datavorous/entrance-exam-dataset huggingface dataset:
The length of the perpendicular from the point $(b, a)$ to the line $\frac{x}{a}-\frac{y}{b}=1$, is

Options:
A $\left|\frac{a^2-a b+b^2}{\sqrt{a^2+b^2}}\right|$
B $\left|\frac{b^2-a b-a^2}{\sqrt{a^2+b^2}}\right|$
C $\left|\frac{a^2+a b-b^2}{\sqrt{a^2+b^2}}\right|$
D None of these

The correct answer is:
$\left|\frac{b^2-a b-a^2}{\sqrt{a^2+b^2}}\right|$

Solution:
Length of perpendicular is
$\left|\frac{\frac{b}{a}-\frac{a}{b}-1}{\sqrt{\left(\frac{1}{a}\right)^2+\left(\frac{1}{b}\right)^2}}\right|=\left|\frac{b^2-a^2-a b}{\sqrt{a^2+b^2}}\right|$

Topic: Mathematics, Straight Lines, JEE Main
The length of the perpendicular (in units) from the point 1,2,4 on the straight line x-21=y-72=z-3-1 lies in the interval

Options:
A 1,32
B 2,3
C 0,2
D 4,5

The correct answer is:
0,2

Solution:
Let, point A is 1,2,4 and any general point on the line is BÎ»+2,2Î»+7,-Î»+3,
ABâ=Î»+1i^+2Î»+5j^+-Î»-1k^
and a vector parallel to the line is bâ=i^+2j^-k^
ABââbâ=0âÎ»+1+4Î»+10+Î»+1=0
âÎ»=-2
The length of the perpendicular =ABâ
=-2+12+-4+52+2-12=3 units

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The length of the portion of the common tangent to x2+y2=16 and 9x2+25y2=225 between the two points of contact is

Options:
A 94 units
B 34 units
C 347 units
D 547 units

The correct answer is:
347 units

Solution:
Equation of the ellipse is x225+y29=1
Any point on the ellipse is 5cosâ¡Î¸,3sinâ¡Î¸ at which the equation of the tangent is
x5cosâ¡Î¸+y3sinâ¡Î¸=1
If it is also a tangent to the circle x2+y2=16, then its distance from the origin is equal to the radius,
â1cos2Î¸25+sin2Î¸9=4
âcos2Î¸25+1-cos2Î¸9=116
âcosâ¡Î¸=5716
If l be the length of the tangent then l2=S'
or l2=25cos2Î¸+9sin2Î¸-16=16cos2Î¸-7
=16.25.7162-7=175-11216=6316
âl=347 units

Topic: Mathematics, Ellipse, JEE Main
The length of the potentiometer wire is 600 cm and a current of 40 mA is flowing in it. When a cell of emf 2 V and internal resistance 10 Î© is balanced on this potentiometer the balance length is found to be 500 cm. The resistance of potentiometer wire will be

Options:
A 20Â Î©
B 40Â Î©
C 60 Î©
D 80Â Î©

The correct answer is:
60 Î©

Solution:
EMF of external circuit=Potential gradient × Balance length
2=IRL×l0
R=2LI×l0=2×60040×10-3×500=12200×10-3=12×103200
= 60 Ω .

Topic: Physics, Current Electricity, JEE Main
The length of the projection of the line segment joining the points (5,-1, 4) and (4,-1, 3) on the plane, x+y+z=7 is

Options:
A 23
B 23
C 23
D 13

The correct answer is:
23

Solution:
DR's of AA' (1, 1, 1), as this will be normal to the plane.
DR's of AB is (1, 0, 1)
⇒cos⁡θ=1×1+1×0+1×112+12+1212+02+12=26=23 , also AB=2
Projection A1B1=2.sin⁡θ
=2.1-23
=23

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2018 (08 Apr)
The length of the radius of the circle which touches the x -axis at the point 1,0 and passes through the point 2,3 is

Options:
A 103 units
B 35 units
C 65 units
D 53 units

The correct answer is:
53 units

Solution:
Circle touches the x -axis at 1,0
So, center of circle is 1,k and radius is k.
Hence, the equation of circle is:
x-12+y-k2=k2
It passes through the point 2,3 ,

â1+9-6k+k2=k2

â´k=53

Topic: Mathematics, Circle, JEE Main
The length of the second's pendulum on the surface of the earth is 1 m. If the acceleration due to gravity on the surface of the moon is one-sixth of the acceleration due to gravity of the earth, then the length of the second's pendulum on the surface of the moon will be

Options:
A 16 m
B 6 m
C 136 m
D 36 m

The correct answer is:
16 m

Solution:
Seconds pendulum is a special pendulum having a time period of 2 s.
Using the formula T=2Ï L g
If the time period remains the same and g is reduced by a factor of 6.
L should be reduced by a factor of 6.
âL=16 m on moon.

Topic: Physics, Oscillations, JEE Main
The length of the segment of the tangent line to the curve $x=a \cos ^3 t, y=\alpha \sin ^3 t$, at any point on the curve cut off by the coordinate axes is

Options:
A $4 a$
B $a$
C $a^2$
D $2 a$

The correct answer is:
$a$

Solution:
Given, curves are
and
$$
\begin{aligned}
& x=a \cos ^3 t \\
& y=a \sin ^3 t
\end{aligned}
$$
On differentiating Eq. (i) and Eq. (ii) w.r.t.t, we get
and
$$
\begin{aligned}
& \frac{d y}{d t}=3 a \sin ^2 t \cos t \\
& \frac{d x}{d t}=-3 a \cos ^2 t \sin t
\end{aligned}
$$
$$
\begin{aligned}
\therefore \quad \frac{d y}{d x} & =\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{3 a \sin ^2 t \cos t}{-3 a \cos ^2 t \sin t} \\
& =\frac{-\sin t}{\cos t}=m_1
\end{aligned}
$$
Equation of tangent at point $\left(\operatorname{acos}^3 t, a \sin ^3 t\right)$ is
$$
\begin{gathered}
\quad y-a \sin ^3 t=\frac{-\sin t}{\cos t}\left(x-a \cos ^3 t\right) \\
\Rightarrow y \cos t-a \cos t \sin ^3 t=-x \sin t+a \cos ^3 t \sin t \\
\Rightarrow x \sin t+y \cos t=a \cos t \sin t\left[\sin ^2 t+\cos ^2 t\right] \\
\Rightarrow x \sin t+y \cos t=a \cos t \sin t
\end{gathered}
$$
For finding the points cut off by the coordinate axes put $y=0$ in Eq. (iii), we get
$$
\begin{aligned}
& \Rightarrow \quad x \sin t=\operatorname{acos} t \sin t \\
& \Rightarrow \quad x=a \cos t \\
& \text { and put } \\
& x=0 \text { in Eq. (iii), we get } \\
& \Rightarrow \quad y \cos t=a \cos t \sin t \\
& \Rightarrow \quad y=a \sin t \\
&
\end{aligned}
$$
Hence, points are $A(a \cos t, 0)$ and $B(0, a \sin t)$,
$$
\begin{aligned}
\therefore \quad \text { Length of } & A B=\sqrt{(a \cos t)^2+(a \sin f)^2} \\
& =\sqrt{a^2\left(\cos ^2 t+\sin ^2 t\right)}=\sqrt{a^2}=a
\end{aligned}
$$

Topic: Mathematics, Application of Derivatives, JEE Main
The length of the shadow of a pole inclined at \( 10^{\circ} \) to the vertical towards the sun is \( 2.05 \) metres, when the elevation of the sun is \( 38^{\circ} \). The length of the pole is

Options:
A \( \frac{2.05 \sin 38^{\circ}}{\sin 42^{\circ}} \)
B \( \frac{2.05 \sin 42^{\circ}}{\sin 38^{\circ}} \)
C \( \frac{2.05 \cos 38^{\circ}}{\cos 42^{\circ}} \)
D None of these

The correct answer is:
\( \frac{2.05 \sin 38^{\circ}}{\sin 42^{\circ}} \)

Solution:
sin38°l=sinSPO2.05

=sin180°-38°-90°-10°2.05 ⇒ l=2.05sin38°sin42° .

Topic: Mathematics, Heights and Distances, JEE Main
The length of the shadows of a vertical pole of height h, thrown by the sun's rays at three different moments are h, 2h and 3h. The sum of the angles of elevation of the rays at these three moments is equal to

Options:
A π2
B π3
C π4
D π6

The correct answer is:
π2

Solution:
Let OA be the vertical pole of height h and OP1, OP2, OP3 be the lengths of shadow.
In âAOP1 we have,
tanâ¡Î¸1=OAOP1=hh=1
â¹ Î¸1=Ï4
In âAOP2 we have,
tanâ¡Î¸2=OAOP2=22h=12
â¹ Î¸2=tan-1â¡12
Similarly, tanâ¡Î¸3=OAOP3=h3h=13
Î¸3=tan-1â¡13
â´ Sum of the angles
=Î¸1+Î¸2+Î¸3
=Ï4+tan-1â¡12+tan-1â¡13
=Ï4+tan-1â¡12+131-12Ã13
=Ï4+tan-1â¡1
=Ï4+Ï4=Ï2

Topic: Mathematics, Heights and Distances, JEE Main
The length of the two rods made up of the same metal and having the same area of cross-section are $0.6 \mathrm{~m}$ and $0.8 \mathrm{~m}$ respectively. The temperature between the ends of first rod is $90^{\circ} \mathrm{C}$ and $60^{\circ} \mathrm{C}$ and that for the other rod is 150 and $110^{\circ} \mathrm{C}$. For which rod the rate of conduction will be greater

Options:
A First
B Second
C Same for both
D None of the above

The correct answer is:
Second

Solution:
The rate of heat flow in a rod is given by the formula $\mathrm{H}=\mathrm{kA}((\mathrm{T} 1-\mathrm{T} 2) / \mathrm{L})$
Rate of heat transfer is directly proportional to difference in temperature.

Topic: Physics, Thermal Properties of Matter, JEE Main
The length of the wire between two ends of sonometer is $100 \mathrm{~cm}$. What should be the positions of two bridges below the wire so that the three segments of the wire have their fundamental frequencies in the ratio $1: 3: 5$.

Options:
A $\frac{1500}{23} \mathrm{~cm}, \frac{500}{23} \mathrm{~cm}$
B $\frac{1500}{23} \mathrm{~cm}, \frac{300}{23} \mathrm{~cm}$
C $\frac{300}{23} \mathrm{~cm}, \frac{1500}{23} \mathrm{~cm}$
D $\frac{1500}{23} \mathrm{~cm}, \frac{2000}{23} \mathrm{~cm}$

The correct answer is:
$\frac{1500}{23} \mathrm{~cm}, \frac{2000}{23} \mathrm{~cm}$

Solution:
According to question
Length of wire $=L=100 \mathrm{~cm}$
And ratio of fundamental frequencies
$$
=1: 3: 5
$$
$$
\begin{aligned}
L_1: L_2: L_3 & =\frac{1}{1}: \frac{1}{3}: \frac{1}{5} \\
& =15: 5: 3
\end{aligned}
$$
Let $x=$ comman factor then
$$
\begin{aligned}
& 15 x+5 x+3 x=100 \\
& \Rightarrow \quad 2 x=100 \\
& \Rightarrow \quad x=\frac{100}{23} \\
& \therefore \quad L_1=15 \times \frac{100}{23} \\
& =\frac{1500}{23} \mathrm{~cm} \\
& \therefore \quad L_2=5 \times \frac{100}{23} \\
& =\frac{1500}{23} \mathrm{~cm} \\
&
\end{aligned}
$$
$$
\begin{aligned}
\therefore \quad L_3 & =3 \times \frac{100}{23} \\
& =\frac{300}{23} \mathrm{~cm}
\end{aligned}
$$
This shows that bridge should be placed from $A$ at $\frac{1500}{23} \mathrm{~cm}$ and $\frac{2000}{23} \mathrm{~cm}$ respectively

Topic: Physics, Waves and Sound, JEE Main
The length of the wire is doubled. Its conductance will

Options:
A remain unchanged
B be halved
C be doubled
D be quadrupled

The correct answer is:
be halved

Solution:
The inverse of resistance of any wire or conductor is defined as conductance.

conductance G=1R

Resistance of a wire is given by

R=ρLA

Here, ρ= Resistivity of wire,

L= Length of the wire,

A= Cross-sectional area.

If the length of the wire is doubled, the resistance will also get doubled.

thus the conductance of the wire will be halved.

Topic: Chemistry, Electrochemistry, JEE Main
The length of two open organ pipes are $l$ and $(l+\Delta l)$ respectively. Neglecting end correction, the frequency of beats between them will be approximately

Options:
A $\frac{v}{2 l}$
B $\frac{v}{4 l}$
C $\frac{v \Delta l}{2 l^2}$
D $\frac{v \Delta l}{l}$

The correct answer is:
$\frac{v \Delta l}{2 l^2}$

Solution:
$\lambda_1=2 l, \lambda_2=2 l+2 \Delta l \Rightarrow n_1=\frac{v}{2 l}$ and $n_2=\frac{v}{2 l+2 \Delta l}$
$\Rightarrow$ No. of beats $=n_1-n_2=\frac{v}{2}\left(\frac{1}{l}-\frac{1}{l+\Delta l}\right)=\frac{v \Delta l}{2 l^2}$

Topic: Physics, Waves and Sound, JEE Main
The length of two opposite edges of a tetrahedron are 12 and 15 units and the shortest distance between them is 10 units. If the volume of the tetrahedron is 200 cubic units, then the angle between the 2 edges is

Options:
A sin-112
B sin-123
C sin-134
D sin-145

The correct answer is:
sin-123

Solution:
Let, ABCD be the tetrahedron and position vectors of A,B,C,D be aâ,bâ,câ,dâ respectively.
Given, bâ-aâ=12 & dâ-câ=15
Shortest distance =dâ-bââbâ-aâÃdâ-câbâ-aâÃdâ-câ=10 ... 1
Also, volume =16dâ-aââbâ-aâÃcâ-aâ=200 ... 2
From 1 and 2, we get,
bâ-aâÃdâ-câÃ10=6Ã200
bâ-aâdâ-câsinâ¡Î¸=120
sinâ¡Î¸=23âÎ¸=sin-123

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The lengths of major and minor axis of an ellipse are 10 and 8 respectively and its major axis along the $y$-axis. The equation of the ellipse referred to its center as origin is

Options:
A $\frac{x^2}{25}+\frac{y^2}{16}=1$
B $\frac{x^2}{16}+\frac{y^2}{25}=1$
C $\frac{x^2}{100}+\frac{y^2}{64}=1$
D $\frac{x^2}{64}+\frac{y^2}{100}=1$

The correct answer is:
$\frac{x^2}{16}+\frac{y^2}{25}=1$

Solution:
Here given that $2 b=10,2 a=8 \Rightarrow b=5, a=4$. Hence the required equation is $\frac{x^2}{16}+\frac{y^2}{25}=1$.

Topic: Mathematics, Ellipse, JEE Main
The lengths of sides of a cuboid are a, 2a and 3a. If the relative percentage error in the measurement of a is 1%, then what is the relative percentage error in the measurement of the volume of the cube.

Options:

The correct answer is:
3

Solution:
V=a×2a×3a

⇒V=6a3

⇒ΔVV=3Δaa=3%

Topic: Physics, Units and Dimensions, JEE Main
The lengths of the perpendiculars from the points m2,2m, mn,m+n and n2,2n to the line x+3y+3=0 are in

Options:
A Arithmetic progression
B Geometric progression
C Harmonic progression
D None of these

The correct answer is:
Geometric progression

Solution:
The given line is x+3y+3=0 â¦ 1
Let p1, p2 and p3 are the lengths of the perpendiculars from the points Am2,2m, Bmn,m+n and Cn2,2n respectively on the given line, then
p1=m2+32m+31+3âm+322
p2=mn+3m+n+31+3âm+3n+32
p3=n2+32n+31+3ân+322
Clearly, p1.p3=p22
â´ p1,p2,p3 are in GP

Topic: Mathematics, Straight Lines, JEE Main
The lengths of the sides of a triangle are 10+x2, 10+x2 and 20-2x2. If for x=k, the area of the triangle is maximum, then 3k2 is equal to

Options:
A 5
B 12
C 10
D 20

The correct answer is:
10

Solution:
Area of triangle, Δ=2020-10-x220-10-x220-20+2x2

Δ=210x10-x2

For maxima, dΔdx=0⇒21010-x2-410x2=0

⇒ x2=103

Here, d2∆dx2<0

Hence 3k2=10

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2022 (27 Jun Shift 1)
The lengths of the tangents from any point on the circle x2+y2+8x+1=0 to the circles x2+y2+7x+1=0 and x2+y2+4x+1=0 are in the ratio

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

The correct answer is:
1:2

Solution:
Let the point on x2+y2+8x+1=0 is h,k

⇒h2+k2+8h+1=0
Now ratio of the lengths of the tangents from h,k to the circles x2+y2+7x+1=0 and x2+y2+4x+1=0 is
h2+k2+7h+1h2+k2+4h+1=7h-8h4h-8h
=12=1:2

Topic: Mathematics, Circle, JEE Main
The lengths of the tangents from any point on the circle \( x^{2}+y^{2}+8 x+1=0 \) to the circles \( x^{2}+y^{2}+7 x+1=0 \) and \( x^{2}+y^{2}+4 x+1=0 \) are in the ratio

Options:
A \( 1: 2 \)
B \( 1: 3 \)
C \( 1: 4 \)
D \( 1: \sqrt{2} \)

The correct answer is:
\( 1: 2 \)

Solution:
Let the point on x2+y2+8x+1=0 is h,k

⇒h2+k2+8h+1=0
Now ratio of the lengths of the tangents from h,k to the circles x2+y2+7x+1=0 and x2+y2+4x+1=0 is
h2+k2+7h+1h2+k2+4h+1=7h-8h4h-8h
=12=1:2

Topic: Mathematics, Circle, JEE Main
The lengths of two adjacent sides of a cyclic quadrilateral are 2 units and 5 units and the angle between them is 60o. If the area of the quadrilateral is 43 sq. units, then the perimeter of the quadrilateral is

Options:
A 12.5 units
B 13 units
C 13.2 units
D 12 units

The correct answer is:
12 units

Solution:
cos⁡ 60 o =4+25-c22.2.5
⇒10=29-c2
⇒c2=19
⇒c=19
Now,
cos120°= a 2 + b 2 − c 2 2ab
⇒ -12=a2+b2-192ab
⇒a2+b2-19=-ab
⇒a2+b2+ab=19
Area of quadrilateral =12×2×5sin⁡60°+12absin⁡120°=43
⇒532+ab34=43
⇒ab4=4-52=32
⇒ab=6
⇒a2+b2=13
⇒a=2, b=3
Perimeter
=2+5+2+3
=12 units

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2017 (09 Apr Online)
The lengths of two adjacent sides of a cyclic quadrilateral are 2 units and 5 units and the angle between them is 60o. If the area of the quadrilateral is 43 sq. units, then the perimeter of the quadrilateral is

Options:
A 12.5 units
B 13 units
C 13.2 units
D 12 units

The correct answer is:
12 units

Solution:
cos⁡ 60 o =4+25-c22.2.5
⇒10=29-c2
⇒c2=19
⇒c=19
Now,
cos120°= a 2 + b 2 − c 2 2ab
⇒ -12=a2+b2-192ab
⇒a2+b2-19=-ab
⇒a2+b2+ab=19
Area of quadrilateral =12×2×5sin⁡60°+12absin⁡120°=43
⇒532+ab34=43
⇒ab4=4-52=32
⇒ab=6
⇒a2+b2=13
⇒a=2, b=3
Perimeter
=2+5+2+3
=12 units

Topic: Mathematics, Trigonometric Equations, JEE Main
The lengths of two adjacent sides of a cyclic quadrilateral are 2 units and 5 units and the angle between them is 60o. If the area of the quadrilateral is 43 sq. units, then the perimeter of the quadrilateral is

Options:
A 12.5 units
B 13 units
C 13.2 units
D 12 units

The correct answer is:
12 units

Solution:
cosâ¡ 60 o =4+25-c22.2.5
â10=29-c2
âc2=19
âc=19
Now,
cos120Â°= a 2 + b 2 â c 2 2ab
â -12=a2+b2-192ab
âa2+b2-19=-ab
âa2+b2+ab=19
Area of quadrilateral =12Ã2Ã5sinâ¡60Â°+12absinâ¡120Â°=43
â532+ab34=43
âab4=4-52=32
âab=6
âa2+b2=13
âa=2, b=3
Perimeter
=2+5+2+3
=12

Topic: Mathematics, Properties of Triangles, JEE Main
The letter $\mathrm{T}$ in T-lymphocyte refers to:

Options:
A Thymus
B Thyroid
C Thalamus
D Tonsil

The correct answer is:
Thymus

Solution:
No Solution Available

Topic: Biology, Human Health and Diseases, NEET, NEET 2009 (Mains)
The letters of the word 'MANKIND' are written in all possible orders and arranged in serial order as in an English dictionary. Then the serial number of the word 'MANKIND' is _____ .

Options:

The correct answer is:
1492

Solution:
M	A	N	K	I	N	D

Arranging the letters alphabetically, we get

ADIKMNN

When the word starts with any of the letters A/D/I/K, the number of possibilities=6!2!×4=1440

Now when the word starts with MA, then the number of possibilities=4!2!×3=36

Now when the word starts with MAN, then the number of possibilities=3!×2=12

Now when the word starts with MANK, then the number of possibilities=4

Hence, rank of the work MANKIND is 1492

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2022 (25 Jul Shift 1)
The letters of the word MODESTY are written in all possible orders and these words are written out as in a dictionary, then the rank of the word MODESTY is

Options:
A $5040$
B $720$
C $1681$
D $2520$

The correct answer is:
$1681$

Solution:
Words start with D are $6!720$, start with E are 720 , start with MD are $5!120$ and start with ME are 120 . Now the first word starts with MO is nothing but MODESTY. Hence rank of MODESTY is 1681.

Topic: Mathematics, Permutation Combination, JEE Main
The letters of the word OUGHT are written in all possible ways and these words are arranged as in a dictionary, in a series. Then the serial number of the word TOUGH is :

Options:
A 89
B 84
C 86
D 79

The correct answer is:
89

Solution:
Lets arrange the letters of OUGHT in alphabetical order.

G,H,O,T,U

Number of words starting with G

=4!=24

Number of words starting with H

=4!=24

Number of words starting with O

=4!=24

Number of words starting with TG

3!=6!

Number of words starting with TH

=3!=6

Number of words starting with TOG

=2!=2

Number of words starting with TOH

=2!=2

Next word is TOUGH.

Hence, the serial number of the word TOUGH is

=24+24+24+6+6+2+2+1=89

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2023 (29 Jan Shift 2)
The letters of the word 'QUESTION' are arranged in a row at random. The probability that there are exactly two letters between $\mathrm{Q}$ and $\mathrm{S}$ is

Options:
A $\frac{1}{14}$
B $\frac{5}{7}$
C $\frac{1}{7}$
D $\frac{5}{28}$

The correct answer is:
$\frac{5}{28}$

Solution:
The given word is 'QUESTION', let E be the event that there are exactly two letters between $Q$ and $S$.
Here, number of sample space,
$n(S)=8 \text { ! }$
number of possible events,
$n(E)={ }^6 P_2 \times 2 ! \times 5 \text { ! }$
$\therefore \quad$ Required probability
$\begin{aligned}
& P(E)=\frac{n(E)}{n(S)}=\frac{{ }^6 P_2 \times 2 ! \times 5 !}{8 !}=\frac{6 ! \times 2 ! \times 5 !}{4 ! \times 8 !} \\
& =\frac{6 ! \times 2 ! \times 5 ! \times 4 !}{4 ! \times 7 \times 8 \times 6 !}=\frac{2 \times 5}{7 \times 8}=\frac{5}{28}
\end{aligned}$

Topic: Mathematics, Permutation Combination, JEE Main
The leucocytes contain which of the following in large quantity?

Options:
A Basophils
B Neutrophils
C Eosinophils
D Monocytes

The correct answer is:
Neutrophils

Solution:
White blood cells are also known as leucocytes. There are two main groups of WBCs
(a) Agranulocytes (30\%)
(b) Granulocytes ( $70 \%)$
Granulocytes are made in the bone marrow but by the cells different from those that make red blood cells. These can be further subdivided into
(i) Neutrophils
(ii) Eosinophils
(iii) Basophils.
Neutrophils (phagocytes) make up about 70\% of the total number of granular WBCs. They commonly squeeze between the cells of the capillary walls and wander through the intercellular spaces. They are actively phagocytic which engulf and digest disease causing bacteria.

Topic: Biology, Body Fluids and Circulation, JIPMER, JIPMER 2009
The level of water in a tank is 5 m high. A hole of the area 10 cm2 is made in the bottom of the tank. The rate of leakage of water from the hole is (g =10 m s-2)

Options:
A 10-2 m3 s-1
B 102 m3 s-1
C 10 m3 s-1
D 10-4 m3 s-1

The correct answer is:
10-2 m3 s-1

Solution:
The velocity of efflux, v=2 gh;
The volume of liquid flowing out per second
=v×A=2gh×A
=2×10×5×10×10-4=10-2 m3 s-1

Topic: Physics, Mechanical Properties of Fluids, JEE Main
The level of water in a tank is 5 m high. A hole of the area 10 cm2 is made in the bottom of the tank. The rate of leakage of water from the hole is

Options:
A 10-2 m3 s-1
B 102 m3 s-1
C 10 m3 s-1
D 10-4 m-3 s-1

The correct answer is:
10-2 m3 s-1

Solution:
The velocity of efflux, v=2 gh;
The volume of liquid flowing out per second
=vÃA=2 ghÃA
=2Ã10Ã5Ã10Ã10-4=10-2 m3 s-1

Topic: Physics, Mechanical Properties of Fluids, JEE Main
The level of water in a tank is \( 5 \mathrm{~m} \) high. A hole of the area \( 10 \mathrm{~cm}^{2} \) is made in the bottom of the tank. The rate of leakage of water from the hole is \( \left(g=10 \mathrm{~m} \mathrm{~s}^{-2}\right) \)

Options:
A \( 10^{-2} \mathrm{~m}^{3} \mathrm{~s}^{-1} \)
B \( 10^{2} \mathrm{~m}^{3} \mathrm{~s}^{-1} \)
C \( 10 \mathrm{~m}^{3} \mathrm{~s}^{-1} \)
D \( 10^{-4} \mathrm{~m}^{3} \mathrm{~s}^{-1} \)

The correct answer is:
\( 10^{-2} \mathrm{~m}^{3} \mathrm{~s}^{-1} \)

Solution:
The velocity of efflux, v=2gh;
The volume of liquid flowing out per second
=v×A=2gh×A
=2×10×5×10×10-4=10-2 m3 s-1

Topic: Physics, Mechanical Properties of Fluids, JEE Main
The level of water in the pond will A man is sitting on a boat which is floating in a pond. If the man drinks some water from the pond.

Options:
A rise a little
B sink a little
C remains stationary
D none of these

The correct answer is:
remains stationary

Solution:
When a man sitting in a boat drinks m gram of water from the pond, the weight of (boat+man) system will increase by mg and so the system will displace m gram more water for floating. Due to removal of water form pond, the water level in pond will fall. But due to water displaced by the floating system the water level in the pond will rise. Since the water removed from the pond is equal to the water displaced by the system, the level of water in the pond remains unchanged.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
The Lewis structure of allene is

Which of the following statements correctly gives answers for all 3 parts:
(a) Is the molecule planar?
(b) Does 1, 3 - dichloro propadiene show geometrical isomerism?
(c) Is the molecule 1, 3 - dichloro propadiene polar?

Options:
A (i) Non-planar
(ii) No geometrical isomerism
(iii) Polar
B (i) Planar
(ii) No geometrical isomerism
(iii) Polar
C (i) Planar
(ii) Yes geometrical isomerism
(iii) Polar
D (i) Non Planar
(ii) No geometrical isomerism
(iii) Not Polar

The correct answer is:
(i) Non-planar
(ii) No geometrical isomerism
(iii) Polar

Solution:
Hybridization of central carbon is sp and two of its unhybridized p -orbitals are involved in Ï - bonding. Also p - orbitals at an atom are orthogonal (at 90 Î¿ angle), the two Ï - bonds at central carbon are also orthogonal. Let these p - orbitals, involved in Ï - bonding, at central carbon be p y and p z, then

Hybridization at the terminal carbons are both sp 2 . The two p - orbitals involved in sp 2 hybridization are p x and p y at left terminal and p x and p z at right terminals. Therefore, the three s p 2 hybrid orbitals at left terminal carbon are in xy plane, while the same at right terminal carbon are in xz
plane. As a result, the two triangular planes at terminals are at right angles as shown below:

The two H - C - H planes at terminals are perpendicular to each other. ( Ï - bonds are also perpendicular to each other.)
Hence, (a) Molecule is non - planar. (b) It doesn't show geometrical isomerism because two H and chlorine are in different planes. (c) polar.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The LH and FSH secretion from the anterior pituitary is stimulated by the

Options:
A gonadotropin releasing hormone.
B testosterone from testis
C melatonin from pineal gland
D thyroxine from thyroid gland

The correct answer is:
gonadotropin releasing hormone.

Solution:
Gonadotropins are hormones synthesized and released by the anterior pituitary, which act on the gonads (testes and ovaries) to increase the production of sex hormones and stimulate production of either sperm or ova. Follicle stimulating hormone (FSH) and luteinizing hormones (LH) are the main gonadotropins.

Topic: Biology, Chemical Coordination and Integration, NEET