datavorous/entrance-exam-dataset huggingface dataset:
Conducting part of the tertiary bronchi and bronchioles end up in

Options:
A Segmental bronchi
B Segmental bronchiole
C Respiratory bronchioles
D Terminal bronchiole

The correct answer is:
Terminal bronchiole

Solution:
Tracheobronchial tree consists of -
I. Conducting part (from the external nostrils up to the terminal bronchioles). It performs the following functions -
A. Transport of the atmospheric air to the alveoli.
B. Removal of foreign particles.
C. Humidification of air.
D. Maintaining the temperature of air equal to body temperature.
II. Exchange part (includes alveoli and their ducts) - It is the site of actual diffusion of O2 and CO2 between blood and atmospheric air.

Topic: Biology, Breathing and Exchange of Gases, NEET
Conductivity of a photodiode starts changing only if the wavelength of incident light is less than 660 nm. The band gap of photodiode is found to be X8 eV. The value of X is:

(Given h=6.6×10-34 J s, e=1.6×10-19 C)

Options:
A 15
B 11
C 13
D 21

The correct answer is:
15

Solution:
Given: Threshold wavelength =660 nm.

Therefore, energy required(or band gap energy) Eg=hcλ=6.6×10-34×3×108660×10-9 J

=6.6×10-34×3×108660×10-9×1.6×10-19 eV

=158eV

So, X=15.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2024 (01 Feb Shift 2)
Conductivity of a solution is directly proportional to

Options:
A Dilution
B Number of ions
C Current density
D Volume of the solution

The correct answer is:
Number of ions

Solution:
Conductivity of a solution is directly proportional to the number of ions.

Topic: Chemistry, Electrochemistry, JEE Main
Conductivity (unit Siemen’s) is directly proportional to area of the vessel and the concentration of the solution in it and is inversely proportional to the length of the vessel then the unit of the constant of proportionality is

Options:
A Sm mol ${ }^{-1}$
B $\mathrm{Sm}^2 \mathrm{~mol}^{-1}$
C $S^{-2} m^2 \mathrm{~mol}$
D $\mathrm{S}^2 \mathrm{~m}^2 \mathrm{~mol}^{-2}$

The correct answer is:
$\mathrm{Sm}^2 \mathrm{~mol}^{-1}$

Solution:
$C=\frac{K[A] A}{l}, K=\frac{C \times l}{[A] A}=\frac{\mathrm{Sm}}{\mathrm{mol} \mathrm{m}^{-3} \mathrm{~m}^2}=\mathrm{Sm}^2 \mathrm{~mol}^{-1}$

Topic: Chemistry, Electrochemistry, JEE Main
Conductivity (unit Siemen’s S) is directly proportional to area of the vessel and the concentration
of the solution in it and is inversely proportional to the length of the vessel then the unit of the
constant of proportionality is

Options:
A
$\mathrm{Sm} \mathrm{~mol} ^{-1}$

B
$\mathrm{Sm}^2 \mathrm{~mol}^{-1}$

C
$\mathrm{S}^{-2} \mathrm{~m}^2 \mathrm{~mol}$

D
$\mathrm{S}^2 \mathrm{~m}^2 \mathrm{~mol}^{-2}$

The correct answer is:
$\mathrm{Sm}^2 \mathrm{~mol}^{-1}$

Solution:
No Solution Available

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2002
Congestion of the lungs is one of the main symptoms of

Options:
A angina pectoris
B heart block
C heart failure
D cardiac arrest

The correct answer is:
heart failure

Solution:
Heart failure means impairement of heart to pump out blood or fill with blood. This causes distension of blood vessels in the lungs and thus results in lung congestion. This results in fluid accumulation in the lungs too

Topic: Biology, Body Fluids and Circulation, NEET
Conjugate of $1+i$ is

Options:
A $i$
B $1$
C $1-i$
D $1+i$

The correct answer is:
$1-i$

Solution:
$1-i$

Topic: Mathematics, Complex Number, JEE Main
Conservation of momentum in a collision between particles can be understood from

Options:
A
Conservation of energy

B
Newton's first law only

C
Newton's second law only

D
both Newton's second and third law

The correct answer is:
both Newton's second and third law

Solution:
By Newton's second law :
$$
\vec{F}_{e x t}=\frac{d p}{d t}
$$
As $\vec{F}_{e x t}$ in law of conservation of momentum is zero.
If $F_{e x t}=0, d p=0 \Rightarrow p=$ constant
Hence, momentum of a system will remain conserve if external force on the system is zero.
In case of collision between particle equal and opposite forces will act on individual particle by Newton's third law.
So, $\vec{F}_{12}=-\vec{F}_{21} \quad\left(\because \vec{F}_{e x t}=0\right)$
Total force on the system will be zero.
$$
\begin{aligned}
&\frac{d p_{12}}{d t}=-\frac{d \vec{p}_{21}}{d t} \text { or } d \vec{p}_{12}=-d \vec{p}_{21} \\
&\Rightarrow\left(d \vec{p}_{12}+d \vec{p}_{21}=0\right)
\end{aligned}
$$
So prove the law of conservation of momentum and verifies the option (d).

Topic: Physics, Laws of Motion
Consider Ι1=∫π4π2esin⁡x+1ecos⁡x+1dx and Ι2=∫π4π2ecos⁡x+1esin⁡x+1dx, then

Options:
A Ι1>Ι2
B Ι1<Ι2
C Ι1=Ι2
D Ι1+Ι2=0

The correct answer is:
Ι1>Ι2

Solution:
As sin⁡x>cos⁡x, ∀x∈π4,π2
⇒esin⁡x>ecos⁡x
∴esin⁡x+1ecos⁡x+1>ecos⁡x+1esin⁡x+1
⇒∫π4π2esin⁡x+1ecos⁡x+1dx>∫π4π2ecos⁡x+1esin⁡x+1
⇒Ι1>Ι2>0

Topic: Mathematics, Definite Integration, JEE Main
Consider ÎÎ±=â«Î±Î±2dxx (where Î±>0), then the value of âr=25Îr+âk=25Î1k is

Options:
A 0
B 1
C ln⁡2
D ln⁡4

The correct answer is:
0

Solution:
Ια=ln⁡xαα2=ln⁡α2-ln⁡α
=ln⁡α2α
=ln⁡α
Thus, ∑r=25Ιr=ln⁡2+ln⁡3+ln⁡4+ln⁡5 and

∑r=25Ι1k=ln⁡12+ln⁡13+ln⁡14+ln⁡15=-ln⁡2+ln⁡3+ln⁡4+ln⁡5
Hence, their sum equals to zero

Topic: Mathematics, Definite Integration, JEE Main
Consider 3 non-zero vectors aâ, bâ and câ such that aââcââ 0. If a vector râ satisfies the relation râÃaâ=bâÃaâ and rââcâ=0, then aââcârâÃbâ+bââcâaâÃrâ is equal to

Options:

The correct answer is:
0

Solution:
Given, r→×a→=b→×a→

⇒r→-b→×a→=0
⇒r→-b→=λa→⇒r→=b→+λa→
∴r→⋅c→=c→⋅b→+λa→⋅c→=0⇒λ=-b→⋅c→a→⋅c→……..i
r→×b→=b→+λa→×b→=λa→×b→……..ii
a→×r→=a→×b→+λa→=a→×b→……..iii
Now from ii & iii,

a→⋅c→r→×b→+b→⋅c→a→×r→
=a→⋅c→λa→×b→+b→⋅c→a→×b→
=a→×b→a→⋅c→λ+b→⋅c→=0 (from i)

Topic: Mathematics, Vector Algebra, JEE Main
Consider∫3x4+2x2+1x4+x2+1dx=fx. If f1=3, then f22 is equal to

Options:

The correct answer is:
84

Solution:
Multiplying the numerator and denominator by x in the given integral, we get,

fx=∫3x5+2x3+xx6+x4+x2dx

Now substitute x6+x4+x2=t2
⇒23x5+2x3+xdx=2tdt
So, fx=∫tdtt=x6+x4+x2+C
As f1=3⇒C=0

Hence, f2=64+16+4⇒f22=84

Topic: Mathematics, Indefinite Integration, JEE Main
Consider 5 independent Bernoulli's trials each with probability of success $\mathrm{p}$. If the probability of at least one failure is greater than or equal to $\frac{31}{32}$, then $\mathrm{p}$ lies in the interval

Options:
A
$\left(\frac{3}{4}, \frac{11}{12}\right]$

B
$\left[0, \frac{1}{2}\right]$

C
$\left(\frac{11}{12}, 1\right]$

D
$\left(\frac{1}{2}, \frac{3}{4}\right]$

The correct answer is:
$\left[0, \frac{1}{2}\right]$

Solution:
$\mathrm{n}=5$
Success $=\mathrm{p}$
Failure $=\mathrm{q}$
$\mathrm{P}$ (at least one failure) $\geq \frac{31}{32}$
$1-\mathrm{P}$ (no failure) $\geq \frac{31}{32}$
$1-\mathrm{P}(\mathrm{x}=5) \geq \frac{31}{32}$
$1-{ }^5 \mathrm{C}_5 \mathrm{p}^5 \geq \frac{31}{32}$
$-\mathrm{p}^5 \geq-\frac{1}{32}$
$\mathrm{p}^5 \leq \frac{1}{32}$
$\mathrm{p} \leq \frac{1}{2}$
$\mathrm{p} \in\left[0, \frac{1}{2}\right]$

Topic: Mathematics, Probability, JEE Main, JEE Main 2011
Consider 10 observation x1,x2,...x10, such that ∑i=110xi−α=2 and ∑i=110xi−β2=40, where α,β are positive integers. Let the mean and the variance of the observations be 65 and 8425 respectively. The βα is equal to:

Options:
A 2
B 32
C 52
D 1

The correct answer is:
2

Solution:
Given observations are x1,x2,...,x10

⇒∑i=110xi−α=2

⇒∑i=110xi−10α=2

Mean, μ=65=∑xi10

⇒∑xi=12

⇒10α+2=12

⇒α=1

Now, ∑i=110xi−β2=40. Let yi=xi−β

⇒σy2=110∑yi2−y¯2

⇒8425=110∑xi−β2−∑i=110xi−β102

⇒8425=4−12−10β102

⇒6−5β52=4−8425

⇒6−5β52=1625

⇒6−5β=±4

⇒β=25 (not possible) or β=2

Hence, βα=2

Topic: Mathematics, Statistics, JEE Main, JEE Main 2024 (01 Feb Shift 2)
Consider A=â«01dx1+x3, then A satisfies

Options:
A A>π4
B A<π4
C A=π4
D A=π6

The correct answer is:
A>π4

Solution:
As, xâ0,1âx2>x3
â1+x2>1+x3 or 11+x2<11+x3
Thus, â«01dx1+x2<â«01dx1+x3
i.e., tan-1x01<A
âA>tan-11-tan10
âA>Ï4

Topic: Mathematics, Definite Integration, JEE Main
Consider A=â«0Ï4sin2xxdx, then

Options:
A A>π2
B A=π2
C A<π2
D A>π

The correct answer is:
A<π2

Solution:
As sinθ<θ, ∀θ∈0,π2
∴sin2x<2x
So, A=2∫0π4sin2x2xdx<2∫0π41dx
⇒A<2π4=π2

Topic: Mathematics, Definite Integration, JEE Main
Consider a 20 kg uniform circular disk of radius 0.2 m. It is pin supported at its center and is at rest initially. The disk is acted upon by a constant force F=20 N through a massless string wrapped around its periphery as shown in the figure.

Suppose the disk makes n number of revolutions to attain an angular speed of 50 rad s-1. The value of n, to the nearest integer, is _______ .

[Given : In one complete revolution, the disk rotates by 6.28 rad]

Options:

The correct answer is:
20

Solution:
α=τI=F.R.mR2/2=2 FmR

α=2×20020×0.2=10 rad/s2

ω2=ω02+2αΔθ

502=02+210Δθ⇒Δθ=250020
Δθ=125 rad

No. of =1252π≈20 revolution

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2021 (16 Mar Shift 1)
Consider a 72 cm long wire AB as shown in the figure. The galvanometer jockey is placed at P on AB at a distance x cm from A. The galvanometer shows zero deflection.

The value of x, to the nearest integer, is

Options:

The correct answer is:
48

Solution:
In balanced conditions,

126=x72-x

⇒x=48 cm.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2021 (18 Mar Shift 2)
Consider A=a11a12a21a22 and B=1121 such that AB=BA, then the value of a12a21+a11a22 is

Options:
A 2
B 4
C 32
D 12

The correct answer is:
32

Solution:
Given:
AB=BA
a11a12a21a221121=1121a11a12a21a22
a11+2a12a11+a12a21+2a22a21+a22=a11+a21a12+a222a11+a212a12+a22
a11+2a12=a11+a21 ⇒2a12=a21
a11+a12=a12+a22⇒a11=a22
a21+2a22=2a11+a21⇒a11=a22
a21+a22=2a12+a22⇒a21=2a12
a12a21=12, a11a22=1

Topic: Mathematics, Matrices, JEE Main
Consider a ΔABC where A1,3,2,B−2,8,0 and C3,6,7. If the angle bisector of ∠BAC meets the line BC at D, then the length of the projection of the vector AD→ on the vector AC→ is:

Options:
A 37238
B 382
C 39238
D 19

The correct answer is:
37238

Solution:
Given,

A1,3,2, B-2,8,0 & C3,6,7

And angle bisector of ∠BAC meets the line BC at D,

Now, plotting the diagram we get,

Now, finding

AC→=2i^+3j^+5k^

⇒AC→=4+9+25=38

And AB→=9+25+4=38

Using angle bisector theorem, we get

BDCD=ABAC

⇒BD:CD=1:1

Using section formula to find coordinates of D,

⇒D≡1×3+1-21+2, 1×6+1×81+2, 1×7+1×01+2

⇒D≡12, 7, 72

⇒AD→=−12i^+4j^+32k^

⇒AD→=12-i^+8j^+3k^

Length of projection of AD→ on AC→ is given by,

⇒d=AD→·AC→AC→

⇒d=37238

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2024 (01 Feb Shift 2)
Consider a binary star system of star A and star B with masses mA and mB revolving in a circular orbit of radii rA and rB, respectively. If TA and TB are the time period of star A and star B, respectively, then:

Options:
A TATB=rArB32
B TA=TB
C TA>TB (if mA>mB)
D TA>TB (if rA>rB)

The correct answer is:
TA=TB

Solution:
1) From the centre of mass, we get

M1R1=M2R2------1

2) Force acting on both starts i.e. gravitational force and centripetal force are balanced.

GM1M2R2=M1V12R1=M1V12R1------2

As we know, V1=2πR1T1&V2=2πR2T2

Substituting equation 3 in equation 2, we get

M1R1T12=M1R2T22------4

From equation 1, we conclude in eqution 4

T12=T22 or T1=T2

Topic: Physics, Gravitation, JEE Main, JEE Main 2021 (20 Jul Shift 2)
Consider a black body radiation in a cubical box at absolute temperature $T$. If the length of each side of the box is doubled and the temperature of the walls of the box and that of the radiation is halved, then the total energy

Options:
A halves
B doubles
C quadruples
D remains the same

The correct answer is:
remains the same

Solution:
Assuming temperature of the body and cubica box is same initially, i.e., $T$ and finally it becomes $T / 2$ Because, temperature of body and surrounding remains same.
Hence, no net loss of radiation occur through the body. This total energy remains constant.

Topic: Physics, Thermal Properties of Matter, WBJEE, WBJEE 2014
Consider a block and trolley system as shown in figure. If the coefficient of kinetic friction between the trolley and the surface is 0.04, the acceleration of the system in m s-2 is: (Consider that the string is massless and unstretchable and the pulley is also massless and frictionless) :

Options:
A 3
B 4
C 2
D 1.2

The correct answer is:
2

Solution:
Maximum kinetic friction acting, fk=μN=0.04×20g=8 Newton

Therefore, acceleration of the system will bea=60−826=2 m s-2

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2024 (01 Feb Shift 1)
Consider a block kept on an inclined plane (inclined at 45°) as shown in the figure. If the force required to just push it up the incline is 2 times the force required to just prevent it from sliding down, the coefficient of friction between the block and inclined plane (µ) is equal to :

.

Options:
A 0.33
B 0.60
C 0.25
D 0.50

The correct answer is:
0.33

Solution:
The forces acting on block is shown in above figure.

Here, normal force, N=mgcos45° and frictional force, f=μN.

Balancing the force along incline, we have

F1=mg sin45°+f=mg sin45°+μN

⇒F1=mg2+μmg cos45°

⇒F1=mg21+μ

Now, the forces acting on block to just push it up on inclined plane is shown below.

Balancing the force along incline, we have

F2=mg sin45°-f=mg sin45°-μN

=mg21-μ

Given, F1=2F2

⇒mg21+μ=2mg21-μ

⇒1+μ=2-2μ

⇒μ=13=0.33

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2023 (25 Jan Shift 2)
Consider a brass rod and a steel rod (80 cm longer than the brass rod) at 0 Â°C. It is observed that on increasing temperatures of the two rods by same amount difference in lengths of the two rods does not change. Given that the thermal coefficient of linear expansion for steel and brass are 11Ã10-6 â-1 and 19Ã10-6 â-1 respectively. The sum of lengths of the two rods at 0o C is

Options:
A 2 m
B 4 m
C 3 m
D 1.5 m

The correct answer is:
3 m

Solution:
The length and coefficient of thermal expansion for the rods are respectively;
Î»1,Î±1â steel rod
Î»2,Î±2â Brass rod
As difference in lengths is constant
âÎÎ»1=ÎÎ»2
âÎ»1Î±1ÎT=Î»2Î±2ÎT
âÎ»1Î±1=Î»2Î±2â¦..i
Also âÎ»1-Î»2=Î»â¦â¦ii
Î»1=1Î±2Î±2-Î±1 and Î»2=1Î±1Î±2-Î±1
â sum =1Î±2+Î±1Î±2-Î±1=80Ã308
=300 cm=3 m

Topic: Physics, Thermal Properties of Matter, JEE Main
Consider a car moving on a straight road with a speed of $100 \mathrm{~m} / \mathrm{s}$. The distance at which car can be stopped is $\left[\mu_{\mathrm{k}}=0.5\right]$

Options:
A
$800 \mathrm{~m}$

B
$1000 \mathrm{~m}$

C
$100 \mathrm{~m}$

D
$400 \mathrm{~m}$

The correct answer is:
$1000 \mathrm{~m}$

Solution:
$\mu_{\mathrm{k}} \mathrm{mgs}=\frac{1}{2} \mathrm{mu}^2$
$\mathrm{~s}=\frac{\mathrm{u}^2}{2 \mu_{\mathrm{k}} \mathrm{g}}=1000 \mathrm{~m}$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2005
Consider a certain reaction A → Products with k = 2.0 × 10–2s–1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L–1

Options:
A 0.756 M
B 0.368 M
C 0.135 M
D 0.956 M

The correct answer is:
0.135 M

Solution:
Correct Option is : (C)
0.135 M

Topic: Chemistry, Chemical Kinetics, NEET
Consider a certain reaction \(\mathrm{A} \longrightarrow\) Products with \(k=\) \(2 \cdot 0 \times 10^{-2} \mathrm{~s}^{-1}\). Calculate the concentration of A remaining after \(100 \mathrm{~s}\) if the initial concentration of \(\mathrm{A}\) is \(1.0 \mathrm{~mol} \mathrm{~L}^{-1}\).

Options:

None

Solution:
\(k=\frac{2 \cdot 303}{t} \log \frac{\left[\mathrm{A}_0\right]}{[\mathrm{A}]}\)
\(2 \cdot 0 \times 10^{-2} \mathrm{~s}^{-1}=\frac{2 \cdot 303}{100 \mathrm{~s}} \log \frac{1 \cdot 0 \mathrm{~mol} \mathrm{~L}^{-1}}{[\mathrm{~A}]}\)
\(\log [\mathrm{A}]=-0.8684\)
\(=\) Antilog \((-0.8684)=0 \cdot 1354 \mathrm{~mol} \mathrm{~L}^{-1}\).

Topic: Chemistry, Chemical Kinetics
Consider a circle C which touches the y- axis at 0,6 and cuts off an intercept 65 on the x- axis. Then the radius of the circle C is equal to :

Options:
A 53
B 9
C 8
D 82

The correct answer is:
9

Solution:
We have, a circle C which touches the y- axis at 0,6 and cuts off an intercept 65 on the x- axis.

We know that, length of the chord of the circle using perpendicular distance from the center is 2r2-d2 where r is the radius of the circle and d is the perpendicular distance from the center to the chord.

From the given data, we have the following diagram

Hence, length of the chord of the circle ⇒2r2-d2=65

⇒r2-d2=35

On squaring on both sides

⇒r2-d2=45

⇒d2=r2-45

Given that circle touches the y-axis at 0, 6. Therefore, d=6.

⇒r2=45-36

r=36+45

r=9.

Topic: Mathematics, Circle, JEE Main, JEE Main 2021 (27 Jul Shift 2)
Consider a circle x-α2+y-β2=50, where α,β>0. If the circle touches the line y+x=0 at the point P, whose distance from the origin is 42 , then (α+β)2 is equal to _______.

Options:

The correct answer is:
100

Solution:
Given: x-α2+y-β2=50

The given equation of circle represent centre as α,β and radius as 52 units.

Now, x+y=0 is tangent to the given circle at P.

We know that, radius is perpendicular to tangent at the point of tangency.

⇒CP⊥x+y=0 and CP=r

⇒r=α×1+β×112+12

⇒52=α+β2

⇒50=α+β22

⇒α+β2=100

Topic: Mathematics, Circle, JEE Main, JEE Main 2024 (27 Jan Shift 2)
Consider a class of 5 girls and 7 boys. The number of different teams consisting of 2 girls and 3 boys that can be formed from this class, if there are two specific boys A and B, who refuse to be the members of the same team, is:

Options:
A 300
B 200
C 500
D 350

The correct answer is:
300

Solution:
7B5G→3B2G
Total number of ways of forming a team of 3 boys and 2 girls = 7C3× 5C2=7.6.51.2.3×5.41.2=350
Total number of ways of forming a team if two specific boys B1,B2 always join the team
= 5C1× 5C2=50
So, the total number of ways of forming a team so that two specific boys never come together
=350-50=300

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2019 (09 Jan Shift 1)
Consider a collection of a large number of particles each moving with speed v. The direction of velocity is randomly distributed in the collection. Find the magnitude of the relative velocity between a pair of particles averaged over all the pairs in the collection.

Options:
A v
B 4v
C 4vπ
D 4πv

The correct answer is:
4vπ

Solution:
Consider any two particles whose velocity vector particles making angle θ .
Relative velocity vAB = vA - vB
⇒ vrel=v2+v2-2v2 cos θ
⇒ vrel=2v sinθ2
Average of vrel=∫02πvrel dθ∫02π dθ=4vπ

Topic: Physics, Motion In Two Dimensions, JEE Main
Consider a compound slab consisting of two different materials having equal thickness and thermal conductivities $K$ and $2 K$, respectively. The equivalent thermal conductivity of the slab is:

Options:
A $\frac{4}{3} K$
B $\sqrt{2} K$
C $3 \mathrm{~K}$
D $\frac{2}{3} K$

The correct answer is:
$\frac{4}{3} K$

Solution:
$$
\therefore \frac{2 l}{K_{e g}(A)}=\frac{l}{2 K A}+\frac{l}{K A}
$$
for series connection $\mathrm{R}=\mathrm{R}_1+\mathrm{P}_2$
$$
\Rightarrow K_{e g}=\frac{4}{3} K
$$

Topic: Physics, Thermal Properties of Matter, JEE Main
Consider a concave mirror of $10 \mathrm{~cm}$ focal length illuminated by an object kept at a distance of $25 \mathrm{~cm}$. The distance at which the image is formed and its magnification respectively are

Options:
A $-16.7 \mathrm{~cm}$ and -0.67
B $7.1 \mathrm{~cm}$ and 0.29
C $-16.7 \mathrm{~cm}$ and 0.67
D $7.1 \mathrm{~cm}$ and -0.29

The correct answer is:
$-16.7 \mathrm{~cm}$ and -0.67

Solution:
Using mirror equation, we have,
$$
\begin{aligned}
\frac{1}{v}+\frac{1}{u}= & \frac{1}{f} \\
\frac{1}{v}= & \frac{1}{f}-\frac{1}{u} \\
\Rightarrow \quad \frac{1}{v}= & \frac{1}{(-10)}-\frac{1}{(-25)} \\
& {[\therefore \text { Given } f=-10 \mathrm{~cm}, u=-25 \mathrm{~cm}] } \\
\Rightarrow \quad v= & -16.7 \mathrm{~cm}
\end{aligned}
$$
or
and magnification,
$$
m=-\left(\frac{v}{u}\right)=\frac{-(-16.7)}{-25}=-0.67
$$

Topic: Physics, Ray Optics, JEE Main
Consider a cubical hall ABCDPQRS with each side 20 m and base as ABCD. If G is the center of the wall BCRQ and T is the midpoint of the side AB, then the angle of elevation of G at the point T is

Options:
A sin-1⁡13
B cos-1⁡13
C tan-1⁡13
D cot-1⁡13

The correct answer is:
sin-1⁡13

Solution:
Let H be the midpoint of BC

Since â TBH=90o,

âTH2=BT2+BH2=102+102=200
Also, since â THG=90o,

âTG2=TH2+GH2=200+100=300
Let Î¸ be the required angle of elevation of G at T.
Then, sinâ¡Î¸=GHTG=10103=13
âÎ¸=sin-1â¡13

Topic: Mathematics, Heights and Distances, JEE Main
Consider a cuboid of sides 2x,4x and 5x and a closed hemisphere of radius r. If the sum of their surface areas is constant k, then the ratio x:r, for which the sum of their volumes is maximum, is

Options:
A 2:5
B 19:45
C 3:8
D 19:15

The correct answer is:
19:45

Solution:
Let sum of surface area

k=3πr2+76x2

∵ k is constant so dkdx=0

⇒ 6πrdrdx+2×76x=0

⇒ drdx=-76x3πr ⋯1

Now total volume V=23πr3+40x3

For maximum volume dVdx=0

⇒2πr2drdx+120x2=0 ⋯2

⇒ 2πr2-76x3πr+120x2=0

⇒ 8x-19r3+15x=0

⇒ xr=1945

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2022 (26 Jun Shift 2)
Consider a curve y=yx in the first quadrant as shown in the figure. Let the area A1 is twice the area A2. Then the normal to the curve perpendicular to the line 2x-12y=15 does NOT pass through the point __

âââââââ

Options:
A 6,21
B 8,9
C 10,-4
D 12,-15

The correct answer is:
10,-4

Solution:
Given that A1=2A2

Now, from the graph

A1+A2=x×y-2×4=xy-8

⇒32 A1=xy-8

⇒A1=23xy-163

⇒∫4xfxdx=23xy-163

⇒fx=23xdydx+y

⇒23xdydx=y3

⇒2∫dyy=∫dxx

⇒2lny=lnx+lnc

⇒y2=cx

As f4=2⇒c=1

So y2=x is the given curve.

Given that the slope of normal =-6

So the equation of normal will be

y=-6x-12-6-14-63

⇒y=-6x+3+54

⇒y+6x=57

Only 10,-4 does not satisfy from the given options.

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2022 (27 Jul Shift 2)
Consider a cycle tyre being filled with air by a pump. Let $V$ be the volume of the tyre (fixed) and at each stroke of the pump $\Delta V(\ll V)$ of air is transferred to the tube adiabatically. What is the work done when the pressure in the tube is increased from $p_1$ to $p_2$ ?

Options:

None

Solution:
When air is transferred into tyre adiabatically then let initial volume of air in tyre $V$, after pumping one stroke, if volume is increased by $\Delta V$ and pressure is increased by $\Delta p$. Then for just before and after an stroke, we can write
$$
\begin{aligned}
&p_1 V_1^\gamma=p_2 V_2^\gamma \\
&p(V+\Delta V)^\gamma=(p+\Delta p) V^\gamma(\because \text { volume is fixed) } \\
&p V^\gamma\left(1+\frac{\Delta V}{V}\right)^\gamma=p\left(1+\frac{\Delta p}{p}\right) V^\gamma \\
&\text { (As volume of tyre } V \text { remains constant) } \\
&p V^\gamma\left(1+\gamma \frac{\Delta V}{V}\right)=p V^\gamma\left(1+\frac{\Delta p}{p}\right)
\end{aligned}
$$
On expanding by binomial theorem and neglecting the higher terms of $\Delta V$ as $(\Delta V \ll V)$
$$
\begin{aligned}
&1+\gamma \frac{d V}{d V}=1+\frac{\Delta p}{p} \\
&\gamma \frac{\Delta V}{V}=\frac{\Delta p}{p} \Rightarrow \Delta V=\frac{1}{\gamma} \frac{V}{p} \Delta p \\
&d V=\frac{1}{\gamma} \frac{V}{p} d p
\end{aligned}
$$
Integrating both sides with work done limits $W_1$ to $W_2$ and pressure limit from $p_1$ to $p_2$.
$$
\begin{aligned}
&\int_{W_1}^{W_2}(d W)=\int_{p_1}^{p_2} p d V=\int_{p_1}^{p_2} p \times \frac{1}{\gamma} \frac{V}{p} d p\left[\because d V=\frac{1}{\gamma} \frac{V}{p} d p\right] \\
&=\frac{V}{\gamma} \int_{p_1}^{p_2} d p=\frac{V}{\gamma}\left(p_2-p_1\right) \\
&W=\frac{\left(p_2-p_1\right)}{\gamma} V
\end{aligned}
$$

Topic: Physics, Thermodynamics
Consider a cylinder of mass M resting on a rough horizontal rug that is pulled out from under it with acceleration 'a' perpendicular to the axis of the cylinder. What is Ffriction at point P ? It is assumed that the cylinder does not slip.

Options:
A Ma
B Ma 2
C Ma 3
D Mg

The correct answer is:
Ma 3

Solution:
For no slipping at P
a 1 + α R = a → (1)
f = Ma 1 = M a - α R → (2)
f. R = I · α = MR 2 2 · α → (3)
Solve the equations (2) and (3)
f = Ma 3 â

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2014 (19 Apr Online)