datavorous/entrance-exam-dataset huggingface dataset:
Dimensional formula for thermal conductivity is (here K denotes the temperature):

Options:
A MLT–2K
B MLT–2K–2
C MLT–3K
D MLT–3K–1

The correct answer is:
MLT–3K–1

Solution:
k=Q/tΔxAΔT

=M1L2T−2LL2KT=M1L1T−3K−1

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2020 (04 Sep Shift 1)
Dimensional formula of magnetic flux is

Options:
A ML2T−2A−1
B ML0T−2A−2
C M0L−2T−2A−3
D ML2T−2A3

The correct answer is:
ML2T−2A−1

Solution:
φ=BA=FI×LA=[MLT−2][L2][A][L]=[ML2T−2A−1]

Topic: Physics, Units and Dimensions, JEE Main
Dimensions of permeability (μ0) are

Options:
A A−2M1L1T−2
B MLT−2
C ML0T−1
D A−1MLT2

The correct answer is:
A−2M1L1T−2

Solution:
F= μ 0 4π 2 I 1 I 2 l r ⇒ μ 0 =[F] [A] −2 =[ML T −2 A −2 ]

Topic: Physics, Units and Dimensions, JEE Main
Dimensions of resistance in an electrical circuit, in terms of dimension of mass $M$, of length $L$, of time $T$ and of current $I$, would be

Options:
A $M L^2 T^{-2}$
B $M L^2 T^{-1} I^{-1}$
C $M L^2 T^{-3} I^{-2}$
D $M L^2 T^{-3} I^{-1}$

The correct answer is:
$M L^2 T^{-3} I^{-2}$

Solution:
We know that
$\begin{aligned}
R & =\frac{V}{I}=\frac{\left[M L^2 T^{-2}\right]}{\left[I^2 T\right]} \\
& =\left[M L^2 T^{-3} I^{-2}\right]
\end{aligned}$

Topic: Physics, Units and Dimensions, JEE Main
Dimethyl ether when heated with excess $\mathrm{HI}$ gives

Options:
A $\mathrm{CH}_3 \mathrm{I}$ and $\mathrm{CH}_3 \mathrm{OH}$
B $\mathrm{CH}_3 \mathrm{I}$ and $\mathrm{H}_2 \mathrm{O}$
C $\mathrm{C}_2 \mathrm{H}_6+\mathrm{CH}_3 \mathrm{I}$ and $\mathrm{CH}_3 \mathrm{OH}$
D $\mathrm{CH}_3 \mathrm{l}$ and $\mathrm{HCHO}$

The correct answer is:
$\mathrm{CH}_3 \mathrm{I}$ and $\mathrm{H}_2 \mathrm{O}$

Solution:
$\mathrm{CH}_3-\mathrm{O}-\mathrm{CH}_3+2 \mathrm{HI} \rightarrow 2 \mathrm{CH}_3 \mathrm{I}+\mathrm{H}_2 \mathrm{O}$

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:
\(\mathrm{N}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_3(\mathrm{~g})\)
(i) Calculate the mass of ammonia produced if \(2.00 \times 10^3 \mathrm{~g}\) dinitrogen reacts with \(1.00 \times 10^3 \mathrm{~g}\) of dihydrogen.
(ii) Will any of the two reactants remain unreacted?
(iii) If yes, which one and what would be its mass?

Options:

None

Solution:
(i) Balanced equation for preparation of ammonia is
\(\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_3(\mathrm{~g})\)
\(1 \mathrm{~mol}\) of \(\mathrm{N}_2\), i.e., \(28 \mathrm{~g}\) react with \(3 \mathrm{~mol}\) of \(\mathrm{H}_2\), i.e., \(6 \mathrm{~g}\) of \(\mathrm{H}_2 .2000 \mathrm{~g}\) of \(\mathrm{N}_2\) will react with
\(\mathrm{H}_2=\frac{6}{28} \times 2000 \mathrm{~g}=428.6 \mathrm{~g}\) but the amount of \(\mathrm{H}_2\) actually present is \(1000 \mathrm{~g}\), thus \(\mathrm{H}_2\) is in excess and remain unreacted. Thus, \(\mathrm{N}_2\) is the limiting reagent while \(\mathrm{H}_2\) is the excess reagent.
1 mol of \(\mathrm{N}_2\), i.e., \(28 \mathrm{~g}\) of \(\mathrm{N}_2\) produce \(\mathrm{NH}_3=2 \mathrm{~mol}=34 \mathrm{~g} 2000 \mathrm{~g}\) of \(\mathrm{N}_2\) will produce \(\mathrm{NH}_3\) \(=\frac{34}{28} \times 2000 \mathrm{~g}=2428.57 \mathrm{~g}\)
(ii) \(\mathrm{H}_2\) will remain unreacted.
(iii) Mass left unreacted \(=1000 \mathrm{~g}-428.6 \mathrm{~g}\)
\(=571.4 \mathrm{~g}\)

Topic: Chemistry, Some Basic Concepts of Chemistry
Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation: N2(g) + H2(g) → 2NH3(g) Will any of the two reactants remain unreacted? and if yes, which one and what would be its mass?

Options:
A Yes, 571g Hydrogen will remain unreacted
B No
C Yes, 571g Nitrogen will remain unreacted
D Yes, 475g Hydrogen will remain unreacted

The correct answer is:
Yes, 571g Hydrogen will remain unreacted

Solution:
Correct Option is : (A)
Yes, 571g Hydrogen will remain unreacted

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
Dinitrogen and dioxygen are main constituents of air but these do not react with each other to form oxides of nitrogen because

Options:
A
the reaction is endothermic and requires very high temperature

B
the reaction can be initiated only in presence of a catalyst

C
oxides of nitrogen are unstable

D
$\mathrm{N}_2$ and $\mathrm{O}_2$ are unreactive

The correct answer is:
the reaction is endothermic and requires very high temperature

Solution:
Nitrogen and oxygen do not react with each other at normal temperature. At high altitude when lightning strikes, they combine to form oxides of nitrogen.
$$
\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \stackrel{3000^{\circ} \mathrm{C}}{\longrightarrow} 2 \mathrm{NO}(\mathrm{g})
$$

Topic: Chemistry, Environmental Chemistry
Dinitrogen and dioxygen the main constituents of air do not react with each other in atmosphere to form oxides of nitrogen because

Options:
A N2 is unreactive in the condition of atmosphere.
B Oxides of nitrogen are unstable.
C Reaction between them can occur in the presence of a catalyst.
D The reaction is endothermic and require very high temperature.

The correct answer is:
The reaction is endothermic and require very high temperature.

Solution:
Dinitrogen and dioxygen are the major constituents of air. These gases do not react with each other to form oxides of nitrogen because the reaction is endothermic and requires very high temperature. The dissociation energy of N2 is very high due to the presence of triple bond and it is very stable.

N2+O2⇄1483-2000 K2NO

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2022 (28 Jul Shift 2)
Dinitrogen is a robust compound, but reacts at hig altitude to form oxides. The oxide of nitrogen the can damage plant leaves and retard photosynthes is

Options:
A NO
B NO3-
C NO2
D NO2-

The correct answer is:
NO2

Solution:
Dinitrogen and dioxygen are the main constituents of air. These gases do not react with each other at a normal temperature. At high altitudes when lightning strikes, they combine to form oxides of nitrogen.

N2g+O2g→2NOg

NO reacts instantly with oxygen to give NO2

2NOg+O2g→2NO2g

Rate of production of NO2 is faster when nitric oxide reacts with ozone in the stratosphere

The irritant red haze in the traffic and congested places is due to oxides of nitrogen. Higher concentrations of NO2 damage the leaves of plants and retard the rate of photosynthesis. It is toxic to living tissues also

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2022 (29 Jul Shift 2)
Dinitrogen is used

Options:
A In manufacture of calcium cyanamide
B In cryosurgery
C As a refrigerant
D All of these

The correct answer is:
All of these

Solution:
Dinotrogen is used in preparation of Ca(CN)2 in cryosurgery and as refrigerant

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Dinosaurs became extinct in

Options:
A Silurian period
B Jurassic period
C Triassic age
D Cretaceous period

The correct answer is:
Cretaceous period

Solution:
The giant reptiles became extinct in cretaceous period mainly because of their lack of adaptability, depleted supply of food, predation and large scale destruction of their eggs.

Topic: Biology, Evolution, NEET
Dipole moment is shown by

Options:
A
1,2-dichlorobenzene

B
trans 2, 3-dichloro-2-butene

C
1,4-chlorobenzene

D
trans-1,2-dinitroethene

The correct answer is:
1,2-dichlorobenzene

Solution:
In 1, 2 dichlorobenzene the two dipoles are at $60^{\circ}$ (i.e. unsymmetric). Thus possesses dipole moment.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2012 (26 May Online)
Dipole moment of $\mathrm{HCl}=1.03 \mathrm{D}, \mathrm{HI}=0.38 \mathrm{D}$.
Bond length of $\mathrm{HCl}=1.3 Å$ and $\mathrm{HI}=1.6 Å$. The ratio of fraction of electric charge, $\delta$, existing on each atom in $\mathrm{HCl}$ and $\mathrm{HI}$ is

Options:
A $12: 1$
B $2.7: 1$
C $3.3: 1$
D $1: 3.3$

The correct answer is:
$3.3: 1$

Solution:
Dipole moment,
$(\mu)=\delta \times d$
where, $\delta=$ magnitude of electric charge $d=$ distance between particles (or bond length)
$$
\begin{array}{l}
\therefore \delta=\frac{\mu}{d} \\
\begin{aligned}
\text { or } & \frac{\delta_{\mathrm{HCl}}}{\delta_{\mathrm{HI}}}=\frac{\mu_{\mathrm{HCl}}}{d_{\mathrm{HCl}}} \times \frac{d_{\mathrm{HI}}}{\mu_{\mathrm{HI}}} \\
&=\frac{1.03 \times 1.6}{1.3 \times 0.38}=3.3: 1
\end{aligned}
\end{array}
$$

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Direction cosines of the normal to the plane containing the lines x2=y3=z5 and x-12=y-13=z-15 are

Options:
A 214,-314,514
B 214,-314,114
C 214,-114,114
D 313,-213,1

The correct answer is:
214,-314,114

Solution:
The normal vector to the plane =i^j^k^111235
=2i^-3j^+k^
Hence, direction cosines are 214,-314,114

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Direction ratios of the normal to the plane passing through the points $(0,1,1),(1,1,2)$ and $(-1,2,-2)$ are

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

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

Solution:
$I(1)+m(0)+n(2-1)=0 \Rightarrow I+n=0$
and $I(-2)+m(1)+n(-4)=0 \Rightarrow 2 l-m+4 n=0$
$\therefore I=-\frac{m}{2}=-n$
Hence direction ratios are $(1,-2,-1)$.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Direction:
The question has a paragraph followed by two statements, Statement $-1$ and statement $-2$. Of the given four alternatives after the statements, choose the one that describes the statements.
A thin air film is formed by putting the convex surface of a plane - convex lens over a plane glass plate. With monochromatic light, this film gives an interference pattern due to light reflected from the top (convex) surface and the bottom (glass plate) surface of the film.
Statement-1 : When light reflects from the air-glass plate interface, the reflected wave suffers a phase change of $\pi$.
Statement-2 : The centre of the interference pattern is dark.

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 false. Statement- 2 is true.

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

The correct answer is:
Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.

Solution:
As light enters from air to glass it suffers a phase change on $\pi$ and therefore at centre there will be destructive interference.

Topic: Physics, Wave Optics, JEE Main, JEE Main 2011
Discuss and compare the dipole moments as well as bond angles in \( \mathrm{NCl}_{3} \) and \( \mathrm{PCl}_{3} \).

Options:
A Bond angle \( \mathrm{NCl}_{3}> \) Bond angle \( \mathrm{PCl}_{3} \)
B Bond angle \( \mathrm{NCl}_{3} < \) Bond angle \( \mathrm{PCl}_{3} \)
C Bond angle \( \mathrm{NCl}_{3}= \) Bond angle \( \mathrm{PCl}_{3} \)
D None of these

The correct answer is:
Bond angle \( \mathrm{NCl}_{3}> \) Bond angle \( \mathrm{PCl}_{3} \)

Solution:
Dipole moment: The magnitude of P-Cl bond moment is much larger than that of N-Cl ( greater electronegativity difference in the former case) bond, therefore, PCl3 is more polar than NCl3
Bond angle: Cl-N Cl bond angle is greater than Cl-P Cl bond angle. This is due to smaller bond pair - bond pair repulsion in PCl3.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Discuss the causes and effects of global warming. What measures need to be taken to control global warming?

Options:

None

Solution:
Increase in atmospheric concentration of green house gases has resulted in rise of atmospheric temperature by $0.6^{\circ} \mathrm{C}$ (global warming) in the 20th century. This has been confirmed by intergovernmental panel on climatic change (IPCC) in its reports of 1991 and 1992. This predictable change in near future may affect climate, sea level, range of species distribution, food production as well as fisheries resources in the oceans.
Causes of global warming :
(i) Increase in concentration of greenhouse gases.
(ii) Increase of automobile and use of fossil fuel.
(iii) Deforestation and change in land use.
(iv) CFC and aerosol emission from refiigerator and aeroplane.
(v) Increased particulate matter in lower atmosphere.
Effects of global warming :
(i) $\mathrm{CO}_2$ fertilisation effect.
(ii) Many species of plants, being sensitive to temperature will die with sudden rise in temperature and their place will be taken over by scrub vegetation.
(iii) Loss of biodiversity.
(iv) Rise in sea level.
(v) Possibilities of drought and floods.
(vi) Erruption of plant disease and pests.
(vii) Change in rainfall pattern.
Methods that can reduce the atmospheric concentration of greenhouses gases are
(i) Reducing the greenhouse gas emission by limiting the use of fossil fuels, and by developing alternative renewable sources of energy (wind energy, solar energy etc.)
(ii) Increasing the vegetation cover, mainly the forests, for photosynthetic utilization of $\mathrm{CO}_2$.
(iii) Minimizing the use of nitrogen fertilizers in agriculture for reducing $\mathrm{N}_2 \mathrm{O}$ emissions.
(iv) Developing substitutes for chlorofluorocar-bons.

Topic: Biology, Environmental Issues
Discuss the factors responsible for ascent of xylem sap in plants.

Options:

None

Solution:
Ascent of xylem sap is supported by following factors.
(i) Root pressure (positive pressure that develops in the xylem sap of root).
(ii) Cohesion (adhesion of water molecules due to hydrogen bonding).
(iii) Adhesion (force between tracheary wall and water molecule).
(iv) Transpiration pull (tension develops due to transpiration) is the main cause of ascent of xylem sap.

Topic: Biology, Transport in Plants
Discuss the main purpose of vulcanisation of rubber.

Options:

None

Solution:
Natural rubber has the following disadvantages:
(a) It is soft and sticky and becomes even more so at high temperatures and brittle at low temperatures. Therefore, rubber is generally used in a narrow temperature range \((283-335 \mathrm{~K})\) where its elasticity is maintained.
(b) It has large water absorption capacity, has low tensile strength and low resistance to abrasion.
(c) It is not resistant to the action of organic solvents.
(d) It is easily attacked by oxygen and other oxidising agents.
To improve all these properties, natural rubber is vulcanised by heating it with about \(5 \%\) sulphur at \(373-415 \mathrm{~K}\). The vulcanized rubber thus obtained has excellent elasticity over a larger range of temperature, has low water absorption tendency and is resistant to the action of organic solvents and oxidising agents.

Topic: Chemistry, Polymers
Discuss the main steps in the digestion of proteins as the food passes through different parts of the alimentary canal.

Options:

None

Solution:
(a) In stomach the proenzyme pepsinogen, on exposure to $\mathrm{HCl}$ converted into active pepsin that converts proteins into proteoses and peptones.
Protein $\stackrel{\text { Pepsin }}{\longrightarrow}$ Protease, peptones, peptides
(b) Proteins, proteoses and peptides in the chyme reaching the intestine are acted upon by proteolytic enzymes of pancreatic juice and converted to dipeptides.
$\stackrel{\text { Large peptides }}{\stackrel{\text { Carboxypeptidase }}{\longrightarrow}}$
Dipeptide + Amino acids
(c) The enzymes in succuss entericus act on the end product to form amino acids.
Dipeptides $\stackrel{\text { Dipeptidase }}{\longrightarrow}$ Amino acids

Topic: Biology, Digestion and Absorption
Discuss the trends in chemical reactivity of group 15 elements.

Options:
A Reactivity towards oxygen increases down the group.
B Reactivity towards hydrogen decreases down the group.
C Reactivity towards halogens increases down the group.
D Reactivity towards metals remains constant down the group.

The correct answer is:
Reactivity towards oxygen increases down the group.

Solution:
Correct Option is : (A)
Reactivity towards oxygen increases down the group.

Topic: Chemistry, p Block Elements (Group 13 & 14), NEET
Discuss the two main limitations of plant hybridisation programme.

Options:

None

Solution:
The limitations of plant hybridisation programme include the following
(i) Limited number of identified disease resistant genes present in various crop varieties or wild relatives are available.
(ii) Parents should be compatible for hybridisation.
(iii) Process is time consuming and tedious.

Topic: Biology, Strategies for Enhancement in Food Production
Discuss the various types of positive interactions between species.

Options:

None

Solution:
The interspecific intractions are of three types-positive or beneficial, negative or antagonistic and neutral.
Some positive interactions are scavenging, commensalism, protocooperation and mutualism.
Mutualism This type of interaction confers benefits on both the species interacting, Examples include.
(i) Lichens represent an intimate mutualistic relationship between a fungus and photosynthesising algae or cyanobacteria.
(ii) The mycorrhizae are associations between fungi and the roots of higher plants. The fungi help the plant in the absorption of essential nutrients from the soil while the plant in turn provides the fungi with carbohydrates for energy yielding.
(iii) Plants offer nectar, juicy and nutritious fruits to animals which in return pollinate their flowers and disperse their seeds.
Commensalism This is the interaction in which one species benefits the and the other is unaffected.
(i) An orchid growing as an epiphyte on a mango branch.
(ii) Barnacles growing on the back of a whale.
(iii) Sea anemone has stinging tentacles so that the clown fish that lives among them gets protection from predators.
Scavenging is the act of feeding by scavenger like bacteria, fungi on the remains of dead animals.
Protocooperation in the type of relationship in which both partners mutually obtain benefits. But they associate purely to benefit from each other and can live without each other.

Topic: Biology, Organisms and Populations
Discuss with your teacher about
(i) haploid insects and lower plants where cell-division occurs and
(ii) some haploid cells in higher plants where celldivision does not occur.

Options:

None

Solution:
(i) Some insects like honey bee drones are haploid. They are not fertile.
(ii) In lower plant, main plant body is haploid produces haploid microspores by mitosis. The gametes of Chlamydomonas fuse to form diploid zygotes. Meiosis take place at this stage in lower plants forming haploid spous which give rise to new plant.

Topic: Biology, Cell Cycle and Cell Division
Dispersion without deviation is produced by two thin (small angled) prisms which are combined together. One prism has angle \( 5^{\circ} \) and refractive index \( 1.56 \). If the other prism has refractive index \( 1.7 \), what is its angle?

Options:
A \( 3^{\circ} \)
B \( 4^{\circ} \)
C \( 5^{\circ} \)
D \( 6^{\circ} \)

The correct answer is:
\( 4^{\circ} \)

Solution:
Dispersion without deviation by combining two prisms of angles A1 and A2 with refractive indices n1 and n2 is given by

(n1−1)A1=(n2−1)A2 ...i

Here n1=1.56, A1=5°, n2=1.7

From equation i,

⇒ 1.56−1×5=1.7−1×A2

⇒ A2=4°

Topic: Physics, Ray Optics, JEE Main
Displacement x versus t2 graph is shown for a particle. The acceleration of the particle is

Options:
A 4 m s-2
B 8 m s-2
C Zero
D 2 m s-2

The correct answer is:
4 m s-2

Solution:
From the graph,
x=Slope(t2)
x=2-01-0(t2)
x=2t2
dxdt=4t
d2xdt2=4
a=4 m s-2 ∵ d2xdt2=a

Topic: Physics, Motion In One Dimension, JEE Main
Dissolution of sodium sulphate in water is exothermic (with evolution of heat). If the temperature is raised for a saturated solution of sodium sulphate, then according to Le-Chatelier principle

Options:
A The solution will become supersaturated
B Some solid will precipitate out from the solution
C More solid will dissolve
D Solution concentration will remain unchanged

The correct answer is:
Some solid will precipitate out from the solution

Solution:
When temperature increases precipitation of sodium sulphate takes place. Because reaction is exothermic so reverse reaction will take place.

Topic: Chemistry, Chemical Equilibrium, NEET
Dissolving 120 g of a compound of (mol. wt. 60) in 1000 g of water gave a solution of density 1.12 g/mL. The molarity of the solution is :

Options:
A 2.50 M
B 1.00 M
C 2.00 M
D 4.00 M

The correct answer is:
2.00 M

Solution:
Mass of compound =120g ;   moles= 120g 60g/mole =2
Mass of solution = 1 0 0 0 + 1 2 0 = 1 1 2 0 g
Volume of solution mass d = 1120g 1.12g/mL =1000mL
Molarity = 2 moles 1 0 0 0 mL Ã 1 0 0 0 mL
= 2 M (no of moles/Litre)

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2014 (09 Apr Online)
Distance between parallel planes $2 x-2 y+z+3=0$ and $4 x-4 y+2 z+5=0$ is

Options:
A $\frac{2}{3}$
B $\frac{1}{3}$
C $\frac{1}{6}$
D 2

The correct answer is:
$\frac{1}{6}$

Solution:
The required distance is given by
$\left|\frac{3}{\sqrt{2^2+2^2+1^2}}-\frac{5}{\sqrt{4^2+4^2+2^2}}\right|=\left|1-\frac{5}{6}\right|=\frac{1}{6}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Distance between two non-intersecting planes P1 and P2 is 5 units, where P1 is 2x-3y+6z+26=0 and P2 is 4x+by+cz+d=0. The point A-3,0,-1 lies between the planes P1 and P2, then the value of 3b+4c-5d is equal to

Options:
A 580
B 120
C -18
D -120

The correct answer is:
120

Solution:
Since both the planes are parallel
P1:4x-6y+12z+52=0
P2:4x-6y+12z+d=0
Therefore, d-5214=5âd-52=70
âd=122,-18
â´P2 is 4x-6y+12z+122=0 or
4x-6y+12z-18=0
Since the point -3,0,-1 is lying between P1 and P2
â´ On substituting the point in both the equations of the plane, both the expressions must be of opposite signs
Hence, 4x-6y+12z-18=0 is the equation of the required plane

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Distance between two parallel planes 2x+y+2z=8 and 4x+2y+4z+5=0 is

Options:
A 7 2
B 9 2
C 3 2
D 5 2

The correct answer is:
7 2

Solution:
Distance between
2x+y+2z=8 and
2x+y+2z= − 5 2 is
d=8+5222+12+22
d=72

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2013 (07 Apr)
Distance between two parallel planes $2 x+y+2 z=8$ and $4 x+2 y+4 z+5=0$ is

Options:
A
$\frac{3}{2}$

B
$\frac{5}{2}$

C
$\frac{7}{2}$

D
$\frac{9}{2}$

The correct answer is:
$\frac{7}{2}$

Solution:
Given planes are
$$
2 x+y+2 z-8=0,4 x+2 y+4 z+5=0 \Rightarrow 2 x+y+2 z+5 / 2=0
$$
Distance between planes $=\frac{\left|d_1-d_2\right|}{\sqrt{a^2+b^2+c^2}}=\frac{|-8-5 / 2|}{\sqrt{2^2+1^2+2^2}}=\frac{7}{2}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2004
Distance of the centre of mass of a solid uniform cone from its vertex is z0. If the radius of its base is R and its height is h then z0 is equal to:

Options:
A 3h28R
B h24R
C 3h4
D 5h8

The correct answer is:
3h4

Solution:
Let the mass of solid cylinder is M and angle with vertical is θ. Now, we take a small element of thickness dy at a distance y from the vertex. The radius of small element is r, as shown in the figure below.

Then, tanθ=Rh=ry which will give, r=ytanθ. The density of the cone is given by,

ρ=M13πR2h=3MπR2h.

The mass of the small element is given by,

dM=ρdV

dM=3MπR2h×πr2dy=3My2tan2θR2hdy

The center of mass of the cone is given by,

z0=1M∫0hydM

z0=1M∫0hy·3My2tan2θR2hdy

z0=3R2hRh2y440h

z0=3h4.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Distance of the centre of mass of a solid uniform cone from its vertex is z0. If the radius of its base is R and its height is h then z0 is equal to:

Options:
A 3h28R
B h24R
C 3h4
D 5h8

The correct answer is:
3h4

Solution:
Let the mass of solid cylinder is M and angle with vertical is θ. Now, we take a small element of thickness dy at a distance y from the vertex. The radius of small element is r, as shown in the figure below.

Then, tanθ=Rh=ry which will give, r=ytanθ. The density of the cone is given by,

ρ=M13πR2h=3MπR2h.

The mass of the small element is given by,

dM=ρdV

dM=3MπR2h×πr2dy=3My2tan2θR2hdy

The center of mass of the cone is given by,

z0=1M∫0hydM

z0=1M∫0hy·3My2tan2θR2hdy

z0=3R2hRh2y440h

z0=3h4.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2015 (04 Apr)