datavorous/entrance-exam-dataset huggingface dataset:
A line passing through the point A(9,0) makes an angle of 30° with the positive direction of x-axis. If this line is rotated about A through an angle of 15° in the clockwise direction, then its equation in the new position is

Options:
A y3-2+x=9
B x3-2+y=9
C x3+2+y=9
D y3+2+x=9

The correct answer is:
y3-2+x=9

Solution:
Given,

A straight line passes through A9,0 and makes a angle 30° with the x-axis,

And it is rotated by 15° degree clockwise

So, the new angle with the x-axis will be θ=30°-15°=15°

Hence, the equation of the line will be,

y-0=tan15°x-9

⇒y=tan45°-30°x-9

⇒y=1-131+13x-9

⇒y=2-3x-9

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2024 (30 Jan Shift 1)
A line passing through the point A2,2 encloses an area of 4 sq. units with coordinate axes. The distance (in units) of the origin from the line is equal to

Options:
A 1
B 22
C 2
D 2

The correct answer is:
2

Solution:
Let, the equation of the line is xa+yb=1⇒2a+2b=1
⇒2b+a=ab….i
Case I: If both intercepts are positive
Then, 12ab=4⇒ab=8⇒b+a=4
⇒a4-a8⇒a2-4a+8=0
⇒ No ‘a’ is possible (Discriminant is negative)
Case II: If intercept are of opposite sign, then -12ab=4⇒ab=-8
⇒b+a=-82=-4⇒a2+b2=16-2-8=32
⇒ the distance of origin from the line is aba2+b2=832=2 units

Topic: Mathematics, Straight Lines, JEE Main
A line passing through the point of intersection of $x+y=4$ and $x-y=2$ makes an angle $\tan ^{-1}\left(\frac{3}{4}\right)$ with the $X$ -axis. It intersects the parabola $y^{2}=4(x-3)$ at points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right),$ respectively. Then $\left|x_{1}-x_{2}\right|$ is equal to

Options:
A $\frac{16}{9}$
B $\frac{32}{9}$
C $\frac{40}{9}$
D $\frac{80}{9}$

The correct answer is:
$\frac{32}{9}$

Solution:
Given equations are
$x+y=4 ...(i)$
and $\quad x-y=2$ ...(ii)
From Eqs. (i) and $(ii),$ we get
$x=3$ and $y=1$
The line through this point making an angle tan $^{-1} \frac{3}{4}$ with the $X$ -axis is
$(y-1)=\frac{3}{4}(x-3) \quad\left[\because m=\frac{3}{4}\right]$
$\Rightarrow \quad y=\frac{3 x}{4}-\frac{5}{4}=\frac{3 x-5}{4}$
Since, this line intersects the parabola
$y^{2}=4(x-3)$ at points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$, respectively.
$\therefore$ Putting $y=\frac{3 x-5}{4}$ in equation of parabola, we get
$$
\left(\frac{3 x-5}{4}\right)^{2}=4(x-3)
$$
$\Rightarrow \quad 9 x^{2}-94 x+217=0$
$\Rightarrow \quad x_{1}+x_{2}=\frac{94}{9}$ and $x_{1} x_{2}=\frac{217}{9}$
$\begin{aligned} \therefore \quad\left|x_{1}-x_{2}\right| &=\sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}} \\ &=\sqrt{\left(\frac{94}{9}\right)^{2}-4 \times \frac{217}{9}} \\ &=\frac{32}{9} \end{aligned}$

Topic: Mathematics, Parabola, JEE Main
A line perpendicular to the line $a x+b y+c=0$ and passes through $(a, b)$. The equation of the line is

Options:
A $b x-a y+\left(a^2-b^2\right)=0$
B $b x-a y-\left(a^2-b^2\right)=0$
C $b x-a y=0$
D None of these

The correct answer is:
$b x-a y=0$

Solution:
The required equation of line which passes through $(a, b)$ and gradient $\frac{b}{a}$ is, $y-b=\frac{b}{a}(x-a)$ $\Rightarrow b x-a y=0$

Topic: Mathematics, Straight Lines, JEE Main
A line segment AB of length λ moves such that the points A and B remain on the periphery of a circle of radius λ. Then the locus of the point, that divides the line segment AB in the ratio 2:3, is a circle of radius

Options:
A 35λ
B 23λ
C 195λ
D 197λ

The correct answer is:
195λ

Solution:
Given,

A line segment AB of length λ moves such that the points A and B remain on the periphery of a circle of radius λ,

Now taking the points on the circle of radius λ as Bλcosθ1,λsinθ1 & Aλcosθ2λsinθ2 and taking the point Ph,k which divides the line segment in AB of length λ in 2:3

Now plotting the diagram we get,

Now, let O be the origin and radius of circle is λ and AB=λ and using distance formula we get,

AB=λ=λcosθ1-λcosθ22+λsinθ1-λsinθ22

⇒1=2-2 cosθ1-θ2

⇒cosθ1-θ2=12

Now using section formula we get,

h=2λ cosθ1+3λ cosθ25 and k=2λ sinθ1+3λ sinθ25

Now squaring and adding above two value we get,

h2+k2=λ2254+9+12cosθ1-θ2

⇒h2+k2=λ225·19

Hence, Radius =λ519

Topic: Mathematics, Circle, JEE Main, JEE Main 2023 (10 Apr Shift 1)
A line through the point $\mathrm{A}(2,0)$ which makes an angle of $30^{\circ}$ with the positive direction of $x$-axis is rotated about A in clockwise direction through an angle $15^{\circ}$. Then the equation of the straight line in the new position is

Options:
A $(2-\sqrt{3}) x+y-4+2 \sqrt{3}=0$
B $(2-\sqrt{3}) x-y-4+2 \sqrt{3}=0$
C $(2-\sqrt{3}) x-y+4+2 \sqrt{3}=0$
D $(2-\sqrt{3})_x+y+4+2 \sqrt{3}=0$

The correct answer is:
$(2-\sqrt{3}) x-y-4+2 \sqrt{3}=0$

Solution:
Hints : Equation of line in new position :
$$
\begin{aligned}
& y-0=\tan 15^{\circ}(x-2) \\
& \Rightarrow y=\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)(x-2) \\
& \Rightarrow y=\frac{(\sqrt{3}-1)^2}{2}(x-2)
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow 2 y=(4-2 \sqrt{3})(x-2) \\
& \Rightarrow y=(2-\sqrt{3})(x-2) \\
& \Rightarrow(2-\sqrt{3}) x-y-4+2 \sqrt{3}=0
\end{aligned}
$$

Topic: Mathematics, Straight Lines, JEE Main
A line with direction cosines proportional to $2,1,2$ meets each of the lines $x=y+a=z$ and $x+a=2 y=2 z$. The co-ordinates of each of the point of intersection are given by

Options:
A
$(3 a, 3 a, 3 a),(a, a, a)$

B
$(3 a, 2 a, 3 a),(a, a, a)$

C
$(3 a, 2 a, 3 a),(a, a, 2 a)$

D
$(2 a, 3 a, 3 a),(2 a, a, a)$

The correct answer is:
$(3 a, 2 a, 3 a),(a, a, a)$

Solution:
Any point on the line $\frac{x}{1}=\frac{y+a}{1}=\frac{z}{1}=t_1 \quad$ (say) is $\left(t_1, t_1-a, t_1\right)$ and any point on the line $\frac{x+a}{2}=\frac{y}{1}=\frac{z}{1}=t_2 \quad($ say $)$ is $\left(2 t_2-a, t_2, t_2\right)$.
Now direction cosine of the lines intersecting the above lines is proportional to $\left(2 t_2-a-t_1, t_2-t_1+a, t_2-t_1\right)$.
Hence $2 \mathrm{t}_2-\mathrm{a}-\mathrm{t}_1=2 \mathrm{k}, \mathrm{t}_2-\mathrm{t}_1+\mathrm{a}=\mathrm{k}$ and $\mathrm{t}_2-\mathrm{t}_1=2 \mathrm{k}$
On solving these, we get $t_1=3 a, t_2=a$.
Hence points are $(3 a, 2 a, 3 a)$ and $(a, a, a)$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2004
A line with direction ratio 2,1,2 meets the lines x=y+2=z and x+2=2y=2z respectively at the point P and Q. if the length of the perpendicular from the point (1,2,12) to the line PQ is l, then l2 is

Options:

The correct answer is:
65

Solution:
Let Pt, t-2, t and Q2s-2, s, s

DR's of PQ are 2, 1, 2

⇒2 s-2-t2=s-t+21=s-t2

⇒2s-2-t2=s-t+21

⇒2s-2-t=2s-2t+4

⇒t=6

⇒s-t+21=s-t2

⇒s-6+21=s-62

⇒2s-8=s-6

⇒s=2

⇒P6,4,6 and Q2, 2, 2

PQ:x-22=y-21=z-22=λ

⇒x=2λ+2, y=λ+2 and z=2λ+2

Let, F(2λ+2,λ+2,2λ+2)

A1,2,12

AF→·PQ→=0

⇒22λ+1+1λ+22λ-10=0

⇒4λ+2+λ+4λ-20=0

⇒λ=2

So F6, 4, 6 and AF=65

⇒l2=65

Topic: Mathematics, Three Dimensional Geometry, JEE Main
A line with direction ratio 2,1,2 meets the lines x=y+2=z and x+2=2y=2z respectively at the point P and Q. if the length of the perpendicular from the point (1,2,12) to the line PQ is l, then l2 is

Options:

The correct answer is:
65

Solution:
Let Pt, t-2, t and Q2s-2, s, s

DR's of PQ are 2, 1, 2

⇒2 s-2-t2=s-t+21=s-t2

⇒2s-2-t2=s-t+21

⇒2s-2-t=2s-2t+4

⇒t=6

⇒s-t+21=s-t2

⇒s-6+21=s-62

⇒2s-8=s-6

⇒s=2

⇒P6,4,6 and Q2, 2, 2

PQ:x-22=y-21=z-22=λ

⇒x=2λ+2, y=λ+2 and z=2λ+2

Let, F(2λ+2,λ+2,2λ+2)

A1,2,12

AF→·PQ→=0

⇒22λ+1+1λ+22λ-10=0

⇒4λ+2+λ+4λ-20=0

⇒λ=2

So F6, 4, 6 and AF=65

⇒l2=65

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2024 (29 Jan Shift 1)
A line with gradient 2 intersects a line with gradient 6 at the point 40,30. The distance between y-intercepts of these lines is

Options:
A 160
B 180
C 108
D 120

The correct answer is:
160

Solution:
Let C1 and C2 be the x -intercept of line with slope 2 and 6 respectively.
y-0=2x-C1
y=2x-2C1â¦â¦.i
Similarly y=6x-6C2â¦â¦.ii
Both i and ii satisfy x=40 and y=30
30=80-2C1âC1=25
And 30=240â6 C 2 â6 C 2 =210â C 2 =35
Hence, 6C2-2C1=160

Topic: Mathematics, Straight Lines, JEE Main
A line with positive direction cosines passes through the point $P(2,-1,2)$ and makes equal angles with the coordinate axes. If the line meets the plane $2 x+y+z=9$ at point $Q$, then the length $P Q$ equals

Options:
A
$\sqrt{2}$

B
2

C
$\sqrt{3}$

D
1

The correct answer is:
$\sqrt{3}$

Solution:
Point $P$ is $(2,-1,2)$
Let this line meet at $Q(h, k, w)$
Direction ratio of this line is
$(h-2, k+1, w-2)$
Since, $d c_s$ are equal $\& d r_s$ are also equal,
So, $h-2=k+1+w-2$
$\Rightarrow \quad k=h-3$ and $w=h$
This line meets the plane
$2 x+y+z=9$ at $Q$, so,
$2 h+k+w=9$ or $2 h+h-3+h=9$
$\Rightarrow 4 h-3=9 \Rightarrow h=3$
and $k=0$ and $w=3$
Distance
$$
\begin{aligned}
P Q & =\sqrt{\left(3-2^2\right)+\left(0-(-1)^2\right)+(3-2)^2} \\
& =\sqrt{1^2+1^2+1^2}=\sqrt{3}
\end{aligned}
$$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2012 (07 May Online)
A line, with the slope greater than one, passes through the point A4,3 and intersects the line x-y-2=0 at the point B. If the length of the line segment AB is 293, then B also lies on the line

Options:
A 2x+y=9
B 3x-2y=7
C x+2y=6
D 2x-3y=3

The correct answer is:
x+2y=6

Solution:
Given point B lie on the line x-y-2=0, so the point will be Bx1,x1-2

Now given distance between AB is 293, so by distance formula,

x1-42+x1-2-32=293

Now on squaring on both side we get,

18x12-162x1+340=0

On solving we get, x1=519 or x1=103

So, y1=339 or y1=43

Now we can see only x+2y=6 will satisfy the point 103,43

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2022 (25 Jul Shift 1)
A line x-a2=y-b3=z-c4 intersects a plane x-y+z=4 at a point where the line x-12=y+35=z+12 meets the plane. Also, a plane ax-2y+bz=3 meet them at the same point, then 11a+b+c is equal to

Options:

The correct answer is:
210

Solution:
Any general point on the line x-12=y+35=z+12 is 2λ+1,5λ-3,2λ-1
It must satisfy x-y+z=4⇒2λ+1-5λ+3+2λ-1=4
⇒-λ+3=4⇒λ=-1
So, the point of intersection of the plane and the line is -1,-8,-3
This point also lies on ax-2y+bz=3⇒-a+16-3b=3
⇒a+3b=13….i
This point also satisfy x-a2=y-b3=z-c4
⇒1+a2=8+b3=3+c4…ii
By taking 1+a2=8+b3⇒3+3a=16+2b
⇒3a-2b=13…iii
From i and iii
a=6511,b=2611
From ii
1+65112=8+26113=3+c4
3811=3811=3+c4⇒c=11911
⇒11a+b+c=210

Topic: Mathematics, Three Dimensional Geometry, JEE Main
A linear aperture whose width is 0.02 cm is placed immediately in front of a lens of focal length 60 cm. The aperture is illuminated normally by a parallel beam of wavelength 5Ã10-5 cm. The distance of the first dark band of the diffraction pattern from the centre of the screen is

Options:
A 0.10  cm
B 0.25  cm
C 0.20  cm
D 0.15  cm

The correct answer is:
0.15  cm

Solution:
By the principle of single slit diffraction pattern
θ=λa=yf
y=fλa=60×5×10-50.02=0.15 cm

Topic: Physics, Wave Optics, JEE Main
A linear object is placed along the axis of a mirror as shown in the figure. If f is the focal length of the mirror, then the length of the image is

Options:
A 2f3
B f
C f3
D none

The correct answer is:
f

Solution:
The image of object at 2f will be formed at 2f. So the object and image will coincide at 2f.
1v-1u=1f
1v-1-3f2=1f
1v=1f-23f=13f
∴ Length of image =3f-2f=f

Topic: Physics, Ray Optics, JEE Main
A linear programming problem is as follows:
Maximize $Z=30 x-18 y$
subject to the constraints,
$\begin{aligned}
& 3 x+4 y \leq 60 \\
& 5 x-3 y \geq 20 \\
& x, y \geq 0
\end{aligned}$
In the feasible region, the maximum value of $Z$ occurs at

Options:
A Two points
B One point
C Infinite number of points
D More than one of the above

The correct answer is:
Infinite number of points

Solution:
- Graph of the constraints in xy-plane needed to be drawn.
- The two constraints are given by $3 x+4 y \leq 60$ and $5 x-3 y \leq 20$.
- To draw $3 x+4 y \leq 60$, by putting $x=0$, we get $y=15$. Subsequently putting $y=0$ gives $x=20$.
- Two points obtained are $(0,15)$ and $(20,0)$. Joining these points and extending the line on both sides will give the line $3 x+4 y \leq 60$.
- Similarly, $5 x-3 y \leq 20$ will be drawn.
- Based on the inequalities, the feasible region is obtained which is shown in red color in the figure below:
- Now the corner points need to be determined.
- First corner point will be the origin $(0,0)$.
- Second corner point is the point where $3 x+4 y \leq 60$ intersects the $y$-axis. Putting $x=0$, we get the first corner point as $(0,15)$.
- Third corner point is the point where $5 x-3 y \leq 20$ intersects the $x$-axis. Putting $y=0$, we get the first corner point as $(4,0)$.
- Fourth corner point will be the intersection of two inequalities. Solving them we get, $y=\frac{240}{29}$ and $x=\frac{260}{29}$.
- Below is the table showing the corner points and the value of the objective functions at those points.
$\begin{array}{|l|l|l|} \hline \text{Sr. No.} & \text{Corner Points} & Z=30 x- 18 y \\ \hline 1 & (0,0) & 0 \\ \hline 2 & (0,15) & -270 \\ \hline 3 & (4,0) & 120 \\ \hline 4 & \left(\frac{260}{29}, \frac{240}{29}\right) & 120 \\ \hline \end{array}$
- The table indicates that the objective function has a maximum value at $(4,0)$ and $\left(\frac{260}{29}, \frac{240}{29}\right)$.
- Hence, the maximum value of the function lies on any point on the line joining the points $(4,0)$ and $\left(\frac{260}{29}, \frac{240}{29}\right.$ ).
- There will infinite number of points on the line where the value of the objective function will be maximum.

Topic: Mathematics, Linear Programming, JEE Main
A linguistic club of a certain Institute consists of 6 girls and 4 boys. A team of 4 members to be selected from this group including the selection of a Captain (from among these 4 members) for the team. If the team has to include atmost one boy, the number of ways of selecting the team is

Options:
A 95
B 260
C 320
D 380

The correct answer is:
380

Solution:
Case I: No boy is included. Selecting 4 girls from 6 girls $={ }^6 \mathrm{C}_4$
Selecting 1 captain from selected members $={ }^4 C_1$ Total number of ways $={ }^6 \mathrm{C}_4 \times{ }^4 \mathrm{C}_1=60$
Case II: One boy is included.
Selecting 3 girls and 1 boy from given members $={ }^6 \mathrm{C}_3 \times{ }^4 \mathrm{C}_1$.
Selecting 1 captain from the selected members $={ }^4 \mathrm{C}_1$.
Total Number of ways $={ }^6 \mathrm{C}_3 \times{ }^4 \mathrm{C}_1 \times{ }^4 \mathrm{C}_1=320$.
Total Number of ways $=320+60=380$.

Topic: Mathematics, Permutation Combination, JEE Main
A liquid A of mass 100 g at 100Â°C is added to 50 g of a liquid B at temperature 75Â°C, the temperature of the mixture becomes 90Â°C. Now if 100 g of liquid A at 100Â°C is added to 50 g of liquid B at 50Â°C, temperature of the mixture will be

Options:
A 80°C
B 60°C
C 70°C
D 85°C

The correct answer is:
80°C

Solution:
Heat loss will always be equal to heat gain.

As per 1stcondition,
100×SA×(100-90)=50×SB×(90-75) ___(i)

As per 2nd condition,
100×SA(100-θ)=50×SB(θ-50) ___(ii)
Dividing (ii) by (i), we get
100-θ100-90=θ-5090-75
300-3θ=2θ-100
θ=80°C

Topic: Physics, Thermal Properties of Matter, JEE Main
A liquid column of height $0.04 \mathrm{~cm}$ balances excess pressure of a soap bubble of certain radius. If density of liquid is $8 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}$ and surface tension of soap solution is $0.28 \mathrm{Nm}^{-1}$, then diameter of the soap bubble is _____ $\mathrm{cm}$. (if $g=10 \mathrm{~m} \mathrm{~s}^{-2}$ )

Options:

The correct answer is:
7

Solution:
$\begin{aligned} & \rho \mathrm{gh}=\frac{4 \mathrm{~S}}{\mathrm{R}} \\ & \Rightarrow \mathrm{R}=\frac{4 \times 0.28}{8 \times 10^3 \times 10 \times 4 \times 10^{-4}} \\ & \Rightarrow \frac{0.28}{8} \mathrm{~m}=\frac{28}{8} \mathrm{~cm} \\ & \Rightarrow \mathrm{R}=3.5 \mathrm{~cm}\end{aligned}$
Diameter $=7 \mathrm{~cm}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2024 (08 Apr Shift 1)
A liquid does not wet the sides of a solid, if the angle of contact is

Options:
A Zero
B Obtuse (More than $90^{\circ}$ )
C Acute (Less than $90^{\circ}$ )
D $90^{\circ}$

The correct answer is:
Obtuse (More than $90^{\circ}$ )

Solution:
When the contact angle is obtuse (more then $90^{\circ}$ ), liquid beads up and does not easily penetrate the solid to wet it. So the correct answer is ' $B$ '.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A liquid does not wet the solid surface if angle of contact is:

Options:
A equal to 45°
B equal to 60°
C greater than 90°
D zero

The correct answer is:
greater than 90°

Solution:
A liquid does not wet the solid surface if the angle of contact is obtuse ie, greater than 90°. In this case cohesive forces will be greater than adhesive forces and so, the liquid does not wet the surface of solid.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A liquid drop having 6 excess electrons is kept stationary under a uniform electric field of $25.5$ $\mathrm{kVm}^{-1}$. The density of liquid is $1.26 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}$. The radius of the drop is (neglect buoyancy).

Options:
A
$4.3 \times 10^{-7} \mathrm{~m}$

B
$7.8 \times 10^{-7} \mathrm{~m}$

C
$0.078 \times 10^{-7} \mathrm{~m}$

D
$3.4 \times 10^{-7} \mathrm{~m}$

The correct answer is:
$7.8 \times 10^{-7} \mathrm{~m}$

Solution:
$\mathrm{F}=\mathrm{qE}=m g\left(\mathrm{q}=6 \mathrm{e}=6 \times 1.6 \times 10^{-19}\right)$
$\operatorname{Density}(\mathrm{d})=\frac{\text { mass }}{\text { volume }}=\frac{\mathrm{m}}{\frac{4}{3} \pi \mathrm{r}^3}$
or $r^3=\frac{m}{\frac{4}{3} \pi d}$
Putting the value of $d$ and $m\left(=\frac{q E}{g}\right)$ and
solving we get $\mathrm{r}=7.8 \times 10^{-7} \mathrm{~m}$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2013 (23 Apr Online)
A liquid drop having surface energy E is spread into 512 droplets of same size. The final surface energy of the droplets is nE. What is the value of n?

Options:

The correct answer is:
8

Solution:
The surface area of the liquid drop is A=4ÏR2
Its surface energy is E
When the drop splits in 512 droplets, the surface area of each droplet is 4Ïr2
â´ Total surface area A2 =512Ã4Ïr2
The volume of bigger drop is 43ÏR3 and volume of small droplets is 512Ã43Ïr3
â´ 43ÏR3=512Ã43Ïr3âr=R8
â´ A2=512Ã4Ïr2=512Ã4ÏR82=8A1
Surface energy E=A.T (T is surface tension and A is area)
â´ EnE=A2âTA1âT=8âA1A1=8
â´ En=8E

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A liquid drop having surface energy E is spread into 512 droplets of same size. The final surface energy of the droplets is

Options:
A 2E
B 4E
C 8E
D 12E

The correct answer is:
8E

Solution:
The surface area of the liquid drop is A=4πR2
Its surface energy is E
When the drop splits in 512 droplets, the surface area of each droplet is 4πr2
∴ Total surface area A2 =512×4πr2
The volume of bigger drop is 43πR3 and volume of small droplets is 512×43πr3
∴ 43πR3=512×43πr3⇒r=R8
∴ A2=512×4πr2=512×4πR82=8A1
Surface energy E=A.T (T is surface tension and A is area)
∴ EnE0=A2⋅TA1⋅T=8⋅A1A1=8
∴ En=8E

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A liquid drop having surface energy E is spread into 512 droplets of the same size. The final surface energy of the droplets is

Options:
A 2E
B 4E
C 8E
D 12E

The correct answer is:
8E

Solution:
The surface area of the liquid drop is A=4πR2
Its surface energy is E
When the drop splits in 512 droplets, the surface area of each droplet is 4πr2
∴ Total surface area A2 =512×4πr2
The volume of the bigger drop is 43πR3 and the volume of small droplets is 512×43πr3
∴ 43πR3=512×43πr3⇒r=R8
∴ A2=512×4πr2=512×4πR82=8A1
Surface energy E=A.T (T is surface tension and A is area)
∴ EnE0=A2⋅TA1⋅T=8⋅A1A1=8
∴ En=8E

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A liquid drop placed on a horizontal plane has a near spherical shape (slightly flattened due to gravity). Let $\mathrm{R}$ be the radius of its largest horizontal section. A small disturbance causes the drop to vibrate with frequency $\mathrm{v}$ about its equilibrium shape. By dimensional analysis the ratio $\frac{\mathrm{v}}{\sqrt{\sigma / \rho \mathrm{R}^{3}}}$ can be (Here $\sigma$ is surface tension, $\rho$ is density, $\mathrm{g}$ is acceleration due to gravity, and $\mathrm{k}$ is arbitrary dimensionless constant)-

Options:
A $\mathrm{k} \rho \mathrm{gR}^{2} / \sigma$
B $\mathrm{k} \rho \mathrm{R}^{2} / \mathrm{g} \sigma$
C $\mathrm{k} \rho \mathrm{R}^{3} / \mathrm{g} \sigma$
D $\mathrm{k} \rho / \mathrm{g} \sigma$

The correct answer is:
$\mathrm{k} \rho \mathrm{gR}^{2} / \sigma$

Solution:
$\frac{\mathrm{v}}{\sqrt{\sigma / \rho \mathrm{R}^{3}}}$ is dimensionless $\mathrm{k} \rho \mathrm{gR}^{2} / \sigma$ is also dimensionless

Topic: Physics, Units and Dimensions, JEE Main
A liquid flows through a horizontal tube. The velocities of the liquid in the two sections, which have areas of crosssection $A_1$ and $A_2$, are ${ }^{v_1}$ and ${ }^{v_2}$ respectively. The difference in the levels of the liquid in the two vertical tubes is $h$

Options:
A The volume of the liquid flowing through the tube in unit time is $A_1 v_1$
B $v_2-v_1=\sqrt{2 g h}$
C $v_2^2-v_1^2=2 g h$
D The energy per unit mass of the liquid is the same in both sections of the tube

The correct answers are:
The volume of the liquid flowing through the tube in unit time is $A_1 v_1$, $v_2^2-v_1^2=2 g h$, The energy per unit mass of the liquid is the same in both sections of the tube

Solution:
According to equation of continuity the volume of liquid flowing through the tube in unit time remains constant i.e. $A_1 v_1=A_2 v_2$, hence option (1) is correct
According to Bernoulli's theorem,
$P_1+\frac{1}{2} \rho v_1^2=P_2+\frac{1}{2} \rho v_2^2$
$\Rightarrow P_1-P_2=\frac{1}{2} \rho\left(v_2^2-v_1^2\right) \Rightarrow h \rho g=\frac{1}{2} \rho\left(v_2^2-v_1^2\right)$
$\therefore v_2^2-v_1^2=2 g h$
Hence option (3) is correct.
Also, according to Bernoulli's theorem option (4) is correct

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A liquid having a coefficient of volume expansion Î³ is filled in a cylindrical vessel made out of glass having coefficient of linear expansion Î±=Î³5. At room temperature, the level of liquid in the vessel is l0 and when the temperature is increased by âT, the level of liquid in the vessel becomes lâl01+nÎ±âT. What is the value of n? [Given, Î±âTâª1]

Options:

The correct answer is:
3

Solution:
Let us assume that the volume of the liquid in the vessel is V0 and the area of cross-section of the vessel is A0, then

l0=V0A0

Similarly,

l=VA=V01+Î³âTA01+2Î±âT

âl=l01+Î³âT1+2Î±âT-1

Using binomial approximation, we get

l=l01+Î³âT1-2Î±âT

âl=l01+Î³-2Î±âT-2Î±Î³âT2

Neglecting 2Î±Î³âT2, since it is insignificant as compared to Î³-2Î±âT, we get

lâl01+Î³-2Î±âTâlâl01+5Î±-2Î±âTâlâl01+3Î±âT

So, n=3

Topic: Physics, Thermal Properties of Matter, JEE Main
A liquid in a beaker has temperature $\theta(t)$ at time $t$ and $\theta_0$ is temperature of surroundings, then according to Newton's law of cooling the correct graph between $\log _e\left(\theta-\theta_0\right)$ and $t$ is

Options:
A

B

C

D

The correct answer is:

Solution:
According to Newtons law of cooling.
$\frac{\mathrm{d} \theta}{\mathrm{dt}} \propto-\left(\theta-\theta_0\right)$
$\Rightarrow \quad \frac{\mathrm{d} \theta}{\mathrm{dt}}=-\mathrm{k}\left(\theta-\theta_0\right)$
$\int \frac{d \theta}{\theta-\theta_0}=\int-k d t$
$\Rightarrow \quad \ln \left(\theta-\theta_0\right)=-k t+c$
Hence the plot of $\ln \left(\theta-\theta_0\right)$ vs $t$ should be a straight line with negative slope.

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2012 (Offline)
A liquid is coming out from a vertical tube. The relation between the weight of the drop $W$, surface tension of the liquid $T$ and radius of the tube $r$ is given by, if the angle of contact is zero

Options:
A $W=\pi r^2 T$
B $W=2 \pi r T$
C $W=2 r^2 \pi T$
D $W=\frac{3}{4} \pi r^3 T$

The correct answer is:
$W=2 \pi r T$

Solution:
Correct option is 2. $W=2 \pi r T$
If the angle of contact is zero
The relation between the weight of the drop $W$, surface tension of the liquid $T$ and radius of the tube $r$ is given by,
$W=2 \pi r T$

Topic: Physics, Mechanical Properties of Fluids, JEE Main