datavorous/entrance-exam-dataset huggingface dataset:
A tuning fork of frequency $512 \mathrm{~Hz}$ makes 4 beats/seconds with the vibrating string of a piano. The beat frequency decreases to 2 beats/s when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was

Options:
A $510 \mathrm{~Hz}$
B $514 \mathrm{~Hz}$
C $516 \mathrm{~Hz}$
D $508 \mathrm{~Hz}$

The correct answer is:
$508 \mathrm{~Hz}$

Solution:
Suppose $n_p=$ frequency of piano $=$ ?
$\left(n_p \propto \sqrt{T}\right)$
$\mathrm{n}_{\mathrm{f}}=$ frequency of tuning fork $=512 \mathrm{~Hz}$
$\mathrm{x}$ = Beat frequency $=4$ beats $/ \mathrm{s}$, which is decreasing $(4 \rightarrow 2)$ after changing the tension of piano wire. Also, tension of piano wire is increasing so $n_p \uparrow$
Hence, $\mathrm{n}_{\mathrm{p}} \uparrow-\mathrm{n}_{\mathrm{f}}=\mathrm{x} \downarrow \longrightarrow$ wrong $\mathrm{n}_{\mathrm{f}}-\mathrm{n}_{\mathrm{p}} \uparrow=\mathrm{x} \downarrow \longrightarrow$ correjct $\Rightarrow \mathrm{n}_{\mathrm{p}}=\mathrm{n}_{\mathrm{f}}-\mathrm{x}=512-4=508 \mathrm{~Hz}$

Topic: Physics, Waves and Sound, NEET, NEET 2010 (Screening)
A tuning fork of frequency n is held near the open end of a tube, which is closed at the other end, and the length of the tube is adjusted until resonance occurs. If the two shortest lengths to produce resonance are L1 and L2, then the speed of the sound is

Options:
A n(L2-L1)
B nL2-L12
C 4n(L2-L1)
D 2n(L2-L1)

The correct answer is:
2n(L2-L1)

Solution:
n= v4L1 and n=3v4L2
â´v=4nL1 and 3v=4nL2
or 2v=4nL2-L1
or v=2nL2-L1

Topic: Physics, Waves and Sound, NEET
A tuning fork of known frequency 256 Hz makes 5 beats s-1 with the vibrating string of a piano. The beat frequency decreases to 2 beats s-1 when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was

Options:
A 258 Hz
B 254 Hz
C 251 Hz
D 261 Hz

The correct answer is:
251 Hz

Solution:
ν1=256 Hz
For tuning for ν2-ν1=±5,
ν2=frequency of piano
ν2=256+5 Hz or 256-5 Hz

ν2=261 Hz or 251 Hz
When tension is increased, the beat frequency decreases to 2 beats s-1.
If we assume that the frequency of the piano string is 261 Hz, then on increasing tension, frequency, more than 261 Hz. But it is given that beat frequency decreases to 2, therefore, 261 is not possible.
Hence, 251 Hz was the frequency of the piano string before increasing tension.

Topic: Physics, Waves and Sound, NEET
A tuning fork of known frequency 256 Hz makes 5 beats/s with the vibrating string of a piano. The beat frequency decreases to 2 beats/s when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was

Options:
A (256+2)Hz
B (256-2)Hz
C (256-5)Hz
D (256+5)Hz

The correct answer is:
(256-5)Hz

Solution:
v1=256 Hz
For tuning for v2-v1=Â±5,
v2=frequency of piano
v2=256+5Hz or 256-5Hz
When tension is increased, the beat frequency decreases to 2 beats/s.
If we assume that the frequency of the piano string is 261 Hz, then on increasing tension, frequency, more than 261 Hz. But it is given that beat frequency decreases to 2, therefore, 261 is not possible.
Hence, 251 Hz i.e., 256-5 was the frequency of the piano string before increasing tension.

Topic: Physics, Waves and Sound, NEET
A tuning fork vibrating with a frequency of 512 Hz is kept close to the open end of a tube filled with water (Fig. 15.4). The water level in the tube is gradually lowered. When the water level is 17 cm below the open end, what is the speed of sound in air at 0°C, if the room temperature is 20°C?

Options:
A 336 m/s
B 330 m/s
C 340 m/s
D 350 m/s

The correct answer is:
336 m/s

Solution:
Correct Option is : (A)
336 m/s

Topic: Chemistry, Waves and Sound, NEET
A tuning fork with frequency $800 \mathrm{~Hz}$ produces resonance in a resonance column tube with upper end open and lower end closed by water surface. Successive resonance are observed at length $9.75 \mathrm{~cm}, 31.25 \mathrm{~cm}$ and $52.75 \mathrm{~cm}$. The speed of sound in air is

Options:
A $500 \mathrm{~m} / \mathrm{s}$
B $156 \mathrm{~m} / \mathrm{s}$
C $344 \mathrm{~m} / \mathrm{s}$
D $172 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$344 \mathrm{~m} / \mathrm{s}$

Solution:
For vibrating tunning fork over a resonance tube, the first resonance is obtained at the length
$$
l_1=\frac{\lambda}{4}
$$
and for second resonance,
$$
\mathrm{l}_2=\frac{\lambda}{4}+\frac{\lambda}{2}=\frac{3 \lambda}{4}
$$
From Eq. (i) and (ii), we get
$$
\begin{array}{cc}
& \mathrm{l}_2-\mathrm{l}_1=\frac{3 \lambda}{4}-\frac{\lambda}{4}=\frac{\lambda}{2} \\
\Rightarrow & \lambda=2\left(l_2-l_1\right) \\
\Rightarrow & \mathrm{v}=2 \mathrm{f}\left(\mathrm{l}_2-\mathrm{l}_1\right)
\end{array}
$$
..(iii) $\left(\because \lambda=\frac{\mathrm{v}}{\mathrm{f}}\right)$
Here, $\mathrm{f}=800 \mathrm{~Hz}, \mathrm{l}_1=9.75 \mathrm{~cm}, \mathrm{l}_2=31.25 \mathrm{~cm}$
Substituting the given values in Eq. (iii), we get
$$
\begin{aligned}
& \Rightarrow \mathrm{v}=2 \times 800(31.25-9.75) \\
& =34400 \mathrm{~cm} / \mathrm{s}=344 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$

Topic: Physics, Waves and Sound, NEET, NEET 2019 (Odisha)
A tunnel is dug along the diameter of the earth. A mass m is dropped into it. How much time does it take to cross the earth?

Options:
A 169.2Â min
B 84.6Â min
C 21.2Â min
D 42.3 min

The correct answer is:
42.3 min

Solution:
Time period to cross the earth once along its diameter =T2
=122ÏRg
=ÏRg
Where, R is the radius of the earth
=3.14Ã6.4Ã1069.8=2537.5 s
=42.3âmin

Topic: Physics, Oscillations, NEET
A tunnel is dug through the centre of the Earth. Show that a body of mass ‘m’ when dropped from rest from one end of the tunnel will execute simple harmonic motion.

Options:
A $2\pi \sqrt{\frac{R}{2g}}$
B $2\pi \sqrt{\frac{2R}{g}}$
C $2\pi \sqrt{\frac{R}{g}}$
D $2\pi \sqrt{\frac{3R}{2g}}$

The correct answer is:
$2\pi \sqrt{\frac{R}{g}}$

Solution:
Correct Option is : (C)
$2\pi \sqrt{\frac{R}{g}}$

Topic: Chemistry, Oscillations, NEET
A tunnel is made across the earth of radius R, passing through its centre. A ball is dropped from a height h in the tunnel, height h is very small. The motion will be periodic with time period:

Options:
A 2πRg+4hg
B 2πRg+42hg
C 2πRg+hg
D 2πRg+2hg

The correct answer is:
2πRg+42hg

Solution:
When the ball is in the tunnel at distance x from the centre of the earth, then gravitational force acting on ball is
F=Gmx2Ã43Ïx3Ï=GÃ43ÏÏmx
Mass of the earth, M=43ÏR3Ï
or 43ÏÏ=MR3
â´ F=GMmxR3 ie, Fâx
As this F is directed towards the centre of earth ie, the mean position so the ball will execute periodic motion about the centre of earth
Here inertia factor=mass of ball=m
Spring factor=GMmR3=gmR
â´ time period of oscillation of ball in the tunnel is
T'=2Ïinertia factorspring factor
=2Ïmgm/R
=2ÏRg
Time spent by ball outside the tunnel on both the sides will be = 4 2 h / g
Therefore, total time period of oscillation of ball is
=2ÏRg+42hg

Topic: Physics, Oscillations, NEET
A two-slit Young's interference experiment is done with monochromatic light of wavelength 6000 Aâ. These slits are 2 mm apart. The fringes are observed on a screen placed 10 cm away from the slits. A transparent plate of thickness of 0.5 mm is placed in front of one of the slits. It is found that the interference pattern shifts by 5 mm. What is the refractive index of the transparent plate?

Options:
A 1.2
B 1.5
C 1.8
D 43

The correct answer is:
1.2

Solution:
Shift = Î¼ - 1 tD d
5 Ã 1 0 - 3 = Î¼ - 1 Ã 0.5 Ã 1 0 - 3 Ã 1 0 Ã 1 0 - 2 2 Ã 1 0 - 3
â Î¼ - 1 = 1 5
â Î¼ = 6 5 = 1.2

Topic: Physics, Wave Optics, NEET
A two-slit Young's interference experiment is done with monochromatic light of wavelength 6000 Aâ. These slits are 2 mm apart. The fringes are observed on a screen placed 10 cm away from the slits. A transparent plate of thickness of 0.5 mm is placed in front of one of the slits. It is found that the interference pattern shifts by 5 mm. What is the refractive index of the transparent plate?

Options:
A 1.2
B 1.5
C 1.8
D 4/3

The correct answer is:
1.2

Solution:
Shift = Î¼ - 1 tD d
5 Ã 1 0 - 3 = Î¼ - 1 Ã 0.5 Ã 1 0 - 3 Ã 1 0 Ã 1 0 - 2 2 Ã 1 0 - 3
â Î¼ - 1 = 1 5
â Î¼ = 6 5 = 1.2

Topic: Physics, Wave Optics, NEET
A type of cell called a lymphocyte makes proteins that are exported from the cell. It is possible to track the path of these proteins within the cell by labeling them with radioactive isotopes. Which of the following might be the path of the protein from the site where its polypeptides are made to the lymphocyte's plasma membrane?

Options:
A RER → Golgi body → Plasma membrane
B Golgi body → RER → Plasma membrane
C SER → Lysosome → Plasma membrane
D Nucleus → Golgi body → RER → Plasma membrane

The correct answer is:
RER → Golgi body → Plasma membrane

Solution:
The surface of the rough endoplasmic reticulum is studded with the ribosome. It is located near the nucleus and continues with the outer layer of the nuclear envelope. This RER is very important in the synthesis and packaging of proteins. These proteins are transported through tubules to the smooth endoplasmic reticulum that lies near to the Golgi apparatus and then these proteins diffuse to the deepest layer of the Golgi apparatus with the help of transport vehicles. Then these proteins are transported to the cell membrane or plasma membrane.

Topic: Biology, Cell - The Unit of Life, NEET
A type of natural selection that preserves existing allelic frequencies is called

Options:
A unidirectional selection
B stabilizing selection
C disruptive selection
D bidirectional selection

The correct answer is:
stabilizing selection

Solution:
There are three types of natural selection: Stabilizing or balancing, directional and disruptive. In the stabilizing type the two extremes of a population are not able to survive and only the mean character value survives. In directional change, more individuals acquire value other than the mean character value. In disruption type more individuals acquire peripheral character value at both ends of the distribution curve.

Topic: Biology, Evolution, NEET
A typical angiosperm embryo sac at maturity is:

Options:
A 8-nucleate and 7-celled
B 7-nucleate and 8-celled
C 7 -nucleate and 7- celled
D 8-nucleate and 8-celled

The correct answer is:
8-nucleate and 7-celled

Solution:
The nucleus of the functional megaspore divides mitotically to form two nuclei which move to the opposite poles, forming the 2-nucleate embryo sac. Two more sequential mitotic nuclear divisions result in the formation of the 4-nucleate and later the 8-nucleate stages of the embryo sac.

There is a characteristic distribution of the cells within the embryo sac.

Three cells are grouped together at the micropylar end and constitute the egg apparatus. The egg apparatus, in turn, consists of two synergids and one egg cell.

Three cells are at the chalazal end and are called the antipodals.

The large central cell, as mentioned earlier, has two polar nuclei.

Thus, a typical angiosperm embryo sac, at maturity, though 8-nucleate is 7-celled.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET, NEET 2021
A typical angiosperm embryo sac contains 8 nucleus and 7 cells. These 8 nucleus are formed by ___________

Options:
A 4 mitotic divisions
B 7 mitotic divisions
C 5 meiotic divisions
D 4 meiotic divisions

The correct answer is:
7 mitotic divisions

Solution:
Number of mitosis = Number of Daughter Nucleus -1. So if there is8 Nucleus then number of Mitosis = (8-1) =7mitosis. It means total 7 mitosis required to produce 7 celled and 8 nucleated embryo sac.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
A typical anther is

Options:
A unilobed
B bilobed
C trilobed
D tetralobed

The correct answer is:
bilobed

Solution:
A typical anther consists of two lobes, so it is called as ditheceous anther.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
A typical fat molecule is made up of

Options:
A Three glycerol molecules and one fatty acid molecules.
B One glycerol and three fatty acid molecules.
C One glycerol and one fatty acid molecule.
D Three glycerol and three fatty acid molecules.

The correct answer is:
One glycerol and three fatty acid molecules.

Solution:
Fat is a triglyceride which is made up of 3 molecules of fatty acids and one molecule of glycerol [tri acyl gleceroid (TAG)].
Fats are stored in the form of TAG.

Topic: Biology, Biomolecules (B), NEET, NEET 2016 (Phase 1)
A typical fat molecule is made up of

Options:
A Three glycerol molecules and one fatty acid molecules
B One glycerol and three fatty acid molecules
C One glycerol and one fatty acid molecule
D Three glycerol and three fatty acid molecules

The correct answer is:
One glycerol and three fatty acid molecules

Solution:
Fat is a triglyceride which is made up of 3 molecules of fatty acids and one molecule of glycerol [tri acyl gleceroid (TAG)].
Fats are stored in the form of TAG.

Topic: Chemistry, Biomolecules, NEET
A typical process of cell division in a human cell approximately lasts for

Options:
A only about an hour
B about 12 hours
C only about 24 hours
D about 6 hours

The correct answer is:
only about an hour

Solution:
In a human eukaryotic cell the whole cell cycle process takes place in about 24 hours. The interphase consumes around 23 hours of the total duration of the cycle. The whole cell division stage is quite fast and short lasting about one hour only.

Topic: Biology, Cell Cycle and Cell Division, NEET
A typical somatic cell of female possess

Options:
A 22 autosomes + one sex chromosome
B 44 autosomes + one sex chromosome
C 22 autosomes + two sex chromosomes
D 44 autosomes + two sex chromosomes

The correct answer is:
44 autosomes + two sex chromosomes

Solution:
Chromosome composition of a normal female = 44 + XX
Chromosome composition of a normal male = 44 + XY

Topic: Biology, Principles of Inheritance and Variation, NEET
A U - shaped wire is dipped in a soap solution and removed. The thin soap film formed between the wire and light slider supports a weight of 1.5Ã10-2 N. The length of the slider is 30 cm. What is the surface tension of the film?

Options:
A 3×10-3 N m-1
B 2×10-5 N m-1
C 4Ã10-4 N m-1
D 2.5×10-2 N m-1

The correct answer is:
2.5×10-2 N m-1

Solution:
A soap film has two free surfaces, so the total length of the film to be supported,
l=2×30 cm=0.60 m
Let the surface tension of the film
If the total on the slider due to surface tension,
then f=T×2l=T×0.6 N
W=1.5×10-2 N
In equilibrium position, the force f on the slider due to surface tension must be balanced by the weight w supported by the slider
i.e., f=w=mg
T×0.6=1.5×10-2
T=1.5×10-20.6
T=2.5×10-2 N m-1

Topic: Physics, Mechanical Properties of Fluids, NEET
A U tube with both ends open to the atmosphere is partially filled with water. Oil, which is immiscible with water, is poured into one side until it stands at a distance of 10 mm above the water level on the other side. Meanwhile, the water rises by 65 mm from its original level (see diagram). The density of the oil is____ρw=1000 kg m-3

Options:
A 650 kg m-3
B 425Â kgÂ m-3
C 800Â kgÂ m-3
D 928 kg m-3

The correct answer is:
928 kg m-3

Solution:
∵ Pressure in a stationary liquid at the same level is the same everywhere. As both ends of the tube are open, the pressure on both free surfaces should be equal.
hoil ρoil g=hwater ρwaterg
140×ρoil=130×ρwater
ρoil=928 kg m-3.

Topic: Physics, Mechanical Properties of Fluids, NEET, NEET 2017
A uniform bar of length 6a and mass 8m lies on a smooth horizontal table. Two point masses m and 2m moving in the same horizontal plane with speeds 2v and v respectively, strike the bar and stick to the bar after collision. Then the angular velocity of the bar (about the centre of mass) is

Options:
A zero
B 2v3a
C v5a
D 3v5a

The correct answer is:
v5a

Solution:
From conservation of angular momentum about point C
Linitial2m+Linitialm=Lfinalsystem
m2v 2a+2mva=1128m6a2+m2a2+2ma2Ï
6 mva=30 ma2Ï
Ï=v5a

Topic: Physics, Rotational Motion, NEET
A uniform bar of mass m is supported by a pivot at its top about which the bar can swing like a pendulum. If a force F is applied perpendicular to the lower end of the bar as shown in figure, what is the value of F in order to hold the bar in equilibrium at an angle θ from the vertical

Options:
A 2mgsin⁡θ
B mgsin⁡θ
C mg2sin⁡θ
D mg2cos⁡θ

The correct answer is:
mg2sin⁡θ

Solution:
Let the length of bar is l , then according to the problem.
For equilibrium of the bar,
Fextnet=0 and Ïextnet=0
taking torque about pivot O,
OBÃMg+OAÃF=0
â -OB Mgsin Î¸+OA.Fsin 90o=0
(clockwise) (anti-clockwise)
â´ lF=l2Mgsin Î¸
â F=Mgsin Î¸2

Topic: Physics, Rotational Motion, NEET
A uniform but time-varying magnetic field B(t) exists in a circular region of radius a and is directed into the plane of the paper as shown. The magnitude of the induced electric field at point P (outside the circular region) at a distance r from the centre of the circular region
â

Options:
A Is zero
B Decreases as 1r
C Increases as r
D Decreases as 1r2

The correct answer is:
Decreases as 1r

Solution:
∫E→.dl→=|dϕdt|=A|dBdt|
⇒E2πr=πa2dBdt for r≥a
Therefore, E=a22rdBdt
Therefore, the Induced electric field ∝ 1 r

For r≤a

E( 2πr )=π r 2 | dB dt |
or
E=r2dBdt or E ∝ r
At r = a, E=a2dBdt
Therefore, the variation of E with r (distance from centre) will be as follows:

Topic: Physics, Electromagnetic Induction, NEET
A uniform but time-varying magnetic field exists in the cylindrical region and directed into the paper. If field decrease with time and a positive charge placed at any point inside the region, then it moves

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

The correct answer is:
along 4

Solution:
â«E.dl=-Ïr2dBdt
EÃ2Ïr=+Ïr2ÃdBdt (as dBdt=-ve)
E=+ ve
â´ assume direction is right
so q charge move along 4

Topic: Physics, Electrostatics, NEET
A uniform chain of length 2 m is kept on a table such that a length of 60 cm hangs freely from the edge of the table. The total mass of the chain is 4 kg. The work done in pulling the entire chain on the table is

Options:
A 7.2 J

B 3.6 J

C 120 J

D 1200 J

The correct answer is:
3.6 J

Solution:
Mass per unit length=ML
=42=2 kg m-1
The mass of 0.6 m of chain
=0.6Ã2=1.2 kg
â´Center of mass of hanging part
h=0.6+02=0.3 m
Hence, work done in pulling the chain on the table =work done against the gravity force
W=mgh=1.2Ã10Ã0.3=3.6 J

Topic: Physics, Work Power Energy, NEET
A uniform chain of length L and mass M is lying on a smooth table and one-third of its length is hanging vertically down over the edge of the table. If g is acceleration due to gravity, the work required to pull the hanging part on to the table is

Options:
A MgL
B MgL3
C MgL9
D MgL18

The correct answer is:
MgL18

Solution:
Mass of hanging portion is M3 (one-third) and centre of mass c is at a distance h=L6 below the table top.
Therefore, the required work done is,
W=mgh=M3gL6=MgL18

Topic: Physics, Work Power Energy, NEET
A uniform chain of mass 4 kg and length 2 m is kept on table such that 310th of the chain hangs freely from the edge of the table. How much work has to be done in pulling the entire chain on the table?

Options:
A 7.2 J
B 120 J
C 1200 J
D 3.6 J

The correct answer is:
3.6 J

Solution:
The center of mass of the hanging part is at 0.3 m from table
Mass of hanging part =4Ã0.â¡62=1.â¡2kg
â´ Wâ¡ = mâ¡ gâ¡ hâ¡
=1.â¡2Ã10Ã0.â¡3 (centre of hanged part distance from top is taken as several points are located at different distance from top)
=3.â¡6J

Topic: Physics, Work Power Energy, NEET
A uniform circular disc of radius 50 cm at rest is free to rotate about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of 2.0 rad s-2. Its net acceleration in m s-2 at the end of 2.0 s is approximately:

Options:
A 8.0
B 7.0
C 6.0
D 3.0

The correct answer is:
8.0

Solution:
At the end of 2ââsec, w=w0+Î±t
w=0+22=4 rad/sec
Particle acceleration towards the center is =ac=rw2
ac=1242=8 m/s 2
Tangential acceleration a t =âÎ±âR
=ââ1ââ m/s 2
a net = a c 2 + a t 2
= 8 2 + 1 2
= 65 â8 (approximately)

Topic: Physics, Rotational Motion, NEET, NEET 2016 (Phase 1)
A uniform circular disc placed on a rough horizontal surface is initially given a velocity v0 and angular velocity ω0 as shown in figure. The disc comes to rest after moving some distance in the direction of motion. Then v0rω0 is:

Options:
A 12
B 2
C 13
D 3

The correct answer is:
12

Solution:
From conservation of angular momentum about a point on the surface,
Initial angular momentum = final angular momentum:
Mv0r-12Mr2ω0=0
Therefore v0rω0=12

Topic: Physics, Rotational Motion, NEET
A uniform circular ring of mass m and radius R is placed freely on a horizontal smooth surface as shown in figure. A particle of mass m is connected to the circumference of the ring with a massless string. The particle is imparted velocity v0 perpendicular to length of string as shown. If T is tension in the string just after the particle imparted velocity, then

Acceleration of point P at this instant, is

Options:
A 4T3m
B Tm
C 2Tm
D 3T2m

The correct answer is:
2Tm

Solution:
Calculate acceleration of each in ground frame.
T=mac
T.R=mR2Î±
T=mRÎ±
aP=(aC+Î±R)
T=macâac=Î±R
ap=2Tm

Topic: Physics, Rotational Motion, NEET
A uniform conducting wire of length 12a and resistance R is wound up as a current-carrying coil in the shape of,

(i) an equilateral triangle of side a.
(ii) a square of side a.
The magnetic dipole moments of the coil in each case respectively are:

Options:
A 3Ia2 and 3Ia2
B 3Ia2 and Ia2
C 3Ia2 and 4Ia2
D 4Ia2 and 3Ia2

The correct answer is:
3Ia2 and 3Ia2

Solution:
Area of equilateral triangle of side a,

A=12×a×3a2=34a2

Magnetic moment of each triangle,

m=IA=34a2I

12 a length of wire will form 4 loops of triangle of side a.

Magnetic moment of triangle,

M1=Nm=4×34a2I=3Ia2

Similarly, magnetic moment of square

M2=Nm=3×Ia2

=3Ia2

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2021
A uniform copper rod of 50 cm length is insulated on the sides and has its ends exposed to ice and steam respectively. If there is a layer of water 1 mm thick at each end, the temperature gradient in Â°C m-1 in the bar is (assume that the thermal conductivity of copper is 400 W m-1 K-1 and water is 0.4 W m-1 K-1 )

Options:
A 60
B 40
C 50
D 65

The correct answer is:
40

Solution:
Given, Kc=400 W M-1 K-1,
Kw=0.4 W M-1 K-1
θ1=100 °C, θ2=0 °C
In steady-state, the flow of heat will be the same throughout the whole system
KwA100-θ10-3=KcAθ-θ′50100
Or 100-θ=KcKw×10-3×θ-θ′0.5
=4000.4×10-3×θ-θ′0.5
100-θ=2θ-θ′
(100−θ) K w A 10 −3 = ( θ ′ −0) K w A 10 −3
100−θ= θ ′
Or 100-θ=4θ-200
⇒ 5θ=300⇒θ=60 °C
∴ θ′=100-θ=100-60=40 °C
Hence, the temperature gradient along the bar
=θ-θ′0.5=60-400.5= 40 °C m-1

Topic: Physics, Thermal Properties of Matter, NEET
A uniform copper wire of length 1 m and cross-section area 5Ã10-7 m2 carries a current of 1 A . Assuming that there are 8Ã1028 free electrons per m3 in copper, how long will an electron take to drift from one end of the wire to the other?

Options:
A 0.8Ã103 s
B 1.6Ã103 s
C 3.2Ã103 s
D 6.4Ã103 s

The correct answer is:
6.4Ã103 s

Solution:
Current in wire i=Anevd
Here,
A=5×10-7 m2
∴1=8×1028×1.6×10-19×5×10-7×vd
or vd=18×1028×1.6×10-19×5×10-7
Now,
t=1vd=8×1028×1.6×10-19×5×10-7
=64×102=6.4×103 s

Topic: Physics, Current Electricity, NEET
A uniform cube of mass m and edge a moves on a horizontal surface along the positive x-axis, with an initial velocity v0

Options:
A During motion, N>mg
B During motion, normal reaction acts on the centre of mass
C During motion, the normal reaction shifts towards positive x-axis from the centre of mass
D During motion, normal reaction shifts in the direction of the forces of friction

The correct answer is:
During motion, the normal reaction shifts towards positive x-axis from the centre of mass

Solution:
In this case, the force of friction has a tendency to rotate the cube in clockwise direction. So normal reaction will shift in forward direction to balance the torque due to force of friction.
Hence, (During motion, the normal reaction shifts towards positive x-axis from the centre of mass) is correct.

Topic: Physics, Rotational Motion, NEET