â° out of phase with each other will be around

Options:
A 70 cm
B 0.7 m
C 12.0 cm
D 120.0 cm

The correct answer is:
12.0 cm

Solution:
Given, frequency f = 500 Hz, velocity v = 350 m/s, phase Ï = 6 0 â
By the formula, Î» = v f = 3 5 0 5 0 0 = 0.7 m
Also we know that Î x = Î Ï Î» 2 Ï
= 6 0 Ã Ï 1 8 0 Ã 0.7 2 Ï
â 0.12 metre â 12 cm

Topic: Physics, Waves and Sound, NEET
A wave of frequency 500 Hz has a velocity of 350 m s-1. The distance between two nearest points on the wave which are 60Â° out of phase with each other will be around

Options:
A 70 cm
B 0.7 m
C 12.0 cm
D 120.0 cm

The correct answer is:
12.0 cm

Solution:
Given, frequency f= 500 Hz, velocity v = 350 m s-1, phase ϕ = 6 0 ∘
By the formula, λ=vf=350500=0.7 m
Also we know that Δ x = Δ ϕ λ 2 π
= 6 0 × π 1 8 0 × 0.7 2 π
≈ 0.12 metre ≈ 12 cm

Topic: Physics, Waves and Sound, NEET
A wave pulse in a string is described by the equation y1=53x-4t2+2 and another wave pulse in the same string is described by y2=-53x+4t-62+2 . The values of y1,y2 and x are in metres and t is in seconds. Which of the following statement is correct?

Options:
A y1 travels along -x -direction and y2 along +x -direction
B Both y1 and y2 travel along +x -direction
C At x=1m, y1 and y2 always cancel
D At time t=1s, y1 and y2 exactly cancel everywhere

The correct answer is:
At x=1m, y1 and y2 always cancel

Solution:
∵ Given,
y1=53x-4t2+2, y2=-53x+4t-62+2
According option d , at x=1
y1=53-4t2+2
y2=-53+4t-62+2
=-53-4t2+2
Both wave pulse equation are existing in same string therefore resultant equation of wave pulse.
y=y1+y2=0
Hence, option At x=1m, y1 and y2 always cancel is correct.

Topic: Physics, Waves and Sound, NEET
A wave pulse is generated in a string that lies along x-axis. At the points A and B, as shown in the figure, if RA and RB are the ratio of wave speed to the particle speed respectively then :

Options:
A RA>RB

B RB>RA

C RA=RB

D Information is not sufficient to decide.

The correct answer is:
RB>RA

Solution:
Vp=-Vslope
⇒ VpV=-slope=1R
⇒ RBRA=slopeAslopeB
as we can see
slope A > slope B
⇒ RB>RA

Topic: Physics, Waves and Sound, NEET
A wave pulse is travelling on a string of linear mass density 6.4Ã10-3 kg m-1 under a load of 80 kgf. Calculate the time taken by the pulse to traverse the string, if its length is 0.7 m

Options:
A 2×10-3 s

B 3Ã10-3 s
C 4Ã10-2 s
D 5×102 s

The correct answer is:
2×10-3 s

Solution:
Here, T = 80 kgf = = 80 Ã 9.8 N ;
m = 6 .4 Ã 10 â3 kg m-1
Now, Ï = T m = 8 0 Ã 9 Â· 8 6 Â· 4 Ã 1 0 - 3 = 3 5 0 m s -1
Therefore, time taken by the wave pulse to traverse 0.7 m length of the string,
t = lâ¡ Ï = 0 . 7 3 5 0 = 2 Ã 1 0 - 3 s

Topic: Physics, Waves and Sound, NEET
A wave pulse on a string has the dimension shown in the diagram. The wave speed is v = 1 cm s-1. If point O is a free end. The shape of the wave at time t = 3 s is

Options:
A
B
C
D

The correct answer is:

Solution:
Time taken for maxima to reach the end is 3 sec and at it's free end the disturbance will add up and get double there.

Topic: Physics, Waves and Sound, NEET
A wave travelling in the positive $x$-direction having displacement along $y$-direction as $1 \mathrm{~m}$, wavelength $2 \pi \mathrm{m}$ and frequency of $\frac{1}{\pi} \mathrm{Hz}$ is represented by

Options:
A $y=\sin (x-2 t)$
B $y=\sin (2 \pi x-2 \pi t)$
C $y=\sin (10 \pi x-20 \pi t)$
D $y=\sin (2 \pi x+2 \pi t)$

The correct answer is:
$y=\sin (x-2 t)$

Solution:
Given $a=1 \mathrm{~m}$
As,
$\begin{aligned}
y & =\operatorname{asin}(k x-\omega t) \\
& =\sin \left(\frac{2 \pi}{2 \pi} x-2 \pi \times \frac{1}{\pi} t\right) \\
& =\sin (x-2 t)
\end{aligned}$

Topic: Physics, Waves and Sound, NEET, NEET 2013 (All India)
A wave travelling in the positive $x$-direction with amplitude $\mathrm{A}=0.2 \mathrm{~m}$, velocity $v=360 \mathrm{~ms}^{-1}$ and wavelength $\lambda$ $=60 \mathrm{~m}$, then the correct expression for the wave is:

Options:
A $y=0.2 \sin \left[2 \pi\left(6 t+\frac{x}{60}\right)\right]$
B $y=0.2 \sin \left[\pi\left(6 t+\frac{x}{60}\right)\right]$
C $y=0.2 \sin \left[2 \pi\left(6 t-\frac{x}{60}\right)\right]$
D $y=0.2 \sin \left[\pi\left(6 t-\frac{x}{60}\right)\right]$

The correct answer is:
$y=0.2 \sin \left[2 \pi\left(6 t-\frac{x}{60}\right)\right]$

Solution:
The general equation of wave travelling in $x$-direction is given as
$$
y=A \sin \left[\frac{2 \pi}{\lambda}(V t-x)\right]
$$
Where $A=0.2 \mathrm{~m}, v=360 \mathrm{~ms}^{-1}, \lambda=$ $60 \mathrm{~m}$
$\therefore$ The equation becomes
$$
\begin{aligned}
& y=0.2 \sin \left[\frac{2 \pi}{60}(360 t-x)\right] \\
& y=0.2 \sin \left[2 \pi\left(6 t-\frac{x}{60}\right)\right]
\end{aligned}
$$

Topic: Physics, Waves and Sound, NEET, NEET 2002
A waveform shown when applied to the following circuit will produce which of the following output waveforms. Assume ideal diode configuration and R1=R2

Options:
A
B
C
D

The correct answer is:

Solution:
The P-N junction will conduct only when it is forward biased i.e. when -5 V is fed to it, so it will conduct only for third quarter part of signal shown and when it conducts potential drop 5 V will be across both the resistors, so output voltage across R2 is 2.5 V.
∴ V 0 = - 2.5 V

Topic: Physics, Semiconductors, NEET
A wavefront is represented by the plane y=3-x. The propagation of wave takes place at

Options:
A 45° with x - direction
B 135° with x - direction
C 60° with x - direction
D No sufficient data

The correct answer is:
45° with x - direction

Solution:
The propagation of ray takes place in
perpendicular direction of wavefront.
Here, m1m2= -1
m1=slope of wavefront=-1
m2=Slope of ray
-1m2= -1 âm2=1
tanâ¡Î¸=1 âÎ¸=45Â°

Topic: Physics, Wave Optics, NEET
A weak acid HX (Ka=10â5) on reaction with NaOH gives NaX. For 0.1 M aqueous solution of NaX, the % hydrolysis is

Options:
A 0.001%
B 0.01%
C 0.15%
D 1%

The correct answer is:
0.01%

Solution:
X - + H 2 O ⇌ HX + OH - 1 - h h h

h=KwKa×C=10-1410-5×0.1=10-4
Or 10-2% i.e., 0.01%

Topic: Chemistry, Ionic Equilibrium, NEET
A weak acid, HA and a $\mathrm{K}_a$ of $1.00 \times 10^{-5}$. If $0.100 \mathrm{~mol}$ of this acid is dissolved in one litre of water, the percentage of acid dissociated at equilibrium is closer to:

Options:
A $1.00 \%$
B $99.9 \%$
C $0.100 \%$
D $99.0 \%$

The correct answer is:
$1.00 \%$

Solution:
Given $\mathrm{K}_a=1.00 \times 10^{-5}, \mathrm{C}=0.100$ mol
for a weak electrolyte, degree of dissociation
$\propto=\sqrt{\frac{\mathrm{K}_a}{\mathrm{C}}}=\sqrt{\frac{1 \times 10^{-5}}{0.100}}=10^{-2}=1 \%$
Related Theory
In any acid-base reaction, the equilibrium will favour the reaction that moves the proton to the stronger base. This equilibrium constant is referred to as the ion-product constant for water, Kw. In pure water, some molecules act as bases and some as acids.

Topic: Chemistry, Ionic Equilibrium, NEET, NEET 2007
A weak acid HX has the dissociation constant 1×10-5 M. It forms a salt NaX on reaction with alkali. The degree of hydrolysis of 0.1 M solution of NaX is

Options:
A 0.0001%
B 0.15 %
C 0.01 %
D 0.1 %

The correct answer is:
0.01 %

Solution:
Hydrolysis reaction is X-+H2O→HX+OH-
For a salt of weak acid with a strong base Kh=KwKa=cα2=10-1410-5=0.1×α2
α2=10-8
α=10-4
so, the degree of hydrolysis =10-4×100=10-2=0.01%

Topic: Chemistry, Ionic Equilibrium, NEET
A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g of carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. What is the (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula?

Options:
A (i) CH , (ii) 20 g/mol and (iii) C2H2
B (i) C2H , (ii) 26 g/mol and (iii) C4H2
C (i) C2H , (ii) 30 g/mol and (iii) C2H2
D (i) CH , (ii) 26 g/mol and (iii) C2H2

The correct answer is:
(i) CH , (ii) 26 g/mol and (iii) C2H2

Solution:
Correct Option is : (D)
(i) CH , (ii) 26 g/mol and (iii) C2H2

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
(a) What happens when Borax is heated strongly?

Options:
A Borax undergoes dehydration and forms sodium metaborate.
B Borax decomposes to form boric oxide.
C Borax melts and forms a glassy bead.
D Borax sublimes and forms boric acid.

The correct answer is:
Borax decomposes to form boric oxide.

Solution:
Correct Option is : (B)
Borax decomposes to form boric oxide.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
A Wheatstone bridge is used to determine the value of unknown resistance X by adjusting the variable resistance Y as shown in the figure. For the most precise measurement of X, the resistances P and Q

Options:
A should be approximately equal and are small
B should be very large and unequal
C do not play any significant role
D should be approximately equal to 2X

The correct answer is:
should be approximately equal and are small

Solution:
For precise measurement of X, value of X should be close to Y. Therefore, resistances P and Q should be equal.
If P and Q are large the current in the galvanometer will be small, and it will be difficult to find the null point. Therefore, P and Q should be small.

Topic: Physics, Current Electricity, NEET, NEET 2022 (Phase 1)
A wheel has angular acceleration of 3.0 $\mathrm{rad} / \mathrm{sec}^2$ and an initial angular speed of $2~ \mathrm{rad}/ \mathrm{sec}$. In a time of $2 \mathrm{sec}$ it has rotated through an angle (in radian) of:

Options:
A 10
B 12
C 4
D 6

The correct answer is:
10

Solution:
Here $\theta=\omega t+\frac{1}{2} t^2$
$\begin{aligned}
& =(2)(2)+\frac{1}{2}(3)(2)^2 \\
& =4+6=10 \mathrm{rad}
\end{aligned}$

Topic: Physics, Motion In Two Dimensions, NEET, NEET 2007
A wheel having moment of inertia $2 \mathrm{~kg}$ $\mathrm{m}^2$ about its vertical axis, rotates at the rate of $60 \mathrm{rpm}$ about the axis. The torque which can stop the wheel's rotation in one minute would be:

Options:
A $\frac{2 \pi}{15} \mathrm{~N}-\mathrm{m}$
B $\frac{\pi}{12} \mathrm{~N}-\mathrm{m}$
C $\frac{\pi 5}{15} \mathrm{~N}-\mathrm{m}$
D $\frac{\pi}{18} \mathrm{~N}-\mathrm{m}$

The correct answer is:
$\frac{\pi 5}{15} \mathrm{~N}-\mathrm{m}$

Solution:
Given, $\mathrm{I}=2 \mathrm{~kg}-m^2$,
$\omega_0=\frac{60}{60} \times 2 \pi \mathrm{rad} / \mathrm{s}$
$\omega_0$ and $t=60 \mathrm{~s}$
The torque required to stop the wheel's rotation is
$\begin{aligned}
\tau & =\mathrm{I} \propto l\left(\frac{\omega_0-\omega}{t}\right) \\
\tau & =\frac{2 \times 2 \pi \times 60}{60 \times 60} \\
& =\frac{\pi}{15} \mathrm{~N}-\mathrm{m}
\end{aligned}$

Topic: Physics, Rotational Motion, NEET, NEET 2004
A wheel is rotating with an angular speed of 20 rad s-1. It is stopped to rest by applying a constant torque in 4 s. If the moment of inertia of the wheel about its axis is 0.20 kg/m2, then the work done by the torque in two seconds will be

Options:
A 10 J
B 20 J
C 30 J
D 40 J

The correct answer is:
30 J

Solution:
ω1=20 rad s-1, ω=0 , t=4 s . So angular retardation α=ω1-ω2t=204=5 rad s-2
Now, angular speed after 2 s

ω2=ω1-αt=20-5×2=10 rad s-1 .

Work done by torque in 2 s = loss in kinetic energy =12lω12-ω12=120.20 202-102
=120.20 400-100
=120.20×300=12×60=30 J

Topic: Physics, Rotational Motion, NEET
A wheel of a bullock cart is rolling on a level road as shown in the figure below. If its linear speed is $v$ in the direction shown, which one of the following options is correct ( $P$ and $Q$ are any highest and lowest points on the wheel, respectively)?

Options:
A Point $P$ moves faster than point $Q$
B Both the points $P$ and $Q$ move with equal speed
C Point $P$ has zero speed
D Point $P$ moves slower than point $Q$

The correct answer is:
Point $P$ moves faster than point $Q$

Solution:
In the case of pure rolling,
The topmost point will have velocity $2 v$ while point $Q$ i.e. lowest point will have zero velocity. Hence point $P$ moves faster than point $Q$.

Topic: Physics, Rotational Motion, NEET, NEET 2024
A wheel of radius 2 m rolls on the ground with uniform velocity 4 m s-1. The relative acceleration of the topmost point of the wheel with respect to the bottom-most point of the wheel is

Options:
A 8 m s-2
B 16 m s-2
C 4 m s-2
D 32 m s-2

The correct answer is:
16 m s-2

Solution:
Given, the wheel's radius =2 m,
velocity =4 m s-1
As velocity is uniform, therefore acceleration experienced by the wheel in a circular motion,
a=v2r=422=8 m s-2
â´ Relative acceleration of the topmost point of the wheel with respect to the bottom-most point of the wheel
= aT- aB
=8--8=16 m s-2

Topic: Physics, Laws of Motion, NEET
A wheel rotating at 12 rev s-1 is brought to rest in 6 s. The average angular deceleration in rad s-2 of the wheel during this process is

Options:
A 4π
B 4
C 72
D π

The correct answer is:
4π

Solution:
ωi=12×2π rad s-1
=24π rad s-1
ω f =0
∆t=6 s
As ω f = ω i +α∆t
We have, α= − ω i Δt
= -24π6
= -4π rad sec-2

Topic: Physics, Rotational Motion, NEET
A wheel which is initially at rest is subjected to a constant angular acceleration about its axis. It rotates through an angle of 15o in time t seconds. The increase in angle through which it rotates in the next 2t seconds is

Options:
A 90o
B 120o
C 30o
D 45o

The correct answer is:
120o

Solution:
If angular acceleration is constant, we have
θ=ω0t+12αt2
The given θ=15o
ω0=0
For the first conditions (time = t sec)
15o=0+12αt2
⇒ 15o=12αt2 ......(i)
For the second conditions (time = t+2t sec)
θ1=12α3t2=12α9t2 .......(ii)
So for 2t sec, Δθ=θ1-12αt2
Δθ=9×12αt2-12αt2
=812αt2
=8×15o=120o (from equation(i))

Topic: Physics, Laws of Motion, NEET
A wheel with 20 metallic spokes each 1 m long is rotated with a speed of 120 rpm in a plane perpendicular to a magnetic field of 0.4 G. The induced emf between the axle and rim of the wheel will be, 1 G=10-4 T

Options:
A 2.51×10-4 V
B 2.51×10-5 V
C 4.0×10-5 V
D 2.51 V

The correct answer is:
2.51×10-4 V

Solution:
Given, the angular velocity, ω=2π60×120 rad s-1.

Recall the formula of induced EMF in terms of angular velocity and radius of the ring, ε=12Bωr2, now put the proper values at proper places,

⇒ε=12×0.4×10-4×2π12060(1)2

⇒ε=0.8π×10-4=2.512×10-4 V.

Topic: Physics, Electromagnetic Induction, NEET, NEET 2020 (Phase 2)
A which stage of meiosis does the genetic constituent of gametes is finally decided?

Options:
A Metaphase-I
B Anaphase-II
C Metaphase-II
D Anaphase-I

The correct answer is:
Anaphase-I

Solution:
When each daughter cell gets a half number of the chromosome from a parent cell through which the genetic constituent of gametes is finally decided. It is done in anaphase I stage of meiosis in which homologous pairs of chromosomes are separated and pulled to opposite ends. Due to which single set of chromosomes is present in each end.

Topic: Biology, Cell Cycle and Cell Division, NEET
A whistle giving out 450 Hz approaches a stationary observer at a speed of 33 m s-1. The frequency heard by the observer (in Hz) is (Speed of sound = 330 m s-1)

Options:
A 409
B 429
C 517
D 500

The correct answer is:
500

Solution:
The source is moving towards the observer
f⁡'=f⁡v⁡v⁡-v⁡1=450330330-33
f⁡'=500 Hz

Topic: Physics, Waves and Sound, NEET
A whistle producing sound waves of frequencies 9500 Hz and above is approaching a stationary person with speed of v m s-1. The velocity of sound in air is 300 m s-1. If the person can hear frequencies up to a maximum of 10000 Hz, the maximum value of v up to which he can hear the whistle is

Options:
A 30 m s -1
B 15 2 m s -1
C 152 m s-1
D 15 m s -1

The correct answer is:
15 m s -1

Solution:
ν'ν=vsvs-v
where vs is the velocity of sound in air.
100009500=300300-v
⇒300-v=285 ⇒v=15 m s-1

Topic: Physics, Waves and Sound, NEET
A whistle producing sound waves of frequency 9500 Hz above is approaching a stationary person with speedv m s-1. The velocity of sound in air is 300 m s-1. If the person can hear frequency up to a maximum of 10,000 Hz, the maximum value of v up to which he can hear the whistle is

Options:
A 152 m s-1
B 15 m s-1
C 30 m s-1
D none of these

The correct answer is:
15 m s-1

Solution:
The velocity of sound in air =300 m s-1
Let v be the maximum value of source velocity for which the person is able to hear the sound, then
10000=fapp=300300-v×9500
⟹ v=15 m s-1

Topic: Physics, Waves and Sound, NEET