datavorous/entrance-exam-dataset huggingface dataset:
Find the least positive integer \( m \) for which function \( f \) defined as \( f(x)=\sin x-m x+k \) is a decreasing function for all \( x \in R \)

Options:

The correct answer is:
2

Solution:
f(x)=sinx-mx+k
f is differentiable. f'(x)=cosx-m
f'(x)<0⇔cosx-m<0
⇔m>cosx
For function f to be decreasing for all x m>1

⇒The least positive integer m=2

Topic: Mathematics, Application of Derivatives, JEE Main
Find the least positive value of k, if the range of 15,14, k, 25,30,35 is 23.

Options:
A 11
B 13
C 12
D 14

The correct answer is:
12

Solution:
Let k be the least value and 35 be the Highest value.
Since, Range = Highest value - Lowest value
⇒23=35-k

⇒k=12.

Topic: Mathematics, Statistics, JEE Main
Find the length of the chord of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$, whose middle point is $\left(\frac{1}{2}, \frac{2}{5}\right)$

Options:
A $\frac{7}{6} \sqrt{40}$
B $\frac{7}{5} \sqrt{41}$
C $\frac{5}{7} \sqrt{41}$
D none of the above

The correct answer is:
$\frac{7}{5} \sqrt{41}$

Solution:
Chord of ellipse with mid-point $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ is $\mathrm{T}=\mathrm{S}_1$
$\begin{aligned}
& \therefore \frac{\mathrm{x}}{50}+\frac{\mathrm{y}}{40}-1=\frac{1}{25 \times 4}+\frac{4}{25 \times 16}-1 \\
& \therefore 4 \mathrm{x}+5 \mathrm{y}=4
\end{aligned}$
This line when solved with the ellipse gives the points of intersection.
$\begin{aligned}
& \therefore \frac{\mathrm{x}^2}{25}+\frac{\left(\frac{4(1-\mathrm{x})}{5}\right)^2}{16}=1 \\
& \therefore \frac{\mathrm{x}^2}{25}+\frac{16(1-\mathrm{x})^2}{16 \times 25}=1 \\
& \therefore \frac{\mathrm{x}^2+(1-\mathrm{x})^2}{25}=1 \\
& \therefore 2 \mathrm{x}^2-2 \mathrm{x}-24=\mathrm{o} \\
& \therefore \mathrm{x}=-3 \text { or } \mathrm{x}=4
\end{aligned}$
Hence, the points of intersection are $\left(-3, \frac{16}{5}\right)$ and $\left(4,-\frac{12}{5}\right)$.
$\begin{aligned}
& \therefore \text { length of chord }=\sqrt{(-3-4)^2+\left(\frac{16}{5}+\frac{12}{5}\right)^2} \\
& =\sqrt{(7)^2+\left(\frac{4 \times 7}{5}\right)^2}
\end{aligned}$
$\begin{aligned}
& =7 \sqrt{ } 1+\frac{16}{25} \\
& =\frac{7}{5} \sqrt{41}
\end{aligned}$
This is the required answer.

Topic: Mathematics, Ellipse, JEE Main
Find the magnetic field at point P due to a straight line segment AB of length 6 cm carrying a current of 5 A. (See figure) μ0=4π×10-7 NA-2

Options:
A 1.5×10-5 T
B 2.0×10-5 T
C 3.0×10-5 T
D 2.5×10-5 T

The correct answer is:
1.5×10-5 T

Solution:
θ=370
r=4 cm
B=μ0i4πr [sin37°+sin37°]
=4π×10-754π×4×10-235+35
=32×10-5=1.5×10-5 T

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2019 (12 Apr Shift 2)
Find the magnetic field at the point P in figure. The curved portion is a semicircle connected to two long straight wires.

Options:
A μ0i2r1+2π
B μ0i2r1+1π
C μ0i2r12+12π
D μ0i2r12+1π

The correct answer is:
μ0i2r12+12π

Solution:
Magnitude of magnetic field due to semi-infinite portion of wire AB at point C is

B1=μ0i4πr

Magnitude of magnetic field due to semicircular arc BD at point C is

B2=μ0i4r

Magnitude of magnetic field due to semi-infinite portion DE at point C is 0.

Using right-hand rule, it can be seen that directions of B1→ and B2→ are both outwards from the plane.

Therefore, the magnitude of net magnetic field at point C is

B=B1+B2

=μ0i2r12+12π

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2023 (01 Feb Shift 1)
Find the magnetic induction at point O, if the current carrying wire is in the shape shown in the figure.

Options:
A μ 0 I 5 π r 3 π 2 + 1 â
B μ 0 I 3 π r 4 π 3 + 1 â
C Î¼ 0 I 2 Ï r 3Ï 4 + 1 â
D Î¼ 0 I 4 Ï r 3 Ï 2 + 1 â

The correct answer is:
μ 0 I 4 π r 3 π 2 + 1 ⊙

Solution:
B1=0,
B2=μ04πIr×3π2⊙,
B3=μ04πIr⊙
B=B2+B3=μ0I4πr3π2+1⊙

Topic: Physics, Magnetic Effects of Current, JEE Main
Find the magnetic moment of \( \left[\mathrm{NiCl}_{4}\right]^{2-} \)

Options:
A \( 1.8 \mathrm{BM} \)
B \( 5.86 \quad \mathrm{BM} \)
C \( 2.82 \mathrm{BM} \)
D \( 3.82 \mathrm{BM} \)

The correct answer is:
\( 2.82 \mathrm{BM} \)

Solution:
4s2 3d8

μ=2×(3)=6=2.83 BM

Topic: Chemistry, Coordination Compounds, JEE Main
find the major product

Options:
A
B
C
D None of these

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
Find the maximum value of the function \( f(x)=3 x^{3}-18 x^{2}+27 x-40 \) in the set \( S=\left\{x \in R: x^{2}+30-11 x \leq 0\right\} \)

Options:
A \( 122 \)
B \( 222 \)
C \( -122 \)
D \( -222 \)

The correct answer is:
\( 122 \)

Solution:
fx=3x3-18x2+27x-40

x2-11x+30≤0⇒x∈5, 6

f′x=9x2-36x+27=9 x2-4x+3=9 x-1 x-3

for x∈5, 6 fx is increasing function

hence maximum value in this interval occurs at x=6

so f6=648-648+162-40=122

Topic: Mathematics, Application of Derivatives, JEE Main
Find the mean deviation about the mean for the data $4,7,8,9,10,12,13,17$

Options:
A 3
B 24
C 10
D 8

The correct answer is:
3

Solution:
Arithmetic mean $\bar{x}$ of $4,7,8,9,10,12,13$, 17 is
$\bar{x}=$ $\frac{4+7+8+9+10+12+13+17}{8}=\frac{80}{8}=10$
$\Sigma\left|x_i-\bar{x}\right|=6+3+2+1+0+2+3+7=24$
$\therefore$ Mean deviation about mean
$=\text { M.D. }(\bar{x})=\frac{\sum\left|x_i-\bar{x}\right|}{n}=\frac{24}{8}=3$

Topic: Mathematics, Statistics, JEE Main
Find the mean deviation from the mean for the following data :
\( 4,7,8,9,10,12,13,17 . \)

Options:

The correct answer is:
3

Solution:
Given that data 4,7,8,9,10,12,13,17

We will find mean deviation from the mean

∴ x¯=Σxin=808=10

x i	x i - x ¯	x i - x ¯

4	- 6	6

7	- 3	3

8	- 2	2

9	- 1	1

10	0	0

12	2	2

13	3	3

17	7	7

80	0 *	24

∴ M.D. x¯=Σi=1nxi-x¯n=248=3

Then M.D x- =3

Topic: Mathematics, Statistics, JEE Main
Find the metallic and ferromagnetic substance.

Options:
A VO2
B MnO2
C TiO2
D CrO2

The correct answer is:
CrO2

Solution:
In d-block elements,
MnO2 and CrO2 exhibit strong attraction to magnetic fields and are able to retain their magnetic properties.
MnO2 is anti-ferromagnetic and CrO2 is ferromagnetic.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2016 (03 Apr)
Find the minimum attainable pressure of an ideal gas in process T= T 0 +Î± V 2, where T 0 and Î± are positive constants and V is the volume of one mole of gas.

Options:
A 2R Î± T 0
B 3R α T 0
C 3R
D 3R Î± T 0 2

The correct answer is:
2R α T 0

Solution:
Given,
T= T 0 +Î± V 2 ...(i)
For 1 mol of a gas, PV=RT
of V= RT P
Substituting this value in Eq. (i), we get
T= T 0 +Î± RT P 2 = T 0 +Î± R 2 T 2 P 2
or T P 2 = T 0 P 2 +Î± R 2 T 2
or P= Î± RT Tâ T 0 â1/2 ...(ii)
After differentiating, we get
dP dT = Î± R Tâ T 0 â1/2 â 1 2 T Tâ T 0 â3/2
For minimum pressure,
dP dT =0
â´ 0 = Î± R T - T 0 - 1 / 2 - 1 2 T T - T 0 - 3 / 2
After solving, T = 2T0
From Eq. (ii),
P min = Î± R2 T 0 2 T 0 â T 0 â1/2
=2R Î± T 0

Topic: Physics, Thermodynamics, JEE Main
Find the mode of the following distribution:
\(\begin{array}{|l|l|}\hline \text { Class interval } & \text { frequency }(f) \\\hline 0-10 & 5 \\\hline 10-20 & 8 \\\hline 20-30 & 7 \\\hline 30-40 & 12 \leftarrow f_{m-1} \\\hline 40-50 & 28 \leftarrow f_{m} \\\hline 50-60 & 20 \leftarrow f_{m+1} \\\hline 60-70 & 10 \\\hline\end{array}\)

Options:

The correct answer is:
46.67

Solution:
We have,

Class interval	frequencyf	cumulative frequencycf

0-10	5	5

10-20	8	13

20-30	7	20

30-40	12 ←fm-1	32

40-50	28 ←fm	60

50-60	20 ←fm+1	80

60-70	10	90

Total	90

Median=902=45 it belongs to a class interval 40-50.

l=40, fm=28, fm-1=12, fm+1=20, h=10

So, mode =ℓ+fm-fm-12 fm-fm-1-fm+1×h

=40+28-122×28-12-20×10

=46.67

Topic: Mathematics, Statistics, JEE Main
Find the modulation index of an AM wave having 8 V variation where maximum amplitude of the AM wave is 9 V.

Options:
A 0.8
B 0.5
C 0.2
D 0.1

The correct answer is:
0.8

Solution:
Modulation index is given by m=AmAc, where, Am is amplitude of modulating signal and Ac is amplitude of carrier signal.

Given: 2Am=8

Am+Ac=9⇒Ac=5

∴m=45=0.8

Topic: Physics, Communication System, JEE Main, JEE Main 2022 (29 Jul Shift 1)
Find the molarity of water. Given: ρ = 1000 kg /m 3
[Report your answer upto one decimal place].

Options:

The correct answer is:
55.5

Solution:
Mass of 1.0 L water = 1000 g

Molarity=wtmol.wtÃ1000VmL
âMolarity=100018=55.5mol L -1

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
Find the most stable form

Options:
A
B
C
D Both B & C

The correct answer is:

Solution:
This is the most stable form because of hydrogen bonding

Topic: Chemistry, General Organic Chemistry, JEE Main
Find the number of 4-digit numbers that can be formed using the digits $1,2,3,4,5$ if no digit is repeated. How many of these will be even?

Options:

None

Solution:
(i) Out of 5 digits 4-digit numbers are to be formed.
Such numbers are
${ }^5 P_4=5 \times 4 \times 3 \times 2=120$
(ii) When 2 is at units place then remaining three place are filled in ${ }^4 P_3$ ways
$=4 \times 3 \times 2=24$
When 4 is at unit's place, then 4-digit numbers are again $=24$
$\therefore$ Even 4- digit numbers $=2 \times 24=48$

Topic: Mathematics, Permutation Combination
Find the number of lone pairs of electrons present in a \( \mathrm{OF}_{2} \) molecule.

Options:

The correct answer is:
8

Solution:
O atom has = 2 lone pairs

F atom has = (3 lone pairs each)

Total lone pairs=2+3+3=8.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Find the number of molecule having two lone e- pairs, on central atom.

I3+,XeF2,XeF4,H2O,NH2-,H2S,SO42-,NF3,NO2-,XeOF2

Options:

The correct answer is:
6

Solution:

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Find the number of optical isomers for this sizes compound.

Options:
A 3
B 2
C 6
D zero

The correct answer is:
zero

Solution:
In the compound POS or COS present so all compounds are optically inactive

Topic: Chemistry, General Organic Chemistry, JEE Main
Find the number of solutions for; $\quad \cos x=x$

Options:
A 1
B 2
C 3
D 4

The correct answer is:
1

Solution:
As, $\quad \cos x=x$
$\therefore$ to plot the curve for $y=\cos x ; \quad y=x$ and find the point of intersection as to obtain number of solutions. Here, the two curves intersect at a point $A$.
So, $\cos x=x$ has only one solution.

Topic: Mathematics, Functions, JEE Main
Find the number of solutions of: $e^x=x^4$.

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
Let; $f(x)=e^x$ and $g(x)=x^4$, which could be shown as;
From the figure, it is clear they intersect at two points, therefore two solutions.

Topic: Mathematics, Functions, JEE Main
Find the number of solutions of; $\log _{10} x=x$.

Options:
A 1
B 2
C 3
D 4

The correct answer is:
1

Solution:
Let; $f(x)=\log _{10} x$ and $g(x)=x$; which could be shown as;
From above figure, it is clear they intersect at one points, therefore $\mathbf{1}$ solution.

Topic: Mathematics, Functions, JEE Main
Find the number of solutions of; $\sin x=\frac{x}{10}$.

Options:
A 7
B 10
C 15
D 12

The correct answer is:
7

Solution:
Here, let $f(x)=\sin x$ and $g(x)=\frac{x}{10}$ also we know; $-1 \leq \sin x \leq 1$ $\therefore \quad-1 \leq \frac{x}{10} \leq 1 \quad \Rightarrow \quad-10 \leq x \leq 10$
Thus, to sketch both curves when $x \in[-10,10]$
From above figure $f(x)=\sin x$ and $g(x)=\frac{x}{10}$ intersect at 7 points. So, numbers of solutions are 7.

Topic: Mathematics, Functions, JEE Main
Find the number of solutions of the equation,
$\sin x=x^2+x+1 \text {. }$

Options:
A No solution
B 100
C 12
D 5

The correct answer is:
No solution

Solution:
Let; $f(x)=\sin x$ and $g(x)=x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}$ which could be shown as;
which do not intersect at any point, therefore no solution

Topic: Mathematics, Functions, JEE Main
Find the number of ways in which \( 20 \) one rupee coins can be distributed among \( 5 \) people such that each person gets atleast \( 3 \) rupees.

Options:

The correct answer is:
105

Solution:
Since, 5 people each get atleast 3 rupees

So, left Rs. 5 is to be distributed x1+x2+x3+x4+x5=5

Required number of ways = 5+5-1C5-1

=C49=105

Topic: Mathematics, Permutation Combination, JEE Main
Find the number of words of four letters that can be formed from the letters of the word 'EXAMINATION'

Options:

The correct answer is:
2454.00

Solution:
(AA), (II), (NN), E, X, M, T, O

Case-1: All 4 different.

C48×4!=1680

Case-2: 2 Alike, 2 different

3×C27×4!2!=756

Case -3: 2 alike of 1 kind, 2 alike of 2nd kind.

C23×1×1×4!2!2!=18

Total =1680+756+18=2454

Topic: Mathematics, Permutation Combination, JEE Main
Find the odd one out with respect to the functions of an ecosystem.

Options:
A Nutrient cycling
B Energy flow
C Decomposition
D Stratification

The correct answer is:
Stratification

Solution:
Four functions of the ecosystem are nutrient cycling, energy flow, productivity, and decomposition. Stratification along with species composition is two main structural features of the ecosystem.

Topic: Biology, Ecosystem, NEET
Find the odd one out with respect to the respiratory organ.

Options:
A Blue whale
B Tadpole of frog
C Bony fish
D Shark

The correct answer is:
Blue whale

Solution:
Respiration by gills is called branchial respiration, which mainly takes place in most of aquatic arthropods, mollusca, all true fishes and tadpole of frogs.
Blue whale being the mammal exclusively respire by using its lungs, called pulmonary respiration.

Topic: Biology, Breathing and Exchange of Gases, NEET
Find the ODD one out.

Options:
A Eye of octopus and mammals
B Flippers of penguin and dolphins
C Sweet potato and potato
D Upper limb of man and forelimb of cheetah.

The correct answer is:
Upper limb of man and forelimb of cheetah.

Solution:
Organisms with similar structures but different functions are homologous organs. While organisms with similar functions but different structures are analogous organs. Homology represents divergent evolution. Here upper limb of man and forelimb of cheetah represent homologous structure. While the remaining three are analogous structures.

Topic: Biology, Evolution, NEET
Find the ordered triplet \( (A, B, \lambda) \), If \( \int \frac{2 \cos x-\sin x+\lambda}{\cos x+\sin x-2} d x=A \log _{e}(|\cos x+\sin x-2|)+B x+C \)

Options:
A \( \left(\frac{1}{2}, \frac{3}{2},-1\right) \)
B \( \left(\frac{3}{2}, \frac{1}{2},-1\right) \)
C \( \left(\frac{1}{2},-1,-\frac{3}{2}\right) \)
D \( \left(\frac{3}{2},-1,-\frac{1}{2}\right) \)

The correct answer is:
\( \left(\frac{3}{2}, \frac{1}{2},-1\right) \)

Solution:
We have,
∫2cosx-sinx+λcosx+sinx-2dx=Alogecosx+sinx-2+Bx+C
Differentiating R.H.S. w.r.t. x, we get

ddxAlogecosx+sinx-2+Bx+C =Acosx-sinxcosx+sinx-2+B

=Acosx-Asinx+Bcosx+Bsinx-2Bcosx+sinx-2
∴2cosx-sinx+λcosx+sinx-2=Acosx-Asinx+Bcosx+Bsinx-2Bcosx+sinx-2

So, A+B=2, B-A=-1, λ=-2B

⇒A=32, B=12, λ=-1

Topic: Mathematics, Indefinite Integration, JEE Main
Find the peak current and resonant frequency of the following circuit (as shown in figure).

Options:
A 2 A and 50 Hz
B 0.2 A and 100 Hz
C 2 A and 100 Hz
D 0.2 A and 50 Hz

The correct answer is:
0.2 A and 50 Hz

Solution:
as given Z=XL-XC2+R2

XL=ωL=100×100×10-3=10 Ω

XC=1ωC=1100×100×10-6=100 Ω

Z=10-1002+R2=902+1202=30×5=150 Ω

ipeak =Vpeakz=30150=15 amp=0.2 amp

& For resonant frequency

⇒ωL=1ωC⇒ω2=1LC⇒ω=1LC

& f=12πLC⇒12π100×10-3×100×10-6

=100102π=100π2π=50 Hz

as 10≈π

Topic: Physics, Alternating Current, JEE Main
Find the peak current and resonant frequency of the following circuit (as shown in figure).

Options:
A 2 A and 50 Hz
B 0.2 A and 100 Hz
C 2 A and 100 Hz
D 0.2 A and 50 Hz

The correct answer is:
0.2 A and 50 Hz

Solution:
as given Z=XL-XC2+R2

XL=ωL=100×100×10-3=10 Ω

XC=1ωC=1100×100×10-6=100 Ω

Z=10-1002+R2=902+1202=30×5=150 Ω

ipeak =Vpeakz=30150=15 amp=0.2 amp

& For resonant frequency

⇒ωL=1ωC⇒ω2=1LC⇒ω=1LC

& f=12πLC⇒12π100×10-3×100×10-6

=100102π=100π2π=50 Hz

as 10≈π

Topic: Physics, Alternating Current, JEE Main, JEE Main 2021 (26 Feb Shift 2)
Find the pressure exerted by \( 6 \times 10^{23} \) hydrogen molecules which will strike per second a wall of area \( 10^{-4} \mathrm{~m}^{2} \) at \( 60^{\circ} \) with normal. The mass of hydrogen molecules and speed are \( 3.32 \times 10^{-27} \mathrm{~kg} \) and \( 10^{3} \mathrm{~m} \mathrm{~s}^{-1} \) respectively

Options:
A \( 19.92 \times 10^{3} \mathrm{~m} \mathrm{~s}^{-1} \)
B \( 18.2 \times 10^{3} \mathrm{~m} \mathrm{~s}^{-1} \)
C 1. \( 992 \times 10^{3} \mathrm{~m} \mathrm{~s}^{-1} \)
D \( 0.1992 \times 10^{3} \mathrm{~m} \mathrm{~s}^{-1} \)

The correct answer is:
\( 19.92 \times 10^{3} \mathrm{~m} \mathrm{~s}^{-1} \)

Solution:
Given,

The hydrogen molecules strike the wall at 60° with the normal of the wall as shown below

Thus the Change in momentum = 2mvcosθ

Force, F=2mvcosθ×n
Pressure, P=FA=2mvncosθA
=2×3.32×10-27×103×12×6×102310-4
=19.92×103 m s-1.

Topic: Physics, Kinetic Theory of Gases, JEE Main
Find the rate law that corresponds to the data shown for the following reaction?

Experiment	Al0=M	[B]0=M	Initial rate

1	0.012	0.035	0.10

2	0.024	0.070	0.80

3	0.024	0.035	0.10

4	0.012	0.070	0.80

Options:
A Rate =k[A]0[B]3
B Rate=k[B]4
C Rate =k[A][B]3
D Rate =kA2[B]2

The correct answer is:
Rate =k[A]0[B]3

Solution:
Expt. 4 Expt. 1=[0.012]x[0.070]y[0.012]x[0.035]y=0.800.10

2y=8∴y=3

Expt. 3 Expt. 1=[0.024]x[0.035]3[0.012]x[0.035]3=0.100.10

2x=1 x=0

Rate =k[A]0[B]3.

Topic: Chemistry, Chemical Kinetics, JEE Main
Find the ratio of energies of photons produced due to the transition of an electron of a hydrogen atom from its
(i) second permitted energy level to the first level, and
(ii) the highest permitted energy level to the first permitted level.

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

The correct answer is:
3:4

Solution:
(a) Since the second permitted energy level
= E 2 â E 1
The energy of the photon produced
=( â13.6âeV 2 2 )â( â13.6âeV 1 2 )=â3.4+13.6=10.2âeV
( b )âThe highest permitted energy level to the first permitted level
= E â â E 1
=0â( â13.6 )=13.6âeV
â´The ratio of energies of photon
= 10.2 13.6 = 3 4 =3:4

Topic: Physics, Atomic Physics, JEE Main
Find the reading of the ideal ammeter connected in the given circuit. Assume that the cells have negligible internal
resistance.

Options:
A \( 0.8 \mathrm{~A} \)
B \( 0.25 \mathrm{~A} \)
C \( 1.95 \mathrm{~A} \)
D \( 1.0 \mathrm{~A} \)

The correct answer is:
\( 1.95 \mathrm{~A} \)

Solution:
By finding potential difference across any resistance we can find current through each resistance directly.

Total current in the circuit is equal to the sum of individual current through branches. This si also known as Kirchoff's current law or junction rule.

∴ i=3+0.25-0.5-0.8=1.95 A

Topic: Physics, Current Electricity, JEE Main
Find the relative error in \( \mathrm{P} \), if the physical quantity \( \mathrm{P} \) is described by the relation \( P=a^{\frac{1}{2}} b^{2} c^{3} d^{-4} \). It is given that the relative errors in the measurement of \( \mathrm{a}, \mathrm{b}, \mathrm{c} \) and \( \mathrm{d} \) respectively, are \( 2 \%, 1 \%, 3 \% \) and \( 5 \% \).

Options:
A \( 12 \% \)
B \( 8 \% \)
C \( 25 \% \)
D \( 32 \% \)

The correct answer is:
\( 32 \% \)

Solution:
P=a12 b2c3d-4
âppmax=12 âaa+2âbb+3âcc+4âdd
100ÃâPP=12 âaa+2âbb+3âcc+4âddÃ100
% error in P=12Ã2+2Ã1+3 Ã3+4 Ã5
=32%

Topic: Physics, Mathematics in Physics, JEE Main
Find the remainder when $x^4-11 x^3+44 x^2-76 x+48$ is divided by $x^2-7 x+12$.

Options:
A 0
B 1
C $x^2-4 x+4$
D $x^2+4 x-4$

The correct answer is:
0

Solution:
No solution. Refer to answer key.

Topic: Mathematics, Quadratic Equation, AP EAMCET, AP EAMCET 2020 (07 Oct Special)
Find the resistor in which the dissipated power is the greatest -

Options:
A 2 Ω
B 9 Ω
C 18Â Î©
D 5 Î©

The correct answer is:
5 Î©

Solution:
Equivalent circuit
i1 : i2 : i3=12: 16: 13=3 : 1 : 2
i1=36Ãi=i2
i2=16Ãi=i6
i3=26Ãi=i3
ÎH in 5Î©=i2 Ã 5
âH in 2Î©=i24Ã 2=i22
âH in 9Î©=4i29Ã9
Maximum heat is in 5Î© .

Topic: Physics, Current Electricity, JEE Main
Find the solution of differential equation given below:

dydx+y·cosec2(x)=cosec2(x)·cot(x)

Options:
A yecotx=(1+cotx)e-cotx+c
B ye-cotx=(1-cotx)e-cotx+c
C yecotx=(1+cotx)ecotx+c
D ye-cotx=(1+cotx)e-cotx+c

The correct answer is:
ye-cotx=(1+cotx)e-cotx+c

Solution:
Given the differential equation is,

dydx+cosec2(x)·y=cosec2(x)·cotx..........(1)

Comparing with (1), we get

dydx+py=Q

Which is a linear differential equation,

P=cosec2xQ=cosec2x·cotxI.F.=e∫Pdx=e∫cosec2x dx=e-cotx

The solution of (1) is given by,

y·I.F.=∫Q·IF+Cy·e-cotx=∫cosec2x·cotx·e-cotx dx+C.

Now, put cotx=t in RHS

-cosec2x dx=dtcosec2x=-dty·e-cotx=∫e-t·t (-dt)=-∫t·e-t dt=-[t∫e-t dt-∫ddtt(∫e-t dt)]=-[-te-t-∫-e-t dt]=-[-te-t-e-t]+C=te-t+e-t+C=e-t(1+t)+C=e-cotx(1+cotx)+C.

Topic: Mathematics, Differential Equations, JEE Main
Find the stress developed inside a tooth cavity filled with copper when hot tea at temperature of \(57^{\circ} \mathrm{C}\) is drunk. You can take body (tooth) temperature to be \(37^{\circ} \mathrm{C}\) and \(\alpha=1.7 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\), bulk modulus for copper \(B=140 \times 10^9 \mathrm{Nm}^{-2}\).

Options:
A \(1.4 \times 10^8 \mathrm{Nm}^{-2}\)
B \(1.9 \times 10^8 \mathrm{Nm}^{-2}\)
C \(2.0 \times 10^8 \mathrm{Nm}^{-2}\)
D \(3.4 \times 10^7 \mathrm{Nm}^{-2}\)

The correct answer is:
\(1.4 \times 10^8 \mathrm{Nm}^{-2}\)

Solution:
Temperature of hot tea, \(t_2=57^{\circ} \mathrm{C}\)
Normal temperature of tooth, \(t_1=37^{\circ} \mathrm{C}\)
\(\alpha=1.7 \times 10^{-5} \mathrm{C}^{-1}\)
Bulk modulus, \(B=140 \times 10^9 \mathrm{Nm}^{-2}\)
Thermal stress in tooth cavity
\(\begin{aligned}
& =\text {Bulk modulus } \times \text { Volumetric strain } \\
& =B \times \frac{\Delta V}{V}=B \times \gamma \cdot \Delta t \\
& {[\because \Delta V=V \gamma \Delta t]} \\
& =B \times 3 \alpha \times \Delta t \\
& {[\because \gamma=3 \alpha]} \\
& =3 B \alpha \Delta t=3 B \alpha\left(t_2-t_1\right) \\
& =3 \times 140 \times 10^9 \times 1.7 \times 10^{-5}(57-37) \\
& =1.4 \times 10^8 \mathrm{Nm}^{-2} \\
\end{aligned}\)

Topic: Physics, Mechanical Properties of Solids, JEE Main
Find the suitable method from the following to prepare primary amines without the loss of carbons.

Options:
A Gabriel method
B Alkylation method
C Hoffmann bromamide method
D Stephen method

The correct answer is:
Gabriel method

Solution:
Gabriel phthalimide synthesis is used for the preparation of aliphatic primary amines without the loss of carbons. It involves nucleophilic substitution (SN2) of alkyl halides by the anion formed by the phthalimide.

Phthalimide on treatment with ethanolic KOH forms potassium salt of phthalimide which on heating with RX followed by either alkaline hydrolysis or hydrazinolysis with hydrazine produces the corresponding 1°amine.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Find the sum of all two digit numbers which when divided by 4 , yields 1 as remainder.

Options:

None

Solution:
The sum of two digit numbers which when divided by 4 yield las remainder is
$13+17+21+\ldots \ldots \ldots+97$
Let the sum be denoted by $S$ and let 97 be the $n^{\text {th }}$ term, then
$\begin{aligned}
& 97=a+(n-1) d=13+(n-1) d \\
\text { since, } & a=13, d=4
\end{aligned}$
$\begin{array}{ll}
\therefore & 97-13=4(n-1) \\
\text { or } & 84=4(n-1) \\
& n=22
\end{array}$
$\therefore \quad$ The sum of the series,
$\begin{aligned}
&S=13+17+21+\ldots \ldots \ldots \ldots \ldots \\
&a=13, d=4, n=22 \\
&S_{22}=\frac{22}{2}[2 \times 13+(22-1) 4]
\end{aligned}$
$=11[26+21 \times 4]=11(26+84)$
$=11 \times 110=1210$

Topic: Mathematics, Sequences and Series
Find the sum of bond order between same bonded atoms in \( \mathrm{Q} \) and \( \mathrm{R} \) compounds.

Options:

The correct answer is:
3.00

Solution:
Bond order of [O-O] in H2O2=1.0
Bond order of [O-O] in O2=2.0
So sum of bond order between same bonded atoms in Q and R compounds = 1 + 2.0 = 3.0.

Topic: Chemistry, s Block Elements, JEE Main
Find the sum of bond order between same bonded atoms in Q and R compounds.

Options:

The correct answer is:
3.00

Solution:
Bond order of [O-O] in H2O2=1.0
Bond order of [O-O] in O2=2.0
So sum of bond order between same bonded atoms in Q and R compounds = 1 + 2.0 = 3.0.

Topic: Chemistry, s Block Elements, JEE Main
Find the sum of maximum number of electrons having +1 and –1 value of ‘m’ in Ti

Options:

The correct answer is:
10

Solution:
Electronic configuration of Ti is 1s2â2s2â2p6â3s2â3p6â3d2â4s2

Topic: Chemistry, Structure of Atom, JEE Main