datavorous/entrance-exam-dataset huggingface dataset:
In an Linear programming problem, the restrictions or limitations under which the objective function is to be optimised are called

Options:
A Constraints
B Objective function
C Decision variables
D More than one of the above

The correct answer is:
Constraints

Solution:
Objective function
- A linear function of two or more variables which has to be maximized or minimized under the given restrictions is called an objective function,
- The variables used in the objective function are called as decision variables.
Constraints:
- These are the restrictions on the variables of an linear programming problem are called as linear constraints.
- The final solution of the objective function must satisfy these constraints.

Topic: Mathematics, Linear Programming, JEE Main
In an LPP, if the objective function $Z=m x+$ ny has the same maximum value on the two corner points of the feasible region and these two corner points are lying on the same line segment of the constraint then the number of points at which Zmax occurs is?

Options:
A 2
B Infinite
C Finite
D More than one of the above

The correct answer is:
Infinite

Solution:
Given: The objective function $Z=m x+n y$ has the same maximum value on the two corner points of the feasible region.
ion drawn with the feasible region shaded as,
- The corner points of the feasible region are:
C(15,15), B(5,5), M(10,0) \text { and } N(60,0)$
- There is no change in corner points occurs due to extra constraints.
$\begin{array}{|c|l|} \hline \text{Corner Points of feasible region} & z \\ \hline C(15,15) & 6 \\ \hline B(5,5) & 0 \\ \hline M(10,0) & 2 \\ \hline N(60,0) & 0 \\ \hline \end{array}$
- The maximum value of $x+3 y$ will occur at two points $C$ and $N$.
- Now check whether there is a possibility of multiple solutions.
- For that join the points $\mathrm{C}$ and $\mathrm{N}$. If the points $\mathrm{C}$ and $\mathrm{N}$ are lying at the same line segment $\mathrm{CN}$ so for all the points on that line segment $\mathrm{CN}$, the value of the objective function will be a maximum of 60 .
- Since corner points $\mathrm{C}$ and $\mathrm{N}$ are lying on the same line segment $\mathrm{CN}$ so, at every point on the line segment $\mathrm{CN}$, we will get the maximum value of the objective function.
- So, the correct answer is option 2.

Topic: Mathematics, Linear Programming, JEE Main
In an $n-p-n$ transistor

Options:
A the emitter has higher degree of doping compared to that of the collector
B the collector has higher degree of doping compared to that of the emitter
C both the emitter and collector have same degree of doping
D the base region is most heavily doped

The correct answer is:
the emitter has higher degree of doping compared to that of the collector

Solution:
In an $n-p-n$ or $p-n-p$ transistor, the left hand side thick layer of the transistor is heavily doped known as emitter and right hand side thick layer of the transistor is moderately doped known as collector.

Topic: Physics, Semiconductors, JEE Main
In an n-p-n common emitter (CE) transistor the collector current changes from 5 mA to 16 mA for the change in base current from 100 μA and 200 μA, respectively. The current gain of transistor is _____.

Options:
A 110
B 210
C 0.9
D 9

The correct answer is:
110

Solution:
The change in collector current for the given transistor can be written as

∆Ic=16 mA-5 mA=11 mA

The change in base current for the given transistor can be written as

∆Ib=200 μA-100 μA= 100 μA

The formula to calculate the current gain in CE configuration is given by

β=∆Ic∆Ib ...1

Substitute the values of the known parameters into equation (1) to calculate the required quantity.

β=∆Ic∆Ib=11 mA100 μA=110

Topic: Physics, Semiconductors, JEE Main, JEE Main 2023 (12 Apr Shift 1)
In an NPN transistor circuit, the collector current is $10 \mathrm{~mA}$. If $90 \%$ of the electrons emitted reach the collector, the emitter current (i,) and base current (iB) are given by

Options:
A $i_F=-1 m A, i_B=9 m A$
B $i B=9 m A, i B=-1 m A$
C $i_B=1 m A, i_B=11 \mathrm{~mA}$
D $i_E=11 \mathrm{~mA}, i_B=1 \mathrm{~mA}$

The correct answer is:
$i_E=11 \mathrm{~mA}, i_B=1 \mathrm{~mA}$

Solution:
$i_C=\frac{90}{100} \times i_E \Rightarrow 10=0.9 \times i_E=1.1 \mathrm{~mA}$
Also $i_E=i_B+i_C \Rightarrow i_B=11-10=1 \mathrm{~mA}$.

Topic: Physics, Semiconductors, JEE Main
In an NPN transistor the collector current is 24 mA. If $80 \%$ of electrons reach collector its base current in mA is

Options:
A 36
B 26
C 1.6
D 6

The correct answer is:
6

Solution:
Given $i_c=\frac{80}{100} \times i_e \Rightarrow 24=\frac{80}{100} \times i_e \Rightarrow i_e=30 \mathrm{~mA}$
By using $i_e=i_b+i_c \Rightarrow i_b=30-24=6 \mathrm{~mA}$.

Topic: Physics, Semiconductors, JEE Main
In an optics experiment, with the position of the object fixed, a student varies the position of a convex lens and for each position, the screen is adjusted to get a clear image of the object. A graph between the object distance $u$ and the image distance $v$, from the lens, is plotted using the same scale for the two axes. A straight line passing through the origin and making an angle of $45^{\circ}$ with the x-axis meets the experimental curve at $P$. The coordinates of $P$ will be

Options:
A
$(2 f, 2 f)$

B
$\left(\frac{f}{2}, \frac{f}{2}\right)$

C
$(f, f)$

D
$(4 f, 4 f)$

The correct answer is:
$(2 f, 2 f)$

Solution:
It is possible when object kept at centre of curvature.
$$
\begin{aligned}
& u=v \\
& u=2 f, v=2 f .
\end{aligned}
$$

Topic: Physics, Ray Optics, JEE Main, JEE Main 2009
In an organ pipe whose one end is at x=0, the pressure is expressed by p = p 0 cos 3 Ï x 2 sin 3 0 0 Ï t where x is in meter and t in seconds. The organ pipe can be

Options:
A Closed at one end, open at another with length = 0.5 m
B Open at both ends, length =1 m

C Closed at both ends, length =2 m

D Closed at end, open at another with length =23 m

The correct answer is:
Closed at both ends, length =2 m

Solution:
For x = Î»
3 Ï x 2 = 2 Ï
â Î» = 4 3
at x = 0
there is a pressure antinode. So first end must be close and in case of close at bot end
lâ¡ = n Î» 2 â lâ¡ = 2 n 3
for close at one end
lâ¡ = 2 n + 1 Î» 4 = 2 n + 1 3
So for close at one end possible lengths are
2 n + 1 3 = 1 3 , 1 , 5 3 , 7 3
Open at both ends is not possible for close at both end possible length are
2 n 3 = 2 3 , 4 3 , 2 , 8 3 ...... so only possibility is (3)

Topic: Physics, Waves and Sound, JEE Main
In an oscillating LC circuit the maximum charge on the capacitor is Q. The charge on the capacitor when the energy is stored equally between the electric and magnetic field is

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

B
$\frac{\mathrm{Q}}{\sqrt{3}}$

C
$\frac{Q}{\sqrt{2}}$

D
$Q$

The correct answer is:
$\frac{Q}{\sqrt{2}}$

Solution:
$q=\frac{Q}{\sqrt{2}}$

Topic: Physics, Alternating Current, JEE Main, JEE Main 2003
In an oscillator, for sustained oscillations, Barkhausen criterion is Aβ equal to (A= voltage gain without feedback, β= feedback factor)

Options:
A Zero
B 12
C 1
D 2

The correct answer is:
1

Solution:
For sustained oscillation according to Barkhausen criteria. A.β=1

Topic: Physics, Semiconductors, JEE Main
In an oxidation process, oxidation number

Options:
A Decreases
B Increases
C Does not change
D First increases then decreases

The correct answer is:
Increases

Solution:
In oxidation process oxidation state always increases.

Topic: Chemistry, Redox Reactions, JEE Main
In an unbiased p - n junction electrons diffuse from n-region to p-region because:

Options:
A electrons travel across the junction due to potential difference
B only electrons move from n to p region and not the vice - versa
C electron concentration in n - region is more as compared to that in p - region
D holes in p - region attract them

The correct answer is:
electron concentration in n - region is more as compared to that in p - region

Solution:
In a p-n junction diffusion occurs due to spontaneous movement of majority charge carrier from the region of high concentration to low concentration.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2015 (10 Apr Online)
In angiosperms, functional megaspore develops into

Options:
A embryo sac
B ovule
C endosperm
D pollen sac

The correct answer is:
embryo sac

Solution:
In angiosperms, the unit of female gametophyte (embryo sac) is functional megaspore.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET, NEET 2011 (Mains)
In angiosperms the correct sequence of events in formation of female gametophyte in the ovule is
(A) 3 successive free nuclear divisions in functional megaspores.
(B) Degeneration of 3 megaspores.
(C) Meiotic division in megaspore mother cell.
(D) Migration of 3 nuclei towards each pole.
(E) Formation of wall resulting in seven celled embryo sac.
Choose the correct answer from the options given below.

Options:
A $(B),(C),(A),(D),(E)$
B $(C),(B),(A),(D),(E)$
C $(A),(B),(C),(D),(E)$
D $(C), (E), (A), (D), (B)$

The correct answer is:
$(C),(B),(A),(D),(E)$

Solution:
In angiosperms the correct sequence of events in formation of female gametophyte in the ovule is
- The megaspore mother cell undergoes meiosis and forms four haploid megaspores.(C)
- One remains functional and three degenerate.(B)
- 3 successive free nuclear divisions in functional megaspore take place.(A)
- Migration of 3 nuclei towards each pole.(D)
- Formation of wall resulting in seven celled embryo sac.(E)
The correct sequence is $(C),(B),(A),(D),(E)$

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET, NEET 2023 (Manipur)
In Antirrhinum (Snapdragon), a red flower was crossed with a white flower and in F1 generation, pink flowers were obtained. When pink flowers were selfed, the F2 generation showed white, red and pink flowers. Choose the incorrect statement from the following.

Options:
A The experiment does not follow the principle of dominance.
B Pink colour in F1 is due to incomplete dominance.
C Ratio of F2 is 14 (Red) : 24 (Pink) : 14 (White)
D Law of Segregation does not apply in this experiment.

The correct answer is:
Law of Segregation does not apply in this experiment.

Solution:
Law of Segregation:

A pair of alleles for a pair of contrasting characters are brought together in a hybrid, they remain associated but not blend, and they segregate into gametes during gamete formation. This law is based on the fact that the two factors of a character present in an individual do not get mixed up (blending) and both the traits are recovered as such in the F2 generation though one of these is not seen at the F1 stage. During gametes or spore formation, factors of a pair separate or segregate from each other, so that a gamete carries only one factor of a character. This ensures the purity of gametes. Of course a homozygous parent produces all gametes that are similar while a heterozygous will produce two type gametes each having one factor with equal proportion. This law is universally applicable.

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2019 (All India)
In Antirrhinum, two plants with pink flowers were hybridized. The $\mathrm{F}_1$ plants produced red, pink and white flowers in the proportion of 1 red, 2 pink and 1 white. What could be the genotype of the two plants used for hybridization? Red flower colour is determined by $\mathrm{RR}$, and white by $\mathrm{rr}$ genes.

Options:
A rrrr
B $\mathrm{RR}$
C $\mathrm{Rr}$
D $\mathrm{rr}$

The correct answer is:
$\mathrm{Rr}$

Solution:

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2010 (Mains)
In any triangle $\mathrm{ABC}$, if $\mathrm{a}=\sqrt{2} \mathrm{~cm}, \mathrm{~b}=\sqrt{3} \mathrm{~cm}$ and $\mathrm{c}=\sqrt{5} \mathrm{~cm}$, find area of triangle.

Options:
A $\sqrt{6} / 2$
B $5 \mathrm{sq} \mathrm{cm}$
C $3 \mathrm{sq} \mathrm{cm}$
D $\sqrt{2}$ $\frac{7}{\sqrt{65}}$

The correct answer is:
$\sqrt{6} / 2$

Solution:
We know that, $\cos C=\frac{a^2+b^2-c^2}{2 a b}=0$
then, $\quad \angle \mathrm{C}=\frac{\pi}{2}$
so, $\quad \Delta=\frac{1}{2} a b \sin C$ $[\because \sin C=1]$
$\Delta=\frac{1}{2} \times(\sqrt{2})(\sqrt{3})(1)$
$\Delta=\frac{\sqrt{6}}{2}$ sq. cm

Topic: Mathematics, Properties of Triangles, JEE Main
In aquatic ecosystem, __(i)__ is the major conduit for energy flow, as against in the terrestrial ecosystem, a much larger fraction of energy flows through the __(ii)___ than ___(iii)_ .

Options:
A (i)-GFC; (ii)-DFC; (iii)-GFC
B (i)-DFC; (ii)-GFC; (iii)-DFC
C (i)-GFC; (ii)-GFC; (iii)-DFC
D (i)-DFC; (ii)-DFC; (iii)-GFC

The correct answer is:
(i)-GFC; (ii)-DFC; (iii)-GFC

Solution:
GFC- grazing food chain and DFC- detritus food chain.
Food chain starting with dead organic matter is DFC where was food chain starting with a producer is GFC. DFC begins with decomposers and in aquatic system, decomposition is very low. Thus, GFC forms the main channel of energy flow in it. Where in land ecosystem, decomposition rate is higher as a result more amount of energy flows through DFC than GFC.

Topic: Biology, Ecosystem, NEET
In aqueous alkaline solution, two electrons reduction of $\mathrm{HO}_{2}^{-}$ gives

Options:
A $\mathrm{HO}^{-}$
B $\mathrm{H}_{2} \mathrm{O}$
C $\mathrm{O}_{2}$
D $O_{2}^{-}$

The correct answer is:
$\mathrm{HO}^{-}$

Solution:
In aqueous alkaline solution, two electron reduction of $\mathrm{HO}_{2}^{-}$ gives OH The reaction is as follows:
$\begin{array}{l}
\mathrm{H}_{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HO}_{2}^{-} \\
\mathrm{HO}_{2}^{-}+\mathrm{H}_{2} \mathrm{O}+2 e^{-} \rightleftharpoons 3 \mathrm{OH}
\end{array}$

Topic: Chemistry, Redox Reactions, JEE Main
In aqueous solution glucose remains as

Options:
A Only in open chain form
B Only in pyranoze form
C Only in furanose forms
D In all three forms in equilibrium

The correct answer is:
In all three forms in equilibrium

Solution:
Hints: \(\begin{aligned} & \beta-\mathrm{D}-\text { glucose } \rightleftharpoons \mathrm{D}-\text { glucose } \rightleftharpoons \alpha-\mathrm{D}-\text { glucose } \\ & (\approx 64 \%) \quad\qquad(\text {open chain } \approx 0.02 \%) \qquad~~(\approx 34 \%)\end{aligned}\)

Topic: Chemistry, Biomolecules, WBJEE, WBJEE 2011
In aqueous solution the ionization constants for carbonic acid are $\mathrm{K}_1=4.2 \times 10^{-7}$ and $\mathrm{K}_2=4.8 \times 10^{-11}$
Select the correct statement for a saturated $0.034 \mathrm{M}$ solution of the carbonic acid.

Options:
A
The concentration of $\mathrm{CO}_3^{2-}$ is $0.034 \mathrm{M}$.

B
The concentration of $\mathrm{CO}_3^{2-}$ is greater than that of $\mathrm{HCO}_3^{-}$.

C
The concentration of $\mathrm{H}^{+}$and $\mathrm{HCO}_3^{-}$are approximately equal.

D
The concentration of $\mathrm{H}^{+}$is double that of $\mathrm{CO}_3^{2-}$.

The correct answer is:
The concentration of $\mathrm{H}^{+}$and $\mathrm{HCO}_3^{-}$are approximately equal.

Solution:
$\begin{array}{lll}\mathrm{A} \rightarrow & \mathrm{H}_2 \mathrm{CO}_3 \rightleftharpoons \mathrm{H}^{+}+\mathrm{HCO}_3^{-} & \mathrm{K}_1=4.2 \times 10^{-7} \\ \mathrm{~B} \rightarrow & \mathrm{HCO}_3^{-} \rightleftharpoons \mathrm{H}^{+}+\mathrm{CO}_3^{-2} & \mathrm{~K}_2=4.8 \times 10^{-11}\end{array}$
As $\mathrm{K}_2 \ll \mathrm{K}_1$
All major $\left[\mathrm{H}^{+}\right]_{\text {todal }} \approx\left[\mathrm{H}^{+}\right]_{\mathrm{A}}$
and from I equilibrium, $\left[\mathrm{H}^{+}\right]_{\mathrm{A}} \approx\left[\mathrm{HCO}_3^{-}\right] \approx\left[\mathrm{H}^{+}\right]_{\text {tocal }}$
$\left[\mathrm{CO}_3^{-2}\right]$ is negligible compared to $\left[\mathrm{HCO}_3^{-}\right]$or $\left[\mathrm{H}^{+}\right]_{\text {total }}$

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2010
In Arcata marshes, cleaning of wastewater involves:

Options:
A Utilization of microorganisms only for pollution control
B Filteration chlorination like chemical process
C Removal of dissolved heavy metals through biological process
D Only conventional method sewage treatment

The correct answer is:
Removal of dissolved heavy metals through biological process

Solution:
Town Arcata is situated along the northern coast of California. Along with Humboldt State University, the towns people created an integrated waste water treatement process within a natural system. The cleaning occurs in two stages -
(i) The conventional sedimentation, filtering and chlorine treatements are given. After this stage, lost of dangerous pollutants like dissolved heavy metals still remains. To combat this, an innovative approach was taken and
(ii) The biologists developed a series of six connected marshes over 60 hectares of marshland. Appropriate plants, algae, fungi and bacteria were seeded into this area, which neutralise, absorb and assimilate the pollutants. Hence, as the water flows through the marshes, it gets purified naturally.

Topic: Biology, Environmental Issues, NEET
In arrangement given in figure, if the block of mass $m$ is displaced, the frequency is given by

Options:
A $n=\frac{1}{2 \pi} \sqrt{\left(\frac{k_1-k_2}{m}\right)}$
B $n=\frac{1}{2 \pi} \sqrt{\left(\frac{k_1+k_2}{m}\right)}$
C $n=\frac{1}{2 \pi} \sqrt{\left(\frac{m}{k_1+k_2}\right)}$
D $n=\frac{1}{2 \pi} \sqrt{\left(\frac{m}{k_1-k_2}\right)}$

The correct answer is:
$n=\frac{1}{2 \pi} \sqrt{\left(\frac{k_1+k_2}{m}\right)}$

Solution:
With respect to the block the springs are connected in parallel combination.
$\therefore$ Combined stiffness $k=k_1+k_2$ and $n=\frac{1}{2 \pi} \sqrt{\frac{k_1+k_2}{m}}$

Topic: Physics, Oscillations, JEE Main
In assigning R-S configuration which among the following groups has highest priority?

Options:
A -SO3H
B –COOH
C –CHO
D -C6H5

The correct answer is:
-SO3H

Solution:
-SO3H (Sulphur atomic number = 16)

Topic: Chemistry, General Organic Chemistry, JEE Main
In astronomical telescope, which of the following statement is NOT applicable?

Options:
A The telescope is light weight.
B The telescope is very heavy and bulky.
C The telescope suffers spherical aberration.
D The telescope suffers chromatic aberration.

The correct answer is:
The telescope is light weight.

Solution:
The astronomical telescope is used to observe district images of heavenly bodies like stars, planets etc. It is two lens system first one is an objective lens which has a large aperture and large focal length because it forms the image of a distant object at its focal length, and the second one is the eyepiece, which behaves as a simple magnifier with which to see the image formed by the objective.

The astronomical telescope is quite heavy and bulky. Aberration is a property of optical systems like lenses that makes light to be spread out over a region of space rather than focused to a point. This consists of lenses so, it will suffer spherical and chromatic aberration.

Topic: Physics, Ray Optics, JEE Main
In bacteria, plasmid is:

Options:
A extra chromosomal material
B main DNA
C non-functional DNA
D repetitive gene

The correct answer is:
extra chromosomal material

Solution:
A plasmid is a small, often circular DNA molecule found in bacteria and other cells. Plasmids are separate from the bacterial chromosome and replicate independently of it.
Related Theory
Each bacterial cell typically produces many copies of a plasmid, in contrast to making only one copy of its own chromosome. The fact that plasmids are smaller and in greater number than the host chromosome make plasmids easier to isolate in pure form, which is why researchers commonly use them for studying DNA in the laboratory. Plasmids are thus a fundamental tool of recombinant DNA technology.

Topic: Biology, Biological Classification, NEET, NEET 2002
In barley stem vascular bundles are -

Options:
A Closed and radial
B Open and scattered
C Closed and scattered
D Open and in a ring

The correct answer is:
Closed and scattered

Solution:
No Solution Available

Topic: Biology, Anatomy of Flowering Plants, NEET, NEET 2009 (Mains)
In base vs. Acid titration, at the end point methyl orange is present as

Options:
A quinonoid form
B heterocyclic form
C phenolic form
D benzenoid form

The correct answer is:
quinonoid form

Solution:
Methyl orange is a popular pH indicator that is used in the acid base titration.Unlike a universal indicator, methyl orange does not have a full spectrum of color change, but it has a sharp end point.The colour of the solution changes from yellow to red when methyl orange is used as an indicator.That means in acidic medium it is red and exists in quinonoid form

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2022 (25 Jul Shift 2)
In basic medium, H2O2 exhibits which of the following reactions?
(A) Mn2+→Mn4+
(B) I2→I-
(C) PbS→PbSO4
Choose the most appropriate answer from the options given below :

Options:
A (A),(C) only
B (A) only
C (B) only
D (A), (B) only

The correct answer is:
(A), (B) only

Solution:
In basic medium, oxidising action of H2O2.

Mn2++H2O2→Mn+4+2OH-

In basic medium, reducing action of H2O2

I2+H2O2+2OH-→2I-+2H2O+O2

In acidic medium, oxidising action of H2O2

PbS(s)+4H2O2(aq)→PbSO4( s)+4H2O(l)

Hence correct option (4).

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2021 (18 Mar Shift 2)
In between two point charges a conducting spherical shell having charge Q is placed as shown in the diagram. Assuming all charges to be positive, mark out the correct statement.

Options:
A the charge on outer surface of shell is uniformly distributed
B the charge on outer surface of shell is non-uniformly distributed
C the information is insufficient to find the nature of charge distribution on the outer surface
D none of these

The correct answer is:
the charge on outer surface of shell is non-uniformly distributed

Solution:
To solve this question we have to use the concept that "the resultant electric field at any point inside a conducting shell due to all outside charges is zero, " and that "E at any internal point due to uniformly charged conducting spherical shell is zero."
Consider any point P inside the shell, at P the resultant electric field due to I, II and charge on outer surface of shell must be zero.

EP=EI+EII+Eouter surface charge=0
But E I + E II â 0 . So E outer surface charge â 0 it means charge distribution is outer surface of shell is non-uniform.

Topic: Physics, Electrostatics, JEE Main
In biprism experiment, $6^{\text {th }}$ bright band with wavelength ' $\lambda_1$ ' coincides with $7^{\text {th }}$ dark band with wavelength ' $\lambda_2$ ' then the ratio $\lambda_1: \lambda_2$ is (other setting remains the same)

Options:
A 7 : 6
B 13 : 12
C 12 : 13
D 6 : 7

The correct answer is:
13 : 12

Solution:
$\begin{aligned} & \text { Distance of } 6^{\text {th }} \text { bright band }=\frac{6 \lambda_1 D}{d} \\ & \text { Distance of } 7^{\text {th }} \text { dark band }=\frac{6.5 \lambda_2 D}{d} \\ & \therefore \frac{6 \lambda_1 D}{d}=\frac{6.5 \lambda_2 D}{d} \\ & \therefore \frac{\lambda_1}{\lambda_2}=\frac{6.5}{6}=\frac{13}{12}\end{aligned}$

Topic: Physics, Wave Optics, MHT CET, MHT CET 2021 (24 Sep Shift 1)
In biprism experiment, the $4^{\text {th }}$ dark band is formed opposite to one of the slits. The wavelength of light used is
( $\mathrm{D}=$ distance between source and screen, $d$ = distance between the slits)

Options:
A $\frac{\mathrm{d}^2}{9 \mathrm{D}}$
B $\frac{d^2}{11 D}$
C $\frac{d^2}{14 D}$
D $\frac{\mathrm{d}^2}{7 \mathrm{D}}$

The correct answer is:
$\frac{\mathrm{d}^2}{7 \mathrm{D}}$

Solution:
Concept: In biprism experiment, location of the nth dark band is the location of the nth dark fringe given by,
$y_n=(2 n-1) \frac{\lambda D}{2 d}$
Given, slit width $=\mathrm{d}$, and $\mathrm{y}_4=\frac{\mathrm{d}}{2}$
$\begin{aligned}
& \mathrm{y}_4=\left(\frac{7 \lambda \mathrm{D}}{2 \mathrm{~d}}\right)=\left(\frac{\mathrm{d}}{2}\right) \\
& \therefore \lambda=\left(\frac{\mathrm{d}^2}{7 \mathrm{D}}\right)
\end{aligned}$
Option (D) is correct.

Topic: Physics, Wave Optics, JEE Main
In blast furnace iron oxide is reduced by

Options:
A Silica
B Co
C Carbon
D Lime stone

The correct answer is:
Co

Solution:
$\mathrm{Fe}_2 \mathrm{O}_3+3 \mathrm{CO} \rightarrow 2 \mathrm{Fe}+3 \mathrm{CO}_2$

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main