datavorous/entrance-exam-dataset huggingface dataset:
An electron micrograph of a plant cell shows two different types of plastids. One of them is known to store oils and fats in them; such a plastid is known as:

Options:
A Chloroplast
B Chromoplast
C Elaioplast
D Amyloplast

The correct answer is:
Elaioplast

Solution:
Based on the type of pigments plastids can be classified into chloroplasts, chromoplasts and leucoplasts. The leucoplasts are the colourless plastids of varied shapes and sizes with stored nutrients: Amyloplasts store carbohydrates (starch), e.g., potato; elaioplasts store oils and fats whereas the aleuroplasts store proteins.

Topic: Biology, Cell - The Unit of Life, NEET
An electron moves along the line AB with constant velocity V which lies in the same plane as a circular loop of conducting wire, as shown in the figure. What will be the direction of current induced in the loop?

Options:
A no current will be induced
B induced current will be clockwise
C induced current will be anticlockwise
D the current will change direction as the electron passes by

The correct answer is:
the current will change direction as the electron passes by

Solution:
The current will change direction as the electron passes by
As electron approaches the loop flux linked with the loop first increases and then decrease as electron passes by. Therefore direction of induced current changes.

Topic: Physics, Electromagnetic Induction, NEET
An electron moves in a circular orbit with a uniform speed $v$. It produces a magnetic field $B$ at the centre of the circle. The radius of the circle is proportional to:

Options:
A $\sqrt{\frac{B}{v}}$
B $\frac{B}{v}$
C $\sqrt{\frac{v}{B}}$
D $\frac{v}{B}$

The correct answer is:
$\sqrt{\frac{v}{B}}$

Solution:
The magnetic field \((B)\) at a point induced by a flowing current \((I)\) is given as,
\(B=\frac{\mu_0 I}{2 r}\)
Where \(\mu_0\) is the permeability of free space, and \(r\) is the radius of the circular path.
Substituting for \(I\) and rearranging,
\(\begin{aligned}
& B=\frac{\mu_0\left(\frac{e v}{2 \pi r}\right)}{2 r} \\
\Rightarrow & B=\frac{\mu_0 e v}{4 \pi r^2} \\
\Rightarrow & r^2=\frac{\mu_0 e v}{4 \pi B} \\
\Rightarrow & r^2 \propto \frac{v}{B} \\
\therefore & r \propto \sqrt{\frac{v}{B}}
\end{aligned}\)
Therefore, when an electron moves in a circular orbit with a uniform speed \(v\) and produces a magnetic field \(B\) at the centre of the circle, the radius of the circle is proportional to \(\sqrt{\frac{v}{B}}\).

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2005
An electron moves on a straight line path XY as shown. The abcd is a coil adjacent to the path of electron. What will be the direction of current, if any, induced in the coil?

Options:
A adcb.
B the current will reverse its direction as the electron goes past the coil.
C no current induced.
D abcd.

The correct answer is:
the current will reverse its direction as the electron goes past the coil.

Solution:
When e- comes closer the induced current will be anticlockwise
When e- comes farther induced current will be clockwise.

Topic: Physics, Current Electricity, NEET, NEET 2015 (Phase 2)
An electron moves with a constant velocity inside a charged parallel plate capacitor as shown in the figure. The space between the plates is filled with a constant magnetic field B. The plates are having surface charge densities +Ï and -Ï respectively. The time in which the electron comes out from the capacitor is

Options:
A eÏÎµ0lâ¡B
B Îµ0lâ¡BÏ
C el⁡σε0B
D Îµ0lâ¡B3Ï

The correct answer is:
ε0l⁡Bσ

Solution:
| F → e | = | F → m |
∴ q E = B q v
∴ v = F B
t=l⁡v=l⁡BE=l⁡Bσ/ε0=ε0l⁡Bσ

Topic: Physics, Magnetic Effects of Current, NEET
An electron moving in a circular orbit of radius r makes n rotations per second. The magnetic field produced at the center has magnitude:

Options:
A μ0ne2πr
B Zero
C μ0n2er
D μ0ne2r

The correct answer is:
μ0ne2r

Solution:
If an electron is doing n revolutions per second then we can say the current is,
i=qT=eT we know that T=1n.
⇒i=eT=ne

Now, magnetic field at the center of a circular loop is
B=μ0i2r=μ0ne2r

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2015 (Phase 1)
An electron moving with velocity v along the axis approaches a circular current carrying loop as shown in the figure. The magnitude of magnetic force on electron at this instant is-

Options:
A μ04eviR2x(x2⁡+R2⁡)3/2
B μ0eviR2x(x2⁡+R2⁡)3/2
C μ04πeviR2x(x2⁡+R2⁡)3/2
D none of these

The correct answer is:
none of these

Solution:
Direction of electron velocity is along magnetic field lines. Hence F=0

Topic: Physics, Magnetic Effects of Current, NEET
An electron moving with velocity 'v' is found to have a certain value of de-Broglie wavelength. The velocity to be possessed by the neutron to have the same de-Broglie wavelength is

Options:
A 1 8 4 0 v
B 1840 v
C v 1 8 4 0
D v

The correct answer is:
v 1 8 4 0

Solution:
Î» e = h m e v Î» m = h m n v n
â´ Î» e = Î» n
â´âââââ 1 m e v â=â 1 m n v n
or V n v = m e m n = 9.1Ã 10 â28 1.67Ã 10 â24 = 1 1840
or v n = 1 1 8 4 0 Ã v = v 1 8 4 0

Topic: Chemistry, Structure of Atom, NEET
An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquired as a result of photon emission will be
( $m$ is the mass of the electron, $R$ Rydberg constant and $h$ Planck's constant)

Options:
A $\frac{24 h R}{25 m}$
B $\frac{25 h R}{24 m}$
C $\frac{25 m}{24 h R}$
D $\frac{24 m}{25 h R}$

The correct answer is:
$\frac{24 h R}{25 m}$

Solution:
Here, $E_5-E_1=\frac{h c}{\lambda}$
$\begin{gathered}
\frac{R h c}{25}-R h c=\frac{h c}{\lambda} \\
\frac{24}{25} R=\frac{1}{\lambda}
\end{gathered}$
But
$\begin{aligned}
P & =\frac{h}{\lambda} \\
v & =\frac{h}{m \lambda} \\
& =\frac{24}{25} \frac{R h}{m}
\end{aligned}$

Topic: Physics, Atomic Physics, NEET, NEET 2012 (Screening)
An electron of a velocity x is found to have a certain value of de Broglie wavelength. The velocity to be possessed by the neutron to have the same de Broglie wavelength is
(Given me=9.1×10-28g,mn=1.674×10-24g )

Options:
A x
B x1840
C 1840 x
D None

The correct answer is:
x1840

Solution:
λ electron =λ neutron
hmeVe=hmnVn
So Vn=meVemn=9.1×10-28×x1.674×10-24=x1840.

Topic: Chemistry, Structure of Atom, NEET
An electron of mass m and a photon have same energy E. The ratio of de-Broglie wavelengths associated with them is:
(c being velocity of light)

Options:
A 1cE2m12
B E2m12
C c2mE12
D 1c2mE12

The correct answer is:
1cE2m12

Solution:
De-Broglie wavelength is given by
λe=hp=h2m.E for electron
De-Broglie wavelength of photon is given by
λp=hp=hEc=hcE
λeλp=12mE.Ec=1cE2m

Topic: Physics, Dual Nature of Matter, NEET, NEET 2016 (Phase 1)
An electron of mass m with an initial velocity V→=V0 i^  (V0>0) enters in an electric field E→=-E0i^(E0>0 and is a constant) at t=0. If λ0 is its de-Broglie wavelength initially, then its de-Broglie wavelength at time t is

Options:
A λ0 t
B λ01+eE0mV0t
C λ01+eE0mV0t
D λ0

The correct answer is:
λ01+eE0mV0t

Solution:
V→=V0i^
E→=-E0i^
a→=F→m=qE→m=-e-E0i^m
a→=eE0mi^
V=u+at
V=V0+eE0mt
∵ De Broglie wavelength= λ=hmv and initially λ 0 = h m v 0 .
= hmV0+eE0tm=hmV01+eE0tmV0=λ01+eE0tmV0

Topic: Physics, Dual Nature of Matter, NEET, NEET 2018
An electron (q=1.6Ã10-19 C) is moving at right angles to a uniform magnetic field of 3.534Ã10-5 T The time taken by the electron to complete a circular orbit is

Options:
A 2 Î¼s
B 4 Î¼s
C 3 Î¼s
D 1 Î¼s

The correct answer is:
1 Î¼s

Solution:
For a perpendicular magnetic field, magnetic force is provided by the force. So,
mv2r=qvB
i.e. r=mvqB ...(i)
As in uniform circular motion v=rω so the angular frequency of circular motion will be given by
ω=vr=qBm [Using Eq. (i)]
and hence, the time period
T=2πω=2πmqB ...(ii)
Given, B=3.534×10-5 T
q=1.6×10-19 C, m=9.1×10-31 kg, T=?
From Eq. (ii), we get
T=2×3.14×9.1×10-313.534×10-5×1.6×10-19=1×10-6 s=1 μs

Topic: Physics, Magnetic Effects of Current, NEET
An electron with kinetic energy 10 eV is incident on a hydrogen atom in its ground state. The collision

Options:
A Must be elastic
B May be partially elastic
C Must be completely inelastic
D May be completely inelastic

The correct answer is:
Must be elastic

Solution:
10 eV electron cannot excite a hydrogen atom Hence collision is elastic as minimum energy required to excite is 10.2 eV. so no change in total K.E of system is possible.

Topic: Physics, Atomic Physics, NEET
An element (atom mass =100 g/mol) having BCC structure has its cell edge 400 pm. The density of the element is

Options:
A 2.144 g/cm3
B 5.188 g/cm3
C 7.289 g/cm3
D 10.376 g/cm3

The correct answer is:
5.188 g/cm3

Solution:
Ï=ZÃMwNAÃV=2Ã1006Ã1023Ã(400Ã10-10)3
=5.188 g/cm3

Topic: Chemistry, Solid State, NEET
An element (atomic mass = 100 g/mol) having bcc structure has unit cell edge 400 pm. Then density of the element is

Options:
A 10.376 g/cm3
B 5.188 g/cm3
C 7.289 g/cm3
D 2.144 g/cm3

The correct answer is:
5.188 g/cm3

Solution:
In bcc structure number of atoms (Z = 2)
Density =ZÃMa3ÃNA
=2Ã100(400Ã10-10)3Ã6.02Ã1023
=5.188 g/cm3

Topic: Chemistry, Solid State, NEET
An element has a body centered cubic (bcc) structure with a cell edge of 288 pm. The atomic radius is

Options:
A 24×288 pm
B 43×288 pm
C 32×288 pm
D 34×288 pm

The correct answer is:
34×288 pm

Solution:
Body-centered cubic (BCC) is the name given to a type of atom arrangement found in nature. A body-centered cubic unit cell structure consists of atoms arranged in a cube where each corner of the cube shares an atom and with one atom positioned at the center.
3a=4r (for bcc lattice)
r=34×288 pm

Topic: Chemistry, Solid State, NEET, NEET 2020 (Phase 1)
An element has a body-centred cubic (bcc) structure with a cell edge of 288 pm. The density of the element is 7.2 g/cm³. How many atoms are present in 208 g of the element?

Options:
A 2.41 x 1024atoms
B 1.44 x 1024atoms
C 4.8 x 1024atoms
D 6.72 x 1023atoms

The correct answer is:
2.41 x 1024atoms

Solution:
Correct Option is : (A)
2.41 x 1024atoms

Topic: Chemistry, Solid State, NEET
An element of mass M has Z protons and N neutrons. Masses of proton and neutron are mp and mn respectively. Choose the correct relation among following options.

Options:
A MÂ >ZmpÂ +Â Nmn
B M=Â Â ZmpÂ +Â Nmn
C M <Zmp + Nmn
D M may be greater than, less than or equal to Zmp + Nmn depending on nature of nucleus

The correct answer is:
M <Zmp + Nmn

Solution:
When a nucleus is formed, then the mass of nucleus is slightly less than the sum of the mass of Z protons and N neutrons.
That difference in energy also known as binding energy needs to be supplied to nucleus to break it into respective number of protons and neutrons.
M<(Zmp +Nmn)

Topic: Physics, Nuclear Physics, NEET
An element 'P' has atomic number 56. What will be the formula of its halide?

Options:
A PX
B PX2
C PX3
D P2X3

The correct answer is:
PX2

Solution:
Z=56;[Xe]6s2
Formula of halide is PX2

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
An element playing important role in nitrogen fixation is

Options:
A molybdenum
B copper
C manganese
D zinc

The correct answer is:
molybdenum

Solution:
Molybdenum is absorbed as molybdate by plants. It is involved in nitrogen metabolism including nitrogen fixation. It is a component of enzyme nitrogenase and acts as enzyme activator. Its deficiency causes chlorosis and necrosis, whiptail of cauliflower and premature leaf fall.
Copper is absorbed by the plant in ionic form. Its deficiency causes necrosis, die back in Citrus, reclamation in legumes.
Manganese is absorbed by the plants as bivalent ion. It acts as enzyme activator. Its deficiency causes interveinal chlorosis as well as yellowing of starch and their subsequent degenerate.
Zinc is needed for biosynthsis of IAA and also acts as enzyme activator. Its deficiency causes chlorosis, little leaf,rosette, white bud of maize and mottling of leaves.

Topic: Biology, Mineral Nutrition, NEET, NEET 2010 (Screening)
An element whose IUPAC name is ununtrium (Uut) belongs to

Options:
A s-block element
B p-block element
C d-block element
D Transition element

The correct answer is:
p-block element

Solution:
UFor atomic number = 113
electronic confifugration is
113 [Rn] 5f14,6d10,7s2,7p1
p-block, 13 column (III A).

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
An element with atomic number 20 will be placed in which period of the periodic table ?

Options:
A 1
B 2
C 3
D 4

The correct answer is:
4

Solution:
Ca20= 1s22s22p63s2sp64s2
Period⇒ 4th

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
An element with mass number 81 contains 31.7% more neutrons as compared to protons. What is the atomic symbol?

Options:
A 81Br29
B 42Br81
C 81Br39
D 81Br35

The correct answer is:
81Br35

Solution:
Correct Option is : (D)
81Br35

Topic: Chemistry, Structure of Atom, NEET
An element with molar mass 2.7×10-2kg mol-1forms a cubic unit cell with edge length 405 pm. If its density is 2.7×103kg m-3, what is the nature of the cubic unit cell?

Options:
A Simple Cubic
B Body-Centred Cubic
C Face-Centred Cubic
D None of the above

The correct answer is:
Face-Centred Cubic

Solution:
Correct Option is : (C)
Face-Centred Cubic

Topic: Chemistry, Solid State, NEET
An element, X has the following isotopic composition

200X:90% 199X:8.0% 202X:2.0%

The weighted average atomic mass of the naturally occurring element X is closest to

Options:
A 204 amu
B 198 amu
C 197 amu
D 200 amu

The correct answer is:
200 amu

Solution:
Average isotopic mass of X
=200Ã90+199Ã8+202Ã290+8+2
=18000+1592+404100=199.96 amu
â200 amu

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
An element, $\mathrm{X}$ has the following isotopic composition:
${ }^{200} \mathrm{X}: 90 \% \quad{ }^{199} \mathrm{X}: 8.0 \quad{ }^{202} \mathrm{X}: 2.0 \%$
The weighted average atomic mass of the naturally-occurring element $\mathrm{X}$ is closest to

Options:
A $201 \mathrm{amu}$
B $202 \mathrm{amu}$
C $199 \mathrm{amu}$
D $200 \mathrm{amu}$

The correct answer is:
$200 \mathrm{amu}$

Solution:
Average isotopic mass of
$\begin{aligned}
X & =\frac{200 \times 90+199+8+202 \times 2}{90+8+2} \\
& =\frac{18000+1892+404}{100} \\
& =\frac{19996}{100}=199.96 \mathrm{amu} .
\end{aligned}$

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET, NEET 2007
An element X which occurs in the second short period has an outer electronic structure s2p1. What are the formula and acid-base character of its oxides?

Options:
A XO3, basic
B X2O3, basic
C X2O3, amphoteric
D XO2, acidic

The correct answer is:
X2O3, amphoteric

Solution:
The outer electronic configuration =s2p1
Thus, valency =2+1=3
Therefore, the formula of the oxide is X2O3
X is A l 2 O 3 , its nature is amphoteric

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
An elevator and its load have a total mass of 800 kg. The elevator is originally moving downwards at 10 m s-1, it slows down to stop with constant acceleration in a distance of 25 m. Find the tension T in the supporting cable while the elevator is being brought to rest. (Take g=10 m s-2 )

Options:
A 8000 N

B 1600 N

C 9600 N

D 6400 N

The correct answer is:
9600 N

Solution:
As the elevator is going down with decreasing speed, so acceleration is upward duration, Let it is a
T-800g=800a,
T=800(g+a)
From v2=u2-2as,
â´ a=2 m s-2
â´ T=80010+2,
â´ T=9600 N

Topic: Physics, Laws of Motion, NEET
An elevator is going upward and the variation in the velocity of the elevator is as given in the graph. What is the height to which the elevator takes the passengers?

Options:
A 3.6 m
B 28.8 m
C 36.0 m
D 32.4 m

The correct answer is:
36.0 m

Solution:
Area under graph will give displacement
Height = 1 2 Ã4Ã3.6+3.6Ã8=36âm

Topic: Physics, Motion In One Dimension, NEET
An elevator weighing 6000 kg is pulled upwards by a cable with an acceleration of 5 m s-2. Taking g to be 10 m s-2, the tension in the cable is

Options:
A 6000 N
B 9000 N
C 60000 N
D 90000 N

The correct answer is:
90000 N

Solution:
T=mg+a=6000×10+5=90000 N

Topic: Physics, Laws of Motion, NEET
An EM wave is propagating in a medium with a velocity V→=V i^. The instantaneous oscillating electric field of this em wave is along +y axis. Then the direction of oscillating magnetic field of the EM wave will be along

Options:
A -y direction
B +z direction
C -z direction
D -x direction

The correct answer is:
+z direction

Solution:
V^=E^×B^ , i^=j^×k^
∴ B^=k^ , +z direction

Topic: Physics, Electromagnetic Waves, NEET, NEET 2018
An emf is generated by an ac generator having 100 turn coil, of loop area $1 \mathrm{~m}^2$. The coil rotates at a speed of one revolution per second and placed in a uniform magnetic field of $0.05 \mathrm{~T}$ perpendicular to the axis of rotation of the coil. The maximum value of emf is

Options:
A $62.8 \mathrm{~V}$
B $6.28 \mathrm{~V}$
C $3.14 \mathrm{~V}$
D $31.4 \mathrm{~V}$

The correct answer is:
$31.4 \mathrm{~V}$

Solution:
Maximum value of emf generated by an ac generator is given by
$\begin{aligned} \varepsilon_0 & =\mathrm{NBA} \omega \\ & =100 \times 0.05 \times 1 \times 2 \pi \\ & =31.415 \mathrm{~V} \\ & \approx 31.4 \mathrm{~V}\end{aligned}$

Topic: Physics, Electromagnetic Induction, NEET, NEET 2023 (Manipur)
An E.M.F of 5 V is produced in an inductor, when the current changes at a steady rate from 3A to 2 A in 1 millisecond. The value of self-inductance is

Options:
A zero
B 5Â H
C 5000Â H
D 5 mH

The correct answer is:
5 mH

Solution:
Magnitude of self inductance is
L=edidt
L=5(3-2)10-3=5Ã10-3
L=5 mH

Topic: Physics, Electromagnetic Induction, NEET
An emf of 15 V is applied in a circuit coil containing 5 H inductance and 10 Î© resistance. The ratio of currents at time t=â and t=1 s is

Options:
A e2e2-1
B 1-e-1
C e-1
D none of these

The correct answer is:
e2e2-1

Solution:
Here, i=i0, at t=â

Let i be the current at t=1 s
From i=i01-e-RLt
= i01-e-105Ã1=i01-1e2
â´ i0i= e2e2-1

Topic: Physics, Alternating Current, NEET
An e.m.f. of 15 V is applied in a circuit containing 5 Henry inductance and 10 Î© resistance. The ratio of the current at time t=â and t=1 second is

Options:
A e 1 / 2 e 1 / 2 - 1
B e 2 e 2 - 1
C 1 - e - 1
D e - 1

The correct answer is:
e 2 e 2 - 1

Solution:
I = v R 1 - e - Rt L
I 1 I 2 = 1 - e - Rt 1 L 1 - e - R t 2 L
= 1 - e - 2 t 1 1 - e - 2 t 2
For t 1 =ââââ&ââ t 2 =1,ââ I 1 I 2 = 1 1 - e - 2 = e 2 e 2 - 1

Topic: Physics, Alternating Current, NEET
An energy of 24.6 eV is required to remove one of the electrons from a helium atom. The energy required to remove both electrons from helium atoms is

Options:
A 38.2 eV
B 49.2 eV
C 51.8 eV
D 79 eV

The correct answer is:
79 eV

Solution:
Required energy =âI .E 1 â+âI .E 2
=24.6âeV
after removal of 1 st â e â He + form ( H -like), hence Bohr's model applicable.
I .E 2  = I .E H  ×  Z 2
=13.6Ã4=54.4âeV
So total energy required =24.6+54.4=79âeV

Topic: Chemistry, Structure of Atom, NEET
An energy of $484 \mathrm{~J}$ is spent in increasing the speed of a flywheel from $60 \mathrm{rpm}$ to $360 \mathrm{rpm}$. The moment of inertia of the flywheel is:

Options:
A $0.7 \mathrm{~kg}-\mathrm{m}^2$
B $3.22 \mathrm{~kg}-\mathrm{m}^2$
C $30.8 \mathrm{~kg}-\mathrm{m}^2$
D $0.07 \mathrm{~kg}-\mathrm{m}^2$

The correct answer is:
$0.7 \mathrm{~kg}-\mathrm{m}^2$

Solution:
Given,
$\begin{aligned}
& \omega_1=60 \mathrm{rpm}=60 \times \frac{2 \pi}{60}=2 \pi \mathrm{rad} / \mathrm{s} \\
& \omega_2=360 \mathrm{rpm}=360 \times \frac{2 \pi}{60}=12 \pi \mathrm{rad} / \mathrm{s}
\end{aligned}$
Change in kinetic energy,
$\begin{aligned}
\Delta \mathrm{K} . \mathrm{E} & =\frac{1}{2} \mathrm{I}\left(\omega_2^2-\omega_1^2\right) \\
484 & =\frac{1}{2} \times \mathrm{I}\left(\omega_2+\omega_1\right)\left(\omega_2-\omega_1\right) \\
\Rightarrow \quad \mathrm{I} & =\frac{484 \times 2}{14 \pi \times 10 \pi}=0.7 \mathrm{~kg} \cdot \mathrm{m}^2
\end{aligned}$
(where, the symbols have their usual meanings)

Topic: Physics, Rotational Motion, NEET, NEET 2022 (Phase 2)
An energy source will supply a constant current into the load, if its internal resistance is

Options:
A Equal to the resistance of the load
B Very large as compared to the load resistance
C Zero
D Non-zero but less than the resistance of the load

The correct answer is:
Very large as compared to the load resistance

Solution:
I=ER+r
I≈Er=constant
where, R=external resistance
r= internal resistance such that r>>R

Topic: Physics, Current Electricity, NEET
An engine can pull 4 coaches at a maximum speed of 20 m s-1. Mass of the engine is twice the mass of every coach. Assuming resistive forces are proportional to the weight of train and velocity too, approximate maximum speeds of the engine when it pulls 12 and 6 coaches are

Options:
A 8.5 m s-1 and 15 m s-1 respectively

B 6.5 m s-1 and 8 m s-1 respectively
C 8.5 m s-1 and 13 m s-1 respectively
D 10.5 m s-1 and 15 m s-1 respectively

The correct answer is:
8.5 m s-1 and 15 m s-1 respectively

Solution:
When 4 coaches (m each) are attached with engine (2m)
According to question, P=K·6mgv .......(1)
(constant power), (K being proportionally constant)
Since resistive force is proportional to weight
â
Now if 12 coaches are attached
P=K·14mg·v1 ......(2)
So by equation (1) and (2)
6mgv=14mgv1 ⇒ v1=614×v
=614×20=6×107=607⇒ v1=8.5 m s-1
Similarly for 6 coaches, K·6mgv = K·8mgv2
⇒v2=68×20=34×20=15 m s-1

Topic: Physics, Work Power Energy, NEET
An engine has an efficiency of $1 / 6$. When the temperature of sink is reduced by $62^{\circ} \mathrm{C}$, its efficiency is doubled. Temperatures of the source is:

Options:
A $37^{\circ} \mathrm{C}$
B $62^{\circ} \mathrm{C}$
C $99^{\circ} \mathrm{C}$
D $124^{\circ} \mathrm{C}$

The correct answer is:
$99^{\circ} \mathrm{C}$

Solution:
Efficiency of an engine, \(\quad \eta=1-\frac{T_2}{T_1}\) where \(T_1\) is the temperature of the source and \(T_2\) is the temperature of the sink.
\(\therefore \quad \frac{1}{6}=1-\frac{T_2}{T_1} \quad\) or, \(\quad \frac{T_2}{T_1}=\frac{5}{6}\)
\(\therefore \quad 2 \times \frac{1}{6}=1-\frac{\left(T_2-62\right)}{T_1}\)
or, \(\frac{1}{3}=1-\frac{\left(T_2-62\right)}{T_1}\)
or, \(\frac{2}{3}=\frac{T_2-62}{T_1}\) or, \(2 T_1=3 T_2-186\)
or, \(2 T_1=3\left[\frac{5}{6}\right] T_1-186 \quad\) [using (i)]
\(\therefore \quad\left[\frac{5}{2}-2\right] T_1=186 \quad\) or, \(\quad \frac{T_1}{2}=186\)
or, \(T_1=372 \mathrm{~K}=99^{\circ} \mathrm{C}\).

Topic: Physics, Thermodynamics, NEET, NEET 2007
An engine is attached to a wagon through a shock absorber of length 1.5 m. The entire system with a total mass of 50000 kg was moving with a speed of 36 kmh-1, when the brakes are applied to bring it to rest. In this process, the spring of the shock absorber gets compressed by 1.0 m. If 90â% of the energy of the wagon is lost due to friction, the spring constant is

Options:
A 5.0× 10 5   N/m
B 4.0× 10 5   N/m
C 1.0× 10 5   N/m
D 2.0× 10 5   N/m

The correct answer is:
5.0× 10 5   N/m

Solution:
Mass of the system ( m )=50000ââkg
Speed of the system ( v )=36ââkm/h
= 3 6 Ã 5 1 8 m/s âµ 1 km/h = 5 1 8 m/s
=10ââm/s
Compression of the spring ( x )=1.0ââm
KE of the system = 1 2 mv 2
= 1 2 Ã 5 0 0 0 0 Ã ( 1 0 ) 2
= 25000Ã100ââJ
=2.5Ã 10 6 ââJ
Since, 90% KE of the system is lost due to friction, therefore kinetic energy transferred to shock absorber,
=â10%,ââK E total = 10 100 Ã2.5Ã 10 6 J=2.5Ã 10 5 J
This K.E. is converted in to spring energy
2.5Ã 10 5 = 1 2 k x 2
2.5Ã 10 5 = 1 2 ÃkÃ 1 2
k=5.0Ã 10 5 ââN/m

Topic: Physics, Work Power Energy, NEET
An engine moving towards a wall with a velocity 50 m s-1 emits a note of 1.2 kHz. Speed of sound in air is 350 m s-1. The frequency of the note after reflection from the wall as heard by the driver of the engine is

Options:
A 2.4 kHz
B 0.24 kHz
C 1.6 kHz
D 1.2 kHz

The correct answer is:
1.6 kHz

Solution:
The reflection sound appears to propagate in a direction opposite to that of moving engine. Thus, the source and the observer can be presumed to approach each other with same velocity.
v′=vv+vov-vs
=vv+vsv-vs ∵vo=vs
⟹v′=1.2350+50350-50
=1.2×400300=1.6 kHz

Topic: Physics, Waves and Sound, NEET