datavorous/entrance-exam-dataset huggingface dataset:
Two parents having two loci incompletely linked and heterozygous in nature are crossed. What would be the distribution of phenotypic features in the F2 generation?

Options:
A Only parental phenotype
B Only recombinant phenotype
C High parental and low recombinant phenotype
D Low parental and high recombinant phenotype

The correct answer is:
High parental and low recombinant phenotype

Solution:
Incomplete linkage: In majority of cases, the homologous chromosomes undergo breakage and reunion during gametogenesis. During reunion the broken pieces of the chromatids are exchanged, producing some nonparental or new combinations. Therefore, the linkage is rendered incomplete. The phenomenon of interchange of chromosome segments between two homologous chromosomes is called crossing over. Incomplete linkage is very common and has been studied in almost all the organisms.
If two genes present at different gene loci are linked than during segregation more parental type is produced and recombinant types are less in number.

Topic: Biology, Principles of Inheritance and Variation, NEET
Two particles A and B are moving, as shown in the figure.

Their total angular momentum about the point O is

Options:
A 9.8â kgâm2âsâ1
B Zero
C 52.7 kgâm2âsâ1
D 37.9 kgâm2âsâ1

The correct answer is:
9.8â kgâm2âsâ1

Solution:
Total angular momentum about O is given as,
L â = L â 1 + L â 2 =â m 1 v 1 r 1 + m 2 v 2 r 2 (anti clockwise)
= -6.5Ã2.2Ã1.5+3.1Ã3.6Ã2.8 (anti clockwise)
=â21.45+31.248=9.8 kgâm2âsâ1 (anti clockwise)

Topic: Physics, Rotational Motion, NEET
Two particles A and B are moving in uniform circular motion in concentric circles of radii rA and rB with speed vA and vB, respectively. Their time period of rotation is the same. The ratio of angular speed of A to that of B will be,

Options:
A rA:rB
B vA:vB
C rB:rA
D 1:1

The correct answer is:
1:1

Solution:
TA=TBωAωB=TBTA
Same time period ⇒ same angular speed.

Topic: Physics, Motion In Two Dimensions, NEET, NEET 2019 (All India)
Two particles A and B execute simple harmonic motion of period T and 5T/4. They start from mean position. The phase difference between them when the particle A completes one oscillation will be

Options:
A π/2
B Zero
C 2π/5
D π/4

The correct answer is:
2π/5

Solution:
Equation of motion of the particles are
X 1 = A 1 sin 2π T 1 t
and X 2 = A 2 sin 2π T 2 t
∴ Phase difference Δϕ= 2π T 1 − 2π T 2 t
= 2π T − 2π 5T/4 t
at t=T
Δϕ= 2π− 4×2π 5 T T = 2π 5

Topic: Physics, Oscillations, NEET
Two particles A and B move with constant velocities v→1 and v→2. At the initial moment their position vectors are r→1 and r→2 respectively. The condition for particles A and B for their collision is:

Options:
A râ1⋅vâ1=râ2⋅vâ2
B râ1Ãvâ1=râ2Ãvâ2
C râ1-râ2=vâ1-vâ2
D râ1-râ2râ1-râ2=vâ2-vâ1vâ2-vâ1

The correct answer is:
r→1-r→2r→1-r→2=v→2-v→1v→2-v→1

Solution:
For two particles to collide, the direction of the relative velocity of one with respect to other should be directed towards the relative position of the other particle
i.e. \(\frac{\vec{r}_1-\vec{r}_2}{\left|\vec{r}_1-\vec{r}_2\right|} \rightarrow\) direction of relative position of 1 w.r.t. 2.
& \(\frac{\vec{v}_2-\vec{v}_1}{\left|\vec{v}_2-\vec{v}_1\right|} \rightarrow\) direction of velocity of 2 w.r.t. 1
So for collision of A & B
\(\frac{\vec{r}_1-\vec{r}_2}{\left|\vec{r}_1-\vec{r}_2\right|}=\frac{\vec{v}_2-\vec{v}_1}{\left|\vec{v}_2-\vec{v}_1\right|}\)

Topic: Physics, Motion In One Dimension, NEET, NEET 2015 (Phase 2)
Two particles A and B of de Broglie wavelengths λAand λBcombine to form a particle C. The process conserves momentum. Find the de Broglie wavelength of the particle C. (The motion is one dimensional).

Options:
A $\frac{2{{\lambda }_{A}}{{\lambda }_{B}}}{\left| {{\lambda }_{A}}-{{\lambda }_{B}} \right|}$
B $\frac{{{\lambda }_{A}}{{\lambda }_{B}}}{\left| {{\lambda }_{A}}-{{\lambda }_{B}} \right|}$
C $\frac{2{{\lambda }_{A}}{{\lambda }_{B}}}{\left| 2{{\lambda }_{A}}-{{\lambda }_{B}} \right|}$
D $\frac{{{\lambda }_{A}}{{\lambda }_{B}}}{\left| 2{{\lambda }_{A}}-{{\lambda }_{B}} \right|}$

The correct answer is:
$\frac{{{\lambda }_{A}}{{\lambda }_{B}}}{\left| {{\lambda }_{A}}-{{\lambda }_{B}} \right|}$

Solution:
Correct Option is : (B)
$\frac{{{\lambda }_{A}}{{\lambda }_{B}}}{\left| {{\lambda }_{A}}-{{\lambda }_{B}} \right|}$

Topic: Chemistry, Dual Nature of Matter, NEET
Two particles A and B of masses mA and mB respectively and having the same charge are moving in a plane. A uniform magnetic field exists perpendicular to this plane. The speeds of the particles are vAand vB respectively and the trajectories are as shown in the figure. Then

Options:
A mA vA < mB vB
B mA vA > mB vB
C mA < mB and vA < vB
D mA = mB and vA = vB

The correct answer is:
mA vA > mB vB

Solution:
Radius of the circle = mâ¡ vâ¡ Bâ¡ qâ¡
or radius â mv if B and q are same.
(Radius)A > (Radius)B
â´ mA vA > mB vB

Topic: Physics, Magnetic Effects of Current, NEET
Two particles A and B separated by a distance 2R are moving counter-clockwise along the same circular path of radius R each with uniform speed v. At the time t=0 , A is given a tangential acceleration of magnitude a=32v225ÏR in the same direction of the initial velocity. Then which of the following is correct

Options:
A The time-lapse for the two bodies to collide is 6πR5V
B The angle covered by A is 9Ï4
C Angular velocity of A is 11V5R
D Radial acceleration of A is 289V25R

The correct answer is:
The angle covered by A is 9π4

Solution:
As when they
Collide 1232 v225 πRt2=πR
t=5πR4V
Angle covered by A
θ=π+vtR
θ=9π4

Topic: Physics, Laws of Motion, NEET
Two particles A, B are moving on two concentric circles of radii R1 and R2 with equal angular speed Ï. At t=0, their positions and direction of motion are shown in the figure:

The relative velocity VAâ-VBâ at t=Ï2Ï is given by:

Options:
A -ωR1+R2i^
B ωR1+R2i^
C ωR1-R2i^
D ωR2-R1i^

The correct answer is:
ωR2-R1i^

Solution:
t=π2ω
V→A=ωR1-i^
V→B=ωR2-i^
VA→-VB→=ωR2-R1i^

Topic: Physics, Laws of Motion, NEET
Two particles are executing simple harmonic motion of the same amplitude A and frequency ω along the x-axis. Their mean positions are separated by distance x0(x0>A). If the maximum separation between them is (x0+A), the phase difference between their motions is

Options:
A π3
B π4
C π6
D π2

The correct answer is:
π3

Solution:
When the distance between the particles is maximum, their relative velocity will be zero. This means the particles are at the same distance from from their respective mean positions. Since the maximum distance is x0+A, it means the particles are at half the amplitude position. Also, if one of the particles is left of its mean position, then the other one would be to the right of the mean position.

x1=Asin⁡(ωt+ϕ1)

At t=0, let x1=A2, then

ωt+ϕ1=π6 and the particle is moving along positive x-axis

x2=Asin⁡(ωt+ϕ2)

At t=0, x2=-A2

ωt+ϕ2=2π-π6 and this particle is also moving towards positive along positive x-axis

⇒ϕ2-ϕ1=2π-π3

But a phase difference of 2π-θ is same as a phase difference of θ.

So, ϕ2-ϕ1=π3

Topic: Physics, Oscillations, NEET
Two particles are projected from a point at the same instant with velocities whose horizontal components and vertical components are u1,v1 and u2,v2 respectively. The time interval between their passing through the other common point of their path (other than origin) is

Options:
A 2gv1u1-v2u2u1+u2
B 2gv12+v22u1+u2
C 2gu12+u22v1+v2
D 2gv1u2-v2u1u1+u2

The correct answer is:
2gv1u2-v2u1u1+u2

Solution:
The time for particle 1 to pass through common point of their path
t1=xu1
The time for particle 2 to pass through common point of their path
t2=xu2
Difference of time t=t1-t2
Time t=xu1-xu2=x1u1-1u2
y=xtanθ-gx22ux2
y=xv1u1-gx22u12=xv2u2-gx22u22
v1u1-v2u2=xg21u12-1u22
v1u1-v2u2=xg21u1+1u21u1-1u2
t=x1u1-1u2=2gv1u2-v2u1u1+u2

Topic: Physics, Motion In Two Dimensions, NEET
Two particles are projected simultaneously in the same vertical plane, from the same point, but with different speeds and at different angles with horizontal. The path followed by one, as seen by the other is

Options:
A a vertical line
B a parabola
C a hyperbola
D a straight line making a constant angle with the horizontal

The correct answer is:
a straight line making a constant angle with the horizontal

Solution:
Let u → 1 and u → 2 be the initial velocities of the two particles and θ 1 and θ 2 be their angles of projection with the horizontal
The velocities of the two particles after time t are,
v → 1 = u⁡ 1 cos θ 1 i ^ + u⁡ 1 sin θ 1 - g t j ^ and
v → 1 = u⁡ 1 cos θ 1 i ^ + u⁡ 2 sin θ 2 - g t j ^
Their relative velocity is v → 1 2 = v → 1 - v → 2
= u⁡ 1 cos θ 1 - u⁡ 2 cos θ 1 i ^ + u⁡ 1 sin θ 1 - u⁡ 2 sin θ 2 j ^
Which is a constant. So the path followed by one, as seen by the other is a straight line, making a constant angle with the horizontal

Topic: Physics, Motion In Two Dimensions, NEET
Two particles are projected simultaneously with same speed V0 in same vertical plane at angles 30o and 60o with the horizontal. The time at which their velocities becomes parallel is

Options:
A V02g
B V03+12g
C 3 V02g
D V03-1g

The correct answer is:
V03+12g

Solution:
V1→=V0cos⁡30oi^+V0sin⁡30o-gt j^
V2→=V0cos⁡60oi^+V0sin⁡60o-gt j^
∵ V1→|| V2→
⇒ V0cos⁡30oV0cos⁡60o=V0sin⁡30o-gtV0sin⁡60o-gt
⇒ t=V03+12g

Topic: Physics, Motion In Two Dimensions, NEET
Two particles, each of mass m and charge q, are attached to the two ends of a light, rigid rod of length 2R. The rod is rotated at a constant angular speed, about a perpendicular axis passing through its centre. The ratio of the magnitudes of the magnetic moment of the system and its angular momentum about the centre of the rod is

Options:
A q2m
B qm
C 2qm
D qπm

The correct answer is:
q2m

Solution:
Current, i = (frequency) (charge)
=ω2π2q=qωπ
Magnetic moment,
M=iA=qωππR2=qωR2
Angular momentum,
L=2Iω=2mR2ω
∴ ML=qωR22mR2ω=q2m

Topic: Physics, Magnetic Effects of Current, NEET
Two particles of mass 5 kg and 10 kg respectively are attached to the two ends of a rigid rod of length 1 m with negligible mass. The centre of mass of the system from the 5 kg particle is nearly at a distance of:

Options:
A 50 cm
B 67 cm
C 80 cm
D 33 cm

The correct answer is:
67 cm

Solution:
mr=constant⇒r∝1m

r1r2=m2m1=105=21

r1=23r=23×100 m=67 cm

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2020 (Phase 1)
Two particles of mass m and 2m have their position vectors as a function of time as r1t=t i^-t3 j^+2t2 k^ and r1t=t i^-t3 j^-t2 k^ respectively (where t is the time). Which one of the following graphs represents the path of the centre of mass?

Options:
A
B
C
D

The correct answer is:

Solution:
Path of the centre of mass in a two particle system,
rt=m1r1t+m2r2tm1+m2
or rt=mt i^-t3 j^+2t2 k^+2mt i^-t3 j^-t2 k^m+2m
⇒ rt=3t i^-3t3 j^3=t i^-t3 j^

Topic: Physics, Center of Mass Momentum and Collision, NEET
Two particles of masses m1, m2 move with initial velocities u1 and u2. On collision, one of the particles get excited to higher level, after absorbing energy ε. If final velocities of particles be v1 and v2 then we must have:

Options:
A m12u1+m22u2-Îµ=m12v1+m22v2
B 12m1u12+12m2u22=12m1v12+12m2v22-Îµ
C 12m1u12+12m2u22-Îµ=12m1v12+12m2v22
D 1 2 m 2 u 1 2 + 1 2 m 1 u 1 2 + ε = 1 2 m 2 v 1 2 + 1 2 m 1 v 1 2

The correct answer is:
12m1u12+12m2u22-ε=12m1v12+12m2v22

Solution:
From conservation of energy initial total energy
Ei=12m1u12+12m2u22
After collision total final energy
Ef=12m1v12+12m2v22+ε

∴12m1u12+12m2u22=12m1v12+12m2v22+ε
12m1u12+12m2u22-ε=12m1v12+12m2v22

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2015 (Phase 1)
Two particles of masses ma and mb and same charge are projected in a perpendicular magnetic field. They travel along circular paths of radius ra and rb such that ra>rb. Then which of the following is true?

Options:
A ma va>mb vb
B ma>mb and va> vb
C ma=mb and va= vb
D mb vb>ma va

The correct answer is:
ma va>mb vb

Solution:
Radius of circular path ra=ma vaqB
And rb=mb vbqB
According to question
ra>rb
ma vaqB> mb vbqB
or mava>mbvb

Topic: Physics, Magnetic Effects of Current, NEET
Two particles of masses m and 9m are separated by a distance r. At a point on the line joining them, the gravitational field is zero. The gravitation potential at that point is ( G = Universal constant of gravitation)

Options:
A -4Gmr
B -8Gmr
C -16Gmr
D -32Gmr

The correct answer is:
-16Gmr

Solution:
For the system described above
F=0, Gm r 1 2 = Gâ(9m) r 2 2 â r 2 r 1 =3
r 2 = 3 4 rââandââ r 1 = 1 4 r V=â Gm r 1 â Gâ(9m) r 2
V=â Gm r 1 â Gâ(9m) r 2
=â 4Gm r â12 Gm r =â 16Gm r

Topic: Physics, Gravitation, NEET
Two particles of masses m and 2m are connected by a string of length L and placed at rest over a smooth horizontal surface. The particles are then given velocities as indicated in the figure shown. The tension developed in the string will be

Options:
A mv22L
B 3mv24L
C 4mv23L
D 8mv23L

The correct answer is:
8mv23L

Solution:
Vc=2mv-mv(2m+m)=v3
Vrel=v-v3=2v3
r1=mL3m=L3
∴ T=2m×2v32L3=8mv23L

Topic: Physics, Laws of Motion, NEET
Two particles of masses 'm' and '2m' have equal kinetic energies. Their de Broglie wavelengths are in the ratio of

Options:
A 1:1
B 1:2
C 1:2
D 2:1

The correct answer is:
2:1

Solution:
λ=h2Em
λmλ2m=h2Em2E(2m)=21

Topic: Chemistry, Structure of Atom, NEET
Two particles of same mass m go around a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is ;

Options:
A V=12R1Gm
B V=Gm2R
C V=12GmR
D V=4GmR

The correct answer is:
V=12GmR

Solution:
the gravitational force provide the necessary centripetal force for circular motion. so
fg=feâGm22R2=mv2Râ v=12GmR

Topic: Physics, Gravitation, NEET
Two particles P and Q are moving on circle. At a certain instant of time both the particles are diametrically opposite and P has tangential acceleration 8 m/s2 and centripetal acceleration 5 m/s2 whereas Q has only centripetal acceleration of 1 m/s2. At that instant acceleration (in m/s2 ) of P with respect to Q is

Options:
A 14
B 80
C 10
D 12

The correct answer is:
10

Solution:
According to the question, we can draw the following diagram.
From the diagram,
aâP=5i^+8j^
And aâQ=-i^
Now, aârel=aâP-aâQ
aârel=5i^+j^--i^
aârel=6i^+8j^
aârel=62+82=36+64
=100
=10 m/s2

Topic: Physics, Laws of Motion, NEET
Two particles whose masses are 10 kg and 30 kg and their position vectors are i^+j^+k^ and -i^-j^-k^ respectively would have the centre of mass at

Options:
A -i^+j^+k^2
B i^+j^+k^2
C -i^+j^+k^4
D i^+j^+k^4

The correct answer is:
-i^+j^+k^2

Solution:
rcm→=m1r1→+m2r2→m1m2
⇒ rcm→=10i^+j^+k^+30-i^-j^-k^40
⇒ rcm→=-20i^+j^+k^40
⇒ rcm→=-i^+j^+k^2

Topic: Physics, Center of Mass Momentum and Collision, NEET
Two particles X and Y having equal charges, are accelerated through the same potential difference. They enter a region of a uniform magnetic field and describe a circular path of radii r1 and r2 respectively. The ratio of the mass of X to that of Y is

Options:
A r1r2
B r1r2
C r2r12
D r1r22

The correct answer is:
r1r22

Solution:
The force acting on the particle inside the magnetic field is FB=qvBsinâ¡Î¸

Î¸=90Â°, FB=qvB
This provides the necessary centripetal force Fc=mv2r
â´ mv2r=qvBâr=mvqB = 2mqV qB
r 1 r 2 = m x m y
mxmy=r1r22

Topic: Physics, Magnetic Effects of Current, NEET
Two particles X and Y having equal charges, are accelerated through the same potential difference. They enter a region of a uniform magnetic field and describe a circular path of radii r1 and r2 respectively. The ratio of the mass of X to that of Y is

Options:
A r1r2
B r1r2
C r2r12
D r1r22

The correct answer is:
r1r22

Solution:
The force acting on the particle inside the magnetic field is FB=qvBsin⁡θ

θ=90°, FB=qvB
This provides the necessary centripetal force Fc=mv2r
∴ mv2r=qvB⇒r=mvqB = 2mqV qB
r 1 r 2 = m x m y
mxmy=r1r22

Topic: Physics, Magnetic Effects of Current, NEET
Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is

Options:
A 9
B 10
C 8
D 11

The correct answer is:
11

Solution:
Time period of a pendulum,

T∝l

⇒T1T2=l1l2

If n is the number of vibrations after which the pendulums are again in phase, the number of vibration of the longer pendulum will be n-1. Therefore,

T1×n-1T2×n=1 ⇒121n-1100n=1

⇒11n-11=10n ⇒n=11

Topic: Physics, Oscillations, NEET, NEET 2022 (Phase 1)
Two perfectly elastic objects A and B of identical mass are moving with velocities 15 m s-1 and 10 m s-1 respectively, collide along the direction of line joining them. Their velocities after the collision are respectively

Options:
A 10 m s-1, 15 m s-1
B 20 m s-1, 5 m s-1
C 0 m s-1, 25 m s-1
D 5 m s-1, 20 m s-1

The correct answer is:
10 m s-1, 15 m s-1

Solution:
Applying momentum conservation:

15m+10m=mv1+mv2
25=v1+v2
And v2-v1u1-u2=1
⇒ v2-v115-10=1
⇒ v2-v1=5
Solving Eqs. (i) and (ii), we have
v2=15 m s-1, v1=10 m s-1

Topic: Physics, Center of Mass Momentum and Collision, NEET
Two periodic waves of intensities $I_1$ and $I_2$ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is

Options:
A $I_1+I_2$
B $\left(\sqrt{I_1}+\sqrt{I_2}\right)^2$
C $\left(\sqrt{I_1}-\sqrt{I_2}\right)^2$
D $2\left(I_1+I_2\right)$

The correct answer is:
$2\left(I_1+I_2\right)$

Solution:
Resultant intensity of two periodic waves is given by
$I=I_1+I_2+2 \sqrt{I_1 I_2} \cos \delta$
where $\delta$ is the phase difference between the waves.
For maximum intensity, $\delta=2 n \pi ; n=0,1,2 \ldots$ etc.
Therefore, for zero order maxima, $\cos \delta=1$
$I_{\text {max }}=I_1+I_2+2 \sqrt{I_1 I_2}=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2$
For minimum intensity, $\delta=(2 n-1) \pi$;
$n=1,2 \ldots \text { etc }$
Therefore, for Ist order minima, $\cos \delta=-1$
$\begin{aligned}
I_{\min } & =I_1+I_2-2 \sqrt{I_1 I_2} \\
& =\left(\sqrt{I_1}-\sqrt{I_2}\right)^2
\end{aligned}$
Therefore,
$\begin{aligned}
I_{\max }+I_{\min } & =\left(\sqrt{I_1}+\sqrt{I_2}\right)^2+\left(\sqrt{I_1}-\sqrt{I_2}\right)^2 \\
& =2\left(I_1+I_2\right)
\end{aligned}$

Topic: Physics, Waves and Sound, NEET, NEET 2008 (Screening)
Two periodic waves of intensities I1 and I2 pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is

Options:
A I1+I2
B I1+I2
C I1-I22
D 2I1+I2

The correct answer is:
2I1+I2

Solution:
Imaxâ¡ =I1+I2+2I1I2
Iminâ¡ =I1+I2-2I1I2
=I1-I22
Therefore, I max + I min =( I 1 + I 2 +2 I 1 . I 2 )+( I 1 + I 2 −2 I 1 . I 2 )
=2I1+I2

Topic: Physics, Waves and Sound, NEET
Two persons A and B start from the same location and walked around a square in opposite directions with constant speeds. The square has a side 60 m. Speeds of A and B are 4 m s-1 and 2 m s-1 respectively. When will they meet for the first time?

Options:
A 30 s

B 40 s

C 10 s

D 20 s

The correct answer is:
40 s

Solution:
Given, speed of the persons A and B around the square = 4 m s-1 and 2 m s-1, respectively. Let the time t, after which they meet each other. So, total distance travelled by the person A and B in time t is equal to the circumference of the square i.e.,
4t+2t=4×60
6t=240
t=40 s

Topic: Physics, Motion In One Dimension, NEET
Two photons having

Options:
A equal wavelengths have equal linear momenta
B equal energies have equal linear momenta
C equal frequencies have equal linear momenta
D equal linear momenta have equal wavelengths

The correct answer is:
equal linear momenta have equal wavelengths

Solution:
The magnitude of momentum and de-Broglie wavelength are related as
λ=hp
Therefore if linear momenta are same then wavelengths are equal but if wavelengths are same then only the magnitudes of momenta are same while direction might be different.

Topic: Physics, Dual Nature of Matter, NEET
Two photons of same frequencies moving in same medium have

Options:
A Same linear momentum and same wavelengths
B Same linear momentum and same speeds
C Same energies and same linear momentum
D None of the above

The correct answer is:
None of the above

Solution:
Momentum is a vector quantity. Its value depends upon magnitude as well as direction. Hence, none of the above is correct.

Topic: Physics, Dual Nature of Matter, NEET
Two pith balls carrying equal charges are suspended from a common point by strings of equal length, the equilibrium separation between them is $r$. Now the strings are rigidly clamped at half the height. The equilibrium separation between the balls now become.

Options:
A $\left(\frac{1}{\sqrt{2}}\right)^2$
B $\left(\frac{r}{\sqrt[3]{2}}\right)$
C $\left(\frac{2 r}{\sqrt{3}}\right)$
D $\left(\frac{2 r}{3}\right)$

The correct answer is:
$\left(\frac{r}{\sqrt[3]{2}}\right)$

Solution:
Given two pith balls carrying equal charges are suspended from a common point by strings of equal length. The equilibrium separation between them is \(r\). Now, we have to find the equilibrium separation when the strings are clamped at half the height.
We can see from the figure that under equilibrium, \(\tan \theta=\frac{\mathrm{F}}{\mathrm{mg}}\)...(i) and
also, \(\tan \theta=\frac{\mathrm{r} / 2}{\mathrm{y}} \ldots\) (ii)
Equating (i) and (ii), we get
\(\frac{\mathrm{F}}{\mathrm{mg}}=\frac{\mathrm{r}}{2 \mathrm{y}}\)
also, Force, \(\mathrm{F}=\frac{\mathrm{kq}^2}{\mathrm{r}^2}\)
where \(\mathrm{k}=\) constant \(=9 \times 10^9\)
So, we get \(\mathrm{r}^3=\left(\frac{\mathrm{kq}^2}{\mathrm{mg}}\right) \mathrm{y} \ldots\) (iii)
Let \(r^{\prime}\) be the equilibrium separation when \(y \rightarrow \frac{y}{2}\) So, equation (iii) then reduces to:
\(\mathrm{r}^{\prime 3}=\left(\frac{\mathrm{kq}^2}{\mathrm{mg}}\right) \frac{\mathrm{y}}{2} \ldots \text { (iv) }\)
Dividing (iv) by (iii) we get,
So, \(\frac{\mathrm{r}^{\prime 3}}{\mathrm{r}^3}=\frac{1}{2}\)
\(r^{\prime}=r\left(\frac{1}{2}\right)^{1 / 3}\)

Topic: Physics, Electrostatics, NEET, NEET 2013 (All India)
Two plane mirrors are inclined at $70^{\circ}$. A ray incident on one mirror at angle, $\theta$ after reflection falls on second mirror and is reflected from there parallel to first mirror. The value of $\theta$ is:

Options:
A $45^{\circ}$
B $30^{\circ}$
C $55^{\circ}$
D $50^{\circ}$

The correct answer is:
$50^{\circ}$

Solution:
The actual situation is shown in the given degree
$$
\begin{aligned}
\theta+40^{\circ} & =90^{\circ} \\
\Rightarrow \quad \theta & =90-40=50^{\circ}
\end{aligned}
$$

Topic: Physics, Ray Optics, NEET, NEET 2013 (Karnataka)
Two plane mirrors are kept parallel at 20 cm from each other. A point object O is placed exactly in between them. Calculate distance between second images formed by two mirrors.

Options:
A 80 cm
B 60Â cm
C 40Â cm
D 10Â cm

The correct answer is:
80 cm

Solution:
From the diagram it is clear that Distance between I2& I 2 â²
=(30 + 20 + 30) cm=80 cm

Topic: Physics, Ray Optics, NEET
Two plane mirrors are placed at some angle. There are five images formed, when an object is placed symmetrically between them. Find the angle between the mirrors.

Options:
A 60°
B 45°
C 30°
D 90°

The correct answer is:
60°

Solution:
Number of images =360°θ-1
5=360°θ-1
θ=60°

Topic: Physics, Ray Optics, NEET
Two plane mirrors kept parallel at 20 cm from each other. A point object is placed exactly in between them. Calculate distance between second images formed by two mirrors.

Options:
A 80 cm
B 60 cm
C 40 cm
D 10 cm

The correct answer is:
80 cm

Solution:
â´ Distance between I2 & I2′
=30+20+30cmâ¡=80Â cm

Topic: Physics, Ray Optics, NEET
Two plane progressive waves having the same wavelength Î» and same frequencies with intensities 9I0 and 4I0 superimpose. Resulting intensity when the path difference between waves becomes Î»4 is

Options:
A 13I0
B 14I0
C 15I0
D 16I0

The correct answer is:
13I0

Solution:
ϕ=2πλ×∆x
=2πλ×λ4=π2
I=I1+I2+2I1I2cos⁡ϕ
=9I0+4I0=13I0

Topic: Physics, Waves and Sound, NEET