datavorous/entrance-exam-dataset huggingface dataset:
Two men are on the opposite sides of a tower. They measure the angles of elevation of the top of the tower as 45o and 30o respectively. If the height of the tower is 40 m, then the distance between the men is

Options:
A 40 m
B 403Â m
C 68.28 m
D 109.28 m

The correct answer is:
109.28 m

Solution:
From ΔO1 AB, tan⁡45o=40x ⇒ x=40 m
From ΔO2AB, cot⁡30o=y40
⇒ y=403 m
Distance between the men =40+403 =109.28 m .

Topic: Mathematics, Heights and Distances, JEE Main
Two men with weights in the ratio 4 : 3 run up a staircase in time in the ratio 12 : 11. The ratio of the power of the first to that of second is

Options:
A 4 3
B 12 11
C 48 33
D 11 9

The correct answer is:
11 9

Solution:
Power=worktime=energytime=mght=Wâ¡hâ¡t( W = weight )
â´Pâ¡1Pâ¡2=Wâ¡1Wâ¡2Ãt2t1=43Ã1112=119 ( h is same in both cases )

Topic: Physics, Work Power Energy, JEE Main
Two metal cubes $A$ and $B$ of same size are arranged as shown in the figure. The extreme ends of the combination are maintained at the indicated temperatures. The arrangement is thermally insulated. The coefficients of thermal conductivity of $A$ and $B$ are $300 \mathrm{~W} / \mathrm{m}^{\circ} \mathrm{C}$ and $200 \mathrm{~W} / \mathrm{m}^{\circ} \mathrm{C}$, respectively. After steady state is reached, the temperature of the interface will be

Options:
A $45^{\circ} \mathrm{C}$
B $90^{\circ} \mathrm{C}$
C $30^{\circ} \mathrm{C}$
D $60^{\circ} \mathrm{C}$

The correct answer is:
$60^{\circ} \mathrm{C}$

Solution:
Temperature of interface $T=\frac{K_1 \theta_1+K_2 \theta_2}{K_1+K_2}$
$=\frac{300 \times 100+200 \times 0}{300+200}=60^{\circ} \mathrm{C}$

Topic: Physics, Thermal Properties of Matter, JEE Main
Two metal P and Q belong to the second group of periodic table. P forms insoluble oxide but soluble sulphate Q forms a soluble oxide but insoluble sulphate. Hydroxide of metal P is soluble in NaOH while that of metal Q is insoluble in NaOH. What are metal P and Q?

Options:
A P = Be, Q = Ba
B P = Mg, Q = Ca
C P = Ca, Q = Sr
D P = Ba, Q = Mg

The correct answer is:
P = Be, Q = Ba

Solution:
BeO-Insoluble, BeSO4 -Soluble
BaO-Soluble, BaSO4 -Insoluble
BeOH2 -Soluble in NaOH
Ba(OH)2 -Insoluble in NaOH

Topic: Chemistry, s Block Elements, JEE Main
Two metal plates having a potential difference of 800 V are 2 cm apart. It is found that a particle of mass 1.96×10-15 kg remains suspended in the region between the plates. The charge on the particle must be (e= elementary charge)

Options:
A 2 e
B 3 e
C 6 e
D 8 e

The correct answer is:
3 e

Solution:
V=800 V

E=Vd=8002×10-2 V m-1

=4×104 V m-1

At equilibrium electrostatic force must be equal to force due to weight.

qE=mg

⇒ q=mgE

=1.96×10-15×9.84×104 coulomb

=1.96×10-15×9.84×104×1.6×10-19 elementary charges

=3 e

Topic: Physics, Electrostatics, JEE Main
Two metal rods 1 and 2 of same lengths have same temperature difference between their ends. Their thermal conductivities are $K_1$ and $K_2$ and crosssectional areas $A_1$ and $A_2$ respectively. If the rate of heat conduction in 1 is four times that in 2 , then:

Options:
A $K_1 A_1=4 K_2 A_2$
B $K_1 A_1=2 K_2 A_2$
C $4 K_1 A_1=K_2 A_2$
D $K_1 A_1=K_2 A_2$

The correct answer is:
$K_1 A_1=4 K_2 A_2$

Solution:
Let $L=$ length of each rod Temperature difference $=\Delta T$
$\therefore$ Rate of flow of rod (i) is $H_1=$ $\frac{K_1 A_1 \Delta T}{L}$ and Rate of flow of rod (2) is $H_2=\frac{K_2 A_2 \Delta T}{L}$ from the equation
$$
\begin{array}{rlrl}
H_1 & =4 H_2 \\
\Rightarrow & \frac{K_1 A_1 \Delta \mathrm{T}}{L} & =4 \frac{K_2 A_2 \Delta T}{L} \\
\Rightarrow & K_1 A_1 & =4 K_2 A_2
\end{array}
$$

Topic: Physics, Thermal Properties of Matter, JEE Main
Two metal rods of the same lengths and area of cross-sections are fixed end to end between rigid supports. The materials of the rods have Young's moduli Y1 and Y2 and coefficients of linear expansion are Î± 1 and Î± 2 . The junction between the rods does not shift if the rods are cooled. Then

Options:
A Y1α1=Y2α2
B Y1α2=Y2α1
C Y1α12=Y2α22
D Y12α1=Y22α2

The correct answer is:
Y1α1=Y2α2

Solution:
Here Y1 and Y2 are Young's moduli of the material of the rods and α 1 and α 2 are the coefficients of thermal expansion. Using Y=stressstrain, T=YAαt, where t is the temperature. Tension must be the same in both the rods for their junction to be in equilibrium.
∴ Y1Aα1t=Y2Aα2t

Topic: Physics, Thermal Properties of Matter, JEE Main
Two metal slabs of same cross-sectional area have thicknesses $d_1$ and ; and thermal conductivities $K_1$ and $K_2$ respectively, are connected in series. The free ends of the two slabs are kept at temperatures $T_1$ and $T_2\left(T_1>T_2\right)$. The temperature $T$ of their common junction is

Options:
A $\frac{K_1 T_1 d_2+K_2 T_2 d_1}{K_1 d_2+K_2 d_1}$
B $\frac{K_1 T_1+K_2 T_2}{K_1+K_2}$
C $\frac{K_1 T_1+K_2 T_2}{T_1+T_2}$
D $\frac{K_1 T_1 d_1+K_2 T_2 d_2}{K_1 d_2+K_2 d_1}$

The correct answer is:
$\frac{K_1 T_1 d_2+K_2 T_2 d_1}{K_1 d_2+K_2 d_1}$

Solution:
Heat current, $\dot{Q}_1=\frac{K_1\left(T_1-T\right) A}{d_1}$
For second slab,
Heat current, $\dot{Q}_2=\frac{K_2\left(T-T_2\right) A}{d_2}$
As slabs are in series same heat current flows through them
$\dot{Q}_1=\dot{Q}_2$
$\therefore \frac{K_1\left(T_1-T\right) A}{d_1}=\frac{K_2\left(T-T_2\right) A}{d_2}$
Therefore, temperature $T$ of their common junction is
$T=\frac{K_1 T_1 d_2+K_2 T_2 d_1}{K_2 d_1+K_1 d_2}$

Topic: Physics, Thermal Properties of Matter, JEE Main
Two metal spheres of radii $R_1$ and $R_2$ are charged to the same potential. The ratio of charges on the spheres is

Options:
A $\sqrt{R_1}: \sqrt{R_2}$
B $R_1: R_2$
C $R_1^2: R_2^2$
D $R_1^3: R_2^3$

The correct answer is:
$R_1: R_2$

Solution:
Spheres have same potential
i.e. $k \frac{Q_1}{R_1}$ $=k \frac{Q_2}{R_2} \Rightarrow \frac{Q_1}{Q_2}=\frac{R_1}{R_2}$

Topic: Physics, Electrostatics, JEE Main
Two metal strips are riveted together at their ends by four rivets, each of diameter 6 mm. Assuming that each of the rivets carries one-quarter of the load, the maximum tension that can be produced in the riveted strip is [Breaking shear stress of the rivet = 6.9×107 Pa]

Options:
A 7.8Ã 10 3 âN
B 6.9Ã 10 3 âN
C 3.14× 10 3  N
D none of these

The correct answer is:
7.8× 10 3  N

Solution:
Diameter of each rivet ( D )=6ââmm
â´ Radius (r) = D 2 = 3 mm = 3 Ã 10 - 3 m
Maximum shearing stress on each rivet = 6.9 Ã 107 Pa
Let w be the maximum load that can be subjected to the riveted strip.
As each rivet carry one-quarter of the load, therefore
Load on each rivet = w 4
Maximum shearing stress = Maximum shearing force Area
â´ 6.9 Ã 10 7 = w / 4 Ï r 2
or w=6.9Ã 10 7 â4 Ïr 2
âor w=6.9Ã 10 7 Ã4Ã3.1â4Ã ( 3Ã 10 â3 ) 2
=6.9Ã4Ã3.14Ã9Ã10
=7.8Ã 10 3 âN

Topic: Physics, Mechanical Properties of Solids, JEE Main
Two metal wires of identical dimensions are connected in series. If $\sigma_{1}$ and $\sigma_{2}$ are the electrical conductivities of the metal wires respectively, the effective conductivity of the combination is

Options:
A $\sigma_{1}+\sigma_{2}$
B $\frac{\sigma_{1} \sigma_{2}}{\sigma_{1}+\sigma_{2}}$
C $\frac{2 \sigma_{1} \sigma_{2}}{\sigma_{1}+\sigma_{2}}$
D $\frac{\sigma_{1}+\sigma_{2}}{2 \sigma_{1} \sigma_{2}}$

The correct answer is:
$\frac{2 \sigma_{1} \sigma_{2}}{\sigma_{1}+\sigma_{2}}$

Solution:
$R_{e q}=R_{1}+R_{2}$
$\frac{1}{\sigma_{e q}} \frac{2 l}{A}=\frac{1}{\sigma_{1}} \frac{1}{A}+\frac{1}{\sigma_{2}} \frac{1}{A}\left[\sigma_{e q}=\frac{2 \sigma_{1} \sigma_{2}}{\sigma_{1}+\sigma_{2}}\right]$

Topic: Physics, Current Electricity, JEE Main
Two metallic spheres of radii $1 \mathrm{~cm}$ and $3 \mathrm{~cm}$ are given charges of $-1 \times 10^{-2} \mathrm{C}$ and $5 \times 10^{-2} \mathrm{C}$, respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is

Options:
A $2 \times 10^{-2} \mathrm{C}$
B $3 \times 10^{-2} \mathrm{C}$
C $4 \times 10^{-2} \mathrm{C}$
D $1 \times 10^{-2} \mathrm{C}$

The correct answer is:
$3 \times 10^{-2} \mathrm{C}$

Solution:
Charge flows from high potential to low potential
$$
\begin{aligned}
\frac{K Q_1}{3} & =\frac{K Q_2}{1} \\
Q_1 & =3 Q_2
\end{aligned}
$$
Also
$$
\begin{aligned}
Q_1+Q_2 & =4 \times 10^{-2} \\
4 Q_2 & =4 \times 10^{-2}
\end{aligned}
$$
$Q_2=10^{-2}$
and $\quad Q_1=3 \times 10^{-2} \mathrm{C}$

Topic: Physics, Electrostatics, NEET, NEET 2012 (Mains)
Two metallic spheres of radii $1 \mathrm{~cm}$ and $3 \mathrm{~cm}$ are given charges of $-1 \times 10^{-2} \mathrm{C}$ and $5 \times 10^{-2} \mathrm{C}$, respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is

Options:
A $2 \times 10^{-2} \mathrm{C}$
B $3 \times 10^{-2} C$
C $4 \times 10^{-2} C$
D $1 \times 10^{-2} C$

The correct answer is:
$3 \times 10^{-2} C$

Solution:
$\quad \frac{k x}{1 \mathrm{~cm}}=\frac{k(Q-x)}{3 \mathrm{~cm}}$
$3 x=Q-x \Rightarrow 4 x=Q$
$x=\frac{Q}{4}=\frac{4 \times 10^{-2}}{4} C=1 \times 10^{-2}$
$Q^{\prime}=\mathrm{Q}-x=3 \times 10^{-2} \mathrm{C}$

Topic: Physics, Electrostatics, JEE Main
Two metallic spheres S1 and S2 are made of the same material and have got the identical surface finish. The mass of S1 is thrice that of S2. Both the spheres are heated to the same high temperature and placed in the same room having lower temperature but are thermally insulated from each other. The ratio of the initial rate of cooling of S1 to that of S2 is

Options:
A 13
B 13
C 3
D (13)1&sol;3

The correct answer is:
(13)1&sol;3

Solution:
The rate at which energy radiates from the object is

Δ Q Δ t

= eσAT4

Since, ΔQ = mc ΔT, we get

ΔTΔl=eσAT4mc

Also, since m = 43 πr3ρ for a sphere, we get

A = 4πr2 = 4π (3m4πρ)23

Hence,

ΔQΔt=eσT4mc&lsqb;4π(3m4πρ)23&rsqb;=K(1m)13

For the given two bodies

Δ T / Δ t 1 Δ T / Δ t 2 = m 2 m 1 1 3 = 1 3 1 / 3

Topic: Physics, Thermodynamics, JEE Main
Two metallic wires of identical dimensions are connected is series. If σ1 and σ2 are the conductivities of the these wires respectively, the effective conductivity of the combination is :

Options:
A σ1σ2σ1+σ2
B 2σ1σ2σ1+σ2
C σ1+σ22σ1σ2
D σ1+σ2σ1σ2

The correct answer is:
2σ1σ2σ1+σ2

Solution:
Let length of wire be l,
Area of wire as A
For equivalent wire length =2l & area will be A

As the wires are connected in series combination, the thermal resistance

Req=R1+R2

⇒2lσeq A=lσ1 A+lσ1 A

⇒2σeq =1σ1+1σ2⇒σeq =2σ1σ2σ1+σ2

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2022 (29 Jul Shift 1)
Two moles of a gas at 8.21 bar and 300 K are expanded at constant temperature up to 2.73 bar against a constant pressure of 1 bar. How much work (in Latm) is done by the gas?

(neglect the sign)

Options:

The correct answer is:
12

Solution:
V1=2Ã0.0821Ã3008.21=6.0 L
V2=2Ã0.0821Ã3002.73=18.0 L
Wirr,isothermal=-PextÃÎV=-1Ã(18-6)
=-12 L atm

Topic: Chemistry, Thermodynamics (C), JEE Main
Two moles of a gas expand reversibly and isothermally at temperature of 300âK. Initial volume of the gas is 1âL while the final pressure is 4.926 atm. The work done by gas is

Options:
A -11488.28 J
B -15036.28 J
C -22488.28 J
D -33488.28 J

The correct answer is:
-11488.28 J

Solution:
To calculate final volume V2=nÃRÃTP2
V 2 = 2×0.0821×300 4.926 =10 L
Now, V 1 =1 L, V 2 =10 L, T=300 K, n=2 mole
W=−2.303 nRT log V 2 V 1
=-2.303Ã2Ã8.314Ã300Ãlogâ¡101
=-11488.28 J

Topic: Chemistry, Thermodynamics (C), JEE Main
Two moles of a monatomic ideal gas are taken through a cyclic process shown on the P-T diagram. The process CA is represented as PT = constant. If the efficiency of the given cyclic process is 1-x12ln2+15, then what is the value of x?

Options:

The correct answer is:
21

Solution:
For process AB,
TA = 300 K, TB = 600 K
W=nRÎT=nRTB-TA=300nR=600R
Q=nCpÎT=2Ã52R300=1500R
For process BC,
W=nRTlnVfVi=nRTlnpipf
=nRTln2=1200Rln2
For process CA,
W=â«P dV=â«600300KT2nRTKdT
=-2nR300=-1200R
Q=nCvÎT+W
=2Ã32R-300-1200R
=-900R-1200R=-2100R
Î·=600R+1200Rln2-1200R1500R+1200Rln2
Î· = 1 - 2 1 1 2 ln 2 + 1 5
Comparing with given equation, x = 21

Topic: Physics, Thermodynamics, JEE Main
Two moles of a monoatomic gas is mixed with six moles of a diatomic gas. The molar specific heat of the mixture at constant volume is :

Options:
A 94R
B 74R
C 32R
D 52R

The correct answer is:
94R

Solution:
For mixture of gases, we can write

Cv=n1Cv1+n2Cv2n1+n2

=2×32R+6×52R2+6

=94R

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2024 (01 Feb Shift 1)
Two moles of an ideal gas is expanded isothermally and reversibly from 1 litre to 10 litre at 300 K. The enthalpy change (in kJ) for the process is

Options:
A 11.4 kJ
B -11.4 kJ
C 0 kJ
D 4.8 kJ

The correct answer is:
0 kJ

Solution:
ΔH=ΔE+PV=ΔE+ΔRT=ΔE+RΔT
For isothermal expansion of ideal gas, ΔE=0
ΔE=CvΔT
So ΔH=0+0=0 ΔT=0
putting ΔE=CvΔT,ΔH=CvΔT+RΔT=Cv+RΔT=CpΔT for 1 mole
for n moles, ΔH=nCpΔT=0

Topic: Chemistry, Thermodynamics (C), JEE Main
Two moles of an ideal gas, with CPCV=53, are mixed with three moles of another ideal gas CPCV=43. The value of CPCV for the mixture is

Options:
A 1.45
B 1.50
C 1.47
D 1.42

The correct answer is:
1.42

Solution:
Let $n_1,n_2$ be the number of moles of gases $1$ and $2$.
We have $\dfrac{C_{p_1}}{C_{v_1}}=\dfrac{5}{3}$ and We know $C_{p_1}=C_{v_1}+R$
$\therefore \dfrac{5C_{v_1}}{3}=C_{v_1}+R,\dfrac{2C_{v_1}}{3}R_1,C_{v_1}=\dfrac{3R}{2}$
$A$ and $C_{p_1}=C_{ v_ 1 }+R=\dfrac{5R}{2}$
Also, $\dfrac{C_{ p_ 2}}{C_{ v_ 2 }}=\dfrac{4}{3}$ and
$C_{p_2}=C_{v_2}+R$
$\therefore \dfrac{4C_{ v_ 2 }}{3}=C_{ v_ 2 }+R, \dfrac{C_{ v_ 2 }}{3}=R,C_{ v_ 2 }=3R$
$C_{ p_ 2}=C_{ v_ 2 }+ R=4R$
Now $C_{ v_ {mix} }=\dfrac{n_1C_{ v_ 1 }+n_2C_{ v_ 2 }}{n_1+n_2}=\dfrac{2\times \dfrac{3R}{2}+3\times 3R}{2+3}$
$=\dfrac{3R+9R}{5}=\dfrac{12R}{5}$
$C_{ p_ {mix} }=\dfrac{n_1C_{ p_ 1 }+n_2C_{ p_ 2 }}{n_1+n_2}=\dfrac{2\times \dfrac{5R}{2}+3\times 4R}{2+3}$
$=\dfrac{17R}{5}$
$l_{mix}=\dfrac{C_{ p_ {mix} }}{C_{ v_ {mix} }}=\dfrac{17}{12}=1.4166=1.42$.

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2020 (07 Jan Shift 1)
Two moles of an ideal monoatomic gas at 5 bar and 300 K are expanded irreversibly up to a final pressure of 1 bar and 240 K against an external pressure of 0.5 bar. The work done by the gas is −xR . The value of ' x ' is
(Here ' R ' is gas constant)

Options:

The correct answer is:
180

Solution:
W=-pextV2â¡-V1
=-pextnRT2P2-nRT1P1
Wirr,adia=-0.52ÃRÃ2401-2ÃRÃ3005
= -180 R

Topic: Chemistry, Thermodynamics (C), JEE Main
Two moles of an ideal monoatomic gas at 5 bar and 300 K are expanded irreversibly up to an external pressure of 1 bar. The work is done by the gas is xR (in bar.Litre). The value of 'x 'is [R is gas constant]

Options:

The correct answer is:
288

Solution:
Given: pfinal=pext so p1=5 , p2=1 and pext=1 , T1=300,T2=?
for the calculation of T2
nCvT2-T1=-pextV2-V1=-pextnRT2P2-nRT1P1
nCvT2-T1=-pextnRT2P2-nRT1P1
nCvT2-T1=-pextnRT2P2-nRT1P1
CvT2-T1=pextRT2P2-RT1P1
3R2(T2⁡-300)=-1R×T21-R×3005
T 2  = 204 K
Wirr,adia=-12×R×2041-2×R×3005
=–288 R

Topic: Chemistry, Thermodynamics (C), JEE Main
Two moles of an ideal monoatomic gas is taken through a cycle ABCA as shown in the p-T diagram. During the process AB, the pressure and temperature of the gas vary such that pT = constant. If T1=300 K and the amount of heat released by the gas during the process AB is Q=-xR, then the value of x is

Options:

The correct answer is:
2100

Solution:
In the process pVÎ³ = constant
Molar heat capacity, C=RÎ³-1+R1-x

Here the process is p2V = constant pV12 = constant

i.e, x=12

C=R53-1+R1-12C=3.5R
QAâB=nCÎT

QAâB=2Ã3.5RÃ300-600
QAâB=-2100R

Topic: Physics, Thermodynamics, JEE Main
Two moles of an ideal monoatomic gas occupies a volume V at 27oC . The gas expands adiabatically to a volume 2V. Calculate (a) the final temperature of the gas and (b) change in its internal energy.

Options:
A (a)195 K (b)2.7 kJ
B (a)189 K (b)2.7 kJ
C (a)195 K (b)-2.7 kJ
D (a)189 K (b)-2.7 kJ

The correct answer is:
(a)189 K (b)-2.7 kJ

Solution:
TVγ-1=Constant
γ=53⇒300V23=T2V23
⇒T=189K⇒ΔU=nCvΔT=2×32×8.314×-11=-2.7 kJ

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2018 (08 Apr)
Two moles of benzaldehyde and one mole of acetone under alkaline conditions using aqueous \(\mathrm{NaOH}\) after heating gives \(x\) as the major product. The number of \(\pi\) bonds in the product \(x\) is _______

Options:

The correct answer is:
9

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2024 (08 Apr Shift 2)
Two moles of helium are mixed with n moles of hydrogen. If CpCv=32 for the mixture then the value of n is,

Options:
A 1
B 3
C 2
D 32

The correct answer is:
2

Solution:
γmix=1+2fmix
CpCv= fmix+2fmix=32
⇒ fmix=4
fmix= n1f1+ nf2n1+ n2
⇒4=2×3+n×52+n
⇒n=2 mole

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2018 (16 Apr Online)
Two moles of helium gas expanded isothermally and irreversibly at 27 o C from volume 1 dm3 to 1 m3 at constant pressure of 100 k Pa . Calculate the work done.

Options:
A 99900 J
B â 99900 J
C 34464.65 J
D â34464.65 J

The correct answer is:
â 99900 J

Solution:
W=-PâV
Given, P = 1 0 0 k Pa = 1 0 5 Pa
V1=1 dm3=10-3m3,V2=1 m3
W=â 10 5 Ã 1â 10 â3 âJ
â´ W=â 99900âJ

Topic: Chemistry, Thermodynamics (C), JEE Main
Two moles of helium gas is mixed with three moles of hydrogen molecules (taken to be rigid). What is the molar specific heat of mixture at constant volume? R=8.3 J/mol K

Options:
A 17.4 J/mol K
B 19.7 J/mol K
C 15.7 J/mol K
D 21.6 J/mol K

The correct answer is:
17.4 J/mol K

Solution:
Cvmix=n1Cv1+n2Cv2n1+n2=232R+352R2+3
=3R+152R5=2110R=17.4 J/mol K

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2019 (12 Apr Shift 1)
Two monochromatic coherent light beams $A$ and $B$ have intensities $L$ and $\frac{L}{4},$ respectively. If these beams are superposed, the maximum and minimum intensities will be

Options:
A $\frac{9 L}{4}, \frac{L}{4}$
B $\frac{5 L}{4}, 0$
C $\frac{5 L}{2}, 0$
D $2 L, \frac{L}{2}$

The correct answer is:
$\frac{9 L}{4}, \frac{L}{4}$

Solution:
For maximum intensity,
$$
I_{\max }=I_{1}+I_{2}+2 \sqrt{I_{1} I_{2}} \cos \delta
$$
$$
\begin{aligned}
\delta &=0 \\
I_{\max } &=L+\frac{L}{4}+2 \sqrt{L \times \frac{L}{4}} \cdot \cos 0 \\
&=\frac{5 L}{4}+\frac{2 L}{2} \\
&=\frac{5 L+4 L}{4} \\
&=\frac{9 L}{4} \\
I_{\min } &=I_{1}+I_{2}-2 \sqrt{I_{1} I_{2}} \cos \theta \\
\theta &=0^{\circ} \\
&=L+\frac{L}{4}-2 \sqrt{L \times \frac{L}{4}} \times 1 \\
&=\frac{5 L}{4}-L=\frac{L}{4}
\end{aligned}
$$
Thus, maximum and minimum intensities will be $\left(\frac{9 L}{4}, \frac{L}{4}\right)$

Topic: Physics, Waves and Sound, JEE Main
Two monochromatic light beams of intensity 16 and 9 units are interfering. The ratio of intensities of bright and dark parts of the resultant pattern is:

Options:
A
$\frac{16}{9}$

B
$\frac{4}{3}$

C
$\frac{7}{1}$

D
$\frac{49}{1}$

The correct answer is:
$\frac{49}{1}$

Solution:
Intensity $\propto(\text { amplitude })^2$
$$
\begin{aligned}
&\frac{I_1}{I_2}=\frac{16}{9}=\frac{a_1^2}{a_2^2} \\
&\Rightarrow \mathrm{a}_1=4 ; \mathrm{a}_2=3
\end{aligned}
$$
Therefore the ratio of intensities of bright and dark parts
$$
\frac{I_{\text {Bright }}}{I_{\text {Dark }}}=\frac{\left(a_1+a_2\right)^2}{\left(a_1-a_2\right)^2}=\frac{(4+3)^2}{(4-3)^2}=\frac{49}{1}
$$

Topic: Physics, Wave Optics, JEE Main, JEE Main 2014 (11 Apr Online)
Two monomers in maltose are:

Options:
A α-D- glucose and β-D- glucose
B α-D- glucose and α-D- galactose
C α-D- glucose and α-D- Fructose
D α-D- glucose and α-D- glucose

The correct answer is:
α-D- glucose and α-D- glucose

Solution:
Maltose on hydrolysis gives 2 moles of α-D- gluscose.

Topic: Chemistry, Polymers, JEE Main, JEE Main 2020 (08 Jan Shift 2)
Two mutually perpendicular tangents of the parabola y2=4ax at the points Q1 and Q2 on it meet its axis in P1 and P2. If S is the focus of the parabola, then the value of 1SP1+1SP2-1 is equal to

Options:
A a4
B a2
C a
D 2a

The correct answer is:
a

Solution:
Let Q1=at12,2at1 & Q2=at22,2at2 are points of contact of the tangents passing through P1 & P2 respectively.
⇒P1=-at12,0 & P2=-at22,0
⇒P1S=a+at12 & P2S=a+at22
⇒1PS1+1PS2=1a+at12+1a+at22=1a+at12+1a+a-1t12
=1a

Topic: Mathematics, Parabola, JEE Main
Two narrow tube of diameters $d_1$ and $d_2$ are joined together to form a Utube open at both ends. If U-tube contains water, the difference in water levels in the limbs is
( $T$ is the surface tension of water, angle of contact is zero and density of water is $\rho, g$ is the acceleration due to gravity)

Options:
A $\frac{4 T}{\rho g}\left[\frac{d_2-d_1}{d_1 d_2}\right]$
B $\frac{4 T}{\rho g}\left[\frac{d_1 d_2}{d_1+d_2}\right]$
C $\frac{2 T}{\rho g}\left[\frac{d_2-d_1}{d_1 d_2}\right]$
D $\frac{2 T}{\rho g}\left[\frac{d_1+d_2}{d_1 d_2}\right]$

The correct answer is:
$\frac{4 T}{\rho g}\left[\frac{d_2-d_1}{d_1 d_2}\right]$

Solution:
Consider the situation below:
When angle of contact is zero degree, the radius of meniscus equals the radius of tube.
Excess pressure in the first tube $P_1=\frac{4 T}{d_1}$
Excess pressure in the second tube, $P_2=\frac{4 T}{d_2}$
Hence, the pressure difference in the two limbs of the tube is given by,
$\Delta P=P_1-P_2=h \rho g$
$\Rightarrow h=\frac{P_1-P_2}{\rho g}=\frac{4 T}{\rho g}\left(\frac{1}{d_1}-\frac{1}{d_2}\right)=\frac{4 T}{\rho g}\left[\frac{d_2-d_1}{d_1 d_2}\right]$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Two natural numbers are randomly chosen and multiplied, then the chance that their product is divisible by 3 is

Options:
A 49
B 59
C 23
D 19

The correct answer is:
59

Solution:
Let E be the event that a natural number randomly chosen is not divisible by 3, then PE=23
P (required event) =1-PEE=1-232=59

Topic: Mathematics, Probability, JEE Main
Two natural numbers are randomly chosen and multiplied, then the chance that their product is divisible by 5 is

Options:
A 425
B 925
C 1625
D 125

The correct answer is:
925

Solution:
Let E be the event that Q natural number randomly chosen is not divisible by 5, then PE=45
P (required event) =1-PEE=1-452=925

Topic: Mathematics, Probability, JEE Main
Two newspapers A and B are published in a city. It is known that 25% of the city population reads A and 20% reads B while 8% reads both A and B. Further, 30% of those who read A but not B look into advertisements and 40% of those who read B but not A also look into advertisements, while 50% of those who read both A and B look into advertisements. Then the percentage of the population who look into advertisements is:

Options:
A 13.5
B 12.8
C 13.9
D 13

The correct answer is:
13.9

Solution:
Let nA and nB be the percentage of population in a city who read newspaper A and B respectively.

Therefore, nA=25, nB=20 and nA∩B=8

Now, we know that

nA∩B¯=nA-nA∩B

⇒nA∩B¯=25-8

⇒nA∩B¯=17

And,

nA¯∩B=nB-nA∩B

⇒nA¯∩B=20-8

⇒nA¯∩B=12

Therefore, percentage of those who read advertisements is

=30% of nA∩B¯+40% of nB∩A¯+50% of nA∩B

=30100×17+40100×12+50100×8

=1390100

Total persons who look into advertisement =13.9%

Topic: Mathematics, Probability, JEE Main, JEE Main 2019 (09 Apr Shift 2)
Two non-ideal identical batteries are connected in parallel. Consider the following statements
(i) The equivalent e.m.f. is smaller than either of the two e.m.f.s
(ii) The equivalent internal resistance is smaller than either of the two internal resistances

Options:
A Both (i) and (ii) are correct
B (i) is correct but (ii) is wrong
C (ii) is correct but (i) is wrong
D Both (i) and (ii) are wrong

The correct answer is:
(ii) is correct but (i) is wrong

Solution:
Because $E_{e q}=E$ and $r e q=r / 2$

Topic: Physics, Current Electricity, JEE Main
Two non-mixing liquids of densities ρ and nρn>1 are put in a container. The height of each liquid is h. A solid cylinder of length L and density d is put in this container. The cylinder floats with its axis vertical and length pLp<1 in the denser liquid. The density d is equal to:

Options:
A 1+n+1pρ
B 2+n+1pρ
C 2+n-1pρ
D 1+n-1pρ

The correct answer is:
1+n-1pρ

Solution:
f0=mg
ρA1-pLg+nρApLg=dALg
ρ1-p+nρp=d
1-p+npρ=d
1+n-1pρ=d

Topic: Physics, Mechanical Properties of Fluids, JEE Main