datavorous/entrance-exam-dataset huggingface dataset:
If the constant term in the binomial expansion of x522-4xl9 is -84 and the coefficient of x-3l is 2αβ where β<0 is an odd number, then αl-β is equal to _____ .

Options:

The correct answer is:
98

Solution:
Given,

The constant term in the binomial expansion of x522-4xl9 is -84 and the coefficient of x-3 is 2αβ

Now General Term of the binomial is given by, Tr+1=Cr9x5229-r-4xlr=-1rCrx452-5r2-lr23-9

Now for constant term 45-5r-2lr=0⇒r=455+2l

So, coefficient of constant term will be,

-19×9Cr23r-9=-84⇒r=3

Now putting the value of r in r=455+2l we get, l=5

Now coefficient of x-3l=x-15⇒452-5r2-lr=-15⇒r=5 when l=5
Now coefficient of x-15=-159 C526=-6327=β2α

Hence, on comparing we get, α=7, β=-63 & l=5

So, the value of α l-β=7×5--63=98

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2023 (31 Jan Shift 2)
If the constant term in the binomial expansion of x-kx210 is 405, then k equals :

Options:
A 9
B 1
C 3
D 2

The correct answer is:
3

Solution:
Tr+1=Cr10-kx2rx10-r

=Cr·10-kr·x5-5r2

for constant term

⇒5-5r2=0 ⇒r=2

⇒T3=C210·k2=405

⇒10(9)2k2=405

⇒k2=9

⇒k=3

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2020 (06 Sep Shift 2)
If the constant term in the expansion of $\left(\frac{\sqrt[5]{3}}{x}+\frac{2 x}{\sqrt[3]{5}}\right)^{12}, x \neq 0$, is $\alpha \times 2^8 \times \sqrt[5]{3}$, then $25 \alpha$ is equal to :

Options:
A 724
B 742
C 639
D 693

The correct answer is:
693

Solution:
$\begin{aligned} & \mathrm{T}_{\mathrm{r}+1}={ }^{12} \mathrm{C}_{\mathrm{r}}\left(\frac{3^{1 / 5}}{\mathrm{x}}\right)^{12-\mathrm{r}}\left(\frac{2 \mathrm{x}}{5^{1 / 3}}\right)^{\mathrm{r}} \\ & \mathrm{T}_{\mathrm{r}+1}=\frac{{ }^{12} \mathrm{C}_{\mathrm{r}}(3)^{\frac{12-\mathrm{r}}{5}}(2)^{\mathrm{r}}(\mathrm{x})^{2 \mathrm{r}-12}}{(5)^{\mathrm{r} / 3}} \\ & \mathrm{r}=6 \\ & \mathrm{~T}_7=\frac{{ }^{12} \mathrm{C}_6(3)^{6 / 5}(2)^6}{5^2}=\left(\frac{9 \times 11 \times 7}{25}\right) 2^8 \cdot 3^{1 / 5} \\ & 25 \alpha=693\end{aligned}$

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2024 (05 Apr Shift 2)
If the coordinates at one end of a diameter of the circle $x^{2}+y^{2}-8 x-4 y+c=0$ are $(-3,2)$, then the coordinates at the other end are

Options:
A $(5,3)$
B $(6,2)$
C $(1,-8)$
D $(11,2)$

The correct answer is:
$(11,2)$

Solution:
The centre of the given circle is $\mathrm{C} \equiv(4,2)$ Let $\mathrm{A} \equiv(-3,2)$

If $(\alpha, \beta)$ are the coordinates of the other end of the diameter, then, as the middle ploint of the diameter is the centre,
$\therefore \quad \frac{\alpha-3}{2}=4$ and $\frac{\beta+2}{2}=2 \Rightarrow \alpha=11, \beta=2$
Thus, the coordinates of the other end of diameter are $(11,2)$

Topic: Mathematics, Circle, JEE Main
If the coordinates of $A$ and $B$ be $(1,1)$ and $(5,7)$, then the equation of the perpendicular bisector of the line segment $A B$ is

Options:
A $2 x+3 y=18$
B $2 x-3 y+18=0$
C $2 x+3 y-1=0$
D $3 x-2 y+1=0$

The correct answer is:
$2 x+3 y=18$

Solution:
Mid point is $(3,4)$
Slope of perpendicular $=\frac{-1}{6 / 4}=\frac{-2}{3}$
Hence the line is $2 x+3 y=18$.

Topic: Mathematics, Straight Lines, JEE Main
If the co-ordinates of the point in which the line joining the points $(3,5,-7)$ and $(-2,1,8)$ is intersected by the plane $y z$ is $\left[\mathrm{a}, \frac{13}{\mathrm{~b}}, \mathrm{c}\right]$, then $(a+b-c)=$

Options:
A
2
B
3
C
4
D
5

The correct answer is:
3

Solution:
Ratio $=-\left(\frac{3}{-2}\right)=\frac{3}{2}$
$\therefore \quad$ Required co-ordinates of the points are
$\begin{aligned}
& {\left[\frac{6-6}{5}, \frac{10+3}{5}, \frac{-14+24}{5}\right]=\left(0, \frac{13}{5}, 2\right) .} \\
& \therefore a+b-c=0+5-2=3
\end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the coordinates of the vertices of the triangle $A B C$ be $(-1,6),(-3,-9)$, and $(5,-8)$ respectively, then the equation of the median through $C$ is

Options:
A $13 x-14 y-47=0$
B $13 x-14 y+47=0$
C $13 x+14 y+47=0$
D $13 x+14 y-47=0$

The correct answer is:
$13 x+14 y+47=0$

Solution:
Let the Median be CD
Point of \(D\) is \((x, y)\)
Now, \(x=\frac{-3-1}{2}, x=-2\) and \(y=\frac{6-9}{2}=-\frac{3}{2}\)
Slope of \(\mathrm{CD}\) is,
\(=\frac{8-\frac{3}{2}}{5+2}=\frac{13}{14}\)
Equation:
\(C D=y+8=\frac{13}{14}(\mathrm{x}-5)\)
So the equation of Median,
\(13 \mathrm{x}-14 \mathrm{y}-47=0\)

Topic: Mathematics, Straight Lines, JEE Main
If the critical angle for total internal reflection from a medium to vacuum is $30^{\circ}$, the velocity of light in the medium is

Options:
A $3 \times 10^8 \mathrm{~m} / \mathrm{s}$
B $1.5 \times 10^8 \mathrm{~m} / \mathrm{s}$
C $6 \times 10^8 \mathrm{~m} / \mathrm{s}$
D $\sqrt{3} \times 10^8 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$1.5 \times 10^8 \mathrm{~m} / \mathrm{s}$

Solution:
$\begin{gathered}\mu=\frac{1}{\sin C}=\frac{1}{\sin 30}=2 \\ \therefore v=\frac{3 \times 10^8}{2}=1.5 \times 10^8 \mathrm{~m} / \mathrm{s}\end{gathered}$

Topic: Physics, Ray Optics, JEE Main
If the cube roots of unity are $1, \omega, \omega^2$ then the roots of the equation $(x-1)^3+8=0$, are

Options:
A
$-1,-1+2 \omega,-1-2 \omega^2$

B
$-1,-1,-1$

C
$-1,1-2 \omega, 1-2 \omega^2$

D
$-1,1+2 \omega, 1+2 \omega^2$

The correct answer is:
$-1,1-2 \omega, 1-2 \omega^2$

Solution:
$(x-1)^3+8=0 \Rightarrow(x-1)=(-2)(1)^{1 / 3}$ $\Rightarrow x-1=-2$ or $-2 \omega$ or $-2 \omega^2$ or $n=-1$ or $1-2 \omega$ or $1-2 \omega^2$.

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2005
If the cubic equation z3+az2+bz+c=0 ∀a,b,c∈R, c≠0 has a purely imaginary root, then (where i2=-1)

Options:
A c=ab
B b=ac
C the imaginary root is equal to ±ic
D the imaginary root is equal to ±ia

The correct answer is:
c=ab

Solution:
Let, z=iα
⇒c-α2a+ibα-α3=0
α≠0⇒α2=b⇒c=ab
And z=±ib

Topic: Mathematics, Complex Number, JEE Main
If the curve satisfies the differential equation xdydx=x2+y-2 and passes through 1,1, then it also passes through the point

Options:
A 4,4
B 3,3
C 2,2
D 0,0

The correct answer is:
2,2

Solution:
The given equation is dydx+y-1x=x-2x
Integration factor =e∫-1xdx=1x
∴ the solution is y1x=∫1xx-2xdx
⇒yx=x+2x+c
As it passes through 1,1
c=-2
∴y=x2+2-2x
∴ it passes through 2,2

Topic: Mathematics, Differential Equations, JEE Main
If the curve satisfying differential equation xdy=y+x3dx passes through 1,1, then the equation of the curve is

Options:
A y2=x3-x
B y=x2-x
C 2y=x3
D 2y=x3+x

The correct answer is:
2y=x3+x

Solution:
The given equation is xdy-ydxx2=xdx
or dyx=xdx
On integrating, we get, yx=x22+C
⇒y= x 3 2 +C⋅x
As it passes through 1,1,
1=12+C⇒C=12
∴ The equation of curve is 2y=x3+x

Topic: Mathematics, Differential Equations, JEE Main
If the curve x2+2y2=2 intersects the line x+y=1 at two points P and Q, then the angle subtended by the line segment PQ at the origin is

Options:
A π2-tan-113
B π2+tan-113
C π2+tan-114
D π2-tan-114

The correct answer is:
π2+tan-114

Solution:
Homogenising

x2+2y2-2x+y2=0

⇒-x2-4xy=0⇒x2+4xy=0

Lines are x=0 and y=-x4

∴ Angle between lines =π2+tan-114

option 3

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2021 (25 Feb Shift 2)
If the curve y=ax2+bx+c,x∈R, passes through the point 1, 2 and the tangent line to this curve at origin is y=x, then the possible values of a, b, c are:

Options:
A a=-1, b=1, c=1
B a=1, b=0, c=1
C a=1, b=1, c=0
D a=12,b=12,c=1

The correct answer is:
a=1, b=1, c=0

Solution:
a+b+c=2 …1

and dydx|0, 0=1

2ax+b|0, 0=1

b=1

Curve passes through origin

So, c=0

and a=1

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2021 (24 Feb Shift 2)
If the curve y=y(x) is the solution of the differential equation 2x2+x5/4dy-yx+x1/4dx=2x9/4dx, x>0 which passes through the point 1, 1-43loge2, then the value of y(16) is equal to

Options:
A 4313+83loge3
B 313+83loge3
C 4313-83loge3
D 313-83loge3

The correct answer is:
4313-83loge3

Solution:
Given differential equation is 2x2+x5/4dy-yx+x1/4dx=2x9/4dx

⇒2x2+x5/4dydx-yx+x1/4=2x9/4

⇒dydx-yx+x1/42x2+x5/4=2x9/42x2+x5/4

⇒dydx-yx+x1/42xx+x1/4=2x9/42x5/4x3/4+1

⇒dydx-y2x=xx3/4+1

This is a linear differential equation of the type dydx+Py=Q, where P=-12x and Q=xx3/4+1

Now, we have integrating factor I.F.=e∫Pdx

=e-∫dx2x=e-12logex

=elogex-12=x-12

Now, the solution of the linear differential equation is y×I.F.=∫Q×I.F.dx+C

⇒y·x-1/2=∫x·x-1/2x3/4+1dx+C

⇒y·x-1/2 =∫x1/2x3/4+1dx+C

Put x=t4⇒dx=4t3dt

⇒y·x-1/2 =∫t2·4t3dtt3+1

⇒y·x-1/2 =4∫t2t3+1-1t3+1dt

⇒y·x-1/2 =4∫t2dt-4∫t2t3+1dt+C

Let, t3+1=u ⇒3t2dt=du

⇒y·x-1/2 =4∫t2dt-43∫duu+C

⇒y·x-1/2 =4t33-43logeu+C

⇒y·x-1/2 =4t33-43loget3+1+C

⇒y·x-1/2=4x3/43-43logex3/4+1+C

Now, the curve passes through 1, 1-43loge2

⇒1-43loge2=43-43loge2+C

⇒C=-13

⇒y·x-1/2=43x3/4-43logex3/4+1-13

⇒y=43x5/4-43xlogex3/4+1-x3

Hence, at x=16, we get

y(16)=43×32-43×4×loge9-43

⇒y16=1243-323loge3

⇒y16=4313-83loge3.

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2021 (17 Mar Shift 2)
If the curve, y=yx represented by the solution of the differential equation 2xy2-ydx+x dy=0, passes through the intersection of the lines, 2x-3y=1 and 3x+2y=8, then |y1| is equal to ___ .

Options:

The correct answer is:
1

Solution:
2xy2-ydx+xdy=0

2xy2dx-ydx+xdy=0

2xdx=ydx-xdyy2=dxy

Now integrate

x2=xy+c

Now point of intersection of lines are 2,1.

4=21+c ⇒c=2

x2=xy+2

Now y1=-1

⇒|y1|=1

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2021 (25 Feb Shift 2)
If the curves $\frac{x^2}{\alpha}+\frac{y^2}{4}=1$ and $y^3=16 x$ intersect at right angles, then a value of $\alpha$ is :

Options:
A
2

B
$\frac{4}{3}$

C
$\frac{1}{2}$

D
$\frac{3}{4}$

The correct answer is:
$\frac{4}{3}$

Solution:
$$
\begin{aligned}
& \text { } \frac{x^2}{\alpha}+\frac{y^2}{4}=1 \Rightarrow \frac{2 x}{\alpha}+\frac{2 y}{4} \cdot \frac{d y}{d x}=0 \\
& \Rightarrow \frac{d y}{d x}=\frac{-4 x}{\alpha y} \\
& y^3=16 x \Rightarrow 3 y^2 \cdot \frac{d y}{d x}=16 \Rightarrow \frac{d y}{d x}=\frac{16}{3 y^2} \ldots
\end{aligned}
$$
Since curves intersects at right angles
$$
\therefore \frac{-4 x}{\alpha y} \times \frac{16}{3 y^2}=-1 \Rightarrow 3 \alpha y^3=64 x
$$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2013 (23 Apr Online)
If the curves, x2a+y2b=1 and x2c+y2d=1 intersect each other at an angle of 90°, then which of the following relations is TRUE?

Options:
A a-c=b+d
B a-b=c-d
C a+b=c+d
D ab=c+da+b

The correct answer is:
a-b=c-d

Solution:
Given curves, x2a+y2b=1 .....i

On differentiating equation i, we get

2xa+2ybdydx=0

⇒m1=dydx=-xbya ...iii

and x2c+y2d=1.....ii

On differentiating equation ii, we get

2xc+2yddydx=0

⇒m2=dydx=-xdyc ...iv

For orthogonal curves, m1×m2=-1

⇒-xbya×-xdyc=-1

x2y2=-acbd....v

From equation i-ii, we get

x2c-aac=y2b-dbd

⇒x2y2=b-dc-aacbd

From equation v, we get

a-c=b-d

⇒ a-b=c-d.

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2021 (25 Feb Shift 1)
If the curves $x^2+p y^2=1$ and $q x^2+y^2=1$ are orthogonal to each other, then

Options:
A $p-q=2$
B $\frac{1}{p}-\frac{1}{q}=2$
C $\frac{1}{p}+\frac{1}{q}=-2$
D $\frac{1}{p}+\frac{1}{q}=2$

The correct answer is:
$\frac{1}{p}+\frac{1}{q}=2$

Solution:
Given curves are
$$
\begin{aligned}
& x^2+p y^2=1 \\
& q x^2+y^2=1
\end{aligned}
$$
and
On differentiating Eq. (i), w.r.t., $x$ we get
$$
\begin{gathered}
2 x+2 y p \frac{d y}{d x}=0 \\
\Rightarrow \quad \frac{d y}{d x}=m_1=-\frac{x}{p y}
\end{gathered}
$$
On differentiating Eq. (ii), w.r.t. to $x$, we get
$$
\begin{array}{r}
2 q x+2 y \frac{d y}{d x}-0 \\
\Rightarrow \quad \frac{d y}{d x}=m_2=\frac{-q x}{y}
\end{array}
$$
Since, both the curves are orthogonal to each other,
Then,
$$
\Rightarrow \quad \frac{-x}{p y} \cdot \frac{-q x}{y}=-1
$$
$\Rightarrow \quad q x^2=-p y^2$
$\Rightarrow \quad q\left(1-p y^2\right)=-p y^2$
[from Eq. (i)]
$\Rightarrow \quad q-p q y^2=-p y^2$
$\Rightarrow \quad q=(p q-p)=y^2$
$\therefore \quad y^2=\frac{q}{p q-p}$
and
$$
x^2=\frac{-p}{p q-p}
$$
On putting $x^2$ and $y^2$ in Eq. (ii) we get
$$
\begin{array}{cc}
& -\frac{p q}{p q-q}+\frac{q}{p q=p}=1 \\
\Rightarrow & -p q+q=p q-p \Rightarrow p+q=2 p q \\
\Rightarrow & \frac{1}{p}+\frac{1}{q}=2
\end{array}
$$

Topic: Mathematics, Application of Derivatives, JEE Main
If the curves x=y4 and xy=k cut at right angles, then 4k6 is equal to ___ .

Options:

The correct answer is:
4

Solution:
Given, x=y4 and xy=k

By solving equations, we get y5=k …1

Also x=y4

⇒1=4y3dydx⇒dydx=14y3

for xy=k⇒x=ky

⇒1=-ky2·dydx

⇒dydx=-y2k

∵ Curve cut orthogonally

⇒14y3×-y2k=-1

⇒y=14k

∴ from 1 y5=k

⇒14k5=k

⇒4=4k6

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2021 (25 Feb Shift 2)
If the curves y2=6x, 9x2+by2=16 intersect each other at right angles, then the value of b is:

Options:
A 92
B 6
C 72
D 4

The correct answer is:
92

Solution:
Given, curves y2=6x and 9x2+by2=16
Let the point of intersection of curves be x1, y1. So,
2y1dydx=6
⇒ dydx=3y1
18x1+2by1dydx=0
⇒ dydx=-9x1by1
Since both the curve intersect at right angle.
3y1×-9x1by1=-1
⇒27x1=by12 (Since y12=6x1)
⇒27x1=b6x1
x1≠0
b=276=92

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2018 (08 Apr)
If the data x1,x2,…x10 is such that the mean of first four of these is 11, the mean of the remaining six is 16 and the sum of squares of all of these is 2000, then the standard deviation of this data is:

Options:
A 22
B 4
C 2
D 2

The correct answer is:
2

Solution:
Let the numbers are x1,x2,…x10N=10
Given x1+x2+x3+x4=4.11=44x5+x6….+x10=6.16=96⇒∑xi=140
⇒ variance σ2=∑xi2N-∑xiN2
⇒σ2=200010-140102
⇒σ2=200-196=4
⇒ Standard deviation σ=2.

Topic: Mathematics, Statistics, JEE Main, JEE Main 2019 (12 Apr Shift 1)
If the de Broglie wavelength of an electron is equal to 10-3 times the wavelength of a photon of frequency 6Ã1014 Hz, then the speed of the electron is equal to (Speed of light =3Ã108 m s-1, Planck's constant =6.63Ã10-34J s, Mass of electron =9.1Ã10-31 kg )

Options:
A 1.45×106 m s-1
B 1.1×106 m s-1
C 1.7×106 m s-1
D 1.8×106 m s-1

The correct answer is:
1.45×106 m s-1

Solution:
λe=10-3λp
fp=6×1014 Hz=Cλp
λp=3×1086×1014=0.5×10-6 m
λe=0.5×10-9=hmeve
ve=6.63×10-349.1×10-31×0.5×10-9
=1.45×106 m s-1

Topic: Physics, Dual Nature of Matter, JEE Main
If the de Broglie wavelength of the electron in nth Bohr orbit in a hydrogenic atom is equal to 1.5 πa0 (a0 is Bohr radius), then the value of nz is:

Options:
A 1.50
B 1.0
C 0.4
D 0.75

The correct answer is:
0.75

Solution:
2πr=nλ
2πa0n2Z=nλ
2πa0n2Z=n ×1.5πa0
nZ=1.52=0.75

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2019 (12 Jan Shift 2)
If the de Broglie wavelengths associated with a proton and an $\alpha$-particle are equal, then the ratio of velocities of the proton and the $\alpha$-particle will be:

Options:
A
$1: 4$

B
$1: 2$

C
$4: 1$

D
$2: 1$

The correct answer is:
$4: 1$

Solution:
According to question, $\lambda_p=\lambda_\alpha$
Using, $\lambda=\frac{h}{p}=\frac{h}{m v}$
$$
\begin{aligned}
&\text { So, } \frac{h}{m_p \times v_p}=\frac{h}{m_\alpha \times v_\alpha} \\
&\Rightarrow \frac{v_p}{v_\alpha}=\frac{m_\alpha}{m_p}=\frac{4 m_p}{m_p}
\end{aligned}
$$
$(\because$ mass of $\alpha$-particle is 4 times of mass of proton)
So, $\frac{v_p}{v_\alpha}=\frac{4}{1}$; i.e., $4: 1$

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
If the deBroglie wavelength of an electron is equal to $10^{-}$ times the wavelength of a photon of frequency $6 \times 10^{14}$ $\mathrm{Hz}$, then the speed of electron is equal to :

(Speed of light $\left.=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)$

Planck's constant $=6.63 \times 10^{-34} \mathrm{~J} . \mathrm{s}$

Mass of electron $\left.=9.1 \times 10^{-31} \mathrm{~kg}\right)$

Options:
A $1.1 \times 10^{6} \mathrm{~m} / \mathrm{s}$
B $1.7 \times 10^{6} \mathrm{~m} / \mathrm{s}$
C $1.8 \times 10^{6} \mathrm{~m} / \mathrm{s}$
D $1.45 \times 10^{6} \mathrm{~m} / \mathrm{s}$

The correct answer is:
$1.45 \times 10^{6} \mathrm{~m} / \mathrm{s}$

Solution:
de-Broglie wavelength,

$$

\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=10^{-3}\left(\frac{3 \times 10^{8}}{6 \times 10^{14}}\right) \quad\left[\because \lambda=\frac{\mathrm{c}}{\mathrm{v}}\right]

$$

$$

\begin{array}{l}

\mathrm{v}=\frac{6.63 \times 10^{-34} \times 6 \times 10^{14}}{9.1 \times 10^{-31} \times 3 \times 10^{5}} \\

\mathrm{v}=1.45 \times 10^{6} \mathrm{~m} / \mathrm{s}

\end{array}

$$

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2019 (11 Jan Shift 1)
If the definite integral I=∫0πsin⁡xdx=∑ε=0naεsin⁡ε. (where, ⋅ is the greatest integer function), then the value of an+nπ is equal to

Options:

The correct answer is:
1

Solution:
The given integral is
I=∫01sin⁡0dx+∫12sin⁡1dx+∫23sin⁡2dx+∫3πsin⁡3dx
⇒I=2-1sin⁡1+3-2sin⁡2+π-3sin⁡3
∴an=π-3
and n=3
⇒an+n=π or an+nπ=1

Topic: Mathematics, Definite Integration, JEE Main
If the definition of the temperature coefficient of the reaction holds good for a reaction between 27°C and 37°C, the activation energy for the reaction in kJ.mol-1 is

Options:
A 102
B 53.5
C ∞
D 141.5

The correct answer is:
53.5

Solution:
Given, T1=27+273=300 K ; T2=37+273=310K

Temperature coefficient of a reaction is ideally = 2.

Let the rate of reaction at 27°C be k i.e. K1=k , then rate of reaction at 37°C be 2k i.e. K2= 2k
From the Arrhenius equation we get ,
⇒logk2k1=Ea2.303 RT2-T1T1T2 , where Ea is the Activation energy & R=8.314×10-3 kJ mol-1 K-1
⇒log⁡2kk=Ea2.303 ×8.314×10-3×310-300310×300
⇒log⁡2=Ea2.303×8.314×10-3×1093000
⇒Ea=2.303×0.3010×8.314×10-3×9300010 kJ mol-1
⇒Ea=53.5 kJ

Topic: Chemistry, Chemical Kinetics, JEE Main
If the degree of dissociation of aqueous solution of weak monobasic acid is determined to be 0.3, then the observed freezing point will be _____ % higher than the expected/theoretical freezing point. (Nearest integer).

Options:

The correct answer is:
30

Solution:
HA1-α⇋H+α+A-α

Van't Hoff factor, i=1+αn-1

=1+0.32-1

=1+0.3

=1.3

∆Tf without considering dissociation =kf× molality

∆Tf without considering dissociation =ikf× molality

=ΔTf'

ΔTf'-ΔTf=0.3 (kf× molality)

ΔTf1-ΔTfΔTf=0.3kf×molalitykf×molality=0.3

% value =30

Topic: Chemistry, Solutions, JEE Main, JEE Main 2023 (10 Apr Shift 1)
If the degree of freedom of a gas are $f$, then the ratio of two specific heats $C_P / C_V$ is given by

Options:
A $\frac{2}{f}+1$
B $1-\frac{2}{f}$
C $1+\frac{1}{f}$
D $1-\frac{1}{f}$

The correct answer is:
$\frac{2}{f}+1$

Solution:
$\frac{C_P}{C_V}=\gamma=1+\frac{2}{f}$

Topic: Physics, Kinetic Theory of Gases, JEE Main
If the density of oxygen is 16 times that of hydrogen, what will be the ratio of their corresponding velocities of sound waves

Options:
A $1: 4$
B $4: 1$
C $16: 1$
D $1: 16$

The correct answer is:
$1: 4$

Solution:
$v=\sqrt{\frac{\gamma P}{\rho}} \Rightarrow \frac{v_{\mathrm{O}_2}}{v_{\mathrm{H}_2}}=\sqrt{\frac{\rho_{\mathrm{H}_2}}{\rho_{\mathrm{O}_2}}}=\sqrt{\frac{1}{16}}=\frac{1}{4}$

Topic: Physics, Waves and Sound, JEE Main
If the density of some lake water is 1.25 g mL-1 and contains 92 g of Na+ ions per kg of water, calculate the molality of Na+ ions in the lake.

Options:

The correct answer is:
4

Solution:
Molality of Na + ions (m)
=â Numberâofâmolesâofâ Na + âions Massâofâwaterâinâkg
= 92 g / 23 g mol â1 1âkg
= 4 mol  kg −1  or 4 m

Topic: Chemistry, Solutions, JEE Main
If the derivative of \( f(\tan x) \) w.r.t. \( g(\sec x) \) at \( x=\frac{\pi}{4} \), where \( f^{\prime}(1)=2 \) and \( g^{\prime}(\sqrt{2})=4 \) is \( \frac{\sqrt{2}}{k} \) then find \( k \) ?

Options:

The correct answer is:
2.00

Solution:
Let u=ftanx and v=gsecx

⇒dudx=f'tanxsec2x

and dvdx=g'secxsec x tan x

⇒dudv=dudx/dvdx=f'tan xsec2xg'sec xsec x tan x

⇒dudvx=π4=f'tanπ4g'secπ4sinπ4=f'12g'2

=224=12

=12×22

=22

=2k

⇒k =2.

Topic: Mathematics, Differentiation, JEE Main
If the derivative of the function $f(x)=\left\{\begin{array}{cc}b x^2+a x+4 ; & x \geq-1 \\ a x^2+b ; & x < -1\end{array}\right.$ is everywhere continuous, then

Options:
A $a=2, b=3$
B $a=3, b=2$
C $a=-2, b=-3$
D $a=-3, b=-2$

The correct answer is:
$a=2, b=3$

Solution:
We have,
$$
\begin{aligned}
f(x) & =\left\{\begin{array}{cc}
a x^2+b ; & x < -1 \\
b x^2+a x+4 ; & x \geq-1
\end{array}\right. \\
\therefore \quad f^{\prime}(x) & =\left\{\begin{array}{cc}
2 a x, & x < -1 \\
2 b x+a, & x \geq-1
\end{array}\right.
\end{aligned}
$$
$\because f(x)$ is differentiable at $x=-1$
$\therefore$ It is continuous at $x=-1$ and hence
$$
\begin{array}{ll}
& \lim _{\left(x \rightarrow-1^{-}\right)} f(x)=\lim _{x \rightarrow-1^{+}} f(x) \\
\Rightarrow \quad & a+b=b-a+4 \\
\Rightarrow \quad & a=2 \\
\text { and also, } & \lim _{\left(x \rightarrow-1^{-}\right)} f^{\prime}(x)=\lim _{x \rightarrow-1^{+}} f^{\prime}(x) \\
\Rightarrow \quad & -2 a=-2 b+a \\
\Rightarrow \quad & 3 a=2 b \Rightarrow b=3 \quad(\because a=2)
\end{array}
$$
Hence, $a=2, b=3$

Topic: Mathematics, Continuity and Differentiability, JEE Main
If the determinant of the adjoint of a (real) matrix of order 3 is 25 , then the determinant of the inverse of the matrix is

Options:
A $0.2$
B $\pm 5$
C $\frac{1}{\sqrt[5]{625}}$
D $\pm 0.2$

The correct answer is:
$\pm 0.2$

Solution:
Given, the determinant of the adjoint of a (real) matrix of order 3 is $25 .$
i.e., $\quad|\operatorname{adj} A|=25...(i)$
We know that,
$|\operatorname{adj} A|=|A|^{n-1} \quad$ (here, $n=3$ )
$\Rightarrow \quad|A|^{3-1}=|A|^{2}=25 \quad$ [from Eq. (i)]
$\Rightarrow \quad|A|=\pm 5$
$\therefore \quad\left|A^{-1}\right|=|A|^{-1}=\frac{1}{|A|}=\pm \frac{1}{5}=\pm 0.2$ (by property)

Topic: Mathematics, Matrices, JEE Main
If the dielectric constant of a substance $K=\frac{4}{3}$, then the electric susceptibility $\chi$ in terms of vacuum permittivity $\varepsilon_0$ is

Options:
A $\frac{\varepsilon_0}{3}$
B $3 \varepsilon_0$
C $\frac{4}{3} \varepsilon_0$
D $\frac{3}{4} \varepsilon_0$

The correct answer is:
$\frac{\varepsilon_0}{3}$

Solution:
Given, $K=\frac{4}{3}$
For a dielectric, the electric displacement,
$\mathbf{D}=\varepsilon_0 \mathbf{E}+\mathbf{P} \quad \text { or } \quad D=\varepsilon_0 E+P$
$K \varepsilon_0 E=\varepsilon_0 E+\chi E \Rightarrow \chi=(K-1) \varepsilon_0$
$\Rightarrow \quad \chi=\varepsilon_0\left(\frac{4}{3}-1\right) \Rightarrow \chi=\frac{\varepsilon_0}{3}$

Topic: Physics, Electrostatics, JEE Main
If the difference between mean and mode is \( 63 \), the difference between mean and median is

Options:
A \( 189 \)
B \( 21 \)
C \( 31.5 \)
D \( 48.5 \)

The correct answer is:
\( 21 \)

Solution:
By the result

Mode = 3 Median-2 mean
⇒ Mean - mode = 3(mean - median)

⇒ 63=3(mean - median)

i.e. mean - median =21

Topic: Mathematics, Statistics, JEE Main
If the difference between the number of subsets of the sets A and B is 120, then choose the incorrect option.

Options:
A Maximum value of nA∩B=3
B Minimum value of nA∩B=0
C Maximum value of nA∪B=21
D Minimum value of nA∪B=7

The correct answer is:
Maximum value of nA∪B=21

Solution:
Let, nA=Î±,nB=Î²Î±>Î²
given, 2Î±-2Î²=120âÎ±=7,Î²=3
â´nAâ©Bâ0,3âµnAâªB=Î±+Î²=nAâ©B
nAâªBâ7,10

Topic: Mathematics, Sets and Relations, JEE Main
If the difference between the number of subsets of two sets A and B is 120, then nA×B is equal to

Options:
A 21
B 25
C 18
D 24

The correct answer is:
21

Solution:
Let, nA=α & nB=β
Given, 2α-2β=120⇒α,β≡7,3,3,7
∴n(A×B)=7×3=21

Topic: Mathematics, Sets and Relations, JEE Main
If the difference between the roots of the equation $x^2+a x+1=0$ is less than $\sqrt{5}$, then the set of possible values of $a$ is

Options:
A
$(-3,3)$

B
$(-3, \infty)$

C
$(3, \infty)$

D
$(-\infty,-3)$

The correct answer is:
$(-3,3)$

Solution:
$x^2+a x+1=0$
$\alpha+\beta=-a \quad \alpha \beta=1$
$|\alpha-\beta|=\sqrt{(\alpha+\beta)^2-4 \alpha \beta}$
$|\alpha-\beta|=\sqrt{a^2-4}$
$\sqrt{a^2-4} < \sqrt{5}$
$a^2-4 < 5$
$a^2-9 < 0$
$a \in(-3,3)$.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2007
If the differential equation 3x13dy+x-23ydx=3xdx is satisfied by kx13y=x2+c (where c is an arbitrary constant), then the value of k is

Options:
A 13
B 23
C 2
D 1

The correct answer is:
2

Solution:
The given equation is x13.dy+13x-23dx.y=xdx
or dx13.y=xdx
Integrating, we get,

x13.y=x22+λ
Or 2x13y=x2+C
⇒k=2

Topic: Mathematics, Differential Equations, JEE Main