datavorous/entrance-exam-dataset huggingface dataset:
If $M=\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right]$ and $M^2-\lambda M-I_2=0$, then $\lambda=$

Options:
A $-2$
B $2$
C $-4$
D $4$

The correct answer is:
$4$

Solution:
$M^2-\lambda M-I_2=0$
$\Rightarrow\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right]-\left[\begin{array}{cc}\lambda & 2 \lambda \\ 2 \lambda & 3 \lambda\end{array}\right]-\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=0$
$\Rightarrow\left[\begin{array}{cc}5 & 8 \\ 8 & 13\end{array}\right]-\left[\begin{array}{cc}\lambda & 2 \lambda \\ 2 \lambda & 3 \lambda\end{array}\right]-\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=0$
$\Rightarrow\left[\begin{array}{cc}5-\lambda & 8-2 \lambda \\ 8-2 \lambda & 13-3 \lambda\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$\Rightarrow 5-\lambda=1,8-2 \lambda=0,13-3 \lambda=1$
$\Rightarrow \lambda=4$, which satisfies|all the three equations.

Topic: Mathematics, Matrices, JEE Main
If $m_{1}$ and $m_{2}$ are slopes of the lines represented by
$\left(\sec ^{2} \theta-\sin ^{2} \theta\right) x^{2}-2 \tan \theta x y+\sin ^{2} \theta y^{2}=0$, then $\left|m_{1}-m_{2}\right|=$

Options:
A 1
B 2
C 4
D 3

The correct answer is:
2

Solution:
$\left(\operatorname{scc}^{2} \theta-\sin ^{2} \theta\right) x^{2}-2 \tan \theta x y+\sin ^{2} \theta y^{2}=0$
$\sqrt{a x^{2}+2 b x y+b y^{2}}=0$
$\left|m_{1}-m_{2}\right|=?$
$\left|m_{1}-m_{2}\right|=\sqrt{\left(m_{1}+m_{2}\right)^{2}-4 m_{1} m_{2}}$
$m_{1}+m_{2}=\frac{-2 h}{b}$
$m_{1} m_{2}=\frac{a}{b}$
$m_{1}-m_{2} \mid=\sqrt{\frac{4 \tan ^{2} \theta-4\left(\sec ^{2} \theta-\sin ^{2} \theta\right)}{\sin ^{2} \theta}}$
$=2$

Topic: Mathematics, Quadratic Equation, JEE Main
If m1 and m2 are the roots of the equation x2+3+2x+3-1=0, then the area (in sq. units) of the triangle bounded by the lines y=m1x, y=m2x and y=2 is

Options:
A 33+11
B 33-11
C 233
D 211

The correct answer is:
33+11

Solution:
Vertices of the triangle formed by the lines y=m1x, y=m2x & y=2 are
0,0, 2m1,2 & 2m2,2
Area of the triangle =120012m1212m221
=2m2-m1m1m2=2Daca=2Dc=2b2-4acc
=27+43-43-13-1=2113-1
=33+11 sq. units

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If $m_{1}$ and $m_{2}$ are the slopes of tangents to the circle $x^{2}+y^{2}=4$ from the point $(3,2)$, then $m_{1}-m_{2}$ is equal to

Options:
A $\frac{5}{12}$
B $\frac{12}{5}$
C $\frac{3}{2}$
D 0

The correct answer is:
$\frac{12}{5}$

Solution:
Equation of pair of tangents is $S S_{1}=T^{2}$
$\Rightarrow\left(x^{2}+y^{2}-4\right)(9+4-4)=(3 x+2 y-4)^{2}$
$\Rightarrow 5 y^{2}+16 y-12 x y+24 x-52=0$
$\therefore \quad m_{1}+m_{2}=\frac{-2 h}{b}=\frac{12}{5}$
and $\quad m_{1} m_{2}=0$
Now, $m_{1}-m_{2}=\sqrt{\left(m_{1}+m_{2}\right)^{2}-4 m_{1} m_{2}}$
$=\sqrt{\left(\frac{12}{5}\right)^{2}-0}$
$=\frac{12}{5}$

Topic: Mathematics, Circle, JEE Main
If m1 and m2 are the slopes of the tangents to the ellipse x216+y29=1 which passes through 5,4, then the value of m1+m2-m1m2 is equal to

Options:
A 479
B -409
C 223
D 113

The correct answer is:
113

Solution:
Let the equation of tangent passing through 5,4 is y=mx±16m2+9
⇒4=5m±16m2+9
⇒4-5m2=16m2+9⇒9m2-40m+7=0
⇒m1+m2=409,m1m2=79
⇒m1+m2-m1m2=339=113

Topic: Mathematics, Ellipse, JEE Main
If $m_1$ and $m_2$ are the slopes of the tangents to the hyperbola $\frac{x^2}{25}-\frac{y^2}{16}=1$ which pass through the point $(6,2)$, then

Options:
A $m_1+m_2=\frac{24}{11}$
B $m_1 m_2=\frac{20}{11}$
C $m_1+m_2=\frac{48}{11}$
D $m_1 m_2=\frac{11}{20}$

The correct answers are:
$m_1+m_2=\frac{24}{11}$, $m_1 m_2=\frac{20}{11}$

Solution:
The line through $(6,2)$ is
$y-2=m(x-6) \Rightarrow y=m x+2-6 m$
Now from condition of tangency, $(2-6 m)^2=25 m^2-16$
$\begin{aligned}
& \Rightarrow 36 m^2+4-24 m-25 m^2+16=0 \\
& \Rightarrow 11 m^2-24 m+20=0
\end{aligned}$
Obviously its roots are $m_1$ and $m_2$, therefore $m_1+m_2=\frac{24}{11}$ and $m_1 m_2=\frac{20}{11}$

Topic: Mathematics, Hyperbola, JEE Main
If mâa, mâb and mâc are 3 unit vectors such that mâaâmâb=0, mâaâmâc=0 and the angle between mâb and mâc is Ï3, then the value of mâaÃmâb-mâaÃmâc is equal to

Options:
A 1
B 2
C 3
D 4

The correct answer is:
1

Solution:
∵m→a⋅m→b=0,m→a⋅m→c=0
⇒m→a⋅m→b-m→c=0⇒m→a is perpendicular to m→b-m→c
Therefore, m→a×m→b-m→c=m→am→b-m→c
=1×1+1-2×1×1×cosπ3
=1

Topic: Mathematics, Vector Algebra, JEE Main
If $m$ and $\sigma^2$ are the mean and variance of the random variable $X$, whose distribution is given by
Then

Options:
A $m=\sigma^2=2$
B $m=1, \sigma^2=2$
C $m=\sigma^2=1$
D $m=2, \sigma^2=1$

The correct answer is:
$m=\sigma^2=1$

Solution:
Given, distribution is
$\begin{aligned}\therefore Mean, m & =\sum_{i=1}^4 p_i x_i \\ & =0 \times \frac{1}{3}+1 \times \frac{1}{2}+2 \times 0+3 \times \frac{1}{6}\end{aligned}$
$=0+\frac{1}{2}+0+\frac{1}{2}=1$
Variance, $\sigma^2=\sum_{i=1}^4 p_i\left(x_i-m\right)^2$
$\begin{aligned}
& =\frac{1}{3}(0-1)^2+\frac{1}{2}(1-1)^2 \\
& \quad+0(2-1)^2+\frac{1}{6}(3-1)^2 \\
& =\frac{1}{3}+0+0+\frac{2}{3}=1 \\
& m=\sigma^2=1
\end{aligned}$
$\therefore \quad m=\sigma^2=1$

Topic: Mathematics, Statistics, JEE Main
If $\mathrm{M}$ and $\sigma^2$ represent respectively the mean deviation from the mean and the variance for the data $1,3,5,7,11$, $13,17,19,23$ then $3\left(\sigma^2-M\right)=$

Options:
A 232
B 112
C 224
D 136

The correct answer is:
136

Solution:
$\begin{aligned} & \text {} \bar{X}=\frac{\Sigma x_i}{n}=\frac{99}{9}=11 \\ & \begin{aligned} M & =\frac{\Sigma\left|x_i^n-\bar{x}\right|^{\prime}}{n} \\ & =\frac{10+8+6+4+0+2+6+8+12}{9}=\frac{56}{9} \\ \sigma^2 & =\frac{\Sigma\left|x_i-\bar{x}\right|^2}{n}\end{aligned}\end{aligned}$
$\begin{aligned} & =\frac{(10)^2+8^2+6^2+4^2+2^2+6^2+8^2+12^2}{9}=\frac{464}{9} \\ & 3\left(\sigma^2-M\right)=3\left(\frac{464}{9}-\frac{56}{9}\right)=\frac{408 \times 3}{9}=136 .\end{aligned}$

Topic: Mathematics, Statistics, JEE Main
If $m$ and $e$ are the mass and charge of the revolving electron in the orbit of radius $r$ for hydrogen atom, the total energy of the revolving electron will be:

Options:
A
$\frac{1}{2} \frac{e^2}{r}$

B
$-\frac{e^2}{r}$

C
$\frac{m e^2}{r}$

D
$-\frac{1}{2} \frac{e^2}{r}$

The correct answer is:
$-\frac{1}{2} \frac{e^2}{r}$

Solution:
Total energy of a revolving electron is the sum of its kinetic and potential energy. Total energy $=$ K.E. $+$ P. E.
$$
=\frac{e^2}{2 r}+\left(-\frac{e^2}{r}\right)=-\frac{e^2}{2 r}
$$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2014 (12 Apr Online)
If M and m are the maximum and minimum values of yx for pair of real numbers x,y which satisfy the equation x-32+y-32=6, then the value of 1M+1m is

Options:

The correct answer is:
6

Solution:
Let yx=m'. So, on putting y= m ′ x in the given circle,
We get x−3 2 + m ′ x−3 2 =6
⇒x21+m'2-6x1+m'+12=0
Putting discriminant =0, we get,
m'=3+_22
So, m = 3-22 and M=3+22⇒mM=1
⇒1M+1m=m+M
=3-22+3+22
=6

Topic: Mathematics, Circle, JEE Main
If m and M are the minimum and the maximum values of 4+12sin2⁡2x-2cos4⁡x, x ∈R, then M-m is equal to:

Options:
A 15 4
B 94
C 74
D 14

The correct answer is:
94

Solution:
Given expression is 4+12sin2⁡2x-2cos4⁡x
= 4+21-cos2⁡xcos2⁡x-2cos4⁡x

Creating perfect squares, we get
=-4 cos2⁡x-142-1716
Since, 0 ≤cos2⁡x ≤1
⇒ -14 ≤cos2⁡x-14 ≤34
⇒ 0 ≤cos2⁡x-142 ≤916
⇒ -1716 ≤cos2⁡x-142-1716 ≤916-1716
⇒ 174 ≥-4 cos2⁡x-142-1716 ≥2
⇒ M=174 and m=2
⇒ M-m=174-2=94

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2016 (09 Apr Online)
If m and M denotes the minimum and maximum value of 2z+1, where z-2i≤1 and i2=-1, then the value of M-m2 is equal to

Options:
A 17
B 34
C 51
D 16

The correct answer is:
16

Solution:
z-2i=1 represents a circle with centre at 0,2 and radius 1 unit

2z+1=2z--12 represents twice the distance from the point -12,0

Hence from the diagram,

m=2AD=2CD-AC
m=2172-1=17-2
M=2BD=2CD+BC
M=2172+1=17+2⇒M-m2=42=16

Topic: Mathematics, Complex Number, JEE Main
If \( m \) and \( n \) are order and degree of the differential equation \( \left(\frac{d^{2} y}{d x^{2}}\right)^{5}+4 \cdot \frac{\left(\frac{d^{2} y}{d x^{2}}\right)^{3}}{\frac{d^{3} y}{d x^{3}}}+\frac{d^{3} y}{d x^{3}}=x^{2}-1 \), then the value of \( m+n \) is

Options:

The correct answer is:
5.00

Solution:
The given differential equation can be written as

d2ydx25d3ydx3+4d2ydx23+d3ydx32

=x2-1d3ydx3

⇒m=3, n=2

Topic: Mathematics, Differential Equations, JEE Main
If m and n respectively are the number of local maximum and local minimum points of the function fx=∫0x2t2-5t+42+etdt, then the ordered pair m,n is equal to

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

The correct answer is:
2,3

Solution:
Given fx=∫0x2t2-5t+42+etdt

We know that,

ddx∫uxvxfvdx=fvxddxvu-fuxddxux

Using above formula we get f'x=x4-5x2+42+ex2·2x

For critical points f'x=0

So, xx2-1x2-42+ex2=0

⇒xx+1x-1x+2x-2=0

Now by first derivative test we get,

Here points of maxima are -1,1 & points of minima are -2,0,2

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2022 (27 Jun Shift 2)
If m and n respectively are the numbers of positive and negative value of θ in the interval -π,π that satisfy the equation cos2θcosθ2=cos3θcos9θ2, then mn is equal to _____ .

Options:

The correct answer is:
25

Solution:
We have,

cos2θ·cosθ2=cos3θ·cos9θ2

⇒2cos2θcosθ2=2cos9θ2cos3θ

⇒ cos5θ2+cos3θ2=cos15θ2+cos3θ2

⇒cos15θ2=cos5θ2

⇒15θ2=2kπ±5θ2; k∈Z

⇒5θ=2kπ or 10θ=2kπ

⇒θ=2kπ5 or θ=kπ5

∴θ=-π,-4π5,-3π5,-2π5,-π5,0,π5,2π5,3π5,4π5,π

So, m=5,n=5

∴ mn=25.

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2023 (25 Jan Shift 2)
If m arithmetic means (A.Ms) and three geometric means (G.Ms) are inserted between 3 and 243 such that 4th A.M. is equal to 2nd G.M., then m is equal to:

Options:

The correct answer is:
39

Solution:
Let 3, A1, A2, A3,......Am, 243 are in arithmetic progression with m arithmetic means.

Common difference d=243-3m+1=240m+1

Let 3, G1, G2, G3, 243 are in geometric progression with 3 geometric means.

Common ratio r=243313+1=8114=3

Given G2=A4

⇒332=3+4240m+1

⇒27=3+960m+1

⇒m+1=40

⇒m=39

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2020 (03 Sep Shift 2)
If m is a non-zero number and ∫x5m-1+2x4m-1x2m+xm+13dx=fx+c, then fx is equal to

Options:
A x5m-x4m2mx2m+xm+12
B 12mx4mx2m+xm+12
C x5m2mx2m+xm+12
D 2mx5m+x4mx2m+xm+12

The correct answer is:
12mx4mx2m+xm+12

Solution:
∫x5m-1+2x4m-1x2m+xm+13dx=fx+c

⇒∫x5m-1+2x4m-1x6m1+x-m+x-2m3dx=fx+c

⇒∫x-m-1+2x-2m-11+x-m+x-2m3dx=fx+c.

If, 1+x-m+x-2m=t

⇒-mx-m-1+2x-2m-1dx=dt

∴∫-dtmt3=fx+c

⇒-1m∫t-3dt=fx+c

Using ∫xndx=xn+1n+1+C, n≠-1

⇒-1mt-2-2=fx+c

⇒12m1+x-m+x-2m2+c=fx+c

⇒12m1+1xm+1x2m2+c=fx+c

⇒12mx4mx2m+xm+12=fx.

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2014 (19 Apr Online)
If m is any natural number, then the value of the integral â«x3m+x2m+xm2x2m +3xm+61/mdx is (where, C is an arbitrary constant)

Options:
A 16m+12x3m+3x2m+6xm1/m+1 +C
B 16m2x3m+3x2m+6xmÂ 1/m+1Â +C
C 16m2x3m+3x2m+6xm1/m+C
D None of the above

The correct answer is:
16m+12x3m+3x2m+6xm1/m+1 +C

Solution:
Put, I=â«x3m+x2m+xm2x3m+3x2m+6xm1/mxdx
=â«x3m-1+x2m-1+xm-12x3m +3x2m +6xm1/mdx
=16mâ«2x3m+3x2m+6xm1/m6mx3m-1 +6mx2m-1 +6mxm-1dx
Now, let 2x3m+3x2m+6xm=tâ6mx3m-1 +6mx2m-1 +6mxm-1dx=dt

âI=16mâ«t1/mdt =16mt1/m+11/m+1
=16m+12x3m+3x2m+6xm(1/m)+1+C

Topic: Mathematics, Indefinite Integration, JEE Main
If m is chosen in the quadratic equation m2+1x2-3x+m2+12=0 such that the sum of its roots is greatest, then the absolute difference of the cubes of its roots is:

Options:
A 43
B 105
C 83
D 85

The correct answer is:
85

Solution:
Let, the roots of m2+1x2-3x+m2+12=0 are α and β

We know that the sum and product of roots of a quadratic equation ax2+bx+c=0, a≠0 are respectively -ba and ca.

∴ α+β=3m2+1 and αβ=m2+12m2+1=m2+1

∵ α+β is maximum, ∴ m=0

∴ the equation is x2-3x+1=0

And α+β=3, αβ=1

Now, α3-β3=α-βα2+β2+αβ

=α+β2-4αβα+β2-αβ

=9-4 9-1

=85.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2019 (09 Apr Shift 2)
If m is the A.M. of two distinct real numbers I and n I, n>1 and G1, G2 and G3 are three geometric means between I and n, then G14+2G24+G34 equals

Options:
A 4l⁡2m2 n2
B 4 l⁡2mn
C 4 l⁡m2 n
D 4l⁡mn2

The correct answer is:
4 l⁡m2 n

Solution:
Given, m is the A.M. of l & n.
⇒m=l+n2 ...1
G1, G2, G3 are G.M. between l & n.
⇒l, G1, G2, G3, n are in G.P.
⇒n=lr4 ⇒r4=nl ...2
⇒G1=l⁡r, G2=l⁡r2, G3=l⁡r3
⇒G14+2G24+G34=l⁡4r4+2×l⁡4r8+l⁡4×r12
=l⁡4r41+2r4+r8
=l4×nl⁡1+r42=l⁡3n×1+nl⁡2
=l⁡3×n×l⁡+n2l⁡2
=l⁡n×2m2 (From (1))
=4l⁡m2n

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2015 (04 Apr)
If 'M' is the mass of water that rises in a capillary tube of radius 'r' , then mass of water which will rise in a capillary tube of radius '2r' is:

Options:
A 4 M
B 2 M
C M2
D M

The correct answer is:
2 M

Solution:
Mass of liquid rising in tube =ρAh
m=ρπr22Tcos⁡θρgr
m∝r
Mass of liquid rise in tube will become 2M

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2019 (09 Apr Shift 1)
If m is the minimum value of k for which the function fx=xkx-x2 is increasing in the interval [0, 3] and M is the maximum value of f in [0, 3] when k=m, then the ordered pair (m, M) is equal to:

Options:
A 4, 33
B 5, 36
C 3,Â 33
D 4,Â 32

The correct answer is:
4, 33

Solution:
fx=xkx-x2 ⇒ f'x=3kx-4x22kx-x2
As per the given condition f'x≥0 for x∈0, 3
⇒3kx-4x2≥0 for x∈0, 3
⇒3k-4x ≥0 for x∈0, 3
⇒k≥4x3 for x∈0, 3
⇒k≥4 . So minimum value of k is m=4.
Now fx=x4x-x2
Since given function in increasing hence maximum value will occur at x=3
⇒f(3)=34×3-32=33, M=33
Hence (m, M) = (4, 33)

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2019 (12 Apr Shift 1)
If m is the slope of a common tangent to the curves x216+y29=1 and x2+y2=12, then 12m2 is equal to

Options:
A 6
B 9
C 10
D 12

The correct answer is:
9

Solution:
Given,

x216+y29=1

we know that equation of tangent to the ellipse is

y=mx±a2 m2+b2

y=mx±16m2+9 ...1

x2+y2=12

Also equation of tangent to the circle is

y=mx±121+m2 ...2

Now, for common tangent equate eq.1 and 2

⇒16m2+9=121+m2

16m2-12m2=3

4m2=3

12m2=9

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2022 (26 Jun Shift 2)
If Mn2+ has 'X' number of unpaired electrons and 'Y' other positive oxidation states than +2. What is the sum of X + Y here?

Options:

The correct answer is:
10

Solution:
Mn2+→Ar3d5
unpaired e-X=5
Mn can show +3, +4, +5, +6 and +7 oxidation states also other than +2.
Y = 5
X + Y = 5 + 5 = 10

Topic: Chemistry, d and f Block Elements, JEE Main
If m→,  n→ are non-parallel unit vectors and r→ is a vector which is perpendicular to m→ and n→ such that r→=5 and m→+n→2=2+4m→×n→, then the value of m→  n→  r→2 is equal to

Options:
A 7
B 215
C 5
D 103

The correct answer is:
5

Solution:
r→ is parallel to m→ × n→
Let θ be angle between m→ and n→
∴ m→+n→2=2+4 m→×n→
1+1+2cos⁡θ=2+4×sin⁡θ
⇒tan⁡θ=12
m→ n→ r→2=r→.m→×n→2
= r→2m→2n→2sin2θ
=25×15=5

Topic: Mathematics, Vector Algebra, JEE Main
If m number of integers greater than 7000 can be formed with the digits 3,5,7,8 and 9, such that no digit is being repeated, then the value of m100 is

Options:

The correct answer is:
1.92

Solution:
A five digit integer is always greater than 7000.
The number of such integers are 5P5=5!=120.
For a four digit integer to be greater than 7000, it must begin with 7,8 or 9.
The number of such integers are 34P3=72.
Hence, the required number of such integers are 120+72=192.
âm=192
âm100=1.92.

Topic: Mathematics, Permutation Combination, JEE Main
If $M_{O}$ is the mass of an oxygen isotope ${ }_{8} \mathrm{O}^{17}$, $M_{p}$ and $M_{n}$ are the masses of a proton and a neutron, respectively, the nuclear binding energy of the isotope is

Options:
A $\left(M_{O}-8 M_{p}\right) c^{2}$
B $\left(M_{O}-8 M_{p}-9 M_{n}\right) c^{2}$
C $M_{O} c^{2}$
D $\left(M_{O}-17 M_{n}\right) c^{2}$

The correct answer is:
$\left(M_{O}-8 M_{p}-9 M_{n}\right) c^{2}$

Solution:
Binding energy
$$
\begin{aligned}
\mathrm{BE} &=\left(M_{\text {nucleus }}-M_{\text {nucleons }}\right) c^{2} \\
&=\left(M_{O}-8 M_{p}-9 M_{n}\right) c^{2}
\end{aligned}
$$

Topic: Physics, Nuclear Physics, JEE Main
If $\mathrm{M_o}$ is the mass of an oxygen isotope $\mathrm{{ }_8 O^{17}, M_p}$ and $\mathrm{M_N}$ are the masses of a proton and a neutron respectively, the nuclear binding energy of the isotope is

Options:
A
$\mathrm{(M_o-8 M_p) C^2}$

B
$\mathrm{(M_o-8 M_p-9 M_N) C^2}$

C
$\mathrm{M_o C^2}$

D
$\mathrm{(M_{\mathrm{o}}-17 \mathrm{M}_{\mathrm{N}}) \mathrm{C}^2}$

The correct answer is:
$\mathrm{(M_o-8 M_p-9 M_N) C^2}$

Solution:
Binding energy $=\left(\mathrm{M_o-8 M_P-9 M_N}\right) \mathrm{C^2}$

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2007
If $M_o$ is the mass of isotope ${ }_5^{12} B, M_P$ and $M_n$ are the masses of proton and neutron, then nuclear binding energy of isotope is :

Options:
A $\left(M_o-5 M_p\right) C^2$
B $\left(5 M_p+7 M_n-M_o\right) C^2$
C $\left(M_0-12 M_n\right) C^2$
D $\left(M_o-5 M_p-7 M_n\right) C^2$

The correct answer is:
$\left(5 M_p+7 M_n-M_o\right) C^2$

Solution:
$\begin{aligned} & \text { B.E. }=\Delta m C^2 \\ & \left(5 M_p+7 M_n-M_o\right) C^2\end{aligned}$

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2024 (08 Apr Shift 2)
If $m^{\text {th }}$ terms of the series $63+65+67+69+\ldots \ldots \ldots$. and $3+10+17+24+\ldots \ldots$ be equal, then $m=$

Options:
A $11$
B $12$
C $13$
D $15$

The correct answer is:
$13$

Solution:
Given series $63+65+67+69+ \ldots(i)$
and $3+10+17+24+ \ldots(ii)$
Now from (i), $m^{\text {th }}$ term $=(2 m+61)$ and $m^{\text {th }}$ term of (ii) series $=(7 m-4)$
Under condition,
$\Rightarrow 7 m-4=2 m+61 \Rightarrow 5 m=65 \Rightarrow m=13$

Topic: Mathematics, Sequences and Series, JEE Main
If Mz= magnetization of a paramagnetic sample, B= external magnetic field, T= absolute temperature, C= curie constant then according to Curie's law in magnetism, the correct realtion is

Options:
A M Z = T CB
B M Z = CB T
C C= M Z B T
D C= T 2 M z B

The correct answer is:
M Z = CB T

Solution:
M z = M ext V
M z = CB T ....( Paramagnetic )
Where, C=curie constant.

Topic: Physics, Magnetic Properties of Matter, JEE Main
If mammalian ovum fails to get fertilized, which one of the following is unlikely?

Options:
A corpus luteum will disintegrate
B progesterone secretion rapidly declines
C estrogen secretion further decreases
D primary follicle starts developing.

The correct answer is:
primary follicle starts developing.

Solution:
A) Estrogen secretion further decreases- Estrogen is secreted by the corpus luteum. Since it disintegrates in the absence of fertilization, estrogen levels fall.
B) Progesterone secretion starts declining- If pregnancy occurs, human chorionic gonadotropin is released maintaining the corpus luteum allowing it to maintain levels of progesterone. In the other way round, (i.e., in case of fertilization does not occur), progesterone levels fall due to the disintegration of the corpus luteum.
C) Corpus luteum will disintegrate- If the egg is not fertilized, the corpus luteum stops secreting progesterone and decays (after approximately 14 days in humans). It then degenerates into a corpus Albicans, which is a mass of fibrous scar tissue.
D) Primary follicle starts developing- Primary follicle is a precursor to the corpus luteum and thus this is most unlikely to happen. At the start of the menstrual cycle, some 12-20 primary follicles begin to develop under the influence of elevated FSH to form secondary follicles. The primary follicles have formed from primordial follicles, which developed in the ovary at around 10-30 weeks after conception.
So, the correct answer is option B.

Topic: Biology, Human Reproduction, NEET, NEET 2005
If mass is written as m=kcPG-1/2 h1/2, then the value of P will be : (Constants have their usual meaning with k a dimensionless constant)

Options:
A 12
B 13
C 2
D -13

The correct answer is:
12

Solution:
Given:

m=kcPG-1/2 h1/2

Writing dimensions on both sides, we get

M1 L0 T0=LT-1PM-1 L3 T-2-1/2ML2 T-11/2

By comparing exponent of L on both the sides, we get

0=P-32+1⇒P=12

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2024 (30 Jan Shift 2)
If mass, speed and radius of the circular path of the particle are increased by \(100 \%\), then the necessary force required to maintain the circular path will have to be increased by

Options:
A \(100 \%\)
B \(250 \%\)
C \(300 \%\)
D \(400 \%\)

The correct answer is:
\(300 \%\)

Solution:
Centripetal force is given as
\(F=\frac{m v^2}{r}\)
where, \(m=\) mass, \(v=\) velocity and \(r=\) radius.
\(\frac{F_2}{F_1}=\frac{m_2}{m_1} \cdot\left(\frac{v_2}{v_1}\right)^2 \cdot\left(\frac{r_1}{r_2}\right)\) ...(i)
When, mass, speed and radius of circular path of particle increases by \(100 \%\), i.e. these become double.
Hence,
\(\begin{aligned}
& m_2=2 m_1 \\
& v_2=2 v_1 \\
& r_2=2 r_1
\end{aligned}\)
\(\therefore\) From Eq. (i), we have
\(\begin{aligned}
\Rightarrow & \frac{F_2}{F_1} & =\frac{2 m_1}{m_1}\left(\frac{2 v_1}{v_1}\right)^2\left(\frac{r_1}{2 r_1}\right)=2 \times 4 \times \frac{1}{2}=4 \\
\Rightarrow \quad & F_2 & =4 F_1
\end{aligned}\)
Percentage increase in centripetal force,
\(=\frac{F_2-F_1}{F_1} \times 100=\frac{4 F_1-F_1}{F_1} \times 100=300 \%\)

Topic: Physics, Motion In Two Dimensions, AP EAMCET, AP EAMCET 2020 (17 Sep Shift 1)
If mass-energy equivalence is taken into account, when water is cooled to form ice, the mass of
water should

Options:
A
increase

B
remain unchanged

C
decrease

D
first increase then decrease

The correct answer is:
increase

Solution:
Because thermal energy decreases, therefore mass should increase

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2002
If masses of all the molecules of a gas are halved and their speeds are doubled, then the ratio of initial and final pressures is

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

The correct answer is:
1:2

Solution:
Using the relation p=13mnv2V ...(i)
and also p′=13m2n2v2V ...(ii)
Dividing Eq.(ii) by Eq. (i), we get
p′p=2
So, p:p′=1:2
The ratio of initial and final pressures is 1:2.

Topic: Physics, Thermodynamics, JEE Main
If \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) and \(\mathbf{r}\) are vectors such that \(\mathbf{a}\) is not perpendicular to \(\mathbf{b} \cdot \mathbf{r} \times \mathbf{b}=\mathbf{c} \times \mathbf{b}\) and \(\mathbf{r} \cdot \mathbf{a}=\mathbf{0}\) then \(\mathbf{r}=\)

Options:
A \(\mathbf{c}+\frac{(\mathbf{c} \cdot \mathbf{a})}{(\mathbf{b} \cdot \mathbf{a})} \mathbf{b}\)
B \(\mathbf{b}-\frac{(\mathbf{c} \cdot \mathbf{a})}{(\mathbf{b} \cdot \mathbf{a})} \mathbf{c}\)
C \(\mathbf{c}-\frac{(\mathbf{c} \cdot \mathbf{a})}{(\mathbf{b} \cdot \mathbf{a})} \mathbf{b}\)
D \(\mathbf{b}+\frac{(\mathbf{c} \cdot \mathbf{a})}{(\mathbf{b} \cdot \mathbf{a})} \mathbf{c}\)

The correct answer is:
\(\mathbf{c}-\frac{(\mathbf{c} \cdot \mathbf{a})}{(\mathbf{b} \cdot \mathbf{a})} \mathbf{b}\)

Solution:
Given,
\(\begin{aligned}
& \mathbf{r} \times \mathbf{b}=\mathbf{c} \times \mathbf{b} \\
& \Rightarrow \quad(\mathbf{r}-\mathbf{c}) \times \mathbf{b}=\mathbf{0} \\
& \Rightarrow \quad \mathbf{r}-\mathbf{c} \text { is parallel to } \mathbf{b} \text {. } \\
& \Rightarrow \quad(\mathbf{r}-\mathbf{c})=\lambda \mathbf{b} \\
& \mathbf{r}=\mathbf{c}+\lambda \mathbf{b} \\
\end{aligned}\)
or \(\mathbf{r}=\mathbf{c}+\lambda \mathbf{b}\) ...(i)
Also, \(\mathbf{r} \cdot \mathbf{a}=0 \Rightarrow(\mathbf{c}+\lambda \mathbf{b}) \cdot \mathbf{a}=0\)
\(\Rightarrow \quad \mathbf{c} \cdot \mathbf{a}+\lambda \mathbf{b} \cdot \mathbf{a}=\mathbf{0}\)
or \(\lambda=-\left(\frac{\mathbf{c} \cdot \mathbf{a}}{\mathbf{b} \cdot \mathbf{a}}\right)\) ...(ii)
So, from Eq. (i) we substitute ' \(\lambda\) ' from Eq. (ii), we get
\(\mathbf{r}=\mathbf{c}-\left(\frac{\mathbf{c} \cdot \mathbf{a}}{\mathbf{b} \cdot \mathbf{a}}\right) \mathbf{b}\)

Topic: Mathematics, Vector Algebra, JEE Main