datavorous/entrance-exam-dataset huggingface dataset:
In the experimental set up of metre bridge shown in the figure, the null point is obtaine data distance of $40 \mathrm{~cm}$ from

A. If a $10 \Omega$ resistor is connected in series with $\mathrm{R}_{1},$ the null point shifts by $10 \mathrm{~cm}$. The resistance that should be connected in parallel with $\left(\mathrm{R}_{1}+10\right) \Omega$ such that the null point shifts back to its initial position is

Options:
A $20 \Omega$
B $40 \Omega$
C $60 \Omega$
D $30 \Omega$

The correct answer is:
$60 \Omega$

Solution:
Initially at null deflection $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{2}{3}$$\ldots(\mathrm{i})$

Finally at null deflection, when null point is shifted $\frac{\mathrm{R}_{1}+10}{\mathrm{R}_{2}}=1 \Rightarrow \mathrm{R}_{1}+10=\mathrm{R}_{2}$...(ii)

Solving equations (i) and (ii) we get $\frac{2 R_{2}}{3}+10=R_{2}$

$10=\frac{\mathrm{R}_{2}}{3} \Rightarrow \mathrm{R}_{2}=30 \Omega$

$\& R_{1}=20 \Omega$

Now if required resistance is $\mathrm{R}$ then $\frac{\frac{30 \times \mathrm{R}}{30+\mathrm{R}}}{30}=\frac{2}{3}$

$\mathrm{R}=60 \Omega$

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2019 (11 Jan Shift 2)
In the extraction of aluminium bauxite cryolite is dissolved because

Options:
A It increases the electrical conductivity of aluminium oxide
B Bauxite becomes active
C It acts as a solvent
D It reduces melting point of aluminium oxide

The correct answer is:
It increases the electrical conductivity of aluminium oxide

Solution:
In the extraction of aluminum, cryolite is dissolved because it lowers the melting point and increases the electrical conductivity of Almina.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
In the extraction of aluminium the electrolyte is

Options:
A Fused cryolite with felspar
B Fused cryolite with fluorspar
C Pure alumina in molten cryolite
D Pure alumina with bauxite and molten cryolite

The correct answer is:
Pure alumina in molten cryolite

Solution:
In the extraction of aluminium by electrolytic process alumina $\mathrm{Al}_2 \mathrm{O}_3$ is used along with fluorspar and cryolite in molten state.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
In the extraction of Cu, the metal is formed in the Bessemer converter due to the reaction:

Options:
A Cu 2 Sâ+â2â Cu 2 Oâ â Î â6âCuâ+â SO 2
B Cu 2 S→2 Cu+S
C Fe+ Cu 2 O→2 Cu+FeO
D 2  Cu 2 O→4 Cu+ O 2

The correct answer is:
Cu 2 S + 2  Cu 2 O  → Δ  6 Cu +  SO 2

Solution:
The following reaction taken place in the formation of blister copper in the Bessemer converter
Cu 2 S + 2  Cu 2 O  → Δ  6 Cu +  SO 2

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
In the extraction of copper from its sulphide ore, metal is finally obtained by the oxidation of cuprous sulphide with:

Options:
A SO2
B Fe2O3
C Cu2O
D CO

The correct answer is:
Cu2O

Solution:
In the extraction of copper from its sulphide ore, metal is finally obtained by the oxidation of cuprous sulphide with Cu2O. â It is an example of auto reduction.

Cu2S+2Cu2O →6Cu+ SO2

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2018 (16 Apr Online)
In the extraction of copper from its sulphide ore, the metal is finally obtained by the reduction of cuprous oxide with

Options:
A copper (I) sulphide $\left(\mathrm{Cu}_2 \mathrm{~S}\right)$
B sulphur dioxide $\left(\mathrm{SO}_2\right)$
C iron sulphide (FeS)
D carbon monoxide (CO)

The correct answer is:
copper (I) sulphide $\left(\mathrm{Cu}_2 \mathrm{~S}\right)$

Solution:
In the extraction of copper from its sulphide ore, when ore is subjected to roasting, some of it is oxidised to $\mathrm{Cu}_2 \mathrm{O}$ which reacts with the remaining $\mathrm{Cu}_2 \mathrm{~S}$ (sulphide ore) to give copper metal.
$\mathrm{Cu}_2 \mathrm{~S}+2 \mathrm{Cu}_2 \mathrm{O} \longrightarrow 6 \mathrm{Cu}+\mathrm{SO}_2 \uparrow$
In this process $\mathrm{Cu}_2 \mathrm{~S}$ behaves as reducing agent.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, NEET, NEET 2012 (Screening)
In the extraction of copper, its sulphide ore is heated in a reverberatory furnace after mixing with silica to:

Options:
A separate CuO as CuSiO3
B remove calcium as CaSiO3
C decrease the temperature needed for roasting of Cu2S
D remove FeO as FeSiO3

The correct answer is:
remove FeO as FeSiO3

Solution:
In actual practice, the ore is heated in reverberatory furnace after mixing with silica. In the furnace, iron oxide 'slags off' as iron silicate and copper is produced in the form copper matte. This contains Cu2S and FeS.

FeO+SiO2⟶FeSiO3

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2023 (30 Jan Shift 1)
In the figure, a conducting rod of length l=1 meter and mass m=1 kg moves with an initial velocity, u =5 m s-1 . On a fixed horizontal frame containing inductor L=2 H and resistance R=1 Î© . PQ and MN are smooth, conducting wires. There is a uniform magnetic field of strength B=1 T. Initially, there is no current in the inductor. Find the total charge in coulomb, flown through the inductor by the time velocity of the rod becomes vf= 1 m s-1 and the rod has travelled a distance x=3 meter.

Options:

The correct answer is:
1

Solution:
Let i1 and i2 be the current through L and R at any time t
∴ i=i1+i2 ⇒ BlvR=i2 and Blv=Ldi1dt
Force on conducting rod =mdvdt=-ilB=-i1+BlvRlB
⇒ mdv= -lB i1dt-B2l2Rv dt
⇒ m∫dv=-lB ∫i1dt-B2l2R ∫v dt
⇒ mvf-u=-lBQ-B2l2Rx
( v1 = velocity, when it has moved a distance 'x')
⇒ Q=-B2 l2R x-mvf-uB l=1C

Topic: Physics, Electromagnetic Induction, JEE Main
In the figure given below, for an angle of incidence 45° at the top surface, what is the minimum refractive index needed for total internal reflection at the vertical face?

Options:
A 2+12
B 12
C 32
D 2+1

The correct answer is:
32

Solution:
At point A , by Snell's law
μ=sin⁡45osin⁡r⇒sin⁡r=1μ2 ... (i)
At point B , total internal reflection with sin⁡i1=1μ
From figure, i1=90o-r .... (ii)
sin⁡90o-r=1μ⇒cos⁡r=1μ .... (iii)
Now, cos⁡r=1-sin2⁡r=1-12μ2=2μ2-1μ2
From equation (ii) and (iii), 1μ=2μ2-12μ2
Squaring both sides and then solving, we get
μ=32

Topic: Physics, Ray Optics, JEE Main
In the figure given below, the back line (upper) indicates action spectrum for photosynthesis and the lighter line (lower) indicates the absorption spectrum of chlorophyll- $a$, answer the following
(a) What does the action spectrum indicate? How can we plot an action spectrum? Explain with an example.
(b) How can we derive an absorption spectrum for any substances?
(c) If chlorophyll-a is responsible for light reaction of photosynthesis, why do the action spectrum and absorption spectrum not overlap?

Options:

None

Solution:
(a) Action spectrum depicts the relative rates of photosynthesis at different wavelenghths of light. Action spectrum of photosynthesis can be plotted by measurement of oxygen evolution at different wavelength. Englemann (1882) by using a green algae plotted action spectrum.
(b) Absorption spectrum of a substance can be derived by calculating amount of energy of different wavelength of light absorbed.
(c) Chlorophyll a is responsible for light reaction of photosynthesis, but the action spectrum and absorption spectrum do not overlap because, though chlorophyll $\mathrm{a}$ is the main pigment responsible for the absorption of light, other thylakoids pigment like chlorophyll b, xanthophylls, carotenoids, which are accessory pigments also absorb and transfer the energy to chlorophyll a. Indeed they not only enable a wider range of wavelength of incoming light to be utilized for photosynthesis but also protect chlorophyll from photo-oxidation.

Topic: Biology, Photosynthesis in Higher Plants
In the figure given below the plant bears two different types of flowers marked 'A' and 'B'. Identify the types of flowers and state the type of pollination that will occur in them.

Options:

None

Solution:
In the figure the plant bears following two types of flowers. A - Chasmogamous flower (the flowers remain open, exposing anthers and stigmas.)
$\mathrm{B}$ - Cleistogamous flowers (the flowers remain closed, so that anthers and stigmas are never exposed) the types of pollination that will occur in these flowers are Autogamy (within same flower), Geitonogamy (different flowers of same plant) and Xenogamy (genetically different plants).
Ceistogamy is a type autogamy in which some plants, like, Commelina Benghalensis possess both chasmogamous and cleistogamous flowers.
In chasmogamous flowers, the flowers may undergo selfpollination or cross-pollination while the flowers undergo only self-pollination, in cleistogamous flowers.

Topic: Biology, Reproduction in Organisms
In the figure, given that VBB supply can vary from 0 to 5.0 V, VCC=5 V, βdc=200, RB=100 kΩ, RC=1 kΩ and VBE=1.0V. The minimum base current and the input voltage at which the transistor will go to saturation, will be, respectively:

Options:
A 25 Î¼A and 2.8 V
B 20 Î¼A and 2.8 V
C 25 Î¼A and 3.5 V
D 20 Î¼A and 3.5 V

The correct answer is:
25 μA and 3.5 V

Solution:
When switched on,
VCE=0
VCC-RC iC=0
ic=VCCRC=51Ã103=5Ã10-3A
ic=Î²iB
iB=iCÎ²=5Ã10-3200=2.5Ã10-5A =25 Î¼A
using KVL at input side,
VBB-iB RB-VBE=0
VBB=VBE+iBRB
=1+100Ã103Ã25Ã10-6
=1+2.5
=3.5 V

Topic: Physics, Semiconductors, JEE Main, JEE Main 2019 (12 Jan Shift 2)
In the figure given, the electric current flowing through the 5 kΩ resistor is x mA.

The value of x to the nearest integer is ________.

Options:

The correct answer is:
3

Solution:
I=215+1+1=3 mA

Topic: Physics, Current Electricity, JEE Main, JEE Main 2021 (16 Mar Shift 1)
In the figure is shown a system of four capacitors connected across a 10 V battery. The charge that will flow from switch S when it is closed is:

Options:
A Zero
B 20 μC from a to b
C 5 μC from b to a
D 5 μC from a to b

The correct answer is:
5 μC from b to a

Solution:
Finally when switch is closed,
q1=2×5=10 μC
q2=3×5=15 μC
The total charge on dotted region =q2-q1
=15-10=5 μC
Charge 5 μC flows from b to a

Topic: Physics, Capacitance, JEE Main
In the figure is shown a system of four capacitors connected across a 10 V battery. The charge that will flow from switch S when it is closed is:

Options:
A Zero
B 20 μC from a to b
C 5 μC from b to a
D 5 μC from a to b

The correct answer is:
5 μC from b to a

Solution:
Finally when switch is closed,
q1=2×5=10 μC
q2=3×5=15 μC
The total charge on dotted region =q2-q1
=15-10=5 μC
Charge 5 μC flows from b to a

Topic: Physics, Capacitance, JEE Main, JEE Main 2015 (11 Apr Online)
In the figure of sigmoid growth curve given below, label segments 1,2 and 3.

Options:

None

Solution:
Growth is the main characteristic that distinguish living organisms from non-living. All living organism grow in number and then accumulate biomass and grow in size as well. Increase in number of cells as well as increase in size and length of each cell, exhibits growth of all living organism. In unicellular organism, the growth is synchronized with reproduction. These organism when divide they produce offspring (reproduction) i.g., each cell accumulate (synthesise) protoplasm and increase in size but at a certain a limit and divide to from two cells.

Topic: Biology, Plant - Growth and Development
In the figure PQ, PO1 and O1Q are the diameters of semicircles C1, C2 and C3 with centres at O1, O2 and O3 respectively and the circle C4 touches the semicircles C1,C2 and C3. If PQ=24 units and the area of the circle C4 is A sq. units, then the value of 8πA is equal to here, PO1=O1Q. If your answer is K, mark 2K as answer after rounding off to nearest integer.

Options:

The correct answer is:
1

Solution:
Let the point of contact of C4 & C1 is A, center of C4 is O4 & radius is equal to r
âAO1=12âO1O4=12-r
Also, O4O3=r+6 and O1O3=6
âr+62=12-r2+36
â36r=144âr=4âA=16Ï
â8ÏA=8Ï16Ï=12=0.5

Topic: Mathematics, Circle, JEE Main
In the figure PQ, PO1 and O1Q are the diameters of semicircles C1, C2 and C3 with centres at O1, O2 and O3 respectively and the circle C4 touches the semicircles C1,C2 and C3. If PQ=24 units and the area of the circle C4 is A sq. units, then the value of 8πA is equal to here, PO1=O1Q

Options:

The correct answer is:
0.5

Solution:
Let the point of contact of C4 & C1 is A, center of C4 is O4 & radius is equal to r
âAO1=12âO1O4=12-r
Also, O4O3=r+6 and O1O3=6
âr+62=12-r2+36
â36r=144âr=4âA=16Ï
â8ÏA=8Ï16Ï=12=0.5

Topic: Mathematics, Circle, JEE Main
In the figure, pendulum bob on left side is pulled a side to a height $h$ from its initial position. After it is released it collides with the right pendulum bob at rest, which is of same mass. After the collision, the two bobs stick ogether and rise to a height

Options:
A $\frac{3 h}{4}$
B $\frac{2 h}{3}$
C $\frac{h}{2}$
D $\frac{h}{4}$

The correct answer is:
$\frac{h}{4}$

Solution:
When bob $A$ strikes to the bob $B$, then
$m u=(m+m) v^{\prime}$
The potential energy of $A$ at height $h$ gets converted into kinetic energy of this mass, at point $O$, i.e.
$\begin{array}{rlrl}
& m g h =\frac{1}{2} m u^2 \\
\Rightarrow \quad u & =\sqrt{2 g h} & [From Eq. (i)] \\
\therefore \quad v^{\prime} & =\frac{\sqrt{2 g h}}{2}=\sqrt{\frac{g h}{2}}
\end{array}$
Let the combined mass moves to a height $h^{\prime}$, then total mass $=2 m$
Then, $2 m g h^{\prime}=\frac{1}{2}(2 m) v^{\prime 2}$
$\Rightarrow \quad g h^{\prime}=\frac{g h}{4} \Rightarrow h^{\prime}=\frac{h}{4}$

Topic: Physics, Work Power Energy, JEE Main
In the figure, potential difference between A and B is:

Options:
A 10 V
B 5 V
C 15 V
D 0 V

The correct answer is:
10 V

Solution:
Diode is in forward bias, so it will behave as simple wire so,
So, Vab=305+10×5=10 V

Topic: Physics, Current Electricity, JEE Main, JEE Main 2020 (07 Jan Shift 2)
In the figure shown ABC is a uniform wire. If the center of mass of the wire lies vertically below point A, then BCAB is close to

Options:
A 1.85
B 1.5
C 1.37
D 3

The correct answer is:
1.37

Solution:
xcm=x2=(ρx)(x2)12+ρy(y2)ρ(x+y)
⇒ 12+yx=y2x2
⇒ yx=1+32=1.37

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2016 (10 Apr Online)
In the figure shown, a circuit contains two identical resistors with resistance R=5 Ω and an inductance with L=2 mH. An ideal battery of 15V is connected in the circuit. What will be the current through the battery long after the switch is closed?

Options:
A 5.5 A
B 6 A
C 3 A
D 7.5 A

The correct answer is:
6 A

Solution:
After a long time, when the growth of current in the circuit has stopped, the inductor will work as a conductor. Therefore the effective resistance of the circuit will be,

R=52=2.5 Ω

Hence, the current through the circuit will be,

I=VR=152.5=6A

Topic: Physics, Alternating Current, JEE Main, JEE Main 2019 (12 Jan Shift 1)
In the figure shown a semicircular area is removed from a uniform square plate of side \( l \) and mass (before removing) \( m \).
The \( x \)-coordinate of centre of mass of remaining portion is (The origin is at the centre of square)

Options:
A \( -\frac{\pi(\pi-2) \ell}{2(8-\pi)} \)
B \( \frac{\pi(\pi-2) \ell}{2(8-\pi)} \)
C \( -\frac{\pi(\pi-2) \ell}{(8-\pi)} \)
D None of these

The correct answer is:
None of these

Solution:
The centre of mass about x-axis is given as,

xcm=m1x1+m2x2 m1+m2

Here, m1= mass of the square plate=m

x1=centre of mass of the square plate =0 (As the origin is at centre of the square)
m2= mass of the removed part=mass density of square×area of the semicircular=-ml2πl222=-π8m

x2=centre of mass of the removed part=l2-43πl2=l21-43π

Now, the centre of mass of the system will be,xcm=-πm8×l21-43πm-π8m

Hence, the centre of mass of remaining portion xcm=-lπ-432(8-π).

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
In the figure shown below reactant A (represented by square) is in equilibrium with product B (represented by circle). The equilibrium constant is (approx):

Options:
A 4
B 8
C 1
D 2

The correct answer is:
2

Solution:
Bonus (no reaction is given)
A⇌B(Assume reaction)
K= [B] [A] = 11 6 ≃2

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2020 (09 Jan Shift 2)
In the figure shown here, \( A \) is a conducting sphere and \( B \) is a closed spherical surface. If a - \( q \) charge is placed at \( C \) near \( \mathrm{A} \), then the electric flux through the closed surface is-

Options:
A zero
B positive
C negative
D can not be predicted

The correct answer is:
negative

Solution:
Given that the figure shown here, A is a conducting sphere and B is a closed spherical surface. If a -q charge is placed at C near A

Charge enclosed in Gaussian surface is – ve, so the electric flux ϕ = – ve

Topic: Physics, Electrostatics, JEE Main
In the figure shown here, what is the equivalent focal length of the combination of lenses (Assume that all layers are thin)?

Options:
A -50 cm
B 40 cm
C -40 cm
D -100 cm

The correct answer is:
-100 cm

Solution:
The focal length fL of the left plano-concave lens can be calculated as follows-

1fL=n-11R1-1R2= 1.6-11∞-120 cm= -0.620 cm ...1

The focal length fM of the middle bi-convex lens can be calculated as follows-

1fM=n-11R1-1R2= 1.5-1120 cm-1-20 cm= 0.510 cm ...2

The focal length fR of the right plano-concave lens can be calculated as follows-

1fR=n-11R1-1R2= 1.6-11-20 cm-1∞= -0.620 cm ...3

Hence, the effective focal length f of the combination of lenses can be calculated as follows-

1f=1fL+1fM+1fR= -0.620 cm+0.510 cm-0.620 cm= -0.110 cm⇒f= -100 cm

Topic: Physics, Ray Optics, JEE Main
In the figure shown mass of both, the spherical body and block is $\mathrm{m}$. Moment of inertia of the spherical body about centre of mass is $2 m R^2$. The spherical body rolls on the horizontal surface. There is no slipping at any surfaces in contact. The ratio of kinetic energy of the spherical body to that of block is

Options:
A $3 / 4$
B $1 / 3$
C $2 / 3$
D $1 / 2$

The correct answer is:
$2 / 3$

Solution:
Let $v$ be the linear velocity of centre of mass of the spherical body and $w$ its angular velocity about centre of mass.
Then $\left.\omega=\frac{v}{2 R} \right\rvert\, \mathrm{KE}$ of spherical body
$K_1=\frac{1}{2} m v^2+\frac{1}{2} I \omega^2\left|K_1=\frac{1}{2} m v^2+\frac{1}{2}(2 m R)^2\left(\frac{v^2}{4 R^2}\right)=\frac{3}{4} m v^2\right|$ ?.
(i) Speed of the block will be $\left.v^{\prime}=(\omega)(3 R)=3 R \omega=(3 R)\left(\frac{v}{2 R}\right)=\frac{3}{2} v \right\rvert\,$
$\therefore K E$ of block
$K_2=\frac{1}{2} m v^2\left|=\frac{1}{2} m\left(\frac{3}{2} v\right)^2=\frac{9}{8} m v^2\right|$
From equations (i) and (ii),
$| \frac{K_1}{K_2}=\frac{2}{3} |$

Topic: Physics, Rotational Motion, JEE Main
In the figure shown, neglecting friction and mass of pulleys, what is the acceleration of mass \( B \) ?

Options:
A \( \frac{g}{3} \)
B \( \frac{5 g}{2} \)
C \( \frac{2 g}{2} \)
D \( \frac{2 g}{5} \)

The correct answer is:
\( \frac{2 g}{5} \)

Solution:
Forces acting on different parts of the system is shown below.

For block A,

mg-2T=ma ...(i).

For block B,

T-mg=mb ...(ii).

And by constraint motion, if A will move by x distance, then B will move by 2x distance.

a-b+a=0 ...(iii).

Solving (i), (ii) and (iii), we get,

a=-g5 and

b=-2g5.

Topic: Physics, Laws of Motion, JEE Main
In the figure shown, OABC is a rectangle with OA=3 units, OC=4 units. If BD=2.5 units, then the slope of the diagonal OB is equal to

Options:
A 12
B 23
C 12
D 13

The correct answer is:
12

Solution:
tan⁡θ=34,tan⁡ϕ=323=12
slope of OB=tan⁡π2-θ+ϕ=cot⁡θ+ϕ
= 1− 3 4 ⋅ 1 2 3 4 + 1 2 = 1 2

Topic: Mathematics, Straight Lines, JEE Main
In the figure shown plane waves are refracted from air to water using Huygen's principle a, b, c, d, e are lengths on the diagram. The refractive index of air with respect to water is in the ratio.

Options:
A ae

B be

C bd

D db

The correct answer is:
db

Solution:
sin isin r=bcdc=bd
μa,w=sin rsin i

Topic: Physics, Wave Optics, JEE Main
In the figure shown, RB=500 kÎ©, RC=8 kÎ©, VBB=10.6 V and VCC=20 V. The current amplification factor Î²=100 and VBE=0.6 V. Mark the incorrect statement.

Options:
A IB=20 Î¼A
B VCB=3.4 V
C VCE=2.8 V
D the transistor is in active region

The correct answer is:
VCE=2.8 V

Solution:
VBB – IBRB= VBE

10.6-5×105×IB=0.6

⇒IB=20 μA

IC=100 IB=2000 μA

VCE= VCC –ICRC=20-0.002×8000=4 V

VCB=20-0.002×8000-0.6=3.4 V

Topic: Physics, Semiconductors, JEE Main
In the figure shown, the plank is being pulled to the right with a constant speed v. If the cylinder does not slip then

Options:
A The speed of the centre of mass of the cylinder is 2v
B The speed of the centre of mass of the cylinder is v
C The angular velocity of the cylinder is vR
D The angular velocity of the cylinder is zero

The correct answer is:
The angular velocity of the cylinder is vR

Solution:
For no slip condition velocity of plank = velocity of bottom point of sphere.
ω = v R and v c = 0

Topic: Physics, Rotational Motion, JEE Main
In the figure shown, the system is at rest, initially. Two persons, \( A \) and \( B \), of masses \( 40 \mathrm{~kg} \) each, move with speeds \( v_{1} \) and \( v_{2} \), respectively, towards each other on a plank lying on a smooth horizontal surface as shown in the figure. The plank travels a distance of \( 20 \mathrm{~m} \) towards right direction in \( 5 \mathrm{~s} \). (Here, \( v_{1} \) and \( v_{2} \) are given with respect to the plank). Then, (i) \( v_{1}=10 \mathrm{~m} \mathrm{~s}^{-1}, v_{2}=0 \mathrm{~m} \mathrm{~s}^{-1} \) (ii) \( v_{1}=15 \mathrm{~m} \mathrm{~s}^{-1}, v_{2}=5 \mathrm{~m} \mathrm{~s}^{-1} \) (iii) \( v_{1}=20 \mathrm{~m} \mathrm{~s}^{-1}, v_{2}=10 \mathrm{~m} \mathrm{~s}^{-1} \)

Options:
A only (i) and (iii) are possible.
B all (i), (ii), (iii) are not possible.
C only (ii) and (iii) are possible.
D All (i), (ii), (iii) are possible.

The correct answer is:
all (i), (ii), (iii) are not possible.

Solution:
Let the distance travelled by plank be s in time t.

So, the velocity of plank, vP=st.

⇒vP=205⇒vP=4 m s-1

∆xCM=0.

Velocity of A with respect to ground =v1+4

Velocity of B with respect to ground =v2-4

As the net external force in x-direction is zero, the linear momentum is conserved.

40v1+4-40v2-4+20×4=0.

40v1-v2+10=0

v1-v2=-10.

If v1=10 m s-1 then v2=20 m s-1

If v1=20 m s-1 then v2=30 m s-1

If v1=15 m s-1 then v2=25 m s-1.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
In the figure shown, what is the current (in Ampere) drawn from the battery? You are given:
R1=15 Ω , R2=10 Ω , R3=20 Ω , R4=5 Ω, R5=25 Ω, R6=30 Ω, E=15 V

Options:
A 9/32
B 7/18
C 13/24
D 20/3

The correct answer is:
9/32

Solution:
The equivalent resistanceReq of this given network is,

Req=15+253+30=45+25+903=1603

By applying ohm's law, the Current i through the battery is
i=151603=15×3160

∴i=932 A

Topic: Physics, Current Electricity, JEE Main, JEE Main 2019 (08 Apr Shift 2)
In the figure, the chamber A contains a gas, movable chamber B is placed on the top of the gas and it contains $\mathrm{n}$ metal balls. The weight of chamber B is supported by the gas. Chamber $\mathrm{C}$ has vacuum. Let the gas be in equilibrium at pressure $\mathrm{P}$. Let $\mathrm{P}^{\prime}$ be the pressure if one of the balls is taken away. Find $\left(\mathrm{P}-\mathrm{P}^{\prime}\right) / \mathrm{P}$.

Options:
A 1
B $n$
C $2 \mathrm{n}$
D $1 / \mathrm{n}$

The correct answer is:
$1 / \mathrm{n}$

Solution:
No solution. Refer to answer key.

Topic: Physics, Kinetic Theory of Gases, JEE Main
In the figure, the mass of a ball is 95 times the mass of the rod. Length of the rod is 1 m. The level of ball is the same as rod level. Find out the time taken by the ball to reach at the upper end of the rod.

Options:
A 1.4 s
B 2.45 s
C 3.25 s
D 5 s

The correct answer is:
1.4 s

Solution:
Let a1 and a2 be accelerations of a ball (upward) and rod (downward), respectively.
Clearly, from the diagram
2a1-a2 ...(i)
Now, for the ball
2T-95mg-95ma ...(ii)
and for the rod, mg-T=ma2 ...(iii)
On solving equations (i) and (iii), we get
a1=g29 m s-2↑ (upward)
a2=2g29 m s-2↓ (downward)
So, acceleration of ball w.r.t rod =a1+a2=3g29
Now, displacement of ball w.r.t. rod when it reaches the upper end of rod is 1m .
Using the equation of motion,
s=ut+12at2
s=0+12×3×1029t2
t=5830=1.4 s approx

Topic: Physics, Laws of Motion, JEE Main