datavorous/entrance-exam-dataset huggingface dataset:
Find the sum of \( n- \) terms of the series \( \frac{1}{2}, \frac{3}{4}, \frac{7}{8}, \ldots \)

Options:
A \( n-1 \)
B \( n-2^{-n} \)
C \( n-2^{n} \)
D None of these

The correct answer is:
\( n-1 \)

Solution:
As the n-term of the series is =2n-12n=1-2-n.
Now apply sigma to the n term it results in =n-∑0n2-n=n-1

Topic: Mathematics, Sequences and Series, JEE Main
Find the sum of the coefficients âa, b, c and dâ,
in the given balanced chemical reaction, IO3-+aâI-+bâH+âcâH2O+dâI2

Options:

The correct answer is:
17

Solution:
IO3â+ aIâ+ bH+âcH2O + dI2
Step 1: Iâ1âI2oxidation
IO3ââI2reduction
Step 2: 2IO3â+ 12H+âI2+ 6H2O
Step 3: 2IO3â+ 12H++ 10eâI2+ 6H2O
2Iâ âI2+ 2e
Step 4: 2IO3â+ 12H++ 10eââI2+ 6H2O
2IââI2+ 2e5
Step 5: 2IO3â+ 10Iâ+ 12H+â6I2+ 6H2O
IO3â+ 5Iâ+ 6H+â3I2+ 3H2O
On comparing, a = 5, b = 6, c = 3, d = 3

Topic: Chemistry, Redox Reactions, JEE Main
Find the time period of the body (mass \( m \) ), shown in the figure, when displaced slightly in the vertical direction.

Options:
A \( 2 \pi \sqrt{\frac{m}{2 k}} \)
B \( 2 \pi \sqrt{\frac{3 m}{k}} \)
C \( 2 \pi \sqrt{\frac{3 m}{2 k}} \)
D \( \pi \sqrt{\frac{3 m}{k}} \)

The correct answer is:
\( 2 \pi \sqrt{\frac{3 m}{2 k}} \)

Solution:
For spring connected in parallel k+k=2k

For spring in series 1keq=1k+12k=32k

From the given spring arrangement, keq=2k3

As the mass m displaced slightly in the vertical direction it will undergo in SHM.

Time period for SHM is, T=2πmkeq

Therefore, the time period of the body of mass m, when displaced slightly in the vertical direction is T=2π3m2k

Topic: Physics, Oscillations, JEE Main
Find the torque about point $\mathrm{O}$ and $\mathrm{A}$.
$\hat{F}=5 \sqrt{3} \hat{i}+5 \hat{j}$

Options:
A $6 \hat{\mathrm{k}}$
B $-8 \hat{\mathrm{k}}$
C $5 \hat{k}$
D $-5 \sqrt{3 \hat{k}}$

The correct answer is:
$-5 \sqrt{3 \hat{k}}$

Solution:
$\begin{aligned}
& \text { Torque about point } O, \bar{\tau}=\overline{r_0} \times \bar{F}, \overline{r_o}=\hat{i}+\hat{j}, \bar{F}=5 \sqrt{3} \hat{i}+5 \hat{\jmath} \\
& \hat{\mathrm{t}}=(\wedge+\hat{\jmath}) \times(5 \sqrt{3} \hat{i}+5 \hat{\jmath})=5(1-\sqrt{3}) \hat{k}
\end{aligned}$
Torque about point $\mathrm{A}, \bar{\tau}=\overline{\mathrm{r}_{\mathrm{a}}} \times \overline{\mathrm{F}}, \overline{\mathrm{r}_{\mathrm{a}}}=\hat{\jmath}, \overline{\mathrm{F}}=5 \sqrt{3} \hat{1}+5 \hat{\jmath}$
$\bar{\tau}=\hat{\jmath} \times(5 \sqrt{3} \hat{\imath}+5 \hat{\jmath})=5(-\sqrt{3}) \hat{k}$

Topic: Physics, Rotational Motion, JEE Main
Find the total number of feasible products (including stereoisomers) in the following reaction. (No carbocation rearrangement is observed)

Options:

The correct answer is:
4

Solution:
Total no. of stereoisomers = 2n
= 22
= 4

Topic: Chemistry, General Organic Chemistry, JEE Main
Find the value of $-197^{\circ} \mathrm{C}$ temperature in Kelvin.

Options:
A $47 \mathrm{~K}$
B $76 \mathrm{~K}$
C $470 \mathrm{~K}$
D $760 \mathrm{~K}$

The correct answer is:
$76 \mathrm{~K}$

Solution:
$$
\text { Temperature in kelvin, } \mathrm{T}=\mathrm{C}+273=-197+273=76 \mathrm{~K}
$$

Topic: Physics, Thermal Properties of Matter, JEE Main
Find the value of $(7.995)^{1 / 3}$ correct to four decimal places.

Options:
A $1.9995$
B $1.9996$
C $1.9990$
D $1.9991$

The correct answer is:
$1.9996$

Solution:
$$
\begin{array}{l}
(7.995)^{1 / 3}=(8-0.005)^{1 / 3} \\
=2\left(1-\frac{0.005}{8}\right)^{1 / 3} \\
=2\left[1+\left(\frac{1}{3}\right)\left(-\frac{0.005}{8}\right)\right]=2[1-0.000208] \\
=1.9996
\end{array}
$$

Topic: Mathematics, Binomial Theorem, JEE Main
Find the value of $a$, if the distance between the points $P(-3,-14)$ and $Q(a,-5)$ is 9

Options:
A - 3
B +3
C 6
D 9

The correct answer is:
- 3

Solution:
Let the given points be:
$\begin{aligned}
& P(-3,-14)=\left(x_1, y_1\right) \\
& Q(a,-5)=\left(x_2, y_2\right)
\end{aligned}$
Using the distance formula,
$\begin{aligned}
& d=\sqrt{\left[\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2\right]} \\
& P Q=\sqrt{\left[(a+3)^2+(-5+14)^2\right]} \\
& V\left[(a+3)^2+81\right]=9\{\text { from the given \}}
\end{aligned}$
Squaring on both sides,
$\begin{aligned}
& (a+3)^2+81=81 \\
& (a+3)^2=0 \\
& a+3=0 \\
& a=-3
\end{aligned}$

Topic: Mathematics, Straight Lines, JEE Main
Find the value of \(\lim _{x \rightarrow 0} \frac{\sin \left(x^m\right)}{(\sin x)^n}\), given that \(n < m\)

Options:
A 2
B 1
C 0
D \(\infty\)

The correct answer is:
0

Solution:
\(\begin{aligned}
\operatorname{Lim}_{x \rightarrow 0} & \frac{\sin \left(x^m\right)}{(\sin x)^n} ;(n < m) \\
& =\operatorname{Lim}_{x \rightarrow 0} \frac{\frac{\sin x^m}{x^m} x^m}{\left(\frac{\sin x}{x} x\right)^n}=\operatorname{Lim}_{x \rightarrow 0} \frac{\frac{\sin x^m}{x^m}}{\left(\frac{\sin x}{x}\right)^n} x^{m-n} \\
& =\frac{1}{1} \times 0 \quad \{\because m > n\} \\
& =0
\end{aligned}\)
Hence, option (c) is correct.

Topic: Mathematics, Limits, JEE Main
Find the value of $m$ if the points $(5,1),(-2,-3)$ and $(8,2 m)$ are collinear.

Options:
A 19 / 6.
B 19/14
C 15/14
D 15/19

The correct answer is:
19/14

Solution:
Let the given points be:
$\begin{aligned}
& A(5,1)=\left(x_1, y_1\right) \\
& B(-2,-3)=\left(x_2, y_2\right) \\
& C(8,2 m)=\left(x_3, y_3\right)
\end{aligned}$
We know that the area of the triangle formed by collinear points is 0.
So, the area of triangle $A B C=0$
$\begin{aligned}
& \Rightarrow(1 / 2)|5(-3-2 m)-2(2 m-1)+8(1+3)|=0 \\
& \Rightarrow|5(-3-2 m)-2(2 m-1)+8(4)|=0 \\
& \Rightarrow|-15-10 m-4 m+2+32|=0 \\
& \Rightarrow|-14 m+19|=0 \\
& \Rightarrow-14 m+19=0 \\
& \Rightarrow 14 m=19 \\
& \Rightarrow m=19 / 14
\end{aligned}$

Topic: Mathematics, Straight Lines, JEE Main
Find the value of p+q3 for given structure

p = degree of unsturation (DU)
q = number of 2o carbon

Options:

The correct answer is:
3

Solution:
p=2 ring +3π bond =5
q=8,2∘ carbon present in the compound
p+q3=2+83=103=3.33

Topic: Chemistry, General Organic Chemistry, JEE Main
Find the value of $\sin 12^{\circ} \sin 48^{\circ} \sin 54^{\circ}$.

Options:
A $\frac{1}{2}$
B $\frac{1}{4}$
C $\frac{1}{6}$
D $\frac{1}{8}$

The correct answer is:
$\frac{1}{8}$

Solution:
$$
\begin{array}{l}
y^{2}=2 a(x-a) \\
\sin 12^{\circ} \sin 48^{\circ} \sin 54^{\circ} \\
=\sin 12^{\circ} \sin \left(60^{\circ}-12^{\circ}\right) \sin \left(90^{\circ}-36^{\circ}\right) \\
=\frac{\sin 12^{\circ} \sin \left(60^{\circ}-12^{\circ}\right) \sin 72^{\circ} \cos 36^{\circ}}{\sin 72^{\circ}} \\
=\frac{\left[\sin 12^{\circ} \sin \left(60^{\circ}-12^{\circ}\right) \sin \left(60^{\circ}+12^{\circ}\right) \cos 36^{\circ}\right]}{\sin 72^{\circ}} \\
=\frac{\sin 12^{\circ}\left(\sin ^{2} 60^{\circ}-\sin ^{2} 12^{\circ}\right) \cos 36^{\circ}}{\sin 72^{\circ}} \\
=\frac{\sin 12^{\circ}\left(\frac{3}{4}-\sin ^{2} 12^{\circ}\right) \cos 36^{\circ}}{\sin 72^{\circ}} \\
=\frac{\left.3 \sin 12^{\circ}-4 \sin ^{3} 12^{\circ}\right)}{4} \cdot \frac{\cos 36^{\circ}}{\sin 72^{\circ}} \\
=\frac{\sin 72^{\circ}}{24 \cos 36^{\circ}}=\frac{1}{4} \frac{1}{8} \\
=\frac{\sin 72^{\circ}}{8}
\end{array}
$$

Topic: Mathematics, Trigonometric Equations, JEE Main
Find the value of \( \tan ^{-1}\left(-\frac{1}{\sqrt{3}}\right)+\cot ^{-1}\left(\frac{1}{\sqrt{3}}\right)+\tan ^{-1}\left[\sin \left(\frac{-\pi}{2}\right)\right] \)

Options:
A \( -\frac{\pi}{12} \)
B \( +\frac{\pi}{12} \)
C \( -\frac{\pi}{15} \)
D \( +\frac{\pi}{10} \)

The correct answer is:
\( -\frac{\pi}{12} \)

Solution:
tan -1 - 1 3 + cot -1 1 3 + tan -1 sin - π 2
= tan -1 - 1 3 + cot -1 1 3 + tan -1 - sin π 2 âµ sin - θ = - sin θ
= - π 6 + π 3 - tan -1 sin π 2 âµ tan -1 - x = - tan -1 x
= π 6 - tan -1 1 = π 6 - π 4 = - π 1 2 .

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
Find the value of $\tan \left[2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right]$

Options:
A $-1 / 3$
B $-7 / 17$
C $-1 / 2$
D $-1 / 4$

The correct answer is:
$-7 / 17$

Solution:
$\begin{aligned} & 2 \tan ^{-1} \frac{1}{5}=\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{5} \\ &=\tan ^{-1} \frac{\frac{1}{5}+\frac{1}{5}}{1-\frac{1}{5} \cdot \frac{1}{5}}=\tan ^{-1} \frac{2 / 5}{24 / 25}=\tan ^{-1} \frac{5}{12} \\ &=\tan \left(2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right)=\tan \left(\tan ^{-1} \frac{5}{12}-\frac{\pi}{4}\right) \\ &=\frac{\tan \left(\tan ^{-1} \frac{5}{12}\right)-\tan \frac{\pi}{4}}{1+\tan \left(\tan ^{-1} \frac{5}{12}\right) \tan \frac{\pi}{4}}=\frac{-\frac{7}{12}}{\frac{17}{12}}=-\frac{7}{17} \end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
Find the value of x+52 where x = total structural isomers with molecular formula C6H12 containing cyclo propane ring.

Options:

The correct answer is:
5.5

Solution:
x=6
⇒6+52=112=05.50

Topic: Chemistry, General Organic Chemistry, JEE Main
Find the value x+y2 (include sterio isomers)

Options:

The correct answer is:
1.5

Solution:
Value=x+y2=2+12=1.50

Topic: Chemistry, General Organic Chemistry, JEE Main
Find the values of $x$ for which the following functions are identical.
$f(x)=\cos x \text { and } g(x)=\frac{1}{\sqrt{1+\tan ^2 x}}$

Options:
A $x \in[-3,3]$
B $x \in(-\infty,-3] \cup(2,3]$
C $x \in(2,3]$
D $x \in \bigcup_{n \in Z}\left(-\frac{\pi}{2}+2 n \pi, \frac{\pi}{2}+2 n \pi\right)$.

The correct answer is:
$x \in \bigcup_{n \in Z}\left(-\frac{\pi}{2}+2 n \pi, \frac{\pi}{2}+2 n \pi\right)$.

Solution:
$f(x)=\cos x$ has domain $R$ and range $[-1,1]$
$g(x)=\frac{1}{\sqrt{1+\tan ^2 x}}=\frac{1}{\sqrt{\sec ^2 x}}=|\cos x|,$
has domain $R-\{(2 n+1) \pi / 2, n \in Z$ as $\tan x$ is not defined for $x=(2 n+1) \pi / 2, n \in Z$
Also range of $g(x)=|\cos x|$ is $[0,1]$
Hence $f(x)$ and $g(x)$ are identical if $x$ lies in first and fourth quadrant.
$\Rightarrow \quad x \in \bigcup_{n \in \mathcal{Z}}\left(-\frac{\pi}{2}+2 n \pi, \frac{\pi}{2}+2 n \pi\right) .$

Topic: Mathematics, Functions, JEE Main
Find the values of $x$ for which the following functions are identical.
$f(x)=\frac{\sqrt{9-x^2}}{\sqrt{x-2}}$ and $g(x)=\sqrt{\frac{9-x^2}{x-2}}$

Options:
A $x \in[-3,3]$
B $x \in(-\infty,-3] \cup (2,3]$
C $x \in(2,3]$
D $x \in(-3,3)$

The correct answer is:
$x \in(2,3]$

Solution:
$\begin{aligned} f(x) & =\frac{\sqrt{9-x^2}}{\sqrt{x-2}} \\ \Rightarrow & x \in[-3,3] \text { and } x\gt2 \Rightarrow x \in(2,3] \\ & g(x)=\sqrt{\frac{9-x^2}{x-2}} \text { is defined if } \frac{9-x^2}{x-2} \geq 0 \\ \Rightarrow & \frac{x^2-9}{x-2} \leq 0\end{aligned}$
From the sign scheme $x \in(-\infty,-3] \cup(2,3]$ Hence, $f(x)$ and $g(x)$ are identical if $x \in(2,3]$

Topic: Mathematics, Functions, JEE Main
Find the values of $x$ for which the following functions are identical.
$f(x)=x$ and $g(x)=\frac{1}{1 / x}$

Options:
A $x \in R$
B $x \in R-\{0\}$
C $x \in R^{+}$
D none of these

The correct answer is:
$x \in R-\{0\}$

Solution:
$f(x)=x$ is defined for all $x$.
but $g(x)=\frac{1}{1 / x}=x$ is not defined for $x=0$ as $1 / x$ is not defined at $x=0$.
Hence both the functions are identical for $x \in R-\{0\}$.

Topic: Mathematics, Functions, JEE Main
Find the values of x, y, z in the given balanced equation H2SO4+x HI→H2S+y I2+z H2O ,

Options:
A x=3,y=5,z=2
B x=4,y=8,z=5
C x=8,y=4,z=4
D x=5,y=3,z=4

The correct answer is:
x=8,y=4,z=4

Solution:
Sulphur gains 8 electron in reduction and iodide ion loses 1 electron in oxidation.
Thus values of x, y, z are 8, 4, 4 respectively hence the reaction is
H2SO4+8HI→H2S+4I2+4H2O

Topic: Chemistry, Redox Reactions, JEE Main
Find the weight of H2SO4 in 1200 mL of a solution of 0.4 N strength.

Options:
A 23.52 g
B 2.52Â g
C 3.52Â g
D 29.52Â g

The correct answer is:
23.52 g

Solution:
Meq. of H2SO4 = 0.4 Ã 1200 = 480
âµMeq.=NÃV in mL
= w49Ã1000=480
w = 23.52 g

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
Find the x coordinate of the centre of mass of the non-uniform rod of length L given below. The origin is taken at the left end of the rod. The density of the rod as a function of its x-coordinate is Ï=ax2+bx+c, where a, b and c are constants.

Options:
A 2aL2+3bL2+6cL2(3aL2+4bL+8c)
B 4aL3+3bL2+2cL2(3aL2+2bL+c)
C 3aL2+4bL2+6cL4aL2+6bL+8c
D 3aL3+4bL2+6cL2(2aL2+3bL+6c)

The correct answer is:
3aL3+4bL2+6cL2(2aL2+3bL+6c)

Solution:
∫0L(ax2+bx+c)x dx∫0L(ax2+bx+c)dx

∫0L(ax3+bx2+cx)dx∫0L(ax2+bx+c)dx

=ax44+bx33+cx220Lax33+bx22+cx0L
=3aL4+4bL3+6cL2122aL2+3bL2+6cL6
=3aL3+4bL2+6cL22aL2+3bL+6c

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Find total number of solutions of \( \sin x=\frac{|x|}{10} \).

Options:

The correct answer is:
6.00

Solution:
Let y=sinx ...1

y=x10 ...2

six solution.

Topic: Mathematics, Trigonometric Equations, JEE Main
Find where the point \( (5 \cos \theta, 5 \sin \theta) \) lies with respect to ellipse \( \frac{x^{2}}{25}+\frac{y^{2}}{16}=1 \)

Options:
A director circle
B auxiliary circle
C focus
D None of these

The correct answer is:
auxiliary circle

Solution:
Given that, x2a2+y2b2=1

Given point 5cosθ,5sinθ

The auxiliary circle of an ellipse is x2+y2=25.

Given point satisfy the auxilary circle of an ellipse.

Topic: Mathematics, Ellipse, JEE Main
Find Work done in increasing the size of a soap bubble of radius \( 3 \mathrm{~cm} \) to \( 5 \mathrm{~cm} \) is nearly (surface tension of soap solution \( =0.03 \mathrm{Nm}^{-1} \) ). Surface area is

Options:
A \( 0.2 \pi \mathrm{mJ} \)
B \( 2 \pi \mathrm{mJ} \)
C \( 0.4 \pi \mathrm{mJ} \)
D \( 4 \pi \mathrm{mJ} \)

The correct answer is:
\( 0.4 \pi \mathrm{mJ} \)

Solution:
Given,

Work done in increasing the size of a soap bubble from radius3 cm 3 cm to 5 cm.

Work done =2T×Change in surface area,

r2=5 cm and r1=3 cm

⇒W=2T×4πr22-r12

=2×0.03×4π(5)2-(3)2×10-4

=0.06 ×4π ×16×10-4

=3.84π×10-4 J

=0.384×10-3 π J

=0.4π mJ

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Find Work done in the irreversible process C→A .

Options:
A 4.51 atm
B Zero
C 8.12 L atm
D unpredictable

The correct answer is:
8.12 L atm

Solution:
PA=nRTV=1Ã0.0821Ã20020=0.821
W=P.ÎV=0.821(20-10)=8.21 L atm

Topic: Chemistry, Thermodynamics (C), JEE Main
Fire extinguishers contain $\mathrm{H}_2 \mathrm{SO}_4$ and which one of the following?

Options:
A
$\mathrm{NaHCO}_3$ and $\mathrm{Na}_2 \mathrm{CO}_3$

B
$\mathrm{Na}_2 \mathrm{CO}_3$

C
$\mathrm{NaHCO}_3$

D
$\mathrm{CaCO}_3^2$

The correct answer is:
$\mathrm{NaHCO}_3$ and $\mathrm{Na}_2 \mathrm{CO}_3$

Solution:
Fire extinguishers contain a mixture of $\mathrm{Na}_2 \mathrm{CO}_3, \mathrm{NaHCO}_3$ and $\mathrm{H}_2 \mathrm{SO}_4$.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2012 (26 May Online)
First, a set of n equal resistors of 10 Ω each are connected in series to a battery of E.M.F. 20 V and internal resistance 10 Ω. A current I is observed to flow. Then, the n resistors are connected in parallel to the same battery. It is observed that the current is increased 20 times, then the value of n is ___________.

Options:

The correct answer is:
20

Solution:
When the resistances of each 10 Ω Ã¢â¬â¹ are connected in series and parallel then,Rs=10n, and Rp=10n. Given, ip=20is,

⇒2010n+10=202010n+10⇒20n10+10n=202010n+10

⇒n=20.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2021 (27 Aug Shift 1)
First action spectrum of photosynthesis based on study of Cladophora was given by-

Options:
A Jan Ingenhousz
B Joseph Priestley
C Von Sachs
D Engelmann

The correct answer is:
Engelmann

Solution:
Using a prism Engelmann split light into its spectral components and then illuminated a green alga, Cladophora, placed in a suspension of aerobic bacteria. The bacteria were used to detect the sites of O2 evolution. He observed that the bacteria accumulated mainly in the region of blue and red light of the split spectrum. A first action spectrum of photosynthesis was thus described. It resembles roughly the absorption spectra of chlorophyll a and b.

Topic: Biology, Photosynthesis in Higher Plants, NEET
First and second ionization energies of magnesium are 7.646 and 15.035 eV respectively. The amount of energy in kJ/mol needed to convert all the atoms of Magnesium into Mg 2+ ions present in 12 mg of magnesium vapours is: (Report your answer by multiplying with 10 and round it upto nearest integer)
[Given: 1 eV = 96.5 kJ  mol −1 ]

Options:

The correct answer is:
11

Solution:
12âmg 24âg =0.5Ã 10 â3 âmoles
Energy per atom = 7.646â+â15.035â=â22.68âeV
22.68âÃâ96.48â=â21.88âÃâ 10 2 âkJ/mol
E (needed) 2 1 . 8 8 Ã 1 0 2 Ã 0 . 5 Ã 1 0 - 3
10.94 ×  10 −1  = 1.094 kJ/mol

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
First diffraction minimum due to a single slit of width 10-4 cm is at Î¸=30o. The wavelength of the light used is

Options:
A 4000 â«
B 5000Â â«
C 6000Â â«
D 6250Â â«

The correct answer is:
5000 Å

Solution:
asinâ¡Î¸=1ÃÎ»
â Î»=aâsinâ¡Î¸=10-4Ã12=10Ã10-52
=5Ã10-5 cm=5000 â«

Topic: Physics, Wave Optics, JEE Main
First secondary minima in case of diffraction due to single slit is obtained at a distance \( x \) from central maximum position,
then

Options:
A \( x=\frac{\lambda a}{D} \)
B \( \frac{x}{a}=\frac{D}{\lambda} \)
C \( x=\lambda D a \)
D \( x=\frac{\lambda D}{a} \)

The correct answer is:
\( x=\frac{\lambda D}{a} \)

Solution:
In the case of Fraunhofer single slit diffraction,

Condition for the minimum, asinθ=mλ.

Since, sinθ≃tanθ and from the figure tanθ=xD, thus mλ=axD ...1, here m=0,1,2,3.., λ is wavelength, a is width of slit and D is the distance between slit and screen.

Now, for the first secondary minima from the central maximum position, the path difference is an integral number of wavelengths.

Thus, for first secondary minima, m=1 and asinθ=mλ.

From equation 1, we get axD=λ⇒x=λDa

Topic: Physics, Wave Optics, JEE Main
Fish die in water bodies polluted by sewage due to

Options:
A Pathogens
B Bad odour of sewage
C Reduction in oxygen
D Release of poisonous substances

The correct answer is:
Reduction in oxygen

Solution:
Fish die in water bodies polluted by sewage because dissolved O2 content of water is reduced.

Topic: Chemistry, Environmental Chemistry, JEE Main
Five balls of different colours are to be placed in three boxes of different sizes. Each box can hold all five balls. In how many ways can we place the balls so that no box remains empty?

Options:
A 50
B 100
C 150
D 200

The correct answer is:
150

Solution:
Let the boxes be marked as A, B and C. We have to ensure that no box remains empty and all five balls have to put in. There will be two possibilities:
(i) Any two box containing one ball each and 3 rd box containing 3 balls. Number of ways
\(\begin{aligned}
& =\mathrm{A}(1) \mathrm{B}(1) \mathrm{C}(3) \\
& ={ }^5 \mathrm{C}_1 \cdot{ }^4 \mathrm{C}_1 \cdot{ }^3 \mathrm{C}_3=5 \cdot 4 \cdot 1=20
\end{aligned}\)
(ii) Any two box containing 2 balls each and third containing 1 ball, the number of ways
\(\begin{aligned}
& =\mathrm{A}(2) \mathrm{B}(2) \mathrm{C}(1)={ }^5 \mathrm{C}_2 \cdot{ }^3 \mathrm{C}_2 \cdot{ }^1 \mathrm{C}_1 \\
& =10 \times 3 \times 1=30
\end{aligned}\)
Since, the box containing 1 ball could be any of the three boxes A, B, C. Hence, the required number of ways \(=30 \times 3=90\).
Hence, total number of ways \(=60+90=150\).

Topic: Mathematics, Permutation Combination, VITEEE, VITEEE 2023
Five boys and four girls sit in a straight line. The number of ways in which they can be seated if two girls are together and the other two are also together but separated from the first two is equal to

Options:
A 5400
B 10800
C 21600
D 43200

The correct answer is:
43200

Solution:
Four girls can be grouped as 2,2 in 4!2!22! ways.
Five boys can be arranged in 5! ways, six gaps will be generated because of these boys. Two groups of the girls can be placed at any of these 6 places in P62 ways.
The girls in each group can be arranged in 2! ways.
Required ways =4!2!22!×5!×P62×2!×2!
=4×32×2×120×6×5×2×2
=360×120=43200

Topic: Mathematics, Permutation Combination, JEE Main
Five charges $+q,+5 q,-2 q,+3 q$ and $-4 q$ are situated as shown in the figure. The electric flux due to this configuration through the surface $S$ is :

Options:
A $\frac{4 q}{\epsilon_0}$
B $\frac{5 q}{\epsilon_0}$
C $\frac{q}{\epsilon_0}$
D $\frac{3 q}{\epsilon_0}$

The correct answer is:
$\frac{4 q}{\epsilon_0}$

Solution:
As per gauss theorem,
$\begin{aligned}& \phi=\frac{\mathrm{q}_{\text {in }}}{\epsilon_0}=\frac{\mathrm{q}+(-2 \mathrm{q})+5 \mathrm{q}}{\epsilon_0} \\
& \frac{4 \mathrm{q}}{\epsilon_0}\end{aligned}$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2024 (09 Apr Shift 2)
Five coins whose faces are marked 2,3 are thrown. What is the probability of obtainining a total of $12 ?$

Options:
A $\frac{1}{16}$
B $\frac{3}{16}$
C $\frac{5}{16} \quad$
D $\frac{7}{16}$

The correct answer is:
$\frac{5}{16} \quad$

Solution:
Let $\mathrm{E}$ be the event of total of 12 . $\mathrm{E}=(2,2,2,3,3),(2,2,3,3,2),(2,3,3,2,2)$,
$(3,3,2,2,2),(3,2,3,2,2),(3,2,2,3,2)$,
$(3,2,2,2,3),(2,3,2,3,2),(2,3,2,2,3)$,
$(2,2,3,2,3)$
$\mathrm{n}(\mathrm{E})=10$
Sample sapce contain total possibility $=2^{5}=32$ Hence, $n(s)=32$
So, $\mathrm{P}(\mathrm{E})=\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{10}{32}=\frac{5}{16}$

Topic: Mathematics, Probability, JEE Main
Five digit number divisible by 3 is formed using 0, 1, 2, 3, 4, 6 and 7 without repetition. Total number of such numbers are

Options:
A
312

B
3125

C
120

D
216

The correct answer is:
216

Solution:
Required numbers are $5 !+5 !-4 !=216$

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2002
Five digit numbers are formed using the digits 1, 2, 3, 5, 7 with repetitions and are written in descending order with serial numbers. For example, the number 77777 has serial number 1. Then the serial number of 35337 is

Options:

The correct answer is:
1436

Solution:
Given,

Five digit numbers are formed using the digits 1, 2, 3, 5, 7 with repetitions and are written in descending order with serial numbers,

Now number of 5 digit numbers starting with digit 1 will be =5×5×5×5=625

Number of 5 digit numbers starting with digit 2 will be =5×5×5×5=625

Number of 5 digit numbers starting with digit 31 will be =5×5×5=125

Number of 5 digit numbers starting with digit 32 will be =5×5×5=125

Number of 5 digit numbers starting with digit 33 will be =5×5=25

Number of 5 digit numbers starting with digit 351 will be=5×5=25

Number of 5 digit numbers starting with digit 352 will be =5×5=25

Number of 5 digit numbers starting with digit 3531 will be =5

And number of 5 digit numbers starting with digit 3532 will be =5

And before 35337 will be 4 numbers, So rank of 35337 will be 1690

So, in descending order serial number will be

3125-1690+1=1436

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2023 (29 Jan Shift 1)
Five equal point charges with q=10 nC are located at x=2, 4, 5, 10 and 20 m. Taking the value of permittivity of free space Îµ0=136ÏÃ10-9 C2 N-1 m-2 and potential at infinity to be zero, calculate the potential (in V) at x=0?

Options:

The correct answer is:
99

Solution:
Given q=10 nC =​10×10-9 C
The potential at the point x=0, will be the sum of potential produced by the charges placed at x=2, 4, 5, 10 and 20 m
Potential,

V=q4πε0×r
Then, V1=q4πε0 × 2
V2=q4πε0×4
V3=q4πε0×5
V4=q4πε0×10
V5=q4π×20
The resultant potential,
V=V1+V2+V3+V4+V5
=q4πε012+14+15+110+120
=q​4πε010+5+4+2+120
=14πε0×10×10-9×2220
=1×36π4π×10-9×10×10-9×2220
=99 V

Topic: Physics, Electrostatics, JEE Main
Five equal resistance each of resistance $R$ are connected as shown in the figure. A battery of $V$ volts is connected between $\mathrm{A}$ and $\mathrm{B}$. The current flowing in AFCEF will be:

Options:
A $\frac{3 V}{R}$
B $\frac{V}{R}$
C $\frac{V}{2 R}$
D $\frac{2 V}{R}$

The correct answer is:
$\frac{V}{2 R}$

Solution:
Given circuit can be reduced to

Topic: Physics, Current Electricity, JEE Main
Five horses are in a race. Mr. A selects two of the horses at random and bets on them. The probability that Mr. A selected the winning horse is

Options:
A
$\frac{2}{5}$

B
$\frac{4}{5}$

C
$\frac{3}{5}$

D
$\frac{1}{5}$

The correct answer is:
$\frac{2}{5}$

Solution:
$\mathrm{n}(\mathrm{S})={ }^5 \mathrm{C}_2 ; \mathrm{n}(\mathrm{E})={ }^2 \mathrm{C}_1+{ }^2 \mathrm{C}_1$
$\mathrm{p}(\mathrm{E})=\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^2 \mathrm{C}_1+{ }^2 \mathrm{C}_1}{{ }^5 \mathrm{C}_2}=\frac{2}{5}$

Topic: Mathematics, Probability, JEE Main, JEE Main 2003