datavorous/entrance-exam-dataset huggingface dataset:
A rectangular parallelopiped is measured as 1 cm×1 cm×100 cm. If its specific resistance is 3×10-7Ω m, then the resistance between its two opposite rectangular faces will be _____×10-7Ω.

Options:

The correct answer is:
3

Solution:
It is given that

l = 1 cm = 0.01 m

A = 100 cm × 1 cm= 100 cm2

⇒A= 0.01 m2

ρ=3×10-7 Ω m

Using the formula of resistance,

R=ρlA=3×10-7×0.010.01=3×10-7Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (10 Apr Shift 2)
A rectangular plate \( 2 \mathrm{~m} \times 3 \mathrm{~m} \) is immersed in water in such a way that its greatest and least depth are \( 6 \mathrm{~m} \) and \( 4 \mathrm{~m} \), respectively, from the water surface. The total thrust on the plate is

Options:
A \( 294 \times 10^{3} \mathrm{~N} \)
B \( 294 \mathrm{~N} \)
C \( 100 \times 10^{3} \mathrm{~N} \)
D \( 400 \times 10^{3} \mathrm{~N} \)

The correct answer is:
\( 294 \times 10^{3} \mathrm{~N} \)

Solution:
Given the size of the plate =2 m×3 m and
greatest and least depths of the plate are 6 m and 4 m.
We know that area of the plate A=2×3=6 m2.
And the depth of center of the plate
x-=6+42=5 m
∴ Total thrust on the plate
=ρωgAx
=103×9.8×6×5
=294×103 N

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A rectangular solid box of length 0.3 m is held horizontally, with one of its sides on the edge of a platform of height 5 m . When released, it slips off the table in a very short time τ=0.01 s , remaining essentially horizontal. The angle by which it would rotate when it hits the ground will be (in radians) close to:

Options:
A 0.5
B 0.3
C 0.02
D 0.28

The correct answer is:
0.5

Solution:
Angular impulse = change in angular momentum
τΔt=ΔL
mgl2×0.01=ml23ω
ω=3g×0.012l=3×10×0.012×0.3=0.5 rad/sec
Time taken by rod to hit the ground = 2hg=2×510=1 s
Angle rotated by rod = ωt=0.5×1=0.5 rad

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2019 (08 Apr Shift 2)
A rectangular tank filled with some liquid is accelerated along a horizontal surface at 403 m s-2. Inside the liquid, a laser pointer is fixed a the centre of the tank which shoots a thin laser beam in the vertically upward direction. If after refraction from the liquid surface, the laser beam moves along the surface of the liquid, then what is the refractive index of the liquid?

Options:

The correct answer is:
1.25

Solution:
Due to the pseudo force, the free surface of the liquid will become inclined. If θ is the angle made by the free surface of the liquid with the horizontal, then

tanθ=ag=43

⇒sinθ=45

Using snell's law we get

μ sinθ=1

μ=1sinθ=54=1.25

Topic: Physics, Ray Optics, JEE Main
A rectangular tank of depth 8 meter is full of water $(\mu=4 / 3)$, the bottom is seen at the depth

Options:
A $6 \mathrm{~m}$
B $8 / 3 \mathrm{~m}$
C $8 \mathrm{~cm}$
D $10 \mathrm{~cm}$

The correct answer is:
$6 \mathrm{~m}$

Solution:
$\mu=\frac{h}{h^{\prime}} \Rightarrow h^{\prime}=\frac{8}{4 / 3}=6 m$

Topic: Physics, Ray Optics, JEE Main
A rectangular wire loop with length a and width b lies in the xy -plane as shown. Within the loop, there is a time dependent magnetic field given by Bâ= cxcosâ¡Ïti^+ysinâ¡Ïtk^.Here, c and Ï are constants. The magnitude of emf induced in the loop as a function of time is

Options:
A ab2c2ωcos⁡ωt
B ab2cωcos⁡ωt
C a2bc2ωsin⁡ωt
D None of the options

The correct answer is:
ab2c2ωcos⁡ωt

Solution:
from diagram only z component of magnetic field gives flux.
Flux through small elemental area
dÏ=ady cysinâ¡Ït
net flux =acsinâ¡Ïtâ«0by dy
Ï=acb22sinâ¡Ït
â´ Induced emf in the loop.
Îµ=ddtÏm=ddt.ab2c2sinâ¡Ït
Hence, Îµ=ab2c2Ïcosâ¡Ït

Topic: Physics, Electromagnetic Induction, JEE Main
A red coloured mixed oxide (X) on treatment with conc. HNO3 gives a compound Y.(Y) with HCl produces a chloride (Z) which is insoluble in cold water but soluble in hot water, (Z) can also be formed by treating (X) with conc. HCl. Compounds (X),(Y) and (Z) are:

Options:
A Pb 3 O 4 , PbNO 3 , PbCl 2
B Mn3O4,MnO2,MnCl2
C Fe3O4,Fe2O3,FeCl3
D Fe3O4,FeO,FeCl2

The correct answer is:
Pb 3 O 4 , PbNO 3 , PbCl 2

Solution:
Pb3O4X is red coloured mixed oxideof PbO and PbO2.
Pb 3 O 4 X +4HNO 3 → Pb NO 3 2 Y +PbO 2 +2H 2 O
Pb NO 3 2 + 2HCl dil → PbCl 2 Z +2HNO 3
PbCl2 in insoluble in cold water but soluble in hot.
Pb3O4(X)+8HCl→3PbCl2+4H2O+Cl2

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
A red LED emits light at 0.1 watt uniformly around it. The amplitude of the electric field of the light at a distance of 1 m from the diode is:

Options:
A 7.75 V m-1
B 1.73 V m-1
C 2.45 V m-1
D 5.48 V m-1

The correct answer is:
2.45 V m-1

Solution:
For a point source of power = P, then intensity at a point at a separation x from the source is, I=PowerArea=P4πx2∵ The average intensity of E.M. wave is given by, I=12cε0Eo2⇒E0=2P4πε0c x2
∵14πε0=9×109, P=0.1 W, x=1 m, c=3×108 m s-1
⇒E0=2×0.1×9×1093×108×12=6=2.45 V m-1

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2015 (04 Apr)
A red solid is insoluble in water. However, it becomes soluble if some KI is added to water. Heating the red solid in a test tube results in liberation of some violet coloured fumes and droplets of a metal appear on the cooler parts of the test tube. The red solid is

Options:
A HgI2
B HgO
C Pb3O4
D NH42Cr2O7

The correct answer is:
HgI2

Solution:
HgI2 is orange-red in appearance and hardly soluble in water. It dissolves in excess of I- giving complexes HgI3- and HgI42- . Violet fumes are liberated due to the release of Iodine gas upon heating.

Topic: Chemistry, Practical Chemistry, JEE Main
A red solid is insoluble in water. However it becomes soluble if some $\mathrm{KI}$ is added to water. Heating the red solid in a test tube results in liberation of some violet coloured fumes and droplets of a metal appear on the cooler parts of the test tube. The red solid is

Options:
A
$\mathrm{HgI}_2$

B
$\mathrm{HgO}$

C
$\mathrm{Pb}_3 \mathrm{O}_4$

D
$\left(\mathrm{NH}_4\right)_2 \mathrm{Cr}_2 \mathrm{O}_7$

The correct answer is:
$\mathrm{HgI}_2$

Solution:
$\underset{\text { (insoluble) }}{\mathrm{HgI}_2}+\mathrm{KI} \rightarrow \underset{\text { (soluble) }}{\mathrm{K}_2 \mathrm{HgI}_4}$
On heating $\mathrm{HgI}_2$ decomposes as $\mathrm{HgI}_2 \rightleftharpoons \mathrm{Hg}+\mathrm{I}_2$

Topic: Chemistry, d and f Block Elements, JEE Main
A red solid is insoluble in water. However it becomes soluble if some $\mathrm{KI}$ is added to water. Heating the red solid in a test tube results in liberation of some violet coloured fumes and droplets of a metal appear on the cooler parts of the test tube. The red solid is

Options:
A
$\mathrm{HgI}_2$

B
$\mathrm{HgO}$

C
$\mathrm{Pb}_3 \mathrm{O}_4$

D
$\left(\mathrm{NH}_4\right)_2 \mathrm{Cr}_2 \mathrm{O}_7$

The correct answer is:
$\mathrm{HgI}_2$

Solution:
$\underset{\text { (insoluble) }}{\mathrm{HgI}_2}+\mathrm{KI} \rightarrow \underset{\text { (soluble) }}{\mathrm{K}_2 \mathrm{HgI}_4}$
On heating $\mathrm{HgI}_2$ decomposes as $\mathrm{HgI}_2 \rightleftharpoons \mathrm{Hg}+\mathrm{I}_2$

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2003
A redox reaction always involves

Options:
A A change in oxidation number.
B A change in phase.
C The transfer of protons.
D The formation of ions.

The correct answer is:
A change in oxidation number.

Solution:
It is a fact that a redox reaction always involves change in oxidation number.

Topic: Chemistry, Redox Reactions, JEE Main
A reduction in atomic size with increase in atomic number is a characteristic of element of

Options:
A
d-block

B
f-block

C
radioactive series

D
high atomic masses

The correct answer is:
f-block

Solution:
f-block elements show a regular decrease in atomic size due to lanthanide/actinide contraction.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2003
A refrigerator absorbs 2000 cal of heat from ice trays. If the coefficient of performance is 4, then work done by the motor is 1 cal=4.2 J

Options:
A 2100 J
B 4200 J
C 8400 J
D 500 J

The correct answer is:
2100 J

Solution:
Q2=2000 cal

Coefficient of performance, COP=Q2W
W=Q2COP=500 calW=500×4.2 J=2100 J

Topic: Physics, Thermodynamics, JEE Main
A refrigerator consumes an average 35 W power to operate between temperature -10°C to 25°C. If there is no loss of energy then how much average heat per second does it transfer?

Options:
A 350 J s-1
B 298 J s-1
C 263 J s-1
D 35 J s-1

The correct answer is:
263 J s-1

Solution:
W+Q2=Q1

W=Q1-Q2

Q2 W=Q2Q1-Q2

Q2W=T2T1-T2

For one second, W=35 J

Q235=263298-263

Q2=263 J

Hence, average heat per second transferred =263 J s-1

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2021 (26 Aug Shift 2)
A refrigerator is to maintain eatables kept inside at $9^{\circ} \mathrm{C}$. If room temperature is $36^{\circ} \mathrm{C}$. Calculate the coefficient of performence.

Options:

None

Solution:
Given $T_1=36^{\circ} \mathrm{C}=36+273=309 \mathrm{~K}$;
$$
T_2=9^{\circ} \mathrm{C}=10+273=282^{\circ} \mathrm{K}
$$
Coefficient of performance
$$
=\frac{T_2}{T_1-T_2}=\frac{282}{309-282}=10.45
$$

Topic: Physics, Thermodynamics
A refrigerator with the coefficient of performance \( \frac{1}{3} \) releases \( 200 \mathrm{~J} \) of heat to a hot reservoir. Then the work done on the working substance is

Options:
A \( \frac{100}{3} \mathrm{~J} \)
B \( 100 \mathrm{~J} \)
C \( \frac{200}{3} \mathrm{~J} \)
D \( 150 \mathrm{~J} \)

The correct answer is:
\( 150 \mathrm{~J} \)

Solution:
A refrigerator's coefficient of performance is expressed as COP=Useful CoolingWork Required

=Q2W=Q2Q1-Q2.

Given here COP=13, Q1=200 J.

Substituting in above relation, we have,

13=Q2200-Q2

200-Q2=3Q2⇒4Q2=200

⇒Q2=50 J.

Now, the work done on the working substance is W=Q1-Q2=200-50=150 J.

Topic: Physics, Thermodynamics, JEE Main
A relation on the set $\mathrm{A}=\{\mathrm{x}:|\mathrm{x}| < 3, \mathrm{x} \in \mathrm{Z}\}$, where $Z$ is the set of integers is defined by $\mathrm{R}=\{(\mathrm{x}, \mathrm{y}): \mathrm{y}=|\mathrm{x}|, \mathrm{x} \neq-1\}$. Then the number of elements in the power set of $R$ is:

Options:
A
32

B
16

C
8

D
64

The correct answer is:
16

Solution:
$$
\begin{aligned}
&\mathrm{A}=\{x:|x| < 3, x \in Z\} \\
&\mathrm{A}=\{-2,-1,0,1,2\} \\
&\mathrm{R}=\{(x, y): y=|x|, x \neq-1\} \\
&\mathrm{R}=\{(-2,2),(0,0),(1,1),(2,2)\}
\end{aligned}
$$
$\mathrm{R}$ has four elements
Number of elements in the power set of $\mathrm{R}$
$$
=2^4=16
$$

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2014 (12 Apr Online)
A relation \( R \) is defined as \( (x, y) \in R \Rightarrow x^{y}=y^{x} \) for \( x, y \in I-\{0\} \), where \( I \) is the set of all integers. Then the relation \( R \) is:

Options:
A reflexive but not symmetric
B svmmetric but not reflexive
C reflexive and symmetric both
D equivalence relation

The correct answer is:
reflexive and symmetric both

Solution:
∵(x, y)∈R⇒xy=yx
∴(x, x)∈R as xx=xx ∀ x∈I-{0}
∴R is reflexive
Now (x, y)∈R⇒xy=yx⇒yx=xy
⇒(y, x)∈R ∴R is symmetric
Now (x, y)∈R⇒xy=yx and (y, z)∈R⇒yz=zy
∵xy=yx⇒y=xyx
⇒yz=zx⇒xyxz=zx⇒xyzx=zx
⇒xyz=zx2⇒(x, z)∉R
R is not transitive.

Topic: Mathematics, Sets and Relations, JEE Main
A relation R is defined from {2, 3, 4, 5} to {3, 6, 7, 10} by x R y⇒x is relatively prime to y, then domain of R is

Options:
A {2,Â 3,Â 5}
B 3,Â 5
C 2,Â 3,Â 4
D 2, 3, 4, 5

The correct answer is:
2, 3, 4, 5

Solution:
Given, x R y⇒x is relatively prime to y.
R={2, 3, 2, 7, 3, 7, 3, 10, 4, 3, 4, 7, 5, 3, 5, 6, 5, 7}
∴ Domain of R={2, 3, 4, 5}

Topic: Mathematics, Sets and Relations, JEE Main
A reservoir is at 827 °C and Carnot's engine takes a thousand kilocalorie of heat from it and exhausts it to a sink at 27 °C. What is the amount of work and the efficiency of the engine?

Options:
A 2.7×105 cal, 70.70 %
B 7.27×105 cal, 72.72 %
C 2.70×105 cal, 80.70 %
D 3.70×105 cal, 70.70 %

The correct answer is:
7.27×105 cal, 72.72 %

Solution:
Given, Q1=106 cal

T1=827 °C=(827+273)=1100 K,

T2=27 °C=(27+273)=300 K.

As, Q1T1=Q2T2

∴ Q2=T2T1Q1=3001100×106=2.72×105 cal
work done, W=Q1-Q2=106-2.72×105=7.27×105 cal

The efficiency of the engine,
η=1-T2T1×100

⇒η=1-3001100×100

=72.72 %

Topic: Physics, Thermodynamics, JEE Main
A resistance is shown in the figure. Its value and tolerance are given respectively by:

Options:
A 27 kΩ, 10%
B 270 Ω, 5%
C 270 Ω, 10%
D 27 kΩ, 20%

The correct answer is:
27 kΩ, 10%

Solution:
Colour band A and B indicate the first two significant figures of resistance in "ohm" , while the C band gives the decimal multiplier i.e., the number of zeros that follows the two significant figures A and B.

The last band D indicates the tolerance in percentage about the indicate value or in another ward it represents the percentage accuracy of the indicated value.

As per color coding of a carbon resistor

Color and its values are
Red2
Violet7
Orange3
Silver10%
∴ R=27×103±10%

Topic: Physics, Current Electricity, JEE Main, JEE Main 2019 (09 Jan Shift 1)
A resistance of 20 ohms is connected to a source of an alternating potential $V=220 \sin (100 \pi t)$. The time taken by the current to change from its peak value to r.m.s value is

Options:
A $0.2 \mathrm{sec}$
B $0.25 \mathrm{sec}$
C $25 \times 10^{-3} \mathrm{sec}$
D $2.5 \times 10^{-3} \mathrm{sec}$

The correct answer is:
$2.5 \times 10^{-3} \mathrm{sec}$

Solution:
Peak value to r.m.s value means, current becomes $\frac{1}{\sqrt{2}}$ times.
So from $i=i_0 \sin 100 \pi t \Rightarrow \frac{1}{\sqrt{2}} \times i_0=i_0 \sin 100 \pi t$
$\Rightarrow \sin \frac{\pi}{4}=\sin 100 \pi t \Rightarrow t=\frac{1}{400} \mathrm{sec}=2.5 \times 10^{-3} \mathrm{sec}$

Topic: Physics, Alternating Current, JEE Main
A resistance of 40 Ω is connected to a source of alternating current rated 220 V,50 Hz. Find the time taken by the current to change from its maximum value to the rms value :

Options:
A 2.5 ms
B 1.25 ms
C 2.5 s
D 0.25 s

The correct answer is:
2.5 ms

Solution:
The current in AC is of the form of i=i0sinωt. The maximum current is i0 and rms current is i02. The change in the phase when current change from maximum to rms value is π4 rad. Therefore, the time taken for the same will be,

t=∆ϕω=π4×2π×50=2.5 ms.

Topic: Physics, Alternating Current, JEE Main, JEE Main 2022 (24 Jun Shift 1)
A resistance of $300 \Omega$ and an inductance of $\frac{1}{\pi}$ henry are connected in series to a ac voltage of 20 volts and $200 \mathrm{~Hz}$ frequency. The phase angle between the voltage and current is

Options:
A $\tan ^{-1} \frac{4}{3}$
B $\tan ^{-1} \frac{3}{4}$
C $\tan ^{-1} \frac{3}{2}$
D $\tan ^{-1} \frac{2}{5}$

The correct answer is:
$\tan ^{-1} \frac{4}{3}$

Solution:
Phase angle $\tan \phi=\frac{\omega L}{R}=\frac{2 \pi \times 200}{300} \times \frac{1}{\pi}=\frac{4}{3}$
$\therefore \phi=\tan ^{-1} \frac{4}{3}$

Topic: Physics, Alternating Current, JEE Main
A resistance $R$ and a capacitance $C$ are connected in series to a battery of negligible internal resistance through a key. The key is closed at $t=$ 0 . If after $t$ sec the voltage across the capacitance was seven times the voltage across $\mathrm{R}$, the value of $\mathrm{t}$ is

Options:
A
$3 R C \ell n 2$

B
$2 R C \ell n 2$

C
$2 R C \ln 7$

D
$3 R C \ell n 7$

The correct answer is:
$3 R C \ell n 2$

Solution:
$t=3 R C \ln 2$

Topic: Physics, Alternating Current, JEE Main, JEE Main 2012 (12 May Online)
A resistance $\mathrm{R}$ and inductance $\mathrm{L}$ and a capacitor $\mathrm{C}$ all are connected in series with an AC supply. The resistance of $\mathrm{R}$ is 16 ohm and for a given frequency, the inductive reactance of $L$ is 24 ohm and capacitive reactance of $\mathrm{C}$ is $12 \mathrm{ohm} .$ If the current in the circuit is 5 amp., find the potential difference across $\mathrm{R}, \mathrm{L}$ and $\mathrm{C}$.

Options:
A 30,20,50 volt
B 40,100,60 volt
C 70,110,60 volt
D 80,120,60 volt

The correct answer is:
80,120,60 volt

Solution:
$\mathrm{V}_{\mathrm{R}}=\mathrm{i} \mathrm{R}=5 \times 16=80 \mathrm{Volt}$
$$
\begin{array}{l}
\mathrm{V}_{\mathrm{L}}=\mathrm{i} \times(\omega \mathrm{L})=5 \times 24=120 \mathrm{Volt} \\
\mathrm{V}_{\mathrm{C}}=\mathrm{i} \times(\mathrm{l} / \omega \mathrm{C})=5 \times 12=60 \mathrm{Volt}
\end{array}
$$

Topic: Physics, Alternating Current, JEE Main
A resistance R whose value is varied from 1 Î© to 5 Î© is connected to a cell of internal resistance 3 Î©. The power consumed by R

Options:
A increases continuously
B decreases continuously
C first decreases then increases
D first increases then decreases

The correct answer is:
first increases then decreases

Solution:
P=ER+32R

dPdR=0 at R=3Î© so power is maximum at

Therefore increase continuously till R=3Î© and then decreases

Topic: Physics, Current Electricity, JEE Main
A resistor and an inductor are connected in series to an AC source of voltage $150 \sin (100 \pi t+\pi)$ volt. If the current in the circuit is $5 \sin \left(100 \pi t+\frac{2 \pi}{3}\right)$ ampere, then the average power dissipated and the resistance of the resistor are respectively

Options:
A $187.5 \mathrm{~W}, 30 \Omega$
B $187.5 \mathrm{~W}, 15 \Omega$
C $375 \mathrm{~W}, 30 \Omega$
D $375 \mathrm{~W}, 15 \Omega$

The correct answer is:
$187.5 \mathrm{~W}, 15 \Omega$

Solution:
Given, peak voltage, $V_0=150 \mathrm{~V}$
Peak current,
$$
I_0=5
$$
Phase difference between voltage and current,
$$
\phi=\pi-\frac{2 \pi}{3}=\frac{\pi}{3}=60^{\circ}
$$
Now, average power in $R-L$ circuit,
$$
\begin{aligned}
P_{\mathrm{av}}=V_{\mathrm{rms}} & \cdot I_{\mathrm{rms}} \cdot \cos \phi \\
& =\frac{V_0}{\sqrt{2}} \cdot \frac{I_0}{\sqrt{2}} \cos 60^{\circ}=\frac{150}{\sqrt{2}} \times \frac{5}{\sqrt{2}} \times \frac{1}{2} \\
P_{\mathrm{av}}= & 187.5 \mathrm{~W}
\end{aligned}
$$
$\begin{gathered}\text { Impendence, } Z=\frac{V_0}{I_0}=\frac{150}{5} \\ Z=30 \Omega \\ \frac{R}{Z}=\cos \phi \\ \frac{R}{30}=\cos 60^{\circ} \Rightarrow R=30 \times \frac{1}{2} \\ R=15 \Omega\end{gathered}$

Topic: Physics, Alternating Current, JEE Main
A resistor develops 300 J of thermal energy in 15 s, when a current of 2 A is passed through it. If the current increases to 3 A, the energy developed in 10 s is _____ J.

Options:

The correct answer is:
450

Solution:
Energy produced is given by H=i2Rt

When i=2 A, t=15 s, energy is H=300=22×R×15

Resistance is R=5 Ω

When i=3 A, t=10 s.

Energy is H'=32×5×10=450 J

Topic: Physics, Current Electricity, JEE Main, JEE Main 2022 (25 Jun Shift 1)
A resistor develops 500 J of thermal energy in 20 s when a current of 1.5 A is passed through it. If the current is increased from 1.5 A to 3 A, what will be the energy developed in 20 s.

Options:
A 1500 J
B 1000 J
C 500 J
D 2000 J

The correct answer is:
2000 J

Solution:
500=1.52×R×20

E=32×R×20

E=2000 J

Topic: Physics, Current Electricity, JEE Main, JEE Main 2021 (16 Mar Shift 2)
A resistor dissipates 192 J of energy in 1 s when a current of 4 A is passed through it. Now, when the current is doubled, the amount of thermal energy dissipated in 5 s is _____________ J.

Options:

The correct answer is:
3840

Solution:
P=i2Rt
192=4×4×R×1
R=12Ω
P'=i'2Rt'
=8×8×12×5=3840 J

Topic: Physics, Current Electricity, JEE Main, JEE Main 2021 (31 Aug Shift 2)
A resistor is constructed as a hollow cylinder with inner and outer radii ra=0.5 cm, rb=1.0 cm respectively and resistivity Ï=3.5Ã10-5 Î© m. The resistance of the configuration for the length of 5 cm cylinder is

Ã10-3 Î©.

Options:
A 7.42
B 10.56
C 14.38
D 16.48

The correct answer is:
7.42

Solution:
Given,
l=5 cm=5Ã10-2 m
Ï=3.5Ã10-5 Î© m
A=Ï1Ã10-4-0.25Ã10-4
=0.75ÏÃ10-4 m2
=7.5ÏÃ10-5 m2
â´ R=ÏlA=3.5Ã10-5Ã5Ã10-27.5ÏÃ10-5
=7.42Ã10-3 Î©

Topic: Physics, Current Electricity, JEE Main
A resistor 'R' and $2 \mu \mathrm{F}$ capacitor in series is connected through a switch to $200 \mathrm{~V}$ direct supply. Across the capacitor is a neon bulb that lights up at $120 \mathrm{~V}$. Calculate the value of $R$ to make the bulb light up $5 \mathrm{~s}$ after the switch has been closed. $\left(\log _{10} 2.5=0.4\right)$

Options:
A
$1.7 \times 10^5 \Omega$

B
$2.7 \times 10^6 \Omega$

C
$3.3 \times 10^7 \Omega$

D
$1.3 \times 10^4 \Omega$

The correct answer is:
$2.7 \times 10^6 \Omega$

Solution:
$V_c=E\left(1-e^{-t / R c}\right)$
$1-e^{-t / R c}=\frac{120}{200}=\frac{3}{5}$
$\Rightarrow R=\frac{5}{1.84 \times 10^{-6}}=2.7 \times 10^6 \Omega$

Topic: Physics, Capacitance, JEE Main, JEE Main 2011
A resonance air column of length $20 \mathrm{~cm}$ resonates with a tuning fork of frequency $250 \mathrm{~Hz}$. The speed of sound in air is

Options:
A 300 m/s
B 200 m/s
C 150 m/s
D 75 m/s

The correct answer is:
200 m/s

Solution:
Length of air column l $=20 \mathrm{~cm}=0.2 \mathrm{~m}$
In resonance air column,
$\lambda=4 l=4 \times 0.2=0.8 \mathrm{~m}$
Now, $\lambda=0.8 \mathrm{~m}$ and $v=250 \mathrm{~Hz}$
Hence, speed of sound in air $v=v \lambda=250 \times 0.8=200 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Waves and Sound, JEE Main
A resonance circuit having inductance and resistance 2×10-4 H and 6.28 Ω respectively oscillates at 10 MHz frequency. The value of quality factor of this resonator is_______. π=3.14

Options:

The correct answer is:
2000

Solution:
Given: L=2×10-4 H

R=6.28 Ω

f=10 MHz=107 Hz

Since quality factor,

Q=ω0LR=2πfLR

∴ Q=2π×107×2×10-46.28

Q=2×103=2000

Topic: Physics, Alternating Current, JEE Main, JEE Main 2021 (24 Feb Shift 1)
A resonance tube is old and has a jagged end. It is still used in the laboratory to determine the velocity of sound in air. A tuning fork of frequency 512 Hz produces first resonance when the tube is filled with water to a mark 11 cm below a reference mark, near the open end of the tube. The experiment is repeated with another fork of frequency 256 Hz which produces first resonance when water reaches a mark 27 cm below the reference mark. The velocity of sound in air, obtained in the experiment, is close to

Options:
A 335 m s-1
B 341 m s-1
C 322 m s-1
D 328 m s-1

The correct answer is:
328 m s-1

Solution:
λ14=11+e ⇒v512 × 4=11+e ...(1)
λ24=27+e ⇒v256 × 4=27+e ...(2)
Subtracting the equations,
v256×4×12=16 cm=0.16 m
⇒v=0.16×2×4×256
=328 m s-1

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2019 (12 Jan Shift 2)
A reverse biased diode is,

Options:
A
B
C
D

The correct answer is:

Solution:
In reverse biased diode connection, the p-side of the diode is connected to the negative terminal of the voltage source and n-type is connected to the negative terminal of the voltage source.

Thus, among the given options, option:

represents the reverse biased mode of connection of diodes.

Topic: Physics, Semiconductors, JEE Main
A reversible engine converts one-sixth of the heat input into work. When the temperature of the sink is reduced by $62^{\circ} \mathrm{C}$, the efficiency of the engine is doubled. The temperatures of the source and sink are

Options:
A $80^{\circ} \mathrm{C}, 37^{\circ} \mathrm{C}$
B $95^{\circ} \mathrm{C}, 28^{\circ} \mathrm{C}$
C $90^{\circ} \mathrm{C}, 37^{\circ} \mathrm{C}$
D $99^{\circ} \mathrm{C}, 37^{\circ} \mathrm{C}$

The correct answer is:
$99^{\circ} \mathrm{C}, 37^{\circ} \mathrm{C}$

Solution:
WE know, Efficiency, $\begin{aligned} \eta & =\frac{\left(T_2-T_1\right)}{T_2}=\frac{1}{6} \\ & \Rightarrow 6 \mathrm{~T}_2-6 \mathrm{~T}_1=\mathrm{T}_2 \\ & \Rightarrow \mathrm{T}_2=1.2 \mathrm{~T}_1 \ldots \ldots . \text { (i) }\end{aligned}$
Where $\mathrm{T}_2$ is the source temperature and $\mathrm{T}_1$ is the sink temperature.
If the sink temperature is reduced by $62^{\circ} \mathrm{C}$, the efficiency gets doubled i.e.
$\begin{aligned} \eta & =\frac{T_2-\left(T_1-62\right)}{T_2}=2 \times \frac{1}{6}=\frac{2}{6} \\ & \Rightarrow 6 \mathrm{~T}_2-6 \mathrm{~T}_1+372=2 \mathrm{~T}_2 \\ & \Rightarrow 6 \mathrm{~T}_2-2 \mathrm{~T}_2-6 \mathrm{~T}_1+372=0 \\ & \Rightarrow 4 \mathrm{~T}_2-6 \mathrm{~T}_1+372=0\end{aligned}$
Substituting the value of $\mathrm{T}_2$ from equation
(i), we will get,
$\begin{aligned} & 4\left(1.2 \mathrm{~T}_1\right)-6 \mathrm{~T}_1+372=0 \\ & \Rightarrow 4.8 \mathrm{~T}_1-6 \mathrm{~T}_1+372=0 \\ & \Rightarrow 372=1.2 \mathrm{~T}_1 \\ & \Rightarrow \mathrm{T}_1=310 \mathrm{~K}=37^{\circ} \mathrm{C}\end{aligned}$
Therefore, $\mathrm{T}_2=1.2 \times 310=372 \mathrm{~K}=99^{\circ} \mathrm{C}$ Hence, the temperature of the source and the sink is $99^{\circ} \mathrm{C}$ and $37^{\circ} \mathrm{C}$ respectively.

Topic: Physics, Thermodynamics, JEE Main