datavorous/entrance-exam-dataset huggingface dataset:
In an experiment with 15 observations on $x$, the following results were available $\sum x^2=2830, \quad \sum x=170$. On observation that was 20 was found to be wrong and was replaced by the correct value 30 . Then the corrected variance is

Options:
A $78.00$
B $188.66$
C $177 \cdot 33$
D $8.33$

The correct answer is:
$78.00$

Solution:
$\sum x=170, \sum x^2=2830$
Increase in $\sum x=10$, then $\sum x^{\prime}=170+10=180$
Increase in $\sum x^2=900-400=500$, then $\sum x^{\prime 2}=2830+500=3330$
$\therefore$ Variance $=\frac{1}{n} \sum x^{\prime 2}-\left(\frac{\sum x^{\prime}}{n}\right)^2$
$=\frac{3330}{15}-\left(\frac{180}{15}\right)^2=222-144=78$.

Topic: Mathematics, Statistics, JEE Main
In an experiment with 15 observations on $x$, the following results were available: $\Sigma x^2=2830, \Sigma x=170$ One observation that was 20 was found to be wrong and was replaced by the correct value 30 . The corrected variance is

Options:
A
$8.33$

B
$78.00$

C
$188.66$

D
$177.33$

The correct answer is:
$78.00$

Solution:
$\Sigma x=170, \Sigma x^2=2830$ increase in $\Sigma x=10$, then
$\Sigma x^{\prime}=170+10=180$
Increase in $\Sigma \mathrm{x}^2=900-400=500$ then
$\Sigma x^{\prime 2}=2830+500=3330$
Variance $=\frac{1}{n} \Sigma x^{\prime 2}-\left(\frac{1}{n} \Sigma x^{\prime}\right)^2$
$=\frac{1}{15} \times 3330-\left(\frac{1}{15} \times 180\right)^2=222-144=78$

Topic: Mathematics, Statistics, JEE Main, JEE Main 2003
In an experiment with a convex lens. The plot of the image distance v' against the object distance μ' measured from the focus gives a curve v'μ'=225. If all the distances are measured in cm. The magnitude of the focal length of the lens is _____ cm.

Options:

The correct answer is:
15

Solution:
Using Newton's formula for lenses,

vu=f2 , here, f is focal length of the lens.

Given here, v'μ'=225

Thus, focal length of the lens is f2=225

Or f=15 cm

Topic: Physics, Ray Optics, JEE Main, JEE Main 2022 (28 Jul Shift 2)
In an experiment with a potentiometer, $\mathrm{V}_{\mathrm{B}}=10 \mathrm{~V}$. $\mathrm{R}$ is adjusted to be $50 \Omega$ (figure). A student wanting to measure voltage $\mathrm{E}_1$ of a battery (approx. $8 \mathrm{~V}$ ) finds no null point possible. He then diminishes $\mathrm{R}$ to $10 \Omega$ and is able to locate the null point on the last (4th) segment of the potentiometer. Find the resistance of the potentiometer wire and potential drop per unit length across the wire in the second case.

Options:

None

Solution:
Consider $R^{\prime}$ be the resistance of the potentiometer wire and given variable resistor $(R=50 \Omega)$
Then the effective resistance of potentiometer
$$
=50 \Omega+\mathrm{R}^{\prime}
$$
Effective voltage applied across potentiometer $=10 \mathrm{~V}$ The current through the main circuit,
$$
\begin{aligned}
\mathrm{I} &=\frac{\mathrm{V}}{50 \Omega+\mathrm{R}^{\prime}} \\
&=\frac{10}{50 \Omega+\mathrm{R}^{\prime}}
\end{aligned}
$$
Potential difference across wire of potentiometer,
$$
I^{\prime}=\frac{10 R^{\prime}}{50 \Omega+R^{\prime}}
$$
As, with $50 \Omega$ resistor, null point can not obtained by 8 volt it's possible only when
$$
\frac{10 \times \mathrm{R}^{\prime}}{50+\mathrm{R}^{\prime}} < 8
$$
So, $\quad 10 \mathrm{R}^{\prime} < 400+8 \mathrm{R}^{\prime}$
$$
2 \mathrm{R}^{\prime} < 400 \text { or } \mathrm{R}^{\prime} < 200 \Omega
$$
Similarly with $10 \Omega$ resistor, null point is obtained at $V^{\prime}>8$ its possible only when
$$
\frac{10 \times R^{\prime}}{10+R^{\prime}}>8
$$
So, $\quad 2 R^{\prime}>80, \quad R^{\prime}>40$
As the null point is obtained on 4th segment or at $\left(\frac{3}{4}\right)$ of total length so at $\left(\frac{3}{4} R^{\prime}\right)$.
$$
\begin{aligned}
\frac{10 \times \frac{3}{4} R^{\prime}}{10+R^{\prime}} & < 8 \\
7.5 R^{\prime} & < 80+8 R^{\prime} \\
R^{\prime} &>160 \\
160 < R^{\prime} < 200
\end{aligned}
$$
So,
(At balance point)
Any R' between $160 \Omega$ and $200 \Omega$ will achieve.
Since, the null point on the last (4th) segment of the potentiometer, therefore potential drop across $400 \mathrm{~cm}$ of wire $>8 \mathrm{~V}$.
This imply that potential gradient
or
$$
\begin{aligned}
\mathrm{k} \times 400 \mathrm{~cm} &>8 \mathrm{~V} \quad \text { (at balance point) } \\
\mathrm{k} \times 4 \mathrm{~m} &>8 \mathrm{~V} \\
\mathrm{k}>2 \mathrm{~V} / \mathrm{m}
\end{aligned}
$$
Similarly, potential drop across $300 \mathrm{~cm}$ wire $ < 8 \mathrm{~V}$
$$
\mathrm{k} \times 300 \mathrm{~cm} < 8 \mathrm{~V}
$$
$$
\begin{aligned}
\mathrm{k} \times 3 \mathrm{~m} & < 8 \mathrm{~V} \\
\mathrm{k} & < \frac{8}{3} \mathrm{~V} / \mathrm{m}
\end{aligned}
$$
Hence, $\frac{8}{3} V / m>k>2 V / m$

Topic: Physics, Current Electricity
In an experiment with sonometer a tuning fork of frequency $256 \mathrm{~Hz}$ resonates with a length of $25 \mathrm{~cm}$ and another tuning fork resonates with a length of $16 \mathrm{~cm}$. Tension of the string remaining constant the frequency of the second tuning fork is

Options:
A $163.84 \mathrm{~Hz}$
B $400 \mathrm{~Hz}$
C $320 \mathrm{~Hz}$
D $204.8 \mathrm{~Hz}$

The correct answer is:
$400 \mathrm{~Hz}$

Solution:
The wavelength of the sound is proportional to the length of wire. Also, the velocity of sound is same in both cases, so
$\begin{aligned}
& \Rightarrow & v_{1} &=v_{2} \\
\Rightarrow & & f_{1} \lambda_{1} &=f_{2} \lambda_{2} \\
\Rightarrow & 256 \times 25 &=f_{2} \times 16 \\
\Rightarrow & & f_{2}=\frac{256 \times 25}{16}=400 \mathrm{~Hz}
\end{aligned}$

Topic: Physics, Waves and Sound, COMEDK, COMEDK 2017
In an expression $a \times 10^{\mathrm{b}}$;

Options:
A $b$ is order of magnitude for $a \geq 5$
B $b$ is order of magnitude for $a \leq 5$
C $a$ is order of magnitude for $b \leq 5$
D $b$ is order of magnitude for $5 < a \leq 10$

The correct answer is:
$b$ is order of magnitude for $a \leq 5$

Solution:
$\begin{aligned} & a \times 10^b \\ & \text { if } a \leq 5 \text { order is } b \\ & a>5 \text { order is } b+1\end{aligned}$

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2024 (08 Apr Shift 1)
In an ice crystal, each water molecule is hydrogen bonded to _____ neighbouring molecules.

Options:

The correct answer is:
4

Solution:
Hydrogen bonding occurs between hydrogen and strongly electronegative atoms like F, N, O. In water or ice hydrogen bonding occurs as there is hydrogen as well as oxygen both are present. Water has 2 hydrogen atoms and 1 oxygen atom. The oxygen of one water molecule has two lone pairs of electrons, each of which can form hydrogen bonds with hydrogen on other two water molecules.

Each water molecule is H-bonded to 4 neighbouring molecules.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2023 (06 Apr Shift 2)
In an ideal double-slit experiment, when a glass - plate Î¼=1.5 of thickness t is introduced in the path of one interfering beams (wavelength Î»), the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass - plate is

Options:
A λ
B λ3
C 2λ3
D 2λ

The correct answer is:
2λ

Solution:
According to given condition,
μ-1t=nλ For minima put n=1
So, μ-1tmin=λ
tmin=λμ-1
= λ1.5-1=2λ

Topic: Physics, Wave Optics, NEET
In an ideal gas at temperature T, the average force that a molecule applies on the walls of a closed container depends on T as Tq . A good estimate for q is:

Options:
A 2
B 12
C 1
D 14

The correct answer is:
1

Solution:
Average linear velocity for collision to occur
t=2du
Change in momentum in 1 collision
Îp=2 mu
â´ average force in collision
= Îpt
(u= root mean square speed)
=2 mu2d Ãu
â Fâu2
â´ u2 âT
â FâT
â q=1

Topic: Physics, Kinetic Theory of Gases, JEE Main
In an ideal gas at temperature T, the average force that a molecule applies on the walls of a closed container depends on T as Tq . A good estimate for q is:

Options:
A 2
B 12
C 1
D 14

The correct answer is:
1

Solution:
Average linear velocity for collision to occur
t=2du
Change in momentum in 1 collision
Îp=2 mu
â´ average force in collision
= Îpt
(u= root mean square speed)
=2 mu2d Ãu
â Fâu2
â´ u2 âT
â FâT
â q=1

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2015 (10 Apr Online)
In an ideal gas at temperature T, the average force that a molecule applies on the walls of a closed container depends on T as Tq. A good estimate for q is

Options:

The correct answer is:
1

Solution:
The pressure exerted by molecules of a gas on the walls p=13nmc2
∴ p∝c2
But C∝T and C2∝T
∴ P∝T
∵ Force =P×A
∴ F∝T1
q=1

Topic: Physics, Kinetic Theory of Gases, JEE Main
In an increasing geometric progression of positive terms, the sum of the second and sixth terms is $\frac{70}{3}$ and the product of the third and fifth terms is 49 . Then the sum of the $4^{\text {th }}, 6^{\text {th }}$ and $8^{\text {th }}$ terms is equal to :

Options:
A 96
B 91
C 84
D 78

The correct answer is:
91

Solution:
$\mathrm{T}_2+\mathrm{T}_6=\frac{70}{3}$
$\mathrm{ar}+\mathrm{ar}^5=\frac{70}{3}$
$\mathrm{~T}_3 \cdot \mathrm{T}_5=49$
$\mathrm{ar}^2 \cdot \mathrm{ar}^4=49$
$\mathrm{a}^2 \mathrm{r}^6=49$
$\mathrm{ar}^3=+7, \mathrm{a}=\frac{7}{\mathrm{r}^3}$
$\mathrm{ar}\left(1+\mathrm{r}^4\right)=\frac{70}{3}$
$\frac{7}{\mathrm{r}^2}\left(1+\mathrm{r}^4\right)=\frac{70}{3}, \mathrm{r}^2=\mathrm{t}$
$\frac{1}{\mathrm{t}}\left(1+\mathrm{t}^2\right)=\frac{10}{3}$
$3 \mathrm{t}^2-10 \mathrm{t}+3=0$
$\mathrm{t}=3, \frac{1}{3}$
Increasing G.P.
$\mathrm{r}^2=3, \mathrm{r}=\sqrt{3}$
$\mathrm{T}_4+\mathrm{T}_6+\mathrm{T}_8$
$=\mathrm{ar}^3+\mathrm{ar}^5+\mathrm{ar}^7$
$=\operatorname{ar}^3\left(1+\mathrm{r}^2+\mathrm{r}^4\right)$
$=7(1+3+9)=91$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2024 (08 Apr Shift 2)
In an increasing geometric progression, the sum of the first and the last term is 99, the product of the second and the last but one term is 288 and the sum of all the terms is 189. Then, the number of terms in the progression is equal to

Options:
A 5
B 6
C 7
D 8

The correct answer is:
6

Solution:
G.P. is increasing, i.e. r>1.
Given, a+arn-1=99, ar⋅arn-2=288 and a1-rn1-r=189 .
a1+rn-1=99 and a2rn-1=288
⇒a1+288a2=99
⇒a2+288=99a⇒a2-99a+288=0
⇒a=3, 96⇒rn-1=288a2=32, 132
As r>1⇒rn-1=32 for a=3
Now, a1-rn1-r=189⇒31-r⋅rn-11-r=189
⇒1-r⋅321-r=63⇒1-32r=63-63r
⇒31r=62⇒r=2.

Now, 2n-1=32⇒n=6.

Topic: Mathematics, Sequences and Series, JEE Main
In an increasing geometric series, the sum of the second and the sixth term is 252 and the product of the third and fifth term is 25. Then, the sum of 4th,6th and 8th terms is equal to:

Options:
A 35
B 32
C 26
D 30

The correct answer is:
35

Solution:
a, ar, ar2....

T2+T6=252⇒ar1+r4=252

a2r21+r42=6254 ⋯1

T3·T5=25⇒ar2ar4=25

a2r6=25 ...2

On dividing 1 by 2

1+r42r4=254

4r8-17r4+4=0

4r4-1r4-4=0

r4=14,4⇒r4=4 (an increasing geometric series)

a2r6=25⇒ar32=25

T4+T6+T8=ar3+ar5+ar7

=ar31+r2+r4

=51+2+4=35

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2021 (26 Feb Shift 1)
In an inflorescence where flowers are borne laterally in an acropetal succession, the position of the youngest floral bud shall be

Options:
A Proximal
B Distal
C Intercalary
D Anywhere

The correct answer is:
Distal

Solution:
In racemose type of inflorescences the main axis continues to grow, the flowers are borne laterally in an acropetal succession, i.e. youngest flower is present at apex and oldest flower is present at the base. In racemose, inflorescence the growth of floral axis is unlimited or indefinite.

Topic: Biology, Morphology of Flowering Plants, NEET
In an interference experiment the ratio of amplitudes of coherent waves is a1a2=13 . The ratio of maximum and minimum intensities of fringes will be:

Options:
A 4
B 9
C 2
D 18

The correct answer is:
4

Solution:
ImaxImin=a1+a2a1-a22=a1/a2+1a1/a2-12=13+113-12
Imax : Imin=4:1

Topic: Physics, Wave Optics, JEE Main, JEE Main 2019 (08 Apr Shift 1)
In an interference experiment, the spacing between successive maxima or minima is

Options:
A $\lambda d / D$
B $\lambda D / d$
C $d D / \lambda$
D $\lambda d / 4 D$

The correct answer is:
$\lambda D / d$

Solution:
In an interference experiment the spacing between successive maxima and minima is called the fringe width and is given by
$$
\beta=D \lambda / d
$$

Topic: Physics, Wave Optics, JEE Main
In an interference experiment, third bright fringe is obtained at a point on the screen with a light of $700 \mathrm{~nm}$. What should be the wavelength of the light source in order to obtain 5 th bright fringe at the same point?

Options:
A $500 \mathrm{~nm}$
B $630 \mathrm{~nm}$
C $750 \mathrm{~nm}$
D $420 \mathrm{~nm}$

The correct answer is:
$420 \mathrm{~nm}$

Solution:
$n_{1} \lambda_{1}=n_{2} \lambda_{2}$
$$
\Rightarrow \quad 3 \times 700=5 \times \lambda_{2}
$$
$$
\Rightarrow \quad \lambda_{2}=420 \mathrm{~nm}
$$

Topic: Physics, Wave Optics, JEE Main
In an interference pattern the ratio of the intensity of light at the bright fringe to that at the dark fringe is $9: 1$. Then, the ratio of the amplitudes of the two interfering waves

Options:
A $3: 1$
B $2: 1$
C $1: 4$
D $5: 4$

The correct answer is:
$2: 1$

Solution:
Given, $\frac{I_{\max }}{I_{\min }}=9$
Since, intensity $\propto(\text { amplitude })^{2}$
$\therefore \quad \frac{A_{\max }}{A_{\min }}=\frac{a_{1}+a_{2}}{a_{1}-a_{2}}=\sqrt{\frac{9}{1}}=\frac{3}{1}$
$\Rightarrow \quad a_{1}+a_{2}=3 a_{1}-3 a_{2}$
$\Rightarrow \quad 2 a_{1}=4 a_{2}$
$\therefore \quad a_{1}: a_{2}=2: 1$

Topic: Physics, Waves and Sound, JEE Main
In an irreversible process taking place at constant $\mathrm{T}$ and $\mathrm{P}$ and in which only pressure-volume work is being done, the change in Gibbs free energy (dG) and change in entropy (dS), satisfy the criteria

Options:
A
$(\mathrm{dS})_{\mathrm{V}, \mathrm{E}}>0,(\mathrm{dG})_{\mathrm{T}, \mathrm{P}} < 0$
B
$(\mathrm{dS})_{\mathrm{V}: \mathrm{E}}=0,(\mathrm{dG})_{\mathrm{TPP}}=0$

C
$(\mathrm{dS})_{\mathrm{V}, \mathrm{E}}=0,(\mathrm{dG})_{\mathrm{T}, \mathrm{P}}>0$

D
$(\mathrm{dS})_{\mathrm{V}, \mathrm{E}} < 0,(\mathrm{dG})_{\mathrm{T}, \mathrm{P}} < 0$

The correct answer is:
$(\mathrm{dS})_{\mathrm{V}, \mathrm{E}}>0,(\mathrm{dG})_{\mathrm{T}, \mathrm{P}} < 0$

Solution:
For spontaneous reaction, $d S>0$ an $\Delta \mathrm{G}$ and $\mathrm{dG}$ should be negative i.e. $ < 0$

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2003
In an isochoric process

Options:
A work done is constant
B volume changes, work done remains same
C volume remains constant and no work is done by the system
D Both volume and work done changes

The correct answer is:
volume remains constant and no work is done by the system

Solution:
In isochoric process, volume of the system remains constant.
i.e. $\quad \Delta V=V_{1}-V_{2}=V-V=0 \quad\left[\because V_{1}=V_{2}=V\right]$
Hence, work done, $W=p \times \Delta V=p \times 0=0$

Topic: Physics, Thermodynamics, JEE Main
In an isochoric process, if \( T_{1}=27{ }^{\circ} \mathrm{C} \) and \( T_{2}=127{ }^{\circ} \mathrm{C} \), then \( \frac{p_{1}}{p_{2}} \) will be equal to

Options:
A \( \frac{9}{59} \)
B \( \frac{2}{3} \)
C \( \frac{3}{4} \)
D \( \frac{1}{3} \)

The correct answer is:
\( \frac{3}{4} \)

Solution:
In an isochoric process, volume of the gas remains constant.

As per ideal gas equation,

pV=nRT.

Thus, at constant volume, p∝T.

⇒p1p2=T1T2

⇒p1p2=27+273127+273

⇒p1p2=300400=34.

Topic: Physics, Thermodynamics, JEE Main
In an isochoric process

Options:
A work done is constant
B volume changes, work done remains same
C volume remains constant and no work is done by the system
D Both volume and work done changes

The correct answer is:
volume remains constant and no work is done by the system

Solution:
In isochoric process, volume of the system remains constant.
i.e. $\quad \Delta V=V_{1}-V_{2}=V-V=0 \quad\left[\because V_{1}=V_{2}=V\right]$
Hence, work done, $W=p \times \Delta V=p \times 0=0$

Topic: Physics, Thermodynamics, COMEDK, COMEDK 2015
In an isolated system, what is the tendency regarding the system's energy and entropy?

Options:
A The system's energy tends to become more ordered, and its entropy decreases.
B The system's energy tends to become more disordered, and its entropy increases.
C The system's energy remains constant, and its entropy remains constant.
D The system's energy decreases, and its entropy remains constant.

The correct answer is:
The system's energy tends to become more disordered, and its entropy increases.

Solution:
In an isolated system, there is a tendency for the system's energy to become more disordered or chaotic, leading to an increase in entropy.

Topic: Chemistry, Thermodynamics (C), NEET
In an isosceles right angled triangle, if the equation of the hypotenuse and its opposite vertex are 3x+4y=4 and 2, 2, then the slopes of the remaining two sides are

Options:
A 17, -7
B -17, 7
C 17, 7
D -17, -7

The correct answer is:
17, -7

Solution:
Given that in an isosceles triangle equation of hypotenuse is 3x+4y=4 i.e. y=-34x+44

∴Slope the hypotenuse is -34.

Let the slope of required line is m.

As, tanθ=m1-m21+m1m2

∵Angle made by the hypotenuse with other two sides is 45°.

∴tan 45°=m+341-3m4

⇒±1=4m+34-3m

⇒4m+3=±4-3m

⇒4m+3=4-3m or 4m+3=-4+3m

⇒m=17,m=-7

Hence, option 1 is correct.

Topic: Mathematics, Straight Lines, TS EAMCET, TS EAMCET 2021 (05 Aug Shift 2)
In an isosceles trapezium, the length of one of the parallel sides, and the leng ths of the non-parallelsides are all equal to 30 . In order to maximize the area of the trapezium, the smallest angle should be

Options:
A $\frac{\pi}{6}$
B $\frac{\pi}{4}$
C $\frac{\pi}{3}$
D $\frac{\pi}{2}$

The correct answer is:
$\frac{\pi}{3}$

Solution:
In $\triangle A B F, y=30 \cos \theta$
Area is $A=\frac{1}{2}[60+60 \cos \theta](30 \sin \theta)$
$\begin{array}{l}
=30(1+\cos \theta)(30 \sin \theta) \\
=900(\sin \theta+\sin \theta \cos \theta)
\end{array}$
For maximum or minimum
$\begin{array}{l}
\frac{d A}{d \theta}=900\left[\cos \theta+\left(-\sin ^{2} \theta+\cos ^{2} \theta\right)\right]=0 \\
\cos \theta-1+\cos ^{2} \theta+\cos ^{2} \theta=0 \\
2 \cos ^{2} \theta+\cos \theta-1=0 \\
(2 \cos \theta-1)(\cos \theta+1)=0
\end{array}$
$\cos \theta=\frac{1}{2}$ or $\cos \theta=-1$ (not possible) $\theta=60^{\circ}$

Topic: Mathematics, Application of Derivatives, JEE Main
In an isosceles triangle ABC, the vertex A is 6,1 and the equation of the base BC is 2x+y=4. Let the point B lie on the line x+3y=7. If α,β is the centroid ∆ABC, then 15α+β is equal to

Options:
A 51
B 39
C 41
D 49

The correct answer is:
51

Solution:
In △ABC, AB=AC

Solving 2x+y=4 & x+3y=7, we get

B≡1,2

Let C≡h,k and as it lies on 2x+y=4

so 2h+k=4

Now, AB2=AC2

26=(h-6)2+k-12⇒26=h-62+3-2h2

⇒26=5h2-24h+45 ⇒h-15h-19=0

⇒h=195 (as h=1rejected)

so k=-185

Hence, centroid =6+1+1953,1+2-1853≡185,-15

i.e. 15α+β=15×175=51

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2022 (27 Jun Shift 1)
In an isothermal expansion of one mole of an ideal gas against vacuum from 10 litre to 100 litre at 27oC , the quantity of heat absorbed by the gas is

Options:

The correct answer is:
0

Solution:
The quality of heat absorbed by the gas, Q=ΔE+W, here W is the work done by the gas. For free expansion i.e., expansion against vacuum, W=0; at constant temperature, the internal energy of an ideal gas is constant and so ΔE=0. Thus Q=0.

Topic: Chemistry, Thermodynamics (C), JEE Main
In an \( L C R \) series AC circuit, the voltage across each of the components is \( 50 \mathrm{~V} \). The voltage across the \( L C \) combination will be

Options:
A \( 50 \mathrm{~V} \)
B \( 50 \sqrt{2} \mathrm{~V} \)
C \( 100 \mathrm{~V} \)
D zero

The correct answer is:
zero

Solution:
In a series LCR circuit, the phase difference between voltages across an inductor and a capacitor is π, therefore, the net voltage gets subtracted.

Hence, voltage across LC combination will be zero.

Topic: Physics, Alternating Current, JEE Main
In an LCR series circuit, an inductor 30 mH and a resistor 1 Ω are connected to an AC source of angular frequency 300 rad s-1. The value of capacitance for which the current leads the voltage by 45° is 1x×10-3 F. Then the value of x is

Options:

The correct answer is:
3

Solution:
tanϕ=xC-xLR
tan45=xC-xLR
xC-xL=R
1ωC-ωL=R
1ωC-300×0.03=1
1ωC=10
C=110ω=110×300
C=13×10-3
X=3

Topic: Physics, Alternating Current, JEE Main, JEE Main 2021 (20 Jul Shift 1)
In an $L C R$ circuit shown in the following figure, what will be the readings of the voltmeter across the resistor and ammeter if an a.c. source of $220 \mathrm{~V}$ and $100 \mathrm{~Hz}$ is connected to it as shown?

Options:
A
$800 \mathrm{~V}, 8 \mathrm{~A}$

B
$110 \mathrm{~V}, 1.1 \mathrm{~A}$

C
$300 \mathrm{~V}, 3 \mathrm{~A}$

D
$220 \mathrm{~V}, 2.2 \mathrm{~A}$

The correct answer is:
$220 \mathrm{~V}, 2.2 \mathrm{~A}$

Solution:
In case of series RLC circuit, Equation of voltage is given by
$$
\mathrm{V}^2=V_R^2+\left(V_L-V_C^2\right)
$$
Here, $V=220 V ; V_L=V_C=300 \mathrm{~V}$
$$
\therefore V_R=\sqrt{V^2}=220 \mathrm{~V}
$$
Current $i=\frac{V}{R}=\frac{220}{100}=2.2 \mathrm{~A}$

Topic: Physics, Alternating Current, JEE Main, JEE Main 2012 (07 May Online)
In an $L R$-circuit, time constant is that time in which current grows from zero to the value (where $I_0$ is the steady state current)

Options:
A $0.63$ $I_0$
B $0.50$ $I_0$
C $0.37$ $I_0$
D $I_0$

The correct answer is:
$0.63$ $I_0$

Solution:
Current at any instant of time $t$ after closing an $L-R$ circuit is given by
$I=I_0\left[1-e^{\frac{-R}{L} t}\right]$
Time constant $t=\frac{L}{R}$
$\therefore I=I_0\left[1-e^{\frac{-R}{L} \times \frac{L}{R}}\right]=I_0\left(1-e^{-1}\right)=I_0\left(1-\frac{1}{e}\right)$
$=I_0\left(1-\frac{1}{2.718}\right)=0.63 I_0=63 \%$ of $I_0$

Topic: Physics, Electromagnetic Induction, JEE Main
In an LC oscillator, if values of inductance and capacitance become twice and eight times, respectively, then the resonant frequency of oscillator becomes x times its initial resonant frequency ω0. The value of x is:

Options:
A 14
B 16
C 116
D 4

The correct answer is:
14

Solution:
Resonance frequency of an LC oscillations circuit is stated as

ω0=1LC, here, L and C represents inductance and capacitance of circuit.

Now, when inductance and capacitance changes, i.e., L'=2L and C'=8C

Then, resonant frequency is ω=12L×8C=14LC

⇒ω=ω04

Thus, the value of x=14.

Topic: Physics, Alternating Current, JEE Main, JEE Main 2023 (25 Jan Shift 1)
In an LCR circuit as shown below both switches are open initially. Now switch S 1 is closed, S 2 kept open. ( q is charge on the capacitor and τ = RC is capacitive time constant). Which of the following statement is correct?

Options:
A At t = 2 τ , q = CV 1 - e - 2
B At t = τ 2 , q = CV 1 - e -1
C Work done by the battery is half of the energy dissipated in the resistor
D At t = τ , q = CV / 2

The correct answer is:
At t = 2 τ , q = CV 1 - e - 2

Solution:
q = CV 1 - e t / τ

At t = 2 τ
q = CV 1 - e - 2

Topic: Physics, Capacitance, JEE Main, JEE Main 2013 (07 Apr)
In an LCR circuit, the capacitance is made one-fourth, when in resonance. What should be the change in inductance, so that the circuit remains in resonance?

Options:
A 4 times
B 14 times
C 8 times
D 2 times

The correct answer is:
4 times

Solution:
For resonance
ω = 1 LC
for ω to remain in resonance LC should remain the same so if C is made one-forth L should be made 4 times.

Topic: Physics, Alternating Current, NEET
In an LCR circuit, the resonant frequency is 600 Hz and half-power points are at 650 Hz and 550 Hz. The quality factor is

Options:

The correct answer is:
6

Solution:
In LCR series, resonant frequency,
fr=600 Hz
∴ωr=2π×600 Hz
Frequencies at half power are given as,
f1=550 Hz and f2=650 Hz
ω1=2π×550 Hz and ω2=2π×650 Hz
Quality factor, Q=ωrω2-ω1
=2π×6002π×650-2π×550
=2π×6002π×100=6

Topic: Physics, Alternating Current, JEE Main
In an LCR series a.c. circuit, the voltage across each of the components, $L, C$ and $R$ is $50 \mathrm{~V}$. The voltage across the LC combination will be

Options:
A
$50 \mathrm{~V}$

B
$50 \sqrt{2} \mathrm{~V}$

C
$100 \mathrm{~V}$

D
$0 \mathrm{~V}$ (zero)

The correct answer is:
$0 \mathrm{~V}$ (zero)

Solution:
In series LCR circuit, the voltage across the inductor $(L)$ and the capacitor $(C)$ are in opposite phase.

Topic: Physics, Alternating Current, JEE Main, JEE Main 2004