datavorous/entrance-exam-dataset huggingface dataset:
A thin circular ring of mass m and radius R is rotating about its axis perpendicular to the plane of the ring with a constant angular velocity ω. Two point particles each of mass M are attached gently to the opposite ends of a diameter of the ring. The ring now rotates, with an angular velocity ω2. Then, the ratio mM is

Options:
A 1
B 2
C 12
D 2

The correct answer is:
2

Solution:
Angular Momentum before particle is attached to the ring, =IÏ=mR2Ï
Angular Momentum after two particle is attached to the ring,
=mR2Ï2+2MR2Ï2
According law of conservation of momentum, (momentum before) = (momentum after)
â mR2Ï=mR2+2MR2Ï2
â 2mR2=mR2+2MR2
So mM=2

Topic: Physics, Rotational Motion, NEET
A thin circular ring of Mass $M$ and radius $r$ is rotating about its axis with a constant angular velocity $\omega$. Four objects each of mass $m$, are kept gently to the opposite ends of two perpendicular diameters of the ring. The angular velocity of the ring will be:

Options:
A $\frac{M \omega}{4 m}$
B $\frac{M \omega}{M+4 m}$
C $\frac{(M+4 m) \omega}{M}$
D $\frac{(M-4 m) \omega}{M+4 m}$

The correct answer is:
$\frac{M \omega}{M+4 m}$

Solution:
According to law of conservation of angular momentum
$$
\begin{aligned}
& M r^2 \omega=\left(M r^2+4 m r^2\right) \omega^{\prime} \\
& \Rightarrow \omega^{\prime}=\frac{M \omega}{M+4 m}
\end{aligned}
$$

Topic: Physics, Rotational Motion, NEET, NEET 2003
A thin circular ring of mass M and radius r is rotating about its axis with a constant angular velocity ω. Two objects, each of mass m, are attached gently to the opposite ends of the diameter of the ring. The wheel now rotates with an angular velocity

Options:
A ωM(M+m)
B ω(M-2m)(M+2m)
C ωM(M+2m)
D ω(M+2m)M

The correct answer is:
ωM(M+2m)

Solution:
I 1 Ï 1 = I 2 Ï 2
â´ Ï 2 = I 1 I 2 Ï = ( M r 2 M r 2 + 2 m r 2 ) Ï = ( M M + 2 m ) Ï

Topic: Physics, Rotational Motion, NEET
A thin circular ring of mass $M$ and radius $R$ is rotating in a horizontal plane about an axis vertical to its plane with a constant angular velocity $\omega$. If two objects each of mass $m$ be attached gently to the opposite ends of a diameter of the ring, the ring will then rotate with an angular velocity

Options:
A $\frac{\omega(M-2 m)}{M+2 m}$
B $\frac{\omega M}{M+2 m}$
C $\frac{\omega(M+2 m)}{M}$
D $\frac{\omega M}{M+m}$

The correct answer is:
$\frac{\omega M}{M+2 m}$

Solution:
Key Idea Apply law of conservation of angular momentum.
$$
\mathrm{I}_1 \omega_1=\mathrm{I}_2 \omega_2
$$
In the given case
$$
\begin{aligned}
\mathrm{I}_1 & =\mathrm{MR}^2 \\
\mathrm{I}_2 & =\mathrm{MR}^2+2 \mathrm{mR}^2 \\
\omega_1 & =\omega
\end{aligned}
$$
Then
$$
\omega_2=\frac{I_1}{I_2} \omega=\frac{M}{M+2 m} \omega
$$

Topic: Physics, Rotational Motion, NEET, NEET 2009 (Screening)
A thin circular ring of mass M and radius R rotates about an axis through its centre and perpendicular to its plane, with a constant angular velocity ω. Four small spheres each of mass m (negligible radius) are kept gently to the opposite ends of two mutually perpendicular diameters of the ring. The new angular velocity of the ring will be

Options:
A M4mω
B M+4mMω
C MM-4mω
D MM+4mω

The correct answer is:
MM+4mω

Solution:
According to conservation of angular momentum,
IÏ=constant,

hence we can write

I1Ï1=I2Ï2
or MR2Ï=M+4mR2Ï2
or Ï2=MM+4mÏ

Topic: Physics, Rotational Motion, NEET
A thin conducting ring of radius $R$ is given a charge $+Q$. The electric field at the centre $O$ of the ring due to the charge on the part $A K B$ of the ring is $E$. The electric field at the centre due to the charge on the part $A C D B$ of the ring is

Options:
A $3 \mathrm{E}$ along $\mathrm{KO}$
B E along $O K$
C $E$ along $K O$
D $3 E$ along $O K$

The correct answer is:
E along $O K$

Solution:
Key Idea: It is given that the ring is conducting. As the ring is conducting, so electric field at its centre is zero, ie,
$\overrightarrow{\mathbf{E}}_{\text {total }}=0$
$\begin{aligned}
\overrightarrow{\mathbf{E}}_{A K B}+\overrightarrow{\mathbf{E}}_{A C D B} & =0 \\
\overrightarrow{\mathbf{E}}_{A C D B} & =-\overrightarrow{\mathbf{E}}_{A K B} \\
\overrightarrow{\mathbf{E}}_{A C D B} & =-\overrightarrow{\mathbf{E}} \text { (along } K O \text { ) }
\end{aligned}$
Therefore, the electric field at the centre due to the charge on the part $A C D B$ of the ring is $E$ along $O K$.
lternative :

The fields at $O$ due to $A C$ and $B D$ cancel each other.
The field due to $C D$ is acting in the direction $O K$ and equal in magnitude to $E$ due to $A K B$.

Topic: Physics, Electrostatics, NEET, NEET 2008 (Screening)
A thin convex lens of focal length 25 cm is cut into two pieces 0.5 cm above the principal axis. The top part is placed at (0,0), and an object is placed at (-50 cm, 0). Find the coordinates of the image.

Options:
A (50 cm, -1 cm)
B (-50 cm, 1 cm)
C (-50 cm, -1 cm)
D (50 cm, 1 cm)

The correct answer is:
(50 cm, -1 cm)

Solution:
Correct Option is : (A)
(50 cm, -1 cm)

Topic: Chemistry, Ray Optics, NEET
A thin copper wire of length L increases in length by 1% when heated from temperature T1 to T2. The percentage change in the area when a thin copper plate having dimensions 2LÃL is heated from T1 to T2 is

Options:
A 2%
B 1%
C 4%
D 3%

The correct answer is:
2%

Solution:
Fractional charge in length
Îll=Î±.ÎÎ¸
Fractional change in area
ÎAA=Î².ÎÎ¸
But Î²=2Î±
â´ ÎAA=2Î±ÎÎ¸
ÎAA=2Îll
ÎAA=21%=2%

Topic: Physics, Thermal Properties of Matter, NEET
A thin diamagnetic rod is placed vertically between the poles of an electromagnet. When the current in the electromagnet is switched on, then the diamagnetic rod is pushed up, out of the horizontal magnetic field. Hence, the rod gains gravitational potential energy. The work required to do this comes from

Options:
A The lattice structure of the material of the rod
B The magnetic field
C The current source
D The induced electric field due to the changing magnetic field

The correct answer is:
The current source

Solution:
The work required to lift must be coming from the current source which is causing the current to blow through electromagnet.

Topic: Physics, Magnetic Properties of Matter, NEET, NEET 2018
A thin flat circular disc of radius 4.5 cm is placed gently over the surface of water. If surface tension of water is $0.07 \mathrm{~N} \mathrm{~m}^{-1}$, then the excess force required to take it away from the surface is

Options:
A 198 N
B 1.98 mN
C 99 N
D 19.8 mN

The correct answer is:
19.8 mN

Solution:
$\begin{aligned} & \text { Excess force }=T \times 2 \pi R \\ & =\frac{7}{100} \times 2 \times 3.14 \times \frac{4.5}{100} \\ & =197.82 \times 10^{-4} \\ & =19.8 \times 10^{-3} \mathrm{~N} \\ & =19.8 \mathrm{mN}\end{aligned}$

Topic: Physics, Mechanical Properties of Fluids, NEET, NEET 2024
A thin flexible wire of length L is connected to two adjacent fixed points and carries a current I in the clockwise direction, as shown in the figure. When the system is put in a uniform magnetic field of strength B going into the plane of the paper, the wire takes the shape of a circle. The tension in the wire is

Options:
A IBL
B IBLπ
C IBL2Ï
D IBL4Ï

The correct answer is:
IBL2π

Solution:
L=2ÏR
â´ R=L2Ï
2Tsinâ¡dÎ¸=Fm
For small angles, sinâ¡(dÎ¸)âdÎ¸
â´ 2TdÎ¸=I dLBsinâ¡90Â°
=I 2R .dÎ¸.B
â´ T=IRB=ILB2Ï

Topic: Physics, Magnetic Effects of Current, NEET
A thin glass prism of μ=1.5 is immersed in water of μ=1.33. The ratio of deviation of the ray in water to that in air for the same prism is

Options:
A 1:4
B 1:2
C 1:8
D 1:3

The correct answer is:
1:4

Solution:
In air δ=1.5-1A=A2
In water δ=μgμw-1A=1.51.33-1A=A8
δin water δin air=14
We know that, δin air=aμg-1A and δin water=wμg-1A.
δwδa=μgμw-1μ9-1
δwδa=3243-132-1
δwδa=98-132-1
δwδa=1812=14

Topic: Physics, Ray Optics, NEET
A thin magnetic needle vibrates in the horizontal plane with a period of 4 s. The needle is cut into two halves by a plane normal to the magnetic axis of the needle. Then, the period of vibration of each half needle is approximately

Options:
A 4 s
B 2 s
C 8 s
D 1 s

The correct answer is:
2 s

Solution:
The moment of inertia of needle when it cuts into two halves by a plane normal to magnetic axis of the needle
I′=112m2l22 or I′=I8
Also, M′=M2
∴T′=2πI′M′B
The time of vibrations
=2πI×28×MB
=12T=12×4
=2 s

Topic: Physics, Magnetic Properties of Matter, NEET
A thin, muscular wall (oval depression) in the heart is seen

Options:
A Inter-atrial septum
B Inter-ventricular septum
C Right auriculo-ventricular septum
D Left auriculo-ventricular septum

The correct answer is:
Inter-atrial septum

Solution:
An oval depression called fossa ovalis is present in the inter auricular septum within the right auricle. Through this foramen, the blood from the right auricle is communicated towards the left auricle in the embryo. Over the course of several months, the foramen ovale closes, leaving a shallow depression known as the fossa ovalis in the adult heart.

Topic: Biology, Body Fluids and Circulation, NEET
A thin plastic ring of radius R carries a charge q uniformly spread over the entire ring. The ring is rotating with angular velocity Ï about its axis. The magnetic induction at the centre of the ring is

Options:
A μ0qω2πR
B μ0qωπR
C μ0qω4πR
D zero

The correct answer is:
μ0qω4πR

Solution:
Due to charge q moving on circular path
Bcentre=µ0I2R=µ0qf2R=µ0qω4πR

Topic: Physics, Magnetic Effects of Current, NEET
A thin prism has different medium on its either side. A light ray is incident almost normally on the first face. What is the angle of deviation if all the angles are very small

Options:
A I1-μ1μ2-A1-μμ2
B I1-μ1μ2+A1-μμ2
C I1+μ1μ2-A1-μμ2
D none of these

The correct answer is:
I1-μ1μ2-A1-μμ2

Solution:
From Snell's law at first refraction,
μ1I=μr1
r1=Iμ1μ ...(1)
From Snell's law at second refraction,
μr2=μ2e
μA-r1=μ2e
∴ From (1),
μA-Iμ1μ=μ2e
e=μμ2A-Iμ1μ ...(2)
Overall deviation,
δ=I+e-A
δ=I+μμ2A-Iμ1μ-A
δ=I1-μ1μ2-A1-μμ2

Topic: Physics, Ray Optics, NEET
A thin prism having refracting angle 10o is made of glass of refractive index 1.42. This prism is combined with another thin prism made of glass of refractive index 1.7. This combination produces dispersion without deviation. The refracting angle of second prism should be

Options:
A 4o
B 6o
C 8o
D 10o

The correct answer is:
6o

Solution:
For dispersion without deviation
μ-1A+μ′-1A′=0
μ-1A=μ′-1A′
1.42-1×10o=1.7-1A′
4.2=0.7A′
A′=6o

Topic: Physics, Ray Optics, NEET, NEET 2017
A thin prism P1 with angle 4° and made from glass of refractive index 1.54 is combined with another thin prism P2 made from glass of refractive index 1.72 to produce dispersion without deviation. The angle of the prism P2 is:

Options:
A 5.33°
B 4Â°
C 3Â°
D 2.6Â°

The correct answer is:
3°

Solution:
Deviation
d=m-1A
Given, dnet=0
∴m1-1A1=m2-1A2 or
A2=μ1-1μ2-1A1=1.54-11.72-14°=3°

Topic: Physics, Ray Optics, NEET
A thin ring of radius $R$ metre has charge $q$ coulomb uniformly spread on it. The ring rotates about its axis with a constant frequency of $f$ revolution/s. The value of magnetic induction in $\mathrm{Wb} \mathrm{m}^{-2}$ at the centre of the ring is

Options:
A $\frac{\mu_0 q \mathrm{f}}{2 \pi \mathrm{R}}$
B $\frac{\mu_0 \mathrm{q}}{2 \pi \mathrm{f} R}$
C $\frac{\mu_0 q}{2 f \mathrm{f}}$
D $\frac{\mu_0 q f}{2 R}$

The correct answer is:
$\frac{\mu_0 q f}{2 R}$

Solution:
The magnetic field at the centre of the circle
\(B=\frac{\mu_0 I}{2 R}=\frac{\mu_0}{2 R}\left(\frac{q}{t}\right)=\frac{\mu_0 q f}{2 R}\)

Topic: Physics, Electromagnetic Induction, NEET, NEET 2010 (Screening)
A thin rod of 5 cm length is kept along the axis of a concave mirror of 10 cm focal length such that its image is real and magnified and one end touches the rod. Its magnification will be

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
End A of the rod acts as an object for mirror and Aâ² will be its image so u=2fâl=20â5=15 cm
âµ 1f=1v+1u â1-10=1v-115âv=-30 cm.
Now m=Length ofâimageLength of object=(30-20)5=2

Topic: Physics, Ray Optics, NEET
A thin rod of length $L$ and mass $M$ is bent at its midpoint into two halves so that the angle between them is $90^{\circ}$. The moment of inertia of the bent rod about an axis passing through the bending point and perpendicular to the plane defined by the two halves of the rod is

Options:
A $\frac{M L^2}{24}$
B $\frac{M L^2}{12}$
C $\frac{M L^2}{6}$
D $\frac{\sqrt{2} M L^2}{24}$

The correct answer is:
$\frac{M L^2}{12}$

Solution:
Since rod is bent at the middle, so each part of it will have same length $\left(\frac{L}{2}\right)$ and mass $\left(\frac{M}{2}\right)$ as shown.

Moment of inertia of each part through its one end
$=\frac{1}{3}\left(\frac{M}{2}\right)\left(\frac{L}{2}\right)^2$
Hence, net moment of inertia through its middle point $O$ is
$\begin{aligned}
I & =\frac{1}{3}\left(\frac{M}{2}\right)\left(\frac{L}{2}\right)^2+\frac{1}{3}\left(\frac{M}{2}\right)\left(\frac{L}{2}\right)^2 \\
& =\frac{1}{3}\left[\frac{M L^2}{8}+\frac{M L^2}{8}\right]=\frac{M L^2}{12}
\end{aligned}$

Topic: Physics, Rotational Motion, NEET, NEET 2008 (Screening)
A thin rod of length L is lying along the x-axis with its ends at x = 0 and x = L. Its linear density ( mass/length ) varies with x as kXLn where n can be zero or any positive number. If the position x C M of the centre of mass of the rod is plotted against n, which of the following graphs best approximates the dependence of x C M on n?

Options:
A
B
C
D

The correct answer is:

Solution:
xCÂ·M=â«0LkLnÂ·xnÂ·dxxâ«0LkLnÂ·xnÂ·dx
âxCÂ·M=â«0Lxn+1dxâ«0Lxndx=Ln+2n+2Â·n+1Ln+1
âxCÂ·M=Ln+1n+2
The variation of the mass with x is given by
dxdn=L{n+21ân+1n+22}=Ln+22
If the rod has the same density as at x = 0 i.e., n = 0, therefore uniform, the cenre of mass would have been at L/2. As the density increases with length, the centre of mass shifts towards the right. Therefore it can only be (b).

Topic: Physics, Center of Mass Momentum and Collision, NEET
A thin semicircular conducting ring of the radius R is falling with its plane vertical in a horizontal magnetic induction Bâ. At the position MNQ, the speed of the ring is v and the potential difference developed across the ring is

Options:
A Zero
B BπR22 and M is at a higher potential
C πBRv and Q is at a higher potential
D 2RBv and Q is at a higher potential

The correct answer is:
2RBv and Q is at a higher potential

Solution:
Induced motional emf in MNQ is equivalent to the motional emf in an imaginary wire MQ i. e.,
eMNQ= eMQ= Bvl = Bv(2R) [l=MQ=2R]
VQ-VM=vB2R
Therefore, the potential difference developed across the ring is 2RBv with Q at a higher potential.

Topic: Physics, Electromagnetic Induction, NEET
A thin semicircular conducting ring PQR of radius r is falling with its plane vertical in a horizontal magnetic field B, as shown in figure. The potential difference developed across the ring when its speed is v, is:

Options:
A Zero
B Bvπr2/2 and P is at higher potential
C πrBV and R is at higher potential
D 2rBv and R is at higher potential

The correct answer is:
2rBv and R is at higher potential

Solution:
Here we have to calculate the emf of the conducting ring when it is falling

So, we have to calculate with the following formulae

emf=VBleq=VB2R

where R is at higher potential and P is at lower potential.

Topic: Physics, Electromagnetic Induction, NEET, NEET 2014
A thin slice is cut out of a glass cylinder along a plane parallel to its axis. The slice is placed on a flat glass plate with the curved surface downwards. Monochromatic light is incident normally from the top. The observed interference fringes from this combination do not follow one of the following statements.

Options:
A The fringes are straight and parallel to the length of the piece.
B The line of contact of the cylindrical glass piece and the glass plate appears dark.
C The fringe spacing increases as we go outwards.
D The fringes are formed due to the interferance of light rays reflected from the curved surface of the cylindrical piece and the top surface of the glass plate.

The correct answer is:
The fringe spacing increases as we go outwards.

Solution:
Here a thin film is formed between curve surface of cylinder and flat surface of glass which results in interference pattern. For a fringe, locus and all point having equal thickness will be straight line parallel to length of the piece. At line of contact, for first reflection, as ray passes from glass to air, no phase changes and as it goes from air to glass, it will change phase. So, both rays will be out of phase resulting in dark fringe.
as t n + 1 - t n = Î» 2 Î¼ = costant
w = t n + 1 - t n Î¸
As we go outward, Î¸ increases
w decreases. So, 3rd option is wrong.

Topic: Physics, Wave Optics, NEET
A thin spherical conducting shell of radius R is given a charge q. Another charge Q is placed at the centre of the shell. The electrostatic potential at the point P at a distance R2 from the centre of the shell is

Options:
A 2Q4πε0R-2q4πε0R
B 2Q4πε0R+q4πε0R
C (q⁡+Q⁡)4πε02R
D 2Q4πε0R

The correct answer is:
2Q4πε0R+q4πε0R

Solution:
Potential due to point charge
V1=Q4πε0R2
Potential due to shell at inside point
V2=VR2=Vsurface
V2=q4πε0R
Vtotal=2Q4πε0R+q4πε0R

Topic: Physics, Electrostatics, NEET
A thin spherical shell is charged by some source. The potential difference between the two points $C$ and $P$ (in V ) shown in the figure is:
(Take $\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9$ SI units)

Options:
A $1 \times 10^5$
B $0.5 \times 10^5$
C Zero
D $3 \times 10^5$

The correct answer is:
Zero

Solution:
For uniformly charged spherical shell,
$\begin{aligned}
& V=\frac{k q}{R} \quad(\text { For } r \leq R) \\
& \therefore \quad V_C=V_P \\
& \\
& V_C-V_P=\text { Zero }
\end{aligned}$

Topic: Physics, Electrostatics, NEET, NEET 2024
A thin uniform circular disc of mass m and radius R is rotating in a horizontal plane about an axis passing through its centre and perpendicular to the plane with an angular velocity ω. Another disc of same dimensions but of mass 14m is placed gently on the first disc co-axially. The angular velocity of the system is

Options:
A 2Ï
B 45ω
C 34ω
D 13ω

The correct answer is:
45ω

Solution:
Using conservation of angular momentum
12mR2ω={12mR2+12m4R2}ω′
12mR2ω=58mR2ω′
ω′=45 ω

Topic: Physics, Rotational Motion, NEET
A thin uniform rod of length l and mass m is swinging freely, about a horizontal axis passing through its end. Its maximum angular speed is Ï. The maximum height, to which its centre of mass rises, is

Options:
A 13Â l2Ï2g
B 16Â lÏg
C 12Â l2Ï2g
D 16 l2Ï2g

The correct answer is:
16 l2Ï2g

Solution:
TEi=TEr
12IÏ2=mgh
12Ã13ml2Ï2=mgh
â h=16 l2Ï2g

Topic: Physics, Rotational Motion, NEET
A thin walled glass sphere of radius 2.5 cm is filled with water. An object O is placed at 7.5 cm from the surface of the sphere. Neglecting the effect of glass wall, at what distance the image I of the object, measured from the centre of sphere is formed? (refractive index of water = 1.333)

Options:
A 20Â cm
B 15Â cm
C 10 cm
D 7.5Â cmÂ

The correct answer is:
10 cm

Solution:
For refraction at first surface,
u=-7.5 cm, R1=2.5 cm
Î¼1=1, Î¼2=43
â´ Î¼2vâ²-Î¼1u=Î¼2-Î¼1R1
â´ 4/3vâ²-1-7.5=4/3-12.5
â 43vâ²=17.5-17.5=0
â vâ²=â
It means the ray is parallel within the sphere.
For refraction at second surface,
u=-â, Î¼1=43
Î¼2=1, R2=-2.5 cm
â´ Î¼2vâ²-Î¼1u=Î¼2-Î¼1R2 gives
â 1vâ²-4/3-â=1-4/3-2.5
â v=+ 7.5 cm
i.e. Image is formed at distance 7.5 cm to the right of P2 . Hence, distance of final image P from centre C=7.5+2.5=10 cm.

Topic: Physics, Ray Optics, NEET
A tightly wound 100 turns coil of radius 10 cm carries a current of 7 A . The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as $4 \pi \times 10^{-7}$ SI units):

Options:
A 4.4 T
B 4.4 mT
C 44 T
D 44 mT

The correct answer is:
4.4 mT

Solution:
The magnitude of magnetic field due to circular coil of $N$ turns is given by
$\begin{aligned}
& B_C=\frac{\mu_0 i N}{2 R} \\
& =\frac{4 \pi \times 10^{-7} \times 7 \times 100}{2 \times 0.1} \\
& =4.4 \times 10^{-3} \mathrm{~T} \\
& =4.4 \mathrm{mT}
\end{aligned}$

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2024
A tire of radius R rolls on a flat surface with angular velocity ω and velocity v as shown in the diagram. If v>ωR, in which direction does friction from the tire act on the road?

Options:
A Towards downwards
B Towards upwards
C Towards the left
D Towards the right

The correct answer is:
Towards the left

Solution:
The solution of this question can simply be understood by taking a situation of car driver, where at higher speed he applies sudden brakes, i.e. v >Ïr and the tire rubs the road and we have rolling with forwards slipping i.e. friction acting in opposite direction of v. (towards left in the above situation.) Friction force on the tire is towards right and on the road is towards left.

Topic: Physics, Rotational Motion, NEET
A tissue is characterized by the presence of thin walls and isodiametric cells that are either closely packed or have intercellular spaces. This tissue is found in

Options:
A Shoot apex
B Wood fibres
C Pith of monocot root
D Bast fibres

The correct answer is:
Pith of monocot root

Solution:
Parenchyma forms the major component within organs. The cells of the parenchyma are generally isodiametric. They may be spherical, oval, round, polygonal or elongated in shape. Their walls are thin and made up of cellulose. They may either be closely packed or have small intercellular spaces. The parenchyma performs various functions including photosynthesis, storage, secretion.

Topic: Biology, Anatomy of Flowering Plants, NEET
A total of 512 seeds are collected from the cross between AaBb × AaBb. Find the number of plants produced with the genotype AAbb and Aabb.

Options:
A 35
B 350
C 96
D 285

The correct answer is:
96

Solution:
For a dihybrid cross, the genotypic ratio will be represented as AABB:AaBB:AABb:AaBb:AAbb:Aabb:aaBB:aaBb:aabb :: 1:2:2:4:1:2:1:2:1.
Therefore, the ratio of plants with genotype AAbb and Aabb will be 316

Hence, the number of plants with genotype AAbb and Aabb will be 316 x 512 = 96

Topic: Biology, Principles of Inheritance and Variation, NEET
A toy car with charge q moves on a frictionless horizontal plane surface under the influence of a uniform electric field E→. Due to the force qE→, its velocity increases from 0 to 6 m s-1 in one second duration. At that instant the direction of the field is reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between 0 to 3 seconds are respectively

Options:
A 1 m s-1, 3.5 m s-1
B 1 m s-1, 3 m s-1
C 2 m s-1, 4 m s-1
D 1.5 m s-1, 3 m s-1

The correct answer is:
1 m s-1, 3 m s-1

Solution:
0<t≤1 s velocity increases from 0 to 6 m/s
1 s≤t≤2 s : velocity decreases from 6 m/s to 0 m/s
But car continues to move forward and then it reverses its direction at t=2 s and after 2 s accelerates backward
Distance undefined
| 1 2 ×6×2 |+| 1 2 ×( −6 )×1 |
=6+3
=9 m
Displacement = Total area under v−t graph
=6−3=3 m
∴ Average velocity =DisplacementTime=33=1m/s
∴ Average speed =Total distanceTime=93=3m/s

Topic: Physics, Motion In One Dimension, NEET, NEET 2018
A train is moving towards East and a car is along North, both are moving with same speed. The observed direction of car to the passenger in the train is

Options:
A East-North
B South-East
C West-North
D North-South

The correct answer is:
West-North

Solution:
According to the situation,
The velocity of car with respect to the train,
vc t=vc-vt
=vj^-vi^=v-i^+j^
Thus,
The velocity of the car w.r.t. the train is in the West-North direction.

Topic: Physics, Motion In One Dimension, NEET
A train moves towards a stationary observer with speed 34 m s-1. The train sounds a whistle and its frequency registered by the observer is f1. If the train's speed is reduced to 17 m s-1, the frequency registered is f2 . If the speed of sound is 340 m s-1 then the ratio f1f2 is

Options:
A 1819
B 12
C 2
D 1918

The correct answer is:
1918

Solution:
f1=fvv-vs
f1=f340340-34=f340306
And f2=f340340-17=f340323
∴f1f2=323306=1918

Topic: Physics, Waves and Sound, NEET