datavorous/entrance-exam-dataset huggingface dataset:
In the Bohr model an electron moves in a circular orbit around the proton. Considering the orbiting electron to be a circular current loop, the magnetic moment of the hydrogen atom, when the electron is in $n^{\text {th }}$ excited state, is :

Options:
A
$\left(\frac{e}{2 m} \frac{n^2 h}{2 \pi}\right)$

B
$\left(\frac{e}{m}\right) \frac{n h}{2 \pi}$

C
$\left(\frac{e}{2 m}\right) \frac{n h}{2 \pi}$

D
$\left(\frac{e}{m}\right) \frac{n^2 h}{2 \pi}$

The correct answer is:
$\left(\frac{e}{2 m}\right) \frac{n h}{2 \pi}$

Solution:
Magnetic moment of the hydrogen atom, when the electron is in $\mathrm{n}^{\text {th }}$ excited state, i.e., $n^{\prime}=(n+1)$
As magnetic moment $\mathrm{M}_{\mathrm{n}}=\mathrm{I}_{\mathrm{n}} \mathrm{A}=\mathrm{i}_{\mathrm{n}}\left(\pi \mathrm{r}_{\mathrm{n}}^2\right)$
$\begin{aligned}
& i_n=e V_n=\frac{m z^2 e^5}{4 \varepsilon_0^2 n^3 h^3} \\
& r_n=\frac{n^2 h^2}{4 \pi^2 k z m e^2}\left(k=\frac{1}{4 \pi \epsilon_0}\right)
\end{aligned}$
Solving we get magnetic moment of the hydrogen atom for $\mathrm{n}^{\text {th }}$ excited state
$M_{n^{\prime}}=\left(\frac{e}{2 m}\right) \frac{n h}{2 \pi}$

Topic: Physics, Atomic Physics, JEE Main
In the Bohr model an electron moves in a circular orbit around the proton. Considering the orbiting electron to be a circular current loop, the magnetic moment of the hydrogen atom, when the electron is in $n^{\text {th }}$ excited state, is :

Options:
A
$\left(\frac{e}{2 m} \frac{n^2 h}{2 \pi}\right)$

B
$\left(\frac{e}{m}\right) \frac{n h}{2 \pi}$

C
$\left(\frac{e}{2 m}\right) \frac{n h}{2 \pi}$

D
$\left(\frac{e}{m}\right) \frac{n^2 h}{2 \pi}$

The correct answer is:
$\left(\frac{e}{2 m}\right) \frac{n h}{2 \pi}$

Solution:
Magnetic moment of the hydrogen atom, when the electron is in $\mathrm{n}^{\text {th }}$ excited state, i.e., $n^{\prime}=(n+1)$
As magnetic moment $\mathrm{M}_{\mathrm{n}}=\mathrm{I}_{\mathrm{n}} \mathrm{A}=\mathrm{i}_{\mathrm{n}}\left(\pi \mathrm{r}_{\mathrm{n}}^2\right)$
$\begin{aligned}
& i_n=e V_n=\frac{m z^2 e^5}{4 \varepsilon_0^2 n^3 h^3} \\
& r_n=\frac{n^2 h^2}{4 \pi^2 k z m e^2}\left(k=\frac{1}{4 \pi \epsilon_0}\right)
\end{aligned}$
Solving we get magnetic moment of the hydrogen atom for $\mathrm{n}^{\text {th }}$ excited state
$M_{n^{\prime}}=\left(\frac{e}{2 m}\right) \frac{n h}{2 \pi}$

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2013 (09 Apr Online)
In the Bohr model an electron of mass $m$ moves in a circular orbit around the proton. Considering the orbiting electron to be a circular current loop, the magnetic moment of the hydrogen atom, when the electron is in $n$th excited state. (Assume, $h=$ Planck's constant)

Options:
A $\left(\frac{e}{2 m} \frac{n^2 h}{2 \pi}\right)$
B $\left(\frac{e}{m}\right) \frac{n h}{2 \pi}$
C $\left(\frac{e}{2 m}\right) \frac{n h}{2 \pi}$
D $\left(\frac{e}{m}\right) \frac{n^2 h}{2 \pi}$

The correct answer is:
$\left(\frac{e}{2 m}\right) \frac{n h}{2 \pi}$

Solution:
If $R$ be the radius of circular path, then magnetic moment,
$M=i \times A=(e \times f) \times\left(\pi R^2\right) \quad\left(\because i=\frac{e}{T}=e f\right)$
$=e \times\left(\frac{v}{2 \pi R}\right) \times\left(\pi R^2\right)$
$\left[\because \omega=2 \pi f \Rightarrow \frac{V}{R}=2 \pi f \Rightarrow f=\frac{V}{2 \pi R}\right]$
$\Rightarrow \quad M=\frac{e v R}{2}$ $\ldots$ (i)
Now, using Bohr's principle,
Angular momentum, $L=m v R=\frac{n h}{2 \pi}$
$\Rightarrow \quad v R=\frac{n h}{2 \pi m}$ $\ldots$ (ii)
From Eqs. (i) and (ii), we get
$M=\frac{e}{2} \times\left(\frac{n h}{2 \pi m}\right)=\left(\frac{e}{2 m}\right) \times\left(\frac{n h}{2 \pi}\right)$

Topic: Physics, Atomic Physics, JEE Main
In the Bohr's model of hydrogen-like atom the force between the nucleus and the electron is modified as $F=\frac{e^2}{4 \pi \varepsilon_0}\left(\frac{1}{r^2}+\frac{\beta}{r^3}\right)$, where $\beta$ is a constant. For this atom, the radius of the $n^{\text {th }}$ orbit in terms of the Bohr radius $\left(a_0=\frac{\varepsilon_0 h^2}{m \pi e^2}\right)$ is :

Options:
A
$r_n=a_0 n-\beta$

B
$r_n=a_0 n^2+\beta$

C
$r_n=a_0 n^2-\beta$

D
$r_n=a_0 n+\beta$

The correct answer is:
$r_n=a_0 n^2-\beta$

Solution:
As $F=\frac{m^2}{r}=\frac{e^2}{4 \pi \epsilon_0}\left(\frac{1}{r^2}+\frac{B}{r^3}\right)$ and $\mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi} \Rightarrow \mathrm{v}=\frac{\mathrm{nh}}{2 \pi \mathrm{mr}}$ $\therefore \mathrm{m}\left(\frac{\mathrm{nh}}{2 \pi \mathrm{mr}}\right)^2 \times \frac{1}{\mathrm{r}}=\frac{\mathrm{e}^2}{4 \pi \epsilon_0}\left(\frac{1}{\mathrm{r}^2}+\frac{\mathrm{B}}{\mathrm{r}^3}\right)$ or, $\frac{1}{\mathrm{r}^2}+\frac{\mathrm{B}}{\mathrm{r}^3}=\frac{\mathrm{mn}^2 \mathrm{~h}^2 4 \pi \epsilon_0}{4 \pi^2 \mathrm{~m}^2 \mathrm{e}^2 \mathrm{r}^3}$
$$
\begin{aligned}
& \text { or, } \frac{\mathrm{a}_0 \mathrm{n}^2}{\mathrm{r}^3}=\frac{1}{\mathrm{r}^2}+\frac{\mathrm{B}}{\mathrm{r}^3} \\
& \left(\because \mathrm{a}_0=\frac{\epsilon_0 \mathrm{~h}^2}{\mathrm{~m} \pi \mathrm{e}^2} \text { Given }\right) \\
& \therefore \mathrm{r}=\mathrm{a}_0 \mathrm{n}^2-\mathrm{B}
\end{aligned}
$$

Topic: Physics, Atomic Physics, JEE Main
In the Bohr's model of hydrogen-like atom the force between the nucleus and the electron is modified as $F=\frac{e^2}{4 \pi \varepsilon_0}\left(\frac{1}{r^2}+\frac{\beta}{r^3}\right)$, where $\beta$ is a constant. For this atom, the radius of the $n^{\text {th }}$ orbit in terms of the Bohr radius $\left(a_0=\frac{\varepsilon_0 h^2}{m \pi e^2}\right)$ is :

Options:
A
$r_n=a_0 n-\beta$

B
$r_n=a_0 n^2+\beta$

C
$r_n=a_0 n^2-\beta$

D
$r_n=a_0 n+\beta$

The correct answer is:
$r_n=a_0 n^2-\beta$

Solution:
As $F=\frac{m^2}{r}=\frac{e^2}{4 \pi \epsilon_0}\left(\frac{1}{r^2}+\frac{B}{r^3}\right)$ and $\mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi} \Rightarrow \mathrm{v}=\frac{\mathrm{nh}}{2 \pi \mathrm{mr}}$ $\therefore \mathrm{m}\left(\frac{\mathrm{nh}}{2 \pi \mathrm{mr}}\right)^2 \times \frac{1}{\mathrm{r}}=\frac{\mathrm{e}^2}{4 \pi \epsilon_0}\left(\frac{1}{\mathrm{r}^2}+\frac{\mathrm{B}}{\mathrm{r}^3}\right)$ or, $\frac{1}{\mathrm{r}^2}+\frac{\mathrm{B}}{\mathrm{r}^3}=\frac{\mathrm{mn}^2 \mathrm{~h}^2 4 \pi \epsilon_0}{4 \pi^2 \mathrm{~m}^2 \mathrm{e}^2 \mathrm{r}^3}$
$$
\begin{aligned}
& \text { or, } \frac{\mathrm{a}_0 \mathrm{n}^2}{\mathrm{r}^3}=\frac{1}{\mathrm{r}^2}+\frac{\mathrm{B}}{\mathrm{r}^3} \\
& \left(\because \mathrm{a}_0=\frac{\epsilon_0 \mathrm{~h}^2}{\mathrm{~m} \pi \mathrm{e}^2} \text { Given }\right) \\
& \therefore \mathrm{r}=\mathrm{a}_0 \mathrm{n}^2-\mathrm{B}
\end{aligned}
$$

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2013 (23 Apr Online)
In the Boolean algebra \( Y=A \cdot B \) indicates that -

Options:
A output \( Y \) exists when either input \( A \) or input \( B \) exists.
B output \( Y \) exists only both inputs \( A \) and \( B \) exist.
C output \( Y \) exists when either input \( A \) exists or input \( B \). exists but not when both inputs \( A \) and \( B \) exist.
D product of \( A \) and \( B \) is \( Y \).

The correct answer is:
output \( Y \) exists only both inputs \( A \) and \( B \) exist.

Solution:
We have Y=A·B

Output exists means that output Y is 1.

Case 1:-A=1 and B=0⇒Y=1·0=0

Case 2:-A=0 and B=1⇒Y=0·1=0

Case 3:-A=1 and B=1⇒Y=1·1=1

Case 4:-A=0 and B=0⇒Y=0·0=0

Hence, output exists only when both inputs exist.

Topic: Physics, Semiconductors, JEE Main
In the Bragg's equation for diffraction of $X$-rays, $n$ represents for

Options:
A Quantum number
B An integer
C Avogadro's numbers
D Moles

The correct answer is:
An integer

Solution:
Bragg's equation is $n \lambda=2 d \sin \theta$
where $n$ is an integer i.e. $1,2,3,4$ etc.

Topic: Chemistry, Solid State, JEE Main
In the C4 cycle, the 4 carbon-containing acids are formed in

Options:
A mesophyll cell of maize leaf.
B mesophyll cell of cactus leaf.
C mesophyll cell of tomato leaf.
D mesophyll cell of banana leaf.

The correct answer is:
mesophyll cell of maize leaf.

Solution:
Plants that are adapted to dry tropical regions have the C4 pathway. Though these plants have the C4 oxaloacetic acid as the first CO2 fixation product they use the C3 pathway or the Calvin cycle as the main biosynthetic pathway. Since this is discovered by Hatch and Slack hence known as Hatch and Slack pathway.

Topic: Biology, Photosynthesis in Higher Plants, NEET
In the carbonyl compound \( \mathrm{Fe}_{2}(\mathrm{CO})_{9} \) what is the EAN of Fe:

Options:
A \( 34 \)
B \( 35 \)
C \( 36 \)
D \( 37 \)

The correct answer is:
\( 36 \)

Solution:
EAN of Fe=26+1+1+1+1+3×2=36

In the above expression, 1 is written for normal covalent bond and ligand donate 2 electrons.

Topic: Chemistry, Coordination Compounds, JEE Main
In the carbylamine reaction, R-X is converted to R-Y via the intermediate Z, R-X,R-Y and Z, respectively are

Options:
A R-NH2,R-NC , carbene
B R-NH2,R-NC , nitrene
C R-NC,R-NH2 , carbene
D R-OH,R-NC , nitrene

The correct answer is:
R-NH2,R-NC , carbene

Solution:
In carbylamine reaction, isocyanide (bad smelling compound) is obtained when primary amine (aliphatic as well as aromatic) reacts with chloroform in the presence of alkali. This reaction proceeds through dichlorocarbene intermediate which is obtained by the reaction of chloroform with alkene.
CHCl3+KOH→KCl+H2O+:CCl2Dichlorocarbene'Z'

The X=RNH2 and Y=RNC .

Topic: Chemistry, Amines, JEE Main
In the Carnot engine when the heat is taken from the source, then the temperature of the source

Options:
A remains constant
B does not remain constant
C decreases
D increases

The correct answer is:
remains constant

Solution:
Both source and sink in a Carnot cycle have
infinite capacity.
So, extracting or expelling heat in them does not change their temperatures.

Topic: Physics, Thermodynamics, TS EAMCET, TS EAMCET 2020 (09 Sep Shift 2)
In the case of a satellite moving along a circular orbit, a larger orbit corresponds to

Options:
A longer period and smaller velocity.
B larger velocity and longer period.
C smaller period and smaller velocity.
D smaller period and larger velocity.

The correct answer is:
longer period and smaller velocity.

Solution:
From Kepler's third law, the time period of satellite is
T2 ∝ r3
And larger the orbit then larger is the value of r, larger will be the time period.
Now the velocity of satellite is

v=GMr⇒v ∝1r.
Hence, larger value of r Corresponds to smaller value of v.

Topic: Physics, Gravitation, JEE Main
In the case of alkali metals, the covalent character decreases in the order :

Options:
A $\mathrm{MI}>\mathrm{MBr}>\mathrm{MCl}>\mathrm{MF}$
B $\mathrm{MCl}>\mathrm{MI}>\mathrm{MBr}>\mathrm{MF}$
C $\mathrm{MF}>\mathrm{MCl}>\mathrm{MBr}>\mathrm{MI}$
D $\mathrm{MF}>\mathrm{MCl}>\mathrm{MI}>\mathrm{MBr}$

The correct answer is:
$\mathrm{MI}>\mathrm{MBr}>\mathrm{MCl}>\mathrm{MF}$

Solution:
$\mathrm{MI}>\mathrm{MBr}>\mathrm{MCl}>\mathrm{MF}$
Down the group increases covalent character

Topic: Chemistry, s Block Elements, NEET, NEET 2009 (Mains)
In the case of amplitude modulation to avoid distortion the modulation index μ should be :

Options:
A μ≤1
B μ≥1
C μ=2
D μ=0

The correct answer is:
μ≤1

Solution:
The modulation index of an amplitude modulated signal is defined as the measure or extent of amplitude variation about an un-modulated carrier.

Modulation index is μ=AmAc

To avoid distortion, μ≤1 because μ>1 will result in interference between career frequency and message frequency, resulting into distortion.

Topic: Physics, Communication System, JEE Main, JEE Main 2022 (28 Jul Shift 1)
In the case of auto catalysis

Options:
A Solvent catalyses
B Product catalyses
C Reactant catalyses
D Heat produced in the reaction catalyses

The correct answer is:
Reactant catalyses

Solution:
In the case of auto catalysis, one of the product of the reaction acts as a catalyst.
For example: when nitric acid is poured on copper, the reaction gradually become faster as the nitrous acid formation increases during the reaction, which acts as an auto catalyst.

Topic: Chemistry, Surface Chemistry, JEE Main
In the case of peppered moth (Biston betularia) the black-coloured from became dominant over the light-coloured form in England during industrial revolution. This is an example of -

Options:
A Inheritance of darker colour character acquired due to the darker environment
B Natural selection whereby the darker forms were selected.
C Appearance of the darker coloured individuals due to very poor sunlight
D Protective mimicry

The correct answer is:
Natural selection whereby the darker forms were selected.

Solution:
No Solution Available

Topic: Biology, Evolution, NEET, NEET 2009 (Mains)
In the case of strongly exothermic reactions, what can be expected regarding the value of the equilibrium constant (K)?

Options:
A K becomes much larger than 1
B K becomes much smaller than 1
C K remains unchanged
D K is not affected by ∆rH°

The correct answer is:
K becomes much larger than 1

Solution:
In the case of strongly exothermic reactions, where ∆rH° is large and negative, the value of K becomes much larger than 1, suggesting that the reaction can go to near completion and form a significant amount of product.

Topic: Chemistry, Thermodynamics (C), NEET
In the Castner's process for the extraction of sodium, the anode is made of.......metal.

Options:
A Copper
B Iron
C Sodium
D Nickel

The correct answer is:
Sodium

Solution:
In castner process Na metal is made of anode.

Topic: Chemistry, s Block Elements, JEE Main
In the cell, PtsH2g, 1bar HCl aqAgClsAgsPts,the cell potential is 0.92 V when a 10-6 molar HCl solution is used. The standard electrode potential of Ag|AgCl|Cl- electrode is:
(Given, 2.303RTF=0.06 V at 298 K)

Options:
A 0.76 V
B 0.20 V
C 0.40 V
D 0.94 V

The correct answer is:
0.20 V

Solution:
A:12H2g→H+ aq.+e-; Eo=0.0 V
C :AgCl s+e-→Ag s+Cl-; Eo=xV
Ecell=Ecello-0.061log⁡H+Cl-
0.92=x-0.061log⁡10-12
0.92=x+0.72
x=0.92-0.72
=0.2V

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2019 (10 Jan Shift 2)
In the cell $\mathrm{Zn|} \mathrm{Zn}^{2+} \| \mathrm{Cu}^{2+} \mid \mathrm{Cu}$, the negative electrode is

Options:
A $\mathrm{Cu}$
B $\mathrm{Cu}^{2+}$
C $\mathrm{Zn}$
D $\mathrm{Zn}^{2+}$

The correct answer is:
$\mathrm{Zn}$

Solution:
In the cell $\mathrm{Zn}\left|\mathrm{Zn}^{2+} \| C u^{2+}\right| \mathrm{Cu}$ the negative electrode (anode) is $\mathrm{Zn}$. In electrochemical cell representation anode is always written on the left side while cathode on the right side.

Topic: Chemistry, Electrochemistry, JEE Main
In the chemical reaction, $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NH}_2-\mathrm{CHCl}_3+3 \mathrm{KOH} \longrightarrow(\mathrm{A})+(\mathrm{B})+3 \mathrm{H}_2 \mathrm{O}$, the compound $\mathrm{(A)}$ and $\mathrm{(B)}$ are respectively

Options:
A
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{CN}$ and $3 \mathrm{KCl}$

B
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CONH}_2$ and $3 \mathrm{KCl}$

C
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{NC}$ and $\mathrm{K}_2 \mathrm{CO}_3$

D
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{NC}$ and $3 \mathrm{KCl}$

The correct answer is:
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{NC}$ and $3 \mathrm{KCl}$

Solution:
It is example of carbylamine reaction. so, the product will be $\mathrm{C}_2 \mathrm{H}_5 \mathrm{NC}$ and $\mathrm{KCl}$.
Hence, (D) is the correct answer.

Topic: Chemistry, Amines, JEE Main, JEE Main 2007
In the chemical reactions given above A and B respectively are :

Options:
A H3PO2 and CH3CH2Cl
B CH3CH2OH and H3PO2
C H3PO2 and CH3CH2OH
D CH3CH2Cl and H3PO2

The correct answer is:
H3PO2 and CH3CH2Cl

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2021 (22 Jul Shift 1)
In the chemical reactions,
the compounds ' $\mathrm{A}$ ' and ' $\mathrm{B}$ ' respectively are

Options:
A
nitrobenzene and fluorobenzene

B
phenol and benzene

C
benzene diazonium chloride and fluorobenzene

D
nitrobenzene and chlorobenzene

The correct answer is:
benzene diazonium chloride and fluorobenzene

Solution:

Topic: Chemistry, Amines, JEE Main, JEE Main 2010
In the chemistry of vision in mammals, one of the photosensitive substances is called

Options:
A Melatonin
B Retinol
C Rhodopsin
D Melanin

The correct answer is:
Rhodopsin

Solution:
Melanin is the pigment that gives colour to the skin. Retinol is the other name for vitamin A. Melatonin is the hormone released from the pituitary gland to maintain diurnal rhythm.

Topic: Biology, Neural Control and Coordination, NEET
In the circle given below, let OA=1 unit, OB=13 unit and PQ⊥OB. Then, the area of the triangle PQB (in square units) is :

Options:
A 243
B 263
C 242
D 262

The correct answer is:
243

Solution:
Since x-axis is normal to the circle at O, and normal to the circle passes through its centre. So, OB is a diameter.

Now, perpendicular from centre of circle to chord bisects the chord, so PA=AQ=λ (Let)

Also, product of segments of chords in a circle is equal i.e. OA·AB=AP·AQ

⇒1·12=λ·λ

⇒λ=23 units

Now, area of ΔPQB=12×PQ×AB=12×2λ×AB

i.e. Δ=12·43×12 sq. units

=243 sq. units

Topic: Mathematics, Circle, JEE Main, JEE Main 2021 (26 Feb Shift 1)
In the circuit below, $A$ and $B$ represent two inputs and $C$ represents the output. The circuit represents

Options:
A
NOR gate

B
AND gate

C
NAND gate

D
OR gate

The correct answer is:
OR gate

Solution:
\begin{array}{l|l|l}
\mathrm{A} & \mathrm{B} & \mathrm{C} \\
\hline 0 & 0 & 0 \\
\hline 0 & 1 & 1 \\
\hline 1 & 0 & 1 \\
\hline 1 & 1 & 1
\end{array}

Topic: Physics, Semiconductors, JEE Main, JEE Main 2008
In the circuit diagram shown in figure given below, the current flowing through resistance 3 Ω is x3 A. The value of x is ________.

Options:

The correct answer is:
1

Solution:
Let the current flowing through the circuit be I A in anti-clockwise direction.

The net resistance is R=4.5 Ω+2 Ω+1.5 Ω=8 Ω

Hence, I=VR=4 V8 Ω=0.5 A

The current flowing through the 3 Ω resistor is

66+3×0.5 =13 A

So, x=1

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (11 Apr Shift 1)
In the circuit diagram shown in the adjoining figure, the resultant capacitance between $P$ and $Q$ is

Options:
A $47 \mu \mathrm{F}$
B $3 \mu \mathrm{F}$
C $60 \mu \mathrm{F}$
D $10 \mu \mathrm{F}$

The correct answer is:
$3 \mu \mathrm{F}$

Solution:
The given circuit can be drawn as $12 \mu \mathrm{F}$ where $C=(3+2) \mu F=5 \mu F$
$\frac{1}{C_{P Q}}=\frac{1}{5}+\frac{1}{20}+\frac{1}{12}=\frac{20}{60}=\frac{1}{3}$
$\Rightarrow C_{P Q}=3 \mu F$

Topic: Physics, Capacitance, JEE Main
In the circuit diagrams (A, B, C and D) shown below, $\mathrm{R}$ is a high resistance and $\mathrm{S}$ is a resistance of the order of galvanometer resistance G. The correct circuit, corresponding to the half deflection method for finding the resistance and figure of merit of the galvanometer, is the circuit labelled as:
(a)
(b)
(C)
(d)

Options:
A
Circuit A with $\mathrm{G}=\frac{\mathrm{RS}}{(\mathrm{R}-\mathrm{S})}$

B
Circuit B with $\mathrm{G}=\mathrm{S}$

C
Circuit $\mathrm{C}$ with $\mathrm{G}=\mathrm{S}$

D
Circuit $D$ with $\left.G=\frac{R S}{(R-S}\right)$

The correct answer is:
Circuit $D$ with $\left.G=\frac{R S}{(R-S}\right)$

Solution:
The correct circuit diagram is $D$ with galvanometer resistance
$$
G=\frac{R S}{R-S}
$$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2014 (11 Apr Online)
In the circuit given below, the value of the current is

Options:
A O$ amp
B $10^{-2} \mathrm{amp}$
C $10^2$ amp
D $10^{-3} \mathrm{amp}$

The correct answer is:
$10^{-2} \mathrm{amp}$

Solution:
current flow is possible and $i=\frac{V}{R}=\frac{(4-1)}{300}=10^{-2} \mathrm{~A}$

Topic: Physics, Semiconductors, JEE Main
In the circuit, given in the figure currents in different branches and value of one resistor are shown. Then potential at point B with respect to the point A is:

Options:
A +2V
B -2V
C –1V
D +1V

The correct answer is:
+1V

Solution:
Using kirchhoff's law in the circuit ACDB, we get

Voltage at B= voltage at A + voltage of battery of 1 V + potential difference across 2 Ω resistance - voltage of battery of 2 V

⇒VA+1+2(1)-2=VB

Now potential at point B with respect to point A is given by,

⇒VB-VA=1 V

Topic: Physics, Current Electricity, JEE Main, JEE Main 2020 (05 Sep Shift 2)
In the circuit here, the steady state voltage across capacitor $C$ is a fraction of the battery e.m.f. The fraction is decided by

Options:
A $R_1$ only
B $R_1$ and $R_2$ only
C $R_1$ and $R_3$ only
D $R_1, R_2$ and $R_3$

The correct answer is:
$R_1$ and $R_2$ only

Solution:
In steady state potential difference across capacitor
$V_2=$ potential difference across resistance
$R_2=\left(\frac{R_2}{R+R_2}\right) V$
Hence $\quad V_2$ depends upon $R_2$ and $R_1$

Topic: Physics, Capacitance, JEE Main
In the circuit of CE amplifier, a silicon transistor is used. The value of VCC=+20 V,RL=3 kÎ© , collector voltage =5 V,Î²=100. Then the base resistance RB should be (Take input voltage drop across Si transistor =0.7 V )

Options:
A 3.86Ã105 Î©
B 2.16×104 Ω
C 1.39×104 Ω
D 5.44×105 Ω

The correct answer is:
3.86Ã105 Î©

Solution:
VCC=VC+ICRL
â´20=5+3Ã103IC
â´IC=153Ã103
=5Ã10-3 A=5 mA ,
Î²=ICIB
â´IB=ICÎ²=5Ã10-3100
=5Ã10-5 A
VCC=IBRB+P.D. across input junction
As silicon transistor is in forward bias, the P.D. across input junction is 0.7 volt.
â´20=5Ã10-5RB+0.7
â´RB=19.35Ã10-5
â´RB=3.86Ã105 Î©

Topic: Physics, Semiconductors, JEE Main
In the circuit of CE amplifier, a silicon transistor is used. The value of VCC=+20 V,RL=3 kÎ© , collector voltage =5 V,Î²=100. Then the base resistance RB should be ------ Ã105 Î© (Take input voltage drop across Si transistor =0.7 V )

Options:

The correct answer is:
3.86

Solution:
VCC=VC+ICRL
∴20=5+3×103IC
∴IC=153×103
=5×10-3 A=5 mA ,
β=ICIB
∴IB=ICβ=5×10-3100
=5×10-5 A
VCC=IBRB+P.D. across input junction
As silicon transistor is in forward bias, the P.D. across input junction is 0.7 volt.
∴20=5×10-5RB+0.7
∴RB=19.35×10-5
∴RB=3.86×105Ω

Topic: Physics, Semiconductors, JEE Main
In the circuit shown, a four-wire potentiometer is made of a 400 cm long wire, which extends between A and B . The resistance per unit length of the potentiometer wire is r=0.01 Ω/cm. If an ideal voltmeter is connected as shown with jockey J at 50 cm from end A , the expected reading of the voltmeter will be:

Options:
A 0.50 V
B 0.25 V
C 0.75Â V
D 0.20Â V

The correct answer is:
0.25 V

Solution:
From the given question we take,

Current in the wire, I=VR=1.5+1.50.5+0.5+1+0.01×400=0.5 A

Resistance of 50 cm long wire, =50×0.01=0.5 Ω

Reading = Current × Resistance
=0.5×0.5
=0.25 V

Topic: Physics, Current Electricity, JEE Main, JEE Main 2019 (08 Apr Shift 2)
In the circuit shown as P≠R and the reading of the galvanometer G is same with the switch open or closed. Then

Options:
A IR=IG
B IP=IG
C IQ=IG
D IQ=IR

The correct answer is:
IR=IG

Solution:
As galvanometer deflection remains unaffected with switch S open or closed, hence the bridge circuit is balanced. Hence, Ip=IQand IR=IG
However, as P≠R, hence IP≠IR

Topic: Physics, Current Electricity, NEET
In the circuit shown below, all the inductors (assumed ideal) and resistors are identical. The current through the resistance on the right is I after the key $\mathrm{K}$ has been switched on for a long time. The currents through the three resistors(in order, from left to right) immediately after the key is switched off are

Options:
A $2 I$ upwards, $I$ downwards and $I$ downwards.
B $2 I$ downwards, $I$ downwards and $I$ downwards.
C $I$ downwards, $I$ downwards and $I$ downwards
D $0, I$ downwards and $I$ downwards

The correct answer is:
$2 I$ upwards, $I$ downwards and $I$ downwards.

Solution:
As current does not change through inductor immediately. It remains I through right \& middle resistance \& 2 i upwards in left resistance?

Topic: Physics, Alternating Current, KVPY, KVPY 2015 (SB/SX)
In the circuit shown below, i1 and i2 are the steady-state values of the current through L1 and L2 respectively, then i1 is

Options:
A i1=EL2RL1+L2
B i1=EL1RL1+L2
C i1=EL2RL1L2
D i1=EL1RL1L2

The correct answer is:
i1=EL2RL1+L2

Solution:
Since the potential difference across the inductors is same, we get

L1di1dt=L2di2dt

L1di1=L2di2

L1∆i1=L2∆i2

L1i1=L2i2 (Since the initial value of current through both the inductors was zero)

i1i2=L2L1

The steady current i passing through R is

i=ER

i1=L2L1+L2i=EL2RL1+L2

Topic: Physics, Electromagnetic Induction, JEE Main