datavorous/entrance-exam-dataset huggingface dataset:
A sealed container with negligible coefficient of volumetric expansion contains helium (a monoatomic gas). When it is heated from $300 \mathrm{~K}$ to $600 \mathrm{~K}$, the average K.E. of helium atoms is

Options:
A Halved
B Unchanged
C Doubled
D Increased by factor

The correct answer is:
Doubled

Solution:
Kinetic energy $\propto$ Temperature. Hence if temperature is doubles, Kinetic energy will also be doubled

Topic: Physics, Kinetic Theory of Gases, JEE Main
A sealed flask with a capacity of 2dm3 contains 11 g of propane gas. The flask is so weak that it will burst if the pressure becomes 2MPa. The minimum temperature at which the flask will burst is C∘. [Nearest integer] (Given: R=8.3 J K-1 mol-1. Atomic masses of C and H are 12u and 1u respectively.) (Assume that propane behaves as an ideal gas.)

Options:

The correct answer is:
1655

Solution:
Real gases behave ideally at high temperature and low pressure.

Number of moles of propane gas, C3H8=1144=0.25 moles

PV=nRT

P=Pressure=2MPa

V=Volume=2dm3

n=Number of moles

T=Temperature

⇒2×106×2×10-3=0.25×8.3×T

⇒T=1927.710 K=1654.56 °C

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2022 (25 Jul Shift 2)
A seconds pendulum clock having steel wire is calibrated at 20 Â°C. When temperature is increased to 30 Â°C, then calculate how much time does the clock lose or gain in one week Î±steel=1.2Ã10-5 âC-1

Options:
A 0.3628 s
B 3.626 s
C 362.8 s
D 36.28 s

The correct answer is:
36.28 s

Solution:
The time period of the second's pendulum = 2 s
Change in the time period, Î T = 1 2 T Î± Î Î¸
Î T = 1 2 2 1 . 2 Ã 1 0 - 5 3 0 â - 2 0 â
= 1 . 2 Ã 1 0 - 4 s
New time period, T â² = T + Î T
T â² = 2 . 0 0 0 1 2 s
Time lost in a week,
Î t = Î T T â² Ã t
= 1 . 2 Ã 1 0 - 4 2 . 0 0 0 1 2 Ã 7 Ã 2 4 Ã 3 6 0 0
= 3 6 . 28 s

Topic: Physics, Thermal Properties of Matter, JEE Main
A sector is removed from a metallic disc and the remaining region is bent into the shape of a circular conical funnel with volume $2 \sqrt{3} \pi$. The least possible diameter of the disc is

Options:
A 4
B 6
C 8
D 12

The correct answer is:
6

Solution:
for a cone
$\mathrm{r}$ will be cone slant height
$$
V=2 \sqrt{3}
$$
Let $x$ radius of cone
$\mathrm{h}$ be height then
$$
\begin{array}{l}
\frac{1}{3} \pi r^{2} h=2 \sqrt{3} \pi \\
x^{2} h=6 \sqrt{3} \Rightarrow x^{2}=\frac{6 \sqrt{3}}{h}
\end{array}
$$
least Diameter $\Rightarrow$ least slant height of cone $\ell^{2}=\mathrm{x}^{2}+\mathrm{h}^{2}$ $\mathrm{r}^{2}=\mathrm{x}^{2}+\mathrm{h}^{2}$
$$
r^{2}=\frac{6 \sqrt{3}}{h}+h^{2}
$$
Diff. w. to $\mathrm{x}$
$$
2 r \frac{\mathrm{dr}}{\mathrm{dh}}=\frac{-6 \sqrt{3}}{\mathrm{~h}^{2}}+2 \mathrm{~h}
$$
for maximum \& minimum $\frac{\mathrm{dr}}{\mathrm{dh}}=0$
$$
\begin{array}{l}
\mathrm{h}=\sqrt{3} \Rightarrow \mathrm{h}^{2}=3 \\
\mathrm{x}^{2}=\frac{6 \sqrt{3}}{\sqrt{3}} \\
\mathrm{x}^{2}=6 \\
\mathrm{r}^{2}=6+3 \\
\mathrm{r}^{2}=9 \\
\mathrm{r}=3 \\
\mathrm{~d}=2 \mathrm{r} \\
\mathrm{d}=6
\end{array}
$$

Topic: Mathematics, Application of Derivatives, JEE Main
A sector is removed from a metallic disc and the remaining region is bent into the shape of a circular conical funnel with volume $2 \sqrt{3} \pi$. The least possible diameter of the disc is

Options:
A 4
B 6
C 8
D 12

The correct answer is:
6

Solution:
for a cone
$\mathrm{r}$ will be cone slant height
$$
V=2 \sqrt{3}
$$
Let $x$ radius of cone
$\mathrm{h}$ be height then
$$
\begin{array}{l}
\frac{1}{3} \pi r^{2} h=2 \sqrt{3} \pi \\
x^{2} h=6 \sqrt{3} \Rightarrow x^{2}=\frac{6 \sqrt{3}}{h}
\end{array}
$$
least Diameter $\Rightarrow$ least slant height of cone $\ell^{2}=\mathrm{x}^{2}+\mathrm{h}^{2}$ $\mathrm{r}^{2}=\mathrm{x}^{2}+\mathrm{h}^{2}$
$$
r^{2}=\frac{6 \sqrt{3}}{h}+h^{2}
$$
Diff. w. to $\mathrm{x}$
$$
2 r \frac{\mathrm{dr}}{\mathrm{dh}}=\frac{-6 \sqrt{3}}{\mathrm{~h}^{2}}+2 \mathrm{~h}
$$
for maximum \& minimum $\frac{\mathrm{dr}}{\mathrm{dh}}=0$
$$
\begin{array}{l}
\mathrm{h}=\sqrt{3} \Rightarrow \mathrm{h}^{2}=3 \\
\mathrm{x}^{2}=\frac{6 \sqrt{3}}{\sqrt{3}} \\
\mathrm{x}^{2}=6 \\
\mathrm{r}^{2}=6+3 \\
\mathrm{r}^{2}=9 \\
\mathrm{r}=3 \\
\mathrm{~d}=2 \mathrm{r} \\
\mathrm{d}=6
\end{array}
$$

Topic: Mathematics, Application of Derivatives, KVPY, KVPY 2017 (19 Nov SB/SX)
A segment of wire vibrates with a fundamental frequency of $450 \mathrm{~Hz}$ under a tension of $9 \mathrm{~kg} \mathrm{wt}$. Then tension at which the fundamental frequency of the same wire becomes $900 \mathrm{~Hz}$ is

Options:
A $36 \mathrm{~kg}-\mathrm{wt}$
B $27 \mathrm{~kg}-\mathrm{wt}$
C $18 \mathrm{~kg}-\mathrm{wt}$
D $72 \mathrm{~kg}-\mathrm{wt}$

The correct answer is:
$36 \mathrm{~kg}-\mathrm{wt}$

Solution:
Fundamental frequency of wire
$\begin{array}{rlrl}f & =\frac{1}{2 \pi} \sqrt{\frac{T}{m}} \\ \text { or } & f & \propto \sqrt{T} \\ \text { or } & \frac{f_2}{f_1} & =\sqrt{\frac{T_2}{T_1}} \\ \text { or } & \frac{900}{450} & =\sqrt{\frac{T_2}{9}} \\ \text { or } & T_2 & =4 \times 9=36 \mathrm{~kg}-\mathrm{wt}\end{array}$

Topic: Physics, Waves and Sound, JEE Main
A semicircular disc of mass M and radius R is free to rotate about its diameter. The moment of inertia of semicircular disc about a line perpendicular to its plane through centre is

Options:
A 34MR2
B MR22
C MR23
D MR24

The correct answer is:
MR22

Solution:
The moment of inertia of disc of radius R about its axis is

I=M'R22

where, M'= mass of complete disc =2M

I=2MR22

I=MR2 ...i

If I1 be the moment of inertia of semicircular disc about its. axis, then disc may be assumed as combination of two semi-circular parts,

Thus, I=I1+I1⇒I1=I2=MR22

Topic: Physics, Rotational Motion, JEE Main
A semi-circular loop of radius $30 \mathrm{~cm}$ wire carries current $6 \mathrm{~A}$. An uniform magnetic field $0.5 \mathrm{~T}$ is present perpendicular to the plane of the loop. What is the magnitude of force exerted on the wire?

Options:
A $0.9 \mathrm{~N}$
B $1.8 \mathrm{~N}$
C $0.8 \mathrm{~N}$
D $1.4 \mathrm{~N}$

The correct answer is:
$1.8 \mathrm{~N}$

Solution:
Given, the radius of a circular loop,
$$
R=30 \times 10^{-2} \mathrm{~m}
$$
current, $I=6 \mathrm{~A}$ and magnetic field, $B=0.5 \mathrm{~T}$.
Force exerted on the wire,
$$
\begin{array}{rlrl}
& F & =I L B \sin \theta \\
\Rightarrow \quad & F & =6 \times 2 \times 30 \times 10^{-2} \times 0.5 \times \sin 90^{\circ} \\
\Rightarrow \quad & & {\left[\because \theta=90^{\circ} \text { and } \sin 90^{\circ}=1\right]} \\
& =1.8 \mathrm{~N}
\end{array}
$$
Hence, the correct option is (2).

Topic: Physics, Magnetic Effects of Current, JEE Main
A semicircular wedge of mass M is placed on a horizontal floor. A man of mass m starts moving on wedge from position A with constant speed v relative to the wedge. Initially, both man & wedge were at rest . Neglect the friction between wedge & horizontal surface and man doesn't slips on the wedge while man moves from A to C relative to wedge. The correct statement is

Options:
A speed of wedge will first increase then decrease
B speed of the wedge will remain constant
C speed of the wedge first increases up to a maximum value and then remains constant
D the wedge doesn't move

The correct answer is:
speed of wedge will first increase then decrease

Solution:
The velocity of the wedge = –mv cosθm+M
θ : Angle made by v→manwith the horizontal

Topic: Physics, Center of Mass Momentum and Collision, NEET
A semicircular wire of radius r rotates in uniform magnetic field B about its diameter with angular velocity ω. If the total resistance of the circuit is R, then the mean power generated per period of rotation is

Options:
A πBr2ω28 R
B πBrω28R
C πBrω28 R
D 0

The correct answer is:
πBr2ω28 R

Solution:
The flux linked with coil (made of wire of radius r) of area A and magnetic field B, is given by

ϕ=BA cos θ=B πr22 cos ωt

ϕ=πBr22 cos ωt [∵ A=πr22 and θ=ωt]

By Faraday's law of electromagnetic induction, induced emf, e=-dϕdt

Hence, after differentiating,

e=-ddtπBr22 cos ωt=πBr2ω2 sin ωt (where, R= resistance in the circuit)

∴ Power P=e2R=π2B2r4ω2 sin2ωt4 R

∴ Mean power P¯=π2B2r4ω24 R·sin2ωt¯

=π2B2r4ω24 R·12 ∵ sin2ωt¯=12

=πBr2ω28 R

Topic: Physics, Electromagnetic Induction, JEE Main
A semiconductor $X$ is made by doping a germanium crystal with arsenic $(Z=33)$. A second semiconductor $Y$ is made by doping germanium with indium $(Z=49)$. The two are joined end to end and connected to a battery as shown. Which of the following statements is correct

Options:
A $X$ is $P$-type, $Y$ is $N$-type and the junction is forward biased
B $X$ is $N$-type, $Y$ is $P$-type and the junction is forward biased
C $X$ is $P$-type, $Y$ is $N$-type and the junction is reverse biased
D $X$ is $N$-type, $Y$ is $P$-type and the junction is reverse biased

The correct answer is:
$X$ is $N$-type, $Y$ is $P$-type and the junction is reverse biased

Solution:
Arsenic has five valence electrons, so it a donor impurity. Hence $X$ becomes $N$-type semiconductor. Indium has only three outer electrons, so it is an acceptor impurity. Hence $Y$ becomes P-type semiconductor. Also $N$ (ie. $X$ ) is connected to positive terminal of battery and P(i.e. $Y$ is connected to negative terminal of battery so $P N$-junction is reverse biased.

Topic: Physics, Semiconductors, JEE Main
A series AC circuit containing an inductor 20 mH, a capacitor 120 μF and a resistor 60 Ω is driven by an AC source of 24 V/50 Hz. The energy dissipated in the circuit in 60 s is:

Options:
A 5.17×102 J
B 3.39×103 J
C 2.26×103 J
D 5.65×102 J

The correct answer is:
5.17×102 J

Solution:
(Power factor), Pf=cos⁡ϕ=RZ

Where R, Z are the resistance and impedance respectively.
(Energy dissipated) E=Vrms2Z×cosϕ×t
⇒E=Vrms2Z×RZ×t

⇒E=Vrms2Z2×R×t

Vrms, t are the RMS voltage and time respectively.

We know that the Impedance of a LCR circuit is given by the formula,
Z2=R2+ωL-1ωC2⇒Z2=602+2π×50×20×10-3-12π×50×120×10-62

Where ω, L, C are the angular frequency, inductance and capacitance in the circuit respectively.
⇒Z2=602+2π-100012π2
⇒Z2=4009.7
∴E=2424009.7×60×60 J

= 517.14 J

=5.17×102 J

Topic: Physics, Alternating Current, JEE Main, JEE Main 2019 (09 Jan Shift 2)
A series combination of $n_1$ capacitors, each of capacity $C_1$ is charged by source of potential difference $4 \mathrm{~V}$. When another parallel combination of $n_2$ capacitors each of capacity $C_2$ is charged by a source of potential difference $V$, it has the same total energy stored in it as the first combination has. The value of $C_2$ in terms of $C_1$ is then

Options:
A
$16 \frac{n_2}{n_1} C_1$

B
$\frac{2 C_1}{n_1 n_2}$

C
$2 \frac{n_2}{n_1} C_1$

D
$\frac{16 C_1}{n_1 n_2}$

The correct answer is:
$\frac{16 C_1}{n_1 n_2}$

Solution:
Equivalent capacitance of $n_2$ number of capacitors each of capacitance $\mathrm{C}_2$ in parallel $=n_2 C_2$
Equivalent capacitance of $n_1$ number of capacitors each of capacitances $\mathrm{C}_1$ in series.
Capacitance of each is $C_1=\frac{C_1}{n_1}$
According to question, total energy stored in both the combinations are same
$$
\begin{aligned}
& \text { i.e., } \frac{1}{2}\left(\frac{C_1}{n_1}\right)(4 V)^2=\frac{1}{2}\left(n_2 C_2 \text { b }^2\right. \\
& \therefore C_2=\frac{16 C_1}{n_1 n_2} \\
&
\end{aligned}
$$

Topic: Physics, Capacitance, JEE Main
A series combination of $n_1$ capacitors, each of capacity $C_1$ is charged by source of potential difference $4 \mathrm{~V}$. When another parallel combination of $n_2$ capacitors each of capacity $C_2$ is charged by a source of potential difference $V$, it has the same total energy stored in it as the first combination has. The value of $C_2$ in terms of $C_1$ is then

Options:
A
$16 \frac{n_2}{n_1} C_1$

B
$\frac{2 C_1}{n_1 n_2}$

C
$2 \frac{n_2}{n_1} C_1$

D
$\frac{16 C_1}{n_1 n_2}$

The correct answer is:
$\frac{16 C_1}{n_1 n_2}$

Solution:
Equivalent capacitance of $n_2$ number of capacitors each of capacitance $\mathrm{C}_2$ in parallel $=n_2 C_2$
Equivalent capacitance of $n_1$ number of capacitors each of capacitances $\mathrm{C}_1$ in series.
Capacitance of each is $C_1=\frac{C_1}{n_1}$
According to question, total energy stored in both the combinations are same
$$
\begin{aligned}
& \text { i.e., } \frac{1}{2}\left(\frac{C_1}{n_1}\right)(4 V)^2=\frac{1}{2}\left(n_2 C_2 \text { b }^2\right. \\
& \therefore C_2=\frac{16 C_1}{n_1 n_2} \\
&
\end{aligned}
$$

Topic: Physics, Capacitance, JEE Main, JEE Main 2012 (12 May Online)
A series combination of $n_1$ capacitors, each of value $\mathrm{C}_1$, is charged by a source of potential difference $4 \mathrm{~V}$. When another parallel combination of $n_2$ capacitors, each of value $C_2$, is charged by a source of potential difference $V$, it has the same (total) energy stored in it, as the first combination has. The value of $\mathrm{C}_2$, in terms of $\mathrm{C}_1$, is then

Options:
A $\frac{2 \mathrm{C}_1}{\mathrm{n}_1 \mathrm{n}_2}$
B $16 \frac{\mathrm{n}_2}{\mathrm{n}_1} \mathrm{C}_1$
C $2 \frac{\mathrm{n}_2}{\mathrm{n}_1} \mathrm{C}_1$
D $\frac{16 C_1}{n_1 n_2}$

The correct answer is:
$\frac{16 C_1}{n_1 n_2}$

Solution:
P.D. across each capacitor is \(V_1=\frac{4 \mathrm{~V}}{\mathrm{n}_1}\)
Energy stored across the combination is \(\mathrm{U}_1=\mathrm{n}_1 \times \frac{1}{2} \mathrm{C}_1 \frac{16 \mathrm{~V}^2}{\mathrm{n}_1^2}=\frac{16 \mathrm{C}_1 \mathrm{~V}^2}{2 \mathrm{n}_1}\)

\(\begin{aligned}
& \text {Energy stored across parallel combination is } \mathrm{U}_2=\mathrm{n}_2 \times \frac{1}{2} \mathrm{C}_2 \mathrm{~V}^2\\
& \frac{16 C_1 v^2}{2 n_1}=\frac{n_2 C_1 v^2}{2} \\
& \Rightarrow C_2=\frac{16 C_1}{n_1 n_2}
\end{aligned}\)

Topic: Physics, Magnetic Properties of Matter, JEE Main
A series combination of resistor of resistance 100 Ω inductor of inductance 1 H and capacitor of capacitance 6.25 μF is connected to an ac source. The quality factor of the circuit will be _____.

Options:

The correct answer is:
4

Solution:
The formula to calculate the quality factor of an LCR circuit is given by

Q=1RLC ...1

Substitute the values of the known parameters into equation (1) to calculate the required quality factor of the circuit.

Q=110016.25×10-6=1100×10002.5=4

Topic: Physics, Alternating Current, JEE Main, JEE Main 2023 (08 Apr Shift 2)
A series \( L C R \) circuit is connected to an ac voltage source. When \( L \) is removed from the circuit, the phase difference between current and voltage is \( \frac{\pi}{3} \). If instead \( C \) is removed from the circuit, the phase difference is again \( \frac{\pi}{3} \) between current and voltage. The power factor of the circuit is:

Options:
A \( 0.5 \)
B \( 1.0 \)
C \( -1.0 \)
D zero

The correct answer is:
\( 1.0 \)

Solution:
When L removed tanϕ=XCR

When L removed tanϕ=XLR

XCR=XLR⇒ Resonance

Z=R

cosϕ=RZ=RR=1

Topic: Physics, Alternating Current, JEE Main
A series LCR circuit with 100 Ω resistance is connected to an ac source of 200 V with an angular frequency of 300 rad s-1. If only the capacitor is removed, the current lags behind the voltage by a phase of 60 ∘. If only the inductor is removed, the current leads the voltage by a phase of 60 ∘. The current and the power dissipated in the circuit is

Options:
A 3 A , 300 W
B 5 A ,  100 W
C 4 A , 200 W
D 2 A , 400 W

The correct answer is:
2 A , 400 W

Solution:
From given condition circuit is at resonance
For LR circuit,
tan60= X L R â X L = 3 R
For CR circuit
tan60= X C R â X C = 3 R â X L = X C = 3 R
I rms = V rms R = 200 100 = 2 Amp
P = I 2 R = 4 Ã 100 = 400 Watt

Topic: Physics, Alternating Current, JEE Main
A series LR circuit connected with an ac source E=(25 sin1000t) V has a power factor of 12. If the source of emf is changed to E=(20 sin2000t) V, the new power factor of the circuit will be :

Options:
A 12
B 13
C 15
D 17

The correct answer is:
15

Solution:
For LR circuit, cosϕ=RXL2+R2=12

⇒R=XL

Now, ω'=2ω

Therefore, XL'=ω'L=2ωL=2XL

Hence, cosϕ'=R2XL2+R2=15

Topic: Physics, Alternating Current, JEE Main, JEE Main 2024 (30 Jan Shift 1)
A series L-R circuit is connected to a battery of emf V. If the circuit is switched on at t=0, then the time at which the energy stored in the inductor reaches 1n times of its maximum value, is :

Options:
A LRlnnn-1
B LRlnn+1n-1
C LRlnnn+1
D LRlnn-1n

The correct answer is:
LRlnnn-1

Solution:
Potential energy stored in inductor is given by U=12L2

∪∝l2

UU0=1I02

1n=1I02

1I0=1-e-RT/L=1n

t=LRlnnn-1

Topic: Physics, Alternating Current, JEE Main, JEE Main 2020 (04 Sep Shift 2)
A series LR circuit is connected to a voltage source with Vt=V0sinωt . After a very large time, current It behaves as t0≫LR:

Options:
A
B
C
D

The correct answer is:

Solution:
When sinusoidal voltage is applied across LR circuit, current is also sinusoidal but it lags by some phase.

Current will be in the form of
I=I0sinωt-ϕ
Graph will be sinusoidal.

Here I0 is the maximum current and ϕ is phase difference between V & I

Topic: Physics, Alternating Current, JEE Main, JEE Main 2016 (09 Apr Online)
A series LCR circuit consists of R=80 Ω. XL=100 Ω, and XC=40 Ω. The input voltage is 2500 cos100πt V. The amplitude of current, in the circuit, is _____ A.

Options:

The correct answer is:
25

Solution:
Comparing input voltage with standard expression, V=V0cosωt, we get ω=100π.

Now, impedance Z=R2+XL-XC2

=802+100-402

=100Ω

Therefore, peak current

i0=V0Z=2500100 A=25 A

Topic: Physics, Alternating Current, JEE Main, JEE Main 2023 (31 Jan Shift 2)
A series LCR circuit is connected to an AC source of 220 V, 50 Hz. The circuit contains a resistance R=80 Ω, an inductor of inductive reactance XL=70 Ω, and a capacitor of capacitive reactance XC=130 Ω. The power factor of circuit is x10. The value of x is:

Options:

The correct answer is:
8

Solution:
Power factor in LCR circuit is given by,

cos ϕ=RZ, where, Z is the total impedance of the circuit.

So, cosϕ=RR2+XC-XL2

Putting the values, we have

⇒cos ϕ=80(80)2+(60)2

⇒cos ϕ=80100⇒810

Thus, the value of x=8.

Topic: Physics, Alternating Current, JEE Main, JEE Main 2023 (25 Jan Shift 2)
A series LCR circuit is designed to resonate at an angular frequency ω0=105 rad s-1. The circuit draws 16 W power from 120 V source at resonance. The value of resistance R in the circuit is ________ Ω.

Options:

The correct answer is:
900

Solution:
At resonance

P=V2R

R=V2P=120216

=900 Ω

Topic: Physics, Alternating Current, JEE Main, JEE Main 2021 (24 Feb Shift 2)
A series LCR circuit is subjected to an ac signal of $200 \mathrm{~V}, 50 \mathrm{~Hz}$. If the voltage across the inductor $(\mathrm{L}=10 \mathrm{mH})$ is $31.4 \mathrm{~V}$, then the current in this circuit is _____ .

Options:
A $68 \mathrm{~A}$
B $63 \mathrm{~A}$
C $10 \mathrm{~A}$
D $10 \mathrm{~mA}$

The correct answer is:
$10 \mathrm{~A}$

Solution:
$\begin{aligned} & \text { Voltage across inductor } \mathrm{V}_{\mathrm{L}}=\mathrm{IX}_{\mathrm{L}} \\ & 31.4=\mathrm{I}[\mathrm{L} \omega] \\ & 31.4=\mathrm{I}[\mathrm{L}(2 \pi \mathrm{f})] \\ & 31.4=\mathrm{I}\left[10 \times 10^{-3}(2 \times 3.14) \times 50\right. \\ & \Rightarrow \mathrm{I}=10 \mathrm{~A}\end{aligned}$

Topic: Physics, Alternating Current, JEE Main, JEE Main 2024 (05 Apr Shift 2)
A series LCR circuit of R=5 Ω, L=20 mH and C=0.5 μF is connected across an AC supply of 250 V, having variable frequency. The power dissipated at resonance condition is ________×102 W.

Options:

The correct answer is:
125

Solution:
XL=XC (due to resonance)

Z=R so irms=VZ=VR

V2R=250×2505=125×102 W

Topic: Physics, Alternating Current, JEE Main, JEE Main 2021 (20 Jul Shift 2)
A series LCR circuit with L=100π mH,C=10-3π F and R=10 Ω, is connected across an AC source of 220 V,50 Hz supply. The power factor of the circuit would be ____.

Options:

The correct answer is:
1

Solution:
The capacitive reactance of the circuit is given by,

Xc=1ωC=π2π×50×10-3=10 Ω

The inductive reactance of the circuit is given by,

XL=ωL=2π×50×100π×10-3=10 Ω

From above calculations, it is clear that

XC=XL

∴ Power factor =RZ=RR=1

Topic: Physics, Alternating Current, JEE Main, JEE Main 2024 (27 Jan Shift 2)
A series $\mathrm{LR}$ circuit is connected to an ac source of frequency $\omega$ and the inductive reactance is equal to $2 \mathrm{R}$. A capacitance of capacitive reactance equal to $\mathrm{R}$ is added in series with $\mathrm{L}$ and $\mathrm{R}$. The ratio of the new power factor to the old one is :

Options:
A
$\sqrt{\frac{2}{3}}$

B
$\sqrt{\frac{2}{5}}$

C
$\sqrt{\frac{3}{2}}$

D
$\sqrt{\frac{5}{2}}$

The correct answer is:
$\sqrt{\frac{5}{2}}$

Solution:
Power factor ${ }_{\text {(old) }}$
$=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+\mathrm{X}_{\mathrm{L}}{ }^2}}=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+(2 \mathrm{R})^2}}=\frac{\mathrm{R}}{\sqrt{5} \mathrm{R}}$
Power factor ${ }_{\text {(new) }}$
$=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+\left(X_L-X_C\right)^2}}=\frac{R}{\sqrt{R^2+(2 R-R)^2}}$
$=\frac{R}{\sqrt{2} R}$
$\therefore \frac{\text { New power factor }}{\text { Old power factor }}=\frac{\frac{\mathrm{R}}{\sqrt{2} \mathrm{R}}}{\frac{\mathrm{R}}{\sqrt{5} \mathrm{R}}}=\sqrt{\frac{5}{2}}$

Topic: Physics, Alternating Current, JEE Main
A series $\mathrm{LR}$ circuit is connected to an ac source of frequency $\omega$ and the inductive reactance is equal to $2 \mathrm{R}$. A capacitance of capacitive reactance equal to $\mathrm{R}$ is added in series with $\mathrm{L}$ and $\mathrm{R}$. The ratio of the new power factor to the old one is :

Options:
A
$\sqrt{\frac{2}{3}}$

B
$\sqrt{\frac{2}{5}}$

C
$\sqrt{\frac{3}{2}}$

D
$\sqrt{\frac{5}{2}}$

The correct answer is:
$\sqrt{\frac{5}{2}}$

Solution:
Power factor ${ }_{\text {(old) }}$
$=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+\mathrm{X}_{\mathrm{L}}{ }^2}}=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+(2 \mathrm{R})^2}}=\frac{\mathrm{R}}{\sqrt{5} \mathrm{R}}$
Power factor ${ }_{\text {(new) }}$
$=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+\left(X_L-X_C\right)^2}}=\frac{R}{\sqrt{R^2+(2 R-R)^2}}$
$=\frac{R}{\sqrt{2} R}$
$\therefore \frac{\text { New power factor }}{\text { Old power factor }}=\frac{\frac{\mathrm{R}}{\sqrt{2} \mathrm{R}}}{\frac{\mathrm{R}}{\sqrt{5} \mathrm{R}}}=\sqrt{\frac{5}{2}}$

Topic: Physics, Alternating Current, JEE Main, JEE Main 2013 (25 Apr Online)
A series \( R-C \) combination is connected to an \( A C \) voltage of angular frequency \( \omega=500 \mathrm{rad} / \mathrm{s} \). If the impedance of the \( R-C \) circuit is \( \mathrm{R} \sqrt{1.25} \), the time constant (in millisecond) of the circuit is.

Options:
A \( 4 \mathrm{~ms} \)
B \( 3 \mathrm{~ms} \)
C \( 2 \mathrm{~ms} \)
D \( 1 \mathrm{~ms} \)

The correct answer is:
\( 4 \mathrm{~ms} \)

Solution:
Z = R 2 + X C 2 = R 1.25
∴ R 2 + X C 2 = 1.25 R 2
or X C = R 2 or 1 ω C = R 2 or C=2Rω
∴ Time constant = CR = 2 ω = 2 500 = 4 ms

Topic: Physics, Alternating Current, JEE Main
A set of $n$ values $x_{1}, x_{2}, \ldots, x_{n}$ has standard deviation $\sigma$. What is the standard deviation of $n$ values $x_{1}+k_{1} x_{2}+k_{1, \ldots,} x_{n}+k ?$

Options:
A 0
B $\mathrm{a}+\mathrm{k}$
C $\sigma-\bar{k}$
D $\underline{k} \in$

The correct answer is:
0

Solution:
We know that, if a number is added in values, then the standard deviation remains unaltered.
$\therefore$ Standard deviation of new values $=\sigma$

Topic: Mathematics, Statistics, NDA, NDA 2010 (Phase 2)
A set of parallel chords of the parabola $y^2=4 a x$ have their mid-point on

Options:
A Any straight line through the vertex
B Any straight line through the focus
C Any straight line parallel to the axis
D Another parabola

The correct answer is:
Any straight line parallel to the axis

Solution:
Let $y=m x+c$ is chord and $c$ is variable
$\Rightarrow x=\left(\frac{y-c}{m}\right)$ by $y^2=4 a x$
For getting points of intersection,
$y^2=4 a\left(\frac{y-c}{m}\right) \Rightarrow$ $y^2-\frac{4 a y}{m}+\frac{4 a c}{m}=0$
$\Rightarrow y_1+y_2=\frac{4 a}{m} \Rightarrow \frac{y_1+y_2}{2}=\frac{2 a}{m}$
which is a constant; independent to $c$.

Topic: Mathematics, Parabola, JEE Main