datavorous/entrance-exam-dataset huggingface dataset:
A biocontrol agent to be a part of an integrated pest management should be

Options:
A species-specific and symbiotic
B free-living and broad spectrum
C narrow spectrum and symbiotic
D species-specific and inactive on non-target organisms

The correct answer is:
species-specific and inactive on non-target organisms

Solution:
A biocontrol agent to be a part of an Integrated Pest Management (IPM) programme should be species-specific and inactive or have no negative impacts on non-target organisms like plants, mammals, birds, fish and even on other non-target insects. It should kill only targeted insects/pests (organisms).

Topic: Biology, Microbes in Human Welfare, NEET, NEET 2019 (Odisha)
A biologist studied the population of rats in a barn. He found that the average natality was 250 , average mortality 240, immigration 20 and emigration 30. The net increase in population is

Options:
A 10
B 15
C 05
D Zero

The correct answer is:
Zero

Solution:
A population has birth rates and death rates. The rates are expressed as change in numbers (increase or decrease) with respect to members of the population.
In this case, the net increases in population will be
Birth rate $(B)+$ Immigration $([)$ - Death rate $(D)$ + Emmigration (l)

Topic: Biology, Organisms and Populations, NEET, NEET 2013 (All India)
A bird is moving back and forth between two cars on a straight road. The first car travels at a speed of 18 m/h, and the second car travels at a speed of 27 km/h. The bird moves from the first car towards the second at a speed of 36 km/h. When the two cars are separated by 36 km, what is the total distance covered by the bird?

Options:
A 25.2 km
B 28.8 km
C 32.4 km
D 36 km

The correct answer is:
28.8 km

Solution:
Correct Option is : (B)
28.8 km

Topic: Chemistry, Motion In One Dimension, NEET
A bivalent of meiosis-I consists of

Options:
A Two chromatids and one centromere
B Two chromatids and two centromeres
C Four chromatids and two centromeres
D Four chromatids and four centromeres

The correct answer is:
Four chromatids and two centromeres

Solution:
A bivalent of meiosis-I consists of four chromatids and two centromeres.

Topic: Biology, Cell Cycle and Cell Division, NEET
A black body at $227^{\circ} \mathrm{C}$ radiates heat at the rate of $7 \mathrm{cal} \mathrm{cm}^{-2} \mathrm{~s}^{-1}$. At a temperature of $727^{\circ} \mathrm{C}$, the rate of heat radiated in the same units will be

Options:
A 60
B 50
C 112
D 80

The correct answer is:
112

Solution:
Key Idea Apply Stefan's law of radiation.
$$
\begin{aligned}
\mathrm{E} & =\sigma \mathrm{T}^4 \\
\frac{\mathrm{E}_1}{\mathrm{E}_2} & =\left[\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right]^4 \\
\Rightarrow \quad \mathrm{E}_2 & =7\left[\frac{273+727}{273+227}\right]^4 \\
& =\left(\frac{1000}{500}\right)^4 \times 7 \\
& =112 \mathrm{cal}-\mathrm{cm}^{-2} \mathrm{~s}^{-1}
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter, NEET, NEET 2009 (Screening)
A black body at $1227^{\circ} \mathrm{C}$ emits radiations with maximum intensity at a wavelength of $5000 Å$. If the temperature of the body is increased by $1000^{\circ} \mathrm{C}$, the maximum intensity will be observed at:

Options:
A $3000 Å$
B $4000 Å$
C $5000 Å$
D $6000 Å$

The correct answer is:
$3000 Å$

Solution:
By wiens displacement law $\lambda_m$
$$
T=B
$$
UCwe have,
$$
\begin{aligned}
(5000)(1500) & =\lambda_m^{\prime}(1500+1000) \\
\Rightarrow \quad \lambda_m^{\prime} & =\frac{5000 \times 1500}{2500} \\
& =3000 Å .
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter, NEET, NEET 2006
A black body at 1373 â emits maximum energy corresponding to a wavelength of 1.78 microns. The temperature of the moon for which Î»m=14 micron would be

Options:
A 62.6 ℃
B -58.9 ℃
C -63.7 ℃
D 64.2 ℃

The correct answer is:
-63.7 ℃

Solution:
T2T1= λm1λm2=1.7814
T2=1.7814 ×T1=1.78141373+273
= 209.3 K= 209.3 -273=-63.7℃

Topic: Physics, Thermal Properties of Matter, NEET
A black body is at $727^{\circ} \mathrm{C}$. It emits energy at a rate which is proportional to:

Options:
A $(1000)^4$
B $(1000)^2$
C $(727)^4$
D $(727)^2$

The correct answer is:
$(1000)^2$

Solution:
According to Stefan's law Rate of radiated energy $E \propto T^4$ where $T=$ absolute temperature of a black body
$\therefore E \propto(727+273)^4 \text { or } E \propto[1000]^4$

Topic: Physics, Thermal Properties of Matter, NEET, NEET 2007
A black body is at a temperature of 2880 K. The energy of radiation emitted by this object with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. Wien's constant b = 2.88 x 106 nm - K, then

Options:
A U1 = 0
B U2 = 0
C U1 = U2
D U2 > U1

The correct answer is:
U2 > U1

Solution:
From Wien's law, λ m T = constant, where T is the temperature of black body and λ m is the wavelength corresponding to maximum energy of emission. Energy distribution of black body radiation is given below :
i. U1 and U2 are not zero because a black body emits radiations of nearly all wavelengths.
ii. Since U1 corresponding to lower wavelength, U3 corresponds to higher wavelength and U2 corresponds to medium wave length hence U2 > U1.

Topic: Physics, Thermal Properties of Matter, NEET
A black body is at a temperature of 5760 K. The energy of radiation emitted by the body at wavelength 250 nm is U1, at wavelength 500 nm is U2 and that at 1000 nm is U3. Wien's constant, b=2.88×106 nm K. Which of the following is correct?

Options:
A U1=0
B U3=0
C U1>U2
D \(\mathrm{U}_3 < \mathrm{U}_2 > \mathrm{U}_1\)

The correct answer is:
\(\mathrm{U}_3 < \mathrm{U}_2 > \mathrm{U}_1\)

Solution:
Given

Temperature =5760 K

Wien's constant b=2.88×106 nm K

According to Wien's displacement law, maximum amount of emitted radiation corresponding to λm=bT

λm=2.88×106 nm K5760K=500 nm

From the graph, U1<U2>U3

Topic: Physics, Thermal Properties of Matter, NEET, NEET 2016 (Phase 1)
A black hole is an object whose gravitational field is so strong that even light cannot escape from it. To what approximate radius would earth mass=5.98×1024kg have to be compressed to be a black hole?

Options:
A 10-9m
B 10-6m
C 10-2m
D 100 m

The correct answer is:
10-2m

Solution:
According to question the light is unable to escape the gravitational pull of the black hole thus the escape velocity is equal to the speed of the light so,
Escape velocity is Ve=C.
We know that escape velocity is given by,

2GMR=3×108

Putting in the values we get,
⇒2×203×10-116×1024R=3×108

Thus, on solving we get,
R≈9 mm≈10-2m

Topic: Physics, Gravitation, NEET, NEET 2014
A blind person after walking 10 steps in one direction each of length 80 cm, turns randomly to the left or to right by 90Â°. After walking a total of 40 steps, the maximum displacement of the person from its starting point can be

Options:
A Zero
B 82 m
C 162 m
D 32 m

The correct answer is:
162 m

Solution:
This person cannot walk straight more than 10 steps.
Distance covered in 10 steps
=10×0.8=8m
As the person has to turn after every 10 steps, the only way to have maximum displacement is walk as shown in the figure.
The maximum displacement =AE=AC+CE
=82+82=162 m

Topic: Physics, Motion In One Dimension, NEET
A block A kept on an inclined surface just begins to slide if the inclination is 30º. The block is replaced by another block B and it is found that it just begins to slide if the inclination is 40º

Options:
A mass of A > mass of B
B mass of A < mass of B
C mass of A = mass of B
D all the three are possible

The correct answer is:
all the three are possible

Solution:
To slide a block
the angle of friction = angle of repose.
μ=tan⁡θ
This will not depend on the mass of the body.

Topic: Physics, Laws of Motion, NEET
A block A of mass 2m is hanging from a vertical massless spring of spring constant k and is in equilibrium. Another block B of mass m strikes the block A with velocity u and sticks to it as shown in the figure. The magnitude of the acceleration of the combined system of the blocks just after the collision is

Options:
A g2
B g3
C g
D zero

The correct answer is:
g3

Solution:
For block A to be in equilibrium, the spring force acting on it should be equal to its weight. This means

Fspring=2mg

Just after the collision, the combined mass of the system is 3m and the spring force is the same, hence

a=3mg-2mg3m=g3

Topic: Physics, Center of Mass Momentum and Collision, NEET
A block A of mass m1 rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of table and from its other end another block B of mass m2 is suspended. The coefficient of kinetic friction between the block and the table is μk. When the block B is sliding on the table, the tension in string is:

Options:
A m2+Î¼km1gm1+m2
B m2-Î¼km1gm1+m2
C m1m21+Î¼kgm1+m2
D m1m21-Î¼kgm1+m2Â

The correct answer is:
m1m21+Î¼kgm1+m2

Solution:
From the figure,
\(\begin{aligned}
& m_2 g-T=m_2 a \ldots \text { (i) } \\
& T-\mu_k m_1 g=m_1 a \ldots \text { (ii) } \\
& \text {multiply eqn (i) with } \mathrm{m}_1 \text { and eqn (ii) with } \mathrm{m}_2 \\
& m_1 m_2 g-T m_1=m_1 m_2 a \ldots \text { (iii) } \\
& T m_2-\mu_k m_1 m_2 g=m_1 m_2 a \ldots \text { (iv) } \\
& \text{From eqn (iii) and (iv), we get,} \\
& m_1 m_2 g-T m_1=T m_2-\mu_k m_1 m_2 g \\
& m_1 m_2 g\left(1+\mu_k\right)=T\left(m_1+m_2\right) \\
& \therefore T=\frac{m_1 m_2 g\left(1+\mu_k\right)}{m_1+m_2}
\end{aligned}\)

Topic: Physics, Laws of Motion, NEET, NEET 2015 (Phase 1)
A block A starts sliding on a smooth track from a height h as shown in the figure. The track smoothly joins into a conveyor belt which is being driven by a pulley of radius r. If the angular velocity ω of the pulley is such that the block A doesn't slip on the belt, then the value of ω is

Options:
A ghr
B gh2r
C 2ghr
D 2ghr

The correct answer is:
2ghr

Solution:
The speed of the block on reaching the conveyor belt is

v=2gh

There will be no slipping between the block and the belt if both have the same linear velocity.

⇒v=2gh=ωr⇒ω=2ghr

Topic: Physics, Laws of Motion, NEET
A block B is pushed momentarily along a horizontal surface with an initial velocity v. If Î¼ is the coefficient of sliding friction between B and the surface, block B will come to rest after a time

Options:
A vgμ
B gμv
C gv
D vg

The correct answer is:
vgμ

Solution:
Block B will come to rest, it force applied to it will vanish due to frictional force acting between block B and surface ie,
force applied=frictional force
ie, μ mg=ma
or μ mg=mvt
or t=vμ g

Topic: Physics, Laws of Motion, NEET
A block $B$ is pushed momentarily along a horizontal surface with an initial velocity $V$. If $\mu$ is the coefficient of sliding friction between B and the surface, block B will come to rest after a time.

Options:
A $\frac{g \mu}{V}$
B $\frac{g}{V}$
C $\frac{V}{g}$
D $\frac{V}{g \mu}$

The correct answer is:
$\frac{V}{g \mu}$

Solution:
By using $v=u+a t$
$\begin{aligned}
0 & =v-\mu g t \\
\Rightarrow \quad t & =\frac{V}{\mu g}
\end{aligned}$

Topic: Physics, Laws of Motion, NEET, NEET 2007
A block B is pushed momentarily along a horizontal surface with an initial velocity V. If Î¼ is the coefficient of sliding friction between B and the surface, then the block B will come to rest after a time

Options:
A gμV
B gV
C Vg
D Vg μ

The correct answer is:
Vg μ

Solution:
Friction is the retarding force for the block
F=ma=μR=μmg
Therefore, from the first equation of motion
v=u-at
0=V-μg×t ⇒ Vμg=t

Topic: Physics, Laws of Motion, NEET
A block is placed on an inclined plane moving towards right horizontally with an acceleration a0=g. The length of the plane AC=1 m. Friction is absent everywhere. The time taken by the block to reach from C to A is g=10 m/s2

Options:
A 1.2Â s
B 0.74 s
C 2.56Â s
D 0.42Â s

The correct answer is:
0.74 s

Solution:
Drawing a free body diagram of the block with respect to the plane.
Acceleration of the block up the plane is
a=mgcosâ¡30Â°-mgsinâ¡30Â°m=3-12g=3.66 m/s2
Applying s=12at2
t=2sa=2Ã13.66=0.74 s

Topic: Physics, Laws of Motion, NEET
A block mass $m$ is placed on a smooth wedge of inclination $\theta$. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block ( $g$ is acceleration due to gravity) will be:

Options:
A $m g \cos \theta$
B $m g \sin \theta$
C $m g$
D $\mathrm{mg} / \cos \theta$

The correct answer is:
$\mathrm{mg} / \cos \theta$

Solution:
Here $N \cos \theta=m g$
$\Rightarrow N=\frac{m g}{\cos \theta}$

Topic: Physics, Laws of Motion, NEET, NEET 2004
A block moves down a smooth inclined plane of inclination Î¸. Its velocity on reaching the bottom is v. If it slides down a rough inclined plane of same inclination, its velocity on reaching the bottom is v / n, where n is a number greater than 1. The coefficient of friction is given by

Options:
A Î¼=tanâ¡Î¸1-1n2
B Î¼=cotâ¡Î¸1-1n2
C μ = tan θ 1 - 1 n 2 1/2
D μ = cot θ 1 - 1 n 2 1/2

The correct answer is:
μ=tan⁡θ1-1n2

Solution:
For smooth, v2=2g sinÎ¸ s . . .(1)
For rough, vn2= 2 g sinÎ¸-Î¼cosÎ¸S. . .(2)
on solving, Î¼=1-1n2tanÎ¸

Topic: Physics, Laws of Motion, NEET
A block of mass 1 kg is placed on a truck which accelerates with acceleration 5 m s-2. The coefficient of static friction between the block and truck is 0.6. The frictional force acting on the block is

Options:
A 5 N
B 6 N
C 5.82 N
D 4.6 N

The correct answer is:
5 N

Solution:
Given, a=5 m s-2, m=1 kg, μ=0.6.

∴ Maximum static frictional force fs=μsmg=0.6×1×10=6 N
Force on a block due to acceleration of truck Fext=ma=1×5=5 N
Hence, Fext<fs
Therefore, force of friction = external force = 5 N

Topic: Physics, Laws of Motion, NEET
A block of mass 1 kg is pushed up a surface inclined to the horizontal at an angle of 30° by a force of 10 N parallel to the inclined surface (Fig. 6.15). The coefficient of friction between the block and the incline is 0.1. If the block is pushed up by 10 m along the incline, what is the work done against gravity?

Options:
A 25 J
B 50 J
C 75 J
D 100 J

The correct answer is:
50 J

Solution:
Correct Option is : (B)
50 J

Topic: Chemistry, Work Power Energy, NEET
A block of mass 1 kg is pushed up a surface inclined to the horizontal at an angle of 30° by a force of 10 N parallel to the inclined surface (Fig. 6.15). The coefficient of friction between the block and the incline is 0.1. If the block is pushed up by 10 m along the incline, what is the increase in potential energy?

Options:
A 50 J
B 100 J
C 25 J
D 75 J

The correct answer is:
50 J

Solution:
Correct Option is : (A)
50 J

Topic: Chemistry, Work Power Energy, NEET
A block of mass 1 kg is pushed up a surface inclined to the horizontal at an angle of 30° by a force of 10 N parallel to the inclined surface (Fig. 6.15). The coefficient of friction between the block and the incline is 0.1. If the block is pushed up by 10 m along the incline, what is the work done against the force of friction?

Options:
A 8.66 J
B 9.81 J
C 10.66 J
D 11.62 J

The correct answer is:
8.66 J

Solution:
Correct Option is : (A)
8.66 J

Topic: Chemistry, Work Power Energy, NEET
A block of mass 1 kg lies on a horizontal surface in a truck. The coefficient of static friction between the block and the surface is 0.6. If the acceleration of the truck is 5âmâsâ2, find the frictional force acting on the block.

Options:
A 5 N
B 2.5 N
C 5.88 N
D 9.8 N

The correct answer is:
5 N

Solution:
The maximum frictional force that can act on the mass of 1 kg will be

=Î¼N=Î¼âmg=0.6âÃâ1âÃ9.8=5.88âN

The truck is accelerating at 5âmâsâ2.

The pseudo force acting on the mass as seen by the observer on the truck is mÃa=1Ã5=5 N. The frictional force will try to oppose the movement of the mass by this force. Therefore the frictional force acting will be 5 N.

Topic: Physics, Laws of Motion, NEET
A block of mass 1 kg lies on a horizontal surface in a truck. The coefficient of static friction between the block and the surface is 0.6. If the acceleration of the truck is 5 m s-2, the frictional force acting on the block is :

Options:
A 5 N
B 6 N
C 10 N
D 15 N

The correct answer is:
5 N

Solution:
Maximum friction = m μg = 0.6 × 1 0 × 1 = 6 N
Pseudo force = ma=5 N
So required friction force is only 5 N although maximum friction force available is 6 N

Topic: Physics, Laws of Motion, NEET
A block of mass 1 kg moving with a speed of 4 m s-1, collides with another block of mass 2 kg which is at rest. The lighter block comes to rest after collision. The loss in the K.E. of the system will be

Options:
A 8Â J
B 4Ã10-7Â J
C 4 J
D 0Â J

The correct answer is:
4 J

Solution:
Let v1 be the initial velocity of 1 kg block and v2 be the final velocity of 2 kg block.
From law of conservation of momentum, we get
1v1=2(v2) ⇒v2=v12
⇒v2=42=2 m s-1 ∵v1=4 m s-1
ΔKinetic energy=12m1v12-12m2v22=121×42-2×22
=12×8=4 J

Topic: Physics, Center of Mass Momentum and Collision, NEET
A block of mass 2 kg is having velocity 45 m s-1 in the positive x-direction at the origin. The only force acting on it is F=3x2-12x N. Its velocity when it is at x=2 m is

Options:
A 8 m s-1
B 4 m s-1
C 1024 m s-1
D 20 m s-1

The correct answer is:
8 m s-1

Solution:
From work energy theorem
W=ÎK
â«02Fdx=12mv2-u2
â«023x2-12xdx=122v2-452
x3-6x202=v2-80
8-24+80=v2
v=8 m s-1

Topic: Physics, Work Power Energy, NEET
A block of mass 2 kg is at the origin at t=0 and is having velocity 45 m s-1 in the positive x-direction. The only force acting on it is a conservative and its potential energy is defind as U=â x3 + 6x2 + 15 (SI units). Its velocity when its acceleration is minimum after t=0 is-

Options:
A 8 m s-1
B 4 m s-1
C 24 m s-1
D 20 m s-1

The correct answer is:
8 m s-1

Solution:
F=–dUdx⇒F=3x2 – 12x
Now F=min. ⇒dFdx=0 ⇒ x=2 m
Ui + Ki=Uf + Kf
15 + 12 × 2 × 80=-23 + 6 × 22 +15 + 12 × 2 × v2
v2=64
v=8 m s-1

Topic: Physics, Work Power Energy, NEET
A block of mass 2 kg is attached to the spring of spring constant 50 N/m. The block is pulled to a distance of 5 cm from its equilibrium position at x=0 on a horizontal frictionless surface and released at t=0 from rest. The expression for its displacement at anytime t is

Options:
A 5sin5t+Ï/2
B sin5t+π/2
C 5sin5t+3π/2
D 5sint+π/2

The correct answer is:
5sin5t+π/2

Solution:
The spring -block system will perform SHM about the mean position with an amplitude 5cm.
Given, spring constant k=50 N/m
m = mass attached = 2 kg
â´ Angular frequency Ï = k mâ¡ = 5 0 2 = 2 5 = 5 rad/s
Assuming the displacement function
y(t) = A sin (Ït + Ï)
where, Ï = initial phase
But given at t = 0, y(t) = +A
â´ y(0) = + A = A sin (Ï Ã 0 + Ï)
or sin Ï = 1 â Ï = Ï 2
&the desired equation is
y t = Aâ¡ sin Ï t + Ï 2
= A cos Ït
Putting A = 5 cm, Ï = 5 rad/s
We get, y(t) = 5 sin (5t + Ï/2)
where, t is in second and y is in centimetre.

Topic: Physics, Oscillations, NEET
A block of mass 2 kg is kept at origin at t=0 and is having velocity 45 m/s in positive x-direction. The only force on it is a conservative and its potential energy is defined as U=-x3+6x2+15 ( SI units). Its velocity when the force acting on it is minimum (after the time t=0 ) is

Options:
A 8 m/s
B 4 m/s
C 1024 m/s
D none of these

The correct answer is:
8 m/s

Solution:
At x=0
K=12Ã2Ã80=80 J and U=15 J
â´ Total energy is,
E=K+U=95 J
Force, F=-dUdx
âF=3x2-12x
For F to be minimum, dFdx=0
âx=2 m
At x=2 m
E=K+U
â95=12Ã2Ãv2+-8+24+15
âv=8 m/s

Topic: Physics, Work Power Energy, NEET
A block of mass 2 kg is kept at origin at t=0 and is having velocity 45 m/s in positive x-direction. The only force acting on it is a conservative and its potential energy is defind as U=â x3 + 6x2 + 15 (SI units). Its velocity when its acceleration is minimum after t=0 is-

Options:
A 8 m/s
B 4Â m/s
C 1024Â m/s
D 20Â m/s

The correct answer is:
8 m/s

Solution:
F=âdUdxâF=3x2 â 12x
Now F=min. âdFdx=0 â x=2 m
Ui + Ki=Uf + Kf
15 + 12 Ã 2 Ã 80=[-23 + 6 Ã 22 +15] + 12 Ã 2 Ã v2
v2=64
v=8 m/sec

Topic: Physics, Work Power Energy, NEET
A block of mass $2 \mathrm{~kg}$ is placed on an inclined rough surface $\mathrm{AC}$ (as shown in figure) of coefficient of friction $\mu$. If $g=10 \mathrm{~ms}^{-2}$, the net force (in $\mathrm{N}$ ) on the block will be

Options:
A $10$
B $20$
C $10 \sqrt{3}$
D Zero

The correct answer is:
Zero

Solution:
$\begin{aligned} \Rightarrow\left(F_{\text {net }}\right) \mathrm{H} & =m g \sin 30^{\circ}-\mu m g \cos 30^{\circ} \\ & =\frac{m g}{2}-\frac{1}{\sqrt{3}} m g \cdot \frac{\sqrt{3}}{2}\end{aligned}$
$F_{\text {net }}=0$

Topic: Physics, Laws of Motion, NEET, NEET 2023 (Manipur)
A block of mass 2 kg slides on an inclined plane which makes an angle of 30Â° with the horizontal. The coefficient of friction between the block and the surface is32. What force along the plane should be applied to the block so that it moves down?

Options:
A 11.21
B 4.5
C 6.7
D 30.14

The correct answer is:
11.21

Solution:
mg sin θ=(2)(10)(12) =10 N = F1 (say)

μmg cos θ= (32) (2)(10) (32) =21. 21 N =F2 (say)
Force required to move the block down the plane with constant velocity.
F1 will be acting downwards, while F2 upwards.
Since F2 > F1, force required

F=F2-F1=11. 21 N

Topic: Physics, Laws of Motion, NEET
A block of mass 4 kg is kept on a rough horizontal surface. The coefficient of static friction is 0.8. If a force of 19 N is applied on the block parallel to the floor, then the force of friction between the block and floor is

Options:
A 32 N
B 18 N
C 19 N
D 9.8 N

The correct answer is:
19 N

Solution:
f=μmg=0.8×4×10=32 N
Applied force F<f therefore, the force acting on the block will be 19 N.

Topic: Physics, Laws of Motion, NEET
A block of mass 10 kg is in contact against the inner wall of a hollow cylindrical drum of radius 1 m. The coefficient of friction between the block and the inner wall of the cylinder is 0.1. The minimum angular velocity needed for the cylinder to keep the block stationary when the cylinder is vertical and rotating about its axis will be g=10  m s-2,

Options:
A 10 rad s-1
B 102π rad s-1
C 10  rad s-1
D 10 π  rad s-1

The correct answer is:
10  rad s-1

Solution:
âââââââ
N=mω2R
f=mg
For minimum angular velocity,
f=μN
mg=μ·mω2R
ω=gμR=100.1×1=10 rad s-1

Topic: Physics, Laws of Motion, NEET, NEET 2019 (All India)
A block of mass 10 kg is in contact with a cart. If the coefficient of static friction is Î¼=0.5, then the minimum acceleration of cart, that will prevent falling of block, is

Options:
A 5 m s-2
B 10 m s-2
C 20 m s-2
D 15 m s-2

The correct answer is:
20 m s-2

Solution:
FBD of block
mg=Fs
mg=μN
mg=μMa
a=gμ=100.5
=20 m s-2

Topic: Physics, Laws of Motion, NEET
A block of mass 10 kg, moving in x direction with a constant speed of 10 m s-1 , is subjected to a retarding force F=(0.1)x J m-1 during its travel from x=20 m to 30 m . Its final kinetic energy will be:

Options:
A 475 J
B 450 J
C 275 J
D 250 J

The correct answer is:
475 J

Solution:
m=10 kg, u=10 m s-1
Final KE will be given by
KEf=KEi+W
=12mu2+∫2030-0.1xdx
=12×10×100+-0.1x222030
=500+-0.12900-400
KEf=500+-0.15002=475 J

Topic: Physics, Work Power Energy, NEET, NEET 2015 (Phase 1)
A block of mass $10 \mathrm{~kg}$ placed on the rough horizontal surface having coefficient of friction $\mu=0.5$, if a horizontal force of $100 \mathrm{~N}$ acting on it then acceleration of the block will be:

Options:
A $10 \mathrm{~ms}^{-1}$
B $5 \mathrm{~ms}^{-2}$
C $15 \mathrm{~ms}^{-2}$
D $0.5 \mathrm{~ms}^{-2}$

The correct answer is:
$5 \mathrm{~ms}^{-2}$

Solution:
Here $f_{\max }=\mu N=\mu m g$ $=0.5 \times 10 \times 10=50 \mathrm{~N}$ $\Rightarrow a=\frac{\text { net force }}{\text { mass }}$ $=\frac{100-50}{10}=5 \mathrm{~m} / \mathrm{sec}^2$

Topic: Physics, Laws of Motion, NEET, NEET 2002
A block of mass 70 kg is kept on a rough horizontal surface and coefficient of static friction between block and surface is 0.4. A man is trying to pull the block by applying a horizontal force. If the surface exerts a force F on the block then

Options:
A F must be 700 N
B F must be 280 N
C 700 N â¤Fâ¤754 N
D F maybe greater than 754 N

The correct answer is:
700 N â¤Fâ¤754 N

Solution:
Normal reaction,
N=mg=700 N
Limiting friction,
fL=μmg=0.4700
fL=280 N
F=N2+f2
When f=0, Fmin=N
When f=fL,Fmin=700 N
Fmax=7002+2802
Fmax=754 N
∴ 700 N⁡≤F⁡≤754 N

Topic: Physics, Laws of Motion, NEET