datavorous/entrance-exam-dataset huggingface dataset:
The false statement regarding alkane is

Options:
A this does not perform polymerization reaction
B this does not gives elimination reaction
C it does not disappear the colour of dilute $\mathrm{KMnO}_{4}$ solution
D it does not decolourise bromine water

The correct answer is:
this does not gives elimination reaction

Solution:
Alkanes show elimination reactions. Higher alkanes eliminate hydrogen and lower alkanes on heating in the absence of air at $500-700^{\circ} \mathrm{C}$.

Topic: Chemistry, Hydrocarbons, JEE Main
The family containing mustard and its main characters are

Options:
A Brassicaceae - Tetramerous flowers, six stamens, bicarpellary gynoecium, siliqua type fruit
B Brassicaceae - Pentamerous flowers, many stamens, pentacarpellary gynoecium, capsule type fruit
C Solanaceae - Pentamerous flowers, five stamens, bicarpellary gynoecium, berry type fruit
D Poaceae - Trimerous flowers, three stamens, monocarpellary gynoecium, caryopsis type of fruit.

The correct answer is:
Brassicaceae - Tetramerous flowers, six stamens, bicarpellary gynoecium, siliqua type fruit

Solution:
Brassica (mustard) belongs to family Brassicaceae. Family brassicaceae is characterised by polypetalous 4 petals, polysepalous 4 sepals, six stamens (which are in tetradynamous condition i.e. 4 +2 or in two whorls, 4 inner long and 2 outer short). Gynoecium is bicarpellary, syncarpous unilocular with parietal placentation. Fruit is usually siliqua (dry dehiscent fruit formed by bicarpellary syncarpous gynoecium with replum) and rarely lomentum.

Topic: Biology, Morphology of Flowering Plants, AIIMS, AIIMS 2005
The family containing mustard and its main characters are

Options:
A Brassicaceae - Tetramerous flowers, six stamens, bicarpellary gynoecium, siliqua type fruit
B Brassicaceae - Pentamerous flowers, many stamens, pentacarpellary gynoecium, capsule type fruit
C Solanaceae - Pentamerous flowers, five stamens, bicarpellary gynoecium, berry type fruit
D Liliaceae - Actinomorphic flowers, six stamens, tricarpellary gynoecium, capsule type of fruit

The correct answer is:
Brassicaceae - Tetramerous flowers, six stamens, bicarpellary gynoecium, siliqua type fruit

Solution:
Mustard (Brassica campestris) is a member of Brassicaceae family which consists of tetramerous flower having tetradynamous stamens (2 + 4), bicarpellary syncarpous gynoecium with false septum called replum and the fruit is siliqua.

Topic: Biology, Morphology of Flowering Plants, NEET
The fatty acid which shows reducing property is

Options:
A Acetic acid
B Ethanoic Acid
C Oxalic acid
D Formic acid

The correct answer is:
Formic acid

Solution:
Formic acid, $\mathrm{HCOOH}$ shows reducing property.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The feeding of cattle should be carried out in a scientific manner with special emphasis on

Options:
A Quality of fodder.
B Quantity of fodder.
C Both the quality and quantity of fodder.
D Neither the quality nor the quantity of fodder.

The correct answer is:
Both the quality and quantity of fodder.

Solution:
In dairy farm management:
a. Processes and systems are involved that increase milk yield and improve quality.
b. Proper care is given to the cattle: well housed, well-fed (A. Clean water, B. Fodder with quality and quantity), and maintaining a disease-free environment.
c. Cleanliness and hygiene of the handler and animal are also maintained while milking storage and transport of the milk and its products.
d. Regular inspections and visits by a veterinary doctor are mandatory.

Topic: Biology, Strategies for Enhancement in Food Production, NEET
The fifth term of the H.P.,$2,2 \frac{1}{2}, 3 \frac{1}{3} \ldots\ldots\ldots$ will be in A.P.

Options:
A $5 \frac{1}{5}$
B $3 \frac{1}{5}$
C $\frac{1}{10}$
D $10$

The correct answer is:
$10$

Solution:
Series, $2,2 \frac{1}{2}, 3 \frac{1}{3}, \ldots \ldots$ are in H.P.
$\Rightarrow$ $\frac{1}{2}, \frac{2}{5}, \frac{3}{10}, \ldots \ldots$will be in A.P.
Now first term $a=\frac{1}{2}$and common difference $d=-\frac{1}{10}$
So, $5^{\text {th }}$term of the A.P. $=\frac{1}{2}+(5-1)\left(-\frac{1}{10}\right)=\frac{1}{10}$
Hence $5^{\text {th }}$ term in H.P. is $10$

Topic: Mathematics, Sequences and Series, JEE Main
The figure below shows a battery of emf Îµ connected to an inductor L and a resistance R in series. If the switch is closed at t=0 , then the total charge that flows from the battery in one time constant of the circuit is

Options:
A εLR21-1e
B εL2eR2
C εL4eR2
D ÎµLeR2

The correct answer is:
εLeR2

Solution:
q=∫0τεR(1-e-t/τ)dt

q=εRt-e-t/τ-1τ0τ=εRτ+τe-1-τ
q=εR×1e×LR=εLR2e

Topic: Physics, Alternating Current, JEE Main
The figure below shows a block of mass 15 kg, kept on a rough inclined plane of angle 30Â° and coefficient of static friction equal to 0.5. It is being acted upon by two forces. What should be the minimum value of P (in N) so that the block doesn't slip downwards? [Take g=10 ms-2 and 3=1.7]

Options:

The correct answer is:
16.25

Solution:
F+mgsinθ=P+μ mg cosθ
∴5+ 15×10× 1 2 =P= 0.5×15×10× 3 2
∴80=P+ 75 3 2
∴P=80− 75 3 2
∴P= 160−75 3 2
∴P=16.25 N

Topic: Physics, Laws of Motion, JEE Main
The figure below shows a block of mass M connected to an ideal string which passes through a thin fixed smooth pipe. On the other end, a particle of mass m is connected which revolves in a vertical circle of radius r. If the coefficient of friction between M and the surface is Î¼=23, then for what minimum value of M, the block of mass m can undergo complete vertical circular motion?

Options:
A Mmin=6m
B Mmin=9m
C Mmin=3m
D Mmin=15m

The correct answer is:
Mmin=9m

Solution:
For the particle to undergo complete vertical circular motion, its speed at the lowest point should be at least v=5gr. Also, the tension in the string is maximum when the particle is at the lowest point. The tension at the lowest position of the particle is

T=mg+mv2r

⇒T=mg+m5grr=6mg

The friction force acting on the block should be able to balance the maximum tension.

f=T=6mg

f=6mg≤μMg

⇒M≥6m23=9m

Mmin=9m

Topic: Physics, Laws of Motion, JEE Main
The figure below shows a wheatstone bridge with resistors P and Q having almost equal resistance. When R=400 Î©, the bridge is in balanced condition. If on interchanging P and Q, the bridge is again balanced for R=405 Î©, then the value of X is

Options:
A 404.5 Ω
B 402.5 Ω
C 403.5 Ω
D 401.5 Ω

The correct answer is:
402.5 Ω

Solution:
Initially,

PQ=400x

Once we interchange P and Q, we get

QP=405X

⇒PQ×QP=400×405X2

⇒X=400×405≈402.5 Ω

Topic: Physics, Current Electricity, JEE Main
The figure below shows an isosceles triangular frame ABC made out of two different materials and D is the midpoint of the side AB. The coefficient of thermal expansion of the rod ADB is α1 and for the rods AC and BC is α2. End C is fixed and the whole system is placed on a smooth horizontal surface. When the temperature of the system increases, it is found that the distance remains fixed, then

Options:
A l1l2=2α2α1
B l1l2=2α1α2
C l1l2=α1α2
D l1l2=α2α1

The correct answer is:
l1l2=2α2α1

Solution:
According to condition of the problem, height of the isosceles triangle ABC is unchanged. The dotted lines show configuration
After a temperature rise. Increase in length of rod AB,
âl1=l1Î±1âT
Thus AA′=1/2Â l1Î±1âT
We draw a normal from A to A′C (the final length of AC ). Increase in length of AC is A′N
A′N=l2Î±2âT
Considering increase in angle Î¸ to be very small.
A′NâAA′cosâ¡Î¸
Where cosâ¡Î¸=l12l2
Thus, we have l2Î±2âT=12l1Î±1âTl12l2
Hence l1l2=2Î±2Î±2

Topic: Physics, Thermal Properties of Matter, JEE Main
The figure below shows currents in a part of electric circuit. The current $i$ is

Options:
A $1.7 \mathrm{amp}$
B $3.7 \mathrm{amp}$
C $1.3 \mathrm{amp}$
D $1 \mathrm{amp}$

The correct answer is:
$1.7 \mathrm{amp}$

Solution:
According to Kirchhoff's first law
At junction $A, i_{A B}=2+2=4 A$
At junction $B, i_{A B}=i_{B C}-1=3 A$
At junction $C, i=i_{B C}-1.3=3-1.3=1.7$

Topic: Physics, Current Electricity, JEE Main
The figure below shows the structure of a mitochondrion with its four parts labelled (A), (B), (C) and (D). Select the part correctly matched with its function.

Options:
A D-Outer membrane-gives-rise to inner membrane by splitting
B B-Inner membrane-forms infolding called cristae
C C-Cristae-possess single circular DNA molecule and ribosomes
D A-Matrix-major site for respiratory chain enzymes

The correct answer is:
B-Inner membrane-forms infolding called cristae

Solution:
The inner membrane of mitochondria forms a number of infoldings called the cristae (sing-crista) towards the matrix. The cristae increase the surface area.

Topic: Biology, Cell - The Unit of Life, NEET, NEET 2011 (Mains)
The figure below shows the top view of a hollow cylindrical room having a frictionless floor. A block of mass m is projected with a speed v, on the smooth horizontal floor, along the wall of the room. The average normal force exerted by the cylindrical walls, on the block, during its motion from A to B is

Options:
A mv2R
B mv2πR
C 2mv2R
D 2mv2πR

The correct answer is:
2mv2πR

Solution:
The change in momentum of the block from A to B is

Δp=2mv

The time it takes to move from A to B is

Δt=πRv

The average force exerted by the wall on the block is

Favg=∆p∆t=2mv2πR

Topic: Physics, Laws of Motion, JEE Main
The figure below shows three steps (I, II, III) of Polymerase Chain Reaction (PCR). Select the option giving correct identification together with what it represents?

Options:
A B-Denaturation at a temperature of about $98^{\circ} \mathrm{C}$ separating the two DNA strands
B A-Denaturation at a temperature of about $50^{\circ} \mathrm{C}$
C C-Extension in the presence of heat stable DNA polymerase
D A-Annealing with two sets of primers

The correct answer is:
C-Extension in the presence of heat stable DNA polymerase

Solution:
There are three steps in polymerase chain reaction namely.
(i) Denaturation-at $94^{\circ} \mathrm{C}$ for DNA strand separation.
(ii) Annealing-binding of primer.
(iii) Extension in presence of DNA polymerase which is heat stable.

Topic: Biology, Biotechnology - Principles and Processes, NEET, NEET 2012 (Mains)
The figure given below is a diagrammatic representation of the response of organisms to abiotic factors. What do a, b, and c represent respectively?

Options:
A Conformer, Regulator, Partial regulator
B Regulator, Partial regulator, Conformer
C Partial regulator, Regulator, Conformer
D Regulator, Conformer, Partial regulator

The correct answer is:
Regulator, Conformer, Partial regulator

Solution:
In graph, the line 'a' represents regulator, line 'b' represent conformer and line 'c' represent partial regulator. Organism that are able to maintain homeostasis by physiological means that ensures constant body temperature are called regulators. Organism that are not able to maintain a constant internal temperature are called conformers.

Topic: Biology, Organisms and Populations, NEET
The figure given below shows a network of four capacitors of capacitance equal to C1=C, C2=2C, C3=3C, C4=4C are connected to a battery. If the ratio of charges on C2 and C4 is pq then what is the value of p+q, where p and q are smallest positive integers.

Options:

The correct answer is:
25

Solution:
The charge flowing through C4 is
q4=C4ÃV=4 CV
The series combination of C1, C2 and C3
1Câ²=1C+12C+13C
1Câ²=6+3+26C=116CâCâ²=6C11
Now, Câ² and C4 form parallel combination giving
C"=Câ²+C4=6C11+4C=50C11
Net Charge, q=C"V=5011CV
qâ²=q-q4=5011 CV-4CV=6CV11
Total Charge flowing through C1,C2 and C3 will be
qâ²=q-q4=5011CV-4CV=6CV11
Since, C1, C2, C3 are in series combination. Hence, charge flowing through these will be same
Hence, q2=q1=q3=qâ²=6 CV11
Thus, q2q4=6 CV114 CV=322

Topic: Physics, Electrostatics, JEE Main
The figure given below shows a small part of human lung where exchange of gas takes place. In which one of the options given below, the one part A, B C or D is correctly identified along with its function.

Options:
A A - Alveolar cavity - main site of exchange of respiratory gases
B D - Capillary wall - exchange of and takes place here
C B - Red blood cell - transport of mainly
D C - Arterial capillary - passes oxygen to tissues

The correct answer is:
A - Alveolar cavity - main site of exchange of respiratory gases

Solution:
Alveoli are the primary sites of exchange of gases. The exchange of gases $\left(\mathrm{O}_2\right.$ and $\left.\mathrm{CO}_2\right)$ between the alveoli and the blood occurs by simple diffusion.

Topic: Biology, Breathing and Exchange of Gases, NEET, NEET 2011 (Screening)
The figure given below shows a small part of human lung where exchange of gas takes place. In which one of the options given below, the one part A, B, C or D is correctly identified along with its function.

Options:
A A - Alveolar cavity - main site of exchange of respiratory gases
B D - Capillary wall - mainly made up of cuboidal epithelium
C B - Red blood cell - transport of mainly carbaminohaemoglobin
D C - Arterial capillary - passes oxygen to tissues

The correct answer is:
A - Alveolar cavity - main site of exchange of respiratory gases

Solution:

Topic: Biology, Breathing and Exchange of Gases, NEET
The figure given below shows a small part of human lung where exchange of gases takes place. In which one of the options given below, the one part A, B, C or D is correctly identified along with its function?

​

Options:
A B : red blood cell â transport of CO2 mainly
B D : capillary wall â exchange of O2 and CO2 takes place here.
C A : alveolar cavity â main site of exchange of respiratory gases
D C : arterial capillary – passes oxygen to tissues

The correct answer is:
A : alveolar cavity – main site of exchange of respiratory gases

Solution:
In the given figure the exchange of respiratory gases occurs in alveolar cavity.
The diffusion membrane is made up of three major layers namely, the thin squamous epithelium of alveoli, the endothelium of alveolar capillaries and the basement substance in between them. However, its total thickness is much less than a millimetre. Therefore, all the factors in our body are favourable for diffusion of O2 from alveoli to tissues and that of CO2 from tissues to alveoli.

Topic: Biology, Breathing and Exchange of Gases, NEET
The figure given below shows the conversion of a substrate into product by an enzyme. In which one of the four options (a-d), the components of reaction labelled as $A, B, C$ and $D$ are identified correctly?
$\begin{array}{|c|c|c|c|c|}
\hline & A & B & C & D \\
\hline \text{(a)} & \text{Potential energy} & \text{Transition state} & \text{Activation energy with enzyme} & \text{Activation energy without enzyme} \\
\hline \text{(b)} & \text{Transition energy} & \text{Potential energy} & \text{Activation energy without enzyme} & \text{Activation energy with enzyme} \\
\hline \text{(c)} & \text{Potential energy} & \text{Transition state} & \text{Activation energy with enzyme} & \text{Activation energy without enzyme} \\
\hline \text{(d)} & \text{Activation energy with enzyme} & \text{Transition state} & \text{Activation energy without enzyme} & \text{Potential energy} \\
\hline
\end{array}$

Options:
A (a)
B (b)
C (c)
D (d)

The correct answer is:
(b)

Solution:
The amount of energy required to raise the energy of molecules at which chemical reaction can occur is called activation energy. Thus, activation energy is actually the energy required to form the transition state. Enzymes dramatically reduce the activation energy of a reaction, so that most molecules can easily get over the activation energy barrier and quickly turn into product. Simply we can say that activation energy of an enzyme catalysed reaction is lower than that of an uncatalysed reaction.

Topic: Biology, Biomolecules (B), NEET, NEET 2010 (Mains)
The figure gives experimentally measured B vs. H variation in a ferromagnetic material. The retentivity, coercivity and saturation, respectively, of the material are:

Options:
A 1.5T,50A/m and 1.0T
B 1.5T,50A/m and 1.0T
C 150A/m,1.0T and 1.5T
D 1.0T,50A/m and 1.5T

The correct answer is:
1.0T,50A/m and 1.5T

Solution:
x=retentivity
y=coercivity
z=saturationmagnetization

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2020 (07 Jan Shift 2)
The figure in the complex plane given by
10zz¯-3z2+z¯2+4iz2-z¯2=0

Options:
A a straight line
B a circle
C a parabola
D an ellipse

The correct answer is:
a straight line

Solution:
10zz¯-3(z+z¯)2-2zz¯+4i((z+z¯)(z-z¯))=0
Let z=x+iy

10x2+y2-34x2-2x2-2y2+4i(2x(2iy))=0
⇒4x2+16y2-16xy=0
⇒x2-4xy+4y2=0
⇒(x-2y)2=0⇒x=2y
∴ straight line

Topic: Mathematics, Complex Number, JEE Main
The figure represents a voltage regulator circuit using a Zener diode. The breakdown voltage of the Zener diode is 6 V and the load resistance is, RL=4 kΩ . The series resistance of the circuit is Ri=1 kΩ . If the battery voltage VB varies from 8 V to 16 V , what are the minimum and maximum values of the current through Zener diode?

Options:
A 0.5 mA; 6 mA
B 1 mA; 8.5 mA
C 0.5 mA; 8.5 mA
D 1.5 mA; 8.5 mA

The correct answer is:
0.5 mA; 8.5 mA

Solution:
For V0=8V
8 V-1000 i0=6 V
i0=2 mA
i0-i1=64000=1.5 mA
i1=0.5 mA
For V0=16 V
16 V-1000 i0=6 V
i0=10 mA
i0-i2=64000=1.5 mA
i2=8.5 mA

Topic: Physics, Semiconductors, JEE Main, JEE Main 2019 (10 Apr Shift 2)
The figure shown below, calculate the net current from the battery and net resistance of the circuit.

Options:
A 1.33 A, 152 Î©
B 1.33 A, 15 Ω
C 1.5 A, 10 Ω
D 1.5 A, 15 Ω

The correct answer is:
1.33 A, 152 Î©

Solution:
In the given circuit, B and E are at the same potential. Points C and F are also at the same potential.
Therefore, the resistance between B and E, C and F will be ineffective and neglected. Now, the circuit becomes
In the above circuit, branch ABCD and AEFD are in the parallel combination.
Thus, the net resistance of the circuit.
1Rnet=15+5+5+15+5+5=115+115=215
âRnet=152 Î©
â´ Current, I=VRnet=1015/2=2015=43=1.33 A

Topic: Physics, Current Electricity, JEE Main
The figure shown below is a screwgauge. When the circular scale division matches the main scale at 43, the 6.5 mm mark on the main scale is just visible. The main scale has 1 2 mm marks. In complete rotation, the screw advances by 1 2 mm and circular scale has 50 divisions. The reading of the screwgauge is

Options:
A 6.93 mm

B 5.93 mm

C 4.93 mm

D 7.93 mm

The correct answer is:
6.93 mm

Solution:
Object thickness = 6.5 mm + 4 3 1 / 2 mm 5 0 = 6.93 mm

Topic: Physics, Experimental Physics, JEE Main
The figure shows a block of mass m resting on a horizontal smooth floor at a distance l from a rigid wall. Block is pushed toward the right by a distance 3l2 and released from rest. When block passes from its mean position, another block of mass m1 is gently placed on it which sticks to it due to friction. The value of m1 so that the combined block just touches the left wall is η4m. Find the value of η.

Options:

The correct answer is:
5

Solution:
When block P is released from rest from a distance 3l2 toward right from mean position, this will be the amplitude of oscillation so velocity of block when passing from its mean position is given as
v=AÏ=3l2km
If mass m1 is added to it and just after if velocity of combined block becomes v1, from momentum conservation, we have
mv = (m + m1)v1
â´ v1=mm+m1â3l2km
If this is the velocity of combined block at mean position, it is given as
v1 =A1Ï1 where Ï1=km+m1
where A1 and Ï1 are the new amplitude and angular frequency of SHM of the block.
It is given that combined block just reaches the left wall.
Thus, the new amplitude of oscillation must be l . So, we have,
mm+m1â3l2km=lkm+m1
On solving, we get
m1=54m
â´ n=5

Topic: Physics, Oscillations, JEE Main
The figure shows a capacitor of capacitance C connected to a battery via a switch, having a total charge Q on it, in steady-state. When the switch S is turned from position A to position B, the energy dissipated in the circuit is

Options:
A 18Q2C
B 38Q2C
C 34Q2C
D 58Q2C

The correct answer is:
38Q2C

Solution:
Q0=Cε
Q1C=Q23C Q1+Q2=Q0
⇒Q1=Q04; Q2=3Q04
Energy dissipated,
E=12Q0 2C-12Q1 2C-12Q2 23C
=12CQ02-Q0216-9Q023×16
=Q0232C16-1-3=38Q02C

Topic: Physics, Capacitance, JEE Main, JEE Main 2019 (12 Jan Shift 1)
The figure shows a capacitor of capacitance C connected to a battery via a switch, having a total charge Q on it, in steady-state. When the switch S is turned from position A to position B, the energy dissipated in the circuit is

Options:
A 18Q2C
B 38Q2C
C 34Q2C
D 58Q2C

The correct answer is:
38Q2C

Solution:
Q0=Cε
Q1C=Q23C Q1+Q2=Q0
⇒Q1=Q04; Q2=3Q04
Energy dissipated,
E=12Q0 2C-12Q1 2C-12Q2 23C
=12CQ02-Q0216-9Q023×16
=Q0232C16-1-3=38Q02C

Topic: Physics, Electrostatics, JEE Main
The figure shows a circuit diagram of a 'Wheatstone Bridge' to measure the resistance $G$ of the galvanometer. The relation $\frac{P}{Q}=\frac{R}{G}$ will be satisfied only when

Options:
A The galvanometer shows a deflection when switch $S$ is closed
B The galvanometer shows a deflection when switch $S$ is open
C The galvanometer shows no change in deflection whether $S$ is open or closed
D The galvanometer shows no deflection

The correct answer is:
The galvanometer shows no change in deflection whether $S$ is open or closed

Solution:
In balance condition, no current will flow through the branch containing $S$

Topic: Physics, Current Electricity, JEE Main
The figure shows a circuit that contains four identical resistors with resistance R=2.0 Ω, two identical inductors with inductance L=2.0 mH and an ideal battery with E.M.F. E=9 V. The current i just after the switch S is closed will be:

Options:
A 9 A
B 3.37 A
C 2.25 A
D 3.0 A

The correct answer is:
2.25 A

Solution:
Just after the switch is closed, inductor will behave like infinite resistance (open circuit) so the circuit will look like

i=9R+R=94=2.25 A

Topic: Physics, Alternating Current, JEE Main, JEE Main 2021 (24 Feb Shift 2)
The figure shows a circular area of tthe radius R where a uniform magnetic field B→ is going into the plane of the paper and increasing in magnitude at a constant rate. In that case, which of the following graphs, drawn schematically, correctly shows the variation of the induced electric field E(r)?

Options:
A
B
C
D

The correct answer is:

Solution:
E=dϕdt=AdBdt

for

r<R, E2πr=πr2dBdt⇒Eαr

for

r>R, E2πr=πR2dBdt⇒Eα1r

Topic: Physics, Electromagnetic Induction, JEE Main
The figure shows a circular area of tthe radius R where a uniform magnetic field B→ is going into the plane of the paper and increasing in magnitude at a constant rate. In that case, which of the following graphs, drawn schematically, correctly shows the variation of the induced electric field E(r)?

Options:
A
B
C
D

The correct answer is:

Solution:
E=dϕdt=AdBdt

for

r<R, E2πr=πr2dBdt⇒Eαr

for

r>R, E2πr=πR2dBdt⇒Eα1r

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2014 (19 Apr Online)
The figure shows a combination of two NOT gates and a NOR gate.
The combination is equivalent to $\mathrm{a}$

Options:
A
NAND gate

B
NOR gate

C
AND gate

D
OR gate

The correct answer is:
AND gate

Solution:
Truth table is as shown :
Thus the combination of two NOT gates and one NOR gate is equivalent to a AND gate.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2012 (26 May Online)
The figure shows a current-carrying loop, some part of which is circular and some part is a line segment. The magnetic induction at the centre is

Options:
A 3μ0Iπ4a
B Î¼0I4Ïa1+Ï
C μ0I8πa
D 3Î¼0I8a

The correct answer is:
3μ0I8a

Solution:
The magnetic field due to circular arc is
â B=Î¼0IÎ±4Ïr=Î¼0I4Ïa(3Ï2)=3Î¼0I8a

Topic: Physics, Magnetic Effects of Current, JEE Main
The figure shows a current-carrying square loop ABCD of side 10 cm and current i=10 A. The magnetic moment M â of the loop is

Options:
A 0.05 i ^ - 3 k ^ A m 2
B 0.05 j ^ + k ^ A m 2
C 0.05 3 i ^ + k ^ A m 2
D i ^ + k ^ A m 2

The correct answer is:
0.05 i ^ - 3 k ^ A m 2

Solution:
Areal vectors will be â¥ to the plane.
Area vector is at 6 0 o angle with x-axis and 3 0 o angle with -z-axis.
M â = NI A â
M â = 1 0 Ã 1 0 - 2 A m 2
M = 0 Â· 1 A m 2
Mâ=Mxi^-Mzk^=Mcos60i^-Mcos30ok^
M â = 0.05 i ^ - 3 k ^

Topic: Physics, Magnetic Effects of Current, JEE Main