datavorous/entrance-exam-dataset huggingface dataset:
If the external forces acting on a system have zero resultant, the center of mass

Options:
A May move but not accelerate
B May Accelerate
C Must not move
D None of the above

The correct answer is:
May move but not accelerate

Solution:
According to the equation of motion of the centre of mass
M aCM=Fext
If Fext=0, aCM=0
â´ vCM=constant
i.e., if no external force acts on a system the velocity of its centre of mass remains constant. Thus, the centre of mass may move but not accelerate.

Topic: Physics, Center of Mass Momentum and Collision, NEET
If the external forces acting on a system have zero resultant, the centre of mass

Options:
A must not move
B must not accelerate
C must move
D may accelerate

The correct answer is:
must not accelerate

Solution:
If external forces acting on a system have zero resultant then acceleration is zero and velocity remains constant.

Topic: Physics, Rotational Motion, NEET
If the extremities of a diameter of a circle are $(0,0)$ and
$\left(\mathrm{a}^{3}, 1 / \mathrm{a}^{3}\right)$, then the circle passes through which one of the
following points?

Options:
A $\left(a^{2}, 1 / a^{2}\right)$
B $(\mathrm{a}, \mathrm{l} / \mathrm{a})$
C $(\mathrm{a},-\mathrm{a})$
D $(1 / a, a)$

The correct answer is:
$(1 / a, a)$

Solution:
Given the extremities of a diameter of a circle as
$(0,0)$ and $\left(\mathrm{a}^{3}, \frac{1}{\mathrm{a}}\right)$, equation of circle is
$(x-0)\left(x-a^{3}\right)+(y-0)\left(y-\frac{1}{a^{3}}\right)=0$
$\Rightarrow \quad x^{2}-x a^{3}+y^{2}-\frac{y}{a^{3}}=0$
$\Rightarrow \quad x^{2}+y^{2}-x a^{3}-\frac{y}{a^{3}}=0$
Putting $\mathrm{x}=\frac{1}{\mathrm{a}}$ and $\mathrm{y}=\mathrm{a}$, the equation is satisfied.
Thus, the circle passes through the point $\left(\frac{1}{\mathrm{a}}, \mathrm{a}\right)$.

Topic: Mathematics, Circle, JEE Main
If the extremities of the base of an isosceles triangle are the points $(2 a, 0)$ and $(0, a)$ and the equation of one of the sides is $x=2 a$, then the area of the triangle, in square units, is :

Options:
A
$\frac{5}{4} a^2$

B
$\frac{5}{2} a^2$

C
$\frac{25 a^2}{4}$

D
$5 a^2$

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

Solution:
$$
\text { Let } y \text {-coordinate of } \mathrm{C}=b
$$
$$
\begin{aligned}
& \mathrm{AB}=\sqrt{4 a^2+a^2}=\sqrt{5} a \\
& \text { Now, } \mathrm{AC}=\mathrm{BC} \Rightarrow b=\sqrt{4 a^2+(b-a)^2} \\
& \Rightarrow b^2=4 a^2+b^2+a^2-2 a b \\
& \Rightarrow 2 a b=5 a^2 \Rightarrow b=\frac{5 a}{2} \\
& \therefore \mathrm{C}=\left(2 a, \frac{5 a}{2}\right)
\end{aligned}
$$
Hence area of the triangle
$$
\begin{aligned}
& =\frac{1}{2}\left|\begin{array}{ccc}
2 a & 0 & 1 \\
0 & a & 1 \\
2 a & \frac{5 a}{2} & 1
\end{array}\right|=\frac{1}{2}\left|\begin{array}{ccc}
2 a & 0 & 1 \\
0 & a & 1 \\
0 & \frac{5 a}{2} & 0
\end{array}\right| \\
& =\frac{1}{2} \times 2 a\left(-\frac{5 a}{2}\right)=-\frac{5 a^2}{2}
\end{aligned}
$$
Since area is always $+\mathrm{ve}$, hence area
$$
=\frac{5 a^2}{2} \text { sq. unit }
$$

Topic: Mathematics, Properties of Triangles, JEE Main, JEE Main 2013 (23 Apr Online)
If the filament of a 100 W bulb has an area 0.25 cm2 and behaves as a perfect black body, find the wavelength corresponding to the maximum in its energy distribution. [Given Ï=5.67Ã10-8 J m-2 s K-4, b=2.89Ã10-3 m K]

Options:
A 8751.23 Ã
B 2898.14 Ã
C 9971.9 Ã
D 7055.5 Ã

The correct answer is:
9971.9 Ã

Solution:
In the bulb filament given, energy radiated per second per m2 of its surface area is given as
E=PA=1000.25Ã10-4=4Ã106J s-1 m-2
If T is the temperature of the filament then according to Stefan's law, we have
E = Ï T 4
or 4 Ã 1 0 6 = 5.67 Ã 1 0 - 8 Ã T 4
or T 4 = 4 Ã 1 0 6 5.67 Ã 1 0 - 8 = 7.055 Ã 1 0 1 3
or T = 7.055 Ã 1 0 1 3 1 / 4 = 2898.14 K
If the filament radiates the maximum energy at a wavelength Î» m , from Wein's displacement law, we have
Î» m T = b
or Î» m = b T
=2.89Ã10-32898.14=9971.9 Ã

Topic: Physics, Thermal Properties of Matter, JEE Main
If the first excitation energy of a hydrogen-like atom is 27.3 eV, then the ionization energy of this atom will be

Options:

The correct answer is:
36.4

Solution:
First excitation energy
=Z2Rhc112â122=Z2Rhc34

â´âââ34Rhc=27.3âeV

â´âââZ2Rhc=4327.3 eV=36.4 eV

Topic: Physics, Atomic Physics, JEE Main
If the first, second and last terms of an A.P. be $a, b, 2a$ respectively, then its sum will be

Options:
A $\frac{a b}{b-a}$
B $\frac{a b}{2(b-a)}$
C $\frac{3 a b}{2(b-a)}$
D $\frac{3 a b}{4(b-a)}$

The correct answer is:
$\frac{3 a b}{2(b-a)}$

Solution:
We have first term $A=a \ldots(i)$
Second term $A+d=b \ldots(ii)$
and last term $I=2 a \ldots(iii)$
From (i), (ii) and (iii),
$d=(b-a)$ and $n=\frac{b}{b-a}$
Then sum
$S=\frac{n}{2}[a+l]=$ $=\frac{b}{2(b-a)}[a+2 a]=\frac{3 a b}{2(b-a)}$
Trick : Let $a=2, b=3$ then the sum $=9$ which is given by option (3).

Topic: Mathematics, Sequences and Series, JEE Main
If the first term of a G.P. be 5 and common ratio be -5 , then which term is 3125

Options:
A $6^{\text {th }}$
B $5^{\text {th }}$
C $7^{\text {th }}$
D $8^{\text {th }}$

The correct answer is:
$5^{\text {th }}$

Solution:
Given that first term $a=5$ and common ratio $r=-5$. Suppose that $n^{\text {th }}$ term is 3125 ,
then $a r^{n-1}=3125 \Rightarrow 5(-5)^{n-1}=3125 \Rightarrow(-5)^{n-1}=5^4$
Hence $n=5$.

Topic: Mathematics, Sequences and Series, JEE Main
If the first term of an A.P. is 3 and the sum of its first 25 terms is equal to the sum of its next 15 terms, then the common difference of this A.P.is

Options:
A 16
B 15
C 14
D 17

The correct answer is:
16

Solution:
Sum of first25 terms=sum of next 15 terms

S25=SNext 15

⇒S25+SNext 15=S40

⇒2×S25=S40

⇒2×2522×3+24d=4022×3+39d

⇒52×3+24d=42×3+39d

⇒30+120d=24+156d

⇒36d=6⇒d=16

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2020 (03 Sep Shift 1)
If the first term of an A.P. be 10 , last term is 50 and the sum of all the terms is 300 , then the number of terms are

Options:
A $5$
B $8$
C $10$
D $15$

The correct answer is:
$10$

Solution:
Given that first term $a=10$, last term $I=50$ and sum $S=300$
$\therefore S=\frac{n}{2}(a+l)$
$\Rightarrow 300=\frac{n}{2}(10+50) \Rightarrow n=10$.

Topic: Mathematics, Sequences and Series, JEE Main
If the foci of a hyperbola are same as that of the ellipse x29+y225=1 and the eccentricity of the hyperbola is 158 times the eccentricity of the ellipse, then the smaller focal distance of the point 2, 14325 on the hyperbola, is equal to

Options:
A 725-83
B 1425-43
C 1425-163
D 725+83

The correct answer is:
725-83

Solution:
Given equation of ellipse is, x29+y225=1

⇒a=3, b=5

We know that, e=1-a2b2

⇒e=1-925=45

Now, foci=0,±be

=0,±4

Since, eccentricity of hyperbola is given as 158 same to that of ellipse, ∴ eH=45×158=32

Let equation of the hyperbola be x2A2-y2B2=-1.

⇒B.eH=4

⇒B=83

⇒A2=B2eH2-1=64994-1

⇒A2=809

⇒x2809-y2649=-1

Directrix: y=±BeH=±169

⇒PS=e·PM

⇒PS=32143·25-169

⇒PS=725-83

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2024 (31 Jan Shift 1)
If the foci of the conics $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{7}=1$ and $\frac{x^{2}}{144}-\frac{y^{2}}{81}=\frac{1}{25}$
were to coincide, then what is the value of a ?

Options:
A 2
B 3
C 4
D 16

The correct answer is:
4

Solution:
The equation of ellipse is given as
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{7}=1$
Eccentricity is given by
$e=\sqrt{1-\frac{7}{a^{2}}}$
Therefore, foci of ellipse are $(\pm \mathrm{ae}, 0) \mathrm{ie}$,
$\left(\pm a \sqrt{1-\frac{7}{a^{2}}}, 0\right)$
Now, the equation of given hyper bola is
$\frac{x^{2}}{144}-\frac{y^{2}}{81}=\frac{1}{25} \Rightarrow \frac{x^{2}}{\frac{144}{25}}-\frac{y^{2}}{\frac{81}{25}}=1$
So, a $=\frac{12}{5}$ and $b=\frac{9}{5}$
$\therefore \quad e^{\prime}=\sqrt{1+\frac{81 / 25}{144 / 25}}=\sqrt{\frac{144+81}{144}}=\sqrt{\frac{225}{144}}$
$\quad=\frac{15}{12}$
Foci of hyper bola are $\left(\pm \frac{12}{5}, \frac{15}{12}, 0\right)$ ie, $(\pm 3,0)$.
Since these foci coincides
$\Rightarrow 3=a \sqrt{1-\frac{7}{a^{2}}}$
$\Rightarrow \frac{3}{\mathrm{a}}=\sqrt{1-\frac{7}{\mathrm{a}^{2}}}$
$\Rightarrow \frac{9}{\mathrm{a}^{2}}=1-\frac{7}{\mathrm{a}^{2}}$
$\Rightarrow \frac{16}{a^{2}}=1 \Rightarrow a=4$

Topic: Mathematics, Ellipse, JEE Main
If the foci of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ coincide with the foci of the hyperbola $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$, then $b^2$ is equal to

Options:
A
8

B
10

C
7

D
9

The correct answer is:
7

Solution:
Given equation of ellipse is
$$
\frac{x^2}{16}+\frac{y^2}{b^2}=1
$$
$$
\begin{aligned}
& \text { eccentricity }=e=\sqrt{1-\frac{b^2}{16}} \\
& \text { foci: } \pm a e=\pm 4 \sqrt{1-\frac{b^2}{16}}
\end{aligned}
$$
Equation of hyperbola is $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$
$$
\begin{aligned}
& \Rightarrow \frac{x^2}{\frac{144}{25}}-\frac{y^2}{\frac{81}{25}}=1 \\
& \text { eccentricity }=e=\sqrt{1+\frac{81}{25} \times \frac{25}{144}}=\sqrt{1+\frac{81}{144}} \\
& =\sqrt{\frac{225}{144}}=\frac{15}{12} \\
& \text { foci: } \pm a e=\pm \frac{12}{5} \times \frac{15}{12}=\pm 3 \\
&
\end{aligned}
$$
Since, foci of ellipse and hyperbola coincide
& \therefore \pm 4 \sqrt{1-\frac{b^2}{16}}=\pm 3 \Rightarrow b^2=7 \\
$$

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2012 (19 May Online)
If the foci of the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{b^{2}}=1$ and the hyperbola $\frac{x^{2}}{144}-\frac{y^{2}}{81}=\frac{1}{25}$ coincide, then the value of $b^{2}$ is

Options:
A 25
B 9
C 16
D 4

The correct answer is:
16

Solution:
Eccentricity for $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is $b^{2}=a^{2}\left(1-e^{2}\right)$ and eccentricity for $\frac{x^{2}}{\frac{144}{25}}-\frac{y^{2}}{\frac{81}{25}}=1$ is
Again, foci $=a_{1} e_{1}=\frac{12}{5} \times \frac{15}{12}=3$
So, focus of hyperbola is $(3,0)=(a e, 0)$
and focus of ellipse is $(a e, 0)=(5 a, 0)$
As, this foci are same, so
$$
\begin{aligned}
& 5 a=3 \\
\therefore & e &=\frac{3}{5} \\
& \text { So, } \quad & e^{2} &=1-\frac{b^{2}}{a^{2}}=1-\frac{b^{2}}{25} \\
\Rightarrow & & \frac{b^{2}}{25} &=1-e^{2}=1-\frac{9}{25} \\
\Rightarrow & & \frac{b^{2}}{25} &=\frac{16}{25} \\
\Rightarrow & & b^{2} &=16
\end{aligned}
$$

Topic: Mathematics, Ellipse, JEE Main
If the focii of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ and the hyperbola $\frac{x^2}{4}-\frac{y^2}{b^2}=1$ coincide, then $b^2$ is equal to

Options:
A $4$
B $5$
C $8$
D $9$

The correct answer is:
$5$

Solution:
Given, equation of ellipse is
$$
\frac{x^2}{25}+\frac{y^2}{16}=1
$$
and equation of hyperbola is
$$
\frac{x^2}{4}-\frac{y^2}{b^2}=1
$$
eccentricity of ellipse
$$
\begin{aligned}
& b^2=a^2\left(1-e^2\right) \\
& \Rightarrow \quad 16=25\left(1-e^2\right) \\
& \Rightarrow \quad e^2=1-\frac{16}{25}=\frac{9}{25} \\
& \Rightarrow \quad e= \pm \frac{3}{5} \\
&
\end{aligned}
$$
Focii of the ellipse $=( \pm a e, 0)=( \pm 3,0)$ which coincide with focii of the hyperbola.
Let $e_1$ be the eccentricity of the hyperbola.
$$
\begin{array}{ll}
\therefore & \pm a e_1= \pm 3 \\
\Rightarrow & e_1=\frac{3}{2}
\end{array}
$$
Now, $\quad b^2=a^2\left(e_1^2-1\right)$
$$
\begin{array}{cc}
\Rightarrow & b^2=4\left(\frac{9}{4}-1\right)=4 \times \frac{5}{4} \\
\Rightarrow & b^2=5
\end{array}
$$

Topic: Mathematics, Hyperbola, JEE Main
If the focus of a hyperbola is (±3,0) and the equation of a tangent is 2x+y-4=0, then the equation of the hyperbola is

Options:
A 4x2-5y2=20
B 5x2-4y2=20
C 4x2-5y2=1
D 5x2-4y2=1

The correct answer is:
4x2-5y2=20

Solution:
Given, ±ae, 0=±3, 0
⇒ ae=3
⇒ a2e2=9
⇒ b2+a2=9 ...(i)
∵ 2x+y-4=0
⇒ y=-2x+4
is the tangent to the hyperbola
∴ 42=a2-22-b2 ∵c2=a2m2-b2
⇒ 4a2-b2=16 ...(ii)
On solving Eqs. (i) and (ii), we get,
a2=5, b2=4
∴ Equation of hyperbola is x25–y24 =1
⇒ 4x2-5y2=20

Topic: Mathematics, Hyperbola, JEE Main
If the following system of linear equations
2x+y+z=5
x-y+z=3
x+y+a z=b
has no solution, then :

Options:
A a=-13,b≠73
B a≠13,b=73
C a≠-13,b=73
D a=13,b≠73

The correct answer is:
a=13,b≠73

Solution:
Given system of equations

2x+y+z=5
x-y+z=3
x+y+a z=b
has no solution,

D=2111-1111a=0⇒-2a-2-a+1+2=0⇒a=13

D1=5113-11b1a=5-a-1 + -13a-b +13+b = -8a+2b-2

D2=2511311ba=23a-b -5 a-1+1b-3 = a-b+3

D3=2151-1311b=2-b-3 - 1 b-3 + 52 = 7-3b

For the system to have no solution ⇒D = 0 , D3 ≠ 0

∴ a = 13 , b ≠73.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2021 (31 Aug Shift 1)
If the foot of perpendicular drawn from the point 2,5,1 on a line passing through α,2α,5 is 15,25,35, then α is equal to

Options:
A 199
B 11570
C 5754
D 125

The correct answer is:
125

Solution:
From diagram PQ→⋅QR→=0
2-15i^+5-25j^+1-35k^·α-15i^+2α-25j^+5-35k^=0
⇒9α5-925+46α5-4625+4425=0
⇒α=125

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the foot of the perpendicular drawn from the point 1,0,3 on a line passing through α,7,1 is 53,73,173, then α is equal to

Options:

The correct answer is:
4

Solution:
Since PQ, is perpendicular to L, therefore
1-53α-53+-737-73+3-1731-173=0
-2α3+109-989+1129=0
2α3=249⇒α=4.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2020 (07 Jan Shift 2)
If the foot of the perpendicular drawn from the point \( (1,0,3) \) on a line passing through \( (\alpha, 7,1) \) is \( \left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right) \), then \( \alpha \) is equal to

Options:

The correct answer is:
4

Solution:
D.R. of BP=53-α,73-7,173-1

D.R. of AP=53-1,73-0,173-3

BP⊥AP

⇒α=4

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the foot of the perpendicular from $(0,0,0)$ to a plane is $(1,2,3)$, then the equation of the plane is

Options:
A $2 x+y+3 z=14$
B $x+2 y+3 z=14$
C $x+2 y+3 z+14=0$
D $x+2 y-3 z=14$

The correct answer is:
$x+2 y+3 z=14$

Solution:
The equation of a plane passing through $(1,2,3)$ is
$a(x-1)+b(y-2)+c(z-3)=0$
$\begin{aligned} & \text { Also, }(a, b, c)=(1-0,2-0,3-0) \\ & -(1,2,3) \\ & \therefore 1(x-1)+2(y-2)+3(z-3)=0 \\ & \Rightarrow \quad x+2 y+3 z-1-4-9=0 \\ & \Rightarrow \quad x+2 y+3 z=14 \\ & \end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the foot of the perpendicular from $(0,0,0)$ to the plane is $(1,2,2)$, then the equation of the plane is

Options:
A $-x+2 y+8 z-9=0$
B $x+2 y+2 z-9=0$
C $x+y+z-5=0$
D $x+2 y-3 z+1=0$

The correct answer is:
$x+2 y+2 z-9=0$

Solution:
Now, we take the equation of plane $x+2 y+2 z-9=0$. Since, it is passing through the point $(1,2,2)$.
$\begin{aligned} & & 1+2(2)+2(2)-9 & =0 \\ \Rightarrow & & 0 & =0\end{aligned}$
Thus, the required equation of the plane is
$x+2 y+2 z-9=0$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the foot of the perpendicular from point 4,3,8 on the line L1:x-al=y-23=z-b4, l≠0 is 3,5,7, then the shortest distance between the line L1 and line L2:x-23=y-44=z-55 is equal to

Options:
A 12
B 16
C

23

D 13

The correct answer is:
16

Solution:
3,5,7 satisfy the line L1

3-aℓ=5-23=7-b4

3-aℓ=1 & 7-b4=1

a+ℓ=3 …1 & b=3 ...2

v→1=4,3,8-3,5,7

v→1=1,-2,1

v→2=ℓ,3,4

v→1·v→2=0⇒ℓ-6+4=0⇒ℓ=2

a+ℓ=3⇒a=1

L1:x-12=y-23=z-34

L2:x-23=y-44=z-55

A=1,2,3

B=2,4,5

AB→=1,2,2

p→=2i^+3j^+4k^

q→=3i^+4j^+5k^

p→×q→=-i^+2j^-k^

Shortest distance =AB→·p→×q→p→×q→=16

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (16 Mar Shift 2)
If the foot of the perpendicular from the point A-1,4,3 on the plane P:2x+my+nz=4, is -2,72,32, then the distance of the point A from the plane P, measured parallel to a line with direction ratios 3,-1,-4, is equal to

Options:
A 1
B 26
C 22
D 14

The correct answer is:
26

Solution:
Plotting the data of the given value we have,

Now let B be foot of ⊥coordinates, so given B=-2,72,32

Now direction ratio of line AB will be -2+1,72-4,32-3≡-1,-12,-32≡2,1,3

Now comparing with 2x+my+nz=4 we get, m=1,n=3

So equation of AC which is passing through -1,4,3 will be

x+13=y-4-1=z-3-4=λ

So point C is 3λ-1,-λ+4,-4λ+3, but given C lies on the plane 2x+y+3z=4, so on satisfying the point we get,

6λ-2-λ+4-12λ+9=4

⇒λ=1⇒C2,3,-1

Now finding the distance between A & C we get,

⇒ AC=26

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2022 (29 Jul Shift 1)
If the force applied to a body produces rotational motion in an anticlockwise sense, the moment of the force is considered
as,

Options:
A negative
B zero
C positive
D infinity

The correct answer is:
positive

Solution:
Sign convention used for moment of couple.

When the forces applied to a body rotates the body in the anticlockwise manner, then the moment of force is assumed to be positive. On the other hand, if the rotation takes place in clockwise sense then the moment of force is assumed to be negative.

Topic: Physics, Rotational Motion, JEE Main
If the force is given by $\mathrm{F}=\mathrm{at}+\mathrm{bt}^{2}$ with $\mathrm{t}$ as time. The dimensions of a and b are

Options:
A $\left[\mathrm{MLT}^{-4}\right]$ and $\left[\mathrm{MLT}^{-2}\right]$
B $\left[\mathrm{MLT}^{-3}\right]$ and $\left[\mathrm{MLT}^{-4}\right]$
C $\left[\mathrm{ML}^{2} \mathrm{~T}^{-3}\right]$ and $\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]$
D $\left[\mathrm{ML}^{2} \mathrm{~T}^{-3}\right]$ and $\left[\mathrm{ML}^{3} \mathrm{~T}^{-4}\right]$

The correct answer is:
$\left[\mathrm{MLT}^{-3}\right]$ and $\left[\mathrm{MLT}^{-4}\right]$

Solution:
Dimension of at = Dimension of F
$[a t]=[F] \Rightarrow[a]=\left[\frac{F}{t}\right]$
$[b]=\left[\frac{M L T^{-2}}{T}\right] \Rightarrow[a]=\left[M L T^{-3}\right]$
Dimension of $\mathrm{bt}^{2}=$ Dimension of $\mathrm{F}$
$\left[\mathrm{bt}^{2}\right]=[\mathrm{F}] \Rightarrow[\mathrm{b}]=\left[\frac{\mathrm{F}}{\mathrm{t}^{2}}\right]$
$[\mathrm{b}]=\left[\frac{\mathrm{ML} \mathrm{T}^{-4}}{\mathrm{~T}^{2}}\right] \Rightarrow[\mathrm{b}]=\left[\mathrm{MLT}^{-4}\right]$

Topic: Physics, Units and Dimensions, JEE Main
If the forest cover is reduced to half, the most likely outcome will be _____.

Options:
A tribals living in these areas will starve to death.
B cattle in these and the adjoining areas will die due to lack of fodder.
C large areas will become deserts.
D crop breeding programmes will suffer due to the reduced availability of a variety of germplasm.

The correct answer is:
large areas will become deserts.

Solution:
If the forest cover is reduced to half, large areas will become deserts. The development of the fertile top soil takes centuries. However, it can be removed very easily due to mainly deforestation, resulting in arid patches of land. When large barren patches extend and meet over time, a desert is created. Internationally, it has been recognised that desertification is a major problem nowadays, particularly due to increased urbanisation.

Topic: Biology, Environmental Issues, NEET
If the formula of basic Beryllium nitrate is BenONO36 . What is the value of 'n' here

Options:

The correct answer is:
4

Solution:
Basic Beryllium nitrate is Be4ONO36
2n−2−6=0
Hence, n=4

Topic: Chemistry, s Block Elements, JEE Main
If the formula of Borax is Na2 B4Ox(OH)y·zH2O, then x+y+z= ______

Options:

The correct answer is:
17

Solution:
The crude name of borax is Tincal. It has the molecular formula Na2B4O7.10H2O and structural formula of borax is Na2 B4O5(OH)4.8H2O

Based on the structural formula, x=5, y= 4 and z=8

(x + y + z) = 5 + 4 + 8 = 17

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2023 (13 Apr Shift 2)
If the four complex numbers z,z¯,z¯-2Rez¯ and z-2Rez represent the vertices of a square of side 4 units in the Argand plane, then z is equal to :

Options:
A 42
B 4
C 22
D 2

The correct answer is:
22

Solution:
Let z=x+iy

∴ length of side =4

then z-z¯=4

2iy=4

y=2

also z-z-2Rez=4

2x=4⇒x=2

z=x2+y2=22

Topic: Mathematics, Complex Number, JEE Main
If the four complex numbers z,z¯,z¯-2Rez¯ and z-2Rez represent the vertices of a square of side 4 units in the Argand plane, then z is equal to :

Options:
A 42
B 4
C 22
D 2

The correct answer is:
22

Solution:
Let z=x+iy

∴ length of side =4

then z-z¯=4

2iy=4

y=2

also z-z-2Rez=4

2x=4⇒x=2

z=x2+y2=22

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2020 (05 Sep Shift 1)
If the four letter words (need not be meaningful) are to be formed using the letters from the word "MEDITERRANEAN" such that the first letter is R and the fourth letter is E, then the total number of all such words is :

Options:
A 110
B 59
C 11!2!3
D 56

The correct answer is:
59

Solution:
M,EEE,D,I,T,RR,AA,NN
R_ _ _ E_
Two empty places can filled with identical letters EE, AA, NN ⇒ 3 ways
Two empty places, can be filled with distinct letters M,E,D,I,T,R,A,N ⇒ 8P2
∴ Number of words 3+8P2=59

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2016 (09 Apr Online)
If the four points, whose position vectors are 3i^-4j^+2k^,i^+2j^-k^,-2i^-j^+3k^ and 5i^-2αj^+4k^ are coplanar, then α is equal to

Options:
A 7317
B -10717
C -7317
D 10717

The correct answer is:
7317

Solution:
Let the four points be A: (3,-4,2), B: (1,2,-1), C: (-2,-1,3) and D: (5,-2α, 4)

A, B, C, D are coplanar points, then

⇒1-32+4-1-2-2-3-1+43-25-3-2α+44-2=0

⇒-26-3-5312-2α+42=0

⇒-3α+146=0

⇒α=7317

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (25 Jan Shift 2)
If the four positive integers are selected randomly from the set of positive integers, then the probability that the number $1,3,7$ and 9 are in the unit place in the product of 4 -digit, so selected is

Options:
A $\frac{7}{625}$
B $\frac{2}{5}$
C $\frac{5}{625}$
D $\frac{16}{625}$

The correct answer is:
$\frac{16}{625}$

Solution:
The number of digits on unit place of any number $=10$
$\therefore$
$$
n(S)=10
$$
The necessary condition for becoming the digits $1,3,5$ or 7 at the unit place of product of four numbers that the digits $1,3,5$ or 7 at unit place of every number.
$$
\begin{array}{ll}
\therefore & n(A)=4 \\
\therefore & P(A)=\frac{4}{10}=\frac{2}{5}
\end{array}
$$
So, required $y=\left(\frac{2}{5}\right)^{4}=\frac{16}{625}$

Topic: Mathematics, Permutation Combination, JEE Main
If the fourth term in the Binomial expansion of 2x+xlog8x6, x>0 is 20×87, then a value of x is

Options:
A 8–2
B 8
C 83
D 82

The correct answer is:
82

Solution:
The general term in the binomial expansion a+bn is Tr+1=Crnan-rbr.

Given, the fourth term in the expansion of 2x+xlog8x6 is

T4=20×87

⇒C362x3xlog8x3=20×87

⇒6!3!·3!2x3xlog8x3=20×237

⇒6×5×4×3!3×2×1·3!2x3xlog8x3=20×221

⇒202x3xlog8x3=20×221

⇒2xxlog8x3=273

⇒2xxlog8x=27

⇒1xxlog8x=26

⇒xlog8x-1=82

Taking log both side to the base 8

⇒log8xlog8x-1=log882

Using logamn=nlogam, we get

⇒(log8x-1)log8x=2

⇒t-1t=2 (Let log8x=t)

⇒t2-t-2=0

⇒t=2 or -1

⇒log8x=2 or -1

If logax=b, ⇒x=ab,

⇒x=82 or 8-1

⇒x=64 or 18.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2019 (09 Apr Shift 1)
If the fourth term in the binomial expansion of x11+log10x+x1126 is equal to 200, and x>1, then the value of x is

Options:
A 100
B 104
C 103
D 10

The correct answer is:
10

Solution:
General term for the given binomial is
Tr+1= 6Crx11+log10x6-rx112r
Given T4=200, putting r=3
⇒T4= 6C3x11+log10x3x1123=200
⇒x11+log10x32⋅x14=10
⇒x321+log10x+14=10
⇒321+log10x+14log10x=1
⇒x=10,10-4
⇒x=10 ( as x>1)

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2019 (08 Apr Shift 2)
If the fourth term in the expansion of x+xlog2x7 is 4480, then the value of x where x∈N is equal to:

Options:
A 2
B 4
C 3
D 1

The correct answer is:
2

Solution:
Given x+xlog2x7.

The general term in the expansion of x+xlog2x7 is Cr7xrxlog2x7-r.

Then, T4=C37x4xlog2x3=4480

⇒7·6·53·2·1x4+log2x=4480

⇒x4+3log2x=128

⇒x4+3log2x=27

Now, from the option only at x=2, LHS=RHS.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2021 (17 Mar Shift 1)
If the fractional part of the number 240315 is k15, then k is equal to

Options:
A 4
B 14
C 8
D 6

The correct answer is:
8

Solution:
2403=23.24100
=8.15+1100
=8100C015100+ 100C11599+....+ 100C9915+ 100C100
=8.15λ+1
=8.15λ+8
⇒240315=8λ+815
Hence, k=8

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2019 (09 Jan Shift 1)
If the freezing point of a 0.01 molal aqueous solution of a cobalt (III) bromide-ammonia complex (which behaves as a strong electrolyte) is -0.0557Â°C , the number of bromide(s) in the coordination sphere of the complex is [ Kf for water =1.86 K kg mol-1 ]

Options:

The correct answer is:
1

Solution:
ÎTf=iKfm
0.0557=iÃ1.86Ã0.01
i=3, means complex will give 3-ions

Hence, formula should be CoNH35BrBr2

Topic: Chemistry, Solutions, JEE Main