datavorous/entrance-exam-dataset huggingface dataset:
Sum of $n$ terms of series $12+16+24+40+\ldots$. will be

Options:
A $2\left(2^n-1\right)+8 n$
B $2\left(2^n-1\right)+6 n$
C $3\left(2^n-1\right)+8 n$
D $4\left(2^n-1\right)+8 n$

The correct answer is:
$4\left(2^n-1\right)+8 n$

Solution:
Let $n^{\text {th }}$ term of series is $T_n$ then
$S_n=12+16+24+40+\ldots . .+T_n$
Again $S_n=12+16+24+\ldots \ldots+T_n$
On subtraction
$0=(12+4+8+16+\ldots+$ upto $n$ terms $)-T_n$
or $T_n=12+[4+8+16+\ldots+\operatorname{upto}(n-1)$ terms $]$
$=12+\frac{4\left(2^{n-1}-1\right)}{2-1}=2^{n+1}+8$
On putting $n=1,2,3 \ldots .$.
$T_1=2^2+8 \quad T_2=2^3+8 \quad T_3=2^4+8 \ldots \ldots .$. etc.
$S_n=T_1+T_2+T_3+\ldots .+T_n$
$=\left(2^2+2^3+2^4+\ldots\right.$ upto $n$ terms $)$
$+(8+8+8+\ldots \ldots$. upto $n$ terms $)$
$=\frac{2^2\left(2^n-1\right)}{2-1}+8 n=4\left(2^n-1\right)+8 n$.

Topic: Mathematics, Sequences and Series, JEE Main
Sum of odd terms is $\mathrm{A}$ and sum of even terms is $\mathrm{B}$ in the expansion $(x+a)^n$, then

Options:
A $A B=\frac{1}{4}(x-a)^{2 n}-(x+a)^{2 n}$
B $2 A B=(x+a)^{2 n}-(x-a)^{2 n}$
C $4 A B=(x+a)^{2 n}-(x-a)^{2 n}$
D None of these

The correct answer is:
$4 A B=(x+a)^{2 n}-(x-a)^{2 n}$

Solution:
$(x+a)^n={ }^n C_0 x^n+{ }^n C_1 x^{n-1} a+{ }^n C_2 x^{n-2} a^2+{ }^n C_3 x^{n-3} a^3+\ldots . .$
But by the condition,
$\begin{aligned}& A={ }^n C_0 x^n+{ }^n C_2 x^{n-2} a^2+{ }^n C_4 x^{n-4} a^4+\ldots . . \\
& B={ }^n C_1 x^{n-1} a+{ }^n C_3 x^{n-3} a^3+\ldots . .\end{aligned}$
Hence $A B=\frac{1}{4}\left\{(x+a)^{2 n}-(x-a)^{2 n}\right\}$
or $4 A B=(x+a)^{2 n}-(x-a)^{2 n}$

Topic: Mathematics, Binomial Theorem, JEE Main
sum of ordinate points of co normal

Options:
A 1
B 0
C 2
D not constant

The correct answer is:
0

Solution:
For a parabola $y=x^2$, the sum of the $y$-coordinates (ordinates) of the points where the normals intersect at a common point is always zero. This is a specific property of the parabola. If the normals intersect at a common point, then the sum of the ordinates of these points on the parabola is zero.

Topic: Mathematics, Ellipse, JEE Main
Sum of oxidation state (magnitude) and coordination number of cobalt in NaCobpyCl4 is

Options:

The correct answer is:
9

Solution:
2,2′-Bipyridine is an organic compound, This colorless solid is an important isomer of the bipyridine family. It is a bidentate chelating ligand, forming complexes with many transition metals.

There is one bidentate ligand and four monodentate ligand attached with Co metal ion. So, Coordination no. will be =6

This bpy ligand has no net charge. and chloride ion has -1 charge. So, Oxidation state of metal ion will be=3

Their sub will be =6+3=9

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2022 (29 Jul Shift 2)
Sum of oxidation states of bromine in bromic acid and perbromic acid is

Options:

The correct answer is:
12

Solution:
Structure of Bromic acid:
HBrO3 (Bromic acid)
Oxidation State of Br=+5
Structure of Per-boric acid:
HBrO4 (per bromic acid)
Oxidation state of Br=+7
Thus, Sum of Oxidation state =12

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2023 (01 Feb Shift 1)
Sum of series of \( \cot ^{-1} \frac{5}{\sqrt{3}}+\cot ^{-1} \frac{9}{\sqrt{3}}+\cot ^{-1} \frac{15}{\sqrt{3}}+\cot ^{-1} \frac{23}{\sqrt{3}}+\ldots \ldots \) is equal to

Options:
A \( \frac{\pi}{4} \)
B \( \frac{\pi}{3} \)
C \( \frac{\pi}{6} \)
D \( \frac{\pi}{12} \)

The correct answer is:
\( \frac{\pi}{3} \)

Solution:
cot-153+cot-193+cot-1153+cot-1233+.....∞

=tan-123-131+23×13+tan-133-231+33×23+.....+....+tan-1n+13-n31+n+13×n3+.....

=tan23−tan−113+tan−133−tan−123+tan−143−tan−133+.....+tan−1n+13−tan−1n3

⇒ Sn=tan-1n+13-tan-113

S∞=π2-π6=π3

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
Sum of the first hundred numbers common to the two arithmetic progressions 12,15,18,……… and 17,21,25,…….. is

Options:
A 56100
B 65100
C 61500
D 51600

The correct answer is:
61500

Solution:
AP1 =12,15,18,â¦â¦. (common difference d1=3)
AP2 =17,21,25,â¦â¦. (common difference d2=4)
First term of the series of common numbers =21 .
Here, a=21, and common difference of the series of common numbers =LCM of d1 and d2=12
â Required sum of the first hundred terms =10022Ã21+100-112
=10021+99Ã6
=61500

Topic: Mathematics, Sequences and Series, JEE Main
Sum of the series P=121+2+132+23+143+34+…+110099+99100 is

Options:
A 110
B 310
C 910
D 12

The correct answer is:
910

Solution:
tr=1r+1r+rr+1
=1rr+1r+1+r×r+1-rr+1-r
=r+1-rrr+1=1r-1r+1
P=∑r=199tr=1-1100=910

Topic: Mathematics, Sequences and Series, JEE Main
Sum of the squares of all integral values of a for which the inequality x 2 + a x + a 2 + 6 a < 0 is satisfied for all x â 1 ,â¡ 2 must be equal to

Options:
A 90
B 89
C 88
D 91

The correct answer is:
91

Solution:
Let, fâ¡ x = x 2 + a x + a 2 + 6 a
â´ f(1)â¤0

â a 2 + 7 a + 1 < 0
or - 7 - 3 5 2 < a < - 7 + 3 5 2 ....(i)
f(2)â¤0
â a 2 + 8 a + 4 < 0
or - 4 - 2 3 < a < - 4 + 2 3 ....(ii)
and D>0
âa2-4Â·1a2+6a>0
â a 2 + 8 a < 0
or - 8 < a < 0 ....(iii)
From Eqs. (i), (ii) and (iii), we get,
- 7 - 3 5 2 â¤ a â¤ - 4 + 2 3
Hence, integral values of a are - 6 ,â¡ - 5 ,â¡ - 4 ,â¡ - 3 ,â¡ - 2 ,â¡ - 1
Required Sum
= - 6 2 + - 5 2 + - 4 2 + - 3 2 + - 2 2 + - 1 2
= 91.

Topic: Mathematics, Quadratic Equation, JEE Main
Sum of the two binary numbers $(1000010)_2$ and $(11011)_2$ is

Options:
A $(111101)_2$
B $(111111)_2$
C $(101111)_2$
D $(111001)_2$

The correct answer is:
$(111101)_2$

Solution:
$\begin{aligned} & (100010)_2=2^5 \times 1+2^4 \times 0+2^3 \times 0+2^2 \times 0+ \\ & 2^1 \times 1+2^0 \times 0=32+0+0+0+2+0=(34)_{10} \\ & \text { and }(11011)_2=2^4 \times 1+2^3 \times 1+2^2 \times 0+2^1 \times 1+2^0 \times 1 \\ & =16+8+0+2+1=(27)_{10} \\ & \therefore \text { Sum }(100010)_2+(11011)_2=(34)_{10}+(27)_{10}=(61)_{10}\end{aligned}$
Now
$\therefore$ Required sum (in binary system)
$(100010)_2+(11011)_2=(111101)_2$

Topic: Physics, Semiconductors, JEE Main
Sum of total number of amphoteric and neutral oxides among the following is:
CO,NO, Al 2 O 3 ,PbO 2 ,CaO, SnO 2 ,ZnO

Options:

The correct answer is:
6

Solution:
CO &NOare neutral oxide.
CaO basic oxide. Rest of them are amphoteric oxide.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Sum to n terms of the series 141.3+243.5+345.7+..... is equal to

Options:
A nn+12n2+182n+1
B nn+1n2+n+162n+1
C n+12n+12+182n+1
D nn+12n+12+1162n+1

The correct answer is:
nn+1n2+n+162n+1

Solution:
The kth term of the given series is
uk=k42k-12k+1=k24+116+13212k-1-12k+1
Therefore the sum of n terms is
Sn=u1+u2+....+un
=14∑k=1nk2+n16+1321-12n+1
=nn+12n+14.6+n16+n162n+1
=nn+1n2+n+162n+1

Topic: Mathematics, Sequences and Series, JEE Main
\( \sum_{r=1}^{n} \frac{r}{r^{4}+r^{2}+1} \) is equal to

Options:
A \( \frac{n^{2}+n}{2\left(n^{2}+n+1\right)} \)
B \( \frac{n^{2}+2 n}{2\left(n^{2}+n+1\right)} \)
C \( \frac{2 n^{2}+n}{2\left(n^{2}+n+1\right)} \)
D \( \frac{n^{2}+n}{\left(n^{2}+n+1\right)} \)

The correct answer is:
\( \frac{n^{2}+n}{2\left(n^{2}+n+1\right)} \)

Solution:
Let, Ιr=rr4+r2+1=rr2+r+1r2-r+1
=12r2+r+1-r2-r+1r2+r+1r2-r+1
=-121r2+r+1-1r2-r+1
=-12Vr-Vr-1
⇒∑r=1nΙr=-12Vn-V0
=-121n2+n+1-1
=121-1n2+n+1=12n2+nn2+n+1

Topic: Mathematics, Sequences and Series, JEE Main
Sun light falls normally on a surface of area 36 cm2 and exerts an average force of 7.2×10-9 N within a time period of 20 minutes. Considering a case of complete absorption, the energy flux of incident light is

Options:
A 25.92×102 W cm-2
B 8.64×10-6 W cm-2
C 6.0 W cm-2
D 0.06 W cm-2

The correct answer is:
0.06 W cm-2

Solution:
The radiation pressure in case of complete absorption is given by,

P=Ic.

Therefore, the force will be,

F=Ic×A⇒I=FcA

⇒I=7.2×10-9×3×10836×10-4=600 W m-2⇒I=0.06 W cm-2

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2022 (28 Jul Shift 2)
Sun subtends an angle of 0.5Â° at the centre of curvature of a concave mirror of radius of curvature 15 m. The diameter of the image of the sun formed by the mirror is

Options:
A 8.55 cm

B 7.55 cm

C 6.55 cm

D 5.55 cm

The correct answer is:
6.55 cm

Solution:
Î¸=12ÃÏ180 rad
diametar of imagefocal length=Î¸
Or diameter of image
=12ÃÏ180Ã152Ã100 cm=6.55 cm

Topic: Physics, Ray Optics, JEE Main
Sunlight of intensity 50 W m-2 is incident normally on the surface of a solar panel. Some part of incident energy (25%) is reflected from the surface and the rest is absorbed. The force exerted on 1 m2 surface area will be close to c=3×108 m s-1

Options:
A 15×10-8 N
B 20×10-8 N
C 10×10-8 N
D 35×10-8 N

The correct answer is:
20×10-8 N

Solution:
The pressure exerted by the photons

P=0.75×Ic +0.25×2Ic ( P = pressure, c = speed of light, I = intensity )
P=1.25 ×Ic
We know the force can be computed using
F=P×A

( F = force, P = pressure, A = area )
F=1.25×50×13×108
F=20.83×10-8 N

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2019 (09 Apr Shift 2)
Suppose $\theta \epsilon\left[0, \frac{\pi}{4}\right]$ is a solution of $4 \cos \theta-3 \sin \theta=1$. Then $\cos \theta$ is equal to :

Options:
A $\frac{4}{(3 \sqrt{6}+2)}$
B $\frac{6+\sqrt{6}}{(3 \sqrt{6}+2)}$
C $\frac{4}{(3 \sqrt{6}-2)}$
D $\frac{6-\sqrt{6}}{(3 \sqrt{6}-2)}$

The correct answer is:
$\frac{4}{(3 \sqrt{6}-2)}$

Solution:
$4\left(\frac{1-\tan ^2 \theta / 2}{1+\tan ^2 \theta / 2}\right)-3\left(\frac{2 \tan \frac{\theta}{2}}{1+\tan ^2 \frac{\theta}{2}}\right)=1$
let $\tan \frac{\theta}{2}=\mathrm{t}$
$\begin{aligned}
& \frac{4-4 t^2-6 t}{1+t^2}=1 \\
& 4-4 t^2-6 t=1+t^2
\end{aligned}$
$\begin{aligned} & \Rightarrow 5 t^2+6 t-3=0 \\ & \Rightarrow t=\frac{-6 \pm \sqrt{36-4(5)(-3)}}{2(5)} \\ & =\frac{-6 \pm \sqrt{96}}{10} \\ & =\frac{-6 \pm 4 \sqrt{6}}{10} \\ & t=\frac{-3+2 \sqrt{6}}{5} \\ & \cos \theta=\frac{1-t^2}{1+t^2}=\frac{1-\left(\frac{2 \sqrt{6}-3}{5}\right)^2}{1+\left(\frac{2 \sqrt{6}-3}{5}\right)^2}=\frac{1-\left(\frac{24+9-12 \sqrt{6}}{25}\right)}{1+\left(\frac{24+9-12 \sqrt{6}}{25}\right)} \\ & =\frac{25-33+12 \sqrt{6}}{25+33-12 \sqrt{6}}=\frac{12 \sqrt{6}-8}{58-12 \sqrt{6}}=\frac{6 \sqrt{6}-4}{29-6 \sqrt{6}} \times \frac{29+6 \sqrt{6}}{29+6 \sqrt{6}} \\ & =\frac{100+150 \sqrt{6}}{625}=\frac{4+6 \sqrt{6}}{25} \times \frac{4-6 \sqrt{6}}{4-6 \sqrt{6}} \\ & =\frac{-200}{25(4-6 \sqrt{6})}=\frac{-8}{4-6 \sqrt{6}}=\frac{4}{3 \sqrt{6}-2}\end{aligned}$

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2024 (05 Apr Shift 1)
Suppose 28-p, p, 70-α, α are the coefficient of four consecutive terms in the expansion of (1+x)n. Then the value of 2α-3p equals

Options:
A 7
B 10
C 4
D 6

The correct answer is:
7

Solution:
Given,

28-p, p, 70-α, α are the coefficient of four consecutive terms in the expansion of (1+x)n

So, Crn=28-p, Cr+1n=p, Cr+2n=70-α & Cr+3n=α

⇒Crn+Cr+1n=28 & Cr+2n+ Cr+3n=70

⇒Cr+1n+1=28 & Cr+3n+1=70

We know that, C28=28 or C128=28,

Now on comparing with Cr+1n+1=28 we get, n,r=7,1 or n,r=27,0

Now, checking with Cr+3n+1=70 we get, n,r=7,1

So, p=C27=21 and α=C47=35

Hence, the value of 2α-3p=70-63=7

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2024 (30 Jan Shift 2)
Suppose 60% w/w aqueous solution of glucose (C6H12O6) and 20% w/w aqueous solution of urea (NH2CONH2) have equal molarity, then which solution has higher density:

Options:
A Both have equal density
B Glucose solution
C Urea solution
D Cannot be predicted

The correct answer is:
Both have equal density

Solution:
Given:
M1=M2
10d1x1Mol. wt.1=10d2x2Mol. wt.2
â´ d1d2=x2x1ÃMol. wt.1 Mol. wt.2=2060Ã18060=11
Thus, d1=d2 Here x 1 , x 2 ââ are percentage weights.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
Suppose a1,a2,2,a3,a4 be in an arithmetico-geometric progression. If the common ratio of the corresponding geometric progression is 2 and the sum of all 5 terms of the arithmetico-geometric progression is 492, then a4 is equal to ______________

Options:

The correct answer is:
16

Solution:
Given,

a1,a2,2,a3,a4 be in an arithmetico-geometric progression,

And the common ratio of the corresponding geometric progression is 2 and the sum of all 5 terms of the arithmetico-geometric progression is 492

So,common ratio Here r = 2 and let common difference = d

And we know that,

a2=a1+d×2

2=a1+2d×22 ..........i

a3=a1+3d×23

a4=a1+4d×24

Now adding all above equations we get,

31a1+98d=492 .........ii

Now from i & ii

a1=0,d=14

Hence, a4=16(0+1)=16

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2023 (10 Apr Shift 2)
Suppose a1,a2,…,an,… be an arithmetic progression of natural numbers. If the ratio of the sum of the first five terms to the sum of first nine terms of the progression is 5:17 and 110<a15<120, then the sum of the first ten terms of the progression is equal to

Options:
A 290
B 380
C 460
D 510

The correct answer is:
380

Solution:
Given the ratio of the sum of the first five terms to the sum of the first nine terms is 5:17,

So, S5S9=517⇒522a+4d922a+8d=517

⇒172a+4d=92a+8d

⇒34a+68d=18a+72d

⇒d=4a

Now a15=a+14d=57a

Also given 110<a15<120

⇒110<57a<120

⇒a=2 ∴ d=8

So, S10=1022×2+9×8=380

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2022 (27 Jul Shift 1)
Suppose A=abcd is a real matrix with non-zero entries, ad-bc=0 and A2=A. Then, a+d equals

Options:
A 1
B 2
C 3
D 4

The correct answer is:
1

Solution:
We have,

A=abcd

A2=abcdabcd

A2=a2+bcab+bdac+cdbc+d2

Given, A2=A and ad-bc=0

∴a2+bcab+bdac+cdbc+d2=abcd

ab+bd=b

ba+d=b

⇒a+d=1

Topic: Mathematics, Matrices, JEE Main
Suppose $A B$ is a focal chord of the parabola $y^2=12 x$ of length $l$ and slope $\mathrm{m} < \sqrt{3}$. If the distance of the chord $\mathrm{AB}$ from the origin is $\mathrm{d}$, then $l \mathrm{~d}^2$ is equal to _______

Options:

The correct answer is:
108

Solution:
$\begin{aligned} & \ell=4 \mathrm{a} \operatorname{cosec}^2 \theta \\ & \ell=12 \times \frac{9}{\mathrm{~d}^2} \\ & \ell \mathrm{d}^2=108\end{aligned}$

Topic: Mathematics, Parabola, JEE Main, JEE Main 2024 (05 Apr Shift 1)
Suppose a class has 7 students. The average marks of these students in the mathematics examination is 62, and their variance is 20. A student fails in the examination if he/she gets less than 50 marks, then in worst case, the number of students can fail is

Options:

The correct answer is:
0

Solution:
Applying the formula of variance we get,

20=∑i=17xi-6227

⇒x1-622+x2-622+….+x7-622=140

If x1=49 (taking the least possibility)

49-622=169

then,

x2-622+….+x7-622=140-169=-29 Negative Number which is not possible, therefore, no student can fail.

Topic: Mathematics, Statistics, JEE Main, JEE Main 2022 (28 Jun Shift 2)
Suppose a differentiable function fx satisfies the identity fx+y=fx+fy+xy2+x2y, for all real x and y. If limx→0fxx=1, then f'3 is equal to :

Options:

The correct answer is:
10

Solution:
∵limx→0fxx=1 ⇒f'0=1

fx+y=fx+fy+xy2+x2y

Differentiate w.r.t. x keeping y constant

f'x+y=f'x+0+y2+2xy

put y=-x

f'0=f'x+x2−2x2

1=f'x−x2

f'x=1+x2

f'3=10.

Topic: Mathematics, Functions, JEE Main, JEE Main 2020 (04 Sep Shift 1)
Suppose A is 3×3 matrix consisting of integer entries that are chosen at random from the set -1000,-999,…999,1000. Let P be the probability that either A2=-I or A is diagonal, where I is the 3×3 identity matrix. Then,

Options:
A P<11018
B P=11018
C 521018≤P≤531018
D P≤541018

The correct answer is:
P<11018

Solution:
We have, A is 3×3 matrix consisting of integer entries from -1000,-999,…999,1000

Total entries =2001

Total number of outcomes =20019

Given, A2=-I,A=-A-1 (which is not possible)

A is diagonal matrix.

A=a000a000a

Total number of favourable outcomes

=20013

Required probability

=20013200192001-6=2000+1-6

p=2000-61+12000-6

P=126×10181+12000-6

P<126×1018 [ ∵1+12000 is decreasing]

P<11018

Topic: Mathematics, Probability, JEE Main
Suppose a is a positive real number such that $a^{5}-a^{3}+a=2 .$ Then

Options:
A $a^{6} < 2$
B $2 < a^{6} < 3$
C $3 < \mathrm{a}^{6} < 4$
D $4 \leq \mathrm{a}^{6}$

The correct answer is:
$3 < \mathrm{a}^{6} < 4$

Solution:
$a^{5}-a^{3}+a=2 ; a \in R^{+}$
Let $\mathrm{f}(\mathrm{a})=\mathrm{a}^{5}-\mathrm{a}^{3}+\mathrm{a}-2 ;$ Note $\left.\mathrm{f}(\mathrm{a})>0 \forall \mathrm{a} \in \mathrm{R}\right\}$
for $a^{6}=3 \Rightarrow a=3^{1 / 6}=1.2$ \{use calculator $\}$
we get $\mathrm{f}(1.2) < 0$ and at $\mathrm{a}=4^{1 / 6} \Rightarrow \mathrm{f}\left(4^{1 / 6}\right)>0$
so one root in $\mathrm{a} \in(3,4)$

Topic: Mathematics, Quadratic Equation, JEE Main
Suppose $A$ is any $3 \times 3$ non-singular matrix and $(A-3 I)(A-5 I)=O$, where $I=I_3$ and $O=O_3$. If $\alpha A+$ $\beta A^{-1}=4 I$, then $\alpha+\beta$ is equal to

Options:
A
8

B
12

C
13

D
7

The correct answer is:
8

Solution:
We have
$$
\begin{aligned}
&(A-3 I)(A-5 I)=O \\
&\Rightarrow A^2-8 A+15 I=O
\end{aligned}
$$
Multiplying both sides by $A^{-1}$, we get;
$$
\begin{aligned}
A^{-1} A \cdot A-8 A^{-1} A+15 A^{-1} I=A^{-1} O \\
\Rightarrow & A-8 I+15 A^{-1}=O \\
& A+15 A^{-1}=8 I \\
& \frac{A}{2}+\frac{15 A^{-1}}{2}=4 I \\
\therefore & \alpha+\beta=\frac{1}{2}+\frac{15}{2}=\frac{16}{2}=8
\end{aligned}
$$

Topic: Mathematics, Matrices, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
Suppose a parabola $y=a x^{2}+b x+c$ has two $x$ intercepts one positive and one negative, and its vertex is $(2,2)$. Then which of the following is true

Options:
A $a b>0$
B $b c>0$
C $\mathrm{ca}>0$
D $a+b+c>0$

The correct answer is:
$b c>0$

Solution:
$y=a x 2+b x+c$
$a>0, b < 0, c < 0$
Also $y(1)=a+b+c < 0$

Topic: Mathematics, Parabola, JEE Main
Suppose a population $A$ has 100 observations $101,102, \ldots, 200$, and another population $B$ has 100 observations $151,152, \ldots, 250$. If $V_A$ and $V_B$ represent the variances of the two populations, respectively, then $\frac{V_A}{V_B}$ is

Options:
A
1

B
9/4

C
4/9

D
2/3

The correct answer is:
1

Solution:
$\sigma_x^2=\frac{\sum d_1^2}{n}$. (Here deviations are taken from the mean)
Since $A$ and $B$ both has 100 consecutive integers, therefore both have same standard deviation and hence the variance.
$\therefore \frac{V_A}{V_B}=1\left(\right.$ As $\sum d_i^2$ is same in both the cases).

Topic: Mathematics, Statistics, JEE Main, JEE Main 2006
Suppose a quantity can be dimensionally represented in terms M, L and T, that is y=Ma Lb Tc. The quantity mass

Options:
A maybe represented in terms of L,T and y if a=0
B maybe represented in terms of L,T and y if a â 0
C Can always be dimensionally represented in terms of L,T and y
D Can never be dimensionally represented in terms of L,T and y

The correct answer is:
maybe represented in terms of L,T and y if a â 0

Solution:
M=y1a.Lâba.Tâca
where a should not be zero.

Topic: Physics, Units and Dimensions, JEE Main
Suppose a smooth tunnel is dug along a straight line joining two points on the surface of the earth and a particle is dropped from rest at its one end. Assume that mass of the earth is uniformly distributed over its volume. Then,

Options:
A the particle will emerge from the other end with velocity \( \sqrt{\frac{G M_{\mathrm{e}}}{2 R_{\mathrm{e}}}} \) where \( M_{\mathrm{e}} \) and \( R_{\mathrm{e}} \) are earth's mass and radius
pectively.
B the overlapping will come to rest at the centre of the tunnel.
C potential energy of the particle will be equal to zero at centre of tunnel, if it is along a diameter.
D acceleration of the particle will be proportional to its distance from mid-point of the tunnel.

The correct answer is:
acceleration of the particle will be proportional to its distance from mid-point of the tunnel.

Solution:
In this case, the particle performs Simple harmonic motion and in the case of simple harmonic motion the acceleration is always directly proportional to distance from the mean position and always tending to reach the mean position which means a restoring force is acted.

⇒F=-kx

⇒ma=-kx

⇒a∝-x.

Topic: Physics, Oscillations, JEE Main
Suppose a train that has \( 360 \mathrm{~Hz} \) frequency sound horn is moving at \( 20 \mathrm{~m} / \mathrm{sec} \) towards an observer, then the frequency observed by a stationary observer is- (Speed of sound in air \( =340 \mathrm{~m} / \mathrm{s} \) )

Options:
A \( 20 \mathrm{~Hz} \)
B \( 340 \mathrm{~Hz} \)
C \( 600 \mathrm{~Hz} \)
D \( 40 \mathrm{~Hz} \)

The correct answer is:
\( 340 \mathrm{~Hz} \)

Solution:
Given,

Actual frequency f = 360 Hz

Velocity of sound sourceVs=20 m/s

Speed of sound wave

V = 340 m/s

As the observer is stationary and source is moving towards an observer then the observed or apparent frequency will be as follows.

f'=fVV+Vs

f'=360340340+20=340 Hz

Topic: Physics, Waves and Sound, JEE Main
Suppose a uniformly charged wall provides a uniform electric field of 2×104 N C-1 normally. A charged particle of mass 2 g being suspended through a silk thread of length 20 cm and remain stayed at a distance of 10 cm from the wall. Then the charge on the particle will be 1xμC where x= ________ [use g=10 m s-2]

Options:

The correct answer is:
3

Solution:
As we can see from geometry, sinθ=1020=12

⇒θ=30°

Vertical component of tension will balance the weight and the horizontal component will balance the electric force. Therefore,

tanθ=qEmg

⇒tan30°=q×2×1042×10-3×10

⇒13=q×106

⇒q=13×10-6 C=13 μC

Hence, x=3.

Topic: Physics, Electrostatics, JEE Main, JEE Main 2024 (01 Feb Shift 2)
Suppose an electron is attracted towards the origin by a force $k / r$ where ' $k$ ' is a constant and ' $r$ ' is the distance of the electron from the origin. By applying Bohr model to this system, the radius of the $n^{\text {th }}$ orbital of the electron is found to be ' $r_n$ ' and the kinetic energy of the electron to be $T_n$. Then which of the following is true?

Options:
A
$T_n \propto 1 / n^2, \quad r_n \propto n^2$

B
$T_n$ independent of $n, r_n \propto n$

C
$T_n \propto 1 / n, \quad r_n \propto n$

D
$T_n \propto 1 / n, \quad r_n \propto n^2$

The correct answer is:
$T_n$ independent of $n, r_n \propto n$

Solution:
$$
\begin{aligned}
& \frac{\mathrm{k}}{\mathrm{r}}=\frac{m v^2}{\mathrm{r}} \\
& \mathrm{mv}=\mathrm{k} \quad \text { (independent or } r \text { ) } \\
& \mathrm{n}\left(\frac{\mathrm{h}}{2 \pi}\right)=m v r \Rightarrow r \propto \mathrm{n} \text { and } \mathrm{T}=\frac{1}{2} m v^2 \text { is independent of } \mathrm{n} .
\end{aligned}
$$

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2008
Suppose $\theta$ and $\phi(\neq 0)$ are such that $\sec (\theta+\phi)$, sec $\theta$ and $\sec (\theta-\phi)$ are in A.P. If $\cos \theta=k \cos \left(\frac{\phi}{2}\right)$ for some $k$, then $k$ is equal to

Options:
A
$\pm \sqrt{2}$

B
$\pm 1$

C
$\pm \frac{1}{\sqrt{2}}$

D
$\pm 2$

The correct answer is:
$\pm \sqrt{2}$

Solution:
Since, $\sec (\theta-\phi), \sec \theta$ and $\sec (\theta+\phi)$ are in A.P.,
$$
\begin{aligned}
& \therefore 2 \sec \theta=\sec (\theta-\phi)+\sec (\theta+\phi) \\
& \Rightarrow \frac{2}{\cos \theta}=\frac{\cos (\theta+\phi)+\cos (\theta-\phi)}{\cos (\theta-\phi) \cos (\theta+\phi)}
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow 2\left(\cos ^2 \theta-\sin ^2 \phi\right)=\cos \theta[2 \cos \theta \cos \phi \\
& \Rightarrow \cos ^2 \theta(1-\cos \phi)=\sin ^2 \phi=1-\cos ^2 \phi \\
& \Rightarrow \quad \cos ^2 \theta=1+\cos \phi=2 \cos ^2 \frac{\phi}{2} \\
& \therefore \quad \cos \theta=\pm \sqrt{2} \cos \frac{\phi}{2} \\
& \text { But given } \cos \theta=\mathrm{k} \cos \frac{\phi}{2} \\
& \therefore \quad \mathrm{k}=\pm \sqrt{2}
\end{aligned}
$$

Topic: Mathematics, Trigonometric Equations, JEE Main