datavorous/entrance-exam-dataset huggingface dataset:
A one rupee coin starting from rest rolls down a distance of 1 m on an inclined plane at an angle of 30Â° with the horizontal. Assuming that g=9.8 m s-2, the time taken is

Options:
A 0.68 s
B 0.6 s
C 0.5 s
D 0.7 s

The correct answer is:
0.7 s

Solution:
l=1 m,θ=30o,g=9.81 m s-2,t=?
K 2 = 1 2 R 2
a = 9 . 8 1 × 1 2 1 + 1 2 = 4 . 9 × 2 3
⇒ l=12at2
⇒ 1=12×4.93×2×t2
⇒t=34.9=0.7 s

Topic: Physics, Rotational Motion, NEET
A P-N junction diode connected to a battery of e.m.f. 4.5 V and an external resistance of 1000 Î© . What is the value of current in the circuit, if potential barrier in the diode =0.5 V

Options:
A 4 A
B 4 mA
C 5 A
D 5 mA

The correct answer is:
4 mA

Solution:
Applying loop equation i=4.5-0.51000=4 mA

Topic: Physics, Semiconductors, NEET
A $p-n$ photodiode is fabricated from a semiconductor with a band gap of $2.5 \mathrm{eV}$. It can detect a signal of wavelength

Options:
A $6000 Å$
B $4000 \mathrm{~nm}$
C $6000 \mathrm{~nm}$
D $4000 Å$

The correct answer is:
$4000 Å$

Solution:
Key Idea Only signals having wavelength less than threshold wavelength will be detected.
Energy
$$
\begin{aligned}
& \mathrm{E}=\mathrm{hv}=\mathrm{h} \frac{\mathrm{c}}{\lambda} \\
& \lambda=\frac{\mathrm{hc}}{\mathrm{E}}
\end{aligned}
$$
$$
\Rightarrow \quad \lambda=\frac{\mathrm{hc}}{\mathrm{E}}
$$
Substituting the values of and in the above equation
$$
\lambda=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{2.5 \times 1.6 \times 10^{-19}}=5000 Å
$$
As $4000 Å < 5000 Å$
Signal of wavelength $4000 Å$ can be detected by the photodiode.

Topic: Physics, Semiconductors, NEET, NEET 2009 (Screening)
A packet of weight w is dropped with the help of a parachute and on striking the ground comes to rest with retardation equal to twice the acceleration due to gravity. The force exerted on the ground is

Options:
A w
B 2w
C 3w
D 4w

The correct answer is:
3w

Solution:
Mass of the packet , M=w/g
For packet F-w=Ma
or F-w=wg×2g
or F=3w

Topic: Physics, Laws of Motion, NEET
A parabolic bowl with its bottom at origin has the shape y=x220 . Here x and y are in metres. The maximum height at which a small mass m can be placed on the bowl without slipping (coefficient of static friction is 0.5 ) is

Options:
A 2.5Â m
B 1.25 m
C 1.0Â m
D 4.0Â m

The correct answer is:
1.25 m

Solution:
Slope, dydx=x10
or tanâ¡Î¸=x10 ... (i)
Equilibrium of mass in horizontal direction gives the equation,
Î¼Ncosâ¡Î¸=Nsinâ¡Î¸
or tanâ¡Î¸=Î¼=12 ... (ii)
From Equations. (i) and (ii)
x10=12
or x=5m
â´ y=x220=2520=1.25 m

Topic: Physics, Laws of Motion, NEET
A parachutist after bailing outfalls 50 m without friction. When the parachute opens, it decelerates at 2 m s-2. He reaches the ground with a speed of 3 m s-1. At what height, did he bailout? ( Given g=9.8 m s-2)

Options:
A 293 m
B 111 m
C 91 m
D 182 m

The correct answer is:
293 m

Solution:
Initially, the parachutist falls under gravity
∴u⁡2=2gh⁡=2×9·8×50=980 m2 s-2
He reaches the ground with speed =3 m s-1 ,a=-2 m s-2
∴32=u2-2×2×h1 , 9=980-4h1or h1=9714
h⁡1=242·75 m
Total height = 5 0 + 2 4 2 · 7 5 =292·75≈293 m

Topic: Physics, Motion In One Dimension, NEET
A parallel beam of electrons, travelling in x -direction, falls on a slit of width d (see the figure). If after passing the slit, an electron acquires momentum Py in the y-direction, then for a majority of electrons passing through the slit, (h is Planck's constant)

Options:
A Pyd>h
B |Py| d>>h
C Pyd<h
D Pyd≃h

The correct answer is:
Pyd>h

Solution:
sin⁡θ=λd
sin⁡θ≤1
λ=hp

Topic: Physics, Dual Nature of Matter, NEET
A parallel beam of fast moving electrons is incident normally on a narrow slit. A fluorescent screen is placed at a large distance from the slit. If the speed of the electrons is increased, then which of the following statements is correct?

Options:
A Diffraction pattern is not observed on the screen in the case of electrons
B The angular width of the central maximum of the diffraction pattern will increase
C The angular width of the central maximum will decrease
D The angular width of the central maximum will be unaffected

The correct answer is:
The angular width of the central maximum of the diffraction pattern will increase

Solution:
No Solution Available

Topic: Physics, Wave Optics, NEET, NEET 2013 (All India)
A parallel beam of light is incident from air at an angle α on the side PQ of a right angled triangular prism of refractive index n=2 . Light undergoes total internal reflection in the prism at the face PR when α has a minimum value of 45o. The angle θ of the prism is:

Options:
A 15o
B 22.5o
C 30o
D 45o

The correct answer is:
15o

Solution:
1sinâ¡45o=2sinâ¡r1
r2-r1=θ
θ=45o-30o
⇒ θ=15o

Topic: Physics, Ray Optics, NEET
A parallel beam of light is incident normally on a plane surface absorbing 40 % of the light and reflecting the rest. If the incident beam carries 60 W of power, the force exerted by it on the surface is

Options:
A 3.2Ã10-8 N
B 3.2Ã10-7 N
C 5.12Ã10-7 N
D 5.12Ã10-8 N

The correct answer is:
3.2Ã10-7 N

Solution:
Momentum of incident light per second
p1=Ec=603Ã108=2Ã10-7
Momentum of reflected light per second
p2=60100ÃEc=603Ã108=1.2Ã10-7
Force on the surface = change in momentum per second
=p2--p1=p2+p1=2+1.2Ã10-7=3.2Ã10-7 N

Topic: Physics, Dual Nature of Matter, NEET
A parallel beam of light of wavelength 600 nm is incident normally on a slit of width a. If the distance between the slit and the screen is 0.8 m and the distance of 2nd order minimum from the centre of the screen is 1.5 mm, calculate the width of the slit.

Options:
A 9.2Ã 10 -4 â mm
B 6.4Ã 10 â4 m
C 6.32Ã 10 â6 mm
D None of these

The correct answer is:
6.4Ã 10 â4 m

Solution:
Given Î»=600 nm =600Ã 10 â9 m.
=6Ã 10 â7 m
D=0.8âm,âx=1.5âmm= 1.5Ã 10 â3 m,ân=2,âa=?
âµ a x D =nÎ»
a= nÎ»D x = 2Ã6Ã 10 â7 Ã0.8 1.5Ã 10 â3 = 9.6Ã 10 â4 1.5
=6.4Ã 10 â4 m

Topic: Physics, Wave Optics, NEET
A parallel beam of light of wavelength 4000 Aâ passes through a slit of width 5Ã10-3 m. The angular spread of the central maxima in the diffraction pattern is

Options:
A 1.6×10-3 rad
B 1.6×10-4 rad
C 3.2×10-3 rad
D 3.2×10-4 rad

The correct answer is:
1.6×10-4 rad

Solution:
Angular spread of the central maxima in the diffraction pattern =2λe ( e = width of the slit)
Here, e=5×10-3m, λ=4000Å
Hence, 2 θ=24000×10-105×10-3
=1600×10-7 rad
or 2θ=1.6×10-4 rad

Topic: Physics, Wave Optics, NEET
A parallel beam of light of wavelength $\lambda$ is incident normally on a narrow slit. A diffraction pattern formed on a screen placed perpendicular to the direction of the incident beam. At the second minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of slit is:

Options:
A $2 \pi$
B $3 \pi$
C $4 \pi$
D $\pi \pi$

The correct answer is:
$4 \pi$

Solution:
We know that path difference for second minimum $=2 \lambda$
$$
\begin{aligned}
\therefore \text { Phase difference } & =\frac{2 \pi}{\lambda} \times 2 x \\
& =4 \pi
\end{aligned}
$$

Topic: Physics, Wave Optics, NEET, NEET 2013 (Karnataka)
A parallel beam of monochromatic light falls normally on a single narrow slit. How does the angular width of the principal maximum in the resulting diffraction pattern varies with the width of the slit?

Options:
A Principal maximum increases on increasing the width of slit
B Principal maximum decreases on decreasing the width of slit
C Principal maximum increases on decreasing the width of slit
D None of these

The correct answer is:
Principal maximum increases on decreasing the width of slit

Solution:
Angular width of central maxima =2Î»b
â´ The angular width of the principal maximum increases on decreasing the width of slit (b).

Topic: Physics, Wave Optics, NEET
A parallel beam of monochromatic light is incident normally on a slit. The diffraction pattern is observed on a screen placed at the focal plane of a convex lens. If the slit width is increased, the central maximum of the diffraction pattern will

Options:
A Become broader and fainter
B Become broader and brighter
C Become narrower and fainter
D Become narrower and brighter

The correct answer is:
Become narrower and brighter

Solution:
Width of central maximum
β=Dλd,d being slit width.
If the slit width is increased the central maximum becomes narrower and more light reaches the region.

Topic: Physics, Wave Optics, NEET
A parallel beam of monochromatic light of wavelength 5000Å is incident normally on a single narrow slit of width 0.001 mm. The light is focused by a convex lens on a screen placed in focal plane. The first minimum will be formed for the angle of diffraction equal to

Options:
A 0o
B 15o
C 30o
D 50o

The correct answer is:
30o

Solution:
For first minimum, asin⁡θ=nλ=1λ
sin⁡θ=λa=5000×10-100.001×10-3=0.5
θ=30o

Topic: Physics, Wave Optics, NEET
A parallel monochromatic beam of light is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of incident beam. For the first minimum of the diffraction pattern the phase difference between the rays coming from the edges of the slit is

Options:
A 2π
B π
C π2
D 0

The correct answer is:
2π

Solution:
As dsinθ=n λ when n=1
dsinθ=λ the phase difference is 2π and we get first minima.

Topic: Physics, Wave Optics, NEET
A parallel plate air capacitor has a capacitance C. When it is half-filled with a dielectric constant 5, the percentage increase in the capacitance will be

Options:
A 400 %
B 66.6 %
C 33.3 %
D 200 %

The correct answer is:
66.6 %

Solution:
Initial capacitance
C=Îµ0Ad
When it is half filled by a dielectric of dielectric constant K, then
C 1 = K Îµ 0 A d / 2 = 2 K Îµ 0 A d
and C 2 = Îµ 0 A d / 2 = 2 Îµ 0 A d
â´ 1C''=1C1+1C2
=d2Îµ0A1K+1=d2Îµ0A15+1
=610 dÎµ0A
â´ C''=5Îµ0A3d
Hence, increase in capacitance
= 53Îµ0Ad-Îµ0AdÎµ0Ad=53-1=23=66.6%

Topic: Physics, Electrostatics, NEET
A parallel plate air capacitor has capacity C, distance of separation between plates is d and potential difference V is applied between the plates. The force of attraction between the plates of the capacitor is

Options:
A CV22d.
B CV2d.
C C2V22d2.
D C2V22d.

The correct answer is:
CV22d.

Solution:
Attraction force between the plates,
F=q22Aε0 where q=CV and C=ε0Ad
⇒F=C2V22Cd=CV22d

Topic: Physics, Capacitance, NEET, NEET 2015 (Phase 2)
A parallel plate air capacitor is charged to a potential difference of $V$ volts. After disconnecting the charging battery the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates:

Options:
A increases
B decreases
C does not change
D becomes zero

The correct answer is:
increases

Solution:
According to the question $\phi=$ Const and decrease in $C$ means increase in $V$.
$$
\begin{aligned}
& C=\frac{t o f}{d} \\
& \text { Capacitance }=\frac{\text { Charge }}{\text { Potential difference }} \\
& \Rightarrow \text { Potential difference } \\
& =\frac{\text { Charge }}{\text { Capacitance }} \text {. } \\
&
\end{aligned}
$$

Topic: Physics, Capacitance, NEET, NEET 2006
A parallel plate capacitor (Fig. 8.6) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad s–1.
What is the rms value of the conduction current?

Options:
A 0.69$\mu $A
B 690$\mu $A
C 69$\mu $A
D 6.9$\mu $A

The correct answer is:
6.9$\mu $A

Solution:
Correct Option is : (D)
6.9$\mu $A

Topic: Chemistry, Electromagnetic Waves, NEET
A parallel plate capacitor has a uniform electric field E→ in the space between the plates. If the distance between the plates is d and the area of each plate is A, the energy stored in the capacitor is: ε0= permittivity of free space)

Options:
A 12ε0E2
B ε0EAd
C 12ε0E2Ad
D E2Adε0

The correct answer is:
12ε0E2Ad

Solution:
Electric energy density x=12ε0E2

Volume of the space between plates of Capacitor is Ad.

∴ Energy Stored = Energy density × Volume

=12ε0E2Ad

Topic: Physics, Capacitance, NEET, NEET 2021
A parallel plate capacitor having cross-sectional area A and separation d has air in between the plates. Now an insulating slab of the same area but the thickness, d2, is inserted between the plates as shown in figure having dielectric constant K(=4). The ratio of new capacitance to its original capacitance will be,

Options:
A 2:1
B 8:5
C 6:5
D 4:1

The correct answer is:
8:5

Solution:
First, recall the formula of capacity of a parallel plate capacitor at free space in terms of cross-section area and separation between plates,C0=ϵ0Ad, when a dielectric slabs of thickness, t and dielectric constant k, inserted between plates then-new capacitance becomes,

Ck=ϵ0Ad-t+tk⇒Ck=ϵ0Ad-d2+d8, here t=d2.

⇒Ck=85ϵ0Ad⇒Ck=85C0⇒CkC0=85

Topic: Physics, Capacitance, NEET, NEET 2020 (Phase 2)
A parallel plate capacitor is charged by a battery after charging the capacitor, the battery is disconnected and if a dielectric plate is inserted between the plates, then which one of the following statements is not correct

Options:
A increase in the stored energy
B decrease in the potential difference
C decrease in the electric field
D increase in the capacitance

The correct answer is:
increase in the stored energy

Solution:
New capacitance,
C=KC0
New electric field,
E=E0K
New potential difference
ΔV=Q0C=Q0KC0
ΔV=ΔV0K

Topic: Physics, Electrostatics, NEET
A parallel plate capacitor is charged by connecting it to a battery through a resistor. If $I$ is the current in the circuit, then in the gap between the plates:

Options:
A Displacement current of magnitude equal to I flows in the same direction as I
B Displacement current of magnitude equal to I flows in a direction opposite to that of $I$
C Displacement current of magnitude greater than I flows but can be in any direction
D There is no current

The correct answer is:
Displacement current of magnitude equal to I flows in the same direction as I

Solution:
According to modified Ampere's law
$\oint B . d l=\mu_0\left(I_C+I_D\right)$
For Loop $L_1 \quad I_C \neq 0$ and $I_D=0$
For Loop $L_2 \quad I_C=0$ and $I_D \neq 0$
Due to $\mathrm{KCL} \quad I_C=I_D$

Topic: Physics, Electromagnetic Waves, NEET, NEET 2024
A parallel plate capacitor of capacitance 20 μF is being charged by a voltage source whose potential is changing at the rate of 3 V s-1. The conduction current through the connecting wires, and the displacement current through the plates of the capacitor would be, respectively:

Options:
A zero, 60 μA
B 60 μA, 60 μA
C 60 μA, zero
D zero, zero

The correct answer is:
60 μA, 60 μA

Solution:
q=CV
dqdt=Cdvdt
=20×10-6×3
Conduction current =60 μA
Displacement current =ε0dϕedt
=ε0AddVdt
=60 μA

Topic: Physics, Capacitance, NEET, NEET 2019 (All India)
A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance 2C is similarly charged to a potential difference 2V. The charging battery is now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is

Options:
A zero
B 256CV2
C 92CV2
D 32CV2

The correct answer is:
32CV2

Solution:
The diagramatic representation of given problem is shown in the figure.

The net charge shared between the two capacitors is,
Q'=Q2-Q1=4CV-CV=3CV
The two capacitors will have the same potential, say V'.
The net capacitance of the parallel combination of the two capacitors will be,
C'=C1+C2=C+2C=3C
The potential difference across the capacitors will be,
V'=Q'C'=3CV3C=V
The electrostatic energy of the capacitors will be,
U'=12C'V'2=123CV2=32CV2

Topic: Physics, Electrostatics, NEET
A parallel plate capacitor with air as a dielectric has capacitance C. A slab of dielectric constant K, having same thickness as the separation between the plates is introduced so as to fill one fourth of the capacitor as shown in the figure. The new capacitance will be

Options:
A K+3C4
B K+2C4
C K+1C4
D KC4

The correct answer is:
K+3C4

Solution:
The two condensers with K and with air are in parallel. With air, C1=Îµ0d3A4=3Îµ0A4d
With medium, C2=Îµ0KdA4=Îµ0AK4d
â´C′=C1+C2
or C′=3Îµ0A4d+Îµ0AK4d=Îµ0Ad34+K4
or C′=C4K+3âµC=AÎµ0d

Topic: Physics, Electrostatics, NEET
A parallel plate condenser has a uniform electric field $E(\mathrm{~V} / \mathrm{m})$ in the space between the plates. If the distance between the plates is $d(\mathrm{~m})$ and area of each plate is $A\left(\mathrm{~m}^2\right)$ the energy (joule) stored in the condenser is

Options:
A $\frac{1}{2} \varepsilon_0 E^2$
B $\varepsilon_0 E A d$
C $\frac{1}{2} \varepsilon_0 E^2 A d$
D $E^2 A d / \varepsilon_0$

The correct answer is:
$\frac{1}{2} \varepsilon_0 E^2 A d$

Solution:
The energy stored in the condenser
$\begin{aligned}
& U=\frac{1}{2} C V^2 \\
& U=\frac{1}{2}\left(\frac{A \varepsilon_0}{d}\right)(E d)^2 \quad\left(\because C=\frac{A \varepsilon_0}{d} \text { and } V=E d\right) \\
& U=\frac{1}{2} \varepsilon_0 E^2 A d
\end{aligned}$

Topic: Physics, Capacitance, NEET, NEET 2011 (Screening)
A parallel-plate capacitor of area A, plate separation d and capacitance C is filled with four dielectric materials having dielectric constants k1, k2, k3 and k4 as shown in the figure below. If a single dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectric constant k is given by

Options:
A k=k1+k2+k3+3k4
B k=23k1+k2+k3+2k4
C 2k=3k1+k2+k3+1k4
D k=2k43k1k1+k4+k2k2+k4+k3k3+k4

The correct answer is:
k=2k43k1k1+k4+k2k2+k4+k3k3+k4

Solution:
Here the capacitance, \(C=\frac{A k \epsilon_0}{d}\)
\(C_{K_1}, C_{K_2}, C_{K_3}\) are in parallel combination and they are connected in series with \(\mathrm{C}_{\mathrm{K}_4}\)

Here, \(\mathrm{C}_{\mathrm{K}_1}=\frac{(\mathrm{A} / 3) \mathrm{K}_1 \epsilon_0}{\mathrm{~d} / 2}=\frac{2 \mathrm{~K}_1}{3} \mathrm{C}\),
\(\mathrm{C}_{\mathrm{k}_2}=\frac{(\mathrm{A} / 3) \mathrm{k}_2 \mathrm{\epsilon}_0}{\mathrm{~d} / 2}=\frac{2 \mathrm{k}_2}{3} \mathrm{C}\)
\(C_{K_3}=\frac{(\mathrm{A} / 3) \mathrm{k}_3 \epsilon_0}{\mathrm{~d} / 2} \frac{2 \mathrm{k}_3 \mathrm{C}}{3} \mathrm{C}\)
\(\mathrm{C}_{\mathrm{K}_4}=\frac{(\mathrm{A}) \mathrm{k}_4 \epsilon_0}{\mathrm{~d} / 2}=2 \mathrm{k}_4 \mathrm{C}\)
Now the equivalent capacitance for the combination of four capacitors is
\(\begin{aligned}
& \frac{1}{C_{c q}}=\frac{1}{\left(C_{k_1}+C_{k_2}+C_{k_3}\right)}+\frac{1}{C_{k_4}} \\
& C_{e q}=k C
\end{aligned}\)
\(\frac{1}{k C}=\frac{3}{2 C}\left[\frac{1}{k_1+k_2+k_3}\right]+\frac{1}{2 k_4 C}\)
or \(\frac{1}{k}=\frac{3}{2\left(k_1+k_2+k_3\right)}+\frac{1}{2\left(k_4\right)}\)
or \(k=\frac{2\left(k_1+k_2+k_3\right)}{3}+2\left(k_4\right)\)

Topic: Physics, Capacitance, NEET, NEET 2016 (Phase 2)
A paramagnetic material has 1028 atoms m-3. Its magnetic susceptibility at temperature 350 K is 2.8Ã10-4. Its susceptibility at 300 K is

Options:
A 3.726×10-4
B 2.672×10-4
C 3.267×10-4
D 3.672×10-4

The correct answer is:
3.267×10-4

Solution:
From Curie's law
χ∝1T
χ2χ1=T1T2
χ2=χ1·T1T2=2.8×10-4×350300=3.267×10-4

Topic: Physics, Magnetic Properties of Matter, NEET
A paramagnetic substance of susceptibility 3Ã10-4 is placed in a magnetic field of 3Ã10-4 A m-1. Then, the intensity of magnetisation in units of A m-1 is

Options:
A 9×10-8
B 1×10-4
C 1×10-3
D 9×10-10

The correct answer is:
9×10-8

Solution:
Given, susceptibility, χ=3×10-4
Magnetic field, B=3×10-4 A m-1
Since, the intensity of magnetization,
I=susceptibility×magnetic field
=3×10-4×3×10-4
=9×10-8 A m-1

Topic: Physics, Magnetic Properties of Matter, NEET
A parent with an S1S2 genotype exhibiting sporophytic self-incompatibility. Which of the following plant genotypes of the pollen grains of the same species it can fertilize?

Options:
A S1S3
B S2S3
C S3S4
D S1S4

The correct answer is:
S3S4

Solution:
Self-incompatibility is a mechanism in which pollens of one flower are prevented to fertilize with the other flowers of the same plant. This process prevents inbreeding and promotes outcrossing. So if a parent with S1S2 genotypes is present they will not fertilize with the same genotype. They will fertilize with the different flowers of the same plant i.e.S3S4.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
A particle A has a charge q and particle B has charge 4q with each of them having the mass m. When they are allowed to fall from rest through the same potential difference, the ratio of their speeds vA:vB will be

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

The correct answer is:
1 : 2

Solution:
Speed obtained by the particle after falling through a potential difference of V volt is
vA=2Vqm â¦(i)
And vB= 2VÃ4qm ...(ii)
Now dividing Eq. (i) by Eq. (ii), we get
vAvB=14 = 12
So, vA:vB=1:2

Topic: Physics, Atomic Physics, NEET
A particle (a) is dropped from a height and another particle (b) is projected in the horizontal direction with a speed of $5 \mathrm{~ms}^{-1}$ from the same height, then the correct statement is:

Options:
A particle (a) will reach at the ground first with respect to particle (b)
B particle (b) will reach at the ground first with respect to particle (a)
C both particle will reach at the ground simultaneously
D both particles will reach at the ground with the same speed

The correct answer is:
both particle will reach at the ground simultaneously

Solution:
For particle A
$\begin{aligned}
& h=\frac{1}{2} g t^2 \\
& t_{\mathrm{A}}=\sqrt{\frac{2 h}{g}}
\end{aligned}$
For particle B. In vertical direction
Using $\mathrm{S}=u t+\frac{1}{2}+a t^2$
$\Rightarrow h=\frac{1}{2} g+t_n^2 \Rightarrow t_{\mathrm{B}}=\sqrt{\frac{2 h}{g}}$

Topic: Physics, Motion In Two Dimensions, NEET, NEET 2002
A particle A is projected from the ground with an initial velocity of 10 ms-1at an angle of 60o with horizontal. From what height h should an another particle B be projected horizontally with velocity 5 ms-1 so that both the particles collide in ground at point C if both are projected simultaneously? (g=10ms-2)

Options:
A 10 m
B 30 m
C 15 m
D 25 m

The correct answer is:
15 m

Solution:
For both the particle to collide at point, their time of fight should be same
T = 2 usin Î¸ g = 2 10 sin 60 10 = 2 3 2
time of fight of particle thrown from above
t = 2h g
equation we get,

2h g = 3
h = 15 m

Topic: Physics, Motion In Two Dimensions, NEET
A particle A of charge 1 μC is held fixed at a point P in free space. Another particle B of same charge and mass 4 mg is kept at a distance of 1 mm from P. If B is released then its velocity at a distance of 9 mm from P is (Take 14πε0=9×109 Nm2C-2)

Options:
A 1.5Ã102 m/s
B 1.0 m/s
C 3.0Ã104 m/s
D 2.0Ã103 m/s

The correct answer is:
2.0Ã103 m/s

Solution:
Loss in PE=Gain in KE

KÃ10-6Ã10-6110-3-19Ã10-3=12 mv2

9Ã109Ã10-6Ã10-610-3Ã89=12Ã4Ã10-6ÃV2

V=2Ã103 m/s

Topic: Physics, Electrostatics, NEET
A particle aimed at a target, projected with an angle 15Â° with the horizontal is short of the target by 10 m. If projected with an angle of 45Â° is away from the target by 15 m, then the angle of projection to hit the target is

Options:
A 12sin-1⁡110
B 12sin-1⁡310
C 12sin-1⁡910
D 12sin-1⁡710

The correct answer is:
12sin-1⁡710

Solution:
R1=R-10
R2=R+15
R-10R+15=sin⁡2θ1sin⁡2θ2=sin⁡30osin⁡90o=12
2R-20=R+15
R=35
Rmax=u2g
R=u2sin⁡2θg
50=u2g
35=50g×sin⁡2θg
u2=50 g
sin⁡2θ=3550=710
2θ=sin-1⁡710
θ=12sin-1⁡710

Topic: Physics, Motion In Two Dimensions, NEET
A particle at end of a spring executes simple harmonic motion with a period t1, while the corresponding period of another spring is t2. If two springs are connected in series then time period is T, then which one is true?

Options:
A T=t1+t2
B T2=t12+t22
C T-1=t1-1+t2-1
D T-2=t1-2+t2-2

The correct answer is:
T2=t12+t22

Solution:
When springs are connected in series,

1keq=1k1+1k2

t1=2πmk1 and t2=2πmk2

T=2πmkeq

T2=2π2m1keq

T2=2π2m1k1+1k2

T2=t12+t22

Topic: Physics, Oscillations, NEET