datavorous/entrance-exam-dataset huggingface dataset:
The plane $2 x-y+z=4$ intersects the line segment joining the points $\mathrm{A}(\mathrm{a},-2,4)$ and $\mathrm{B}(2, \mathrm{~b},-3)$ at the point $\mathrm{C}$ in the ratio $2: 1$ and the distance of the point $\mathrm{C}$ from the origin is $\sqrt{5}$. If $a b<0$ and $\mathrm{P}$ is the point $(\mathrm{a}-\mathrm{b}, \mathrm{b}, 2 \mathrm{~b}-\mathrm{a})$ then $\mathrm{CP}^2$ is equal to:

Options:
A 173
B 163
C 733
D 973

The correct answer is:
173

Solution:
Given points are:

A(a,-2,4) and B(2, b,-3)

Point C divides the line segment joining points A and B in the ratio 2:1, so

C≡a+43,2b-23,-23

Also, C lies on 2x-y+2=4, therefore

2a+83-2b-23-23=4

⇒2a+8-2b+2-2=12

⇒a-b=2 ....1

Also, OC=5

⇒a+432+2b-232+49=5

⇒a2+8a+16+4b2+4-8b=45

⇒a2+8a+4b2-8b=25 ...2

Solving 1 and (2), we have

(b+6)2+(2b-2)2=41

⇒5b2+4b-1=0

⇒b=-1 or 15

⇒a=1 or 115

But ab<0⇒a,b≡1,-1

C≡53,-43,-23,P≡(2,-1,-3)

CP2=19+19+499=519=173

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (29 Jan Shift 2)
The plane 4x+7y+4z+81=0 is rotated through a right angle about its line of intersection with the plane 5x+3y+10z=25. If the equation of the plane in its new position is x-4y+6z=K, then the value of K is

Options:
A 106
B -89
C 73
D 37

The correct answer is:
106

Solution:
The equation of the plane through the line of intersection of the planes is
4x+7y+4z+81+λ5x+3y+10z-25=0
4+5λx+7+3λy+4+10λz+81-25λ=0
Which is parallel to x-4y+6z=K
⇒4+5λ1=7+3λ-4=4+10λ6=81-25λ-K
So, -16-20λ=7+3λ
23λ=-23
λ=-1
and 81-25λ-K=4+10λ6
-K=81-25λ64+10λ=-106
Hence, K=106

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The plane containing the line $\frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}-2}{2}=\frac{\mathrm{z}-3}{3}$ and parallel to the line $\frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{1}=\frac{\mathrm{z}}{4}$ passes through the point:

Options:
A
$(1,-2,5)$

B
$(1,0,5)$

C
$(0,3,-5)$

D
$(-1,-3,0)$

The correct answer is:
$(1,0,5)$

Solution:
Equation of the plane containing the line $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$ is
$$
a(x-1)+b(y-2)+c(z-3)=0
$$
where $a .1+b .2+c .3=0$
i.e., $\quad a+2 b+3 c=0$
(ii)
Since the plane (i) parallel to the line
$$
\begin{array}{ll}
& \frac{x}{1}=\frac{y}{1}=\frac{z}{4} \\
\therefore \quad & a .1+b .1+c .4=0 \\
\text { i.e., } \quad & \mathrm{a}+\mathrm{b}+4 \mathrm{c}=0
\end{array}
$$
(iii)
From (ii) and (iii),
$$
\begin{gathered}
\frac{a}{8-3}=\frac{b}{3-4}=\frac{c}{1-2}=k \quad \text { (let) } \\
\therefore \quad a=5 k, b=-k, c=-k
\end{gathered}
$$
On putting the value of $a, b$ and $c$ in equation (i),
$$
\begin{aligned}
&5(x-1)-(y-2)-(z-3)=0 \\
&\Rightarrow 5 x-y-z=0
\end{aligned}
$$
when $x=1, y=0$ and $z=5$; then
L.H.S. of equation (iv) $=5 x-y-2$
$$
\begin{aligned}
&=5 \times 1-0-5 \\
&=0 \\
&=\text { R.H.S. of }
\end{aligned}
$$
equation (iv)
Hence coordinates of the point $(1,0,5)$ satisfy the equation plane represented by equations (iv),
Therefore the plane passes through the point $(1,0,5)$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2014 (11 Apr Online)
The plane containing the line $\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z-1}{3}$ and also containing its projection on the plane $2 \mathrm{x}+3 \mathrm{y}-\mathrm{z}=5$, contains which one of the following points?

Options:
A (2,2,0)
B (-2,2,2)
C (0,-2,2)
D (2,0,-2)

The correct answer is:
(2,0,-2)

Solution:
Let normal to the required plane is $\vec{n}$

$\Rightarrow \vec{n}$ is perpendicular to both vector $2 \hat{i}-\hat{j}+3 \hat{k}$ and

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

$\Rightarrow \quad \vec{n}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 3 \\ 2 & 3 & -1\end{array}\right|=-8 \hat{i}+8 \hat{j}+8 \hat{k}$

$\Rightarrow$ Equation of the required plane is $\Rightarrow \quad(x-3)(-8)+(y+2) \times 8+(z-1) \times 8=0$

$\Rightarrow \quad(x-3)(-1)+(y+2) \times 1+(z-1) \times 1=0$

$\Rightarrow x-3-y-2-z+1=0$

$\because \quad x-y-z=4$ passes through (2,0,-2)

$\therefore \quad$ plane contains (2,0,-2)

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2019 (11 Jan Shift 1)
The plane containing the line x-32=y-b4=z-33 passes through the points a,1,2,2,1,4,2,3,5, then 3a+5b is equal to

Options:
A 4
B 16
C -16
D -4

The correct answer is:
-16

Solution:
If the plane contains the given line then it must pass through 3,b,3 and should be parallel to 2i^+4j^+3k^
Also, the plane passes through the points a,1,22,1,42,3,5.
Equation of the plane is
x-2y-1z-4021243=0
âx-22-y-1-2+z-4-4=0
âx-2+y-1-2z+8=0
âx+y-2z+5=0
â´ points 3,b,3a,1,2 satisfy the equation of the plane, if
3+b-6+5=0âb=-2
a+1-4+5=0âa=-2

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The plane of a dip circle is set in the geographic meridian and the apparent dip is δ1. It is then set in a vertical plane perpendicular to the geographic meridian. The apparent dip angle is δ2. The declination θ at the place is

Options:
A tan−1⁡tan⁡δ1tan⁡δ2
B tan−1⁡tan⁡δ1+tan⁡δ2
C tan−1⁡tan⁡δ1tan⁡δ2
D tan−1⁡tan⁡δ1−tan⁡δ2

The correct answer is:
tan−1⁡tan⁡δ1tan⁡δ2

Solution:
Apparent dip angle if the plane of a dip circle is set in the geographic meridian, tanδ1=tanϕcosα⇒cosα=tanϕtanδ1 ⋯1

Apparent dip angle if the vertical plane perpendicular to the geographic meridian,

tanδ2=tanϕsinα⇒sinα=tanϕtanδ2 ⋯2

From equations (1) and (2) we have

⇒sinαcosα=tanα=tanδ1tanδ2⇒α=tan-1tanδ1tanδ2

Topic: Physics, Magnetic Properties of Matter, JEE Main
The plane passing through the point (4, -1, 2) and parallel to the lines x+23=y-2-1=z+12 and x-21=y-32=z-43 also passes through the point

Options:
A 1, 1, -1
B -1, - 1, -1
C -1, -1, 1
D 1,Â 1,Â 1

The correct answer is:
1, 1, 1

Solution:
Let n→ be the normal vector to the plane passing through (4, -1, 2) and parallel to the given lines, then we know that n→ is perpendicular to both the lines.

Also, we know that a vector perpendicular to two vectors lie along the cross product of the two vectors.

Hence, n→=i^j^k^3-12123

⇒n→=-7i^-7j^+7k^

Thus, the direction ratios of the normal to the plane are <-7, -7, 7>.

The equation of a plane having direction ratios of the normal to the plane as a, b, c and passing through a point x1, y1, z1 is ax-x1+by-y1+cz-z1=0

Hence, the equation of the plane having direction ratios the normal to the plane as <-7, -7, 7> and passing through the point4, -1, 2 is

-7x-4-7y+1+7z-2=0

⇒-7x-7y+7z=-7

⇒x+y-z=1

From the given options only the point 1, 1, 1 satisfy the equation of the plane.

Hence, the plane passes through the point 1, 1, 1.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2019 (10 Jan Shift 1)
The plane, passing through the points (0, –1, 2) and (–1, 2, 1) and parallel to the line passing through (5, 1, –7) and (1, –1, –1), also passes through the point

Options:
A -2, 5, 0
B 1, -2, 1
C 2, 0, 1
D 0, 5, -2

The correct answer is:
-2, 5, 0

Solution:
Let A0,-1,2 & B-1,2,1 and given plane is parallel to line passing through 5,1,-7 & 1,-1,-1

So, normal vector to the plane is given by,

n→=i^j^k^-1-02+11-21-5-1-1-1+7

⇒n→=i^j^k^-13-1-4-26=i^16-j^-10+k^14

⇒n→=16i^+10j^+14k^

∴Equation of plane

16(x – 0) + 10(y + 1) + 14 (z – 2) = 0

⇒ 8x+5y+7z=9

Now satisfying the points from option in the plane we get,

-2,5,0 only point which satisfy the plane.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (13 Apr Shift 2)
The plane passing through the points 1, 2, 1, 2, 1, 2 and parallel to the line, 2x=3y, z=1 also passes through the point

Options:
A (0,6,-2)
B (-2,0,1)
C 0, -6, 2
D 2, 0,- 1

The correct answer is:
(-2,0,1)

Solution:
Equation of plane passes through 2, 1, 2 is ax – 2+by – 1+cz – 2= 0

It also passes through 1, 2, 1

∴–a + b – c = 0 ⇒ a – b + c = 0 ...(1)

Given line can be written as x3=y2=z-10

Now this line is parallel to 1

⇒3a + 2b + c0 = 0 ...2

Solve 1 & 2

a0-2=b3-0=c2+3

a-2=b3=c2+3

a2=b-3=c-5

Plane is 2x – 4 – 3y + 3 – 5z + 10 = 0

2x – 3y – 5z + 9 = 0

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2020 (02 Sep Shift 1)
The plane through the intersection of the planes x+y+z=1 and 2x+3y-z+4=0 and parallel to y- axis also passes through the point

Options:
A 3,3,-1
B -3,1,1
C 3,2,1
D -3,0,-1

The correct answer is:
3,2,1

Solution:
Let required plane is (using concept of family of plane)

x+y+z-1+t2x+3y-z+4=0

⇒1+2tx+1+3ty+1-tz-1+4t=0

Again DR of y- axis is 0,1,0

Since plane is parallel to y- axis

Hence 1+2t×0+1+3t×1+1-t×0=0

⇒t=-13

∴ required plane is 13x+43z-73=0

⇒x+4z-7=0

Which satisfy 3,2,1

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2019 (09 Jan Shift 1)
The plane which bisects the line joining the points 4, -2, 3 and 2, 4, -1 at right angles also passes through the point :

Options:
A 0, −1, 1
B 4, 0, -1
C 4, 0, 1
D (0, 1, –1)

The correct answer is:
4, 0, -1

Solution:
Mid point P≡3, 1, 1

Normal of plane is along the line AB.

Direction ratio's of normal is <4-2, -2 -4, 3--1>

=<2, -6, 4>

=<1, -3, 2>

Equation of plane is 1x-3-3y-1+2z-1=0

⇒x-3y+2z-2=0

Only option B satisfies and is the answer.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2020 (03 Sep Shift 2)
The plane which bisects the line segment joining the points -3, -3, 4 and 3, 7, 6 at right angles, passes through which one of the following points?

Options:
A 2, 1, 3
B 4, 1, -2
C 4, -1, 7
D -2, 3, 5

The correct answer is:
4, 1, -2

Solution:
The midpoint of line joining the given points is 0, 2, 5 and it lies on the required plane.
Normal to the plane, the line joining the given points, has direction ratio <6, 10, 2>≡<3, 5, 1>
⇒ equation of plane is 3x-0+5y-2+z-5=0
⇒3x+5y+z=15
Hence plane passes through point 4, 1, -2

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The plane which bisects the line segment joining the points -3, -3, 4 and 3, 7, 6 at right angles, passes through which one of the following points?

Options:
A 2, 1, 3
B 4, 1, -2
C 4, -1, 7
D -2, 3, 5

The correct answer is:
4, 1, -2

Solution:
The midpoint of line joining the given points is 0, 2, 5 and it lies on the required plane.
Normal to the plane, the line joining the given points, has direction ratio <6, 10, 2>≡<3, 5, 1>
⇒ equation of plane is 3x-0+5y-2+z-5=0
⇒3x+5y+z=15
Hence plane passes through point 4, 1, -2

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2019 (10 Jan Shift 2)
The plane x1+y2+z3=1 intersect x-axis, y-axis and z-axis at A, B and C respectively. If the distance between the origin and the centroid of âABC is k1 units and the volume of the tetrahedron OABC is k2 cubic units, then the value of k12k2 is equal to (where O is the origin)

Options:
A 21
B 149
C 63
D 143

The correct answer is:
149

Solution:
x1+y2+z3=1 cuts x -axis, y -axis, z -axis at A1,0,0,B0,2,0,C0,0,3 respectively
Centroid of âABC is 13,23,1
Distance between centroid and origin is k1=19+49+1
âk1=143 units
Volume of tetrahedron OABC=16ÃOAÃOBÃOC
k2=16Ã1Ã2Ã3=1 cubic unit
k12k2=1491=149

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The plane $x+2 y-z=4$ cuts the sphere $x^2+y^2+z^2-x+z-2=0$ in a circle of radius

Options:
A
3

B
1

C
2

D
$\sqrt{2}$

The correct answer is:
1

Solution:
Perpendicular distance of centre $\left(\frac{1}{2}, 0,-\frac{1}{2}\right)$ from $x+2 y-2=4$
$$
\begin{aligned}
& \frac{\left|\frac{1}{2}+\frac{1}{2}-4\right|}{\sqrt{6}}=\sqrt{\frac{3}{2}} \\
& \text { radius }=\sqrt{\frac{5}{2}-\frac{3}{2}}=1 .
\end{aligned}
$$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2005
The plane \( x+y+z=5 \) is rotated by an angle \( 90^{\circ} \) along the line of intersection with the plane \( x-y-z=2 . \) If the equation of plane in new position is \( a x+b y+c z+d=0 \), where \( a, b, c, d \in I \) then least value of \( |a+b+c+d| \) is equal to

Options:
A \( 10 \)
B \( 9 \)
C \( 4 \)
D \( 11 \)

The correct answer is:
\( 11 \)

Solution:
The required plane is (x+y+z-5)+λ(x-y-z- 2)=0
⇒(1+λ)x+(1-λ)y+(1-λ)z+(-5-2λ)=0
As, (1+λ)+(1-λ)+(1-λ)=0⇒λ=3
∴ The required plane is 4x-2y-2z-11=0⇒|a+b+c+d|=|4+(-2)+(-2)+(-11)|=|-11|=11

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The planet Mars has two moons, if one of them has a period 7 hours, 30 minutes and an orbital radius of 9.0×103 km. Find the mass of Mars.
Given 4π2G=6×1011 N-1 m-2 kg2

Options:
A 5.96×1019 kg
B 3.25×1021 kg
C 7.02×1025 kg
D 6.00×1023 kg

The correct answer is:
6.00×1023 kg

Solution:
Option D is correct.

T2=4π2GM·r3

M=4π2G·r3 T2

By putting values

M=6×1023

Topic: Physics, Gravitation, JEE Main, JEE Main 2021 (27 Jul Shift 2)
The plant body of Funaria is:

Options:
A Sporophyte
B Gametophyte
C Predominantly sporophyte with independent gametophyte
D Predominantly gametophyte with dependent sporophyte.

The correct answer is:
Predominantly gametophyte with dependent sporophyte.

Solution:
Funaria is a bryophyta in which the main plant body is haploid independent gametophyte, and sporophyte is dependent on the gametophyte for nutrition, partially or fully.

Topic: Biology, Plant Kingdom, NEET
The plant hormone that induces cell proliferation, plant growth and development, triggers callus differentiation, embryogenesis, maintenance of root and shoot meristems, keeps the cut flowers and vegetable fresh is

Options:
A Gibberellins
B Cytokinins
C Auxins
D Ethylene

The correct answer is:
Cytokinins

Solution:
Cytokinins induce cell proliferation, trigger callus differentiation to shoot when applied with auxins, plant growth and development, embryogenesis, maintenance of root and shoot meristems. Cytokinins are sprayed to increase the shelf life of vegetables and cut flowers. The main reason behind this is that the cytokinin increases the retention of chlorophyll and delayed senescence in leaves.

Topic: Biology, Plant - Growth and Development, NEET
The plasma membrane consists mainly of

Options:
A phospholipids embedded in a protein bilayer
B proteins embedded in a phospholipid bilayer
C proteins embedded in a polymer of glucose molecules
D proteins embedded in a carbohydrate bilayer

The correct answer is:
proteins embedded in a phospholipid bilayer

Solution:
According to fluid mosaic model given by Singer and Nicholson (1972), plasma membrane consists of a continuous bilayer of phospholipid molecules, in which globular proteins are embedded. This arrangement corresponds to icebergs (proteins) floating in a sea of phospholipids. Proteins stay in the membrane because they have regions of hydrophobic amino acids which interact with fatty acid tails to exclude water. Rest of the molecule is hydrophilic, which faces into or outward, both of which are aqueous environment.

Topic: Biology, Cell - The Unit of Life, NEET, NEET 2010 (Screening)
The plasmids present in the bacterial cells are

Options:
A Circular double stranded helical DNA molecules
B Circular double stranded helical RNA molecules
C Linear double stranded helical RNA molecules
D Linear double stranded helical DNA molecules

The correct answer is:
Circular double stranded helical DNA molecules

Solution:
The plasmid is extrachromosomal circular double-stranded DNA present in the cytoplasm in addition to the main genome. They are autonomously replicating segments and their number varies from cell to cell in bacteria.

Topic: Biology, Biotechnology - Principles and Processes, NEET
The plates of a parallel plate capacitor are charged to a potential difference of 117 V and then connected across a resistor. The potential difference across the capacitor decreases exponentially with respect to time. After 1 s the potential difference between the plates is 39 V, then after 2 s from the start, the potential difference ( in V ) between the plates is

Options:

The correct answer is:
13

Solution:
During discharging VC=Ve-t/RC
t=1, V1=V e-1/RC=V3
â´e-1/RC=13 â¦â¦1
after 2 s, V2=Ve-2/RC=V132=1179=13

Topic: Physics, Current Electricity, JEE Main
The plates of a parallel plate capacitor are charged to a potential of 200 V. Now, a dielectric slab of thickness 4 mm is inserted between its plates and to maintain the same potential difference between the plates of the capacitor, the distance between the plates is increased by 3.2 mm. The dielectric constant of the slab is

Options:
A 1
B 4
C 5
D 6

The correct answer is:
5

Solution:
Îµ0Ad = Îµ0Adâ²-t+tK
âd=dâ²-t+tK
âdâ²-d=t1-1K
Here, dâ²-d=3.2 mm, t=4 mm
â´3.2=41-1K
â3.24=1-1K
â1-1K=45
â´K=5

Topic: Physics, Electrostatics, JEE Main
The plates of a parallel plate capacitor are charged upto $100 \mathrm{~V}$. A $2 \mathrm{~mm}$ thick insulator sheet is inserted between the plates. Then to maintain the same potential difference, the distance between the plates is increased by $1.6 \mathrm{~mm}$. The dielectric constant of the insulator is

Options:
A 6
B 8
C 5
D 4

The correct answer is:
5

Solution:
Potential of capacitor before inserting insulator,
$$
V=\frac{\sigma d}{\varepsilon_0}
$$
where, $\sigma=$ surface charge density, $d=$ distance between plates
and $\varepsilon_0=$ permittivity of vacuum.
Given, $\frac{\sigma d}{\varepsilon_0}=100$
Now, potential after inserting $2 \mathrm{~mm}$ of insulator.
Let its dielectric constant is $k$.
$$
\begin{aligned}
& V_1=E_1 \cdot 2+E_2(d-2) \\
& V_1=\frac{\sigma}{\varepsilon_0 k} 2+\frac{\sigma}{\varepsilon_0}(d-2) \Rightarrow \quad V_1=\frac{\sigma}{\varepsilon_0}\left(\frac{2}{k}+d-2\right)
\end{aligned}
$$
To maintain potential difference, same distance is increased by $1.6 \mathrm{~mm}$
So, extra potential difference, $V_2=\frac{\sigma}{\varepsilon_0}(1.6)$
Clearly, $V_1+V_2=V$
$$
\begin{array}{lc}
& \frac{\sigma}{\varepsilon_0}\left(\frac{2}{k}+d-2\right)+\frac{\sigma}{\varepsilon_0} 1.6=\frac{\sigma d}{\varepsilon_0} \\
\Rightarrow & \frac{\sigma d}{\varepsilon_0}-\frac{2 \sigma}{\varepsilon_0 k}-\frac{\sigma}{\varepsilon_0}(d-2)=\frac{\sigma}{\varepsilon_0} 1.6 \\
\Rightarrow & \frac{2 \sigma}{\varepsilon_0}-\frac{2 \sigma}{\varepsilon_0 k}=\frac{\sigma}{\varepsilon_0} 1.6 \Rightarrow 2-\frac{2}{k}=1.6 \\
\Rightarrow & 2 k-2=1.6 k \\
\Rightarrow & \quad k=5
\end{array}
$$

Topic: Physics, Capacitance, AP EAMCET, AP EAMCET 2018 (24 Apr Shift 1)
The plates of a parallel plate capacitor are charged upto 200 volts. A dielectirc slab of thickness $4 \mathrm{~mm}$ is inserted between its plates. Then, to maintain the same potential difference between the plates of the capacitor, the distance between the plates is increased by $3.2 \mathrm{~mm}$. The dielectric constant of the dielectric slab is

Options:
A $1$
B $4$
C $5$
D $6$

The correct answer is:
$5$

Solution:
$$
\begin{aligned}
& \frac{\varepsilon_0 A}{d}=\frac{\varepsilon_0 A}{d^{\prime}-t+\frac{t}{K}} \\
& \Rightarrow \quad d=d^{\prime}-t+\frac{t}{K} \Rightarrow d^{\prime}-d=t\left(1-\frac{1}{K}\right)
\end{aligned}
$$
Here, $d^{\prime}-d=3.2 \mathrm{~mm}, t=4 \mathrm{~mm}$
$$
\begin{aligned}
& \therefore \quad 3.2=4\left(1-\frac{1}{K}\right) 1-\frac{1}{K} \\
& \Rightarrow \quad \frac{3.2}{4}=1-\frac{1}{K} \Rightarrow 1-\frac{1}{K}=\frac{4}{5} \\
& \therefore \quad K=5 \\
&
\end{aligned}
$$

Topic: Physics, Capacitance, AP EAMCET, AP EAMCET 2004
The plates of a parallel plate capacitor are charged upto 200 volts. A dielectirc slab of thickness $4 \mathrm{~mm}$ is inserted between its plates. Then, to maintain the same potential difference between the plates of the capacitor, the distance between the plates is increased by $3.2 \mathrm{~mm}$. The dielectric constant of the dielectric slab is

Options:
A $1$
B $4$
C $5$
D $6$

The correct answer is:
$5$

Solution:
$$
\begin{aligned}
& \frac{\varepsilon_0 A}{d}=\frac{\varepsilon_0 A}{d^{\prime}-t+\frac{t}{K}} \\
& \Rightarrow \quad d=d^{\prime}-t+\frac{t}{K} \Rightarrow d^{\prime}-d=t\left(1-\frac{1}{K}\right)
\end{aligned}
$$
Here, $d^{\prime}-d=3.2 \mathrm{~mm}, t=4 \mathrm{~mm}$
$$
\begin{aligned}
& \therefore \quad 3.2=4\left(1-\frac{1}{K}\right) 1-\frac{1}{K} \\
& \Rightarrow \quad \frac{3.2}{4}=1-\frac{1}{K} \Rightarrow 1-\frac{1}{K}=\frac{4}{5} \\
& \therefore \quad K=5 \\
&
\end{aligned}
$$

Topic: Physics, Capacitance, JEE Main
The plates of a parallel plate capacitor are charged upto 200 volts. A dielectirc slab of thickness $4 \mathrm{~mm}$ is inserted between its plates. Then, to maintain the same potential difference between the plates of the capacitor, the distance between the plates is increased by $3.2 \mathrm{~mm}$. The dielectric constant of the dielectric slab is

Options:
A $1$
B $4$
C $5$
D $6$

The correct answer is:
$5$

Solution:
$$
\begin{aligned}
& \frac{\varepsilon_0 A}{d}=\frac{\varepsilon_0 A}{d^{\prime}-t+\frac{t}{K}} \\
& \Rightarrow \quad d=d^{\prime}-t+\frac{t}{K} \Rightarrow d^{\prime}-d=t\left(1-\frac{1}{K}\right)
\end{aligned}
$$
Here, $d^{\prime}-d=3.2 \mathrm{~mm}, t=4 \mathrm{~mm}$
$$
\begin{aligned}
& \therefore \quad 3.2=4\left(1-\frac{1}{K}\right) 1-\frac{1}{K} \\
& \Rightarrow \quad \frac{3.2}{4}=1-\frac{1}{K} \Rightarrow 1-\frac{1}{K}=\frac{4}{5} \\
& \therefore \quad K=5 \\
&
\end{aligned}
$$

Topic: Physics, Capacitance, TS EAMCET, TS EAMCET 2004
The plot given below is of the average power delivered to an LRC circuit versus frequency. The quality factor of the circuit is :

Options:
A
$5.0$

B
$2.0$

C
$2.5$

D
$0.4$

The correct answer is:
$2.0$

Solution:
Quality factor of the circuit
$$
=\frac{\omega_0}{\omega_2-\omega_1}=\frac{5}{2.5}=2.0
$$

Topic: Physics, Alternating Current, JEE Main
The plot given below is of the average power delivered to an LRC circuit versus frequency. The quality factor of the circuit is :

Options:
A
$5.0$

B
$2.0$

C
$2.5$

D
$0.4$

The correct answer is:
$2.0$

Solution:
Quality factor of the circuit
$$
=\frac{\omega_0}{\omega_2-\omega_1}=\frac{5}{2.5}=2.0
$$

Topic: Physics, Alternating Current, JEE Main, JEE Main 2013 (23 Apr Online)
The plot of 1 Y A Vs 1 X A ( 1 Y A on y-axis) where A and B form a ideal solution. Y is mole fraction in vapour phase and X is mole fraction in liquid phase, is linear with slope and intercept respectively

Options:
A pA0pB0 and pA0-pB0pB0
B pA0pB0 and pB0-pA0pB0
C pB0pA0 and pA0-pB0pA0
D pB0pA0 and pB0-pA0pB0

The correct answer is:
pB0pA0 and pA0-pB0pA0

Solution:
1yA=pB0+(pA0-pB0)XApA0XA
1yA=pB0pA01XA+pA0-pB0pA0
y=mx+c
Slope =pB0pA0
Intercept =pA0-pB0pA0

Topic: Chemistry, Solutions, JEE Main
The plot of total vapour pressure as a function of mole fraction of the components of an ideal solution formed by mixing liquids X and Y is

Options:
A
B
C
D

The correct answer is:

Solution:
According to Raoult's law,

p=p0χ

According to Dalton's law of partial pressure,

pT=px+py

pT=p0xχx+p0yχy

=p0xχx+py01-χx

pT=p0Y+p0x-p0yχx

Thus, total vapour pressure of solution varies linearly with mole fraction of component X.

Case I, at χX=0

pT=p0Y

Case II, at XX=1

pT=p0Y+p0X-p0Y

pT=pX0

Topic: Chemistry, Solutions, JEE Main
The plot shows the variation of -ln⁡KP Versus temperature for the two reactions.
Ms+12O2g→MO(s) and
Cs+12O2g→CO(s)
Identify the correct statement:

Options:
A At T<1200 K, oxidation of carbon is unfavourable
B Oxidation of carbon is favourable at all temperatures
C At T<1200 K, the reaction MOs+Cs→Ms+CO(g) is spontaneous
D At T>1200 K, carbon will reduce MOs to Ms.

The correct answer is:
At T<1200 K, the reaction MOs+Cs→Ms+CO(g) is spontaneous

Solution:
ΔG=Δ G 0 +RTlnQ
At Equilibrium ΔG=0 & Q=K
∆Go=-RTln⁡Kp
∴-ln⁡Kp=∆GoRT
∆Go∝ Temperature. At high temperature ∆Go value is very high so non-feasible or non-spontaneous process takes place.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2016 (09 Apr Online)
The plots of intensity versus wavelength for three black bodies at temperatures \( T_{1}, T_{2} \) and \( T_{3} \), respectively, are as shown.
Their temperature are such that

Options:
A \( T_{1}>T_{2}>T_{3} \)
B \( T_{1}>T_{3}>T_{2} \)
C \( T_{2}>T_{3}>T_{1} \)
D \( T_{3}>T_{2}>T_{1} \)

The correct answer is:
\( T_{1}>T_{3}>T_{2} \)

Solution:
Wavelengths corresponding to temperatures T1,T2 and T3 are λ1,λ2 and λ3, respectively.

From given graph, we can write,

λ2>λ3>λ1.

According to Wein's law,

λmT=constant

Wavelength is inversely proportional to temperature.

So, above relation can be written as T1>T3>T2.

Topic: Physics, Thermal Properties of Matter, JEE Main