datavorous/entrance-exam-dataset huggingface dataset:
An oxidation-reduction reaction in which 3 electrons are transferred has a ΔG0 of 17.37 kJ mol-1 at 25°C. The value of Ecello (in V) is _______×10−2.

1F=96500 C mol−1

Options:

The correct answer is:
-6

Solution:
ΔG°=−nF Ecello

17.37×103=−3×96500×Ecello

Ecell=−0.06 V

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2020 (05 Sep Shift 1)
An oxide of the element contains $20 \% \mathrm{O}_{2}$ by weight. Calculate the equivalent weight of the element.

Options:
A 8
B 16
C 32
D 12

The correct answer is:
32

Solution:
Gram equivalent of element $=$ Gram equivalent of $\mathrm{O}_{2}$ $\frac{80}{E}=\frac{20}{8}$
$$
E=32
$$

Topic: Chemistry, Redox Reactions, JEE Main
An α particle, a proton and an electron have the same kinetic energy. Which one of the following is correct in case of their de-Broglie wavelength:

Options:
A λα>λp>λe
B λα<λp<λe
C λα=λp=λe
D λα>λp<λe

The correct answer is:
λα<λp<λe

Solution:
As we know, de-Broglie wavelength is given by,

λ=hp

and p=2mK

Therefore,

λ=h2mK

As kinetic energy is the same for all particles,

∴λ∝1m

∵mα>mp>me

λe>λp>λα

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2023 (24 Jan Shift 2)
An α-particle after passing through potential difference of V volt collides with a nucleus. If the atomic number of the nucleus is Z, then distance of closest approach is

Options:
A 14.4ZV Å
B 14.4ZV m
C 14.4VZ m
D 14.4VZ Å

The correct answer is:
14.4ZV Å

Solution:
Given KE of α-particle =2 eV

r=Zee4πε0·KE

=2Ze×9×1092V

⇒ r=2×Z×16×10-19×9×1092V

=14.4·ZV Å

Topic: Physics, Atomic Physics, JEE Main
An α particle and a carbon 12 atom has same kinetic energy K. The ratio of their de-Broglie wavelengths λα:λC12 is

Options:
A 3:1
B 1:3
C 2:3
D 3:1

The correct answer is:
3:1

Solution:
De-Broglie wavelength is stated as λ=hp=h2mK, where, K is kinetic energy and m is mass.

Here, Kα=KC‐12.

Now, the ratio λαλC12=12m4m=3:1

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2022 (27 Jun Shift 1)
An α particle and a proton are accelerated from rest by a potential difference of 200 V. After this, their de Broglie wavelengths are λα and λp respectively. The ratio λpλα is :

Options:
A 8
B 3.8
C 2.8
D 7.8

The correct answer is:
2.8

Solution:
λ=hp=h2mqV

⇒λpλα=mαqαmpqp=4mp×2emp×e=8=22=22

⇒λpλα=2×1.4=2.8

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2021 (25 Feb Shift 1)
An $\alpha$ particle and a proton travel with same velocity in a magnetic field perpendicular to the direction of their velocities, find the ratio of the radii of their circular path

Options:
A $4: 1$
B $1: 4$
C $2: 1$
D $1: 2$

The correct answer is:
$2: 1$

Solution:
As \(\mathrm{r}=\frac{\mathrm{mv}}{\mathrm{Bq}}\)
Hence, \(\frac{r_\alpha}{r_p}=\frac{m_\alpha}{m_p} \times \frac{q_p}{q_a}\)
We know, \(\mathrm{m}_\alpha=4 \mathrm{~m}_{\mathrm{p}}\) and \(\mathrm{q}_\alpha=2 \mathrm{q}_{\mathrm{p}}\)
\(\frac{r_\alpha}{r_p}=\left(\frac{4}{1}\right) \times\left(\frac{1}{2}\right)=\frac{2}{1}=2: 1\)

Topic: Physics, Magnetic Effects of Current, JEE Main
An α -particle moves in a circular path of radius 1 cm in a uniform magnetic field of 0.125 T. The de Broglie wavelength associated with the α -particle is

Options:
A 1.65×10-12 m
B 3.3×10-12 m
C 4.95×10-12 m
D 6.6×10-12 m

The correct answer is:
1.65×10-12 m

Solution:
The expression for the circular path in a uniform magnetic
fields is,

R=mvqB

mv=qRB

We know that de-Broglie wavelength of a charged particle is given by,

λ=hmv

Substitute the expression in the above equation.

λ=hRqαB
λ=h(+2e)BR
λ=6.6×10-342×1.6×10-19×0.125×1×10-2=1.65×10-12 m

Topic: Physics, Dual Nature of Matter, JEE Main
An $\alpha$-particle moves in a circular path of radius $0.83 \mathrm{~cm}$ in the presence of a magnetic field of $0.25 \mathrm{~Wb} / \mathrm{m}^2$. The de-Broglie wavelength associated with the particle will be

Options:
A $1 Å$
B $0.1 Å$
C $10 Å$
D $0.01 Å$

The correct answer is:
$0.01 Å$

Solution:
We knows
$R=\frac{m v}{q B}$
$\begin{aligned}
\lambda & =\frac{h}{m v} \\
\lambda & =\frac{h}{q B R} \\
& =\frac{6.6 \times 10^{-34}}{1.6 \times 10^{-19} \times 0.83 \times 10^{-2} \times 0.25} \\
& =0.01 Å
\end{aligned}$

Topic: Physics, Dual Nature of Matter, JEE Main
An $\alpha$-particle of energy $5 \mathrm{MeV}$ is scattered through $180^{\circ}$ by a fixed uranium nucleus. The distance of the closest approach is of the order of

Options:
A
$1 Å$

B
$10^{-10} \mathrm{~cm}$

C
$10^{-12} \mathrm{~cm}$

D
$10^{-15} \mathrm{~cm}$

The correct answer is:
$10^{-12} \mathrm{~cm}$

Solution:
At closest approach, all the kinetic energy of the $\alpha$-particle will converted into the potential energy of the system, K.E. = P.E.
$$
\begin{aligned}
& 5 \mathrm{MeV}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}} \\
& 5 \times 10^6 \times \mathrm{e}=9 \times 10^9 \frac{\mathrm{Z}_1 \times \mathrm{Z}_2 \mathrm{e}^2}{\mathrm{r}} \\
& \mathrm{r}=\frac{9 \times 10^9 \times 92 \times 2 \times 1.6 \times 10^{-19}}{5 \times 10^6} \\
& \therefore \mathrm{r}=5.3 \times 10^{-14} \mathrm{~m}=5.3 \times 10^{-12} \mathrm{~cm}
\end{aligned}
$$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2004
An α -paticle approaches the target nucleus of copper (Z = 29) in such a way that the value of impact parameter is zero. The distance of closest approach will be

Options:
A 2πε0(K.E.)α29e2
B 29e22πε0(K.E.)α
C 4πε0(K.E.)α29e2
D (K.E.)α

The correct answer is:
29e22πε0(K.E.)α

Solution:
Distance of closest approach
Kq1q2r=(K.E.)α
K=14πε0
14πε0Ze×2er=(K.E.)α
14πε02×29e2r=(K.E.)α
r=12πε0×29e2(K.E.)α

Topic: Chemistry, Structure of Atom, JEE Main
An p-n junction (D) shown in the figure can act as a rectifier. An alternating current source (V) is connected in the circuit. The current (I) in the resistor $R$ can be shown by:

Options:
A

B

C

D

The correct answer is:

Solution:
When postive cycle enters into diode it is forward bias while for negative half cycle it is reversed bias.

Topic: Physics, Semiconductors, JEE Main
An unbanked circular highway curve on the level ground makes a turn of 90Â°. The highway carries traffic at 108 km h-1, and the centripetal force on a vehicle is not to exceed 1 1 0 of its weight. What is the approximate minimum length of the curve, in km?

Options:
A 1.41 km

B 1 km

C 0.6 km

D None of these

The correct answer is:
1.41 km

Solution:
mv2R=mg10

R=10v2g=108×5182=900 m,
Length πR2=1.413 km

Topic: Physics, Laws of Motion, JEE Main
An unbiased coin is tossed 5 times. Suppose that a variable X is assigned the value k when k consecutive heads are obtained for k=3,4,5, otherwise X takes the value -1. Then the expected value of X, is

Options:
A 316
B 18
C -316
D -18

The correct answer is:
18

Solution:
If k represents the number of consecutive heads and Pk is the corresponding probability while an unbiased coin is tossed 5 times, then

k	0	1	2	3	4	5

Pk	125	1225	1125	525	225	125

Now, the expected value of variable X is

-1125+-11225+-11125+3525+4225+5125

=125-1-12-11+15+8+5

=28-2425=425=123=18

Topic: Mathematics, Probability, JEE Main, JEE Main 2020 (07 Jan Shift 1)
An unbiased coin is tossed \( 5 \) times. Suppose that a variable \( X \) is assigned the value \( k \) when \( k \) consecutive heads are
obtained for \( k=3,4,5 \), otherwise \( X \) takes the value \( -1 . \) Then the expected value of \( X \), is

Options:
A \( \frac{3}{16} \)
B \( \frac{1}{8} \)
C \( \frac{1}{16} \)
D \( \frac{3}{8} \)

The correct answer is:
\( \frac{1}{8} \)

Solution:
If k represents the number of consecutive heads and Pk is the corresponding probability while an unbiased coin is tossed 5 times, then

k	0	1	2	3	4	5

Pk	125	1225	1125	525	225	125

Now, the expected value of variable X is

-1125+-11225+-11125+3525+4225+5125

=125-1-12-11+15+8+5

=28-2425=425=123=18

Topic: Mathematics, Probability, JEE Main
An unbiased coin is tossed eight times. The probability of obtaining at least one head and at least one tail is:

Options:
A 127128
B 6364
C 255256
D 12

The correct answer is:
127128

Solution:
Probability of at least one head and at least one tail can be calculated as,

= 1-{PAll Head+PAll Tail}
=1-128+128
=1-127
=1-1128=127128

Topic: Mathematics, Probability, JEE Main, JEE Main 2017 (08 Apr Online)
An unbiased coin is tossed. If the outcome is a head then a pair of unbiased dice is rolled and the sum of the numbers obtained on them is noted. If the toss of the coin results in tail then a card from a well-shuffled pack of nine cards numbered 1, 2, 3,…, 9 is randomly picked and the number on the card is noted. The probability that the noted number is either 7 or 8 is

Options:
A 1336
B 1972
C 1572
D 1936

The correct answer is:
1972

Solution:
Let, A be the event that the noted number is either 7 or 8.

And, E1 be the event that the toss of coin results in head and E2 be the event that the toss of coin results in tail.

Then, we have PE1=PE2=12

And, PA|E1=P(getting the sum of the numbers on the pair of dice as 7 or 8)

The total number of cases, when a pair of dice is thrown are =6×6=36 and the cases of getting sum 7 or 8 are 1, 6, 2, 5, 3, 4, 4, 3, 5, 2, 6, 1, 2, 6, 3, 5, 4, 4, 5, 3, 6, 2=11 cases.

⇒PA|E1=1136

Similarly, PA|E2=(getting the number 7 or 8 from the numbers 1, 2, 3,.., 9)

⇒PA|E2=29

Now, using the total probability theorem, we get

PA=PE1PA|E1+PE1PA|E2

⇒PA=12×1136+12×29

⇒PA=1172+19=1972.

Topic: Mathematics, Probability, JEE Main, JEE Main 2019 (10 Jan Shift 1)
An unbiased die is rolled n times. Let PA,PB and PC be the probability of occurrence of an odd number exactly one, two and three times respectively in n trials. If PA,PB,PC are in arithmetic progression, then n is equal to

Options:
A 4
B 5
C 6
D 7

The correct answer is:
7

Solution:
PA=Cn112112n-1
PB=Cn212212n-2
PC=Cn312312n-3
⇒PA,PB,PC are in AP ⇒2×Cn2=Cn1+Cn3
⇒2×nn-12=n+nn-1n-26
⇒2×nn-12=n+nn-1n-26
⇒n-1=1+n-1n-26⇒6n-6=6+n2-3n+2
⇒n2-9n+14=0⇒n=2 or 7
∴n=2 is not possible ⇒n=7

Topic: Mathematics, Probability, JEE Main
An unbiased die is rolled until a number greater than 4 appears. The probability that an even number of trials are needed, is

Options:
A 1 2
B 2 5
C 1 5
D 2 3

The correct answer is:
2 5

Solution:
p = Probability of success s = 2 6 = 1 3
q = Probability of failure f = 1 - 1 3 = 2 3
The probability that success occurs in even number of trials
=p(fs)+p(fffs)+p(fffffs)+.....
=qp+q3p+q5p+â¦=qp1-q2
= 2 3 Ã 1 3 1 - 2 3 2 = 2 9 1 - 4 9 = 2 9 Ã 9 5 = 2 5

Topic: Mathematics, Probability, JEE Main
An uncharged capacitor of capacitance 4 Î¼ F and a resistance of 2.5 MÎ© are connected in series with 12 V battery at t=0. Find the time in seconds after which the potential difference across the capacitor is 3 times the potential difference across the resistor. [Given, ln(2) = 0.693]

Options:

The correct answer is:
13.86

Solution:
Given : VC = 3VR = 3(V - VC)
Here, V is the applied potential.

∴   V C = 3 4  V
or  V(1−e−t/T)=34 V
∴  e −t/τ  =  1 4
⇒ t = τ ln 4 = 2τ ln 2.
Here τ = CR =10 s
⇒ t = 2 × 10 × 0.693 = 13.86 s

Topic: Physics, Current Electricity, JEE Main
An undeformed spring of spring constant k is connected to a bead of mass m which can move along a frictionless rod as shown in the figure. If the particle strikes the bead at an angle of 45Â° with the horizontal and sticks to it, then the maximum elongation of the spring after the collision is

Options:
A v4m2k
B v2mk
C v2m2k
D v4mk

The correct answer is:
v2mk

Solution:
Conserving the linear momentum of the system along the rod

mv cos 45∘=2mv1

v 1 = v 2 2

Applying work-energy theorem
- 1 2 kx 2 = - 1 2 2 m v 2 2 2
⇒ x=v2mk

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
An unknown alcohol is treated with the Lucas reagent to determine whether the alcohol is primary, secondary or tertiary. Which alcohol reacts fastest and by what mechansim?

Options:
A Tertiary alcohol by SN2
B Tertiary alcohol by SN1
C Secondary alcohol by SN2
D Secondary alcohol by SN1

The correct answer is:
Tertiary alcohol by SN1

Solution:
The reaction of alcohol with Lucas reagent is mostly an SN1 mechanism reaction and the rate of reaction is directly proportional to the carbocation stability formed in the reaction, hence tertiary alcohol will react fastest.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
An unknown alcohol is treated with the "Lucas reagent" to determine whether the alcohol is primary, secondary or tertiary. Which alcohol reacts fastest and by what mechanism?

Options:
A Tertiary alcohol by \( S_{N} 2 \)
B Secondary alcohol by \( \mathrm{S}_{\mathrm{N}} 1 \)
C Tertiary alcohol by \( S_{N} 1 \)
D Secondary alcohol by \( \mathrm{S}_{\mathrm{N}} 2 \)

The correct answer is:
Tertiary alcohol by \( S_{N} 1 \)

Solution:
Lucas reagent is a mixture of conc. HCl and dry anhydrous ZnCl2.
In this reaction, a carbocation is formed as an intermediate and thus the stability and ease of carbocation formation determine the rate of the reaction as that's the slow step of the mechanism. Hence, the order of reactivity is tertiary alcohol > secondary alcohol by SN1 mechanism.
Thus, with Lucas reagent, tertiary alcohol reacts fastest and by SN1mechanism.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
An unknown alcohol is treated with the "Lucas reagent" to determine whether the alcohol is primary, secondary or tertiary. Which alcohol reacts fastest and by what mechanism:

Options:
A Secondary alcohol by SN2
B Tertiary alcohol by SN2
C Secondary alcohol by SN1
D Tertiary alcohol by SN1

The correct answer is:
Tertiary alcohol by SN1

Solution:
3° alcohol reacts fastest giving white turbidity immediately on adding Lucas reagent.
Reaction takes place through carbocation formation involving SN1 mechanism. Rate of reaction in this reaction is directly proportional to the stability of the carbocation formed.
3° R-OH(tertiary alcohol) forms 3° carbocation and therefore, the reaction is fastest.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2013 (07 Apr)
An unknown alcohol is treated with the "Lucas reagent" to determine whether the alcohol is primary, secondary or tertiary. Which alcohol reacts fastest and by what mechanism:

Options:
A Secondary alcohol by SN2
B Tertiary alcohol by SN2
C Secondary alcohol by SN1
D Tertiary alcohol by SN1

The correct answer is:
Tertiary alcohol by SN1

Solution:
3° alcohol reacts fastest giving white turbidity immediately on adding Lucas reagent.
Reaction takes place through carbocation formation involving SN1 mechanism. Rate of reaction in this reaction is directly proportional to the stability of the carbocation formed.
3° R-OH(tertiary alcohol) forms 3° carbocation and therefore, the reaction is fastest.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
An unknown chlorohydrocarbon has 3.55 % of chlorine. If each molecule of the hydrocarbon has one chlorine atom only; chlorine atoms present in 1 g of chlorohydrocarbon are :(Atomic wt. of Cl=35.5 u; Avogadro constant =6.023×1023 mol–1)

Options:
A 6.023 ×109
B 6.023 ×1023
C 6.023 ×1021
D 6.023 ×1020

The correct answer is:
6.023 ×1020

Solution:
CxHyCl
% Cl=3.55
Weight of Cl=1 ×3.55100
nCl-=1 ×3.55100 ×35.5
No. of Cl-ion= 1 × 3.55100 ×35.5×6.023 ×1023= 6.023 ×1020

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2018 (16 Apr Online)
An unknown compound A C8H10O3 on acetylation with CH3COCl/Py forms acetyl derivative of A whose MW is 280. A on treat with CH2N2 gives methyl ether of B having MW 182. If the number of phenolic hydroxyls and alcoholic hydroxyls in the compound A are X and Y respectively. Find the sum of X + Y here?

Options:

The correct answer is:
3

Solution:
Number of hydroxyl group by acetylation method =â 280âMWâofâ C 8 H 10 O 3 42
=280-15442=12642=3
Number of phenolic hydroxyl =  MW of methyl ether−MW 14
=182-15414=2814=2
Hence, X = 2
Number of alcoholic -OH groups =3-2=1
Hence, Y = 1
So X + Y = 2 + 1 = 3.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
An unknown metal of mass 192 g heated to a temperature of 100°C was immersed into a brass calorimeter of mass 128 g containing 240 g of water at a temperature of 8.4oC. Calculate the specific heat of the unknown metal if water temperature stabilizes at 21.5oC. ( Specific heat of brass is 394J kg-1K-1)

Options:
A 916 J kg-1K-1
B 458 J kg-1K-1
C 654 J kg-1K-1
D 1232 J kg-1K-1

The correct answer is:
916 J kg-1K-1

Solution:
Conservation of energy:
msΔT=mbsb+mwswΔTw
192×10-3s100-21.5=128×394+240×4182×10-3×21.5-8.4
s=50432+100368013.115072=916.19≃916 J kg-1 K-1

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2019 (10 Jan Shift 2)
An unknown metal of mass \( 192 g \) heated to a temperature of \( 100^{\circ} \mathrm{C} \) was immersed into a brass calorimeter of mass \( 128 \mathrm{~g} \) containing \( 240 \mathrm{~g} \) of water at a temperature of \( 8.4^{\circ} \mathrm{C} \). Calculate the specific heat of the unknown metal if water temperature stabilizes at \( 21.5^{\circ} \mathrm{C} . \) ( Specific heat of brass is \( 394 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1} \) )

Options:
A \( 916 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1} \)
B \( 458 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1} \)
C \( 654 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1} \)
D \( 1232 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1} \)

The correct answer is:
\( 916 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1} \)

Solution:
Conservation of energy:
msΔT=mbsb+mwswΔTw
192×10-3s100-21.5=128×394+240×4182×10-3×21.5-8.4
s=50432+100368013.115072=916.19≃916 J kg-1 K-1

Topic: Physics, Kinetic Theory of Gases, JEE Main
An unknown transistor needs to be identified as a npn or pnp type. A multimeter, with +ve and -ve terminals, is used to measure resistance between different terminals of transistor. If terminal 2 is the base of the transistor then which of the following is correct for a pnp transistor?

Options:
A +ve terminal 2, -ve terminal 3, resistance low
B +ve terminal 2, -ve terminal 1, resistance high
C +ve terminal 1, -ve terminal 2, resistance high
D +ve terminal 3, -ve terminal 2, resistance high

The correct answer is:
+ve terminal 2, -ve terminal 1, resistance high

Solution:
When pn junction is forward biased resistance is low. When pn junction reverse biased resistance is high. Because of this reason, we can amplify currents.
For this reason, Transistors can be used as:-

An amplifier

A switch

Logic gates

Topic: Physics, Semiconductors, JEE Main, JEE Main 2016 (09 Apr Online)
An unpolarised light beam of intensity 2I0 is passed through a polaroid P and then through another polaroid Q which is oriented in such a way that its passing axis makes an angle of 30° relative to that of P. The intensity of the emergent light is

Options:
A I04
B I02
C 3I04
D 3I02

The correct answer is:
3I04

Solution:
When unpolarized light passes from Polaroid, then it becomes polarised light and intensity becomes half of the original, therefore, I1=2I02=I0.

Given here, θ=30°.

Using law of Malus for the second polariser, I=I0cos2θ

We have, intensity of emergent light, I2=I1cos230°

=I034=3I04

Topic: Physics, Wave Optics, JEE Main, JEE Main 2022 (29 Jul Shift 2)
An unpolarized light of intensity I0 passes through three polarizers, such that the transmission axis of last polarizer is perpendicular to that of first. If the intensity of emergent light is 3I032 and the angle between the transmission axis of first two polarizers is θ, then

Options:
A θ=45°
B θ=37°
C Î¸=30Â°
D θ=53°

The correct answer is:
θ=30°

Solution:
3I032=I02cos2Î¸ sin2Î¸

4cos2Î¸sin2Î¸=34

or sin2Î¸=34=32

Î¸=30Â°

Topic: Physics, Wave Optics, JEE Main
An unstable heavy nucleus at rest breaks into two nuclei which move away with velocities in the ratio of $8: 27$. The ratio of the radii of the nuclei (assumed to be spherical ) is:

Options:
A
$8: 27$

B
$2: 3$

C
$3: 2$

D
$4: 9$

The correct answer is:
$3: 2$

Solution:
Let heavy nucleus breaks into two nuclei of mass $m_1$ and $m_2$ and move away with velocities $V_1$ and $V_2$ respectively.
According to question, $\frac{V_1}{V_2}=\frac{8}{27}$ $m_1 V_1=m_2 V_2$ (Law of momentum conservation)
$$
\begin{aligned}
&\Rightarrow \frac{m_1}{m_2}=\frac{V_2}{V_1}=\frac{27}{8} \\
&\frac{\rho \times \frac{4}{3} \pi R_1^3}{\rho \times \frac{4}{3} \pi R_2^3} \quad\left(\because \text { density } \rho=\frac{\text { mass }}{\text { volume }}\right)
\end{aligned}
$$
$$
\begin{aligned}
&\Rightarrow\left(\frac{\mathrm{R}_1}{\mathrm{R}_2}\right)=\left(\frac{27}{8}\right)^{\frac{1}{3}}=\left(\frac{3}{2}\right)^{3 \times \frac{1}{3}} \\
&\therefore \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{3}{2}
\end{aligned}
$$

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
An upright object is placed at a distance of 40 cm in front of a convergent lens of focal length 20 cm. A convergent mirror of focal length 10 cm is placed at a distance of 60 cm on the other side of the lens. The position and size of the final image will be:

Options:
A 40 cm from the convergent lens, twice the size of the object
B 20 cm from the convergent mirror, twice the size of the object
C 40 cm from the convergent lens, same size of the object
D 20 cm from the convergent mirror, same size of the object

The correct answer is:
40 cm from the convergent lens, same size of the object

Solution:
First refraction, u=-40 cm; f=+20 cm
From lens formula, 1v-1u=1f
1v-1-40=120⇒1v=120-140
v=+40 cm
m1=-1
First reflection, u=-20 cm;f=-10 cm
From mirror formula, 1v+1u=1f
1v-120=1-10
1v=1-10+120⇒v=-20 cm
m2=-1
Second refraction, u=-40 cm,  f=+20 cm
From lens formula, 1v-1-40=120⇒v=+40 cm
m3=-1
mtotal=m1×m2×m3=-1
Finally, image will be at distance 40 cm from lens and is of same size as object.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2019 (08 Apr Shift 1)
An urn contains 5 red and 2 green balls. A ball is drawn at random from the urn. If the drawn ball is green, then a red ball is added to the urn and if the drawn ball is red, then a green ball is added to the urn; the original ball is not returned to the urn. Now, a second ball is drawn at random from it. The probability that the second ball is red, is:

Options:
A 2149
B 2649
C 3249
D 2749

The correct answer is:
3249

Solution:
Let A is event that 2nd drawn ball is red.
∴PA=PR×PA/R+PG×P A/G
=57×47+27×67
=3249

Topic: Mathematics, Probability, JEE Main, JEE Main 2019 (09 Jan Shift 2)
An urn contains 5 red marbles, 4 black marbles and 3 white marbles. Then, the number of ways in which 4 marbles can be drawn so that at the most three of them are red is ___________.

Options:

The correct answer is:
490

Solution:
We have,5 red,4 black, 3 white marbles
Required number of ways

=C05×C47+C15×C37+C25×C27+C35×C17 ​

=35+175+210+70

=490

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2020 (08 Jan Shift 1)
An urn contains 6 white and 9 black balls. Two successive draws of 4 balls are made without replacement. The probability, that the first draw gives all white balls and the second draw gives all black balls, is :

Options:
A 5256
B 5715
C 3715
D 3256

The correct answer is:
3715

Solution:
Total number of balls in the bag =6+9=15.

Number of white balls =6.

Number of black balls =9.

Four balls are drawn at random at a time.

The probability that first draw give four white balls

C46C415=191

The balls are not replaced.

Now, the probability that second draw give four black balls

C49C411=2155

The probability that the first draw give four white balls and second draw give four black balls

=191×2155=3715

Hence, the required probability is 3715.

Topic: Mathematics, Probability, JEE Main, JEE Main 2024 (27 Jan Shift 2)