datavorous/entrance-exam-dataset huggingface dataset:
A wooden block of mass $5 \mathrm{~kg}$ rests on a soft horizontal floor. When an iron cylinder of mass $25 \mathrm{~kg}$ is placed on the top of the block, the floor yields and the block and the cylinder together go down with an acceleration of $0.1 \mathrm{~ms}^{-2}$. The action force of the system on the floor is equal to:

Options:
A $196 \mathrm{~N}$
B $291 \mathrm{~N}$
C $294 \mathrm{~N}$
D $297 \mathrm{~N}$

The correct answer is:
$291 \mathrm{~N}$

Solution:
Taking $\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2$
$\begin{aligned} & 294-\mathrm{N}=30 \times 0.1 \\ & \mathrm{~N}=291\end{aligned}$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2024 (05 Apr Shift 1)
A wooden block of mass 10 gm is dropped from the top of a tower 100 m high. Simultaneously, a bullet of mass 10 gm is fired from the foot of the tower vertically upwards with a velocity of 100 m s-1 . If the bullet is embedded in it, how high will it rise above the tower before it starts falling? g=10 m s-2

Options:

The correct answer is:
75

Solution:
Block and bullet will collide after 1 s at height 95 m from ground.
Velocity of block before collision=gt=10 m s-1
Velocity of bullet before collision=u–gt=90 m s-1
from conservation of linear momentum,
m–10+m90=2 mv′ ⇒ v′=40 m s-1
Thus height raised by the combined mass
h=v′22g=40×402×10=80 m
height from tower=75 m

Topic: Physics, Motion In One Dimension, JEE Main
A wooden block performs SHM on a frictionless surface with frequency Ï0. The block carries a charge +Q on its surface. If now a uniform electric field E â is switched-on as shown, then the SHM of the block will be

Options:
A of the same frequency and with shifted mean position
B of the same frequency and with the same mean position
C of changed frequency and with shifted mean position
D of changed frequency and with the same mean position

The correct answer is:
of the same frequency and with shifted mean position

Solution:
Frequency or time period of SHM is dependent on variable forces. It does not depend on constant external force. Constant external force can only change the mean position. For example, in the given question, mean position is at natural length of spring in the absence of electric field. Whereas in the presence of electric field, mean position will be obtained after a compression of x0. Where x0 is given by
Kx0 = QE
Or x 0 = QE K
Correct answer is (Of the same frequency and with shifted mean position)
.

Topic: Physics, Oscillations, JEE Main
A wooden box excavated from Indus valley had an activity of $1.18 \times 10^{13}$ disintegration per minute per $g$ of carbon. What is the approximate age of this civilisation?

Options:
A $4000$ years
B $5700$ years
C $8100$ years
D $6000$ years

The correct answer is:
$8100$ years

Solution:
$\begin{aligned} \frac{d N}{d t} &=\lambda N \\ \frac{d N}{d t} &=\frac{1}{T} \frac{6.023 \times 10^{23}}{12} \\ T &=\frac{6.023 \times 10^{23}}{12 \times 1.18 \times 10^{13}} \quad\left\{\frac{d N}{d t}=1.18 \times 10^{13}\right\} \\ &=42530 \times 10^{6} \mathrm{~min} \\ &=\frac{4253 \times 10^{6}}{60 \times 24 \times 365} \text { years } \quad\left\{\because \lambda=\frac{1}{T}\right\} \\ &=8092 \text { years } \\ & \approx 8100 \text { years } \end{aligned}$

Topic: Chemistry, Chemical Kinetics, JEE Main
A wooden cylinder of mass 20 g and area of cross-section 1â cm 2 , having a piece of lead of mass 60âg attached to its bottom, floats in water. The cylinder is depressed and then released. The frequency of oscillations is N Ï s â1. Find the value of N. [Neglect the volume of water displaced by the lead piece, take g=9.8âm/s2, density of water Ïw=1 g cm-3] If your answer is K, mark 4K as answer after rounding off to nearest integer.

Options:

The correct answer is:
7

Solution:
Suppose that the loaded wooden block sinks upto a height h. Then,
weight of water displaced by the block
= weight of the block with lead
If a is area of cross-section of the block and Ï, the density of water, then
a hâ¡ Ï g = M + m g
or hâ¡ = M + m a Ï
= 2 0 + 6 0 1 Ã 1 = 8 0 cm âµ density of water = 1gcm - 3
Now for small displacement ' x' from equilibrium
Ïga( x+h )â( M+m )g=-( M+m )A
Ïgax=-( M+m )A
A=- Ïga M+m x
f= 1 2Ï g h
= 1 2Ï 980 80 = 1.75 Ï s â1

Topic: Physics, Oscillations, JEE Main
A wooden cylinder of mass 20 g and area of cross-section 1â cm 2 , having a piece of lead of mass 60âg attached to its bottom, floats in water. The cylinder is depressed and then released. The frequency of oscillations is N Ï s â1. Find the value of N. [Neglect the volume of water displaced by the lead piece, take g=9.8âm/s2, density of water Ïw=1 g cm-3]

Options:

The correct answer is:
1.75

Solution:
Suppose that the loaded wooden block sinks upto a height h. Then,
weight of water displaced by the block
= weight of the block with lead
If a is area of cross-section of the block and Ï, the density of water, then
a hâ¡ Ï g = M + m g
or hâ¡ = M + m a Ï
= 2 0 + 6 0 1 Ã 1 = 8 0 cm âµ density of water = 1gcm - 3
Now for small displacement ' x' from equilibrium
Ïga( x+h )â( M+m )g=-( M+m )A
Ïgax=-( M+m )A
A=- Ïga M+m x
f= 1 2Ï g h
= 1 2Ï 980 80 = 1.75 Ï s â1

Topic: Physics, Oscillations, JEE Main
A wooden stick of length L, radius R and density ρ has a small metal piece of mass m (of negligible volume) attached to its one end. Find the minimum value for the mass m (in terms of given parameters) that would make the stick float vertically in stable equilibrium in a liquid of density σ .

Options:
A 2πR2Lρσρ-1
B πR2Lρ2σρ-1
C πR2Lρσρ-1
D πR2Lρσ2ρ-1

The correct answer is:
πR2Lρσρ-1

Solution:
M+mg=ÏR2hÏg .... (i)
M=ÏR2LÏ
The height of center of mass from bottom
ML/2+mÃ0m+M=ML2m+M
For rotatory equilibrium and for minimum m, thus should be equal to h/2
â´ h2=ML2m+M .... (ii)
â´ h=ML2m+M;m=ÏR2LÏÏ-ÏR2LÏ =ÏR2LÏÏÏ-1

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A wooden wedge of mass M and inclination angle α rests on a smooth floor. A block of mass m is kept on wedge. A force P→ is applied on the wedge as shown in figure, such that a block remains stationary with respect to wedge. The magnitude of force P→ is

Options:
A M+mg tan Î±
B gÂ tanÂ Î±
C mgcosâ¡Î±
D (M+m)gcosecâ¡Î±

The correct answer is:
M+mg tan Î±

Solution:
Since, P=(M+m)a
Now as in free body diagram of block,
macosâ¡Î±=mgsinâ¡Î±
â´ a=gsinâ¡Î±cosâ¡Î±=gtanâ¡Î± or P=M+mgtanâ¡Î±

Topic: Physics, Laws of Motion, JEE Main
A wooden wheel of radius $\mathrm{R}$ is made of two semicircular parts (see figure); The two parts are held together by a ring made of a metal strip of cross sectional area S and length $L$. $L$ is slightly less than $2 \pi R$. To fit the ring on the wheel, it is heated so that its temperature rises by $\Delta \mathrm{T}$ and it just steps over the wheel. As it cools down to surrounding temperature, it presses the semicircular parts together. If the coefficient of linear expansion of the metal is $\alpha$, and its Youngs' modulus is $Y$, the force that one part of the wheel applies on the other part is :

Options:
A
$\mathrm{2 \pi S Y \alpha \Delta T}$

B
$\mathrm{S Y \alpha \Delta T}$

C
$\mathrm{\pi S Y \alpha \Delta T}$

D
$\mathrm{2 S Y \alpha \Delta T}$

The correct answer is:
$\mathrm{2 S Y \alpha \Delta T}$

Solution:
If temperature increases by $\Delta \mathrm{T}$,
Increase in length $\mathrm{L}, \Delta \mathrm{L}=\mathrm{L} \alpha \Delta \mathrm{T}$
$\therefore \quad \frac{\Delta \mathrm{L}}{\mathrm{L}}=\alpha \Delta \mathrm{T}$
Let tension developed in the ring is $\mathrm{T}$.
$\therefore \quad \frac{\mathrm{T}}{\mathrm{S}}=\mathrm{Y} \frac{\Delta \mathrm{L}}{\mathrm{L}}=\mathrm{Y} \alpha \Delta \mathrm{T}$
$\therefore \quad \mathrm{T}=\mathrm{S} \mathrm{Y} \alpha \Delta \mathrm{T}$
From FBD of one part of the wheel,
$\mathrm{F}=2 \mathrm{T}$
Where, $F$ is the force that one part of the wheel applies on the other part.
$\therefore \mathrm{F}=2 \mathrm{SY} \alpha \Delta \mathrm{T}$

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2012 (Offline)
A word has 4 identical letters and rest all are distinct letters. If the total number of words that can be made with the letters of the word be 210, then the total number of different letters in the word is equal to

Options:
A 3
B 5
C 4
D 7

The correct answer is:
4

Solution:
Let the number of distinct letters other than the identical letters be n.
Total letters =n+4
Identical =4, Distinct =n
Number of ways in which they can be arranged =n+4!4!
⇒n+4!4!=210
⇒n+4n+3n+2…6.5
=210=7×6×5
⇒n+4=7⇒n=3
So, the required number of total different letters =3+1=4

Topic: Mathematics, Permutation Combination, JEE Main
A work of 146 kJ is performed on the gas, in order to compress one kilomole of gas adiabatically and in this process the temperature of the gas increases by 7 Â°C. The gas is ( R = 8.3 J mol-1 K-1 )

Options:
A monoatomic
B diatomic
C triatomic
D a mixture of monoatomic and diatomic

The correct answer is:
diatomic

Solution:
According to first law of thermodynamics
ΔQ=ΔU+ΔW
For an adiabatic process, ΔQ=0
∴ 0=ΔU+ΔW
or ΔU=-ΔW
or nCVΔT=-ΔW
or CV=-ΔWnΔT=--146×1031×103×7

Cv=20.8 J mol-1 K-1
For diatomic gas,
CV=52R=52×8.3 = 20.8 J mol-1 K-1
Hence the gas is diatomic.

Topic: Physics, Thermodynamics, JEE Main
A working transistor with its three legs marked P, Q and R is tested using a multimeter. No conduction is found between P and Q. By connecting the common (negative) terminal of the multimeter to R and the other (positive) terminal to P or Q, some resistance is seen on the multimeter. Which of the following is true for the transistor?

Options:
A It is an n-p-n transistor with R as base
B It is an p-n-p transistor with R as collector
C It is an p-n-p transistor with R as emitter
D It is an n-p-n transistor with R as collector

The correct answer is:
It is an n-p-n transistor with R as base

Solution:
Since no conduction is found when multimeter is connected across P and Q it means either both P and Q are n-type or p-type. So, it means R is base. When R is connected to common terminal and conduction is seen when other terminal is connected to P and Q, so it means transistor is in p-n with R as base.

Topic: Physics, Semiconductors, JEE Main
A working transistor with its three legs marked $P, Q$ and $R$ is tested using a multimeter. No conduction is found between $P$ and $Q$. By connecting the common (negative) terminal of the multimeter to $R$ and the other (positive) terminal to $P$ or $Q$, some resistance is seen on the multimeter. Which of the following is true for the transistor?

Options:
A
It is an npn transistor with $R$ as base

B
It is a pnp transistor with $\mathrm{R}$ as collector

C
It is a pnp transistor with $\mathrm{R}$ as emitter

D
It is an npn transistor with $R$ as collector

The correct answer is:
It is a pnp transistor with $\mathrm{R}$ as collector

Solution:
No Solution Available

Topic: Physics, Semiconductors, JEE Main, JEE Main 2008
A working transistor with its three legs marked P, Q and R is tested using a multimeter. No conduction is found between P and Q. By connecting the common (negative) terminal of the multimeter to R and the other (positive) terminal to P or Q, some resistance is seen on the multimeter. Which of the following is true for the transistor?

Options:
A It is an n-p-n transistor with R as base
B It is an p-n-p transistor with R as collector
C It is an p-n-p transistor with R as emitter
D It is an n-p-n transistor with R as collector

The correct answer is:
It is an n-p-n transistor with R as base

Solution:
Since no conduction is found when the multimeter is connected across P and Q it means either both P and Q are n-type or p-type. So, it means R is base. When R is connected to the common terminal and conduction is seen when another terminal is connected to P and Q, so it means transistor is in n-p-n configuration with R as the base.

Topic: Physics, Semiconductors, JEE Main
$\int\left(\sqrt{\frac{a+x}{a-x}}+\sqrt{\frac{a-x}{a+x}}\right) d x$ is equal to

Options:
A $2 \sin ^{-1}\left(\frac{x}{a}\right)+C$
B $2 a \sin ^{-1}\left(\frac{x}{a}\right)+C$
C $2 \cos ^{-1}\left(\frac{x}{a}\right)+C$
D $2 a \cos ^{-1}\left(\frac{x}{a}\right)+C$

The correct answer is:
$2 a \sin ^{-1}\left(\frac{x}{a}\right)+C$

Solution:
$\begin{aligned} & \text { Let } I=\int\left(\sqrt{\frac{a+x}{a-x}}+\sqrt{\frac{a-x}{a+x}}\right) d x \\ & \text { Put } x=a \cos 2 \theta \Rightarrow d x=-2 a \sin 2 \theta d \theta \\ & \therefore I=-\int\left(\sqrt{\frac{a+a \cos 2 \theta}{a-a \cos 2 \theta}}+\sqrt{\frac{a-a \cos 2 \theta}{a+a \cos 2 \theta}}\right) \\ & \qquad \times 2 a \sin 2 \theta d \theta\end{aligned}$
$\begin{aligned} & =-\int\left(\sqrt{\frac{2 \cos ^2 \theta}{2 \sin ^2 \theta}}+\sqrt{\frac{2 \sin ^2 \theta}{2 \cos ^2 \theta}}\right) 2 a \sin 2 \theta d \theta \\ & =-2 a \int\left(\frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta}\right) \sin 2 \theta d \theta \\ & =-2 a \int \frac{1}{\sin \theta \cos \theta} \sin 2 \theta d \theta \\ & =-4 a \int d \theta=-4 a \theta+C_1 \\ & =-2 a \cos ^{-1}\left(\frac{x}{a}\right)+C_1 \\ & =-2 a\left[\frac{\pi}{2}-\sin ^{-1}\left(\frac{x}{a}\right)\right]+C_1 \\ & =2 a \sin ^{-1}\left(\frac{x}{a}\right)+C_1-\pi a \\ & =2 a \sin ^{-1}\left(\frac{x}{a}\right)+C, \text { where } C=C_1-\pi a\end{aligned}$

Topic: Mathematics, Indefinite Integration, JEE Main
$\int a^x d a=$

Options:
A $\frac{a^x}{\log _e a}+c$
B $a^x \log _e a+c$
C $\frac{a^{x+1}}{x+1}+c$
D $x a^{x-1}+c$

The correct answer is:
$\frac{a^x}{\log _e a}+c$

Solution:
\(\int a^x \cdot d x=\frac{a^x}{\log _e a}+c\)
(Direct formula based question).

Topic: Mathematics, Indefinite Integration, JEE Main
A xenon compound A upon partial hydrolysis gives XeO2F2. The number of lone pair of electrons present in compound A is ___ (Round off to the Nearest integer)

Options:

The correct answer is:
19

Solution:
XeF6(A)+2H2O(Limited water)⟶XeO2 F2+4HF

Structure of 'A'

Total l.p. on (A)=19

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2021 (18 Mar Shift 2)
a young infant may be feeding entirely on mother's milk which is white in colour but the stools which the infant passes out is quite yellowish. What is this yellow colour due to ?

Options:
A Pancreatic juice poured into duodenum
B Intestinal juice
C Bile pigments passed through bile juice
D Undigested milk protein casein

The correct answer is:
Bile pigments passed through bile juice

Solution:
No Solution Available

Topic: Biology, Digestion and Absorption, NEET, NEET 2009 (Mains)
A young man of mass \( 60 \mathrm{~kg} \) stands on the floor of a lift which is accelerating downwards at \( 1 \frac{\mathrm{m}}{\mathrm{s}^{2}} \) then the reaction of the
floor of the lift on the man is (Take \( g=9.8 \frac{\mathrm{m}}{\mathrm{s}^{2}} \) )

Options:
A \( 528 \mathrm{~N} \)
B \( 540 \mathrm{~N} \)
C \( 546 \mathrm{~N} \)
D None

The correct answer is:
\( 528 \mathrm{~N} \)

Solution:
R=mg-a

Topic: Physics, Motion In One Dimension, JEE Main
A Young’s double slit experiment uses a monochromatic source. The shape of the
interference fringes formed on a screen is

Options:
A
hyperbola

B
circle

C
straight line

D
parabola

The correct answer is:
straight line

Solution:
Straight line
Note: If instead of young's double slit experiment, young's double hole experiment was given shape would have been hyperbola.

Topic: Physics, Wave Optics, JEE Main, JEE Main 2005
A Young's double slit experiment uses a monochromatic source. The shape of the interference fringes formed on a screen

Options:
A Straight line
B Parabola
C Hyperbola
D Circle

The correct answer is:
Hyperbola

Solution:

Topic: Physics, Wave Optics, JEE Main
A zener diode has a breakdown voltage of \( 5 \mathrm{~V} \), with a maximum power dissipation of \( 240 \mathrm{~mW} \). The maximum current the
diode can handle will be

Options:
A \( 50 \mathrm{~mA} \)
B \( 48 \mathrm{~mA} \)
C \( 46 \mathrm{~mA} \)
D \( 44 \mathrm{~mA} \)

The correct answer is:
\( 48 \mathrm{~mA} \)

Solution:
Given, zener breakdown voltage of zener diode, Vz=5 V.

Maximum power dissipation by the zener diode, Pmax=240 mW.

Let the maximum current the diode can handle be Imax.

The power dissipated in the diode is given as Pmax=VzImax⇒Imax=PmaxVz=2405=48 mA

Topic: Physics, Semiconductors, JEE Main
A zener diode, having breakdown voltage equal to $15 \mathrm{~V}$, is used in a voltage regulator circuit shown in figure. The current through the diode is

Options:
A $10 \mathrm{~mA}$
B $15 \mathrm{~mA}$
C $20 \mathrm{~mA}$
D $5 \mathrm{~mA}$

The correct answer is:
$5 \mathrm{~mA}$

Solution:
For I $1 \mathrm{k} \Omega$
$$
i_1=\frac{15}{1}=15 \mathrm{~mA}
$$
For II $250 \Omega$
$$
\begin{aligned}
& i_{250 \Omega}=\frac{20-15}{250}=\frac{5}{250} \\
&=\frac{20}{1000}=20 \mathrm{~mA} \\
& \therefore \quad i_{\text {zener }}=20-15=5 \mathrm{~mA}
\end{aligned}
$$

Topic: Physics, Semiconductors, NEET, NEET 2011 (Mains)
A zener diode having zener voltage 8 V and power dissipation rating of 0.5 W is connected across a potential divider arranged with maximum potential drop across zener diode is as shown in the diagram. The value of protective resistance Rp is _____Ω.

Options:

The correct answer is:
192

Solution:
P=Vi

0.5=8i

i=116 A

E=20=8+iRP

RP=12×16=192 Ω

Topic: Physics, Semiconductors, JEE Main, JEE Main 2021 (20 Jul Shift 2)
A Zener diode is always ______ connected.

Options:
A Reverse
B Forward
C Either reverse or forward
D None of the above

The correct answer is:
Reverse

Solution:
A zener diode is always connected in reverse bias mode. When the Zener diode is reverse biased the junction potential increases.

Topic: Physics, Semiconductors, JEE Main
A Zener diode is connected to a battery and a load as show below:
The currents, $\mathrm{I}, \mathrm{I}_{\mathrm{Z}}$ and $\mathrm{I}_{\mathrm{L}}$ are respectively.

Options:
A
$15 \mathrm{~mA}, 5 \mathrm{~mA}, 10 \mathrm{~mA}$

B
$15 \mathrm{~mA}, 7.5 \mathrm{~mA}, 7.5 \mathrm{~mA}$

C
$12.5 \mathrm{~mA}, 5 \mathrm{~mA}, 7.5 \mathrm{~mA}$

D
$12.5 \mathrm{~mA}, 7.5 \mathrm{~mA}, 5 \mathrm{~mA}$

The correct answer is:
$12.5 \mathrm{~mA}, 7.5 \mathrm{~mA}, 5 \mathrm{~mA}$

Solution:
Here, $R=4 \mathrm{k} \Omega=4 \times 10^3 \Omega$
$$
V_i=60 \mathrm{~V}
$$
Zener voltage $V_z=10 \mathrm{~V}$ $R_L=2 \mathrm{k} \Omega=2 \times 10^3 \Omega$
Load current, $I_{\mathrm{L}}=\frac{V_Z}{R_L}=\frac{10}{2 \times 10^3}=5 \mathrm{~mA}$
Current through $R, I=\frac{V_i-V_Z}{R}$ $=\frac{60-10}{4 \times 10^3}=\frac{50}{4 \times 10^3}=12.5 \mathrm{~mA}$
Fom circuit diagram,
$$
\begin{aligned}
& I=I_Z+I_L \\
\Rightarrow & 12.5=I_Z+5 \\
\Rightarrow & I_7=12.5-5=7.5 \mathrm{~mA}
\end{aligned}
$$

Topic: Physics, Semiconductors, JEE Main, JEE Main 2014 (11 Apr Online)
A Zener diode is connected to a battery and a load as shown below: The currents I, IZ and IL are respectively

Options:
A 12.5 mA, 7.5 mA, 5 mA
B 15 mA, 7.5 mA, 7.5 mA
C 12.5 mA, 5 mA, 7.5 mA
D 15 mA, 5 mA, 10 mA

The correct answer is:
12.5 mA, 7.5 mA, 5 mA

Solution:
I L = V AB R L = 1 0 2 0 0 0 = 5 mA
I = 6 0 - 1 0 4 0 0 0 = 5 0 4 0 0 0 = 1 2 Â· 5 mA
I = I L + I Z â´ I Z = I - I L = 7 Â· 5 mA
Hence, 12.5 mA, 7.5 mA, 5 mA

Topic: Physics, Semiconductors, JEE Main
A Zener diode is operating in its breakdown region. What will be the change in output voltage if the input voltage is increased?

Options:
A Increase proportionally
B Decrease proportionally
C Remain nearly constant
D Increase exponentially

The correct answer is:
Remain nearly constant

Solution:
When a Zener diode is operating in its breakdown region, it maintains a nearly constant output voltage despite changes in the input voltage, making it suitable for voltage regulation.

Topic: Physics, Semiconductors, JEE Main
A zener diode of power rating 2 W is to be used as a voltage regulator. If the zener diode has a breakdown of 10 V and it has to regulate voltage fluctuated between 6 V and 14 V, the value of Rs for safe operation should be _____ Ω .

Options:

The correct answer is:
20

Solution:
For diode P=Vi

i=PV=210=0.2 A

Now for Vs=14 V

Voltage drop across Rs=14-10=4 V

Thus, V=iRs

Rs=40.2=20 Ω

Topic: Physics, Semiconductors, JEE Main, JEE Main 2021 (27 Aug Shift 2)
A Zener diode of Zener break-down voltage 10 V is connected as shown in the figure. Current through Zener diode is

Options:
A 4 mA
B 0.6 mA
C 6 mA
D Zero

The correct answer is:
Zero

Solution:
If we consider break down in Zener diode, then potential across R3 will be 10 V and R1 will be 2 V .
So current in R3 will be i3=101500=2300 A
and current in R1 will be i1=2500 A
⇒i1<i3 , which is not possible
⇒ The potential difference across the Zener diode does not reach to break down voltage. So no current will flow through the reverse-biased Zener diode.

Topic: Physics, Semiconductors, JEE Main
A Zener diode undergoes breakdown for an electric field of 106 V m-1 at the depletion region of p-n junction. If the width of the depletion region is 2.5 μm , what should be the reverse-biased potential for the breakdown to occur?

Options:
A 3.5 V
B 2.5 V
C 1.5 V
D 0.5 V

The correct answer is:
2.5 V

Solution:
Reverse biased potential for the zener breakdown
Vr=Ed=106×2.5×10-6=2.5 volt

Topic: Physics, Semiconductors, JEE Main
A Zener diode with a breakdown voltage of 4 V is connected in series with a resistance R to a battery of emf 10 V. The maximum power dissipation rating for the Zener diode is 1 W. The value of R to ensure maximum power dissipation across the diode is

Options:
A 12 Ω
B 24 Ω
C 36 Ω
D 6 Ω

The correct answer is:
24 Ω

Solution:
Pzener=4i=1

⇒i=14 A
i=6R=14

⇒R=24 Ω

Topic: Physics, Semiconductors, JEE Main, JEE Main 2016 (09 Apr Online)
A Zener diode with a breakdown voltage of 4 V is connected in series with a resistance R to a battery of emf 10 V. The maximum power dissipation rating for the Zener diode is 1 W. The value of R to ensure maximum power dissipation across the diode is

Options:
A 12 Ω
B 24 Ω
C 36 Ω
D 6 Ω

The correct answer is:
24 Ω

Solution:
Pzener=4i=1

⇒i=14 A
i=6R=14

⇒R=24 Ω

Topic: Physics, Semiconductors, JEE Main
A Zener of breakdown voltage VZ=8 V and maximum Zener current, IZM=10 mA is subjected to an input voltage Vi=10 V with series resistance R=100 Ω. In the given circuit RL represents the variable load resistance. The ratio of maximum and minimum value of RL is _____ .

Options:

The correct answer is:
2

Solution:
Minimum value of RL for which the diode is shorted is

RLRL+100×10=8⇒RL=400 Ω

For maximum value of RL, current through diode is 10 mA

So, iR=iRL+IZM

⇒2100=8RL+10×10-3⇒10×10-3=8RL⇒RL=800 Ω

So, RLmaxRLmin=2

Topic: Physics, Semiconductors, JEE Main, JEE Main 2022 (28 Jun Shift 2)
AB2 is 10% dissociated in water to A2+ and B-. The boiling point of 10.0 molal aqueous solution of AB2 is-- C∘. (Round off to the Nearest Integer).

[Given : Molal elevation constant of water Kb=0.5 K kg mol-1 boiling point of pure water=100°C]

Options:

The correct answer is:
106

Solution:
AB2→A2++2 B-t=0a00t=0a-aαaα2aα

nT=a-aα+aα+2aα

=a1+2α

so i=1+2α

Now ΔTb=i×m×Kb

ΔTb=1+2α×m×Kb

α=0.1 m=10 Kb=0.5

ΔTb=1.2×10×0.5

=6

So boiling point =106

Topic: Chemistry, Solutions, JEE Main, JEE Main 2021 (16 Mar Shift 1)
AB3 is an interhalogen T-shaped molecule. The number of lone pairs of electrons on A is (Integer answer)

Options:

The correct answer is:
2

Solution:
In interhalogen compound AB3 Central atom A is halogen so it contain 7 electrons in its outer most shell. so its structure is

So central atom A contain 2 lone pairs of electrons.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2021 (26 Aug Shift 1)