datavorous/entrance-exam-dataset huggingface dataset:
An iron bar 10 cm in length is kept at 20 Â°C. If the coefficient of linear expansion of iron is Î±=11Ã10-6 Â°C-1, then at 19 Â°C it will be

Options:
A 11×10-6 cm longer
B 11×10-6 cm shorter
C 11×10-5 cm shorter
D 11×10-5 cm longer

The correct answer is:
11×10-5 cm shorter

Solution:
L=L01+αΔθ
⇒ L1L2=1+αΔθ11+αΔθ2
⇒ 10L2=1+11×10-6×201+11×10-6×19
⇒ L2=9.99989
⇒ Length is shorten by
10-9.99989=0.00011=11×10-5 cm

Topic: Physics, Thermal Properties of Matter, NEET
An iron bar of length 10 m is heated from 0Â°C to 100Â°C. If the coefficient of linear thermal expansion of iron is 10Ã10-6 Â°C-1, then increase in the length of bar (in cm) is

Options:
A 0.5 cm
B 1.0 cm
C 1.5 cm
D 2.0 cm

The correct answer is:
1.0 cm

Solution:
The change in length âl is proportional to l and âT. Stated mathematically as
âl=Î±lâT
Where Î± is called the coefficient of linear thermal expansion for the material.
Given: âT=100 Â°C
l=10 m
â´ âl=10Ã100Ã10Ã10-6
= 10-2 m=1 cm

Topic: Physics, Thermal Properties of Matter, NEET
An iron bar of length $L$ has magnetic moment $M$. It is bent at the middle of its length such that the two arms make an angle $60^{\circ}$ with each other. The magnetic moment of this new magnet is :

Options:
A $\frac{M}{2}$
B $2 M$
C $\frac{M}{\sqrt{3}}$
D $M$

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

Solution:
$\begin{aligned} & \Delta I=2 \frac{I}{2} \sin 30^{\circ} \\ & =\frac{1}{2} \\ & M^{\prime}=m / / 2 \\ & =M / 2\end{aligned}$

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2024
An iron load of 2 kg is suspended in the air from the free end of a sonometer wire of length 1 m. A tuning fork of frequency 256 Hz is in resonance with time the length of the sonometer wire. If the load is immersed in water that will be in resonance with the same tuning fork is (specified gravity of iron=8)

Options:
A 8
B 6
C 16
D 18

The correct answer is:
18

Solution:
We know lâT
â´lairÎ¹water=TairTwater
But specific gravity 8=
TairTair-Twater
â¹Twater=78Tair
â´ lairlwater=87
But
lair=17l Given
â´ lwater=18l

Topic: Physics, Waves and Sound, NEET
An iron nail is dropped from a height h from the level of a sand bed. If it penetrates through a distance x in the sand before coming to rest, then the average force exerted by the sand on the nail is

Options:
A mghx+1
B mgxh+1
C mghx-1
D mgxh-1

The correct answer is:
mghx+1

Solution:
The nail has fallen through a total vertical distance of h+x. Hence loss in its potential energy=mgh+x. If average retarding force exerted by sand on the nail is f, then work done W=-Fx
For equilibrium, mgh+x=Fx or F=mgh+xx=mghx+1.

Topic: Physics, Laws of Motion, NEET
An iron rod is placed parallel to magnetic field of intensity 2000 A m. The magnetic flux through the rod is 6Ã10-4 Wb and its cross-sectional area is 3 cm2. The magnetic permeability of the rod in Wb A-1m-1 is

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

The correct answer is:
10-3

Solution:
μ=ϕA×1H=6×10−43×10−4×12×103=10−3 Wb A-1m-1

Topic: Physics, Magnetic Properties of Matter, NEET
An iron rod of length 50 cm is joined at an end to an aluminium rod of length 100 cm. All measurements refer to 20 Â°C. The coefficients of linear expansion of iron and aluminium are 12Ã10-6 Â°C-1 and 24Ã10-6 Â°C-1 respectively. The average coefficient of expansion of the composite system is

Options:
A 36×10-6 °C-1
B 12×10-6 °C-1
C 20×10-6 °C-1
D 48×10-6 °C-1

The correct answer is:
20×10-6 °C-1

Solution:
Length of iron rod at 100oC,
L1 = 50[1 + 12 x 10-6 x (100 - 20)] = 50.048 cm
Length of aluminum rod at 100oC,
L2 = 100[1 + 24 x 10-6 x (100 - 20)] = 100.192 cm
The length of composite system at 20oC
= 50 + 100 = 150 cm
and length of composite system at 100oC
= 50.048 + 100.192 = 150.24 cm
â´ A verage Î± = 0.24 cm 1 5 0 cm Ã 1 0 0 - 2 0
= 20 x 10-6/oC

Topic: Physics, Thermal Properties of Matter, NEET
An iron rod of susceptibility 599 is subjected to a magnetising field of 1200 A m–1. The permeability of the material of the rod is:
μ0=4π×10−7 T m A−1

Options:
A 8.0×10−5 T m A−1
B 2.4π×10−5 T m A−1
C 2.4π×10−7 T m A−1
D 2.4π×10−4 T m A−1

The correct answer is:
2.4π×10−4 T m A−1

Solution:
μr=xm+1=599+1=600

μ=μoμr=4π×10−7×600

=2.4π×10−4TmA

Topic: Physics, Magnetic Properties of Matter, NEET, NEET 2020 (Phase 1)
An iron tyre of diameter 2 m is to be fitted on to a wooden wheel of diameter 2.01 m. The temperature to which the tyre must be heated, if Î±=11Ã10â6 Â°C-1 and room temperature is 20Â°C, will be

Options:
A 474.5 Â°C
B 490.5 Â°C
C 440.5 Â°C
D 460.5 Â°C

The correct answer is:
474.5 Â°C

Solution:
ÎÎ¸=ÎllÎ±
ÎÎ¸=0.012Ã11Ã10-6
ÎÎ¸=10422=454.5
Î¸=454.5+20=474.5 oC

Topic: Physics, Thermal Properties of Matter, NEET
An iron wire AB has diameter of 0.6 mm and length 3 m at 0 â C .The wire is now stretched between the opposite walls of a brass casing at 0 â C . What is the extra tension that will be set up in the wire when the temperature of the system is raised to 40 â C ?

Given Î± brass =18Ã 10 â6 /K

Î± iron =12Ã 10 â6 /K

Y iron =21Ã 10 10 N/m 2

Options:
A 14.2â N
B 13.8 N
C 16.3 N
D 21.7 N

The correct answer is:
14.2â N

Solution:
Increase in length of the iron wire = L α Δ t = L × 12 × 10 - 6 × 40 = 48L × 10 - 5 m
Increase in length of the brass tube = L α Δ t = L × 18 × 10 - 6 × 40 = 72L × 10 - 5 m
Since the increase in length of the brass tube is greater than that of the iron wire, the latter will be tightened. Let T be the tension in the wire. Then
Stress = T⁡ π r⁡ 2 = T⁡ π 3 × 1 0 - 4 2 = T⁡ 9 π × 1 0 - 8
Strain = 7 2 L ⁡ - 4 8 L⁡ × 1 0 - 5 L⁡ = 2 4 × 1 0 - 5
Since stress=Y×strain,T⁡9π×10-8=21×1010×24×10-5
T = 2 1 × 1 0 10 × 2 4 × 1 0 - 5 × 9 π × 1 0 - 8 = 14.2 N

Topic: Physics, Mechanical Properties of Solids, NEET
An isolated particle of mass m is moving in the horizontal plane x, y along the x-axis, at a certain height above the ground. It suddenly explodes into two fragments of masses m4 and 3m4. An instant later, the smaller fragment is at y=+15 cm. The larger fragment at this instant is at

Options:
A y=-5 cm
B y=+20 cm
C y=+5 cm
D y=-20 cm

The correct answer is:
y=-5 cm

Solution:
Before explosion, the particle was moving along x-axis i.e., it has no y-component of velocity. Therefore, the centre of mass will not move in y-direction or we can say ycm=0.
Now, ycm=m1y1+m2y2m1+m2
Therefore, 0=m/4+15+3m/4ym/4+3m/4
or, y=-5 cm

Topic: Physics, Center of Mass Momentum and Collision, NEET
An isomer of C6H14 forms three monochloro derivatives. The isomer may be -
(Excluding stereo isomer)

Options:
A neo-pentane
B n-hexane
C 2, 3-dimethylbutane
D iso-hexane

The correct answer is:
n-hexane

Solution:
In n-hexane, there is a symmetry of the structure at third and fourth carbon. So, chlorine can attach to first, second, and third carbon to give 3 monochloro derivatives. These monochloro derivatives are:

Topic: Chemistry, General Organic Chemistry, NEET
An isotope decays to 1/ 16 th of its mass in 1  h. What is the half-life period of the isotope?

Options:
A 15 min
B 30 min
C 12 min
D 10 min

The correct answer is:
15 min

Solution:
N = N 0 1 2 n
N 0 1 6 = N 0 1 2 n
1 2 4 = 1 2 n
â´ n = 4
T T 1 2 = 4
â 1 T 1 2 = 4
â T 1 2 = 1 4 h = 1 5 min

Topic: Physics, Nuclear Physics, NEET
An L.C.R circuit contains R=50 Î©, L=1 mH and C=0.1 Î¼F . The impedance of the circuit will be minimum for a frequency of

Options:
A 1052 π Hz
B 1062 π Hz
C 2π×105 Hz
D 2π×106 Hz

The correct answer is:
1052 π Hz

Solution:
The impedance of L-C-R circuit will be minimum for a resonant frequency so,
v0=12ÏLC
=12Ï1Ã10-3Ã0.1Ã10-6
=1052Ï Hz

Topic: Physics, Alternating Current, NEET
An LED is constructed from a p-n junction diode using GaAsP. The energy gap is $1.9 \mathrm{eV}$. The wavelength of the light emitted will be equal to

Options:
A $10.4 \times 10^{-26} \mathrm{~m}$
B $654 \mathrm{~nm}$
C $654 Å$
D $654 \times 10^{-11} \mathrm{~m}$

The correct answer is:
$654 \mathrm{~nm}$

Solution:
The energy of light of wavelength $\lambda$ is given by
$$
\mathrm{E}=\mathrm{hv}=\frac{\mathrm{hc}}{\lambda} \Rightarrow \lambda=\frac{\mathrm{hc}}{\mathrm{E}}
$$
Here, $\mathrm{h}=$ Planck's constant $=6.63 \times 10^{-34} \mathrm{~J}-\mathrm{s}$
$\mathrm{c}=$ speed of light $=3 \times 10^8 \mathrm{~m} / \mathrm{s}$
$\mathrm{E}=$ energy gap $=1.9 \mathrm{eV}=1.9 \times 1.6 \times 10^{-19} \mathrm{~J}$
Substituting the given values in Eq. (i), we get
$$
\begin{aligned}
\Rightarrow \quad \lambda & =\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.9 \times 1.6 \times 10^{-19}} \\
& =6.54 \times 10^{-7} \mathrm{~m} \approx 654 \mathrm{~nm}
\end{aligned}
$$
Thus, the wavelength of light emitted from LED will be $654 \mathrm{~nm}$.

Topic: Physics, Dual Nature of Matter, NEET, NEET 2019 (Odisha)
An L.P.G. cylinder contains 15 kg of butane gas at 27°C and 10 atmospheric pressure. It was leaking and its pressure fell down to 8 atmospheric pressure after one day. The gas leaked is

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

The correct answer is:
3 kg

Solution:
From ideal gas equation, PV=nRT
V∝n (Hence V and T are constant)
and further, number of mol of gas (n) ∝ weight of the gas (w)
So P1P2=w1w2
or 10atm8atm=15kgx
or x = 12 kg
Hence, the gas leaked = 15 – 12 = 3 kg.

Topic: Chemistry, States of Matter, NEET
An N-P-N transistor in a common emitter mode is used as a simple voltage amplifier with a collector current of 4 mA. The terminal of a 8 V battery is connected to the collector through a load resistance RL and to the base through a resistance RB. The collector-emitter voltage VCE=4V base-emitter voltage VBE=0.6 V and base current amplification factor Î²d.c.=100. Find the values of RL and RB.

Options:
A RL=1kÎ©, RB=185 kÎ©
B RL=2kÎ©, RB=150 kÎ©
C RL=1 kÎ©, RB=240 kÎ©
D RL=3kÎ©, RB=185kÎ©

The correct answer is:
RL=1kÎ©, RB=185 kÎ©

Solution:
See figure. Potential difference across RL

IcRL=8V-VCE

=8V-4V=4V

Now ICRL=4V

RL=44Ã10-3=103Î©=1kÎ©

Further, for base-emitter equation,

VCC=IBRB+VBE

or IBRB=Potential difference across RB

=VCC-VBE=8-0.6=7.4V

Again IB=ICÎ²=4Ã10-3100=4Ã10-5A

RB=7.44Ã10-5=1.85Ã105Î©=185 kÎ©

Topic: Physics, Semiconductors, NEET
An n-p-n transistor conducts when

Options:
A both collector and emitter are negative with respect to the base potential
B both collector and emitter are positive with respect to the base potential
C collector is positive and emitter is negative with respect to the base potential
D collector is positive and emitter is at same potential as the base potential

The correct answer is:
collector is positive and emitter is negative with respect to the base potential

Solution:
In normal mode of operation of n-p-n transistor, Base-Emitter junction is forward biased and emitter-collector junction is reversed biased.

Topic: Physics, Semiconductors, NEET
An object comprises of a uniform ring of radius R and its uniform chord AB (not necessarily made of the same material) as shown. Which of the following cannot be the centre of mass of the object?

Options:
A R3,R3
B R2,R2
C R4,R4
D none of these

The correct answer is:
R2,R2

Solution:
From figure R2, R2 is the coordinate of the rod only in absence of the ring,
When the ring is also introduced then rcm=mchord rchord+mring rringmchord+mring
Hence coordinate of the center of mass of system rcm=mchord rchordmchord+mring
Hence the value of coordinate will decrease and will not be R2, R2, hence the answer is B

Topic: Physics, Center of Mass Momentum and Collision, NEET
An object falls from a height of 10 m above the ground. After striking the ground it loses $50 \%$ of its kinetic energy. The height upto which the object can rebounce from the ground is:

Options:
A 7.5 m
B 10 m
C 2.5 m
D 5 m

The correct answer is:
5 m

Solution:
K.E. just before striking the ground $K_1=m g h_1=m g(10)$
K.E. just after striking the ground $=K_1-\frac{50}{100} K_1=\frac{50}{100} K_1=\frac{K_1}{2}=K_2$
Now, $K_2=m g h_2$
$\Rightarrow \frac{K_1}{2}=m g h_2$
$\Rightarrow \frac{m g(10)}{2}=m g h_2 \Rightarrow h_2=5 \mathrm{~m}$

Topic: Physics, Work Power Energy, NEET, NEET 2024 (Re-NEET)
An object, initially at rest, explodes into three fragments of equal mass. The momenta of two parts are 2pi^ and pj^ where p is a positive number. The momentum of the third part is

Options:
A Will be of the magnitude 3p inclined at tan-1â¡12 with x -axis
B Will be of the magnitude 5p inclined at tan-1â¡2 with x -axis
C Will be of magnitude 3p inclined at Ï-tan-1â¡12 with x -axis
D Will be of magnitude 5p inclined at Ï-tan-1â¡12 will x -axis

The correct answer is:
Will be of magnitude 5p inclined at π-tan-1⁡12 will x -axis

Solution:
From the figure, the momentum of third part can be represented by vector OA. Using the law of conservation of linear momentum.
Magnitude of OA= 2p2+p2=5p
tanâ¡Î¸=12
Î¸=tan-1â¡12
Therefore, option will be of magnitude 5p inclined at Ï-tan-1â¡12 with x -axis is correct.

Topic: Physics, Center of Mass Momentum and Collision, NEET
An object is approaching a thin convex lens of focal length 0.3 m with a speed of 0.01 m/s. The magnitude of the rate of change of lateral magnification of image when the object is at a distance of 0.4 m from the lens is

Options:
A 0.3
B 0.6
C 0.15
D -0.3

The correct answer is:
0.3

Solution:
Differentiating the lens formula 1v-1u=1f with respect to time, we get
-1v2Â·dÏdt+1u2Â·dudt=0 as f = constant
â´ dÏdt=v2u2Â·dudt
Further, substituting proper values in lens formula, we
have
1v+10Â·4=10Â·3u=-0Â·4m,f=0Â·3m
or, v=1Â·2m
Putting the values in Eq.(i) Magnitude of rate of change of position of image
= 0 Â· 0 9 m / s
Lateral magnification,m=vu
â´ dmdt=uÂ·dvdt-vdudtu2=-0Â·40Â·09-1Â·20Â·010Â·42
= - 0 Â· 3 / s
â´ Magnitude of rate of change of lateral magnification = 0 Â· 3 / s

Topic: Physics, Ray Optics, NEET
An object is dragged along a horizontal surface by the machine which delivers a constant power. How does the distance moved by the object is dependent on time t?

Options:
A t34
B t32
C t14
D t12

The correct answer is:
t32

Solution:
Power =WorkTime⁡=Force×distanceTime=Force×velocity
∴Force×velocity=constant(K⁡)
or ( m⁡ a ) ( a t⁡ ) = K⁡
or a = ( K⁡ m⁡ t⁡ ) 1 / 2
∵ s⁡ = 1 2 a t⁡ 2
∴ s⁡ = 1 2 ( K⁡ m⁡ t⁡ ) 1 / 2 t⁡ 2 = 1 2 ( K⁡ m⁡ ) 1 / 2 t 3 / 2
or s is proportional to t3/2

Topic: Physics, Work Power Energy, NEET
An object is mounted on a wall. Its image of equal size is to be obtained on a parallel wall with the help of a convex lens placed between these walls. The lens is kept at distance $x$ in front of the second wall. The required focal length of the lens will be

Options:
A $\frac{x}{2}$
B $\frac{x}{4}$
C Less than $\frac{x}{4}$
D More than $\frac{x}{4}$ but less than $\frac{x}{2}$

The correct answer is:
$\frac{x}{2}$

Solution:
Height of image of object equal to height of object for convex lens.
If object is placed at $2 f$
$2 f=x$
$f=\frac{x}{2}$

Topic: Physics, Ray Optics, NEET, NEET 2023 (Manipur)
An object is placed asymmetrically between two plane mirrors inclined at an angle of 72°. The number of images formed is

Options:
A 5
B 4
C 2
D Infinite

The correct answer is:
5

Solution:
n=360Â°72Â°=5
Note that 360Î¸ is odd and object lies asymmetrically

Topic: Physics, Ray Optics, NEET
An object is placed at a distance 20 cm from the pole of a convex mirror of focal length 20 cm. The image is produced at

Options:
A 13.3 cm
B 20 cm
C 25 cm
D 10 cm

The correct answer is:
10 cm

Solution:
u = -20 cm, f= 20 cm
From mirror formula,
1f=1v+1u
120=1v+1-20
1v=120+120
1v=220âv = 10 cm

Topic: Physics, Ray Optics, NEET
An object is placed at a distance of 20 cm from the pole of a concave mirror of focal length 10 cm. The distance of the image formed is

Options:
A + 20 cm
B +Â 10Â cm
C - 20 cm
D -Â 10Â cm

The correct answer is:
- 20 cm

Solution:
Given, f=-10 cm, u=-20 cm
1f=1u+1v
1v=1f-1u=1-10+120=-120
v=-20 cm

Topic: Physics, Ray Optics, NEET
An object is placed at a distance of 40 cm from a concave mirror of focal length 15 cm. If the object is displaced through a distance of 20 cm towards the mirror, the displacement of the image will be

Options:
A 30 cm towards the mirror
B 36 cm away from the mirror
C 30 cm away from the mirror
D 36 cm towards the mirror

The correct answer is:
36 cm away from the mirror

Solution:
By mirror formula,v1=ufu-f=-40-15-40+15=600-25=-24 cm
When object is displaced by 20 cm towards mirror v2=ufu-f=-20 -15-20+15=-60 cm
Displacement of image =v2-v1=-36 cm
= 36 cm away from the mirror

Topic: Physics, Ray Optics, NEET, NEET 2018
An object is placed at a distance of f2from a convex lens of focal length f. The image will be

Options:
A At one of the foci, virtual and double its size
B Is greater than 1.5 but less than 2.0
C At 2f,virtual and erect
D None of the above

The correct answer is:
At one of the foci, virtual and double its size

Solution:
1v-1-f/2=1f
1v=1f-2f=-1f
or v=-f
Again, m=vu=-f-f/2=2
Clearly, the image is virtual and double the size

Topic: Physics, Ray Optics, NEET
An object is placed on the principal axis of a concave mirror at a distance of 1.5f (f is the focal length). The image will be at,

Options:
A -3 f
B 1.5 f
C -1.5 f
D 3 f

The correct answer is:
-3 f

Solution:
Given, u=-1.5f, now use the mirror formula for the concave mirror, in terms of distance between object and pole, the distance between image and pole and focal length,

1u+1v=1f⇒1-1.5f+1v=1-f

⇒1v=-1f+11.5f⇒1v=-1.5+11.5f=-0.51.5f

⇒v=-3f.

Topic: Physics, Ray Optics, NEET, NEET 2020 (Phase 2)
An object kept in a large room having air temperature of $25^{\circ} \mathrm{C}$ takes 12 minutes to cool from $80^{\circ} \mathrm{C}$ to $70^{\circ} \mathrm{C}$.
The time taken to cool for the same object from $70^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$ would be nearly

Options:
A $10 \mathrm{~min}$
B 12 min
C 20 min
D $15 \mathrm{~min}$

The correct answer is:
$15 \mathrm{~min}$

Solution:
Key Idea From Newton's law of cooling, the time taken $(t)$ by a body to cool from $T_1$ to $T_2$ when placed in a medium of temperature $T_0$ can be calculated from relation
$$
\frac{T_1-T_2}{t}=\frac{1}{K}\left(\frac{T_1+T_2}{2}-T_0\right)
$$
When the object cool from $80^{\circ} \mathrm{C}$ to $70^{\circ} \mathrm{C}$ in 12 minutes, then from Newton's law of cooling,
$$
\begin{array}{r}
\frac{80-70}{12}=\frac{1}{\mathrm{~K}}\left(\frac{80+70}{2}-25\right) \quad\left[\because \mathrm{T}_0=25^{\circ} \mathrm{C}\right] \\
\frac{5}{6}=\frac{1}{\mathrm{~K}} 50
\end{array}
$$
Similarly, when object cool from $70^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$ we get
$$
\begin{gathered}
\frac{70-60}{\mathrm{t}}=\frac{1}{\mathrm{~K}}\left(\frac{70+60}{2}-25\right) \\
\frac{10}{\mathrm{t}}=\frac{1}{\mathrm{~K}} 40
\end{gathered}
$$
Divide Eq. (i) and (ii), we get
$$
\begin{gathered}
\frac{5}{6} \times \frac{\mathrm{t}}{10}=\frac{50}{40} \\
\Rightarrow \quad \mathrm{t}=\frac{5}{4} \times 12=15 \text { minutes }
\end{gathered}
$$

Topic: Physics, Thermal Properties of Matter, NEET, NEET 2019 (Odisha)
An object moving along horizontal $x$-direction with kinetic energy 10 J is displaced through $x=(3 \hat{i}) \mathrm{m}$ by the force $\vec{F}=(-2 \hat{i}+3 \hat{j}) \mathrm{N}$. The kinetic energy of the object at the end of the displacement $x$ is

Options:
A 10 J
B 16 J
C 4 J
D 6 J

The correct answer is:
4 J

Solution:
Work energy theorem,
$W_{\mathrm{al}}=\Delta \mathrm{K} . \mathrm{E}\left(W_{\mathrm{al}}=\right.$ work done by all forces $)$
$\Rightarrow K_f-K_l=\vec{F} \cdot \Delta \overline{\mathrm{x}}$
$\Rightarrow K_f-10=(-2 \hat{i}+3 \hat{j}) \cdot(3 \hat{i})$
$K_f-10=-6$
$K_{\mathrm{f}}=4 \mathrm{~J}$

Topic: Physics, Work Power Energy, NEET, NEET 2024 (Re-NEET)
An object, moving with a speed of 6.25 m s-1, is decelerated at a rate given by dvdt=-2.5v where v is the instantaneous speed. The time taken by the object, to come to rest would be

Options:
A 2 s

B 4 s
C 8 s
D 1 s

The correct answer is:
2 s

Solution:
dvdt=-2.5 v
â dvv=-2.5dt
â â« 6.25 0 vâ¡ - 1 / 2 d vâ¡ = - 2.5 â« 0 t dt
â - 2.5 tâ¡ 0 tâ¡ = 2 vâ¡ 1 / 2 6.25 0
â t=2 s

Topic: Physics, Motion In One Dimension, NEET
An object O is placed at a distance of 20 cm from a thin Plano-convex lens of focal length 15 cm. The plane surface of the lens is silvered as shown in Fig. The image is formed at a distance of

Options:
A 60 cm to the right of the lens
B 30 cm to the left of the lens
C 24 cm to the right of the lens
D 12 cm to the left of the lens

The correct answer is:
12 cm to the left of the lens

Solution:
The effective focal length of the silvered lens is given by
1F=-2f+1fm=-215+1∞=-215, which gives
F=-152cm. The silvered lens behaves like a concave mirror. Using the spherical mirror formula 1v+1u=1F, we have ∵u=-20 cm and F=-152cm
1v+1-20=2-15
Which gives v=-12 cm. The negative sing indicates that the image is formed to the left of the lens.

Topic: Physics, Ray Optics, NEET
An object O is placed at a height h = 12 cm above water Î¼=43 surface in a beaker. The lower surface of the beaker is polished, as shown in the figure. The location of final image will be

Options:
A 10 cm below water surface
B 10 cm above water surface
C 30 cm below water surface
D 30 cm above water surface

The correct answer is:
30 cm below water surface

Solution:
Consider refraction at water surface, reflection at mirror and then refraction at water surface again.

Topic: Physics, Ray Optics, NEET
An object of mass 1 kg executes simple harmonic oscillations along the x-axis with a frequency of 2Ï Hz. At the position x=1 m, the object has a kinetic energy of 24 J. The amplitude of the oscillation is

Options:
A 32 m
B 2Â m
C 4Â m
D 8Â m

The correct answer is:
2 m

Solution:
K.E.=12mω2A2-x2=24 J
12×1×16A2-1=24
A=2 m

Topic: Physics, Oscillations, NEET
An object of mass 1kg executes simple harmonic oscillations along the x-axis with a frequency of 2πHz . At the position x = 1m , the object has kinetic energy of 24 J. The amplitude of oscillation is :

Options:
A 32m
B 2 m
C 4 m
D 8 m

The correct answer is:
2 m

Solution:
K.E.=12mω2A2-x224 J
12×1×16A2-1=24
A=2 m

Topic: Physics, Oscillations, NEET
An object of mass 3 kg at rest in space suddenly explodes into three parts of the same mass. The momenta of the two parts are 4i^ and 2j^ respectively. Then the energy released in the explosion is

Options:
A 60 J
B 40 J
C 20 J
D 10 J

The correct answer is:
20 J

Solution:
Let vâ be the velocity of the third part. Therefore from conservation of momentum,
0=14i^+12j^+1vâ
â´ vâ=-4i^-2j^
The energy released in the explosion is the sum of the kinetic energy of all three parts,
â´ E=12142+12122+12142+22
E=20 J

Topic: Physics, Center of Mass Momentum and Collision, NEET
An object of mass $3 \mathrm{~kg}$ is at rest. Now a force $F=6 t^2 \hat{i}+4 \hat{j}$ is applied on the object then the velocity of the object at $t=3 \mathrm{~s}$ is:

Options:
A $18 \hat{i}+3 \hat{j}$
B $18 \hat{i}+6 \hat{j}$
C $3 \hat{i}+6 \hat{j}$
D $18 \hat{i}+4 \hat{j}$

The correct answer is:
$18 \hat{i}+6 \hat{j}$

Solution:
From question $\vec{a}=\frac{\vec{F}}{m}=2 t^2 \hat{i}+\frac{4}{3} t \hat{j}$
$\therefore d \vec{V}=\left(2 t^2 \hat{i}+\frac{4}{3} t \hat{j}\right) d t$
Integrated both sides, We have,
$\vec{V}=2\left[\frac{t^3}{3}\right] \hat{i}+\frac{4}{3}\left[\frac{t^2}{2}\right] \hat{j}$
at $t=3 \mathrm{sec}, \vec{V}=\frac{2}{3}(3)^3 i+\frac{4}{6}(3)^2 j$
$=18 \hat{i}+6 \hat{j}$

Topic: Physics, Motion In One Dimension, NEET, NEET 2002
An object of mass 4 kg and another object of mass 1 kg are moving such that they have equal kinetic energies. The ratio of the magnitudes of their linear momenta is

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

The correct answer is:
2:1

Solution:
K.E1 of 4 kg mass = 1 2 (4)â v 1 2

K.E2 of 1 kg mass = 1 2 (1)â v 2 2

Since, K. E 1 =K. E 2 (given)

We have,

1 2 (4)â v 1 2 = 1 2 (1)â v 2 2

âââ v 1 v 2 = 1 4 = 1 2

Now, momentum | P Â¯ | of 4 kg mass =(4)â(1)âkgâ m s

and, momentum | P Â¯ 2 | of 1 kg =(1)â(2)âkgâ m s

â´ââ | P Â¯ 1 | | P Â¯ 2 | = 4 2

= 2 1

Topic: Physics, Work Power Energy, NEET