datavorous/entrance-exam-dataset huggingface dataset:
A uniformly charged solid sphere of radius R has potential V0 (measured with respect to ∞ ) on its surface. For this sphere the equipotential surfaces with potential 3V02,5V04,3V04 and V04 have radius R1 , R2, R3 and R4 respectively. Then

Note : This question had two option correct at the time of examination. Proper corrections are made in the question to avoid it.

Options:
A 2R>R4
B R1=0 and R2>R4-R3
C R1â 0 and R2-R1> (R4-R3)
D R1=0 and R2<R4-R3

The correct answer is:
R1=0 and R2<R4-R3

Solution:
Potential for uniformly charged solid sphere
v = 1 4 π ε 0 Q r outside i.e r > R
v = 1 4 π ε 0 Q R on the surface
v = 1 4 π ε 0 Q R 3 2 - 1 2 r 2 R 2 inside i.e. r < R
Clearly
3 v 0 2 , 5 v 0 4 are inside potentials ∵ > v 0
3 v 0 4 , v 0 4 are outside potentials ∵ < v 0
To get R 1 : 3 v 0 2 = 1 4 π ε 0 Q R 3 2 - 1 2 R 1 2 R 2
where, v 0 = 1 4 π ε 0 Q R
3 2 × 4 π ε 0 Q R = 1 4 π ε 0 Q R 3 2 - 1 2 R 1 2 R 2
3 2 = 3 2 - 1 2 R 1 2 R 2 ⇒ R 1 = 0
To get R 2 : 5 4 v 0 = 1 4 π ε 0 Q R 3 2 - 1 2 R 2 2 R 2
5 4 1 4 π ε 0 Q R = 1 4 π ε 0 Q R 3 2 - 1 2 R 2 2 R 2
5 4 = 3 2 - 1 2 R 2 2 R 2
1 2 R 2 2 R 2 = 1 4 ⇒ R 2 = R 2
To get R 3 : 3 v 0 4 = 1 4 π ε 0 Q R 3
3 4 1 4 π ε 0 Q R = 1 4 π ε 0 Q R 3
3 4 R = 1 R 3
R 3 = 4 3 R
To get R 4 : v 0 4 = 1 4 π ε 0 Q R 4
1 4 × 1 4 π ε 0 Q R = 1 4 π ε 0 Q R 4
R 4 = 4 R
R 4 - R 3 = 4 R - 4 R 3 = 8 R 3 > R 2
R 1 = 0 and R 2 < R 4 - R 3

Topic: Physics, Electrostatics, JEE Main, JEE Main 2015 (04 Apr)
A uniformly tapering conical wire is made from a material of Young's modulus Y and has a normal, unextended length L. The radii, at the upper and lower ends of this conical wire, have values R and 3R, respectively. The upper end of the wire is fixed to a rigid support and a mass M is suspended from its lower end. The equilibrium extended length, of this wire, would equal:

Options:
A L1+29MgπYR2
B L1+19MgπYR2
C L1+13MgπYR2
D L1+23MgπYR2

The correct answer is:
L1+13MgπYR2

Solution:
r=2RLx+R
For a small section
FA=Ydldx
⟹∫0LdL=∫0LMgdxπ2RLx×R2Y
∆L=MgπY-12RxL+R0L×L2R
=MgL3πR2Y
L f =L+ΔL=L[ 1+ mg 3π R 2 Y ]

Topic: Physics, Mechanical Properties of Solids, JEE Main
A uniformly tapering conical wire is made from a material of Young's modulus Y and has a normal, unextended length L. The radii, at the upper and lower ends of this conical wire, have values R and 3R, respectively. The upper end of the wire is fixed to a rigid support and a mass M is suspended from its lower end. The equilibrium extended length, of this wire, would equal:

Options:
A L1+29MgπYR2
B L1+19MgπYR2
C L1+13MgπYR2
D L1+23MgπYR2

The correct answer is:
L1+13MgπYR2

Solution:
r=2RLx+R
For a small section
FA=Ydldx
⟹∫0LdL=∫0LMgdxπ2RLx×R2Y
∆L=MgπY-12RxL+R0L×L2R
=MgL3πR2Y
L f =L+ΔL=L[ 1+ mg 3π R 2 Y ]

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2016 (09 Apr Online)
A uniformly tapering vessel is filled with a liquid of density $900 \mathrm{~kg} / \mathrm{m}^3$. The force that acts on the base of the vessel due to the liquid is $\left(g=10 \mathrm{~ms}^{-2}\right)$

Options:
A $3.6 \mathrm{~N}$
B $7.2 \mathrm{~N}$
C $9.0 \mathrm{~N}$
D $14.4 \mathrm{~N}$

The correct answer is:
$7.2 \mathrm{~N}$

Solution:
Force acting on the base
$F=P \times A=h d g A=0.4 \times 900 \times 10 \times 2 \times 10^{-3}=7.2 \mathrm{~N}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A uniformly wound solenoidal coil of self-inductance 1.8Ã10-4 H and resistance 6 Î© is broken up into two identical coils. These identical coils are then connected in parallel across a 12 V battery of negligible resistance. The time constant of the circuit is

Options:
A 3 Ã10-5 s
B 1.5 ×10-5 s
C 0.75 ×10-5 s
D 6 ×10-5 s

The correct answer is:
3 Ã10-5 s

Solution:
1Lp=1L+1L=2L ⇒LP=L2
Where L is inductance of each part,
= 1.8 ×10-42=0.9×10-4 H
∴ LP=L2= 0.9×10-42=0.45×10-4 H
Resistance of each part, r=62=3 Ω
Now, 1rP=13+13=23

Time constant of circuit,

=LPrP=0.45×10-41.5=3×10-5 s

Topic: Physics, Electromagnetic Induction, JEE Main
A unit negative charge with mass $M$ resides at the mid-point of the straight line of length $2 a$ adjoining two fixed charges of magnitude $+Q$ each. If it is given a very small displacement $x(x< < a)$ in a direction perpendicular to the straight line, it will

Options:
A come back to its original position and stay there
B execute oscillations with frequency
$$
\frac{1}{2 \pi} \sqrt{\frac{Q}{4 \pi \varepsilon_{0} M a^{3}}}
$$
C None of the above
D execute oscillations with frequency
$$
\frac{1}{2 \pi} \sqrt{\frac{Q}{4 \pi \varepsilon_{0} M a^{2}}}
$$

The correct answer is:
None of the above

Solution:
From figure the net force
$\begin{aligned} F_{\text {bet }} &=-F \cos \theta+(-F \cos \theta) \\ &=-2 F \cos \theta \\ &=-2 \times \frac{k Q \times 1}{\left(x^{2}+a^{2}\right)} \times \frac{x}{\sqrt{x^{2}+a^{2}}} \\ &=-\frac{2 k Q}{\left(x^{2}+a^{2}\right)^{3 / 2}} \cdot x \\ F_{\text {net }} &=-\left(\frac{2 k Q}{a^{3}}\right) \cdot x \end{aligned}$
Frequency of oscillation $=\frac{1}{2 \pi} \sqrt{\frac{2 k Q}{M a^{3}}}$
$=\frac{1}{2 \pi} \sqrt{\frac{2 \times \frac{1}{4 \pi \varepsilon_{0}} \cdot Q}{M a^{3}}}$
$=\frac{1}{2 \pi} \sqrt{\frac{Q}{2 \pi e_{0} M a^{3}}}$
$\Rightarrow\left(^{*}\right.$ None of the option matches)

Topic: Physics, Oscillations, JEE Main
A unit vector a makes an angle $\frac{\pi}{4}$ with z-axis. If $\mathbf{a}+\mathbf{i}+\mathbf{j}$ is a unit vector, then $\mathbf{a}$ is equal to

Options:
A $\frac{\mathbf{i}}{2}+\frac{\mathbf{j}}{2}+\frac{\mathbf{k}}{\sqrt{2}}$
B $\frac{\mathbf{i}}{2}+\frac{\mathbf{j}}{2}-\frac{\mathbf{k}}{\sqrt{2}}$
C $-\frac{\mathbf{i}}{2}-\frac{\mathbf{j}}{2}+\frac{\mathbf{k}}{\sqrt{2}}$
D None of these

The correct answer is:
$-\frac{\mathbf{i}}{2}-\frac{\mathbf{j}}{2}+\frac{\mathbf{k}}{\sqrt{2}}$

Solution:
Let $\mathrm{a}=l \hat{i}+m \hat{j}+n \hat{k}$, where $l^2+m^2+n^2=1$.
a makes an angle $\frac{\pi}{4}$ with $z-$ axis.
$\begin{aligned}
& \therefore n=\frac{1}{\sqrt{2}}, \quad I^2+m^2=\frac{1}{2} \\
& \therefore \mathrm{a}=l \mathrm{i}+m \mathrm{j}+\frac{\mathrm{k}}{\sqrt{2}} \\
& \mathrm{a}+\mathrm{i}+\mathrm{j}=(l+1) \mathrm{i}+(m+1) \mathrm{j}+\frac{\mathrm{k}}{\sqrt{2}} \\
& \text {Its magnitude is } 1 \text {, hence }(l+1)^2+(m+1)^2=\frac{1}{2}
\end{aligned}$
Its magnitude is 1, hence
$(l+1)^2+(m+1)^2=\frac{1}{2}$
From (i) and (ii),
$2 I m=\frac{1}{2} \Rightarrow I=m=-\frac{1}{2}$
Hence
$a=-\frac{\mathrm{i}}{2}-\frac{\mathrm{j}}{2}+\frac{\mathrm{k}}{\sqrt{2}}$

Topic: Mathematics, Vector Algebra, JEE Main
A unit vector in the xy-plane that makes an angle of 45° with the vector i˙^+j˙^ and an angle of 60° with the vector 3i˙^-4j˙^ is

Options:
A i˙^+j˙^2
B i˙^-j˙^2
C 2i˙^-j˙^2
D None of these

The correct answer is:
None of these

Solution:
Let the required unit vector be r→=ai˙^+bj˙^
Then, r→=1
⟹ a2+b2=1 ...(i)
Since, r→ makes an angle of 45° with i˙^+j˙^ and an angle of 60° with 3i˙^-4j˙^, therefore
cos⁡π4=r→⋅(i˙^+j˙^)r→|i˙^+j˙^|
and cos⁡π3=r→⋅3i˙^-4j˙^r→3i˙^-4j˙^
⟹12=a+b2
and12=3a-4b5
⟹a+b=1
and 3a-4b=52
⟹a=1314,b=114
∴ r→=1314i˙^+114j˙^

Topic: Mathematics, Vector Algebra, JEE Main
A unit vector which is coplanar to vector $\mathbf{i}+\mathbf{j}+2 \mathbf{k}$ and $\mathbf{i}+2 \mathbf{j}+\mathbf{k}$ and perpendicular to $\mathbf{i}+\mathbf{j}+\mathbf{k}$, is

Options:
A $\frac{\mathbf{i}-\mathbf{j}}{\sqrt{2}}$
B $\pm\left(\frac{\mathbf{j}-\mathbf{k}}{\sqrt{2}}\right)$
C $\frac{\mathbf{k}-\mathbf{i}}{\sqrt{2}}$
D $\frac{\mathbf{i}+\mathbf{j}+\mathbf{k}}{\sqrt{3}}$

The correct answer is:
$\pm\left(\frac{\mathbf{j}-\mathbf{k}}{\sqrt{2}}\right)$

Solution:
Let the vector be given as $a \mathbf{i}+b \mathbf{j}+c \mathbf{k}$. For this vector to be coplanar with $\mathbf{i}+\mathbf{j}+2 \mathbf{k}$ and $\mathbf{i}+2 \mathbf{j}+\mathbf{k}$, we will have $a \mathbf{i}+b \mathbf{j}+c \mathbf{k}=p(\mathbf{i}+\mathbf{j}+2 \mathbf{k})+r(\mathbf{i}+2 \mathbf{j}+\mathbf{k})$
This gives, $a=p+r$
$\begin{aligned}
& b=p+2 r \\
& c=2 p+r
\end{aligned}$
For the vector $a \mathbf{i}+b \mathbf{j}+c \mathbf{k}$ to be perpendicular to $\mathbf{i}+\mathbf{j}+\mathbf{k}$, we will have $(a \mathbf{i}+b \mathbf{j}+c \mathbf{k}) .(\mathbf{i}+\mathbf{j}+\mathbf{k})=0$
$\Rightarrow a+b+c=0$
Adding equation (i) to (iii), we get $4 p+4 r=a+b+c$
$\Rightarrow 4(p+r)=0 \Rightarrow p=-r$
Now with the help of (i), (ii) and (iii), we get
$a=0, \quad b=r, \quad c=p=-r$
Hence the required vector is $r(\mathbf{j}-\mathbf{k})$
To be its unit vector
$r^2+r^2=1 \Rightarrow r= \pm \frac{1}{\sqrt{2}}$
Hence the required unit vector is,
$\pm \frac{1}{\sqrt{2}}(\mathbf{j}-\mathbf{k})$
Trick : Check for option(a) $\frac{\mathbf{i}-\mathbf{j}}{\sqrt{2}}$ is a unit vector and perpendicular to $\mathbf{i}+\mathbf{j}+\mathbf{k}$ $\left|\begin{array}{ccc}\frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} & 0 \\ 1 & 1 & 2 \\ 1 & 2 & 1\end{array}\right|=-\frac{4}{\sqrt{2}} \neq 0$
So it is not coplanar with the given vector.
(b), $\pm\left(\frac{\mathbf{j}-\mathbf{k}}{\sqrt{2}}\right)$
Check for option
(b), $\pm\left(\frac{\mathbf{j}-\mathbf{k}}{\sqrt{2}}\right)$ is a unit vector and also perpendicular to $\mathbf{i}+\mathbf{j}+\mathbf{k},\left|\begin{array}{ccc}0 & \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} \\ 1 & 1 & 2 \\ 1 & 2 & 1\end{array}\right|=0$
So, it is also coplanar with the given vectors.

Topic: Mathematics, Vector Algebra, JEE Main
A unit vector which is perpendicular to $\mathbf{i + 2} \mathbf{j}-2 \mathbf{k}$ and $-\mathbf{i + 2 j + 2 \mathbf { k }}$ is

Options:
A $\frac{1}{\sqrt{5}}(2 \mathbf{i}-\mathbf{k})$
B $\frac{1}{\sqrt{5}}(-2 \mathbf{i}+\mathbf{k})$
C $\frac{1}{\sqrt{5}}(2 \mathbf{i}+\mathbf{j}+\mathbf{k})$
D $\frac{1}{\sqrt{5}}(2 \mathbf{i}+\mathbf{k})$

The correct answer is:
$\frac{1}{\sqrt{5}}(2 \mathbf{i}+\mathbf{k})$

Solution:
Unit vector is equal to $\frac{\mathbf{a} \times \mathbf{b}}{|\mathbf{a} \times \mathbf{b}|}=\frac{1}{\sqrt{5}}(2 \mathbf{i}+\mathbf{k})$.

Topic: Mathematics, Vector Algebra, JEE Main
A unit vector which is perpendicular to the vector $2 \hat{i}-\hat{j}+2 \hat{k}$ and is coplanar with the vectors $\hat{i}+\hat{j}-\hat{k}$ and $2 \hat{i}+2 \hat{j}-\hat{k}$ is

Options:
A
$\frac{2 \hat{j}+\hat{k}}{\sqrt{5}}$

B
$\frac{3 \hat{i}+2 \hat{j}-2 \hat{k}}{\sqrt{17}}$

C
$\frac{3 \hat{i}+2 \hat{j}+2 \hat{k}}{\sqrt{17}}$

D
$\frac{2 \hat{i}+2 \hat{j}-\hat{k}}{3}$

The correct answer is:
$\frac{2 \hat{i}+2 \hat{j}-\hat{k}}{3}$

Solution:
Let $x \hat{i}+y \hat{j}+z \hat{k}$ be the required unit vector.
Since $\hat{a}$ is perpendicular to $(2 \hat{i}-\hat{j}+2 \hat{k})$.
$$
\therefore \quad 2 x-y+2 z=0
$$
Since vector $x \hat{i}+y \hat{j}+z \hat{k}$ is coplanar with the vector $\hat{i}+\hat{j}-\hat{k}$ and $2 \hat{i}+2 \hat{j}-\hat{k}$.
$$
\begin{aligned}
\therefore \quad x \hat{i} & +y \hat{j}+z \hat{k} \\
& =p(\hat{i}+\hat{j}-\hat{k})+q(2 \hat{i}+2 \hat{j}-\hat{k}),
\end{aligned}
$$
where $p$ and $q$ are some scalars.
$$
\begin{aligned}
& \Rightarrow x \hat{i}+y \hat{j}+z \hat{k} \\
& =(p+2 q) \hat{i}+(p+2 q) \hat{j}-(p+q) \hat{k} \\
& \Rightarrow x=p+2 q, y=p+2 q, z=-p-q
\end{aligned}
$$
Now from equation (i),
$$
\begin{aligned}
& 2 p+4 q-p-2 q-2 p-2 q=0 \\
\Rightarrow & -p=0 \Rightarrow p=0 \\
\therefore & x=2 q, y=2 q, z=-q
\end{aligned}
$$
Since vector $x \hat{i}+y \hat{j}+z \hat{k}$ is a unit vector, therefore
$$
\begin{aligned}
& |x \hat{i}+y \hat{j}+z \hat{k}|=1 \\
\Rightarrow & \sqrt{x^2+y^2+z^2}=1 \\
\Rightarrow & x^2+y^2+z^2=1 \\
\Rightarrow & 4 q^2+4 q^2+q^2=1
\end{aligned}
$$
$$
\Rightarrow 9 q^2=1 \Rightarrow q=\pm \frac{1}{3}
$$
When $q=\frac{1}{3}$, then $x=\frac{2}{3}, y=\frac{2}{3}$,
$$
z=-\frac{1}{3}
$$
When $q=-\frac{1}{3}$, then $x=-\frac{2}{3}, y=-\frac{2}{3}$,
$$
z=\frac{1}{3}
$$
Here required unit vector is $\frac{2}{3} \hat{i}+\frac{2}{3} \hat{j}-\frac{1}{3} \hat{k}$ or $-\frac{2}{3} \hat{i}-\frac{2}{3} \hat{j}+\frac{1}{3} \hat{k}$.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2012 (12 May Online)
A U-tube having a horizontal arm of length 20 cm, has a uniform cross-sectional area 1 cm2. It is filled with water of volume 60 cc. The volume of a liquid of density 4 g cc-1 required to be poured in one arm of the U-tube so that no water is left in the horizontal arm of the tube is (take g=9.8 m s-2)

Options:
A 60 cc
B 45 cc
C 50 cc
D 35 cc

The correct answer is:
35 cc

Solution:
h ρ g = 6 0 × 1 × g
h = 1 5 cm
g=980 cm sec-2
Volume = 1 5 + 2 0 × 1 = 3 5 cc

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A U-tube in which the cross-sectional area of the limb on the left is one quarter, the limb on the right contains mercury (density $13.6 \mathrm{~g} / \mathrm{cm}^3$ ). The level of mercury in the narrow limb is at a distance of $36 \mathrm{~cm}$ from the upper end of the tube. What will be the rise in the level of mercury in the right limb if the left limb is filled to the top with water

Options:
A $1.2 \mathrm{~cm}$
B $2.35 \mathrm{~cm}$
C $0.56 \mathrm{~cm}$
D $0.8 \mathrm{~cm}$

The correct answer is:
$0.56 \mathrm{~cm}$

Solution:
If the rise of level in the right limb be $x \mathrm{~cm}$. the fall of level of mercury in left limb be $4 \times \mathrm{cm}$ because the area of cross section of right limb is 4 times as that of left limb.
$\therefore$ Level of water in left limb is $(36+4 x) \mathrm{cm}$.
Now equating pressure at interface of Fig and water $r\left(\right.$ at $\left.A^{\prime} B^{\prime}\right)$
$(36+4 x) \times 1 \times g=5 x \times 13.6 \times g$
By solving we get $x=0.56 \mathrm{~cm}$.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A value of θ∈0,π3, for which 1+cos2θsin2θ4 cos6θcos2θ1+sin2θ4 cos6θcos2θsin2θ1+4 cos6θ=0, is

Options:
A π9
B 7π24
C 7π36
D π18

The correct answer is:
π9

Solution:
R1→R1-R2
1-10cos2θ1+sin2θ4cos6θcos2θsin2θ1+4cos6θ=0
R2→R2-R3
1-1001-1cos2θsin2θ1+4cos6θ=0
⇒1+4cos6θ+sin2θ+1cos2θ=0
1+2cos6θ=0⇒cos6θ=-12
6θ=2π3⇒θ=π9

Topic: Mathematics, Determinants, JEE Main, JEE Main 2019 (12 Apr Shift 2)
A value of \( \alpha \) such that \( \int_{\alpha}^{\alpha+1} \frac{d x}{(x+\alpha)(x+\alpha+1)}=\log _{e}\left(\frac{9}{8}\right) \) is

Options:
A \( -\frac{1}{2} \)
B \( \frac{1}{2} \)
C \( -2 \)
D \( 2 \)

The correct answer is:
\( -2 \)

Solution:
I=∫αα+1dx(x+α)(x+α+1)=∫αα+11x+α –1x+α+1 dx
Using partial fractions
=lnx+α-lnx+α+1αα+1
=lnx+αx+α+1 αα+1
=ln2α+12α+2-ln2α2α+1
=ln2α+122α+12-1=ln⁡98
⇒(2α+1)2=9
⇒2α+1=±3
⇒α=1 or -2.

Topic: Mathematics, Definite Integration, JEE Main
A value of $C$ for which the conclusion of Mean Value Theorem holds for the function $f(x)=$ $\log _{\mathrm{e}} x$ on the interval $[1,3]$ is

Options:
A
$2 \log _3 e$

B
$\frac{1}{2} \log _{\mathrm{e}} 3$

C
$\log _3 e$

D
$\log _e 3$

The correct answer is:
$2 \log _3 e$

Solution:
Using mean value theorem
$\begin{aligned}
& f^{\prime}(c)=\frac{f(3)-f(1)}{3-1} \\
& \Rightarrow \frac{1}{c}=\frac{\log 3-\log 1}{2} \\
& \Rightarrow c=\frac{2}{\log _e 3}=2 \log _3 e.
\end{aligned}$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2007
A value of θ for which 2+3isinθ1-2isinθ is purely imaginary, is

Options:
A sin-134
B sin-113
C π3
D π6

The correct answer is:
sin-113

Solution:
Let, z=2 + 3isin θ1 - 2isin θ

Multiplying the denominator by its conjugate, we get,

z=2+3 isin θ 1+2 isin θ1+4 sin2 θ

=2-6 sin2 θ+i7sin θ1+4 sin2 θ

To make it purely imaginary. Its real part should be 0.

Hence, 2=6 sin2⁡θ

⇒sin⁡θ=±13

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2016 (03 Apr)
A value of α such that ∫αα+1dx(x+α)(x+α+1)=loge98 is

Options:
A -12
B 12
C -2
D 2

The correct answer is:
-2

Solution:
I=∫αα+1dx(x+α)(x+α+1)=∫αα+11x+α –1x+α+1 dx
Using partial fractions
=lnx+α-lnx+α+1αα+1
=lnx+αx+α+1 αα+1
=ln2α+12α+2-ln2α2α+1
=ln2α+122α+12-1=ln⁡98
⇒(2α+1)2=9
⇒2α+1=±3
⇒α=1 or -2.

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2019 (12 Apr Shift 2)
A value of $\tan ^{-1}\left(\sin \left(\cos ^{-1}\left(\sqrt{\frac{2}{3}}\right)\right)\right)$ is

Options:
A
$\frac{\pi}{4}$

B
$\frac{\pi}{2}$

C
$\frac{\pi}{3}$

D
$\frac{\pi}{6}$

The correct answer is:
$\frac{\pi}{6}$

Solution:
Consider $\tan ^{-1}\left[\sin \left(\cos ^{-1} \sqrt{\frac{2}{3}}\right)\right]$
$$
\begin{aligned}
& \text { Let } \cos ^{-1} \sqrt{\frac{2}{3}}=\theta \Rightarrow \cos \theta=\sqrt{\frac{2}{3}} \\
& \Rightarrow \quad \sin \theta=\sqrt{1-\cos ^2 \theta}=\sqrt{1-\frac{2}{3}}=\sqrt{\frac{1}{3}} \\
& \therefore \quad \tan ^{-1}\left[\sin \left(\cos ^{-1} \sqrt{\frac{2}{3}}\right)\right]=\tan ^{-1}[\sin \theta] \\
& =\tan ^{-1}\left[\sqrt{\frac{1}{3}}\right]=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right) \\
& =\frac{\pi}{6}
\end{aligned}
$$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2012 (19 May Online)
A value of $x$ for which $\sin \left(\cot ^{-1}(1+x)\right)=\cos$ $\left(\tan ^{-1} x\right)$, is :

Options:
A
$-\frac{1}{2}$

B
1

C
0

D
$\frac{1}{2}$

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

Solution:
$\sin \left(\cot ^{-1}(1+x)\right)=\cos \left(\tan ^{-1} x\right)$
$\begin{aligned} & \Rightarrow \operatorname{cosec}^2\left(\cot ^{-1}(1+x)\right)=\sec ^2\left(\tan ^{-1} x\right) \\ & \Rightarrow 1+\left[\cot \left(\cot ^{-1}(1+x)\right)\right]^2 \\ & \Rightarrow=1+\left[\sec \left(\tan ^{-1} x\right)\right]^2 \\ & \Rightarrow(1+x)^2=x^2 \Rightarrow x=-\frac{1}{2}\end{aligned}$

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2013 (09 Apr Online)
A value of x satisfying the equation sin⁡cot-1⁡1+x=cos⁡tan-1⁡x, is:

Options:
A -12
B 0
C -1
D 12

The correct answer is:
-12

Solution:
sincot-1(1+x)=costan-1x

Let, cotλ=1+x

tanβ=x

⇒ sinλ=cosβ

⇒ 1x2+2x+2=111+x2

⇒ x2+2x+2=x2+1

⇒ x=-1/2

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2017 (09 Apr Online)
A van der Waal's gas obeys the equation of state $\left(\mathrm{P}+\frac{\mathrm{n}^{2} \mathrm{a}}{\mathrm{V}^{2}}\right)(\mathrm{V}-\mathrm{nb})=\mathrm{nRT}$. Its internal energy is given by $\mathrm{U}=\mathrm{CT}-\frac{\mathrm{n}^{2} \mathrm{a}}{\mathrm{V}} .$ The equation of a quasistatic ad iabat for this gas is given by-

Options:
A $\mathrm{T}^{\mathrm{ChR}} \mathrm{V}=$ constant
B $\mathrm{T}^{(\mathrm{C}+\mathrm{nR}) \mathrm{nn} \mathrm{R}} \mathrm{V}=$ constant
C $\mathrm{T}^{\mathrm{ChR}}(\mathrm{V}-\mathrm{nb})=$ constant
D $\mathrm{P}^{(\mathrm{C}+\mathrm{nR}) \mathrm{ln} \mathrm{R}}(\mathrm{V}-\mathrm{nb})=$ constant

The correct answer is:
$\mathrm{T}^{\mathrm{ChR}}(\mathrm{V}-\mathrm{nb})=$ constant

Solution:
For adiabatic process
$\mathrm{dQ}=0$ and $-\mathrm{dU}=\mathrm{dW} \Rightarrow-\mathrm{nC}_{\mathrm{V}} \Delta \mathrm{T}=\mathrm{P} \Delta \mathrm{V}$ or $-\mathrm{nC}_{\mathrm{V}} \mathrm{dT}=\mathrm{PdV}$
when change is very small
now given $\mathrm{U}=\mathrm{CT}-\frac{\mathrm{n}^{2} \mathrm{a}}{\mathrm{V}} \quad \therefore \mathrm{d} \mathrm{U}=\mathrm{CdT}+\frac{\mathrm{n}^{2} \mathrm{a}}{\mathrm{V}^{2}} \mathrm{~d} \mathrm{~V}$
put this value of $\mathrm{dU} \mathrm{in}-\mathrm{dU}=\mathrm{d} \mathrm{W}$
$$
\therefore-\left(\mathrm{CdT}+\frac{\mathrm{n}^{2} \mathrm{a}}{\mathrm{V}^{2}} \mathrm{~d} \mathrm{~V}\right)=\mathrm{PdV}
$$
also $\mathrm{P}=\left(\frac{\mathrm{nRT}}{\mathrm{V}-\mathrm{nb}}\right)-\frac{\mathrm{n}^{2} \mathrm{a}}{\mathrm{V}^{2}}$ replace it in (1)
$$
-\left(C d T+\frac{n^{2} a}{V^{2}} d V\right)=\left(\left(\frac{n R T}{V-n b}\right)-\frac{n^{2} a}{V^{2}}\right) d V
$$
$$
\begin{array}{l}
\therefore-\mathrm{CdT}=\left(\frac{\mathrm{nRT}}{\mathrm{V}-\mathrm{nb}}\right) \mathrm{dV} \\
\therefore-\frac{\mathrm{C}}{\mathrm{nR}} \frac{\mathrm{dT}}{\mathrm{T}}=\frac{\mathrm{dV}}{\mathrm{V}-\mathrm{nb}}
\end{array}
$$
Integrating we get
$-\ell \mathrm{n} \mathrm{T}^{\mathrm{C} \mathrm{n} \mathrm{R}}=\ell \mathrm{n}(\mathrm{V}-\mathrm{nb})+\mathrm{k} \quad(\mathrm{k} \rightarrow$ constant of integration)
$\therefore \ln \left(\mathrm{T}^{\mathrm{C} / \mathrm{nR}}\right)(\mathrm{V}-\mathrm{nb})=-\mathrm{k}$
or $\left(\mathrm{T}^{\mathrm{C}^{\mathrm{Tn} \mathrm{R}}}\right)(\mathrm{V}-\mathrm{nb})=$ constant

Topic: Physics, Kinetic Theory of Gases, JEE Main
A variable circle passes through the fixed point $A(p, q)$ and touches $x$-axis. The locus of the other end of the diameter through $\mathrm{A}$ is

Options:
A
$(x-p)^2=4 q y$

B
$(x-q)^2=4 p y$

C
$(y-p)^2=4 q x$

D
$(y-q)^2=4 p x$

The correct answer is:
$(x-p)^2=4 q y$

Solution:
Let the other end of diameter is $(\mathrm{h}, \mathrm{k})$ then equation of circle is
$$
(x-h)(x-p)+(y-k)(y-q)=0
$$
Put $y=0$, since $x$-axis touches the circle
$$
\begin{aligned}
& \Rightarrow x^2-(h+p) x+(h p+k q)=0 \Rightarrow(h+p)^2=4(h p+k q) \quad(D=0) \\
& \Rightarrow(x-p)^2=4 q y .
\end{aligned}
$$

Topic: Mathematics, Parabola, JEE Main, JEE Main 2004
A variable line passes through the fixed point $(\alpha, \beta) .$ The locus of the foot of the perpendicular from the origin on the line is

Options:
A $x^{2}+y^{2}-\alpha x-\beta y=0$
B $x^{2}-y^{2}+2 a x+2 \beta y=0$
C $\alpha x+\beta y \pm \sqrt{\left(\alpha^{2}+\beta^{2}\right)}=0$
D $\frac{x^{2}}{\alpha^{2}}+\frac{y^{2}}{\beta^{2}}=1$

The correct answer is:
$x^{2}+y^{2}-\alpha x-\beta y=0$

Solution:
Let $(\alpha, \beta)$ be the given point, let $Q(x, y)$ be the foot of the perpendicular, and let $O$ be the origin. The line can have any direction
$$
\angle P Q O=90^{\circ}
$$
Point $Q$ lies on the circle having diameter $O P$.
The locus of point $Q$.
$$
(x-0)(x-\alpha)+(y-0)(y-\beta)=0
$$
$\Rightarrow \quad x^{2}-x \alpha+y^{2}-y \beta=0$
$$
\Rightarrow \quad x^{2}+y^{2}-\alpha x-\beta y=0
$$

Topic: Mathematics, Straight Lines, JEE Main
A variable line through the point 65,65 cuts the coordinate axes at the points A and B respectively. If the point P divides AB internally in the ratio 2:1, then the equation of the locus of P is

Options:
A 5xy=2x+y
B 5xy=22x+y
C 5xy=x+2y
D 5xy=2x+2y

The correct answer is:
5xy=22x+y

Solution:
Let, A=a,0 and B=0,b, then the equation of AB is xa+yb=1 which passes through 65,65

⇒1a+1b=56
Let, P=h,k, then

h=20+1a3=a3 & k=2b+103=2b3
⇒a=3h & b=3k2
⇒13h+23k=56
⇒ locus of P is 2y+2x=5xy

Topic: Mathematics, Straight Lines, JEE Main
A variable plane passes through a fixed point ( 3 , $2,1)$ and meets $x, y$ and $z$ axes at $A, B$ and $C$ respectively. A plane is drawn parallel to $y z$ - plane through $A$, a second plane is drawn parallel $z x$ plane through $B$ and a third plane is drawn parallel to $x y$ - plane through $C$. Then the locus of the point of intersection of these three planes, is

Options:
A
$(x+y+z=6)$

B
$\frac{x}{3}+\frac{y}{2}+\frac{z}{1}=1$

C
$\frac{3}{x}+\frac{2}{y}+\frac{1}{z}=1$

D
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{11}{6}$

The correct answer is:
$\frac{3}{x}+\frac{2}{y}+\frac{1}{z}=1$

Solution:
If $a, b, \mathrm{c}$ are the intercepts of the variable plane on the $x, y, z$ axes respectively, then the equation of the plane is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ And the point of intersection of the planes parallel to the $x y, y z$ and $z x$ planes is ( $a, b$, c).
As the point $(3,2,1)$ lies on the variable plane, so $\frac{3}{a}+\frac{2}{b}+\frac{1}{c}=1$
Therefore, the required locus is $\frac{3}{x}+\frac{2}{y}+\frac{1}{z}=1$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
A variable plane passes through a fixed point 3, 2, 1 and meets x, y and z-axes at A, B & C respectively. A plane is drawn parallel to the yz– plane through A, a second plane is drawn parallel to the zx- plane through B and a third plane is drawn parallel to the xy- plane through C. Then the locus of the point of intersection of these three planes, is

Options:
A 3x+2y+1z=1
B 1x+1y+1z= 116
C x+y+z=6
D x3+y2+z1= 1

The correct answer is:
3x+2y+1z=1

Solution:
Let plane be xa+yb+zc=1

It passes through 3, 2, 1.

∴3a+2b+1c=1

Now, A(a, 0, 0), B(0, b, 0), C(0, 0, c)

∴ Locus of point of intersection of planes x=a, y=b & z=c is 3x+2y+1z=1

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2018 (15 Apr)