datavorous/entrance-exam-dataset huggingface dataset:
If $\mathbf{A}=\mathbf{B}+\mathbf{C}$ have scalar magnitudes of 5,4,3 units respectively, then the angle between $A$ and $\mathbf{C}$ is

Options:
A $\cos ^{-1}(3 / 5)$
B $\cos ^{-1}(4 / 5)$
C $\pi / 2$
D $\sin ^{-1}(3 / 4)$

The correct answer is:
$\cos ^{-1}(3 / 5)$

Solution:
Here, triangle is

$\theta=\cos ^{-1}\left(\frac{|\mathbf{C}|}{|\mathbf{A}|}\right)$
$\theta=\cos ^{-1}\left(\frac{3}{5}\right)$

Topic: Physics, Mathematics in Physics, JEE Main
If (a, b, c) is the image of the point (1,2,-3) in the line, x+12=y-3-2=z-1, then a+b+c is equal to:

Options:
A 2
B -1
C 3
D 1

The correct answer is:
2

Solution:
R(-1+2r,3-2r,-r)

dr's of PR are (2-2 r,-1+2 r,-3+r)

Then, 2(2-2 r)+2(1-2 r)+1(3-r)=0

9-9r=0 ⇒ r=1

R(1,1,-1)

then a+1=2 ,b+2=2 , c-3=-2 i.e., a=1,b=0, c=1

∴ a+b+c=2

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2020 (05 Sep Shift 1)
If a,b,c∈N and D=abccabbca, then the least positive value of D is

Options:
A 4
B 6
C 3
D 8

The correct answer is:
4

Solution:
abccabbca=a3Ãb3Ãc3-3abc
=a+b+c2a-b2+b-c2+c-a2
For least positive value, when a,b,câN, two of them are 1 and one of them is 2
So, Dmin=4

Topic: Mathematics, Determinants, JEE Main
If a,b,c,Î»âN, then the least possible value of a2+Î»abacbab2+Î»bccacbc2+Î» is

Options:

The correct answer is:
4

Solution:
Multiplying first, second, third rows by a,b,c and then taking a,b,c common from first, second, third columns, we get,
D=a2+λa2a2b2b2+λb2c2c2c2+λ

Now, R1↔R1+R2+R3
=a2+b2+c2+λ111b2b2+λb2c2c2c2+λ

Now, C2↔C2-C1 , C3→C3-C1
=a2+b2+c2+λ100b2λ0c20λ=λ2a2+b2+c2+λ
Hence, Dmin=4 when a=b=c=λ=1

Topic: Mathematics, Determinants, JEE Main
If $a, b, c \in \mathrm{R}$ and 1 is a root of equation $a x^2+b x$ $+c=0$, then the curve $y=4 a x^2+3 b x+2 c, a \neq 0$ intersect $x$-axis at

Options:
A
two distinct points whose coordinates are always rational numbers

B
no point

C
exactly two distinct points

D
exactly one point

The correct answer is:
exactly one point

Solution:
Given $a x^2+b x+c=0$
$$
\begin{aligned}
& \Rightarrow a x^2=-b x-c \\
& \text { Now, consider } \\
& y=4 a x^2+3 b x+2 c \\
& =4[-b x-c]+3 b x+2 c \\
& =-4 b x-4 c+3 b x+2 c \\
& =-b x-2 c
\end{aligned}
$$
Since, this curve intersects $x$-axis
$\therefore$ put $y=0$, we get
$$
\begin{aligned}
& -b x-2 c=0 \Rightarrow-b x=2 c \\
& \Rightarrow x=\frac{-2 c}{b}
\end{aligned}
$$
Thus, given curve intersects $x$-axis at exactly one point.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2012 (26 May Online)
If a,b,c∈R+ such that a+b+c=27, then the maximum value of a2b3c4 is equal to

Options:
A 2 8 ⋅ 3 10
B 2 9 ⋅ 3 11
C 2 10 â 3 12
D 2 11 ⋅ 3 13

The correct answer is:
2 10 ⋅ 3 12

Solution:
Consider nine numbers a2,a2,b3,b3,b3,c4,c4,c4,c4

Using AM≥GM, we get,
a2+a2+b3+b3+b3+c4+c4+c4+c49≥a22b33c4419
⇒a+b+c9≥a2b3c422334419
⇒2799≥a2b3c4223328
⇒ a 2 b 3 c 4 ≤ 2 10 ⋅ 3 12

Topic: Mathematics, Sequences and Series, JEE Main
If $a \neq b \neq c$, the value of $x$ which satisfies the equation$\left|\begin{array}{ccc}0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0\end{array}\right|=0$ is

Options:
A $x=0$
B $x=a$
C $x=b$
D $x=c$

The correct answer is:
$x=0$

Solution:
Obviously, on putting $x=0$, we observe that the determinant becomes
$\Delta_{x=0}=\left|\begin{array}{ccc}0 & -a & -b \\ a & 0 & -c \\ b & c & 0\end{array}\right|=a(b c)-b(a c)=0$
$\therefore \quad x=0$ is a root of the given equation.
Aliter : Expanding $\Delta$, we get
$\Delta=-(x-a)[-(x+b)(x-c)]+(x-b)[(x+a)(x+c)]=0$
$\Rightarrow \quad 2 x^3-(2 \Sigma a b) x=0$
$\Rightarrow$ Either $x=0$ or $x^2=\sum a b$ (i.e., $x= \pm \sum a b$ )
Again $x=0$ satisfies the given equation

Topic: Mathematics, Determinants, JEE Main
If $A+B+C=\pi$, then $\cos 2 A+\cos 2 B+\cos 2 C=$

Options:
A $1+4 \cos A \cos B \sin C$
B $-1+4 \sin A \sin B \cos C$
C $-1-4 \cos A \cos B \cos C$
D None of these

The correct answer is:
$-1-4 \cos A \cos B \cos C$

Solution:
$\begin{aligned} \text { L.H.S. } & =2 \cos (A+B) \cos (A-B)+\left(2 \cos ^2 C-1\right) \\ & =-1-2 \cos C \cos (A-B)+2 \cos ^2 C \\ & =-1-2 \cos C[\cos (A-B)+\cos (A+B)] \\ & =-1-4 \cos A \cos B \cos C .\end{aligned}$

Topic: Mathematics, Trigonometric Equations, JEE Main
If AB is an isothermal, BC is an isochoric and AC is an adiabatic, which of the graphs correctly represents given in figure?

Options:
A

B

C

D

The correct answer is:

Solution:
As slope of adiabatic AC is more than the slope of isothermal AB, and BC is isochoric (ie at constant volume), therefore, figure (b) represents the curves correctly

Topic: Physics, Thermodynamics, JEE Main
If a,b∈R satisfy the equation a2+4b2-4=0, then the minimum value of 2a+3b will be

Options:
A -4
B -5
C -6
D -10

The correct answer is:
-5

Solution:
Given equation isa24+b2=1

Putting a=2cos⁡θ, b=sin⁡θ, we get,

2a+3b=22cos⁡θ+3sin⁡θ=4cosθ+3sinθ
⇒ Minimum value =-42+32=-5

Topic: Mathematics, Ellipse, JEE Main
If $(a+b x)^{-3}=\frac{1}{27}+\frac{1}{3} x+\ldots$, then the ordered pair $(a, b)$ equals to

Options:
A $(3,-27)$
B $\left(1, \frac{1}{3}\right)$
C $(3,9)$
D $(3,-9)$

The correct answer is:
$(3,-9)$

Solution:
$$
\begin{aligned}
\text { Now, }(a+b x)^{-3} & =a^{-3}\left(1+\frac{b x}{a}\right)^{-3} \\
& =\frac{1}{a^3}\left(1-{ }^3 C_1\left(\frac{b x}{a}\right)+\ldots\right) \\
\Rightarrow \quad \frac{1}{a^3} & -{ }^3 C_1 \frac{1}{a^3}\left(\frac{b x}{a}\right)+\ldots \\
& =\frac{1}{27}+\frac{1}{3} x
\end{aligned}
$$
(given)
On equating constant and $x$, we get
$$
\frac{1}{a^3}=\frac{1}{27}
$$
$\begin{array}{rlrl}\text { and } & -{ }^3 C_1 \frac{1}{a^3}\left(\frac{b}{a}\right) & =\frac{1}{3} \\ \Rightarrow & r l r l \\ \text { and } & -{ }^3 C_1\left(\frac{1}{3^4}\right) \times b & =\frac{1}{3} \\ \Rightarrow & -3 \times \frac{1}{3^4} b & =\frac{1}{3} \\ & \Rightarrow & -b & =3^2 \Rightarrow b=-9 \\ & \therefore & (a, b) & =(3,-9)\end{array}$

Topic: Mathematics, Binomial Theorem, JEE Main
If a+bxeyx=x, then d2ydx2=

Options:
A 1x3xy'+y2
B 1x3xy'+y2
C 1x3xy'-y
D 1x3xy'-y2

The correct answer is:
1x3xy'-y2

Solution:
Given, a+bx eyx=x

⇒eyx=xa+bx

⇒logeyx=logxa+bx

⇒yx=logx-loga+bx

⇒y=xlogx-xloga+bx

Differentiate both sides

y'=xx+logx-bxa+bx-loga+bx

y'=x1x-ba+bx+logx-loga+bx

y'=x1x-ba+bx+yx

xy'=x21x-ba+bx+y

xy'=x2a+bx-bxxa+bx+y

xy'=axa+bx+y

xd2ydx2+y'=aa+bx·1-x·ba+bx2+y'

x3d2ydx2=a2x2a+bx2

x3d2ydx2=axa+bx2

x3d2ydx2=xy'-y2

d2ydx2=1x3xy'-y2

Topic: Mathematics, Differentiation, JEE Main
If a ball is thrown vertically with speed $u$, the distance covered during the last $t$ seconds of its ascent is:

Options:
A $u t$
B $\frac{1}{2} g t^2$
C $u t-\frac{1}{2} g t^2$
D $(u+g t) t$

The correct answer is:
$\frac{1}{2} g t^2$

Solution:
Let time of flight be $T$, then $T=\frac{u}{g}$
Again Let $h$ be the distance covered during last ' $t$ ' second of its ascent.
$\therefore$ Velocity at point $\mathrm{B}=V_B=u-g$ $(T-t)$
$=u-g\left(\frac{u}{g}-t\right)=g t$

$\Rightarrow h=V_B t-\frac{1}{2} g t^2$

Topic: Physics, Motion In One Dimension, JEE Main
If a bar magnet of pole strength $m$ and magnetic moment $M$ is cut equally 5 times parallel to its axis and again 3 times perpendicular to its axis, then the pole strength and magnetic moment of each piece are respectively

Options:
A $\frac{m}{20}, \frac{M}{4}$
B $\frac{m}{5}, \frac{M}{20}$
C $\frac{m}{6}, \frac{M}{24}$
D $\frac{m}{5}, \frac{M}{24}$

The correct answer is:
$\frac{m}{6}, \frac{M}{24}$

Solution:
Given, pole strength $=m$
and magnetic moment $=M$
When it is cut 5 times parallel to its axis, pole strength, $m^{\prime}=\frac{m}{6}$
When it is cut 3 times perpendicular to its axis, then length $2 l^{\prime}=\frac{2 l}{4}$
$\therefore$ pole strength, $m^{\prime}=\frac{m}{6}$
and magnetic moment,
$M^{\prime}=\left(\frac{m}{6}\right)\left(\frac{2 l}{4}\right)=\frac{M}{24}$

Topic: Physics, Magnetic Properties of Matter, JEE Main
If $A B C D E F$ is a regular hexagon and $\overrightarrow{A B}+\overrightarrow{A C}+\overrightarrow{A D}+\overrightarrow{A E}+\overrightarrow{A F}=\lambda \overrightarrow{A D}$, then $\lambda=$

Options:
A 2
B 3
C 4
D 6

The correct answer is:
3

Solution:
By triangle law, $\overrightarrow{A B}=\overrightarrow{A D}-\overrightarrow{B D}, \quad \overrightarrow{A C}=\overrightarrow{A D}-\overrightarrow{C D}$
Therefore, $\overrightarrow{A B}+\overrightarrow{A C}+\overrightarrow{A D}+\overrightarrow{A E}+\overrightarrow{A F}$
$=3 \overrightarrow{A D}+(\overrightarrow{A E}-\overrightarrow{B D})+(\overrightarrow{A F}-\overrightarrow{C D})=3 \overrightarrow{A D}$
Hence $\lambda=3$, [Since $\overrightarrow{A E}=\overrightarrow{B D}, \overrightarrow{A F}=\overrightarrow{C D}]$

Topic: Mathematics, Vector Algebra, JEE Main
If $A B C D$ is a parallelogram, $\overrightarrow{A B}=2 \mathbf{i}+4 \mathbf{j}-5 \mathbf{k}$ and $\overrightarrow{A D}=\mathbf{i}+2 \mathbf{j}+3 \mathbf{k}$, then the unit vector in the direction of $B D$ is

Options:
A $\frac{1}{\sqrt{69}}(\mathbf{i}+2 \mathbf{j}-8 \mathbf{k})$
B $\frac{1}{69}(\mathbf{i}+2 \mathbf{j}-8 \mathbf{k})$
C $\frac{1}{\sqrt{69}}(-\mathbf{i}-2 \mathbf{j}+8 \mathbf{k})$
D $\frac{1}{69}(-\mathbf{i}-2 \mathbf{j}+8 \mathbf{k})$

The correct answer is:
$\frac{1}{\sqrt{69}}(-\mathbf{i}-2 \mathbf{j}+8 \mathbf{k})$

Solution:
$\begin{aligned}
& \text {Since } \overrightarrow{A B}+\overrightarrow{B D}=\overrightarrow{A D} \Rightarrow \overrightarrow{B D}=\overrightarrow{A D}-\overrightarrow{A B} \\
& =(\mathbf{i}+2 \mathbf{j}+3 \mathbf{k})-(2 \mathbf{i}+4 \mathbf{j}-5 \mathbf{k})=-\mathbf{i}-2 \mathbf{j}+8 \mathbf{k}
\end{aligned}$
Hence unit vector in the direction of $\overrightarrow{B D}$ is
$\frac{-\mathbf{i}-2 \mathbf{j}+8 \mathbf{k}}{\left|-\mathbf{i}-2 \mathbf{j}+8 \mathbf{k}\right|}=\frac{-\mathbf{i}-2 \mathbf{j}+8 \mathbf{k}}{\sqrt{69}}$

Topic: Mathematics, Vector Algebra, JEE Main
If $A B C D$ is a parallelogram and the position vectors of $A, B, C$ are $\mathbf{i}+3 \mathbf{j}+5 \mathbf{k}, \mathbf{i}+\mathbf{j}+\mathbf{k}$ and $7 \mathbf{i}+7 \mathbf{j}+7 \mathbf{k}$, then the position vector of $D$ will be

Options:
A $7 \mathbf{i}+5 \mathbf{j}+3 \mathbf{k}$
B $7 \mathbf{i}+9 \mathbf{j}+11 \mathbf{k}$
C $9 \mathbf{i}+11 \mathbf{j}+13 \mathbf{k}$
D $8 \mathbf{i}+8 \mathbf{j}+8 \mathbf{k}$

The correct answer is:
$7 \mathbf{i}+9 \mathbf{j}+11 \mathbf{k}$

Solution:
Let position vector of $D$ is $x \mathbf{i}+y \mathbf{j}+\mathbf{z} \mathbf{k}$, then $\overrightarrow{A B}=\overrightarrow{D C} \Rightarrow-2 \mathbf{j}-4 \mathbf{k}=(7-x) \mathbf{i}+(7-y) \mathbf{j}+(7-z) \mathbf{k}$ $\Rightarrow x=7, y=9, z=11$.
Hence position vector of $D$ will be $7 \mathbf{i}+9 \mathbf{j}+11 \mathbf{k}$.

Topic: Mathematics, Vector Algebra, JEE Main
If A be a square matrix of order 3, such that A=5, then Adj-3A-2 is equal to

Options:

The correct answer is:
29.16

Solution:
Adj-3A2=-3A-22=36A-22=36×1A4
=36×125=72925=29.16

Topic: Mathematics, Determinants, JEE Main
If $A$ be an arithmetic mean between two numbers and $S$ be the sum of $n$ arithmetic means between the same numbers, then

Options:
A $S=n A$
B $A=n S$
C $A=S$
D None of these

The correct answer is:
$S=n A$

Solution:
Let the two quantities be $a$ and $b$ and let $A_1, A_2 \ldots \ldots ., A_n$ be the $n$ A.M.'s between them. Then a, $A_1, A_2 \ldots \ldots . . A_n, b$ are in A.P. and let $d$ be the common difference.
Now $T_{n+2}=b=a+(n+2-1) d \Rightarrow d=\frac{b-a}{n+1}$
Also $A_1+A_2+\ldots \ldots+A_n=S_{n+1}-a$
$=\frac{1}{2}(n+1)\left[2 a+(n+1-1) \frac{(b-a)}{(n+1)}\right]-a$
$=\frac{n}{2}[2 a+(b-a)]=\frac{n}{2}(a+b)=n\left(\frac{a+b}{2}\right)=n A$
Trick: Let $1,3,5,7,9$ is in A.P. In this series $A=5, n=3, S=15 \Rightarrow S=n A$

Topic: Mathematics, Sequences and Series, JEE Main
If a body looses half of its velocity on penetrating $3 \mathrm{~cm}$ in a wooden block, then how much will it penetrate more before coming to rest

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

The correct answer is:
1 cm

Solution:
For first condition
Initial velocity $=u$, Final velocity $=u / 2, s=3 \mathrm{~cm}$
From $v^2=u^2-2 a s_{\Rightarrow}\left(\frac{u}{2}\right)^2=u^2-2 a s \quad a=\frac{3 u^2}{8 s}$
Second condition
Initial velocity $=u / 2$, Final velocity $=0$
$\begin{aligned} & \text { From } v^2=u^2-2 a x_{\Rightarrow} 0=\frac{u^2}{4}-2 a x \\ & \quad x=\frac{u^2}{4 \times 2 a}=\frac{u^2 \times 8 s}{4 \times 2 \times 3 u^2}=s / 3=1 \mathrm{~cm}\end{aligned}$

Topic: Physics, Work Power Energy, JEE Main
If a body looses half of its velocity on penetrating 3 cm in a wooden block, then how much will it
penetrate more before coming to rest?

Options:
A
1 cm

B
2 cm

C
3 cm

D
4 cm

The correct answer is:
1 cm

Solution:
Let the initial velocity of the body be v. Hence the final velocity = v/2
Applying $v^2=u^2-2 a s \Rightarrow\left(\frac{v}{2}\right)^2=v^2-2 \cdot a \cdot 3 \Rightarrow a=v^2 / 8$
In IInd case when the body comes to rest, final velocity = 0, initial velocity = v/2
Again, $(0)^2=\left(\frac{v}{2}\right)^2-2 \cdot \frac{v^2}{8} \cdot s ;$ or $s=1 \mathrm{~cm}$
So the extra penetration will be $1 \mathrm{~cm}$

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2002
If a body moving in a circular path maintains constant speed of 10 m s-1 , then which of the following correctly describes the relation between acceleration and radius?

Options:
A
B
C
D

The correct answer is:

Solution:
V = constant
⟹ No tangential acceleration
⟹ Only centripetal acceleration
a=v2R
⟹ aR=constant
⟹ a∝1R

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2015 (10 Apr Online)
If a body moving in a circular path maintains constant speed of $10 \mathrm{~ms}^{-1}$, then which of the following correctly describes the relation between acceleration $(a)$ and radius $(r)$ ?

Options:
A

B

C

D

The correct answer is:

Solution:
Acceleration of a body moving over a circular path is
$\begin{aligned}
& a & =\frac{v^2}{r}=\frac{100}{r} \quad(\because v=10 \mathrm{~m} / \mathrm{s})
\end{aligned}$
$\Rightarrow \quad a \propto \frac{1}{r}$

Topic: Physics, Motion In One Dimension, TS EAMCET, TS EAMCET 2020 (10 Sep Shift 1)
If A=cosθ-sinθsinθcosθ , then the matrix A-50 when θ=π12, is equal to:

Options:
A 3212-1232
B 1232-3212
C 32-121232
D 12-323212

The correct answer is:
3212-1232

Solution:
A=cosθ-sinθsinθcosθ⇒A=cosθ-sinθsinθcosθ=cos2θ+sin2θ=1
And adjA=cosθsinθ-sinθcosθ
⇒A-1=1AadjA=cosθsinθ-sinθcosθ
⇒A-2=A-12=A-1A-1
=cosθsinθ-sinθcosθcosθsinθ-sinθcosθ
=cos2θ-sin2θ2sinθcosθ-2sinθcosθcos2θ-sin2θ
=cos2θsin2θ-sin2θcos2θ
It is visible that
A-1n=cosnθsinnθ-sinnθcosnθ
⇒A-50=A-150=cos50θsin50θ-sin50θcos50θ
Now at θ=π12,sin50θ=sin50π12=sinπ6=12
cos50θ=cosπ6=32
⇒A-50=3212-1232

Topic: Mathematics, Matrices, JEE Main
If A=cosθ-sinθsinθcosθ , then the matrix A-50 when θ=π12, is equal to:

Options:
A 3212-1232
B 1232-3212
C 32-121232
D 12-323212

The correct answer is:
3212-1232

Solution:
A=cosθ-sinθsinθcosθ⇒A=cosθ-sinθsinθcosθ=cos2θ+sin2θ=1
And adjA=cosθsinθ-sinθcosθ
⇒A-1=1AadjA=cosθsinθ-sinθcosθ
⇒A-2=A-12=A-1A-1
=cosθsinθ-sinθcosθcosθsinθ-sinθcosθ
=cos2θ-sin2θ2sinθcosθ-2sinθcosθcos2θ-sin2θ
=cos2θsin2θ-sin2θcos2θ
It is visible that
A-1n=cosnθsinnθ-sinnθcosnθ
⇒A-50=A-150=cos50θsin50θ-sin50θcos50θ
Now at θ=π12,sin50θ=sin50π12=sinπ6=12
cos50θ=cosπ6=32
⇒A-50=3212-1232

Topic: Mathematics, Matrices, JEE Main, JEE Main 2019 (09 Jan Shift 1)
If a carrier wave $\mathrm{c}(\mathrm{t})=\mathrm{A} \sin \omega_{\mathrm{c}} \mathrm{t}$ is amplitude modulated by a modulator signal $\mathrm{m}(\mathrm{t})=\mathrm{A}$ sin $\omega_{\mathrm{m}} \mathrm{t}$ then the equation of modulated signal $\left[\mathrm{C}_{\mathrm{m}}(\mathrm{t})\right]$ and its modulation index are respectively

Options:
A
$C_m(t)=A\left(1+\sin \omega_m t\right) \sin \omega_c t$ and 2

B
$C_m(t)=A\left(1+\sin \omega_m t\right) \sin \omega_m t$ and 1

C
$C_m(t)=A\left(1+\sin \omega_m t\right) \sin \omega_c t$ and 1

D
$C_m(t)=A\left(1+\sin \omega_c t\right) \sin \omega_m t$ and 2

The correct answer is:
$C_m(t)=A\left(1+\sin \omega_m t\right) \sin \omega_c t$ and 1

Solution:
Modulation index
$$
\mathrm{m}_{\mathrm{a}}=\frac{\mathrm{E}_{\mathrm{m}}}{\mathrm{E}_{\mathrm{c}}}=\frac{\mathrm{A}}{\mathrm{A}}=1
$$
Equation of modulated signal $\left[\mathrm{C}_{\mathrm{m}}(\mathrm{t})\right]$
$$
=\mathrm{E}_{(\mathrm{C})}+\mathrm{m}_{\mathrm{a}} \mathrm{E}_{(\mathrm{C})} \sin \omega_{\mathrm{m}}^{\mathrm{t}}
$$
$=\mathrm{A}\left(1+\sin \mathrm{w}_{\mathrm{C}} \mathrm{t}\right) \sin \omega_{\mathrm{m}} \mathrm{t}$
$\left(\right.$ As $\left.\mathrm{E}_{(\mathrm{C})}=\mathrm{A} \sin \omega_{\mathrm{c}} \mathrm{t}\right)$

Topic: Physics, Communication System, JEE Main, JEE Main 2013 (09 Apr Online)
If a carrier wave of 1000 kHz is used to carry the signal, the length of transmitting antenna will be at least equal to

Options:
A 3 m
B 75 m
C 600 m
D 3000 m

The correct answer is:
75 m

Solution:
Î» = 3 Ã 10 8 1 Ã 10 6 = 300 m
lmin = Î» 4 = 7 5 m

Topic: Physics, Communication System, JEE Main
If a cell receives a signal to contract, then its neighbours are also stimulated to contract. This is due to the presence of -

Options:
A tight junctions
B sarcoplasm
C desmosomes
D intercalated disc

The correct answer is:
intercalated disc

Solution:
Unique junctions called intercalated discs (communication junctions) link the cells together and define their borders. Intercalated discs are the major portal for cardiac cell-to-cell communication, which is required for coordinated muscle contraction and maintenance of circulation. Communication junctions at some fusion points allow the cells to contract as a unit, i.e., when one cell receives a signal to contract, its neighbours are also stimulated to contract.

Topic: Biology, Structural Organisation in Animals, NEET
If a charge particle enters perpendiculary in the uniform magnetic field, then :

Options:
A energy and momentum both remains constant
B energy remains constant but momentum changes
C both energy and momentum changes
D energy changes but momentum remains constant

The correct answer is:
energy remains constant but momentum changes

Solution:
When charge particle enters in a magnetic field perpendicularly, then the magnitude of its velocity remains constant, but direction changes, so its energy (scalar quantity) remains constant, but its momentum (vector quantity) changes.

Topic: Physics, Magnetic Effects of Current, JIPMER, JIPMER 2006
If a charge $q$ is placed at the centre of the line joining two equal charges $Q$ such that the system is in equilibrium then the value of $q$ is

Options:
A
$Q / 2$

B
$-Q / 2$

C
$\mathrm{Q} / 4$

D
$-Q / 4$

The correct answer is:
$-Q / 4$

Solution:
Apply the condition for equilibrium of each charge

Topic: Physics, Electrostatics, JEE Main
If a charge $q$ is placed at the centre of the line joining two equal charges $Q$ such that the system is in equilibrium then the value of $q$ is

Options:
A
$Q / 2$

B
$-Q / 2$

C
$\mathrm{Q} / 4$

D
$-Q / 4$

The correct answer is:
$-Q / 4$

Solution:
Apply the condition for equilibrium of each charge

Topic: Physics, Electrostatics, JEE Main, JEE Main 2002
If a chord of the parabola $y^2=4 x$ passes through its focus and makes an angle $\theta$ with the $X$-axis, then its length is

Options:
A $4 \cos ^2 \theta$
B $4 \sin ^2 \theta$
C $4 \operatorname{cosec}^2 \theta$
D $4 \sec ^2 \theta$

The correct answer is:
$4 \operatorname{cosec}^2 \theta$

Solution:
Let $P\left(t^2, 2 t\right)$ be the one end of a focal chord $P Q$ of the parabola $y^2=4 x$, the coordinate of the other end $Q$ are $\left(\frac{1}{t^2}, \frac{-2}{t}\right)$.
$\left[\because t t^{\prime}=-1\right]$
Given, the chord makes a $\theta$ with positive direction of $x$-axis
$\begin{aligned}
\Rightarrow \quad \tan \theta & =\frac{-2 / t-2 t}{1 / t^2-t^2}=\frac{-2}{(1 / t-t)} \\
& =(t-1 / t)=2 \cot \theta
\end{aligned}$
Now from Eq. (i)
$\begin{aligned}
P Q & =(t+1 / t)^2 \\
P Q & =(t-1 / t)^2+4 \\
& =(2 \cot \theta)^2+4 \\
& =4\left(\cot ^2 \theta+1\right) \\
& =4 \operatorname{cosec}^2 \theta
\end{aligned}$

Topic: Mathematics, Parabola, JEE Main
If a circle and ellipse is given by the equations \( x^{2}+y^{2}=16 \) and \( \frac{x^{2}}{25}+\frac{y^{2}}{4}=1 \), then the equation of common tangent between circle and ellipse in the first quadrant is

Options:
A \( y=-\frac{2 x}{\sqrt{3}}+4 \sqrt{\frac{7}{3}} \)
B \( y=\frac{2 x}{\sqrt{3}}+4 \sqrt{\frac{7}{3}} \)
C \( y=\frac{-2 x}{\sqrt{3}}-4 \sqrt{\frac{7}{3}} \)
D None of these

The correct answer is:
\( y=-\frac{2 x}{\sqrt{3}}+4 \sqrt{\frac{7}{3}} \)

Solution:
Let equation of tangent to x2+y2=16 and x225+y24=1 be simultaneously
y=mx+41+m2 ...(i)
and y=mx+25m2+4 ...(ii)
Since, equations (i) and (ii) are same tangent.
∴ 41+m2=25m2+4
⇒ 161+m2=25m2+4
⇒ 9m2=12⇒m=±23
As m<0
⇒ m=-23
So, the equation of the common tangent is,
y=-2x3+473.

Topic: Mathematics, Circle, JEE Main
If a circle $\mathrm{C}$ passing through $(4,0)$ touches the circle $x^2+y^2+4 x-6 y-12=0$ externally at a point $(1,-1)$, then the radius of the circle $\mathrm{C}$ is :

Options:
A
5

B
$2 \sqrt{5}$

C
4

D
$\sqrt{57}$

The correct answer is:
5

Solution:
Let A be the centre of given circle and $\mathrm{B}$ be the centre of circle C.
$$
\begin{aligned}
& x^2+y^2+4 x-6 y-12=0 \\
& \therefore A=(-2,3) \text { and } B=(g, f)
\end{aligned}
$$
Now, from the figure, we have
$$
\frac{-2+g}{2}=1 \text { and } \frac{3+f}{2}=-1
$$
(By mid point formula)
$$
\Rightarrow g=4 \text { and } f=-5
$$
Now, required radius
$$
=\mathrm{OB}=\sqrt{9+16}=\sqrt{25}=5
$$

Topic: Mathematics, Circle, JEE Main, JEE Main 2013 (22 Apr Online)
If a circle C passing through the point 4, 0 touches the circle x2+y2+4x-6y=12 externally at the point 1, -1, then the radius of C is:

Options:
A 4 units
B 5 units
C 25 units
D 57 units

The correct answer is:
5 units

Solution:
The equation of the tangent to the circle x2+y2+2gx+2fy+c=0 at a point x1, y1 is xx1+yy1+gx+x1+fy+y1+c=0.

Thus, the equation of the tangent to x2+y2+4x-6y-12=0at 1, -1 is

x1+y-1+2x+1-3y-1-12=0

⇒3x-4y-7=0.

Also, we know that the equation of a circle touching a circle x2+y2+2gx+2fy+c=0 and line lx+my+n=0 at x1, y1 is x-x12+y-y12+λlx+my+n=0

Hence, the required circle touching the line 3x-4y-7=0 at 1, -1 is

x-12+y+12+λ3x-4y-7=0

Given, this circle passes through 4, 0

⇒9+1+λ12-7=0

⇒λ=-2

Hence, the equation of the circle is x-12+y+12-23x-4y-7=0

⇒x2+y2-8x+10y+16=0

Also, we know that the radius of the circle x2+y2+2gx+2fy+c=0 is g2+f2-c

Hence, radius =-42+52-16=16+25-16=5 units.

Topic: Mathematics, Circle, JEE Main, JEE Main 2019 (10 Jan Shift 1)
If a circle C, whose radius is 3, touches externally the circle x2+y2+2x-4y-4=0 at the point 2,2, then the length of the intercept cut by this circle C on the x-axis is equal to

Options:
A 23
B 5
C 32
D 25

The correct answer is:
25

Solution:
The equation of a tangent to the given circle at 2,2 is

x2+y2+x+2−2y+2−4=0⇒3x−6=0⇒x=2

Family of the circle touching a given line at a given point

x-22+y-22+λx-2=0.
⇒x2+y2+λ−4x−4y+8−2λ=0
Radius=3=λ−422+4−8−2λ
⇒9=λ2−8λ+164−4+2λ
⇒36=λ2−8λ+16−16+8λ
⇒λ=±6

Hence, equation of the circle is x2+y2−10x−4y+20=0.

So, x-intercept=225−20=25.

Topic: Mathematics, Circle, JEE Main, JEE Main 2018 (16 Apr Online)
If a circle drawn by assuming a chord parallel to the transverse axis of hyperbola x2a2-y2b2=1 as diameter always passes through 2,0, then

Options:
A a=b=2
B b≠a
C b=a=1
D b=a=3

The correct answer is:
a=b=2

Solution:
Let, the end points of the diameter of circle are asecâ¡Î¸,btanâ¡Î¸ & -asecâ¡Î¸,btanâ¡Î¸
â Equation of circle is x-asecâ¡Î¸x+asecâ¡Î¸+y-btanâ¡Î¸2=0
âx2-a2sec2â¡Î¸+y2+b2tan2â¡Î¸-2btanâ¡Î¸y=0
Because 2,0 satisfies above equation,
hence, 4-a21+tan2â¡Î¸+0+b2tan2â¡Î¸=0
â4-a2+b2-a2tan2â¡Î¸=0
Which is always true if a=2 & b=2

Topic: Mathematics, Hyperbola, JEE Main