datavorous/entrance-exam-dataset huggingface dataset:
Figure shows charge (q) versus voltage (V) graph for series and parallel combination of two given capacitors. The capacitances are:

Options:
A 60 μF and 40 μF
B 50 μF and 30 μF
C 20 μF and 30 μF
D 40 μF and 10 μF

The correct answer is:
40 μF and 10 μF

Solution:
Q=CV
The slope of graph will give the equivalent capacitance of the combination.
From graph A: C1+C2=50010=50 …(i)
From graph B: C1C2C1+C2=8010=8 …(ii)
From equation (i) & (ii)
C1C2=400 …(iii)
C1+400C1=50 ⇒C12-50C1+400=0
C1=50±2500-16002=50±302
=10 μF,  40 μF

Topic: Physics, Capacitance, JEE Main, JEE Main 2019 (10 Apr Shift 1)
Figure shows electric field lines in which an electric dipole \( p \) is placed as shown. Which of the following statements is
correct?

Options:
A The dipole will not experience any force.
B The dipole will experience a force towards right.
C The dipole will experience a force towards left.
D The dipole will experience a force upwards.

The correct answer is:
The dipole will experience a force towards left.

Solution:
Here, the space between the electric field lines is increasing from left to right, so the strength of electric field decreases with the increase in the space between electric field lines. As a result force on charges also decreases from left to right.

Let the electric field at the points of +q and -q charges be E1 and E2 respectively. From above discussion, E2>E1.

Force on the point charges is given as F→=qE→

Since E2>E1 ⇒F2>F1

Hence, the net force on the dipole is left side.

Topic: Physics, Electrostatics, JEE Main
Figure shows the position-time graph of a particle of mass $4 \mathrm{~kg}$. What is the (a) force on the particle for $t < 0, t>4 s$, $0 < t < 4 s$ ? (b) impulse at $t=0$ and $t=4 s$ ? (Consider onedimensional motion only).

Options:

None

Solution:
(a) When $t < 0$. As this part is horizontal, thus it can be concluded that distance covered by the particle is zero and hence force on the particle is zero.
When $0 < t < 4$ s. As $\mathrm{OA}$ has a constant slope, hence in this intervel, particle moves with constant velocity $\left(\frac{3}{4}=0.75 \mathrm{~ms}^{-1}\right)$. Hence force on the particle is zero. When $t>4$ s. As this portion shows that particle always remains at a distance of $3 \mathrm{~m}$ from the origin i.e., the particle is at rest. Hence force on the particle is zero.
(b) Impulse at $t=0$. Here $u=0, v=0.75 \mathrm{~ms}^{-1}$, $M=4 \mathrm{~kg}$
$\therefore$ Impulse $=$ total change in momentum
$=M v-m u$
$=M(v-u)=4(0.75-0)=3 \mathrm{~kg} \mathrm{~ms}^{-1}$
Impulse at $t=4 s$. Here $u=0.75 \mathrm{~ms}^{-1}, v=0$
$\therefore$ Impulse $=M(v-u)=4(0-0.75)$
$=-3 \mathrm{~kg} \mathrm{~ms}^{-1}$.

Topic: Physics, Laws of Motion
Figure shows the variation in temperature $(\Delta \mathrm{T})$ with the amount of heat supplied $\mathrm{(Q)}$ in an isobaric process corresponding to a monoatomic $\mathrm{(M)}$, diatomic $\mathrm{(D)}$ and a polyatomic $\mathrm{(P)}$ gas. The initial state of all the gases are the same and the scales for the two axes coincide. Ignoring vibrational degrees of freedom, the lines $a, b$ and $c$ respectively correspond to :

Options:
A
$\mathrm{P}$, $\mathrm{M}$ and $\mathrm{D}$

B
$\mathrm{M}$, $\mathrm{D}$ and $\mathrm{P}$

C
$\mathrm{P}$, $\mathrm{D}$ and $\mathrm{M}$

D
$\mathrm{D}$, $\mathrm{M}$ and $\mathrm{P}$

The correct answer is:
$\mathrm{M}$, $\mathrm{D}$ and $\mathrm{P}$

Solution:
On giving same amount of heat at constant pressure, there is no change in temperature for mono, dia and polyatomic.
$(\Delta \mathrm{Q})_{\mathrm{P}}=\mu \mathrm{C}_{\mathrm{p}} \Delta \mathrm{T}\left(\mu=\frac{\text { No. of molecules }}{\text { Avogedro's no. }}\right)$
or $\quad \Delta \mathrm{T} \propto \frac{1}{\text { no. of molecules }}$

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2013 (09 Apr Online)
Figure shows the variation of the internal energy U with the density Ï of an ideal monoatomic gas for a thermodynamic process AB. Process AB is a part of a rectangular hyperbola. Find the work done (in joule) by gas in process AB.

Options:

The correct answer is:
2

Solution:
Suppose final internal energy is x
Process AB is isobaric process
Uρ = constant
2×x=5 ×2
x=5 J
In process AB
ΔU=nfRΔT2
n = number of mole
f = degree of freedom
ΔT = Change in temperature
5-2=n ×3 ×R ×ΔT2
63=nRΔT
2=nRΔT
Work done = ∫PdV
=PΔV
=nRΔT
=2 J

Topic: Physics, Thermodynamics, JEE Main
Figure shows three regions of magnetic field, each of area \( A \) and in each region, a magnitude of magnetic field decreases at a constant rate \( \alpha \). If \( \vec{E} \) is induced electric field, then value of line integral \( \oint \vec{E} \cdot d r \) along the given loop is equal to

Options:
A \( \alpha A \)
B \( -\alpha A \)
C \( 3 \alpha A \)
D \( -3 \alpha A \)

The correct answer is:
\( -\alpha A \)

Solution:
From Faraday's law, induced emf is given by, e=-dϕdt.
For the case of changing magnetic field, an induced electric field is generated and is given by,

⇒∮E→·dr→=-dBAdt
⇒∮E→·dr→=-AdBdt
⇒∮E→·dr→∝Aα,

where -dBdt=α

Topic: Physics, Electromagnetic Induction, JEE Main
Figure shows two spherical surface of radii R and 2R separating three transparent media of refractive index Î¼=4, Î¼=2 and Î¼=1. A ray of light travelling in a medium Î¼=1 is incident on outer sphere tangentially. The net deviation suffered by light ray will be

Options:
A 0°
B 90°
C 240°
D 280°

The correct answer is:
240°

Solution:
Path of light ray for minimum deviation will be
sin⁡90°=2sin⁡θ
⇒θ=30°
Deviation=60°
There are 4 such deviations therefore total deviation=240°.

Topic: Physics, Ray Optics, JEE Main
Figures (a), (b), (c) and (d) show variation of force with time.

The impulse is highest in figure.

Options:
A Fig (c)
B Fig (b)
C Fig (a)
D Fig (d)

The correct answer is:
Fig (b)

Solution:
As we know, impulse is the area under the curve of the force vs. time graph. So, analysing each figure area under curve.

a A1=12×1.0×0.5=14 N‐s

b A2=0.5×2=1 N‐s

c A3=12×1×0.75=38 N‐s

d A4=12×2×0.5=12 N‐s

From above calculations, area under curve in figure (b) is maximum. Thus, fig (b) impulse is highest.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2023 (01 Feb Shift 2)
Filament, hook and basal body are the parts of

Options:
A Flagella in prokaryotes
B Flagella in eukaryotes
C Fimbriae
D Flagella in both eukaryotes and prokaryotes

The correct answer is:
Flagella in prokaryotes

Solution:
Motility in Bacteria is due to presence of thin extensions called as Flagella. The number and arrangement of flagella varies in different types of bacteria. Flagella is made up of three main parts: Filament: Longest and main part that extends from cell surface to the outside and is composed of flagellin protein. Hook: Structure near the cell surface. Basal body: A system of rings embedded in cell envelope. The eukaryotes also possess flagella but these are structurally different from that of the prokaryotic flagella. Their core called the axoneme, possesses a number of microtubules running parallel to the long axis. The axoneme usually has nine pairs of doublets of radially arranged peripheral microtubules, and a pair of centrally located microtubules. Such an arrangement of axonemal microtubules is referred to as the 9+2 array. The central tubules are connected by bridges and is also enclosed by a central sheath, which is connected to one of the tubules of each peripheral doublets by a radial spoke. Thus, there are nine radial spokes. The peripheral doublets are also interconnected by linkers.

Topic: Biology, Biological Classification, NEET
Fill in the blanks
(i) Ca3P2+6HCl→3CaCl2+....p....
(ii) P4+3NaOH+3H2O→.....q...+3NaH2PO2
(iii) PH4I+KOH→KI+H2O+....r....
p, q and r respectively are

Options:
A PH3,H3PO3,PI3
B PH3,PH3,PH3
C PCl3,H3PO4,PH3
D PCl5,PH3,P4O6

The correct answer is:
PH3,PH3,PH3

Solution:
(i) Ca3P2+6HCl→3CaCl2+2PH3(p)
(ii) P4+3NaOH+3H2O→PH3(q)+3NaH2PO2
(iii) PH4I+KOH→KI+H2O+PH3(r)

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Fill in the reagents for the given conversion
CH3COCl→XCH3CHO→YCH3-C|HOH-CH2CHO→ZCH3CH=CHCHO

Options:
A

X	Y	Z
Pd/BaSO4	dil. NaOH	heat

B

X	Y	Z
NaOH	Hydrolysis	heat

C

X	Y	Z
I2/NaOH	LiAlH4	H3O+

D

X	Y	Z
CrO3	Warm	CO2

The correct answer is:

X	Y	Z
Pd/BaSO4	dil. NaOH	heat

Solution:
CH3COCl→Pd/BaSO4(X)CH3CHO→ dil. NaOH (Y)CH3-C|HOH-CH2CHO(Aldol)→ heat (Z)CH3CH=CHCHOCrotonaldehyde

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Final product D in the sequence of reaction is

Options:
A
B
C
D

The correct answer is:

Solution:
Here propan-2-ol is converted into 4-methyl-pentan-2-ol.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Final product (p) in the sequence of reaction is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
Find A, B and C in the following reactions:

NH3+A+CO2→NH42CO3

NH42CO3+H2O+B→NH4HCO3

NH4HCO3+NaCl→NH4Cl+C

Options:
A A-H2O; B-CO2; C-NaHCO3
B A-H2O; B-O2; C-Na2CO3
C A-H2O; B-O2; C-NaHCO3
D A-O2; B-CO2; C-Na2CO3

The correct answer is:
A-H2O; B-CO2; C-NaHCO3

Solution:
Ammonia reacts with carbon dioxide and water to form ammonium bicarbonate.

2NH3+H2O(A)+CO2⟶NH42CO3

NH42CO3+H2O+CO2(B)⟶2NH4HCO3

Ammonium carbonate reacts with sodium chloride to form sodium bicarbonate and ammonium chloride.

NH4HCO3+NaCl⟶NaHCO3(C)+NH4Cl

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2021 (26 Feb Shift 1)
Find correct order of acidic strength in the following?

Options:
A P > Q > R > S
B P > R > Q > S
C R > P > Q > S
D R > Q > P > S

The correct answer is:
P > R > Q > S

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
Find empirical formula of the compound containing a metal, $M=68 \%$ (atomic mass =34) and remaining $32 \%$ oxygen.

Options:
A $M O$
B $M_2 \mathrm{O}$
C $\mathrm{MO}_2$
D $M_2 \mathrm{O}_3$

The correct answer is:
$M O$

Solution:
$\therefore \quad$ Empirical formula $=M O$

Topic: Chemistry, Some Basic Concepts of Chemistry, AIIMS, AIIMS 2019 (25 May)
Find number of possible triplets \( (\alpha, \beta, \gamma) \),if \( A=\left|\begin{array}{ccc}0 & \alpha & \alpha \\ 2 \beta & \beta & -\beta \\ \gamma & -\gamma & \gamma\end{array}\right| \) is an orthogonal matrix.

Options:
A \( 8 \)
B \( 6 \)
C 4
D \( 2 \)

The correct answer is:
\( 8 \)

Solution:
AAT=I
0αα2ββ-βγ-γγ02βγαβ-γα-βγ=100010001
⇒ 2a2α006β20003γ2=100010001
⇒ 2a2=1, 6β2=1, 3γ2=1
∴ α±12, β=±16, γ=±13

Topic: Mathematics, Matrices, JEE Main
Find out EÂ° for F - → 1 2 F 2  +  e − ; if F 2  +  2e − → 2F − ; E o  = + 2.7 V .

Options:
A 1.35 V
B -1.35 V
C 2.7 V
D -2.7 V

The correct answer is:
-2.7 V

Solution:
E° values do not depend upon stoichiometry of change but depends upon the direction,
i.e., EOPo=-ERPo

Topic: Chemistry, Electrochemistry, JEE Main
Find out the compounds that will disproportionate in their aqueous solution.
IClO4-⁡→+0.36VClO3-⁡→+0.33VClO2→+0.66VOCl-→+0.40V12Cl2→+1.36VCl-
IIMnO4-⁡→+0.56VMnO42-⁡→+2.22VMnO2→+0.95VMn3+→+1.55VMn2+→-0.19VMn

Options:
A

(I)	(II)
ClO2, Cl2	MnO42-,Mn3+

B

(I)	(II)
ClO3-,OCl-	MnO2,Mn2+

C

(I)	(II)
OCl- only	MnO2,Mn2+

D

(I)	(II)
ClO3- only	MnO2 only

The correct answer is:

(I)	(II)
ClO2, Cl2	MnO42-,Mn3+

Solution:
If the reaction sequence is written in reduction order then if right side reduction potential is higher than left side reduction potential then that compound will disproportionate.

Topic: Chemistry, Redox Reactions, JEE Main
Find out the correct match from the following table and select the option with all the correct one.

Act	Year

(i)	The National Forest policy	1981

(ii)	The Montreal Protocol	1987

(iii)	The water (Prevention & Control of Pollution) Act	1974

(iv)	The air (Prevention & Control of Pollution) Act	1986

Options:
A (i), (ii) and (iii)
B (ii) and (iv)
C (ii) and (iii) only
D (i), (iii) and (iv)

The correct answer is:
(ii) and (iii) only

Solution:
National forest policy (1988) of India has recommended 33% forest cover for the plains and 67% for the hills. Realising the importance of maintaining the cleanliness of the water bodies, the government of India has passed the Water (prevention and control of pollution) Act, 1974. In India, The air (prevention & Control of Pollution) Act came into force in 1981, but amended in 1987 to include noise as a pollutant. Recognising the deleterious affects of ozone depletion, an international treaty, known as the Montreal Protocol, was signed at Montreal (Canada) in 1987 (effective in 1989) to control the emission of ozone-depleting substances.

Topic: Biology, Environmental Issues, NEET
Find out the double bond equivalent (DBE) value of the given following compound:

Options:

The correct answer is:
11.00

Solution:
The double bond equivalent (DBE) value of the given compound is 11

Topic: Chemistry, General Organic Chemistry, JEE Main
Find out the major product for the following reaction

Options:
A
B
C
D

The correct answer is:

Solution:
In the first step, alcohol group accepts a proton from the hydronium ion, and thus leaves forming a carbocation. As there is a chance for formation of a six-membered ring, a six membered ring is formed with a carbocation. Hydrogen next to the carbocation leaves forming a cycloalkene which is also the saytzeff product. Chain of reactions are as follows:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2023 (29 Jan Shift 2)
Find out the major product from the following reaction.

Options:
A

B

C

D

The correct answer is:

Solution:
Alcohol forms carbocation inthe presence of acidic medium. The carbocation formed from given alcohol undergo rearranegement to get stability. Final it loses proton to get an alkene.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2023 (25 Jan Shift 2)
Find out the major products from the following reaction sequence.

Options:
A
B
C
D

The correct answer is:

Solution:
Cyanide ion acts as a nucleophile and attacks the carbonyl group of the methyl ketone forming a cyanohydrin ion, which on attack by ethanol ion leads to formation of carboxyl group and on further hydrolysis forms an ester. The alkanide ion from the grignard reagent then attacks the carbonyl of ester twice to form a diol with two alcohol groups attached to a 3o carbon. The sequence of reactions are given below:

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2023 (29 Jan Shift 2)
Find out the major products from the following reactions.

Options:
A
B
C
D

The correct answer is:

Solution:
Hydroboration–oxidation is an anti-Markovnikov reaction, with the hydroxyl group attaching to the less-substituted carbon. In an oxymercuration–demercuration reaction, an alkene is treated with mercury(II) acetate, and the product is treated with sodium borohydride. The net result is a Markovnikov addition product in which the OH group bonds to the more substituted carbon atom of the alkene.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2023 (24 Jan Shift 2)
Find out the normality of a solution, when \( 9.8 \mathrm{~g} \) of \( \mathrm{H}_{2} \mathrm{SO}_{4} \) is dissolved in \( 500 \mathrm{~mL} \) of solution.

Options:
A \( 0.4 \)
B \( 0.2 \)
C \( 0.8 \)
D 4. \( 0 \)

The correct answer is:
\( 0.4 \)

Solution:
Normality, N=WeightEquivalent weight×1000V in mL
Equivalent weight is defined as the weight of a compound that contains one equivalent of a proton (for acid) or one equivalent of a hydroxide (for base).
H2SO4+2OH-⇌2H2O+SO42-
Equivalent weight=982=49 g/mol
∴ N=9.8491000500=0.2 2=0.4N.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
Find out the number of chromosomes and the amount of DNA in a spore mother cell of Bryophytes in G2 stage (If the spore contains 5 chromosomes and 5pg of DNA) -

Options:
A 5 Chromosome, 5pg DNA
B 10 Chromosome, 20pg DNA
C 10 Chromosome, 10pg DNA
D 20 Chromosomes, 20pg DNA

The correct answer is:
10 Chromosome, 20pg DNA

Solution:
Spore mother cell is diploid and spore is haploid. If spore has 5 chromosomes and 5pg DNA than the spore mother cell has 10 chromosomes and 10pg DNA in the G1 phase. Since DNA replication occurs during S phase but chromosome number remains unchanged. Thus chromosome in the G2 phase is 10 and the quantity of DNA is 20 pg.

Topic: Biology, Cell Cycle and Cell Division, NEET
Find out the % of oxalate ion in given sample of oxalate salt of which 0.3 g is present in 100 mL of solution required 90 mL of N/20 KMnO4 for complete oxidation.

Options:

The correct answer is:
66.00

Solution:
The redox changes are
Mn +7 +5 e â â Mn 2+
C 3+ 2 â2 C 4+ +2 e â
âµmeq. of oxalate ion=meq. of KMnO4
w/EÃ1000=90Ã1/20
Or w 88/2 ×1000= 9 2 ∵ E C 2 O 4 −2 = 88 2
or w C 2 O 4 â2 =0.198g
â´ % of oxalate in sample =(0.198Ã100)/0.3=66%

Topic: Chemistry, Redox Reactions, JEE Main
Find out the percentage dissociation of an acid having conc. of 10 M and dissociation constant 1.0×10-3 .

Options:
A 0.1
B 0.5
C 1.0
D 2.0

The correct answer is:
1.0

Solution:
α=10-310=10-2
%α=10-2×100=1%

Topic: Chemistry, Ionic Equilibrium, JEE Main
Find out the role of microbes in the following and discuss it with your teacher.
(a) Single cell protein (SCP)
(b) Soil

Options:

None

Solution:
SCP - single cell protein is microbial yield/cell crop of bacterial, yeast, algae rich in protein. The protein content of microbial cell is very high. Dried cell of Pseudomonas grown on petroleum product has $69 \%$ protein and these proteins have all essential amino acids.
Soil : Microbes take part in formation and maintenance of soil. They add organic matter to freshly formed soil. The process is called humification. Some microbes act as biofertilizers and biopesticides.

Topic: Biology, Microbes in Human Welfare
Find out the surface charge density at the intersection of point x=3 m plane and x-axis, in the region of uniform line charge of 8 nC m-1 lying along the z-axis in free space.

Options:
A 0.424 nC m-2
B 47.88 nC m-2
C 0.07 nC m-2
D 4.0 nC m-2

The correct answer is:
0.424 nC m-2

Solution:
2 Kλr=σε0 x=3 m

⇒σ=0.424×10-9 C m-2

Topic: Physics, Electrostatics, JEE Main, JEE Main 2021 (16 Mar Shift 2)
Find out torque about points $A, O$ and $B$ if $\vec{F}=10 \hat{i}$

Options:
A $(0,50 \hat{k},-50 \hat{k})$
B $(0,-50 \hat{k}, 50 \hat{k})$
C $(0,-50 \hat{k},-50 \hat{k})$
D $(0,50 \hat{k}, 50 \hat{k})$

The correct answer is:
$(0,-50 \hat{k},-50 \hat{k})$

Solution:
The correct option is $\mathbf{C}$
$(0,-50 \hat{k},-50 \hat{k})$
Torque about point $A, \vec{r}_A=\vec{r}_A \times \vec{F}, \vec{r}_A=3 \hat{i} \vec{F}=10 \hat{i}$
$\vec{r}_A=3 \hat{i} \times 10 \hat{i}=0$
Torque about point $B, \vec{r}_B=\vec{r}_B \times \vec{F}, \vec{r}_B=5 \hat{j}, \vec{F}=10 \hat{j}$
$\vec{r}_B=5 \hat{j} \times 10 \hat{i}=-50 \hat{k}$
Torque about point $O, \vec{r}_0=\vec{r}_0 \times \vec{F}, \vec{r}_0=3 \hat{i}+5 \hat{j}, \vec{F}=10 \hat{i}$
$\vec{r}_0=(3 \hat{i}+5 \hat{j}) \times 10 \hat{i}=-50 \hat{k}$

Topic: Physics, Rotational Motion, JEE Main
Find out total number of compound(s) in which at least half of Cl- are ionizable
CrCl3.6NH3,CrCl3.5NH3,CrCl3.4NH3,CrCl3.3NH3,PtCl4.6NH3,PtCl4.5NH3,
PtCl4.4NH3,PtCl4.3NH3,PtCl4.2NH3 .

Options:

The correct answer is:
5

Solution:
Total number of compound(s) in which at least half of Cl- are ionizable
CrCl 3 .6NH 3 ,â CrCl 3 .5NH 3 ,â PtCl 4 .6NH 3 ,â PtCl 4 .5NH 3 ,â PtCl 4 .4NH 3

Topic: Chemistry, Coordination Compounds, JEE Main
Find out total number of compounds which are more stable in its ionic form

Options:

The correct answer is:
4

Solution:
, ,
,

Topic: Chemistry, General Organic Chemistry, JEE Main
Find the angle between the diagonals of parallelogram $P Q R S$, if $\mathbf{P Q}=3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $\mathbf{P S}=\hat{\mathbf{i}}-2 \hat{\mathbf{k}}$

Options:
A Only $\cos \theta=-\sqrt{\frac{3}{10}}$
B Both $\cos \theta= \pm \sqrt{\frac{3}{10}}$
C $\tan \theta=-\sqrt{\frac{3}{10}}$
D $\tan \theta=-\sqrt{\frac{11}{10}}$

The correct answer is:
Only $\cos \theta=-\sqrt{\frac{3}{10}}$

Solution:
$$
\text { 60. (a) } \begin{aligned}
\mathbf{P Q} & =3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\
\mathbf{P S} & =\hat{\mathbf{i}}-2 \hat{\mathbf{k}}
\end{aligned}
$$
$$
\begin{aligned}
& \text { Eqs. (i) }+ \text { (ii) } \Rightarrow \mathbf{P Q}+\mathbf{P S}=4 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\mathbf{0} \hat{\mathbf{k}} \\
& \mathbf{P R}=4 \hat{\mathbf{i}}-2 \hat{\mathbf{j}} \\
& \text { Eqs. (i) }- \text { (ii) } \Rightarrow \mathbf{P Q}-\mathbf{P S}=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \\
& -\mathbf{Q S}=\mathbf{2} \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \\
& \Rightarrow \quad \mathbf{Q S}=-2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}} \\
&
\end{aligned}
$$
Let $\boldsymbol{\theta}$ be the angle between diagonal $\mathbf{P R}$ and $\mathbf{Q S}$
$$
\begin{aligned}
\therefore \quad \cos \theta & =\frac{\mathbf{P R} \cdot \mathbf{Q S}}{|\mathbf{P R}||\mathbf{Q S}|} \\
& =\frac{(4 \hat{\mathbf{i}}-2 \hat{\mathbf{j}})(-2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}})}{\sqrt{16+4} \sqrt{4+4+16}} \\
& =\frac{-8-4}{\sqrt{20} \cdot \sqrt{24}}=\frac{-12}{2 \sqrt{5} \cdot 2 \sqrt{6}} \\
& =\frac{-3}{\sqrt{30}}=-\sqrt{\frac{9}{30}}=-\sqrt{\frac{3}{10}} \\
\therefore \quad \cos \theta & =-\sqrt{\frac{3}{10}}
\end{aligned}
$$
Hence, option (1) is correct.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Find the angle between the two planes $2 \mathrm{x}+\mathrm{y}-2 \mathrm{z}=5$ and $3 x-6 y-2 z=7$

Options:
A $\cos ^{-1}(4 / 21)$
B $\cos ^{-1}(2 / 21)$
C $\cos ^{-1}(1 / 21)$
D $\cos ^{-1}(5 / 21)$

The correct answer is:
$\cos ^{-1}(4 / 21)$

Solution:
The angle between two planes is the angle between their normals. From the equation of the planes, the normal vectors are $\overrightarrow{\mathrm{N}}_{1}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{N}}_{2}=3 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}$

Therefore, $\cos \theta=\left|\frac{\overrightarrow{\mathrm{N}}_{1} \cdot \overrightarrow{\mathrm{N}}_{2}}{\left|\overrightarrow{\mathrm{N}}_{1}\right|\left|\overrightarrow{\mathrm{N}}_{2}\right|}\right|$

$$

=\left|\frac{(2 \hat{i}+\hat{j}-2 \hat{k}) \cdot(3 \hat{i}-6 \hat{j}-2 \hat{k})}{\sqrt{4+1+4} \sqrt{9+36+4}}\right|=\left(\frac{4}{21}\right)

$$

Hence, $\theta=\cos ^{-1}\left(\frac{4}{21}\right)$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Find the angle between the vectors $\mathbf{A}=2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$ and $\mathbf{B}=4 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}$.

Options:
A $0^{\circ}$
B $45^{\circ}$
C $60^{\circ}$
D $90^{\circ}$

The correct answer is:
$90^{\circ}$

Solution:
Given, $\mathbf{A}=2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$ and $\mathbf{B}=4 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}$
Let $\theta$ be the angle between $\mathbf{A}$ and $\mathbf{B}$.
By using dot product,
$\mathbf{A} \cdot \mathbf{B}=A B \cos \theta \text {, }$
where, $A$ and $B$ are magnitudes of $\mathbf{A}$ and $\mathbf{B}$ respectively.
Now,
$\begin{aligned}
& A=\sqrt{2^2+4^2+4^2}=6 \\
& B=\sqrt{4^2+2^2+(-4)^2}=6
\end{aligned}$
Substituting the values, in dot product, we get
$\begin{aligned}
& (2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \cdot(4 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}})=(6)(6) \cos \theta \\
& 8+8-16=36 \cos \theta \\
& \Rightarrow \quad \cos \theta=0 \\
& \Rightarrow \quad \theta=90^{\circ}
\end{aligned}$

Topic: Physics, Mathematics in Physics, JEE Main
Find the angle in degree between two lines whose direction cosines are given by \( \ell+\mathrm{m}+n=0, \ell^{2}+\mathrm{m}^{2}-n^{2}=0 \)

Options:

The correct answer is:
60

Solution:
Eliminating n between the given relations, we get
â 2 + m 2 - - â - m 2 = 0
or â m = 0 â â = 0 or m = 0
Now â = 0
â 1 . â + 0 . m + 0 . n = 0 and â + m + n = 0
60
Solving these we get
â 0 = m - 1 = n 1 ....(i)
Similarly m = 0 â 0 . â + 1 . m + 0 . n = 0 and â + m + n = 0
Solving these we get â 1 = m 0 = n - 1 ...... (ii)
thus dr's of given lines are 0, -1, 1 and 1, 0, -1
respectively.
Let Î¸ is angle between them
cos Î¸ = 0 . 1 + - 1 . 0 + 1 - 1 0 2 + - 1 2 + 1 2 1 2 + 0 2 + - 1 2 = 1/2

Topic: Mathematics, Three Dimensional Geometry, JEE Main