datavorous/entrance-exam-dataset huggingface dataset:
In the circult shown in the figure all the resistance are identical and each has the value $r \Omega$. The equivalent resistance of the combination between the points $A$ and $B$ will remain unchanged even when the followring pairs of points marked in the figure are connected through a resistance $\mathbb{R}$.

Options:
A 2 and 6
B 3 and 6
C 4 and 7
D 4 and 6

The correct answers are:
2 and 6, 4 and 7

Solution:
The circuit diagram is as shown below,
According to question, when given pair of point in options are connected through a resistance $R$, then equivalent resistance between point $A$ and $B\left(R_{A}\right)$ remains unchanged. It is only possible when current. does not flows through resistance $R$ and circuit become Wheatstone bridge as shown in the figure below.
At the balanced condition, $\frac{P}{Q}=\frac{R}{S}$
The current flow in $C D$ branch will be zero.
Now by checking each option, from option (a), circuit is,
Now, $\because \frac{P}{Q}=\frac{R}{S} \Rightarrow \frac{2 r}{4 r}=\frac{r}{2 r} \Rightarrow \frac{1}{2}=\frac{1}{2}$
Hence, option is correct. From option (b), circuit is
Now, $\frac{P}{Q}=\frac{R}{S} \Rightarrow \frac{3 r}{3 r} \neq \frac{r}{2 r}$
So, option (b) is also incorrect. From option (c) circuit is,
Now, $\frac{P}{Q}=\frac{R}{S} \Rightarrow \frac{4 r}{2 r}=\frac{2 r}{r} \Rightarrow \frac{2}{1}=\frac{2}{1}$
So, option (c) is also correct. From option (d), circuit is
Now, $\frac{P}{Q}=\frac{R}{S} \Rightarrow \frac{4 r}{2 r} \neq \frac{r}{2 r}$
So, option (d) is also incorrect. Hence, option (a) and (c) are correct.

Topic: Physics, Current Electricity, WBJEE, WBJEE 2019
In the Claisen-Schmidt reaction to prepare \(351 \mathrm{~g}\) of dibenzalacetone using \(87 \mathrm{~g}\) of acetone, the amount of benzaldehyde required is ______ g. (Nearest integer)

Options:

The correct answer is:
318

Solution:
Claisen Schmidt reaction
mw of benzaldehyde $=106$
$106 \times 3=318 \mathrm{gm}$. Benzaldehyde is required to give 1.5 mole (or $351 \mathrm{gm}$ ) product

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2024 (05 Apr Shift 2)
In the closest packed structure of a metallic lattice, the number of nearest nejghbours of a metallic atom is

Options:
A Twelve
B Four
C Eight
D Six

The correct answer is:
Twelve

Solution:
Co-ordination number in $H C P$ and $C C P$ arrangement is 12 while in bcc arrangement is 8 .

Topic: Chemistry, Solid State, JEE Main
In the co-axial system of circle $x^2+y^2+2 g x+c=0$, where $g$ is a parameter, if $c\gt0$ then the circles are

Options:
A Orthogonal
B Touching type
C Intersecting type
D Non-intersecting type

The correct answer is:
Touching type

Solution:
Given, equation of the circle $x^2+y^2+2 g x+c=0$, where $c$ is constant and $g$ represents the parameter of a coaxial system and $c\gt0$. We know that the standard equation of a circle is $x^2+y^2+2 g x+2 f y+c=0$. Comparing the given equation with the standard equation, we get centre $\equiv(-g, 0)$ and radius $\sqrt{g^2-c}$. Therefore radius becomes zero, when $g^2-c=0$ or $g= \pm \sqrt{c}$. Therefore $(\sqrt{c}, 0)$ and $(-\sqrt{c}, 0)$ are the limiting points of the coaxial system of circles. Since $c\gt0$, therefore $\sqrt{c}$ is real and limiting points are real and distinct. Thus the co-axial system is said to be touching type.

Topic: Mathematics, Circle, JEE Main
In the common-base configuration, a transistor has current amplification factor 0.95 . If the transistor is used in common-emitter configuration and base current changes by $2 \mu \mathrm{A}$, then the change in the collector current is

Options:
A $19 \mu \mathrm{A}$
B $0.91 \mu \mathrm{A}$
C $1.9 \mu \mathrm{A}$
D $38 \mu \mathrm{A}$

The correct answer is:
$38 \mu \mathrm{A}$

Solution:
Given, $\alpha=0.95$
So,
$$
\begin{aligned}
\beta & =\frac{\alpha}{1-\alpha}=\frac{0.95}{1-0.95}=19 \\
& =\beta=\frac{\Delta I_C}{\Delta I_B}=19
\end{aligned}
$$
or $\Delta I_C=19 \times 2=38 \mu \mathrm{A}$

Topic: Physics, Semiconductors, JEE Main
In the complex K2FeFeCN6

Options:
A the complex is high spin complex
B both Fe atoms are in the same oxidation state
C the coordination number of iron is 4
D both Fe atoms are in different oxidation state

The correct answer is:
both Fe atoms are in the same oxidation state

Solution:
In the complex K2Fe[FeCN6] both the iron atoms are present in same oxidation state.

Topic: Chemistry, Coordination Compounds, JEE Main
In the complex K4[ThC2O42H2O2] . If coordination number is X and oxidation number of Th is Y. The sum of X + Y is?

Options:

The correct answer is:
8

Solution:
K4ThC2O42H2O
4+Y+2-2+0=0
Y=+2 (oxidation number)
Coordination number (X)=2×2+2=6
X+Y=6+2=8

Topic: Chemistry, Redox Reactions, JEE Main
In the complex $\left[\mathrm{SbF}_5\right]^{2-}, s p^3 d$ hydridisation is present. Geometry of the complex is

Options:
A Square pyramidal
B Square bipyramidal
C Tetrahedral
D Square

The correct answer is:
Square pyramidal

Solution:
Complex with $s p^3 d-$ hybridisation show square pyramidal geometry.

Topic: Chemistry, Coordination Compounds, JEE Main
In the compound $\mathrm{HC} \equiv \mathrm{C}-\mathrm{CH}=\mathrm{CH}_2$ the hybridization of $\mathrm{C}-2$ and $\mathrm{C}-3$ carbons are respectively:

Options:
A $\mathrm{sp}^3 \& \mathrm{sp}^3$
B $\mathrm{sp}^2 \& \mathrm{sp}^3$
C $\mathrm{sp}^2 \& \mathrm{sp}$
D $\mathrm{sp}^3 \& \mathrm{sp}$

The correct answer is:
$\mathrm{sp}^2 \& \mathrm{sp}$

Solution:
Hints:
(Double bond is preferred)

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
In the connections shown in the adjoining figure, the equivalent capacity between $A$ and $B$ will be

Options:
A $10.8 \mu \mathrm{F}$
B $69 \mu F$
C $15 \mu \mathrm{F}$
D $10 \mu F$

The correct answer is:
$10 \mu F$

Solution:
Given circuit can be drawn as follows. It is a balance whetstone bridge type network, hence $24 \mu F$ capacitor can be neglected
Equivalent capacitance between $A$ and $B=4+6$ $=10 \mu \mathrm{F}$.

Topic: Physics, Capacitance, JEE Main
In the context of the Hall-Heroult process for the extraction of Al, which of the following statements is false?

Options:
A Na3AlF6 serves as the electrolyte.
B CO and CO2 are produced in this process.
C Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity.
D Al3+ is reduced at the cathode to form Al.

The correct answer is:
Na3AlF6 serves as the electrolyte.

Solution:
Hall - Heroult process for extraction of Al. Al2O3 is electrolyte Na3AlF6 reduces the fusion temperature & provides good conductivity.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2015 (04 Apr)
In the context of the Hall-Heroult process for the extraction of Al, which of the following statements is false ?

Options:
A CO and CO2 are produced in this process
B Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity
C Al3+ is reduced at the cathode to form Al
D Na3AlF6 helps in increasing the melting point of the mixture

The correct answer is:
Na3AlF6 helps in increasing the melting point of the mixture

Solution:
Hall heroult process:
In the Hall-HÃ©roult process the following simplified reactions take place at the carbon electrodes:
Cathode: Al3++3e-1âAl
Anode: O2-+CâCO+2e-
Overall:
Al2O3+3Câ2Al+3CO
And 2 Al2O3+3Câ4Al+3CO2
In reality, much more CO2 is formed at the anode than CO:
Cryolite (Na3AlF6) lowers the melting point of the mixture and make it more conductive in nature so as to carry out electrolysis of the ore in Hall-Heroult process.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
In the context of the rearrangement of an oxime of a ketone to an amide (represented below)
Which of the following statement is/are correct?

Options:
A It is the cis hydrocarbon radical \( (R) \) with respect to the \( \mathrm{OH} \) group that migrates
B The group that migrates never gets completely detached from the remainder of the molecule during the transformation
C The rearrangement is intermolecular
D None of the above

The correct answer is:
The group that migrates never gets completely detached from the remainder of the molecule during the transformation

Solution:
This reaction is an example of Backmann rearrangement. It is an intramolecular rearrangement. Beckmann Rearrangement is a reaction where oxime is changed to an amide. Beckmann rearrangement consists of an alkyl group migration, that is anti-periplanar to the expulsion of a leaving group and form nitrilium ion. This is followed by attack of water molecule to an imidate and then tautomerization to the amide:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
In the Coolidge tube experiment, if the applied voltage is increased to three times, the short wavelength limit of continuous X-ray spectrum shifts by 20 pm. What is the initial voltage (in kV) applied to the tube?

Options:

The correct answer is:
41

Solution:
âµ cut-off wavelength for continuous x-ray is given as : Î» min = hc e V 0
â´ hc eV 0 = Î» & hc 3 eV 0 = Î» - Î Î»
( V 0 =acceleratingâvoltage )
â V 0 = 2 hc 3 e Î Î» = 4 1 kV

Topic: Physics, Atomic Physics, JEE Main
In the Coolidge tube experiment, if the applied voltage is increased to three times, the short wavelength limit of continuous X-ray spectrum shifts by 20 pm. What is the initial voltage (in kV) applied to the tube to the nearest integer value? [Take hc = 12400 eV A∘, e = 1.6 × 10-19 C]

Options:

The correct answer is:
41

Solution:
∵ the cut-off wavelength for continuous x-ray is given as : λ min = hc e V 0
∴ hc eV 0 = λ & hc 3 eV 0 = λ - Δ λ
( V 0 =accelerating voltage )
⇒ V0=2hc3eΔλ≈41 kV

Topic: Physics, Atomic Physics, JEE Main
In the coordination compound, $\mathrm{K}_4\left[\mathrm{Ni}(\mathrm{CN})_4\right]$, the oxidation state of nickel is

Options:
A
0

B
$+1$

C
$+2$

D
$-1$

The correct answer is:
0

Solution:
$4(+1)+x+(-1) \times 4=0$ $\Rightarrow 4+x-4=0$
$x=0$

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2003
In the cumene to phenol preparation in presence of air, the intermediate is

Options:
A

B

C

D

The correct answer is:

Solution:
Cumene (isopropylbenzene) is oxidised in the presence of air to cumene hydroperoxide. It is converted to phenol and acetone by treating it with dilute acid.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2023 (25 Jan Shift 1)
In the cyclotron, as radius of the circular path of the charged particle increases

Ï=angular velocity,v=linear velocity

Options:
A Both ω and v increase
B Only ω increases, v remains constant
C Only v increases, ω remains constant
D v increases, ω decreases

The correct answer is:
Only v increases, ω remains constant

Solution:
In a cyclotron
R=mvqB
v=qBRm and Ï=qBm
â´ Râv.
Ï,q,m and B are constant.

Topic: Physics, Magnetic Effects of Current, NEET
In the cyclotron, as radius of the circular path of the charged particle increases $(\omega=$ angular velocity, $v=$ linear velocity)

Options:
A both $\omega$ and $v$ increases
B only $\omega$ increases, $v$ remains constant
C $v$ increases, $\omega$ remains constant
D $v$ increases, $\omega$ decreases

The correct answer is:
$v$ increases, $\omega$ remains constant

Solution:
No Solution Available

Topic: Physics, Magnetic Effects of Current, AIIMS, AIIMS 2018 (27 May)
In the cyclotron, as radius of the circular path of the charged particle increases (to = angular
velocity, \( v= \) linear velocity)

Options:
A both \( \omega \) and \( v \) increases
B \( \omega \) only increases, \( v \) remains constant
C \( v \) increases, \( \omega \) remains constant
D \( v \) increases, \( \omega \) decreases

The correct answer is:
\( v \) increases, \( \omega \) remains constant

Solution:
The radius of circular path is given as
\[
\begin{array}{l}
r=\frac{m v \sin \theta}{q} B \\
\Rightarrow r \propto v
\end{array}
\]
Thus in a cyclotron, as radius of the circular path of charged particle increases \( v \) increases and \( \omega \) remains unchanged.

Topic: Physics, Magnetic Effects of Current, KCET, KCET 2016
In the Davisson and Germer experiment, the velocity of electron emitted from the electron gun can be increased by

Options:
A Decreasing the potential difference between the anode and filament
B Increasing the potential difference between the anode and filament
C Increasing the filament current
D Decreasing the filament current

The correct answer is:
Increasing the potential difference between the anode and filament

Solution:
- The velocity of an electron accelerated through a potential is given by
$V=\sqrt{\frac{2 e V_0}{m}}$
Where $e=$ Charge of electron and $V_0=$ potential, $m=$ mass of the particle
- From the above equation it is clear that
$\Rightarrow V \propto \mathrm{V}_0$
- The velocity of the electron is directly proportional to the square root of the potential difference between the anode and filament. Hence, option (2) is the answer

Topic: Physics, Dual Nature of Matter, JEE Main
In the density measurement of a cube, the mass and edge length are measured as 10.00Â±0.10 kg and 0.10Â±0.01 m, respectively. The error in the measurement of density is:

Options:
A 0.31
B 0.10
C 0.07
D 0.01

The correct answer is:
0.31

Solution:
Density ρ=massvolume
∆ρρ=∆mm+∆VV
=∆mm+3∆aa ∵V=a3
=0.110+30.010.1
=0.01+30.1
=0.31

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2019 (09 Apr Shift 1)
In the depression of freezing point experiment
A. Vapour pressure of the solution is less than that of pure solvent
B. Vapour pressure of the solution is more than that of pure solvent
C. Only solute molecules solidify at the freezing point
D. Only solvent molecules solidify at the freezing point

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

The correct answer is:
A and D only

Solution:
Since the solute particles do not evaporate, the vapour pressure of the solution is lower than that of the pure solvent. The lowering of the vapour pressure depends on the number of solute particles that have been dissolved.

In the depression in freezing point experiment, it is found that the vapour pressure of a solution is less than that of pure solvent. Only solvent molecules solidify at freezing point.

Topic: Chemistry, Solutions, JEE Main, JEE Main 2023 (24 Jan Shift 1)
In the Dewar's method of separation of noble gases, the mixture of noble gases is kept in contact with coconut charcoal at $173 \mathrm{~K}$ which one of the following gaseous mixtures is not adsorbed on to the charcoal?

Options:
A $\mathrm{Ar}, \mathrm{Kr}$
B $\mathrm{Xe}, \mathrm{Ar}$
C $\mathrm{He}, \mathrm{Ne}$
D $\mathrm{Xe}, \mathrm{Kr}$

The correct answer is:
$\mathrm{He}, \mathrm{Ne}$

Solution:
Only $\mathrm{He}$ and $\mathrm{Ne}$ are remained unadsorbed on the coconut charcoal at $-100^{\circ} \mathrm{C}(173 \mathrm{~K})$ as their boiling points are less than $-100^{\circ} \mathrm{C} .(\mathrm{He}=4 \mathrm{~K}, \mathrm{Ne}=27 \mathrm{~K})$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), AP EAMCET, AP EAMCET 2004
In the Dewar's method of separation of noble gases, the mixture of noble gases is kept in contact with coconut charcoal at $173 \mathrm{~K}$ which one of the following gaseous mixtures is not adsorbed on to the charcoal?

Options:
A $\mathrm{Ar}, \mathrm{Kr}$
B $\mathrm{Xe}, \mathrm{Ar}$
C $\mathrm{He}, \mathrm{Ne}$
D $\mathrm{Xe}, \mathrm{Kr}$

The correct answer is:
$\mathrm{He}, \mathrm{Ne}$

Solution:
Only $\mathrm{He}$ and $\mathrm{Ne}$ are remained unadsorbed on the coconut charcoal at $-100^{\circ} \mathrm{C}(173 \mathrm{~K})$ as their boiling points are less than $-100^{\circ} \mathrm{C} .(\mathrm{He}=4 \mathrm{~K}, \mathrm{Ne}=27 \mathrm{~K})$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), TS EAMCET, TS EAMCET 2004
In the diagram shown, a rod of mass M has been fixed on a ring of the same mass. The whole system has been placed on a perfectly rough surface. The system is gently displaced so that the ring starts rolling. The velocity of the centre of the ring when the rod becomes horizontal is (the length of the rod is equal to the radius of the ring)

Options:
A 3gR10
B 5gR3
C 3gR7
D 2gR9

The correct answer is:
3gR10

Solution:
The moment of inertia of the system about instantaneous axis of rotation is given by,
I=Iring+Irod
⇒I=(MR2+MR2)+112MR2+MR12 (From parallel axis theorem)
⇒I=2MR2+112MR2+M5R22 ∵R1=R2+R22=5R2
⇒I=2MR2+112MR2+54MR2=2MR2+43MR2=103MR2
Hence, Isystem=103MR2.
Now from work - energy theorem, we get,
MgR2=Rotational kinetic energy of the system about instantaneous axis of rotation
(here R2 is the distance through which the center of mass of the rod descends)
⇒MgR2=12 Isystem ω2
⇒MgR2=12×103MR2×ω2
⇒ω=3g10R
Now velocity of center of mass is given as, v=Rω.
⇒v=R×3g10R
⇒v=3gR10

Topic: Physics, Rotational Motion, JEE Main
In the diagram shown, I1, I2 are the magnitudes of current in the square loop and infinite long straight conductor respectively. If the net magnetic field at the centre of the loop is zero, then the relation between I1 and I2 is

Options:
A I1=I262
B I1=I242
C I1=I222
D I1=I22

The correct answer is:
I1=I262

Solution:
B→=B→1+B→2
B1=μ0I22π3a2
B2=4μ0I124πa2
μ0I23πa=2μ0I12πa
I2=6I12

Topic: Physics, Magnetic Effects of Current, JEE Main
In the diagram shown, no friction at any contact surface. Initially, the spring has no deformation. What will be the maximum deformation in the spring? Consider all the strings to be sufficiently large. Consider the spring constant to be K.

Options:
A 4F3K
B 8F3K
C F3K
D None

The correct answer is:
8F3K

Solution:
aCOM=F3M
with respect to the Center of mass
4F3X1+4F3X2=12KX1+X22
8F3K=X1+X2

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
In the diagram shown, the difference in the two tubes of the manometer is 5 cm, the cross-section of the tube at A and B is 6 mm2 and 10 mm2 respectively. The rate at which water flows through the tube is g=10 m s-2

Options:
A 7.5 cc s-1
B 12.5 cc s-1
C 8.0 cc s-1
D 10.0 cc s-1

The correct answer is:
7.5 cc s-1

Solution:
6vA=10vB ......1

12ρvA2-ρgh=12ρvB2 .........(2)

⇒vA=10.8 m s-1

dVdt=AvA

=6×10-2 cm2×1000.8 cm s-1=7.5 cc s-1

OR

By the concept of the venturi meter tube

rate of flow=Q=A1A22g(h1−h2)(A22−A12)

=6×10×10−42×100×5(−36+100)×10−2

=7.5 cc s-1

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2014 (19 Apr Online)
In the diagram shown, the normal reaction force between $2 \mathrm{~kg}$ and $1 \mathrm{~kg}$ is (Given $g=10 \mathrm{~ms}^{-2}$) (Consider the surface, to be smooth):

Options:
A $25 \mathrm{~N}$
B $39 \mathrm{~N}$
C $6 \mathrm{~N}$
D $10 \mathrm{~N}$

The correct answer is:
$25 \mathrm{~N}$

Solution:
Total force along inclined plane,
$\begin{aligned}
f & =-\mathrm{F}_1+6 g \sin 30^{\circ}+\mathrm{F}_2 \\
f & =-60+6 g \sin 30^{\circ}+18 \\
f & =-60+6 \times 10 \times \frac{1}{2}+18 \\
f & =12 \mathrm{~N} \text { (downwards) } \\
\text {acceleration, } a & =\frac{f}{m}=\frac{12}{6}=2 \mathrm{~m} / \mathrm{s}^2
\end{aligned}$

As per question, the normal force between $2 \mathrm{~kg}$ and $1 \mathrm{~kg}$ is,
$\begin{aligned}
\mathrm{N}-18-10 \sin 30^{\circ} & =m a \\
\mathrm{~N}-18-10 \times \frac{1}{2} & =1 \times 2 \\
\mathrm{~N} & =2+23 \\
\mathrm{~N} & =25 \mathrm{~N}
\end{aligned}$

Topic: Physics, Laws of Motion, JEE Main
In the diagram, the input is across the terminals $A$ and $C$ and the output is across the terminals $B$ and $D$, then the output is

Options:
A Zero
B Same as input
C Full wave rectifier
D Half wave rectifier

The correct answer is:
Full wave rectifier

Solution:
During half cycle when $\mathrm{V}_{\mathrm{M}} \gt \mathrm{V}_{\mathrm{N}}, \mathrm{D}_1$ andD $\mathrm{D}_3$ are forward biased. The path of current is MABPQDCNM. In second half cycle, $V_N\gtV_M, D_2$ and $D_4$ are forward biased while $D_1$ and $D_3$ are reverse biased. Hence, path of current is, NCBPQDAMN.
Therefore, in both half cycles, current flows from P toQ from load resistance $R_L$. It is full wave rectifier.

Topic: Physics, Semiconductors, JEE Main
In the diode-based rectifier circuit given below, if $V_s=V_m \sin \omega t$ and the diode is ideal, then the average value of $V_L$ is

Options:
A $\frac{R_L}{\left(R_L+R_S\right)} \frac{V_m}{\pi}$
B $R_L V_m \sin \omega t$
C $\frac{R_L}{\left(R_L+R_S\right)} V_m$
D $\frac{R_L}{\left(R_L+R_S\right)} V_m \sin \omega t$

The correct answer is:
$\frac{R_L}{\left(R_L+R_S\right)} \frac{V_m}{\pi}$

Solution:
Given, AC voItage, $V_s=V_m \sin \omega t$ and $V_m=$ maximum value of voltage In the given circuit diode will only conduct current in forward bias, so output across $R_L$ will be
Average voltage, $V_{\mathrm{av}}=\frac{V_m}{\pi}$ average current $I_{\mathrm{av}}=\frac{V_{\mathrm{av}}}{R}$ Total resistance, $R=R_S+R_L$
$$
I_{\mathrm{av}}=\frac{V_m}{\left(R_S+R_L\right) \pi}
$$
Now, voltage across $R_L$,
$$
\begin{aligned}
& V_L=I_{\mathrm{av}} \cdot R_L \\
& V_L=\frac{R_L}{R_S+R_L} \cdot \frac{V_m}{\pi}
\end{aligned}
$$

Topic: Physics, Semiconductors, JEE Main
In the DNA molecules

Options:
A There are two strands which run antiparallel-one in $5^{\prime} \rightarrow 3^{\prime}$ direction and other in $3^{\prime} \rightarrow$
$5+$
B The total amount of purine nucleotides and pyrimidine nucleotides is not always equal
C There are two strands which run parallel in the $5^{\prime} \rightarrow 3^{\prime}$ direction
D The proportion of Adenine in relation to thymine varies with the organism

The correct answer is:
There are two strands which run antiparallel-one in $5^{\prime} \rightarrow 3^{\prime}$ direction and other in $3^{\prime} \rightarrow$
$5+$

Solution:
The 2 strands of DNA are antiparallel one in $5^{\prime}-3^{\prime}$ direction and the other in $3^{\prime}-5^{\prime}$ direction.

Topic: Biology, Molecular Basis of Inheritance, NEET, NEET 2008 (Mains)
In the double slit experiment, 5th dark fringe is formed opposite to one of the slits. The wavelength of light is

Options:
A d26D
B d25D
C d215D
D d29D

The correct answer is:
d29D

Solution:
For dark fringe,
YdD=2m-1λ2

Here, m=5, Y=d2
∴d2dD=2×5-1λ2
Or d2D=9λ
∴λ=d29D

Topic: Physics, Wave Optics, JEE Main
In the double slit experiment when a glass plate (refractive index 1.5) of thickness t is introduced in the path of one of the interfering beams (wavelength λ ), intensity at the position where the central maximum occurred previously remains unchanged. Minimum thickness t of the glass plate is

Options:
A 2λ
B 2λ3
C λ3
D λ

The correct answer is:
2λ

Solution:
For calculating the minimum thickness, we use Î¼-1 t=Î»
Or t=Î»Î¼-1=Î»1.5-1=t
Or t=Î»1/2=2Î»

Topic: Physics, Wave Optics, NEET
In the electric field of a point charge $q$, a certain charge is carried from point $A$ to $B, C, D$ and
$E$. Then the work done

Options:
A Is least along the path $A B$
B Is least along the path $A D$
C Is zero along all the paths $AB, AC, AD$ and $AE$
D Is least along $AE$

The correct answer is:
Is zero along all the paths $AB, AC, AD$ and $AE$

Solution:
ABCDE is an equipotential surface, on equipotential surface no work is done in shifting a charge from one place to another.

Topic: Physics, Electrostatics, JEE Main
In the electric network shown, when no current flows through the 4 Ω resistor in the arm EB, the potential difference between the points A and D will be:

Options:
A 5 V
B 3 V
C 4 V
D 6 V

The correct answer is:
5 V

Solution:
Assuming potential at B = 0, then w.r.t. B the potential of other points are as shown in figure then
So, VA=9 V and VD=4 V
∴VA-VD=9-4=5 V

Topic: Physics, Current Electricity, JEE Main, JEE Main 2015 (11 Apr Online)
In the electrical network shown in the figure, What will be the potential difference (in volt) across 3 Î© resistance?

Options:

The correct answer is:
24

Solution:
As 3 Ω and 6 Ω resistances are
In parallel, their equivalent resistance will be 2 Ω. Here 2 Ω and 4 Ω are in series, their equivalent resistance will be 6 Ω. From current distribution law
i1=12×1818=12 A
i2=6×1818=6 A
Now, 12 A current is entering in the parallel combination of 3 Ω and 6 Ω again from current distribution law
i1′′=6×129=8 A
i2′′=3×129=4 A.
∴ The potential difference across 3 Ω resistance
=8×3=24 V

Topic: Physics, Current Electricity, JEE Main