datavorous/entrance-exam-dataset huggingface dataset:
A perpendicular is drawn from a point on the line x-12= y+1-1=z1 to the plane x+y+z=3 such that the foot of the perpendicular Q also lies on the plane x-y+z=3. Then the coordinates of Q are

Options:
A 2, 0, 1
B –1, 0, 4
C 4, 0, –1
D 1, 0, 2

The correct answer is:
2, 0, 1

Solution:
The given line is x-12=y+1-1=z1=λ, (let)

⇒x-12=λ, y+1-1=λ, z1=λ

⇒x=2λ+1, y=-λ-1, z=λ.

Hence, any point on the line is 2λ+1, -λ-1, λ.

Thus, the point P on the line is 2λ+1, -λ-1, λ.

We know that the foot of perpendicular from a point x1, y1, z1 on the plane ax+by+cz+d=0 is x-x1a=y-y1b=z-z1c=-ax1+by1+cz1a2+b2+c2.

Thus, the foot of perpendicular Q from P on the plane x+y+z-3=0 is given by

x-2λ-11=y+λ+11=z-λ1=-2λ-33

∵ Q lies on x+y+z=3  &  x-y+z=3

On adding the two equations, we get y=0

⇒λ+11=-2λ+33

⇒λ=0

Hence, we have x-11=z1=--33

⇒x=2, z=1

So, the coordinate of Q are 2, 0, 1.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2019 (10 Apr Shift 2)
A perpendicular is drawn from a point on the line x-12= y+1-1=z1 to the plane x+y+z=3 such that the foot of the perpendicular Q also lies on the plane x-y+z=3. Then the coordinates of Q are

Options:
A 2, 0, 1
B –1, 0, 4
C 4, 0, –1
D 1, 0, 2

The correct answer is:
2, 0, 1

Solution:
The given line is x-12=y+1-1=z1=λ, (let)

⇒x-12=λ, y+1-1=λ, z1=λ

⇒x=2λ+1, y=-λ-1, z=λ.

Hence, any point on the line is 2λ+1, -λ-1, λ.

Thus, the point P on the line is 2λ+1, -λ-1, λ.

We know that the foot of perpendicular from a point x1, y1, z1 on the plane ax+by+cz+d=0 is x-x1a=y-y1b=z-z1c=-ax1+by1+cz1a2+b2+c2.

Thus, the foot of perpendicular Q from P on the plane x+y+z-3=0 is given by

x-2λ-11=y+λ+11=z-λ1=-2λ-33

∵ Q lies on x+y+z=3  &  x-y+z=3

On adding the two equations, we get y=0

⇒λ+11=-2λ+33

⇒λ=0

Hence, we have x-11=z1=--33

⇒x=2, z=1

So, the coordinate of Q are 2, 0, 1.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
A perpendicular is drawn from a point on the line x-12= y+1-1=z1 to the plane x+y+z=3 such that the foot of the perpendicular Q also lies on the plane x-y+z=3. Then the coordinates of Q are

Options:
A 2, 0, 1
B â1,Â 0,Â 4
C 4,Â 0,Â â1
D 1,Â 0,Â 2Â

The correct answer is:
2, 0, 1

Solution:
Let point P on the line is 2λ+1, -λ-1, λ

The foot of perpendicular Q is given by
x-2λ-11=y+λ+11=z-λ1=-2λ-33
∵ Q Lies on x+y+z=3  &  x-y+z=3
⇒x+z=3 & y=0
y=0⇒λ+1=-2λ+33⇒λ=0
⇒Q is 2, 0, 1

Topic: Mathematics, Three Dimensional Geometry, JEE Main
A person can see clearly object only when they lie between 50 cm to 400 cm from his eyes. In order to increase the maximum distance of distinct vision to infinity, the type and power of the correcting lens, the person has to use, will be

Options:
A Convex, +2.25 D

B Concave, -0.25 D

C Concave, -0.2 D

D Convex, + 0.15 D

The correct answer is:
Concave, -0.25 D

Solution:
The maximum distance of distinct vision = 400 cm. So the image of the object at infinity is to be formed at 400 cm.
Use lens formula
1v-1u=1f
1-400-1∞=1f
P=-0.25 D

Topic: Physics, Ray Optics, JEE Main
A person can throw a ball upto a maximum range of 100 m. How high above the ground he can throw the same ball?

Options:
A 25 m
B 50 m
C 100 m
D 200 m

The correct answer is:
50 m

Solution:
Horizontal range is maximum if the angle of projection is 45° and is given by Rmax=u2g

Here, Rmax=100 m

Then, 100=u2g ⋯1

If the ball is thrown vertically upward, then the ball will attain the maximum height from the ground as Hmax=u22g

Hmax=1002=50 m

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2022 (29 Jun Shift 2)
A person climbs up a stalled escalator in $60 \mathrm{~s}$. If standing on the same but escalator running with constant velocity he takes $40 \mathrm{~s}$. How much time is taken by the person to walk up the moving escalator?

Options:
A
$37 \mathrm{~s}$

B
$27 \mathrm{~s}$

C
$24 \mathrm{~s}$

D
$45 \mathrm{~s}$

The correct answer is:
$24 \mathrm{~s}$

Solution:
Person's speed walking only is $\frac{1}{60} \frac{\text { "escalator" }}{\text { second }}$
Standing the escalator without walking the
speed is $\frac{1}{40} \frac{\text { "escalator" }}{\text { second }}$
Walking with the escalator going, the speed add.
So, the person's speed is $\frac{1}{60}+\frac{1}{40}=\frac{15}{120}$
$\frac{\text { "escalator" }}{\text { second }}$
So, the time to go up the escalator ("seconds per escalator") is
$$
t=\frac{120}{5}=24 \text { second }
$$

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2014 (12 Apr Online)
A person forgets his 4-digit ATM pin code. But he remembers that in the code all the digits are different, the greatest digit is 7 and the sum of the first two digits is equal to the sum of the last two digits. Then the maximum number of trials necessary to obtain the correct code is________.

Options:

The correct answer is:
72

Solution:
Let the 4-digit pin be abcd

Now given sum of first two is equal to last two, so a+b=c+d

And number available are 0, 1, 2, 3, 4, 5, 6, 7

Now Let a+b=c+d=λ

Also the greatest digit that he has used is 7

So, a+b=c+d=λ≥7

Now taking all cases starting from λ=7 we get,

λ=7, a,b→0,7 or 7,0, then c,d→1,6,2,5,3,4,4,3,5,2,6,1 and vice-versa, so 2×12=24 numbers

λ=8, a,b→1,7 or 7,1, then c,d→2,6,3,5,6,2,5,3 and vice versa so total 2×8=16 numbers,

λ=9, a,b→2,7 or 7,2 then c,d→3,6,4,5,5,4,6,3 and vice versa, so total 2×8=16 numbers,

λ=10, a, b→3, 7,7, 3, so c, d→6, 4,4, 6→8 numbers

λ=11, a, b→4, 7,7, 4, so c, d→6, 5,5, 6→8 numbers

λ=12, 13 & 14 are not possible,

So, Total numbers =24+16+16+8+8=72

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2023 (15 Apr Shift 1)
A person goes $10 \mathrm{~km}$ north and $20 \mathrm{~km}$ east. What will be displacement from initial point

Options:
A $22.36 \mathrm{~km}$
B $2 \mathrm{~km}$
C $5 \mathrm{~km}$
D $20 \mathrm{~km}$

The correct answer is:
$22.36 \mathrm{~km}$

Solution:
$\overrightarrow{A C}=\overrightarrow{A B}+\overrightarrow{B C}$
$A C=\sqrt{(A B)^2+(B C)^2}$
$=\sqrt{(10)^2+(20)^2}$
$=\sqrt{100+400}=\sqrt{500}=22.36 \mathrm{~km}$

Topic: Physics, Mathematics in Physics, JEE Main
A person goes to the office either by a car, or scooter, or bus, the probability of which being 27,37,27 respectively. The probability that he reaches office late if he takes a car, or scooter, or bus is 29,19,49 respectively. If he reaches office in time, the probability that he travelled by car is k, then the value of 24k+7 is equal to

Options:
A 7
B 14
C 212
D 312

The correct answer is:
14

Solution:
Let, C,B,S be the events such that a person goes to the office by car, bus or scooter respectively.
PC=27,PS=37,PB=27
Let, T be the event that he reaches office in time
PTC=79,PTS=89,PTB=59
PCT=PCPTCPCPTC+PBPTB+PSPTS
=27Ã7927Ã79+37Ã89+27Ã59=1448=724

Topic: Mathematics, Probability, JEE Main
A person has been using spectacles of power -1.0 diopter for distant vision and a separate reading glass of power 2.0 diopters. What is the least distance of distinct vision for this person:

Options:
A 10 cm
B 40 cm
C 30 cm
D 50 cm

The correct answer is:
50 cm

Solution:
Given, P=2 D=2 m-1

Focal length is 1f=2100 cm-1

Using lens equation 1v-1u=1f

As we know, u=-25 cm

Then, 1v--125=2100

⇒1v=150-125

⇒v=-50 cm

Topic: Physics, Ray Optics, JEE Main, JEE Main 2023 (30 Jan Shift 1)
A person holding a bag starts sliding freely from a point A on a frictionless inclined plane while his bag starts falling down vertically from the same point. The final speeds of the man VM and the bag VB on reaching the ground after falling through the same height should be such that

Options:
A VM<VB
B VM=VB
C They depend on the masses
D VM>VB

The correct answer is:
VM=VB

Solution:
As there is only a gravitational field which is a conservative field, the change in kinetic energy of both the objects will be proportional to their masses and their vertical displacements. Hence, their speeds will be only proportional to their vertical displacements.
so, VM=VB

Topic: Physics, Laws of Motion, JEE Main
A person holding a rifle (mass of person and rifle together is $100 \mathrm{~kg}$ ) stands on a smooth surface and fires 10 shots horizontally, in $5 \mathrm{~s}$. Each bullet has a mass of $10 \mathrm{~g}$ with a muzzle velocity of $800 \mathrm{~ms}^{-1}$. The final velocity acquired by the person and the average force exerted on the person are:

Options:
A $-0.08 \mathrm{~ms}^{-1}, 16 \mathrm{~N}$
B $-0.08 \mathrm{~ms}^{-1}, 16 \mathrm{~N}$
C $-0.08 \mathrm{~ms}^{-1}, 16 \mathrm{~N}$
D $-0.08 \mathrm{~ms}^{-1}, 16 \mathrm{~N}$

The correct answer is:
$-0.08 \mathrm{~ms}^{-1}, 16 \mathrm{~N}$

Solution:
Using law of conservation of moments
$\begin{aligned}
& \text {Initial momentum }=m u_1+m u_2 \\
& \text {final momentum }=n m v_1+ \\
& (M-n m) v_2 \\
& \therefore m u_1+m u_2=n m v_1+(M-n m) v_2 \\
& 0=\frac{10 \times 10}{1000} \times 800 +\left(100-\frac{10 \times 10}{1000}\right) r^2
\end{aligned}$
$\begin{aligned}
-80 & =\frac{999}{10} v_2 \\
\Rightarrow v_2 & =-0.8 \mathrm{~ms}^{-1}
\end{aligned}$
And average force exerted on the person
$\begin{array}{l}
=\frac{n \times m \times u}{\Delta t} \\
=\frac{n \times m \times u}{\Delta t} \\
=\frac{10 \times \frac{1 \mathrm{G}}{1000} \times 800}{5} \\
=\frac{80}{5}=16 \mathrm{~N}
\end{array}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A person invites 8 guests to a dinner and places 5 of them at one table and the remaining 3 at another, both the tables being round. The number of ways in which the guests can be arranged is

Options:
A 40320
B 2688
C 8064
D 13440

The correct answer is:
2688

Solution:
The number of ways to select 5 person out of 8 is ${ }^8 C_5$. Now, number of ways to place 5 on round table is 4 ! and number of ways to place 3 on round table is 2 !.
So, required number of ways $={ }^8 C_5 \times 4 ! \times 2 !$
$$
=\frac{8 \times 7 \times 6}{3 \times 2} \times 4 \times 3 \times 2 \times 2=42 \times 64=2688
$$
Hence, option (b) is correct.

Topic: Mathematics, Permutation Combination, AP EAMCET, AP EAMCET 2019 (20 Apr Shift 2)
A person invites a party of 10 friends at dinner and place them so that 4 are on one round table and 6 on the other round table. The number of ways in which he can arrange the guests is

Options:
A $\frac{(10) !}{6 !}$
B $\frac{(10) !}{24}$
C $\frac{(9) !}{24}$
D None of these

The correct answer is:
$\frac{(10) !}{24}$

Solution:
Selection of 6 guests $=10 \mathrm{C} 6$

Permutation of 6 on round table $=5 !$

Permutation of 4 on round table $=3 !$

Then, total number of arrangements $=10 \mathrm{C} 6.5 ! .3 !$

$=\frac{(10) !}{6 ! 4 !} \cdot 5 ! .3 !=\frac{(10) !}{24}$

Topic: Mathematics, Permutation Combination, JEE Main
A person is having problems with calcium and phosphorus metabolism in his body. Which one of the following glands may not be functioning properly?

Options:
A parotid
B pancreas
C thyroid
D parathyroid.

The correct answer is:
parathyroid.

Solution:
Parathyroid hormone causes the lowering of blood calcium level. This increases the excitability of nerves and muscles causing cramps and convulsions. Parotid gland is the largest salivary gland, involved in the digestion process. Pancreas is a mixed gland, and it secrete hormones - Insulin and glucagon, maintaining blood sugar levels. Thyroid gland secrete thyroxine $\left(\mathrm{T}_4\right)$ and triiodothyronine $\left(\mathrm{T}_3\right)$, which regulate the basal metabolism of the body.
Related Theory
PTH has effects antagonistic to those of calcitonin by increasing blood calcium levels by stimulating osteoclasts to break down bone and release calcium. secreted by the parafollicular $C$-cells of the thyroid gland.

Topic: Biology, Chemical Coordination and Integration, NEET, NEET 2007
A person is observing two trains, one is coming towards him with a velocity of \( 4 \mathrm{~m} / \mathrm{s} \) and the other is going away from him with the same velocity. If the trains sound which of their natural frequency \( 240 \mathrm{~Hz} \), then the person will hear a beat- frequency of :

Options:
A \( 6 \)
B \( 3 \)
C zero
D \( 12 \)

The correct answer is:
\( 6 \)

Solution:
Given:

Velocity of sound vs= 4 ms-1

Velocity of observer vo= 0 ms-1

Frequency fo= 240 Hz

For first train,

From Doppler effect in sound waves,

⇒f=fov-vov+vs

f1=240×320320+4 =240×320324f1=237 Hz

In second train coming towards observer,

f2=320320-4×240 =320316×240f2 =243 Hz

Therefore,

f= f2-f1f= 243-237f= 6 Hz

A beat-frequency f= 6 Hz

Topic: Physics, Waves and Sound, JEE Main
A person is standing in an elevator. In which situation, he experiences weight loss ?

Options:
A When the elevator moves upward with constant acceleration
B When the elevator moves downward with constant acceleration
C When the elevator moves upward with uniform velocity
D When the elevator moves downward with uniform velocity

The correct answer is:
When the elevator moves downward with constant acceleration

Solution:
When the pseudo force acting on the person is in opposite direction to the weight then the person will experience weight loss. The direction of the pseudo force is opposite to the direction of the frame of reference.

Hence, B is the correct answer.

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2022 (26 Jun Shift 1)
A person is swimming with a speed of 10 m s-1 at an angle of 120° with the flow and reaches to a point directly opposite on the other side of the river. The speed of the flow is x m s-1. The value of x to the nearest integer is ______.

Options:

The correct answer is:
5

Solution:
10 sin 30°=x
x=5 m/s

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2021 (18 Mar Shift 1)
A person is to count 4500 currency notes. Let $a_n$ denote the number of notes he counts in the $\mathrm{n}^{\text {th }}$ minute. If $\mathrm{a}_1=\mathrm{a}_2=\ldots \ldots=\mathrm{a}_{10}=150$ and $\mathrm{a}_{10}, \mathrm{a}_{11}, \ldots \ldots$ are in A.P. with common difference $-2$, then the time taken by him to count all notes is

Options:
A
34 minutes

B
125 minutes

C
135 minutes

D
24 minutes

The correct answer is:
34 minutes

Solution:
Till $10^{\text {th }}$ minute number of counted notes $=1500$
$$
3000=\frac{n}{2}[2 \times 148+(n-1)(-2)]=n[148-n+1]
$$
$$
\begin{aligned}
& n^2-149 n+3000=0 \\
& n=125,24 \\
& n=125 \text { is not possible. }
\end{aligned}
$$
Total time $=24+10=34$ minutes.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2010
A person is to count 4500 notes. Let $a_{n}$ denote the number of notes he counts in the nth minute. If $a_{1}=a_{2}=a_{3}=\ldots .$ $=a_{10}=150$, and $a_{10}, a_{11}, a_{12}, \ldots$ are in APwith the common difference $-2$, then the time taken by him to count all the notes is

Options:
A 24 minutes
B 34 minutes
C 125 minutes
D 135 minutes

The correct answer is:
34 minutes

Solution:
Given, $\mathrm{a}_{1}=\mathrm{a}_{2}=\mathrm{a}_{3}=\ldots \ldots . .=\mathrm{a}_{10}=150$
Also, $a_{10}, a_{11}, a_{12}, \ldots$ are in A.P. and $d=-2$ since, $\mathrm{a}_{10}=150$, A.P. is $150,148,146, \ldots$
For the first 10 minutes, he has counted $150 \times 10=1500$ notes. Time taken to count remaining 3000 notes
$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow 3000=\frac{\mathrm{n}}{2}[2 \times 148+(\mathrm{n}-1)(-2)]$
$\Rightarrow 3000=\frac{\mathrm{n}}{2} \times 2(148-\mathrm{n}+1)$
$\Rightarrow 3000=148 n-n^{2}+n$
$\Rightarrow n^{2}-149 n+3000=0$
$\Rightarrow(\mathrm{n}-24)(\mathrm{n}-125)$
$\Rightarrow \mathrm{n}=24$, or $\mathrm{n}=125 .$
Since he has taken 10 minutes to count 1500 notes, he will not take 125 min to count 3000 notes. $\mathrm{So}, \mathrm{n}=24$
$\therefore$ Total time $=10+24=34$ minutes.

Topic: Mathematics, Sequences and Series, NDA, NDA 2017 (Phase 2)
A person likely to develop tetanus is immunized by administering :

Options:
A Weakned germs
B Dead germs
C Preformed antibodies
D Wide spectrum antibiotics

The correct answer is:
Preformed antibodies

Solution:
No Solution Available

Topic: Biology, Human Health and Diseases, NEET, NEET 2009 (Mains)
A person lives in a high-rise building on the bank of a river $50 \mathrm{~m}$ wide. Across the river is a well lit tower of height $40 \mathrm{~m}$. When the person, who is at a height of $10 \mathrm{~m}$, looks through a polarizer at an appropriate angle at light of the tower reflecting from the river surface, he notes that intensity of light coming from distance $\mathrm{X}$ from his building is the least and this corresponds to the light coming from light bulbs at height ' $\mathrm{Y}$ ' on the tower. The values of $\mathrm{X}$ and $\mathrm{Y}$ are respectively close to (refractive index of water $\simeq \frac{4}{3}$ )

Options:
A
$25 \mathrm{~m}, 10 \mathrm{~m}$

B
$13 \mathrm{~m}, 27 \mathrm{~m}$

C
$22 \mathrm{~m}, 13 \mathrm{~m}$

D
$17 \mathrm{~m}, 20 \mathrm{~m}$

The correct answer is:
$13 \mathrm{~m}, 27 \mathrm{~m}$

Solution:
No Solution is Given.

Topic: Physics, Wave Optics, JEE Main, JEE Main 2013 (09 Apr Online)
A person moved from $A$ to $B$ on a circular path as shown in figure. If the distance travelled by him is $60 \mathrm{~m}$, then the magnitude of displacement would be: $\left(\right.$ Given $\left.\cos 135^{\circ}=-0.7\right)$

Options:
A $42 \mathrm{~m}$
B $47 \mathrm{~m}$
C $19 \mathrm{~m}$
D $40 \mathrm{~m}$

The correct answer is:
$47 \mathrm{~m}$

Solution:
$\begin{aligned} & \text { Distance travelled }=60 \mathrm{~m} \\ & \Rightarrow \text { Angle covered }=135^{\circ} \\ & \text { Displacement }=2 \mathrm{R} \sin \left(135^{\circ} / 2\right) \\ & =2\left(\frac{60}{135} \times \frac{180}{\pi}\right)\left[\frac{1-\cos \left(135^{\circ}\right)}{2}\right]^{1 / 2} \\ & =2\left(\frac{80}{\pi}\right)(0.85)^{1 / 2} \\ & \approx 47 \mathrm{~m}\end{aligned}$

Topic: Physics, Rotational Motion, JEE Main
A person moved from A to B on a circular path as shown in figure. If the distance travelled by him is 60 m, then the magnitude of displacement would be : (Given cos135°=-0.7)

Options:
A 42 m
B 47 m
C 19 m
D 40 m

The correct answer is:
47 m

Solution:
The distance travelled by the person d will be equal to the arch length.

⇒d=Rθ

⇒60=R3π4

⇒R=60×43π=80πm

Now the magnitude of displacement can be found using subtraction of vectors. Therefore,

s=R22+R12-2R1R2cosθ⇒s=R2+R2-2R2cos135°

⇒2R2-2R2-0.7

⇒3.4R2=3.480π2

⇒R=47 m

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2022 (25 Jul Shift 1)
A person observes two moving trains, A reaching the station and B leaving the station with equal speed of 30 m s-1. If both trains emit sounds with frequency 300 Hz, (Speed of sound : 330 m s-1) approximate difference of frequencies heard by the person will be :

Options:
A 33 Hz
B 55 Hz
C 80 Hz
D 10 Hz

The correct answer is:
55 Hz

Solution:
Using Doppler effect, apparent frequency is given by f=v±v0v∓vsf0, where f0=actual frequency, v0=velocity of observer and vs=velocity of source

For train B leaving the station,

f1=300330-0330--30=275 Hz ∵v0=0

For train A reaching the station,

f2=300330-0330-30=330 Hz

The difference of frequencies heard by the person∆f=330-275=55 Hz

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2023 (29 Jan Shift 1)
A person of 80 kg mass is standing on the rim of a circular platform of mass 200 kg rotating about its axis at 5 revolutions per minute (rpm). The person now starts moving towards the centre of the platform. What will be the rotational speed (in rpm ) of the platform when the person reaches its centre....

Options:

The correct answer is:
9

Solution:
Applying conservation of angular momentum,

mR2+MR22ω=MR22ω'

80+2002×5=2002ω'

ω'=9rev/sec.

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2020 (03 Sep Shift 1)
A person of mass M is sitting on a swing of length L and swinging with and an angular amplitude θ0. If the person stands up when the swing passes through its lowest point, the work done by him, assuming that his centre of mass moves by a distance l l<<L, is close to:

Options:
A Mgl (1-θ02)
B Mgl(1+θ022)
C Mgl
D Mgl(1+θ02)

The correct answer is:
Mgl(1+θ02)

Solution:
From angular momentum conservation
MV0L=MV(L-l)V=V0LL-l
Now
W=Mgl+12MV02LL-l2-V02
Since l≪L
W=Mgl+12MV021-lL-2-V02=Mgl+12MV021+2lL-V02=Mgl+MV02lL
From SHM
V0=ωA=gL×Lθ0=gLθ0
Using this
W=Mgl+MgL×θ02×lLW=Mgl1+θ02

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2019 (12 Apr Shift 1)
A person of mass M is sitting on a swing of length L and swinging with and angular amplitude θ0. If the person stands up when the swing passes through its lowest point, the work done by him, assuming that his centre of mass moves by a distance l l<<L, is close to:

Options:
A Mgl (1-θ02)
B Mgl(1+θ022)
C Mgl
D Mgl(1+θ02)

The correct answer is:
Mgl(1+θ02)

Solution:
From angular momentum conservation
MV0L=MV(L-l)V=V0LL-l
Now
W=Mgl+12MV02LL-l2-V02
Since l≪L
W=Mgl+12MV021-lL-2-V02=Mgl+12MV021+2lL-V02=Mgl+MV02lL
From SHM
V0=ωA=gL×Lθ0=gLθ0
Using this
W=Mgl+MgL×θ02×lLW=Mgl1+θ02

Topic: Physics, Work Power Energy, JEE Main
A person predicts the outcome of 20 cricket matches of his home team. Each match can result either in a win, loss or tie for the home team. The total number of ways in which he can make the predictions such that exactly 10 predictions are correct, are equal to

Options:
A 20C10⋅210
B 20C10⋅320
C 20C10⋅310
D 20C10⋅220

The correct answer is:
20C10⋅210

Solution:
In order to predict exactly 10 correct predictions,
⇒ we have to select 10 matches out of 20 in 20C10 ways
⇒ ways of making a correct prediction for each match is 1, so for 10 matches it is equal to 110=1.
⇒ Remaining 10 predictions must be wrong so for each incorrect prediction we have 2 choices out of 3, this can be done in 210 ways.
Hence, the required number of ways =20C101⋅210=20C10×210.

Topic: Mathematics, Permutation Combination, JEE Main
A person pushes a box on a rough horizontal plateform surface. He applies a force of 200 N over a distance of 15 m. Thereafter, he gets progressively tired and his applied force reduces linearly with distance to 100 N. The total distance through which the box has been moved is 30 m. What is the work done by the person during the total movement of the box?

Options:
A 3280 J
B 2780 J
C 5690 J
D 5250 J

The correct answer is:
5250 J

Solution:
W=area=(200×15)+12(100+200)×15

=3000+2250

W=5250 J

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2020 (04 Sep Shift 2)
A person sees his virtual image by holding a mirror very close to the face. When he moves the mirror away from his face, the image becomes inverted. What type of mirror he is using

Options:
A Plane mirror
B Convex mirror
C Concave mirror
D None of these

The correct answer is:
Concave mirror

Solution:
Plane mirror and convex mirror always forms erect images. Image formed by concave mirror may be erect or inverted depending on position of object.

Topic: Physics, Ray Optics, JEE Main
A person showing unpredictable moods. outbursts of emotion, quarrelsome behaviour and conflicts with others is suffering from:

Options:
A addictive disorders
B schizophrenia
C borderline personality disorder (BPD)
D mood disorders

The correct answer is:
borderline personality disorder (BPD)

Solution:
Borderline personality disorder is a serious psychiatric illness. It is a serious emotionally unstable personality disorder. The main feature of this disorder is a pervasive pattern of instability in interpersonal relationships, self-image and emotions. It is characterized by an unpredictable mood or mood swings, quarrelsome behaviour and conflicts.
So, the correct answer is 'Borderline personality disorder (BPD)'.

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2006
A person sitting firmly over a rotating stool has his arms stretched. If he folds his arms, his angular momentum about the axis of the rotation

Options:
A increases
B decreases
C remains unchanged
D doubles

The correct answer is:
remains unchanged

Solution:
According to conservation of angular momentum, when no external torque is applied on the rotating system, then its angular momentum remains unchanged.

Topic: Physics, Rotational Motion, COMEDK, COMEDK 2015