datavorous/entrance-exam-dataset huggingface dataset:
In the figure, the reverse breakdown voltage of a Zener diode is 5.6 V, then the current Iz through the diode is

Options:
A 17 mA
B 15 mA
C 10 mA
D 7 mA

The correct answer is:
10 mA

Solution:
For zener break down potential difference across 800Ω resistor will be 5.6V

Vz=5.6V

i2=V2800=5.6800=7mA
ΔV across 200Ω=9-5.6=3.4V
i1=3.4200=17mA
i1=i2+iz

iz=17mA-7mA=10mA

Topic: Physics, Semiconductors, JEE Main
In the figure, what is the magnetic field at the point

Options:
A $\frac{\mu_0 I}{4 \pi r}$
B $\frac{\mu_0 l}{4 \pi r}+\frac{\mu_0 l}{2 \pi r}$
C $\frac{\mu_0 l}{4 r}+\frac{\mu_0 l}{4 \pi r}$
D $\frac{\mu_0 l}{4 r}-\frac{\mu_0 l}{4 \pi r}$

The correct answer is:
$\frac{\mu_0 l}{4 r}+\frac{\mu_0 l}{4 \pi r}$

Solution:
Magnetic field due to different parts are
$\begin{aligned} & B_1=0 \\ & B_2=\frac{\mu_0}{4 \pi} \cdot \frac{\pi i}{r} \\ & B_3=\frac{\mu_0}{4 \pi} \cdot \frac{i}{r} \\ & \therefore B_{\text {net }}=B_2+B_3=\frac{\mu_0 i}{4 r}+\frac{\mu_0 i}{4 \pi r}\end{aligned}$

Topic: Physics, Magnetic Effects of Current, JEE Main
In the first order reaction, the concentration of the reactants is reduced to \( 25 \% \) in one hour. The half-life period of the reaction is

Options:
A \( 2 \mathrm{~h} \)
B \( 4 \mathrm{~h} \)
C \( 1 / 2 \mathrm{~h} \)
D \( 1 / 4 \mathrm{~h} \)

The correct answer is:
\( 1 / 2 \mathrm{~h} \)

Solution:
For first order reaction,
k=2.303t1/2logaa-a2=2.303tloga(a-x)
2.303t1/2log⁡2=2.3031logaa/4
2.303×0.3010t1/2=2.303×log⁡4
t1/2=0.3010log⁡4=0.30100.6020=12h

Topic: Chemistry, Chemical Kinetics, JEE Main
In the five-kingdom classification, Chlamydomonas and Chlorella have been incluced in

Options:
A Protista
B Algae
C Plantae
D Monera

The correct answer is:
Protista

Solution:
RH Whittaker (1969) divided all the living organisms into five kingdoms namely Monera, Protista, Fungi, Plantae and Animalia.

Topic: Biology, Biological Classification, NEET, NEET 2012 (Mains)
In the flame test of a mixture of salts, a green flame with blue centre was observed. Which one of the following cations may be present?

Options:
A Calcium
B Copper
C Barium
D Strontium

The correct answer is:
Copper

Solution:
Metal	Flame colour

Cu	Green flame with blue centre

Ca	Brick red

Ba	Apple green

Sr	Crimson

Topic: Chemistry, Practical Chemistry, JEE Main, JEE Main 2022 (24 Jun Shift 2)
In the following arrangement, the system is initially at rest. The 5 kg block is now released. Assuming the pulleys and string to be massless and smooth. If the acceleration of block C is x10 m s-2, then find the value of x. Take g=9.8 m s-2.

Options:

The correct answer is:
7

Solution:
Block B will not move
5g-T=5a ...(i)
2T-8g=8a2 ...(ii)
2g=14a
a=g7
â´a2=g14=9.814=710 m s-2=0.7
â´x=7

Topic: Physics, Laws of Motion, JEE Main
In the following arrangement, the system is initially held at rest by an external agent. At t=0, the 5 kg block is released. If the acceleration of block C is x10 ms-2, then find the value of x. Assume that the pulleys and strings are massless and smooth. [g=9.8 ms-2]

Options:

The correct answer is:
7

Solution:
Block B will not move
5g-T=5a ...(i)
2T-8g=8a2 ...(ii)
2g=14a
a=g7
â´a2=g14=9.814=710ms-2=0.7
â´x=7

Topic: Physics, Laws of Motion, JEE Main
In the following balanced reaction,
values of $X, Y$ and $Z$ respectively are

Options:
A
2,5,16

B
$8,2,5$

C
$5,2,16$

D
$5,8,4$

The correct answer is:
2,5,16

Solution:
$X \mathrm{MnO}_4^{-}+Y \mathrm{C}_2 \mathrm{O}_4^{-}+Z \mathrm{H}^{+} \leftrightharpoons$
$$
X \mathrm{Mn}^{++}+2 Y \mathrm{CO}_2+\frac{Z}{2} \mathrm{H}_2 \mathrm{O}
$$
First half reaction
$$
\mathrm{MnO}_4^{-} \longrightarrow \mathrm{Mn}^{++}
$$
On balancing
$$
\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{++}+4 \mathrm{H}_2 \mathrm{O}
$$
Second half reaction
$$
\mathrm{C}_2 \mathrm{O}_4^{-} \longrightarrow 2 \mathrm{CO}_2
$$
On balancing
$$
\mathrm{C}_2 \mathrm{O}_4^{-} \longrightarrow 2 \mathrm{CO}_2+2 \mathrm{e}^{-}
$$
On multiplying eqn. (ii) by 5 and (iv) by 2 and then adding we get
$$
\begin{aligned}
& 2 \mathrm{MnO}_4^{-}+5 \mathrm{C}_2 \mathrm{O}_4^{--}+16 \mathrm{H}^{+} \longrightarrow \\
& 2 \mathrm{Mn}^{++}+10 \mathrm{CO}_2+8 \mathrm{H}_2 \mathrm{O}
\end{aligned}
$$

Topic: Chemistry, Redox Reactions, JEE Main
In the following balanced reaction,
values of $X, Y$ and $Z$ respectively are

Options:
A
2,5,16

B
$8,2,5$

C
$5,2,16$

D
$5,8,4$

The correct answer is:
2,5,16

Solution:
$X \mathrm{MnO}_4^{-}+Y \mathrm{C}_2 \mathrm{O}_4^{-}+Z \mathrm{H}^{+} \leftrightharpoons$
$$
X \mathrm{Mn}^{++}+2 Y \mathrm{CO}_2+\frac{Z}{2} \mathrm{H}_2 \mathrm{O}
$$
First half reaction
$$
\mathrm{MnO}_4^{-} \longrightarrow \mathrm{Mn}^{++}
$$
On balancing
$$
\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{++}+4 \mathrm{H}_2 \mathrm{O}
$$
Second half reaction
$$
\mathrm{C}_2 \mathrm{O}_4^{-} \longrightarrow 2 \mathrm{CO}_2
$$
On balancing
$$
\mathrm{C}_2 \mathrm{O}_4^{-} \longrightarrow 2 \mathrm{CO}_2+2 \mathrm{e}^{-}
$$
On multiplying eqn. (ii) by 5 and (iv) by 2 and then adding we get
$$
\begin{aligned}
& 2 \mathrm{MnO}_4^{-}+5 \mathrm{C}_2 \mathrm{O}_4^{--}+16 \mathrm{H}^{+} \longrightarrow \\
& 2 \mathrm{Mn}^{++}+10 \mathrm{CO}_2+8 \mathrm{H}_2 \mathrm{O}
\end{aligned}
$$

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2012 (12 May Online)
In the following carbocation, \( \mathrm{H} / \mathrm{CH}_{3} \) that is most likely to migrate to the positively charged carbon is,

Options:

The correct answer is:
2

Solution:
Rearrangement can take place by three ways:

1. By group shift (hydride shift, phenyl shift, methyl shift, etc.)

2. By ring expansion, if carbocation is adjacent of five and four-membered ring.

3. By ring contraction, if carbocation is on four-membered ring.

This rearrangement takes place for carbonation stability:

The factors affecting carbocation stability:

1. If carbocation is on less percentage s character carbon, it will be more stable.

2. If carbocation is attached to any electron donating group (groups those show positive inductive, mesomeric and hyperconjugation effect), its stability increases.

3. If carbocation is busy in aromatic stability, it will still be more stable than any other factor.

Preference of effects:

Aromaticity > resonance > mesomeric > hyperconjugation > inductive

So here we can see if we shift H2 then carbonation will be stabilised by +M of -OH group which is more stabilisation to carbocation.

Hence answer is 2

Topic: Chemistry, General Organic Chemistry, JEE Main
In the following circuit, \( 5 \Omega \) resistor develops \( 45 \mathrm{~J} \mathrm{~s}^{-1} \) due to current flowing through it. The power developed per second across \( 12 \Omega \) resistor is:

Options:
A \( 15 \mathrm{~W} \)
B \( 192 \mathrm{~W} \)
C \( 36 \mathrm{~W} \)
D \( 64 \mathrm{~W} \)

The correct answer is:
\( 192 \mathrm{~W} \)

Solution:
From the formula of power in 5 Ω,

P=i12R⇒i1=455=3 A

In parallel combination, potential difference across resistors remain same. Hence,

I1R1=I2R2

⇒3×5=i2×(9+6)⇒i2=1 A

From KCL,

i=i1+i2

⇒i=3+1=4 A
Hence, power developed in 12 Ω resistance is,

P=I2R=(4)2×12=192 W

Topic: Physics, Current Electricity, JEE Main
In the following circuit, a voltmeter V is connected across a lamp L . What change would occur in voltmeter reading if the resistance R is reduced

Options:
A Increases
B Decreases
C Remains same
D None of these

The correct answer is:
Increases

Solution:
Here, the emitter-base junction of the N-P-N transistor is forward biased with battery VBB through resistance R . When the value of R is reduced, then the base current ib will increase. As a result, the collector current will also increase. ( i c =β i b ). Due to the increase in ic, the potential difference across L increases and hence the reading of voltmeter will increase.

Topic: Physics, Semiconductors, JEE Main
In the following circuit, the battery has an emf of 2 V and an internal resistance of 23 Ω. The power consumption in the entire circuit is ______ W.

Options:

The correct answer is:
3

Solution:
As we can see, two resistances of 2 Ω in the middle portion(in both left and right side) are in parallel, so their combined resistance will be 2×22+2+2×22+2=2 Ω. Now this resistance along with the topmost & bottommost resistance are in parallel again, so total resistance of the circuit will be,

Req=23+23=43 Ω

∴P=V2Req=443=3 W

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (31 Jan Shift 2)
In the following circuit, the magnitude of current I1, is ______ A.

Options:

The correct answer is:
2

Solution:
The figure above shows the current and voltage at different points in circuit.

Applying Kirchhoff's Junction law at A,

x-(y+5)1+x-22+x-02=0 ...1

Applying Kirchhoff's Junction law at B,

y+5-x1+y-01+y-21=0 ...2

Solving equation (1) and (2), we get

x=3 and y=0

Now, at D junction

I1=i1+i2

⇒I1=y-01+x-02

=0-01+3-02

⇒I1=1.5 A.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (30 Jan Shift 1)
In the following circuit the switch Sis closed at t =0. The charge on the capacitor C1 as a function of time will be given by Ceq= C1C2C1+ C2

Options:
A C1EÂ 1-expâ¡-tRC1
B CeqEÂ exp-tRCeq
C CeqE 1-expâ¡-tRCeq
D C2E 1-expâ¡-tRC2

The correct answer is:
CeqE 1-expâ¡-tRCeq

Solution:
In given circuit capacitors C1 and C2 are connected in series , so equivalent capacitance will be

1Ceq=1C1 +1C2 ⇒Ceq=C1C2C1+C2

at t = 0, the switch is closed and capacitors start charged, at time t

charge on the capacitor is given by

q= q0 1-e-t/τ

where q0 is the maximum charge on the capacitor at steady state and τ is the time constant.

and q0 =CeqE and τ = RCeq

Then from the above equation

q= CeqE 1-e-t/RCeq

∴q= CeqE 1-exptRCeq
Both capacitor will have same charge as they are connected in series

Topic: Physics, Capacitance, JEE Main
In the following circuit the switch Sis closed at t =0. The charge on the capacitor C1 as a function of time will be given by Ceq= C1C2C1+ C2

Options:
A C1EÂ 1-expâ¡-tRC1
B CeqEÂ exp-tRCeq
C CeqE 1-expâ¡-tRCeq
D C2E 1-expâ¡-tRC2

The correct answer is:
CeqE 1-expâ¡-tRCeq

Solution:
In given circuit capacitors C1 and C2 are connected in series , so equivalent capacitance will be

1Ceq=1C1 +1C2 ⇒Ceq=C1C2C1+C2

at t = 0, the switch is closed and capacitors start charged, at time t

charge on the capacitor is given by

q= q0 1-e-t/τ

where q0 is the maximum charge on the capacitor at steady state and τ is the time constant.

and q0 =CeqE and τ = RCeq

Then from the above equation

q= CeqE 1-e-t/RCeq

∴q= CeqE 1-exptRCeq
Both capacitor will have same charge as they are connected in series

Topic: Physics, Capacitance, JEE Main, JEE Main 2018 (16 Apr Online)
In the following circuits, PN-junction diodes D1, D2 and D3 are ideal for the following potential of A and B, the correct increasing order of resistance between A and B will be

(i) - 10 V, - 5 V (ii) - 5 V, - 10 V (iii) - 4 V, - 12 V

Options:
A (i) < (ii) < (iii)

B (iii) < (ii) < (i)

C (ii) = (iii) < (i)

D (i) = (iii) < (ii)

The correct answer is:
(ii) = (iii) < (i)

Solution:
i) In this case of current will flow from B top A. No current will flow through branch C, E (as marked in figure)
⇒ Req=R4+R+R4=3R2
ii) Here current will flow from A to B so no current will flow in branch D
⇒ Req=R4+R2+R4=R
iii) Here also current will flow from A to B, so Req=R same as (ii)
⇒ Rii=Riii < Ri

Topic: Physics, Semiconductors, JEE Main
In the following common emitter configuration a NPN transistor with current gain Î²=100 is used. The output voltage of the amplifier will be

Options:
A 10Â mV
B 0.1Â V
C 1.0 V
D 10Â V

The correct answer is:
1.0 V

Solution:
Voltage gain =Output voltageInput voltage
Vout = Vin × Voltage gain
Vout = Vin × Current gain × Resistance gain
= Vin×β×RoRi=10-3×100×101=1V

Topic: Physics, Semiconductors, JEE Main
In the following common emitter configuration, an NPN transistor with current gain Î²=100 is used. The output voltage of the amplifier will be

Options:
A 10Â mV
B 0.1Â V
C 1.0 V
D 10Â V

The correct answer is:
1.0 V

Solution:
Voltage gain =Output voltageInput voltage
Vout = Vin × Voltage gain
Vout = Vin × Current gain × Resistance gain
= Vin×β×RoRi=10-3×100×101=1V

Topic: Physics, Semiconductors, JEE Main
In the following common emitter configuration, an NPN transistor with current gain β=100 is used. The output voltage of the amplifier will be

Options:
A 10 mV
B 0.1 V
C 1.0 V
D 10 V

The correct answer is:
1.0 V

Solution:
Voltage gain =Output voltageInput voltage
Vout = Vin × Voltage gain
Vout = Vin × Current gain × Resistance

= Vin×β×RoRi=10-3×100×101=1 V

Topic: Physics, Semiconductors, JEE Main
In the following compound, designated as A, B, C will be numbered as

Options:
A 3, 2, 1
B 6, 1, 2
C 3, 1, 2
D 1, 2, 3

The correct answer is:
1, 2, 3

Solution:
According to the hierarchy of functional groups, OH will be given top priority, followed by the secondary amine. Hence, the numbering will be

Topic: Chemistry, General Organic Chemistry, JEE Main
In the following compound, least number of monochlorination is possible

Options:
A \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3} \)
B
C
D

The correct answer is:

Solution:
Write chlorination reaction for all of them to find which gives of the minimum number of monochlorination product.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
In the following compound,

the favourable site/s for protonation is/are :

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

The correct answer is:
(b), (c) and (d)

Solution:
(b), (c) and (d) are the favourable sites for protonation because these are localilsed lone pair electron.

Topic: Chemistry, Amines, JEE Main, JEE Main 2019 (11 Jan Shift 2)
In the following compounds, the decreasing order of basic strength will be:

Options:
A C2H5NH2>NH3>(C2H5)2NH
B (C2H5)NH>NH3>C2H5NH2
C NH3>C2H5NH2>(C2H5)2NH
D (C2H5)2NH>C2H5NH2>NH3

The correct answer is:
(C2H5)2NH>C2H5NH2>NH3

Solution:
Basic character ∝ +I effect
-CH2CH3 exert +I effect so, more the number of -CH2CH3 groups more is the basic character.
NH3<CH3CH2NH21o Amine<(CH3CH2)2NH2o Amine

Topic: Chemistry, Amines, JEE Main, JEE Main 2019 (08 Apr Shift 1)
In the following compounds:
the order of basicity is as follows

Options:
A
IV $>$ III $>$ II $>$ I

B
III $>$ I $>$ II $>$ IV

C
II $>$ III $>$ I $>$ IV

D
I $>$ III $>$ II $>$ IV

The correct answer is:
I $>$ III $>$ II $>$ IV

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2012 (26 May Online)
In the following, digital circuit, what will be the output a 'Z', when the input A,B are 1,0, 0,0, 1,1, 0,1

Options:
A 0,0,1,0
B 1,0,1,1
C 1,1,0,1
D 0,1,0,0

The correct answer is:
0,0,1,0

Solution:
A	B	AB	A+B	P=AB.A+B	Q=P+AB	Q=z

1	0	1	0	0	1	0

0	1	1	0	0	1	0

1	1	0	0	0	0	1

0	0	1	1	1	0	0

Topic: Physics, Semiconductors, JEE Main, JEE Main 2020 (02 Sep Shift 2)
In the following fig. is shown the flow of liquid through a horizontal pipe. Three tubes $A, B$ and $C$ are connected to the pipe. The radii of the tubes $A, B$ and $C$ at the junction are respectively $2 \mathrm{~cm}, 1 \mathrm{~cm}$ and $2 \mathrm{~cm}$. It can be said that the

Options:
A Height of the liquid in the tube $A$ is maximum
B Height of the liquid in the tubes $A$ a $\mathrm{d} B$ is the same
C Height of the liquid in all the three tubes is the same
D Height of the liquid in the tubes $A$ and $C$ is the same

The correct answer is:
Height of the liquid in the tubes $A$ and $C$ is the same

Solution:
As cross-section areas of both the tubes $A$ and $C$ are same and tube is horizontal. Hence according to equation of continuity $v_A=v_C$ and therefore according to Bernoulli's theorem $P_A=P_C$ i.e. height of liquid is same in both the tubes $A$ and $C$.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
In the following figure, the diodes which are forward biased, are
A.
B.
C.
D.

Options:
A C only
B C and A
C B and D
D A, B and D

The correct answer is:
C and A

Solution:
For forward bias p-type should be higher potential and $n$-type at lower potential. Hence (A) and (C) are in forward bias.

Topic: Physics, Semiconductors, NEET, NEET 2011 (Mains)
In the following given reaction 'A' is

Options:
A
B
C
D

The correct answer is:

Solution:
In the positive part of the addendum added to the alkene according to Markonikov's rule. A tertiary carbocation is generated undergo ring expansion as five membered rings is more stable than four membered rings.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2023 (24 Jan Shift 1)
In the following graph, distance travelled by the body in metres is

Options:
A 200
B 250
C 300
D 400

The correct answer is:
200

Solution:
Area under the graph gives distance
Distance $=\frac{1}{2} \times 10 \times 10+10 \times 10+\frac{1}{2} \times 10 \times 10=50+100+50=200$

Topic: Physics, Motion In One Dimension, JEE Main
In the following I refers to current and other symbols have their usual meaning. Choose the option that corresponds to the dimensions of electrical conductivity:

Options:
A M-1L-3T3I
B M-1L-3T3I2
C M-1L3T3I
D ML-3T-3I2

The correct answer is:
M-1L-3T3I2

Solution:
R=LσA
σ=LRA= LIAV
=LIAwIt
=LL2I2TML2T-2
=M-1L-3T3I2

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2016 (09 Apr Online)
In the following indicator diagram, the net amount of work done will be

Options:
A Positive
B Negative
C Zero
D Infinity

The correct answer is:
Negative

Solution:
The cyclic process 1 is clockwise where as process 2 is anticlockwise. Clockwise area represents positive work and anticlockwise area represents negative work. Since negative area (2) > positive area (1), hence net work done is negative.

Topic: Physics, Thermodynamics, JEE Main
In the following List $A$ consists of wave length range of different of EM waves and List $B$ consists of ways to produce them. Match the following Lists.

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

The correct answer is:
4 3 2 1

Solution:
(A) Waves of range $1 \mathrm{~nm}$ to $400 \mathrm{~nm}$ are produced when inner shell electrons in atoms moving from one energy level to a lower level.
(B) The waves of range of wavelength greater the $0.1 \mathrm{~nm}$ is produced when rapid acceleration and deceleration of electrons in aerials takes place.
(C) The waves of wavelength range $1 \mathrm{~nm}$ to $700 \mathrm{~nm}$ are produced by the vibration of atom and molecules in a substance.
(D) The waves of range of wavelength of less than $10^{-3} \mathrm{~nm}$ are produced by the radioactive decay of the nucleus.
The correct match is given by $\mathrm{A} \rightarrow 4, \mathrm{~B} \rightarrow 3, \mathrm{C} \rightarrow 2$ and $\mathrm{D} \rightarrow 1$

Topic: Physics, Electromagnetic Waves, JEE Main
In the following logic circuit the sequence of the inputs A, B are (0,0),(0,1),(1,0) and (1,1). The output Y for this sequence will be :

Options:
A 0,1,0,1
B 0,0,1,1
C 1,1,1,0
D 1,0,1,0

The correct answer is:
1,1,1,0

Solution:
Look at the above circuit, the value of P can be found by the idea of AND gate, P=A·B and the value of Q, can be calculated by the idea of OR gate, Q=A+B. Then expression for Y can be taken by the idea of NAND gate, ⇒Y=P·Q¯ by the idea of Demorgan's theorem, Y=P¯+Q¯=A¯+B¯+A¯·B¯

ABY=A¯+B¯+A¯·B¯001011101110

Topic: Physics, Semiconductors, JEE Main, JEE Main 2021 (31 Aug Shift 1)
In the following metals which one has lowest probable interatomic forces

Options:
A Copper
B Silver
C Zinc
D Mercury

The correct answer is:
Mercury

Solution:
Mercury has very weak interatomic forces so it remains in liquid state.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
In the following molecules,

Hybridisation of carbon a,b and c respectively are :

Options:
A sp3,sp,sp
B sp3,sp2,sp
C sp3,sp2,sp2
D sp3,sp,sp2

The correct answer is:
sp3,sp2,sp2

Solution:
Hybridisation is found by steric number which is generally the summation of sigma bonds and lone pairs of electrons.

So as we can see a has four sigma bonds so it has sp3 hybridisation.

b,c has three sigma bonds so it has sp2 hybridisation.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2021 (18 Mar Shift 2)
In the following monobromination reaction, the number of possible chiral products are

Options:

The correct answer is:
5

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
In the following monobromination reaction, the number of possible chiral products is
(1.0 mole)
(enantiomerically pure)

Options:

The correct answer is:
5

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
In the following network, the potential at O is

Options:
A 4 V
B 3 V
C 6 V
D 4.8 V

The correct answer is:
4.8 V

Solution:
Let the potential at O is V0.
Application of Kirchhoff's first law at junction O gives
8-V02=V0-44+V0-22
=V0-4+2V0-44
48-V02=V0-4+2V0-4
16-2V0=3V0-8
16+8=5V0
V0=245=4.8 V

Topic: Physics, Current Electricity, JEE Main
In the following nuclear reaction,
D→αD1→β-D2→αD3→γD4
Mass number of D is 182 and atomic number is 74 . Mass number and atomic number of D4 respectively will be

Options:
A 174 and 71
B 174 and 69
C 172 and 69
D 172 and 71

The correct answer is:
174 and 71

Solution:
Equivalent reaction can be written as D→D4+2α+β-+γ

So, mass number of D4=mass number of D-2×4=182-8=174

Atomic number of D4=atomic number of D-2×2+1=74-4+1=71

The reaction is

D74182⟶αD172178⟶β-D273178⟶αD371174→γD471174

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2022 (29 Jun Shift 2)
In the following nuclear reaction $\mathrm{x}$ is ${ }_{13} \mathrm{Al}^{27}+{ }_2 \mathrm{He}^4 \rightarrow$ ${ }_0^{n^1+\mathrm{X}}$

Options:
A ${ }_{15} \mathrm{P}^{31}$
B ${ }_{14} \mathrm{Si}^{30}$
C ${ }_{15} \mathrm{P}^{30}$
D ${ }_{15} \mathrm{Si}^{31}$

The correct answer is:
${ }_{15} \mathrm{P}^{30}$

Solution:
${ }_{13} \mathrm{Al}^{27}+{ }_2 \mathrm{He}^4 \longrightarrow{ }_0 \mathrm{n}^1+\mathrm{x}$
Mass no. remain conserve in nuclear reaction.
$A_x=27+4-1=30$
Atomic number remain conserve in nuclear reaction,
$\mathrm{Z}_{\mathrm{x}}=13+2-0=15$
$\therefore{ }_{15} \mathrm{X}^{30}$ the atom is $\mathrm{P}$.
Hence option (3) is correct.

Topic: Physics, Nuclear Physics, TS EAMCET, TS EAMCET 2023 (12 May Shift 2)
In the following options, each one has a pair of names. They are the same hormone with different names. Which one of these does not follow the rule?

Options:
A Parathormone - Collip's hormone
B Tetra-iodotyrosine - thyroxine
C Prolactin - Lactating hormone
D Vasopressin - Anti diuretic hormone

The correct answer is:
Tetra-iodotyrosine - thyroxine

Solution:
The parathyroid hormone was first discovered by Phillip Collip. Hence, it is also called Collip's hormone. It is a hypercalcemic hormone. Prolactin is a hormone produced by the adenohypophysis. It helps in the growth of the mammary gland and formation of milk. After childbirth, it also maintains corpus luteum for about three to six months leading to lactation amenorrhoea. Hence, prolactin is also called luteotrophic hormone. Vasopressin (increases blood pressure) is secreted by the hypothalamus but stored and released by neurohypophysis. It also acts on the kidney stimulating it to absorb more water. Hence, it is also called anti-diuretic hormone. The thyroid gland produces two hormones tri-iodothyronine T3 and tetra-iodothyronine T4. Since a major portion of the hormone produced is T4, it is called thyroxine. In the example, note the spelling. It is written tetra-iodotyrosine. It should have been tetra-iodothyroxine.

Topic: Biology, Chemical Coordination and Integration, NEET
In the following reaction:

A+2B+3C⇌AB2C3
6.0 g of A, 6.0×1023 atoms of B and 0.036 mol of C reacted and formed 4.8 g of compound AB2C3. If the atomic mass of A and C are 60 and 80 amu, respectively. What is the atomic mass of B in amu? (Avogadro number=6×1023)

Options:
A 40 amu
B 60 amu
C 70 amu
D 50 amu

The correct answer is:
50 amu

Solution:
A + 2 B + 3 C → AB 2 C 3 6 g 6 × 1 0 2 3 atoms 0.036 moles Moles 6 6 0 6 × 1 0 2 3 6 × 1 0 2 3 0.1 mole 1.0 mole 0.036 moles
0.11 ,1.02 ,0.0363; the least value is 0.012, i.e., C is the limiting reagent
0.036 moles of C provides 0.012 moles of AB2C3.
0.012 moles of AB2C3 weighs 4.8 g
1 mole of AB2C3 weighs 4.80.012=400 g mol-1
400=60+2B+3×80
B=50 amu

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2015 (11 Apr Online)