datavorous/entrance-exam-dataset huggingface dataset:
This question has Statement-1 and Statement-2. Of the four choices given after the Statements, choose the one that best describes the two Statements.
Statement 1: The internal energy of a perfect gas is entirely kinetic and depends only on absolute temperature of the gas and not on its pressure or volume.
Statement 2: A perfect gas is heated keeping pressure constant and later at constant volume. For the same amount of heat the temperature of the gas at constant pressure is lower than that at constant volume.

Options:
A
Statement-1 is true, Statement- 2 is true and Statement $-2$ is the correct explanation of Statement-1.

B
Statement-1 is true, Statement-2 is false.

C
Statement-1 is true, Statement-2 is true but Statement-2 is not the correct explanation of Statement-1.

D
Statement-1 is false, Statement-2 is true.

The correct answer is:
Statement-1 is true, Statement-2 is true but Statement-2 is not the correct explanation of Statement-1.

Solution:
Internal energy of a system is the energy possessed by the system due to molecular motion and molecular configuration. Internal energy $u=u_k+u_p$ For an ideal gas there is no molecular attraction $\mathrm{u}_{\mathrm{p}}=0$
$\therefore$ Internal energy $\mathrm{u}=\mathrm{u}_{\mathrm{k}}$ (i.e., internal kinetic energy)
and $\mathrm{u}=\mathrm{u}_{\mathrm{k}}=\frac{3}{2} \mu \mathrm{RT}$
At constant pressure $(\Delta \mathrm{Q})_{\mathrm{P}}=\mu \mathrm{C}_{\mathrm{p}} \Delta \mathrm{T}$
At constant volume $(\Delta \mathrm{Q})_{\mathrm{v}}=\mu \mathrm{C}_{\mathrm{v}} \Delta \mathrm{T}$
As C $_{\mathrm{P}}>\mathrm{C}_{\mathrm{V}} \quad \therefore \quad(\Delta \mathrm{T})_{\mathrm{P}} < (\Delta \mathrm{T})_{\mathrm{V}}$

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2013 (23 Apr Online)
This question has Statement-1 and Statement-2. Of the four choices given after the Statements, choose the one that best describes the two Statements.
Statement-1: In Young's double slit experiment, the number of fringes observed in the field of view is small with longer wavelength of light and is large with shorter wavelength of light.
Statement-2: In the double slit experiment the fringe width depends directly on the wavelength of light.

Options:
A
Statement-1 is true, Statement-2 is true and the Statement-2 is correct explanation of the Statement-1.

B
Statement-1 is false and the Statement-2 is true.

C
Statement-1 is true Statement-2 is true and the Statement-2 is not correct explanation of the Statement-1.

D
Statement-1 is true and the Statement-2 is false.

The correct answer is:
Statement-1 is true Statement-2 is true and the Statement-2 is not correct explanation of the Statement-1.

Solution:
Fringe width $\mathrm{B}=\frac{\mathrm{D}}{\mathrm{d}} \lambda$
And number of fringes observed in the field of view is obtained by $\frac{\mathrm{d}}{\lambda}$

Topic: Physics, Wave Optics, JEE Main, JEE Main 2013 (22 Apr Online)
This question has Statement-1 and Statement-2. Of the four choices given after the Statements, choose the one that best describes the two Statements.
Statement 1: No work is required to be done to move a test charge between any two points on an equipotential surface.
Statement 2 : Electric lines of force at the equipotential surfaces are mutually perpendicular to each other.

Options:
A
Statement 1 is true, Statement 2 is true, Statement 2 is the correct explanation of Statement 1.

B
Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation of Statement 1.

C
Statement 1 is true, Statement 2 is false.

D
Statement 1 is false, Statement 2 is true.

The correct answer is:
Statement 1 is true, Statement 2 is false.

Solution:
The work done is moving a charge along an equipotential surface is always zero.
The direction of electric field is perpendicular to the equipotential surface or lines.

Topic: Physics, Electrostatics, JEE Main, JEE Main 2013 (25 Apr Online)
This question has Statement-1 and Statement-2. Of the four choices given after the Statements, choose the one that best describes the two Statements.
Statement 1: Very large size telescopes are reflecting telescopes instead of refracting telescopes.
Statement 2: It is easier to provide mechanical support to large size mirrors than large size lenses.

Options:
A
Statement-1 is true and Statement-2 is false.

B
Statement-1 is false and Statement-2 is true.

C
Statement-1 and statement-2 are true and Statement-2 is correct explanation for statement-1.

D
Statements-1 and statement-2 are true and Statement-2 is not the correct explanation for statement-1.

The correct answer is:
Statement-1 and statement-2 are true and Statement-2 is correct explanation for statement-1.

Solution:
One side of mirror is opaque and another side is reflecting this is not in case of lens hence, it is easier to provide mechanical support to large size mirrors than large size lenses. Reflecting telescopes are based on the same principle except that the formation of images takes place by reflection instead of refraction.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2013 (23 Apr Online)
This question has Statement-1 and Statements2. Of the four choices given after the Statements, choose the one that best describes the two Statements.
Statement-1 : Out of radio waves and microwaves, the radio waves undergo more diffraction.
Statement-2 : Radio waves have greater frequency compared to microwaves.

Options:
A
Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation of Statement-1

B
Statement-1 is false, Statement-2 is true.

C
Statement-1 is true, Statement-2 is false.

D
Statement-1 is true, Statement-2 is true but Statement-2 is not the correct explanation of Statement-1

The correct answer is:
Statement-1 is true, Statement-2 is false.

Solution:
Wavelength of radio waves is greater than microwaves hence frequency of radio waves is less than microwaves.
The degree of diffraction is greater whose wavelength is greater.

Topic: Physics, Wave Optics, JEE Main, JEE Main 2013 (25 Apr Online)
This reduction reaction is known as:

Options:
A Rosenmund reduction
B Wolff-Kishner reduction
C Stephen reduction
D Etard reduction

The correct answer is:
Rosenmund reduction

Solution:
The Rosenmund reaction is a hydrogenation process where molecular hydrogen reacts with the acyl chloride in the presence of a catalyst – palladium on barium sulphate. Barium sulphate reduces the activity of palladium due to its low surface area, thereby preventing over reduction.

It is known as rosenmund reduction, which is the partial reduction of acid chloride to aldehyde.

Hence, the answer is option A.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2024 (30 Jan Shift 1)
This secondary pollutant has caused significant damage to the beauty of historical monuments like the Taj Mahal and Red fort. Identify the correct match of the responsible pollutant and its characteristic feature.

Options:
A Acid rain - ratio of H2SO4 and HNO3 is 3 : 7
B Photochemical Smog - PAN + O3 + Nitrogen oxides
C Acid rain – pH of < 5.6
D Sulphur smog - formed from vapour of H2SO4

The correct answer is:
Acid rain – pH of < 5.6

Solution:
On the basis of interaction of these substances with the environment they can be classified as primary or secondary pollutants. Substances which remain in the same form as they are after being added to the environment are called primary pollutants. Certain pollutants are formed by chemical reactions with the primary pollutants these are called secondary pollutants. When NO2 and SO2 are released from different sources they undergo chemical reactions and form secondary pollutants. They get dissolved in atmospheric water vapor to form sulphuric acid and nitric acid. These come down on earth with rain water this is called acid rain. The pH of acid rain is lesser than 5.6 . In acid rain the ratio of H2SO4 and HNO3 is 7 : 3 . Due to acid rain acidity of soil and water increases and it is also responsible for the damage of historical monuments like Taj Mahal and Red fort.

Topic: Biology, Environmental Issues, NEET
Three balls, A,B and C, are released and all reach the point X (shonw in the figure). Balls A and B are released from two identical structures, one kept on the ground and the other at height, h, from the ground as shown in the figure. They take time tA and tB respectively to reach X (time starts after they leave the end of the horizontal portion of the structure). The ball C is released from a point at height, h, vertically above X and reaches X in time tC. Choose the correct statement.

Options:
A tC<tA<tC
B ta = tb = tc
C tC=tA<tC
D tB<tA=tC

The correct answer is:
ta = tb = tc

Solution:
Work done by gravity on A and B is same.

∴ Horizontal velocity of A= horizontal velocity of B as they leave the horizontal portion of the structure.

∴ tA=tB …(i)

Also vertical velocity of A and vertical velocity of C when released are both zero

∴ They both will cover same vertical distance in same time.

∴ tA=tC ...(ii)

From (i) and (ii)

tA=tB=tC

Note : Correction in option option B should be tA=tB=tC

Topic: Physics, Motion In One Dimension, JEE Main
Three balls are drawn at random from a bag containing 5 blue and 4 yellow balls. Let the random variables $X$ and $Y$ respectively denote the number of blue and yellow balls. If $\bar{X}$ and $\bar{Y}$ are the means of $X$ and $Y$ respectively, then $7 \bar{X}+4 \bar{Y}$ is equal to________

Options:

The correct answer is:
17

Solution:
$\begin{aligned} & 7 \overline{\mathrm{x}}=\frac{{ }^5 \mathrm{C}_1{ }^4 \mathrm{C}_2+{ }^5 \mathrm{C}_2 \cdot{ }^4 \mathrm{C}_1 \times 2+{ }^5 \mathrm{C}_3 \cdot{ }^4 \mathrm{C}_0 \times 3}{{ }^9 \mathrm{C}_3} \times 7 \\ & \frac{30+80+30}{84} \times 7 \\ & =\frac{140}{12}=\frac{70}{6}=\frac{35}{3}\end{aligned}$
$\begin{array}{|l|l|l|l|l|l|}
\hline yellow & 0 & 1 & 2 & 3 & 4 \\
\hline & & { }^5 \mathrm{C}_2{ }^4 \mathrm{C}_1 & { }^5 \mathrm{C}_1{ }^4 \mathrm{C}_2 & { }^5 \mathrm{C}_0{ }^4 \mathrm{C}_3 & 0 \\
\hline
\end{array}$
$4 \bar{y}=\frac{40+60+12}{84} \times 4=\frac{112}{21}=\frac{16}{3}$

Topic: Mathematics, Probability, JEE Main, JEE Main 2024 (08 Apr Shift 1)
Three balls are marked with 2,4 and 6. They are placed in a box and a ball is drawn, its number is noted and the ball drawn is returned to the box. The process is repeated two more times. If the probability of the sum of all the three numbers is 12 is Î», then 108Î» is equal to

Options:
A 28
B 14
C 20
D 10

The correct answer is:
28

Solution:
For the sum of all three numbers to be 12, the possible cases are 2,4,6, 4,4,4
Hence, required probability =3!+133=727

Topic: Mathematics, Probability, JEE Main
Three basic steps are required for making GMO are:

I. Identification of DNA with a desirable gene.
II. Introduction of identified DNA into the host.
III. Maintenance of introduced DNA in the host and transfer of the DNA to its progeny.

What is the correct sequence of involvement?

Options:
A II → I → III
B I → III → II
C II → III → I
D I → II → III

The correct answer is:
I → II → III

Solution:
For genetically modifying an organism a specific desired gene is to be isolated, this desired gene should be introduced in specific host cell where it can multiply and get transferred to all the progeny cells.

Topic: Biology, Biotechnology and Its Applications, NEET
Three blocks A, B and C are lying on a smooth horizontal surface, as shown in the figure. A and B have equal masses, m while C has mass M. Block A is given an initial speed v towards B due to which it collides with B perfectly inelastically. The combined mass collides with C, also perfectly inelastically . 56th of the initial kinetic energy is lost in the whole process. What is the value of M/m?

Options:
A 3
B 4
C 5
D 2

The correct answer is:
4

Solution:
mv=m+mvâ²
m+mvâ²=m+m+Mvf
â´ final velocity vf=m2m+Mv
Initial energy =12mv2
Final energy =122m+Mmv2m+M2
Given that 5612mv2=12mv2-122m+Mmv2m+M2
112mv2=12m2v22m+M
6m=2m+M
Mm=4

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2019 (09 Jan Shift 1)
Three blocks A, B and C are lying on a smooth horizontal surface, as shown in the figure. A and B have equal masses, m while C has mass M. Block A is given an initial speed v towards B due to which it collides with B perfectly inelastically. The combined mass collides with C, also perfectly inelastically . 56th of the initial kinetic energy is lost in the whole process. What is the value of M/m?

Options:
A 3
B 4
C 5
D 2

The correct answer is:
4

Solution:
mv=m+mvâ²
m+mvâ²=m+m+Mvf
â´ final velocity vf=m2m+Mv
Initial energy =12mv2
Final energy =122m+Mmv2m+M2
Given that 5612mv2=12mv2-122m+Mmv2m+M2
112mv2=12m2v22m+M
6m=2m+M
Mm=4

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Three blocks A, B and C are pulled on a horizontal smooth surface by a force of 80 N as shown in figure. The tensions T1 and T2 in the string are respectively:

Options:
A 40N, 64N
B 60N, 80N
C 88N, 96N
D 80N, 100N

The correct answer is:
40N, 64N

Solution:
Since, the length of the string will remain same. Hence, all blocks will have the same acceleration.

aA=aB=aC=F5+3+2=8010=8 m s-2

Therefore, T1=5×8=40 N.

Now,

T2-T1=3×8⇒T2=64 N

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2024 (30 Jan Shift 2)
Three blocks A,B and C each of mass 2 kg, are hanging over a fixed pulley as shown in the figure. The tension in the string connecting B and C is

Options:
A Zero
B 3.3 N
C 13.3 N
D 19.6Â N

The correct answer is:
13.3 N

Solution:
The combination of blocks A, B and C is shown in figure.
a=m+mg-mgm+m+m=g3
mg-T=ma=mg3
T=2mg3=23Ã2Ã10
=13.3 N

Topic: Physics, Laws of Motion, JEE Main
Three blocks A, B and C of equal mass m are placed one over the other on a smooth horizontal ground as shown in the figure. Coefficient of friction between any two blocks is 0.5. The maximum value of the mass of block D so that the blocks A, B and C move without slipping over each other is

Options:
A 6 m
B 5 m
C 3 m
D 4 m

The correct answer is:
3 m

Solution:
Blocks A and C both move due to friction. But less friction is available to A as compared to C because the normal reaction between A and B is less. Maximum friction between A and B can be
fmax=Î¼ mAg=12mg
â´ Maximum acceleration of A can be
amax=fmaxm=g2
Further amax=mDg3m+mD
Or g2=mDg3m+mD
Or mD=3m

Topic: Physics, Laws of Motion, JEE Main
Three blocks are arranged on a horizontal table ABCD as shown in the figure. The strings and pulleys are massless and both the pulleys stand vertical. The strings connecting blocks m1 and m2 are also vertical and are perpendicular to faces AB and BC which are mutually perpendicular to each other. If m1 and m2 are 3 kg and 4 kg respectively. Coefficient of friction between the block m3=10 kg and the surface is Î¼=0.6 then, frictional force on m3 is

Options:
A 30 N
B 40 N
C 50 N
D 60 N

The correct answer is:
50 N

Solution:
Net force on m3=302+402=50 N
And limiting friction on m3=Î¼m3g=60 N
â´ System remain in equilibrium and friction on m3=50 N

Topic: Physics, Laws of Motion, JEE Main
Three blocks are pushed with a force $\mathrm{F}$ across a frictionless table as shown in figure. Let $\mathrm{N}_{1}$ be the contact force between the left two blocks and $\mathrm{N}_{2}$ be the contact force between the right two blocks. Then

Options:
A $F>N_{1}>N_{2}$
B $F>N_{2}>N_{1}$
C $\quad F>N_{1}=N_{2}$
D $F=N_{1}=N_{2}$

The correct answer is:
$F>N_{1}>N_{2}$

Solution:
$\mathrm{F}=6 \mathrm{a}, \mathrm{N}_{1}=5 \mathrm{a}, \mathrm{N}_{2}=3 \mathrm{a}$
$\mathrm{F}>\mathrm{N}_{1}>\mathrm{N}_{2}$

Topic: Physics, Laws of Motion, JEE Main
Three blocks are suspended as shown in the figure. The acceleration of the 500 g block is

Options:
A 6g13 downwards
B 7g13downwards
C 8g13downwards
D 9g13 upwards

The correct answer is:
8g13downwards

Solution:
500g-T=500a ……(1)
T-100gsin⁡30°-T′′=100a
Or, T-T′′-50g=100a ……(2)
Again, T′′-50g=50a ……(3)
From equations (2) and (3),
T-100g=150a ……(4)

Adding equations (1) and (4), 400g=650a or a=400g650=8g13
This acceleration is downwards.

Topic: Physics, Laws of Motion, JEE Main
Three blocks of masses $2 \mathrm{~kg}, 3 \mathrm{~kg}$ and $5 \mathrm{~kg}$ are connected to each other with light string and are then placed on a frictionless surface as shown in the figure. The system is pulled by a force $F=10 \mathrm{~N}$, then tension $T_1=$

Options:
A $1 \mathrm{~N}$
B $5 \mathrm{~N}$
C $8 \mathrm{~N}$
D $10 \mathrm{~N}$

The correct answer is:
$8 \mathrm{~N}$

Solution:
$T_1=\left(\frac{m_2+m_3}{m_1+m_2+m_3}\right) g=\frac{3+5}{2+3+5} \times 10=8 \mathrm{~N}$

Topic: Physics, Laws of Motion, JEE Main
Three blocks with masses $m, 2 m$ and $3 m$ are connected by strings, as shown in the figure. After an upward force $F$ is applied on block $m$, the masses move upward at constant speed $v$. What is the net force on the block of mass $2 m ?$

Options:
A Zero
B $2 \mathrm{mg}$
C $3 \mathrm{mg}$
D $6 \mathrm{mg}$

The correct answer is:
Zero

Solution:
Since all the blocks are moving with constant velocity, then the net force on the all blacks will be zero.

Topic: Physics, Laws of Motion, JEE Main
Three bodies A, B and C have equal kinetic energies and their masses are $400 \mathrm{~g}$. $1.2 \mathrm{~kg}$ and $1.6 \mathrm{~kg}$ respectively. The ratio of their linear momenta is :

Options:
A $\sqrt{2} ; \sqrt{3} ; 1$
B $1: \sqrt{3}: 2$
C $1: \sqrt{3}: \sqrt{2}$
D $\sqrt{3}: \sqrt{2}: 1$

The correct answer is:
$1: \sqrt{3}: 2$

Solution:
$\begin{aligned} & \mathrm{KE}=\frac{\mathrm{P}^2}{2 \mathrm{~m}} \\ & \mathrm{P} \propto \sqrt{\mathrm{m}}\end{aligned}$
Hence, $\mathrm{P}_{\mathrm{A}}: \mathrm{P}_{\mathrm{B}}: \mathrm{P}_{\mathrm{C}}$
$=\sqrt{400}: \sqrt{1200}: \sqrt{1600}=1: \sqrt{3}: 2$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2024 (08 Apr Shift 1)
Three bodies, a ring, a solid disc and a solid sphere roll down the same inclined plane without slipping. The radii of the bodies are Identical and they start from rest. If $V_S, V_R$ and $V_D$ are the speeds of the sphere, ring and disc, respectively when they reach the bottom, then the correct option is

Options:
A $V_S>V_R>V_D$
B $V_D>V_S>V_R$
C $V_R>V_D>V_S$
D $V_S>V_D>V_R$

The correct answer is:
$V_S>V_D>V_R$

Solution:
Given, radii of ring, solid disc and solid sphere are identical, say $R$
When a body of mass $m$ starting from rest rolls down an incined plane, then the velocity of body when it reaches at the ground is given by
$$
v=\sqrt{\frac{2 g h}{1+\frac{k^2}{R^2}}}
$$
where, $h=$ height of inclined plane,
$R=$ radius of body,
and $k=$ radius of gyration
For a ring, $k^2=R^2$
$\therefore$ Speed of the ring, $v_R=\sqrt{\frac{2 g h}{1+\frac{R^2}{R^2}}}=\sqrt{g h}$
For a solid sphere, $k^2=\frac{2 R^2}{5}$
$$
\therefore \text { Speed of the solid sphere, } v_s=\sqrt{\frac{2 g h}{1+\frac{2}{5}}}
$$
From Eqs. (i), (ii) and (iii), it is clear that
$$
V_S>V_D>V_R \text {. }
$$

Topic: Physics, Rotational Motion, JEE Main
Three boxes are labeled as A, B and C and each box contains 5 balls numbered 1, 2, 3, 4 and 5. The balls in each box are well mixed and one ball is chosen at random from each of the 3 boxes. If α, β and γ are the number on the ball from the boxes A, B and C respectively, then the probability that α=β+γ is equal to

Options:
A 125
B 225
C 425
D 15

The correct answer is:
225

Solution:
The total cases=53

∵α=β+γ

The possible cases are 2,1,1,3,1,2,3,2,1,4,1,34,3,1,4,2,2,5,3,25,2,3,5,4,15,1,4

So, the number of favourable cases =10
Hence, the required probability =1053=225

Topic: Mathematics, Probability, JEE Main
Three bulbs are filled with CH4, CO2 and Ne as shown in the picture. The bulbs are connected through pipes of zero volume. When the stopcocks are opened and the temperature is kept constant throughout, the pressure of the system is found to be______ atm. (Nearest integer).

Options:

The correct answer is:
3

Solution:
The total number of moles remains constant.

n1+n2+n3=nf

n1,n2, n3 moles of gases in the individual containers. nf is resultant moles.

Using ideal gas equation PV=nRT, we van write as

P1V1RT+P2V2RT+P3V3RT=PfVfRT

P1V1+P2V2+P3V3=PfVf

2×2+4×3+3×4=Pf×9

Pf=289≈3

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2023 (08 Apr Shift 1)
Three candidates solve a question. Odds in favour of the correct answer are $5: 2,4: 3$ and $3: 4$ respectively for the three candidates. What is the probability that at least two of them solve the question correctly?

Options:
A $209 / 343$
B $134 / 343$
C $149 / 343$
D $60 / 343$

The correct answer is:
$209 / 343$

Solution:
Odd in fav. for student $(\mathrm{A})=\frac{5}{2}=\frac{\mathrm{P}(\mathrm{A})}{\mathrm{P}\left(\mathrm{A}^{\prime}\right)}$
Odd in fav. for student $(\mathrm{B})=\frac{4}{3}=\frac{\mathrm{P}(\mathrm{B})}{\mathrm{P}\left(\mathrm{B}^{\prime}\right)}$
Odd in fav. for student $(\mathrm{C})=\frac{3}{4}=\frac{\mathrm{P}(\mathrm{C})}{\mathrm{P}\left(\mathrm{C}^{\prime}\right)}$
$\Rightarrow \quad \mathrm{P}\left(\mathrm{A}^{\prime}\right)=\frac{2}{5} \mathrm{P}(\mathrm{A}), \mathrm{P}\left(\mathrm{B}^{\prime}\right)=\frac{3}{4} \mathrm{P}(\mathrm{B}), \mathrm{P}\left(\mathrm{C}^{\prime}\right)=\frac{4}{3} \mathrm{P}(\mathrm{C})$
Now $\mathrm{P}(\mathrm{A})+\mathrm{P}\left(\mathrm{A}^{\prime}\right)=1 \Rightarrow \mathrm{P}(\mathrm{A})+\frac{2}{5} \mathrm{P}(\mathrm{A})=1 \Rightarrow \mathrm{P}(\mathrm{A})=\frac{5}{7}$
Also $\mathrm{P}(\mathrm{B})+\mathrm{P}\left(\mathrm{B}^{\prime}\right)=1 \Rightarrow \mathrm{P}(\mathrm{B})+\frac{3}{4} \mathrm{P}(\mathrm{B})=1 \Rightarrow \mathrm{P}(\mathrm{B})=\frac{4}{7}$
And $\mathrm{P}(\mathrm{C})+\mathrm{P}\left(\mathrm{C}^{\prime}\right)=1 \Rightarrow \mathrm{P}(\mathrm{C})+\frac{4}{3} \mathrm{P}(\mathrm{C})=1 \Rightarrow \mathrm{P}(\mathrm{C})=\frac{3}{7}$
$\therefore \quad \mathrm{P}(\mathrm{A})=\frac{2}{5} \times \frac{5}{7}=\frac{2}{7}, \quad \mathrm{P}\left(\mathrm{B}^{\prime}\right)=\frac{3}{4} \times \frac{4}{7}=\frac{3}{7}$,
$\mathrm{P}\left(\mathrm{C}^{\prime}\right)=\frac{4}{3} \times \frac{3}{7}=\frac{4}{7}$
Req. Prob. $=\mathrm{P}(\mathrm{A}) \times \mathrm{P}(\mathrm{B}) \times \mathrm{P}\left(\mathrm{C}^{\prime}\right)+\mathrm{P}(\mathrm{A}) \times \mathrm{P}\left(\mathrm{B}^{\prime}\right) \times \mathrm{P}(\mathrm{C})$
$+\mathrm{P}\left(\mathrm{A}^{\prime}\right) \times \mathrm{P}(\mathrm{B}) \times \mathrm{P}(\mathrm{C})+\mathrm{P}(\mathrm{A}) \times \mathrm{P}(\mathrm{B}) \times \mathrm{P}(\mathrm{C})$
$=\frac{5}{7} \times \frac{4}{7} \times \frac{4}{7}+\frac{5}{7} \times \frac{3}{7} \times \frac{3}{7}+\frac{2}{7} \times \frac{4}{7} \times \frac{3}{7}+\frac{5}{7} \times \frac{4}{7} \times \frac{3}{7}$
$=\frac{209}{343}$

Topic: Mathematics, Probability, JEE Main
Three capacitances, each of 3 μ F , are provided. These cannot be combined to provide the resultant capacitance of :

Options:
A 2 μ F
B 6 Î¼ F
C 1.5 Î¼ F
D 1 μ F

The correct answer is:
6 μ F

Solution:
Equivalent capacitance will be maximum for parallel combination and Minimum for series combination
â´ Possibilities are
1)

C = 1 Î¼ F
2) C = 9 Î¼ F
3) 1 C = 1 6 + 1 3 = 3 6
â´ C = 2 Î¼ F
4) â C = C 2 + C = 3 2 Ã C
= 3 2 = 1.5 Î¼ F

Topic: Physics, Capacitance, JEE Main, JEE Main 2014 (09 Apr Online)
Three capacitances of capacity $10 \mu F, 5 \mu F$ and $5 \mu F$ are connected in parallel. The total capacity will be

Options:
A $10 \mu \mathrm{F}$
B $5 \mu F$
C $20 \mu F$
D None of the above

The correct answer is:
$20 \mu F$

Solution:
In parallel, $C=C+{ }_1 C_2+C_3=20 u \mathrm{~F}$

Topic: Physics, Capacitance, JEE Main
Three capacitors $3 \mu \mathrm{F}, 6 \mu \mathrm{F}$ and $6 \mu \mathrm{F}$ are connected in aeries to a source of $120 \mathrm{V}$. The potential difference, in volt, across the $3 \mu \mathrm{F}$ capacitor will be

Options:
A 24
B 30
C 40
D 60

The correct answer is:
60

Solution:
The combination of three charges in series
$$
\begin{array}{l}
\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}} \\
\frac{1}{C}=\frac{1}{3}+\frac{1}{6}+\frac{1}{6} \Rightarrow C=\frac{6}{4}=1.5 \mu \mathrm{F}
\end{array}
$$
The charge of this circuit
$$
\begin{array}{l}
q=C V=1.5 \times 120 \\
q=180 \mu \mathrm{C}
\end{array}
$$
The potential difference across the $3 \mu \mathrm{F}$
$$
q=C V
$$
$$
V=\frac{q}{C}=\frac{180}{3}=60 \mathrm{V}
$$

Topic: Physics, Capacitance, JEE Main