datavorous/entrance-exam-dataset huggingface dataset:
The chemiosmotic coupling hypothesis of oxidative phosphorylation proposes that adenosine triphosphate (ATP) is formed because

Options:
A There is a change in the permeability of the inner mitochondiral membrane toward adenosine diphosphate (ADP)
B High energy bonds are formed in mitochondrial proteins
C ADP is pumped out of the matrix into the intermembrane space
D A proton gradient forms across the inner membrane

The correct answer is:
A proton gradient forms across the inner membrane

Solution:
Proton motive force develop between perimitochondrial space and matrix across inner mitochondrial membrane.

Topic: Biology, Photosynthesis in Higher Plants, NEET, NEET 2008 (Mains)
The chief component of the middle lamella in plant cell is:

Options:
A Potassium
B Calcium
C Magnesium
D Phosphorus

The correct answer is:
Calcium

Solution:
Middle lamella is the outer most layer of cell wall which functions as cementing layer present between two adjacent cells. It is made up of calcium pectate and magnesium pectate, but mainly calcium pectate which provides rigidity and hardness to the cell wall.

Topic: Biology, Cell - The Unit of Life, NEET
The chloride of a metal contains 71% chlorine by weight and the vapour density of it is 50. The atomic weigh of the metal will be:

Options:
A 29
B 58
C 35.5
D 71

The correct answer is:
29

Solution:
Mol. wt. of metal chloride =50Ã2=100
Let metal chloride be metal chlorine then
Equivalent of metal = Equivalents of chlorine =â W m E m â=â W Cl E Cl âorâ 29 E â=â 71 35.5
E=292
Now a+35.5n=100
or nâÃâEâ+â35âÃâ5nâ=â100
â´n=2
Therefore, a=2ÃE=2Ã292=29

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The chloride that CANNOT get hydrolysed is:

Options:
A $\mathrm{PbCl}_{4}$
B $\mathrm{CCl}_{4}$
C $\mathrm{SnCl}_{4}$
D $\mathrm{SiCl}_{2}$

The correct answer is:
$\mathrm{CCl}_{4}$

Solution:
$\mathrm{CCl}_{4}$ cannot be hydrolysed due to absence of $d$ orbitals at carbon atom.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2019 (11 Jan Shift 1)
The chlorine atom of the following compound

that reacts most readily with AgNO3 to give a precipitate is

Options:
A Cla
B Clb
C Cl
D Cld

The correct answer is:
Cla

Solution:
The reaction between haloalkane and AgNO gives carbocation intermediate. So, more easily it will be formed, more readily it will react to give precipitate.
In the given compound

Cla is easily lost from this compound and carbocation formation takes place readily. This is because this Cla is closest to electronegative atom 0, which will attract the electron density towards itself and readily leaves C-Clabond. Hence, it will most readily react with AgNO3 to give precipitate.

Topic: Chemistry, General Organic Chemistry, JEE Main
The chlorine end of the chlorine monoxide radical carries a charge of +0.167 e. The bond length is 154.6 pm. Calculate the dipole moment of the radical in Debye units.

Options:
A 2.35 D
B 1.24 D
C 1.59 D
D 2.05 D

The correct answer is:
1.24 D

Solution:
Î¼Ì=e.d
e = 1.167e 1.602 ×  10 −19  C 1e −
=2.675× 10 −20  C
d=154.6 pm=154.6 × 10 −12  m
So μ ¯ =(2.675× 10 −20 C)(154.6× 10 −12 )
=4.136 × 10 −30  Cm
= 4.136× 10 −30 3.34× 10 −30 =1.24 D

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The chord $P Q$ of the parabola $y^2=x$, where one end $P$ of the chord is at point $(4,-2)$, is perpendicular to the axis of the parabola. Then the slope of the normal at $Q$ is

Options:
A
$-4$

B
$-\frac{1}{4}$

C
4

D
$\frac{1}{4}$

The correct answer is:
$-4$

Solution:
Point $P$ is $(4,-2)$ and $P Q \perp x$-axis So, $Q=(4,2)$
Equation of tangent at $(4,2)$ is
$$
\begin{aligned}
& y y_1=\frac{1}{2}\left(x+x_1\right) \\
& \Rightarrow 2 y=\frac{1}{2}(x+2) \Rightarrow 4 y=x+2 \\
& \Rightarrow y=\frac{x}{4}+\frac{1}{2}
\end{aligned}
$$
So, slope of tangent $=\frac{1}{4}$
$\therefore$ Slope of normal $=-4$

Topic: Mathematics, Parabola, JEE Main, JEE Main 2012 (26 May Online)
The chords passing through 2,1 intersect the hyperbola x216-y29=1 at A and B. The locus of the point of intersection of tangents at A and B on the hyperbola is

Options:
A x-y=1
B x+y=3
C 9x-8y=72
D 9x+8y=7

The correct answer is:
9x-8y=72

Solution:
Let, the point of intersection of the tangents at A & B is h,k.
Then, the equation of AB is xh16-yk9=1 { âµT=0 is the chord of contact}
Now, xh16-yk9=1 passes through 2,1
Hence, 2h16-k9=1
â9h-8k=72
â the locus is 9x-8y=72

Topic: Mathematics, Hyperbola, JEE Main
The ciliated columnar epithelial cells in humans are known to occur in

Options:
A bronchioles and fallopian tubes
B bile duct and oesophagus
C fallopian tubes and urethra
D eustachian tube and stomach lining

The correct answer is:
bronchioles and fallopian tubes

Solution:
The columnar epithelium is composed of a single layer of tall and slender cells. If the columnar cells bear cilia on their free surface they are called ciliated columnar epithelium. They are mainly present in the inner surface of hollow organs like bronchioles and fallopian tubes. Their function is to move particles or mucus in a specific direction over the epithelium.

Topic: Biology, Structural Organisation in Animals, NEET, NEET 2011 (Screening)
The ciliated epithelial cells are required to move particles or mucus in a specific direction. In humans, these cells are mainly present in:

Options:
A Bile duct and Bronchioles
B Fallopian tubes and Pancreatic duct
C Eustachian tube and Salivary duct
D Bronchioles and Fallopian tubes

The correct answer is:
Bronchioles and Fallopian tubes

Solution:
The human tracheobronchial tree is lined by pseudostratified, ciliated epithelium from the larynx down to the level of the terminal bronchiole. The nasal cavity (apart from its most anterior portion) and the paranasal sinuses are also lined by ciliated epithelium.
The epithelium of the human fallopian tube is a mixture of ciliated and nonciliated cells. Scattered ciliated cells are also found in the endometrium. The function of these ciliated cells is generally believed to be the transport of eggs, zygotes, and secretions. The cilia may further help to maintain the patency of the oviductal lumen

Topic: Biology, Structural Organisation in Animals, NEET, NEET 2019 (All India)
The circle passing through (1,-2) and touching the axis of x at (3,0) also passes through the point

Options:
A 5,-2
B -2, 5
C -5, 2
D 2, -5

The correct answer is:
5,-2

Solution:
Alt : Equation of circle
(x−3) 2 + (y+r) 2 = r 2
Passes through 1, -2
∴ r=2
∴ Circle (x−3) 2 + (y+2) 2 =4

Passes through (5,-2).

Topic: Mathematics, Circle, JEE Main, JEE Main 2013 (07 Apr)
The circle passing through the intersection of the circles, x2+y2-6x=0 and x2+y2-4y=0 having its centre on the line, 2x-3y+12=0, also passes through the point :

Options:
A (–1,3)
B (–3,6)
C (–3,1)
D (1,–3)

The correct answer is:
(–3,6)

Solution:
By family of circle, circle passing through intersection of given circles will be member of

S1+λS2=0 family (λ≠1)

x2+y2-6x+λx2+y2-4y=0

(λ+1)x2+(λ+1)y2-6x-4λy=0

x2+y2-6λ+1x-4λλ+1y=0

Centre 3λ+1,2λλ+1

Centre lies on 2x-3y+12=0

23λ+1-32λλ+1+12=0

6λ+18=0

λ=-3

Now circle is:x2+y2+3x-6y=0

Now check each option

Topic: Mathematics, Circle, JEE Main, JEE Main 2020 (04 Sep Shift 2)
The circle represented by the equation $x^2+y^2+2 g x+2 f y+c=0$ will be a point circle, if

Options:
A $g^2+f^2=c$
B $g^2+f^2 \gt c$
C $g^2+f^2+c=0$
D None of these

The correct answer is:
$g^2+f^2=c$

Solution:
Using condition of point circle $R=\sqrt{g^2+f^2-c}=0 \Rightarrow g^2+f^2=c$

Topic: Mathematics, Circle, JEE Main
The circle $x^2+y^2-3 x-4 y+2=0$ cuts $x$-axis at

Options:
A $(2,0),(-3,0)$
B $(3,0),(4,0)$
C $(1,0),(-1,0)$
D $(1,0),(2,0)$

The correct answer is:
$(1,0),(2,0)$

Solution:
(d) Given, equation of circle is $x^2+y^2-3 x-4 y+2=0$ and it cuts the $x$-axis. $\therefore x^2+0-3 x+2=0$ or $x^2-3 x+2=0$ or $(x-1)(x-2)=0$ or $x=1,2$. Therefore the points are $(1,0)$ and $(2,0)$.

Topic: Mathematics, Circle, JEE Main
The circle $x^2+y^2=4 x+8 y+5$ intersects the line $3 x-4 y=m$ at two distinct points if

Options:
A
$-35 < m < 15$
B
$15 < m < 65$
C
$35 < m < 85$
D
$-85 < m < -35$

The correct answer is:
$-35 < m < 15$

Solution:
Circle $x^2+y^2-4 x-8 y-5=0$
Centre $=(2,4)$, Radius $=\sqrt{4+16+5}=5$
If circle is intersecting line $3 x-4 y=m$
at two distinct points.
$\Rightarrow$ length of perpendicular from centre $ < $ radius
$\Rightarrow \frac{|6-16-\mathrm{m}|}{5} < 5$
$\Rightarrow|10+\mathrm{m}| < 25$
$\Rightarrow-25 < m+10 < 25$
$\Rightarrow-35 < \mathrm{m} < 15$.

Topic: Mathematics, Circle, JEE Main, JEE Main 2010
The circle $x^2+y^2-8 x+4 y+4=0$ touches

Options:
A $x$-axis only
B $y$ - axis only
C Both $x$ and $y$-axis
D Does not touch any axis

The correct answer is:
$y$ - axis only

Solution:
The given equation can be written as $(x-4)^2+(y+2)^2=4^2$. We know that the standard equation of the circle with centre $(h, k)$ is $(x-h)^2+(y-k)^2=r^2$. Comparing the given equation with the standard equation, we get centre $\equiv(4,-2)$ and radius $=4$. Since co-ordinates of the centre of the circle are $(4,-2)$, therefore the given circle touches $y$-axis only

Topic: Mathematics, Circle, JEE Main
The circuit below is used to heat water kept in a bucket.

Assuming heat loss only by Newton's law of cooling, the variation in the temperature of the water in the bucket as a function of time is depicted by

Options:
A
B
C
D

The correct answer is:

Solution:
In given condition,

Heat gained by water heater - Heat lost governed by Newton's law of cooling = Net heat gained by water

⇒i2R1-KT-T0=mSΔTΔt

Here, T= temperature and t= time.

⇒dTdt=i2R1mS-KmST-T0

=i2R1mS+KT0mS-KmST

Above equation can be written as,

dTdt=A+BT

or dTA+BT=dt

Integrating, we get

⇒ 1B.logeA+BT=t

⇒A+BT=etB

⇒T=1BetB-A

Hence, temperature T depends on time t exponentially.

∴ Best suitable graph is

Topic: Physics, Thermal Properties of Matter, JEE Main
The circuit has two oppositely connect ideal diodes in parallel. What is the current flowing in the circuit?

Options:
A 1.33 A

B 1.71 A

C 2.00 A

D 2.71 A

The correct answer is:
2.00 A

Solution:
Since diode D 1 is reverse biased, therefore it will act as an open circuit.
The effective resistance of the circuit is R=4+2=6 Ω .
Current in the circuit is I = E / R = 12 / 6 = 2 A.

Topic: Physics, Semiconductors, JEE Main
The circuit has two oppositely connect ideal diodes in parallel. What is the current following in the circuit?

Options:
A
$1.33 \mathrm{~A}$

B
$1.71 \mathrm{~A}$

C
$2.00 \mathrm{~A}$

D
$2.31 \mathrm{~A}$

The correct answer is:
$2.00 \mathrm{~A}$

Solution:
$D_1$ is reverse biased therefore it will act like an open circuit.
$$
\mathrm{i}=\frac{12}{6}=2.00 \mathrm{~A}
$$

Topic: Physics, Semiconductors, JEE Main, JEE Main 2006
The circuit in figure consists of wires at the top and bottom and identical springs as the left and right sides. The wire at the bottom has a mass of $10 \mathrm{~g}$ and is $5 \mathrm{~cm}$ long. The wire is hanging as shown in the figure. The springs stretch $0.5 \mathrm{~cm}$ under the weight of the wire and the circuit has a total resistance of $12 \Omega$. When the lower wire is subjected to a static magnetic field, the springs, stretch an additional $0.3 \mathrm{~cm}$. The magnetic field is

Options:
A
$0.6 \mathrm{~T}$ and directed out of page

B
$1.2 \mathrm{~T}$ and directed into the plane of page

C
$0.6 \mathrm{~T}$ and directed into the plane of page

D
$1.2 \mathrm{~T}$ and directed out of page

The correct answer is:
$0.6 \mathrm{~T}$ and directed out of page

Solution:
No Solution Available

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2012 (12 May Online)
The circuit
is equivalent to

Options:
A OR gate
B AND gate
C NAND gate
D NOR gate

The correct answer is:
NOR gate

Solution:
Hence NOR gate

Topic: Physics, Semiconductors, NEET, NEET 2008 (Mains)
The circuit is equivalent to

Options:
A AND gate
B NAND gate
C NOR gate
D OR gate

The correct answer is:
NOR gate

Solution:
The gate circuit can be shown by giving two inputs $A$ and $B$.

Output of NOR gate,
$Y_1=\overline{A+B}$
Output of NAND gate,
$\begin{aligned}
Y_2 & =\overline{Y_1 \cdot Y_1} \\
& =\overline{\overline{A+B} \cdot \overline{A+B}} \\
& =\overline{\overline{A+B}}+\overline{\overline{A+B}} \\
& =(A+B)+(A+B) \\
& =A+B
\end{aligned}$
Output of NOT gate,
$\begin{aligned}
Y & =\overline{Y_2} \\
& =\overline{A+B}
\end{aligned}$
which is the output of NOR gate.
Altemative :

Same as NOR Gate
NOR Gate
$\begin{array}{lll}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 0\end{array}$
Note : NOR and NAND gates are said to be universal building blocks.

Topic: Physics, Semiconductors, JEE Main
The circuit shown below contains two ideal diodes, each with a forward resistance of $50 \Omega$. If the battery voltage is $6 \mathrm{~V},$ the current through the $100 \Omega$ resistance (in Amperes) is:

Options:
A 0.036
B 0.02
C 0.027
D 0.03

The correct answer is:
0.02

Solution:
As $D_{2}$ is reversed biased, so no current through $75 \Omega$ resistor.

now $R_{e q}=150+50+100$

$=300 \Omega$

So, required current $I=\frac{\text { BatteryVoltage }}{300}$ $\mathrm{I}=\frac{6}{300}=0.02$

Topic: Physics, Semiconductors, JEE Main, JEE Main 2019 (11 Jan Shift 2)
The circuit shown here has two batteries of $8.0 \mathrm{~V}$ and $16.0 \mathrm{~V}$ and three resistors $3 \Omega, 9 \Omega$ and $9 \Omega$ and a capacitor of $5.0 \mu \mathrm{F}$.
How much is the current $\mathrm{I}$ in the circuit in steady state?

Options:
A
$1.6 \mathrm{~A}$

B
$0.67 \mathrm{~A}$

C
$2.5 \mathrm{~A}$

D
$0.25 \mathrm{~A}$

The correct answer is:
$0.67 \mathrm{~A}$

Solution:
In steady state capacitor is fully charged hence no current will flow through line 2.
By simplyfing the circuit
Hence resultant potential difference across resistances will be $8.0 \mathrm{~V}$.
Thus current $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{8.0}{3+9}=\frac{8}{12}$
or, $\mathrm{I}=\frac{2}{3}=0.67 \mathrm{~A}$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2014 (12 Apr Online)
The circuit shown in the figure consists of a charged capacitor of capacity 3 μF and a charge of 30 μC. At time t=0, when the key is closed, the value of current flowing through the 5 MΩ resistor is x μA. The value of x to the nearest integer is

Options:

The correct answer is:
2

Solution:
i0=VR=30/35×106=2×10-6
∴ =2.00

Topic: Physics, Current Electricity, JEE Main, JEE Main 2021 (18 Mar Shift 1)
The circuit shown in the figure consists of two resistances R1 & R2 connected to two ideal voltmetres V1 & V2. Assume that a voltmeter reads âV=-â«abEâ.dlâ between its terminals. A time-varying magnetic field B(t)=B0t (where B0 is a positive constant of proper dimensions and t is time) exists in a circular region of radius a and it is directed into the plane of the figure. The reading of voltmeter V2 is

Options:
A πa2B0R1R1+R2
B -πa2B0R2R1+R2
C -πa2B0R1R1+R2
D πa2B0R2R1+R2

The correct answer is:
-πa2B0R2R1+R2

Solution:
ϕ=πa2B0t
ε=-dϕdt
⇒ε=πa2B0 (indeed emf)
i=πa2B0R1+R2 (induced current)
∆V1=-∫abE→.dl→=--iR1=πa2B0R1R1+R2
∆V2=-∫abE→.dl→=--iR2=-πa2B0R2R1+R2

Topic: Physics, Electromagnetic Induction, JEE Main
The circuit shown in the figure contains three resistors R1=100 Î©, R2=50 Î© and R3=20 Î© and cells of E.M.F.s E1=2 V and E2. The ammeter indicates a current of 50 mA. Find the E.M.F (in V) of the second cell. (The internal resistance of the ammeter and of the cell should be neglected)

Options:

The correct answer is:
4

Solution:
Using KVL :
1 0 0 i + 0.05 + 5 0 i - 2 = 0
1 5 0 i + 3 = 0
i = - 3 1 5 0 = - 0.02 A
= - 2 0 mA
So, the direction of i is opposite to what we have assumed. If we redraw the circuit.
Using KVL :
2 0 × 0.05 + 1 0 0 × 0.03 - V 2 = 0
1 + 3 - V 2 = 0
V 2 = 4 V

Topic: Physics, Current Electricity, JEE Main
The circular motion of a particle with constant speed is:

Options:
A periodic but not simple harmonic
B simple harmonic but not periodic
C period and simple harmonic
D neither periodic nor simple harmonic

The correct answer is:
periodic but not simple harmonic

Solution:
Motion is periodic but not simple harmonic.

Topic: Physics, Motion In Two Dimensions, JEE Main
The circulation of the essential nutrients between living and non-living components of the biosphere is termed as:

Options:
A Biogeographic cycle
B Cycling of minerals
C Biological cycle
D Biogeochemical cycle

The correct answer is:
Biogeochemical cycle

Solution:
The movement and exchange of organic and inorganic matter between living and non-living component in a cyclical manner refers to nutrient cycle. These are also called as biogeochemical cycles as there is involvement of biotic and abiotic factors and chemical reactions. There are many important cycles which help in maintaining the biosphere like carbon cycle, nitrogen cycle etc.

Topic: Biology, Ecosystem, NEET
The circumcentre of a triangle lies at the origin and its centroid is the midpoint of the line segment joining the points (a2+1, a2+1) and 2a, - 2a, a≠0. Then for any a, the orthocentre of this triangle lies on the line

Options:
A y-a2+1x=0
B y-2ax=0
C y+x=0
D a-12x-a+12y=0

The correct answer is:
a-12x-a+12y=0

Solution:
The mid-point of a line segment joining the points x1, y1 and x2, y2 is x1+x22, y1+y22

Given, the centroid G is the mid-point of the line segment joining the points a2+1, a2+1 and 2a, -2a, thus G≡a2+1+2a2, a2+1-2a2

⇒G≡a+122, a-122.

Also, given circumcentre C is origin ⇒C≡0, 0.

We know that, for a triangle circumcentre, orthocentre and centroid are collinear.

Thus, C, G and orthocentre H are collinear.

Again, we know that the equation of a line passing through the points x1, y1 and x2, y2 is y-y1=y2-y1x2-x1x-x1

Thus, the orthocentre lies on the line joining the points 0, 0 and a+122, a-122 and its equation, is

y-0=a-122-0a+122-0x-0

⇒y=a-12a+12x

⇒a-12x-a+12y=0.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2014 (19 Apr Online)
The circumference of the second orbit of an atom or ion having a single electron is 4Ã10-9 m. The de Broglie wavelength of an electron revolving in this orbit should be

Options:
A 2×10-9 m
B 4×10-9 m
C 8×10-9 m
D 1×10-9 m

The correct answer is:
2×10-9 m

Solution:
2πr=nλ
∴ λ=2πrn=4×10-92=2×10-9 m

Topic: Physics, Atomic Physics, JEE Main
The Cl-Co-Cl bond angle values in a fac- CoNH33Cl3 complex is/are:

Options:
A 90° & 180°
B 90°
C 180°
D 90° & 120°

The correct answer is:
90°

Solution:
In facial isomer, three Cl atoms present on one face of the octahedron. Similarly, three ammonia molecules are present on one face of the octahedron. All 3 same ligands have 90° bond angle between them.

The Cl-Co-Cl bond angle in above octahedral complex is 90°

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2023 (30 Jan Shift 2)
The class of drug to which chlordiazepoxide with above structure belongs is :

Options:
A Analgesic
B Tranquillizer
C Antacid
D Antibiotic

The correct answer is:
Tranquillizer

Solution:
The class of drug to which chlordiazepoxide belongs is tranquillizer drug.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2021 (26 Aug Shift 2)
The class of materials having the largest band gap in the following is

Options:
A Metals
B Semi-metals
C Semiconductors
D Insulators

The correct answer is:
Insulators

Solution:
Insulators have very large energy band gaps. Its value is generally more than $3 \mathrm{eV}$.

Topic: Physics, Semiconductors, JEE Main
The classification of drugs based on _______ is the most useful one for medicinal chemists.

Options:
A pharmacological effect
B drug action
C chemical structure
D molecular targets

The correct answer is:
molecular targets

Solution:
Medicinal chemists are interested in the location at which a particular drug will create action and assign specific drugs to people based on the requirements. This also involves the mechanism of drug action on target based on structure.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
The cleansing action of soap is due to the

Options:
A the presence of Na+ ions in water
B its dissociation into ions in water
C the formation of associated colloids
D its action as emulifying agent

The correct answer is:
its action as emulifying agent

Solution:
the main function of a soap is to stabilize the emulsion formed by oil or greasy dirt with water. Soap forms negatively charged grease micelles. These grease or oil micelles are removed by rinsing with water.

Topic: Chemistry, Chemistry in Everyday Life, NEET
The C-O bond length in $\mathrm{CO}, \mathrm{CO}_{2}$ and $\mathrm{CO}_{3}{ }^{2-}$ follows the order -

Options:
A $\mathrm{CO} < \mathrm{CO}_{2} < \mathrm{CO}_{3}{ }^{2-}$
B $\mathrm{CO}_{2} < \mathrm{CO}_{3}{ }^{2-} < \mathrm{CO}$
C $\mathrm{CO}>\mathrm{CO}_{2}>\mathrm{CO}_{3}^{2-}$
D $\mathrm{CO}_{3}{ }^{2-} < \mathrm{CO}_{2} < \mathrm{CO}$

The correct answer is:
$\mathrm{CO} < \mathrm{CO}_{2} < \mathrm{CO}_{3}{ }^{2-}$

Solution:
Bond Length $\propto \frac{1}{\text { Bond Order }}$
$\mathrm{CO}, \mathrm{B} . \mathrm{O} .=3$
$\mathrm{O}=\mathrm{C}=\mathrm{O}, \mathrm{B} . \mathrm{O} .=2$

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The coagulating power of electrolytes having ions Na+, Al3+ and Ba2+ for arsenic sulphide sol increases in the order,

Options:
A Ba2+<Na+<Al3+
B Al3+<Na+<Ba2+
C Al3+<Ba2+<Na+
D Na+<Ba2+<Al3+

The correct answer is:
Na+<Ba2+<Al3+

Solution:
As2S3 /S2- is negatively charged sol.

According to the Hardy Schulze rule, the greater the charge on the cation (oxidation state), the greater is its coagulating power for the negatively charged sol. Hence, the increasing order is:

Na + < Ba 2 + < Al 3 +

Al 3 + (highest coagulating power).

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2013 (07 Apr)
The coagulation of 10 cm3 of Gold sol by 1 ml 10% NaCl solution is completely prevented by addition of 0.25 g of starch to it. The gold number of starch is-

Options:
A 0.025
B 0.25
C 2.5
D 250

The correct answer is:
250

Solution:
0.25 gm prevents coagulation of 10 ml of gold sol by 1 ml of 10% NaCl
∴ milligrams at starch =250
∴ Gold number =250

Topic: Chemistry, Surface Chemistry, JEE Main
The coagulation values in millimoles per litre of the electrolytes used for coagulation of As2S2 are given below.
I. NaCl=52
II. BaCl2=0.69
III. MgSO4=0.22
The correct order of their coagulation is:

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

The correct answer is:
III > II > I

Solution:
Coagulation value ∝1coagulating  power
∴ Coagulating power As2S3 is
MgSO4>BaCl2>NaCl
III > II > I

Topic: Chemistry, Surface Chemistry, JEE Main
The coefficient of friction between the two blocks shown in the figure is Î¼=0.4 and between the lower block and the ground is zero. The blocks are given velocities of 2 m s-1 and 8 m s-1 respectively in the direction shown in the figure. In how much time (in seconds) the slipping between the blocks will stop?

Options:

The correct answer is:
1

Solution:
The common velocity of the blocks can be calculated as follows

3v=1×2+2×8

v=6 m s-1

The acceleration of the 1 kg block is

a=μg=4 m s-2

v=2+4t

⇒t=1 s

Topic: Physics, Laws of Motion, JEE Main
The coefficient of friction between the tyres and the road is 0.25 . The maximum speed with which a car can be driven round a curve of radius $40 \mathrm{~m}$ without skidding is (assume $g=10 \mathrm{~ms}^{-2}$ )

Options:
A $40 \mathrm{~ms}^{-1}$
B $20 \mathrm{~ms}^{-1}$
C $15 \mathrm{~ms}^{-1}$
D $10 \mathrm{~ms}^{-1}$

The correct answer is:
$10 \mathrm{~ms}^{-1}$

Solution:
$v=\sqrt{\mu \mathrm{rg}}=\sqrt{0.25 \times 40 \times 10}=10 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Motion In Two Dimensions, JEE Main
The coefficient of linear expansion of an inhomogeneous rod changes linearly from Î± 1 to Î± 2 from one end to the other end of the rod. the effective coefficient of linear expansion of rod is

Options:
A α 1 + α 2
B Î± 1 + Î± 2 2
C α 1 α 2
D α 1 - α 2

The correct answer is:
α 1 + α 2 2

Solution:
The value of α at a distance x from the left end is

α x = α 1 + α 2 - α 1 L x

ΔLΔT=∫0Lαxdx

∆L∆T=α1L+α2-α1LL22

∆L∆T=α1+α22L

α eff = α 1 + α 2 2

Topic: Physics, Thermal Properties of Matter, JEE Main