datavorous/entrance-exam-dataset huggingface dataset:
Two bullets are fired horizontally and simultaneously towards each other from roof tops of two buildings $100 \mathrm{~m}$ apart and of same height of $200 \mathrm{~m}$ with the same velocity of $25 \mathrm{~m} / \mathrm{s}$. When and where will the two bullets collides. $\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)$

Options:
A After $2 \mathrm{~s}$ at a height $180 \mathrm{~m}$
B After $2 \mathrm{~s}$ at a height of $20 \mathrm{~m}$
C After $4 \mathrm{~s}$ at a height of $120 \mathrm{~m}$
D They will not collide

The correct answer is:
After $2 \mathrm{~s}$ at a height $180 \mathrm{~m}$

Solution:
Given, distance between the two buildings $\mathrm{d}=100 \mathrm{~m}$
height of each tower, $\mathrm{h}=200 \mathrm{~m}$
speed of each bullet, $\mathrm{v}=25 \mathrm{~ms}^{-1}$
The situation can be shown as below
where, $\mathrm{x}$ be the vertical distance travelled from the top of the building and t be the time at which they collide.
As two bullets are fired toward each other,
So, their relative velocity will be
$$
\mathrm{v}_{\text {rel }}=25-(-25)=50 \mathrm{~ms}^{-1}
$$
Then, time $\mathrm{t}=\frac{\mathrm{d}}{\mathrm{v}_{\text {rel }}}=\frac{100}{50}=2 \mathrm{~s}$
The distance or height at which they collide is calculated from equation of motion,
$$
\mathrm{x}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^2
$$
The bullet is initially a rest i.e. $\mathrm{u}=0$ and $\mathrm{as}$ it is moving under the effect of gravity $a=-g$, so
$$
\begin{aligned}
& x=-\frac{1}{2} g^2 \\
& x=-\frac{1}{2} \times 10(2)^2=-20 \mathrm{~m}
\end{aligned}
$$
The negative sign shows that the bullets will collide $20 \mathrm{~m}$ below the top of tower i.e. at a height of $(200-20)=180 \mathrm{~m}$ from the ground after $2 \mathrm{~s}$.

Topic: Physics, Motion In Two Dimensions, NEET, NEET 2019 (Odisha)
Two bullets are fired simultaneously, horizontally and with different speeds from the same place. Which bullet will hit the ground first?

Options:
A The faster bullet
B The slower bullet
C Both will hit simultaneously
D Depends on the masses

The correct answer is:
Both will hit simultaneously

Solution:
When two bullets are fired simultaneously, horizontally with different speeds, then they cover different horizontal distance because there is no acceleration in this direction.
Since, horizontal distanceR=velocity ×time.
But there is a vertical acceleration towards the earth (g), so the vertical distance covered by both bullet is given by:
y=12gt2, which independent of the initial velocity.

So, both the bullets will hit the ground simultaneously.

Topic: Physics, Motion In Two Dimensions, NEET
Two buses A and B are moving around concentric circular paths of radii rA and rB . If the two buses complete the circular paths at the same time, the ratio of their linear speeds is

Options:
A 1
B rArB
C rBrA
D None of these

The correct answer is:
rArB

Solution:
In this case, time period of both buses are same.
Here, T1=2πω1
T2=2πω2
∵ ω1=ω2
∵T1=T2
∴vAvB=rAωrBω=rArB

Topic: Physics, Laws of Motion, NEET
Two butterfly species are competing for the same nectar of a flower in a garden. To survive and coexist together, they may avoid competition in the same garden by:

Options:
A feeding at the same time
B choosing different foraging patterns
C increasing time spent on attacking each other
D predating on each other

The correct answer is:
choosing different foraging patterns

Solution:
Two competing species can co-exist for the same type of resource by changing their foraging pattern. These butterflies can obtain nectar at different times and chase different floral parts to avoid competition.

Topic: Biology, Organisms and Populations, NEET, NEET 2022 (Phase 2)
Two capacitors of capacitance 6 Î¼F and 12 Î¼F are connected in series with a battery. The voltage across the 6 Î¼F capacitor is 2 V. The total battery voltage is

Options:
A 4 V
B 6 V
C 9 V
D 3 V

The correct answer is:
3 V

Solution:
Charge on 6 Î¼F capacitor = Charge on 12 Î¼F capacitor.

â6Ã10â6Ã2 V=12Ã10â6 C

â´ Potential difference across 12 Î¼F capacitor is

V=12Ã10â1212Ã10â12=1 V

Battery voltage=2 V+1 V=3 V

Topic: Physics, Electrostatics, NEET
Two capacitors of capacitance C are connected in series. If one of them is filled with dielectric substance K, the effective capacitance is

Options:
A KC(1+K)
B C(K+1)
C 2KC(1+K)
D None of these

The correct answer is:
KC(1+K)

Solution:
When plates of capacitor are separated by a dielectric medium of dielectric constant K, its capacity
Cm=KÎµ0Ad=KC0
ie,Â Â Â Â Â Â Â Â Cm=KC0
Here,Â Â Â Â C0=C
â´Â Â Â Â Â Â Â Â Â Cm=KC
Now, two capacitors of capacities KC and C are in series, their effective capacitance
1C′=1KC+1C
or Â Â Â Â Â 1C′Â =1+KKC
â´Â C′=KCK+1

Topic: Physics, Electrostatics, NEET
Two capacitors of capacity 6 Î¼F and 12 Î¼F in series are connected by potential of 150 V. The potential of the capacitor of capacity 12 Î¼F will be

Options:
A 25 V

B 50 V

C 100 V
D 150 V

The correct answer is:
50 V

Solution:
Given, C1=6 μF, C2=12 μF, V=150 volt.
Total capacity, 1C=1C1+1C2=16+112
1C=312 ⇒C=4 μF
Potential of 12μF capacitor
V=qC
V= 4×15012
V=50 V

Topic: Physics, Electrostatics, NEET
Two Carnot engines $\mathrm{A}$ and $\mathrm{B}$ are operated in series. The engine $A$ receives heat from the source at temperature $T_1$ and rejects the heat to the sink at temperature $T$. The second engine $\mathrm{B}$ receives the heat at temperature $\mathrm{T}$ and rejects to its sink at temperature $T_2$. For what value of $T$ the efficiencies of the two engines are equal:

Options:
A $\frac{T_2-T_2}{2}$
B $T_1 T_2$
C $\sqrt{T_1 T_2}$
D $\frac{T_1+T_2}{2}$

The correct answer is:
$\sqrt{T_1 T_2}$

Solution:
We know that efficiency of carnot engine $\eta=1-\frac{T_2}{T_1}$
Where $T_1=$ temperature of source $T_2=$ temp of $\operatorname{sink}$ For engine $A, n_{\mathrm{A}}=1-\frac{T}{T_1}$ an engine $B, n_{\mathrm{B}}=1-\frac{T_2}{T}$ From question
$\begin{array}{l}
n_A =n_B \\
1-\frac{T}{T_1} =1-\frac{T_2}{T} \\
\Rightarrow \frac{T}{T_1}=\frac{T_2}{T} =\mathrm{T}^2 \\
\Rightarrow \mathrm{T}^2 =\mathrm{T}_1 \mathrm{~T}_2 \\
\Rightarrow \mathrm{T} =\sqrt{T_1 T_2}
\end{array}$

Topic: Physics, Thermodynamics, NEET, NEET 2013 (Karnataka)
Two cars moving in opposite directions approach each other with speed of 22 m s-1 and 16.5 m s-1 respectively. The driver of the first car blows a horn having a frequency 400  Hz. The frequency heard by the driver of the second car is [velocity of sound 340 m s-1]

Options:
A 350  Hz
B 361  Hz
C 411  Hz
D 448  Hz

The correct answer is:
448  Hz

Solution:
Due to Doppler’s effect, when both observer and source are moving towards each other, then, the frequency measured by the observer differs from the actual frequency generated by the source.

Thus, the apparent frequency, fA=fv+v0v-vs
=400 340+16.5340-22
fA=448 Hz

Topic: Physics, Waves and Sound, NEET, NEET 2017
Two cars P and Q start from a point at the same time in a straight line and their positions are represented by XPt=at+bt2 and XQt=ft-t2. At what time do the cars have the same velocity?

Options:
A a-f1+b
B a+f2b-1
C a+f21+b
D f-a21+b

The correct answer is:
f-a21+b

Solution:
vP=dXPdt=a+2bt
vQ=dXQdt=f-2t
Now for, vP=vQ
⇒a+2bt=f-2t
⇒2t+2bt=f-a
⇒t=f−a2(b+1)

Topic: Physics, Motion In One Dimension, NEET, NEET 2016 (Phase 2)
Two cells A and B are contiguous. Cell A has osmotic pressure 10 atm, turgor pressure 7 atm and diffusion pressure deficit 3 atm. Cell B has osmotic pressure 8 atm, turgor pressure 3 atm and diffusion pressure deficit 5 atm. The result will be

Options:
A No movement of water
B Equilibrium between the two
C Movement of water from cell A to B
D Movement of water from cell B to A

The correct answer is:
Movement of water from cell A to B

Solution:
Diffusion pressure deficit is the reduction in the diffusion pressure of water in a system over its pure state. It is given by DPD = O.P - W.P (T.P). DPD determines the direction of the net movement of water. It is always from an area or cell of lower DPD to the area or cell of higher DPD. So, cell A having lower DPD, water will move from cell A to B.

Topic: Biology, Transport in Plants, NEET
Two cells, having the same e.m.f. are connected in series through an external resistance $R$. Cells have internal resistances $r_1$ and $r_2\left(r_1>r_2\right)$ respectively. When the circuit is closed, the potential difference across the first cell is zero. The value of $R$ is:

Options:
A $r_1+r_2$
B $r_1-r_2$
C $\frac{r_1+r_2}{2}$
D $\frac{r_1-r_2}{2}$

The correct answer is:
$r_1-r_2$

Solution:
According to the question
$E-I r_1=0 \text { and } \mathrm{I}=\frac{E+E}{r_1+r_2+R}$
$\begin{array}{l}
\therefore \frac{E}{r_1} =\frac{2 E}{r_1+r_2+R} \\
\Rightarrow r_1+r_2+R =2 r_1-r_2 \\
\Rightarrow R =r_1-r_2
\end{array}$

Topic: Physics, Current Electricity, NEET, NEET 2006
Two cells I and II are contiguous. Cell I has osmotic pressure $10 \mathrm{~atm}$. turgor pressure $7 \mathrm{~atm}$ and diffusion pressure deficit $3 \mathrm{~atm}$. Cell II has osmotic pressure $8 \mathrm{~atm}$, turgor pressure $3 \mathrm{~atm}$ and diffusion pressure deficit $5 \mathrm{~atm}$. The result will be:

Options:
A no movement of water
B equilibrium between the two
C movement of water from cell I to II
D movement of water from cell II to I.

The correct answer is:
movement of water from cell I to II

Solution:
The water moves from a lower diffusion pressure deficit to a higher diffusion, pressure deficit i.e., from A to $B$. The cell-to-cell movement of water depends on the DPD, and not on the osmotic pressure and turgor pressure.
Related Theory
Even though the osmotic pressures are same on both sides of the semi-permeable membrane, the movement of water will take place to equate the DPD on both sides of the membrane. Osmosis takes place from lower $D P D$ (higher water potential) to higher DPD (lower water potential).
Therefore, $D P D=O P-T P$

Topic: Biology, Transport in Plants, NEET, NEET 2007
Two charged capacitors have their outer plates fixed and inner plates connected by a spring of force constant k. The magnitude of charge on each capacitor is q and sign of charge is shown in figure. Find the extension in the spring at equilibrium.

Options:
A q22Aε0k
B q24Aε0k
C q2Aε0k
D zero

The correct answer is:
q22Aε0k

Solution:
We know that spring force F=kx will balance electrostatic force of attraction.
Therefore, q22Aε0=kx
⇒x=q22Aε0k

Topic: Physics, Electrostatics, NEET
Two charged particle M and N are projected with same velocity in a uniform magnetic field. Then M and N respectively

Options:
A an electron and a proton
B a deuteron and a proton
C a deuteron and an electron
D a proton and Î± particle

The correct answer is:
a proton and Î± particle

Solution:
The radius of charged particle in a uniform magnetic field
r=mvqB
therefore r â mq ----(i)
according to the given figure, the centre of both the curve is on the same side, hence nature of both the charge is the same.
from figure RM<RN
option 1 is incorrect as of opposite charge
option 2 is incorrect as deuteron has a greater mass hence greater radius of M
option 3 is incorrect as of opposite charge
option 4 is correct as of proton has a less mass hence smaller radius of M

Topic: Physics, Magnetic Effects of Current, NEET
Two charged spherical conductors of radius R1 and R2 are connected by a wire. Then the ratio of surface charge densities of the spheres σ1/σ2 is :

Options:
A R1R2
B R2R1
C R1R2
D R2R1

The correct answer is:
R2R1

Solution:
V1=V2

kq1R1=kq2R2

σ14πR12R1=σ24πR22R2

σ1σ2=R2R1

Topic: Physics, Electrostatics, NEET, NEET 2021
Two charges 4q and q are placed 30 cm apart. At what point the value of the electric field will be zero?

Options:
A 10 cm away from q and between the charges
B 20 cm away from q and between the charges
C 10 cm away from q and outside the line joining the charges
D 10 cm away from 4q and outside the line joining them

The correct answer is:
10 cm away from q and between the charges

Solution:
Position of null point is given by
x1=rq2q1+1
x1=3041+1
x1=10 cm

Topic: Physics, Electrostatics, NEET
Two charges ±10 mC are placed 5.0 mm apart. Determine the electric field at a point P on the axis of the dipole 15 cm away from its centre O on the side of the positive charge, as shown in Fig.

Options:
A 5.2 × ${{10}^{4}}$N ${{C}^{-1}}$
B 5.2 × ${{10}^{5}}$ N ${{C}^{-1}}$
C 2.6 × ${{10}^{5}}$N ${{C}^{-1}}$
D 2.6 × ${{10}^{4}}$N ${{C}^{-1}}$

The correct answer is:
2.6 × ${{10}^{5}}$N ${{C}^{-1}}$

Solution:
Correct Option is : (C)
2.6 × ${{10}^{5}}$N ${{C}^{-1}}$

Topic: Chemistry, Electrostatics, NEET
Two charges ±10 mC are placed 5.0 mm apart. Determine the electric field at a point Q, 15 cm away from O on a line passing through O and normal to the axis of the dipole, as shown in Fig

Options:
A 2.66 ×${{10}^{6}}$ N ${{C}^{-1}}$
B 1.33 ×${{10}^{5}}$ N ${{C}^{-1}}$
C 1.33 ×${{10}^{6}}$ N ${{C}^{-1}}$
D 2.66 ×${{10}^{5}}$ N ${{C}^{-1}}$

The correct answer is:
1.33 ×${{10}^{5}}$ N ${{C}^{-1}}$

Solution:
Correct Option is : (B)
1.33 ×${{10}^{5}}$ N ${{C}^{-1}}$

Topic: Chemistry, Electrostatics, NEET
Two charges q 1 and q 2 are placed 30 cm apart, as shown in the figure. A third charge q 3 is moved along the arc of a circle of radius 40 cm from C to D . The change in the potential energy of the system is q 3 4Ïâ â o k where k is -

Options:
A 8 q 1
B 6 q 1
C 8 q 2
D 6 q 2

The correct answer is:
8 q 2

Solution:
U1 =kq1q34πε0×0.4+kq1q24πε0×0.3+kq2q34πε0×0.5
U2=kq1q24πε0×0.3+kq2q34πε0×0.1+kq1q34πε0×0.4
∴∆U=8kq2q34πε0=8q2q34πε0⇒k=8q2

Topic: Physics, Electrostatics, NEET
Two charges q and –3q are placed fixed on x-axis separated by distance ‘d’. Where should a third charge 2q be placed such that it will not experience any force?

Options:
A

B $\frac{d}{2}\left( 1+\sqrt{3} \right)$
C $\frac{d}{3}\left( 2+\sqrt{3} \right)$
D $\frac{d}{3}\left( 1+\sqrt{3} \right)$

The correct answer is:
$\frac{d}{2}\left( 1+\sqrt{3} \right)$

Solution:
Correct Option is : (B)
$\frac{d}{2}\left( 1+\sqrt{3} \right)$

Topic: Chemistry, Electrostatics, NEET
Two children Ramesh (on path ARB ) and Sohan (on path ASB ), travel down slides of identical height h but different shapes as shown in the figure. Assuming they start down the frictionless slides at the same time with zero initial velocity, which of the following statements is true?

Options:
A Ramesh reaches the bottom first with the same average velocity as Sohan.
B Ramesh reaches the bottom first with a larger average acceleration than Sohan.
C Ramesh reaches the bottom first with the same average acceleration as Sohan.
D They reach the bottom at the same time with the same average acceleration.

The correct answer is:
Ramesh reaches the bottom first with a larger average acceleration than Sohan.

Solution:
Ramesh takes lesser time to reach bottom as velocities of both of them are same at the bottom average acceleration of Ramesh is larger compared to Sohan.

Topic: Physics, Motion In One Dimension, NEET
Two circular coils 1 and 2 are made from the same wire but the radius of the $1^{\text {st }}$ coil is twice that of the $2^{\text {nd }}$ coil. What potential difference in volts should be applied across them so that the magnetic field at their centres is the same?

Options:
A 2
B 3
C 4
D 6

The correct answer is:
4

Solution:
Let $r_1$ and $r_2$ be the radius of coil 1 and coil 2 . If $B_1$ and $B_2$ are magnetic induction at their centre then $B_1=\frac{\mu_0 I_1}{2 r_1}$ and $B_2=\frac{\mu_0 I_2}{2 r_2}$ $B_1=B_2$ and $r_1=2 r_2$ there $I_1=2 I_2$ Again if $R_1$ and $R_2$ are resistance of the coil 2 then $R_1=2 R_2$ (as $R \propto$ length $=2 \pi r)$ and if $v_1$ and $v_2$ are the potential difference across them respectively then
$$
\begin{aligned}
\frac{V_1}{V_2} & =\frac{I_1 R_1}{I_2 R_2} \\
& =\frac{\left(2 I_2\right)\left(2 R_2\right)}{I_2 R_2}=4
\end{aligned}
$$

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2006
Two circular coils A and B are facing each other as shown in the diagram. The current i through A can be altered, then

Options:
A there will be repulsion between A and B if i is increased
B there will be attraction between A and B if i is increased
C There will be neither attraction nor repulsion when i is changed
D Attraction or repulsion between A and B depends on the direction of current. It does not depend whether the current is increased or decreased

The correct answer is:
there will be repulsion between A and B if i is increased

Solution:
When current is increased in A, magnetic flux passing through B will increase, according to Lenz's law direction of induced current in B will be such as there is repulsion between A and B.

Topic: Physics, Electromagnetic Induction, NEET
Two circular coils can be arranged in any of the three following situations as shown in the figure. Their mutual inductance will be

Options:
A Maximum B
B Maximum A
C Maximum C
D Same in all conditions

The correct answer is:
Maximum A

Solution:
The mutual inductance between two coils depends on their degree of flux linkage i.e. the fraction of flux linked with one coil which is also linked to the other coil.
Here, the two coils in arrangement (A) are placed with their planes parallel which allows maximum flux

Topic: Physics, Magnetic Effects of Current, NEET
Two circular coils mounted parallel to each other on the same axis carry steady currents. If an observer between the coils reports that one coil is carrying a clockwise current i1 while the other is carrying a counter-clockwise current i2, then the two coils experience

Options:
A a steady repulsive force
B zero force
C a repulsive force
D a steady attractive force

The correct answer is:
a steady attractive force

Solution:
The coil carrying a current i in clockwise sense coil has south polarity on that face of coil and other coil having current i2 in counter clockwise sense has north polarity on that face of the coil. As south and north poles will attract each other, hence a steady attractive force acts between the two coils.

Topic: Physics, Magnetic Effects of Current, NEET
Two circular coils X and Y have the equal number of turns and carry equal currents in the same sense and subtend same solid angle at the point O. If the smaller coil X is midway between O and Y , then if we represent the magnetic induction due to bigger coil Y at O as BY and that due to smaller coil X at O as Bx , then :

Options:
A BYBX=1
B BYBX=2
C BYBX=12
D BYBX=14

The correct answer is:
BYBX=12

Solution:
[ For same angle α ]

B= μ 0 4π 2M ( R 2 + d 2 ) 3 2
∝R2(R2+d2)32
B y B x = R y 2 [ R y 2 + ( 2d ) 2 ] 3 2 R x 2 [ R x 2 + d 2 ] 3 2
R y =2 R x
⇒ B y B x = 1 2

Topic: Physics, Magnetic Effects of Current, NEET
Two circular coils X and Y having equal number of turns and carry equal currents in the same sense and subtend same solid angle at point O. If the smaller coil X is midway between O and Y, then if we represent the magnetic induction due to bigger coil Y at O as BY and due to smaller coil X at O as BX then

Options:
A BYBX=1
B BYBX=2
C BYBX=1/2
D BYBX=1/4

The correct answer is:
BYBX=1/2

Solution:
By=µ0NI2r222r2+d23/2
Bx=µ0NIr22r2+d/223/2⇒ByBx=12

Topic: Physics, Magnetic Properties of Matter, NEET
Two cities are 150 km apart. Electric power is sent from one city to another city through copper wires. The fall of potential per km is 8 volts and the average resistance per km is 0.5 Ω. The power loss in the wire is:

Options:
A 19.2Â W
B 19.2 kW
C 19.2Â J
D 12.2Â kW

The correct answer is:
19.2 kW

Solution:
Distance between the two cities, 150 km.

Voltage drop per kilometre is 8 V km-1.

Resistance per kilometre 0.5 Ω km-1

Total voltage drop will be,
Vdrop=150 km×8 V km-1=1200 V

Total resistance of wire,
R=0.5 Ω km-1×150 km=75 Ω

Therefore, the power loss will be, P=V2R=150×820.5×150=19.2 kW

Topic: Physics, Current Electricity, NEET, NEET 2014
Two coherent beams of wavelength 5000 Aâ reaching a point would individually produce intensities 1.44 and 4.00 units. If they reach there together, the intensity is 10.24 units. Calculate the lowest phase difference with which the beams reach that point.

Options:
A zero
B π4
C π2
D π

The correct answer is:
zero

Solution:
Resultant intensity at the point where the two coherent beam reach together is given by
I=I1+I2+2I1 I2cos⁡ϕ
⇒ 10.24=1.44+4.00+2 1.44×4.00cos⁡ϕ
⇒ 10.24=5.44+4.8cos⁡ϕ
⇒ cos⁡ϕ=10.24-5.444.8=1
⇒ ϕ=0

Topic: Physics, Wave Optics, NEET
Two coherent point sources S1 and S2 are separated by a small distance d as shown in the figure. The fringes obtained on the screen will be

Options:
A semi-circles
B concentric circles
C points
D straight lines

The correct answer is:
concentric circles

Solution:
Path difference on circles around 'P' is same. So, The fringes obtained on the screen in the given condition will be concentric circles.

Topic: Physics, Wave Optics, NEET
Two coherent point sources S1 and S2 vibrating in phase emit light of wavelength λ . The separation between the sources is 2λ . The smallest distance from S2 on a line passing through S2 and perpendicular to S1S2 where a minimum intensity occurs is

Options:
A 7λ12
B 15λ4
C λ2
D 3λ4

The correct answer is:
7λ12

Solution:
S1P-S2P=3λ2
or 4λ2+x2-x=3λ2
On solving
x=7λ12

Topic: Physics, Wave Optics, NEET
Two coherent point sources S1 and S2 vibrating in phase emit light of wavelength λ. The separation between them is 2λ. The light is collected on a screen placed at a distance D>>λ from the slit S1 as shown. The minimum distance, so that intensity at P is equal to the intensity at O

Options:
A 2D
B 3D
C 8D
D 5D

The correct answer is:
3D

Solution:
Path difference between the light waves form S1 and S2, reaching at P is,
=2Î»cosÎ¸
Since, there is a bright maximum at O. Thus for maximum at P,
2Î»cosÎ¸=nÎ»
For x to be minimum n=1
cosÎ¸=12âÎ¸=60Â°
â´x=D tan60Â°=3D

Topic: Physics, Wave Optics, NEET
Two coherent sources of light interfere and produce fringe patterns on a screen. For central maximum, the phase difference between the two waves will be

Options:
A zero
B π
C 3π2
D π2

The correct answer is:
zero

Solution:
For central maximum, path difference is zero hence phase difference for coherent sources of light will be zero. And the relation between phase difference and path difference is, ∆ϕ=2πλ∆x, where Δx is path difference and ∆ϕ is phase difference. So due to zero path difference phase difference is also zero.

Topic: Physics, Wave Optics, NEET, NEET 2020 (Phase 2)
Two coils A and B have a coefficient of mutual inductance M=2 H. The magnetic flux passing through coil A changes by 4 Wb in 10 s due to change in current in B. Then

Options:
A Change in current in B in this time interval is 0.5 A
B Change in current in B in this time interval is 8 A
C The change in current in B in this time interval is 2 A
D A change in the current of 1 A in coil A will produce a change in flux passing through B by 4 Wb

The correct answer is:
The change in current in B in this time interval is 2 A

Solution:
Here, it is given that M=2H, dϕ=4 Wb, dt=10 s
As ϕ =M i
dϕ=M di
Or di=dϕM=42=2 A

Topic: Physics, Electromagnetic Induction, NEET
Two coils are at fixed locations. when coil 1 has no current and the current in coil 2 increases at the rate 15.0 A s-1 the emf in coil 1 is 25.0 mV, when coil 2 has no current and coil 1 has a current of 3.6 A, the flux linkage in coil 2 is

Options:
A 16 mWb

B 10 mWb

C 4 mWb

D 6 mWb

The correct answer is:
6 mWb

Solution:
Coefficient of mutual induction M is given by
M = e 1 di 2 / dt = ϕ 2 i 1
∴ ϕ 2 = e 1 i 1 di 2 / dt = 25.0 × 10 - 3 3.6 1 5
= 6 × 1 0 - 3 Wb
= 6 mWb

Topic: Physics, Electromagnetic Induction, NEET
Two coils are placed close to each other. The mutual inductance of the pair of coils depends upon-

Options:
A Relative position and orientation of the two coils
B The materials of the wires of the coils
C The currents in the two coils
D The rates at which currents are changing in the two coils

The correct answer is:
Relative position and orientation of the two coils

Solution:
Mutual inductance depends upon the geometrical shapes of the two circuits and their placing.

Topic: Physics, Current Electricity, NEET