datavorous/entrance-exam-dataset huggingface dataset:
A man is walking on a straight line. The arithmetic mean of the reciprocals of the intercepts of this line on the coordinate axes is 14. Three stones A,B and C are placed at the points 1,1,2,2 and 4,4 respectively. Then which of these stones is/are on the path of the man?

Options:
A C only
B All the three
C B only
D A only

The correct answer is:
B only

Solution:
Let the line be y=mx+c
x - intercept : -cm
y - intercept : c

A.M of reciprocals of the intercepts :

-mc+1c2=14 ⇒ 21-m=c

line : y=mx+21-m=c

⇒ y-2-mx-2=0

⇒ line always passes through 2,2

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2021 (24 Feb Shift 1)
A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is 30°. After walking for 10 minutes from A in the same direction, at a point B, he observes that the angle of elevation of the top of the pillar is 60° . Then the time taken (in minutes) by him, from B to reach the pillar, is

Options:
A 20
B 5
C 6
D 10

The correct answer is:
5

Solution:
ΔQPA :hx+y=tan30°

⇒3h=x+y ......i
ΔQPB :hy=tan60°

⇒h= 3y ......ii
From equation i & ii, we get

3y=x+y

⇒y=x2
âµ Speed is uniform
Therefore, distance x walked by him in 10 min.
and distance x2 walked by him in 5 min.

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2016 (03 Apr)
A man is walking towards a vertical pillar in a straight path at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is 30o. After walking for 53+1 minutes from A in the same direction, at a point B, he observes that the angle of elevation of the top of the pillar is 45o. Then the time taken (in minutes) by him, to reach from B to the pillar, is (take 3=1.73)

Options:

The correct answer is:
18.65

Solution:
In ∆QPA: hx+y=tan⁡30o⇒3h=x+y…….(i)
In ∆QPB: hy=tan⁡45o⇒h=y…….(ii)
From (i) and (ii), we get,

3 y=x+y⇒y= x 3 −1
Given that the distance x is covered in 5 3 +1 minutes
Hence, the distance x 3 −1 is covered in 53+13−1=52+3=18.65 minutes

Topic: Mathematics, Heights and Distances, JEE Main
A man makes attempts to hit the target. The probability of hitting the target is $\frac{3}{5}$. Then the probability that $A$ hit the target exactly 2 times in 5 attempts, is

Options:
A $\frac{144}{625}$
B $\frac{72}{3125}$
C $\frac{216}{625}$
D None of these

The correct answer is:
$\frac{144}{625}$

Solution:
Probability of success
$(p)=\frac{3}{5} \Rightarrow q=1-p=\frac{2}{5}$
Hence the probability of 2 hits in 5 attempts
$={ }^5 C_2\left(\frac{3}{5}\right)^2\left(\frac{2}{5}\right)^3=\frac{144}{625}$.

Topic: Mathematics, Probability, JEE Main
A man (mass =50 kg ) and his son (mass =20 kg ) are standing on a frictionless surface facing each other. The man pushes his son so that he starts moving at a speed of 0.70 m s-1 with respect to the man. The speed of the man with respect to the surface is:

Options:
A 0.20 m s-1
B 0.14 m s-1
C 0.47 m s-1
D 0.28 m s-1

The correct answer is:
0.20 m s-1

Solution:
By conservation of linear momentum,

P→i=P→f

Where P→i and P→f are initial and final momentum of system (boy + man)

Now, 0=mbv→b+mmv→m......(i)

vb= velocity of boy w.r.t. ground

vm= velocity of man w.r.t. ground

v→bm= velocity of boy w.r.t. man

So, v→bm=v→b-v→m

or v→b=v→bm+v→m

or v→b=0.7+v→m ........(ii)

From equation (i) and (ii),

0=200.7+v→m+50 v→m

v→m=-1470=-0.20 m s-1

Negative sign shows velocity of man is opposite to boy.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A man (mass =50 kg ) and his son (mass =20 kg ) are standing on a frictionless surface facing each other. The man pushes his son so that he starts moving at a speed of 0.70 m s-1 with respect to the man. The speed of the man with respect to the surface is:

Options:
A 0.20 m s-1
B 0.14 m s-1
C 0.47 m s-1
D 0.28 m s-1

The correct answer is:
0.20 m s-1

Solution:
By conservation of linear momentum,

P→i=P→f

Where P→i and P→f are initial and final momentum of system (boy + man)

Now, 0=mbv→b+mmv→m......(i)

vb= velocity of boy w.r.t. ground

vm= velocity of man w.r.t. ground

v→bm= velocity of boy w.r.t. man

So, v→bm=v→b-v→m

or v→b=v→bm+v→m

or v→b=0.7+v→m ........(ii)

From equation (i) and (ii),

0=200.7+v→m+50 v→m

v→m=-1470=-0.20 m s-1

Negative sign shows velocity of man is opposite to boy.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2019 (12 Apr Shift 1)
A man of $50 \mathrm{~kg}$ is standing at one end on a boat of length $25 \mathrm{~m}$ and mass $200 \mathrm{~kg}$. If he starts running and when he reaches the other end, he has a velocity $2 \mathrm{~ms}^{-1}$ with respect to the boat. The final velocity of the boat is : (in $\mathrm{ms}^{-1}$ )

Options:
A $\frac{2}{5}$
B $\frac{2}{3}$
C $\frac{8}{5}$
D $\frac{8}{3}$

The correct answer is:
$\frac{2}{5}$

Solution:
Let $v$ be the velocity of the boat with respect to the water, then from conservation of linear momentum.
$(200+50) v+50 \times 2=50 \times 0+200 \times 0$
$250 v=-100$
$v=-\frac{100}{250}=-\frac{2}{5} \mathrm{~m} / \mathrm{s}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A man of 60 kg is running on the road and suddenly jumps into a stationary trolly car of mass 120 kg. Then the trolly car starts moving with velocity 2 m s-1. The velocity of the running man was _____ m s-1, when he jumps into the car.

Options:

The correct answer is:
6

Solution:
Applying conservation of the momentum,

mmu=mm+mtv⇒60u=60+120×2⇒u=6 m s-1

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2022 (28 Jun Shift 1)
A man of length $h$ requires a mirror, to see his own complete image of length at least equal to

Options:
A $\frac{h}{4}$
B $\frac{h}{3}$
C $\frac{h}{2}$
D $h$

The correct answer is:
$\frac{h}{2}$

Solution:
To see his own complete image, a man of height $\mathbf{h}$ would require a mirror of length at least equal to half his height, i.e., $=\mathrm{h} / \mathbf{2}$
. This is because light rays from the top and bottom of a person reflect off the mirror at equal angles. So, a mirror half the person's height will suffice to see the full reflection. However, the mirror must be positioned appropriately, with its top and bottom aligned with the person's height. Please note that this is a simplified explanation and actual results may vary depending on the observer's distance from the mirror and the angle of observation.

Topic: Physics, Ray Optics, JEE Main
A man of mass m starts falling towards a planet of mass $\mathrm{M}$ and radius $\mathrm{R}$. As he reaches near to the surface, he realizes that he will pass through a small hole in the planet. As he enters the hole, he sees that the planet is really made of two pieces a spherical shell of negligible thickness of mass $2 \mathrm{M} / 3$ and a point mass $\mathrm{M} / 3$ at the centre. Change in the force of gravity experienced by the man is

Options:
A $\frac{2}{3} \frac{\mathrm{GMm}}{\mathrm{R}^{2}}$
B 0
C $\frac{1}{3} \frac{\mathrm{GMm}}{\mathrm{R}^{2}}$
D $\frac{4}{3} \frac{\mathrm{GMm}}{\mathrm{R}^{2}}$

The correct answer is:
$\frac{2}{3} \frac{\mathrm{GMm}}{\mathrm{R}^{2}}$

Solution:
$\quad$ Change in force of gravity
$$
=\frac{\mathrm{GMm}}{\mathrm{R}^{2}}-\frac{\mathrm{G} \frac{\mathrm{M}}{3} \mathrm{~m}}{\mathrm{R}^{2}}
$$
(only due to mass $\mathrm{M} / 3$ due to shell gravitational field is zero (inside the shell))
$$
=\frac{2 \mathrm{GMm}}{3 \mathrm{R}^{2}}
$$

Topic: Physics, Gravitation, JEE Main
A man on the top of a vertical tower observes a car moving at a uniform speed towards the tower on a horizontal road. If it takes 18 min for the angle of depression of the car to change from 30° to 45°, then the time taken (in min) by the car to reach the foot of the tower is

Options:
A 923-1
B 181+3
C 183-1
D 91+3

The correct answer is:
91+3

Solution:
Let length of tower be h.

⇒AC′=AB=h

And AC=ABcot30°=3h

⇒CC′=3-1h

Time taken by car form C C'=18 min

⇒Time take by car to reach the foot of the tower

=183-1 min

=93+1 min

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2018 (16 Apr Online)
A man $P$ has 7 friends, 4 of them are ladies and 3 are men, His wife $Q$ also has 7 friends, 3 of them are ladies and 4 are men. Assume $P$ and $Q$ have no common friends. Then the total number of ways in which $P$ and $Q$ together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of $P$ and $Q$ are in this party, is

Options:
A 468
B 485
C 484
D 469

The correct answer is:
485

Solution:

Topic: Mathematics, Permutation Combination, JEE Main
A man projects a coin upwards from the gate of a uniformly moving train. The path of coin for the man will be

Options:
A Parabolic
B Inclined straight line
C Vertical straight line
D Horizontal straight line

The correct answer is:
Vertical straight line

Solution:
Because horizontal velocity is same for coin and the observer. So relative horizontal displacement will be zero.
\(\Rightarrow\) So the path will be vertical straight line.

Topic: Physics, Motion In Two Dimensions, JEE Main
A man running a race-course notes that the sum of the distances from the two flag posts from him is always $10 \mathrm{~m}$ and the distance between the flag posts is $8 \mathrm{~m}$. The equations of the path traced by the man is given by

Options:
A $\frac{x^{2}}{9}+\frac{y^{2}}{25}=1$
B $\frac{x^{2}}{9}+\frac{y^{2}}{16}=1$
C $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$
D $\frac{x^{2}}{16}+\frac{y^{2}}{25}=1$

The correct answer is:
$\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$

Solution:
Given that, $S P+S^{\prime} P=10$
$\Rightarrow \quad 2 a=10 \Rightarrow a=5$
and $\quad 2 a e=8$
$a e=4$
$b^{2}=a^{2}\left(1-e^{2}\right)$
$=a^{2}-(a e)^{2}$
$=25-16=9$
Hence, required equation is
$$
\frac{x^{2}}{25}+\frac{y^{2}}{9}=1
$$

Topic: Mathematics, Ellipse, JEE Main
A man runs towards mirror at a speed of $15 \mathrm{~m} / \mathrm{s}$. What is the speed of his image

Options:
A $7.5 \mathrm{~m} / \mathrm{s}$
B $15 \mathrm{~m} / \mathrm{s}$
C $30 \mathrm{~m} / \mathrm{s}$
D $45 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$15 \mathrm{~m} / \mathrm{s}$

Solution:
$15 \mathrm{~m} / \mathrm{s}$, in case of plane mirror speed of object =speed of image

Topic: Physics, Ray Optics, JEE Main
A man saves Rs. 200 in each of the first three months of his service. In each of the subsequent months his saving increases by Rs. 40 more than the saving of immediately previous month. His total saving from the start of service will be Rs. 11040 after

Options:
A
19 months

B
20 months

C
21 months

D
18 months

The correct answer is:
21 months

Solution:
$\begin{array}{ccccccc}1 & 2 & 3 & 4 & 5 & 6 & \ldots \ldots . \\ 200 & 200 & 200 & 240 & 280 & \ldots \ldots . & \ldots \ldots \ldots\end{array}$
Sum $=11040$
$120+80+160+40+200+240+\ldots=11040$
$\frac{n}{2}[2 a+(n-1) d]+80+40=11040$
$\frac{n}{2}[240+(n-1) 40]=10920$
$n[6+n-1]=546$
$n(n+5)=546$
$n=21$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2011
A man standing in a swimming pool looks at a stone lying at the bottom. The depth of the swimming pool is $h$. At what distance from the surface of water is the image of the stone formed (Line of vision is normal; Refractive index of water is $n$ )

Options:
A $h / n$
B $n / h$
C $h$
D $h n$

The correct answer is:
$h / n$

Solution:
$\mu=\frac{h}{h^{\prime}}$
$\Rightarrow \quad h^{\prime}=\frac{h}{n}$

Topic: Physics, Ray Optics, JEE Main
A man standing on a road has to hold his umbrella at $30^{\circ}$ with the vertical to keep the rain away. He throws the umbrella and starts running at $10 \mathrm{~km} / \mathrm{h}$. He finds that raindrops are hitting his head vertically. The actual speed of raindrops is :

Options:
A $20 \mathrm{~km} / \mathrm{h}$
B $10 \sqrt{3} \mathrm{~km} / \mathrm{h}$
C $20 \sqrt{3} \mathrm{~km} / \mathrm{h}$
D $10 \mathrm{~km} / \mathrm{h}$

The correct answer is:
$20 \mathrm{~km} / \mathrm{h}$

Solution:
When the man is at rest with respect to the ground, the rain comes to him at an angle $30^{\circ}$ with the vertical. This is the direction of the velocity of raindrops with respect to the ground.
Here, $\overrightarrow{\mathbf{v}}_{r, g}=$ velocity of the rain with respect to the ground
$\overrightarrow{\mathbf{v}}_{m, g}=$ velocity of the man with respect to the ground
and $\overrightarrow{\mathbf{v}}_{r, m}=$ velocity of the rain with respect to the man.
We have, $\quad \overrightarrow{\mathbf{v}}_{r, g}+\overrightarrow{\mathbf{v}}_{r, m}+\overrightarrow{\mathbf{v}}_{m, g}$ $\ldots$ (i)
Taking horizontal components, Eq. (i) gives
$v_{r, g} \sin 30^{\circ}=v_{m, g}=10 \mathrm{~km} / \mathrm{h}$
or $\quad v_{r, g}=\frac{10}{\sin 30^{\circ}}=20 \mathrm{~km} / \mathrm{h}$.

Topic: Physics, Motion In Two Dimensions, JEE Main
A man stands at one end of a boat which is stationary in water. Neglect water resistance. The man now moves to the other end of the boat and again becomes stationary. The centre of mass of the man plus boat system will remain stationary with respect to water

Options:
A In all cases
B Only when the man is stationary initially and finally
C Only if the man moves without acceleration on the boat
D Only if the man and the boat have equal masses.

The correct answer is:
In all cases

Solution:
There are no external horizontal forces acting on the man plus boat system. (The forces exerted by the man and the boat on each other are internal forces for the system) Therefore, the centre of mass of the system, which is initially at rest, will always be at rest.

Topic: Physics, Center of Mass Momentum and Collision, NEET
A man throws a ball of mass m from a moving plank (A). The ball has a horizontal component of velocity u=334 m s-1 with respect to ground towards another man standing on stationary plank (B), as shown in the figure. The combined mass of plank A with man as well as that of plank B with the other man is 2m each. the x-y plane represents the horizontal plane which is frictionless. If initially, plank A was moving with speed ui^ with respect to ground, then the relative velocity of plank B (in m s-1) with respect to plank A (after the man on plank B catches the ball) minus 10 m s-1 is

Options:

The correct answer is:
7

Solution:
By conservation of momentum,
3mui^=m(ucos37oi^+usin37oj^)+2mVAâ
=VAâ=11u10i^-3u10j^
Similarly =VBâ=4u15i^+3u10j^
â´VBA=17 m s-1

Topic: Physics, Motion In Two Dimensions, JEE Main
A man throws balls with the same speed vertically upwards one after the other at an interval of 2 seconds. What should be the speed of the throw so that more than two balls are in the sky at any time?

Options:
A more than $19.6 \mathrm{~m} / \mathrm{s}$
B at least $9.8 \mathrm{~m} / \mathrm{s}$
C any speed less than $19.6 \mathrm{~m} / \mathrm{s}$
D only with speed $19.6 \mathrm{~m} / \mathrm{s}$

The correct answer is:
more than $19.6 \mathrm{~m} / \mathrm{s}$

Solution:
From question Interval of ball thrown $=2 \mathrm{sec}$
If we want that minimum three balls remain in air then time of flight of first ball must be greater than 4 sec.
$\begin{aligned}
& T>4 \mathrm{sec} \\
& \therefore \frac{2 u}{g}>4 \mathrm{sec} \\
& \Rightarrow \quad u>19.6 \mathrm{~m} / \mathrm{s}
\end{aligned}$

Topic: Physics, Motion In One Dimension, JEE Main
A man throws balls with the same speed vertically upwards one after the other at an interval of 2 seconds. What should be the speed of the throw so that more than two balls are in the sky at any time?

Options:
A more than $19.6 \mathrm{~m} / \mathrm{s}$
B at least $9.8 \mathrm{~m} / \mathrm{s}$
C any speed less than $19.6 \mathrm{~m} / \mathrm{s}$
D only with speed $19.6 \mathrm{~m} / \mathrm{s}$

The correct answer is:
more than $19.6 \mathrm{~m} / \mathrm{s}$

Solution:
From question Interval of ball thrown $=2 \mathrm{sec}$
If we want that minimum three balls remain in air then time of flight of first ball must be greater than 4 sec.
$\begin{aligned}
& T>4 \mathrm{sec} \\
& \therefore \frac{2 u}{g}>4 \mathrm{sec} \\
& \Rightarrow \quad u>19.6 \mathrm{~m} / \mathrm{s}
\end{aligned}$

Topic: Physics, Motion In One Dimension, NEET, NEET 2003
A man tosses a coin 10 times, scoring 1 point for each head and 2 points for each tail. Let $\mathrm{P}(\mathrm{K})$ be the probability of scoring at least $\mathrm{K}$ points. The largest value of $\mathrm{K}$ such that $\mathrm{P}(\mathrm{K})>1 / 2$ is $-$

Options:
A 14
B 15
C 16
D 17

The correct answer is:
16

Solution:
Ways to make the sum $K$ is coefficient of $x^{K}$ in $\left(x+x^{2}\right)^{10}$
Coefficient of $x^{K}$ in $x^{10}(1+x)^{10}$
Coefficient of $x^{K-10}$ in $(1+x)^{10}$
Which is ${ }^{10} \mathrm{C}_{\mathrm{K}-10}$
So ways to make sum minimum $\mathrm{K}$ is
${ }^{10} \mathrm{C}_{\mathrm{K}-10}+{ }^{10} \mathrm{C}_{\mathrm{K}-9}+{ }^{10} \mathrm{C}_{\mathrm{K}-8}+\ldots . .+{ }^{10} \mathrm{C}_{10}$
Probability
$\begin{aligned}
\mathrm{P}(\mathrm{K}) &=\frac{{ }^{10} \mathrm{C}_{\mathrm{K}-10}+{ }^{10} \mathrm{C}_{\mathrm{K}-9}+\ldots \ldots \ldots \ldots . .+{ }^{10} \mathrm{C}_{10}}{2^{10}} \\
\mathrm{P}(\mathrm{K}) &=\frac{2^{10}-\left({ }^{10} \mathrm{C}_{0}+\ldots \ldots+{ }^{10} \mathrm{C}_{\mathrm{K}-11}\right)}{2^{10}} \\
&=1-\frac{{ }^{10} \mathrm{C}_{0}+\ldots \ldots . .+{ }^{10} \mathrm{C}_{\mathrm{K}-11}}{2^{10}}>\frac{1}{2}
\end{aligned}$
But $\mathrm{K}$ should be maximum so ${ }^{10} \mathrm{C}_{\mathrm{K}-11}={ }^{10} \mathrm{C}_{5}$ (middle value) So that ${ }^{10} \mathrm{C}_{0}+\ldots .+{ }^{10} \mathrm{C}_{\mathrm{K}-11}$ is $\max$ So $\mathrm{K}=16$

Topic: Mathematics, Probability, JEE Main
A man wants to distribute 101 coins of a rupee each, among his 3 sons with the condition that no one receives more money than the combined total of the other two. The number of ways of doing this is equal to

Options:
A 103C2-352C2
B 103C23
C 103C26
D 103C2-350C3

The correct answer is:
103C2-352C2

Solution:
Let, the amount received by the sons be Rs. x, Rs. y, and Rs z respectively, then

xâ¤y+z=101-xâ2xâ¤101 .
âxâ¤50, yâ¤50, zâ¤50.
x+y+z=101
So, the required number of ways is the coefficient of x101 in 1+x+x2+....+x513
or coefficient of x101 in the expansion of 1-x501-x3

or coefficient of x101 in the expansion of 1-x5131-x-3
or coefficient of x101 in the expansion of 1-3C1x51+.....1-x-3
or coefficient of x101 in the expansion of1-x-3-3C1x511-x-3 =1Ã103C101-352C50=103C2-352C2

Topic: Mathematics, Permutation Combination, JEE Main
A man wants to reach a certain destination. One-sixth of the total distance is muddy while half the distance is tar road. For the remaining distance he takes a boat. His speed of traveling in mud, in water, on tar road is in the ratio $3: 4: 5$. The ratio of the durations he requires to cross the patch of mud, stream and tar road is

Options:
A $\frac{1}{2}: \frac{4}{3}: \frac{5}{2}$
B $3: 8: 15$
C $10: 15: 18$
D $1: 2: 3$

The correct answer is:
$10: 15: 18$

Solution:
Let distance is $6 \mathrm{x}$
$\begin{array}{llll} & \text { mud }: & \text { tar } & \text { : } & \text { stream } \\ \text { distance } & x & : & 3 x & : & 2 x \\ \text { speed } & 3 v & : & 5 v & : & 4 v \\ \text { time } & \frac{x}{3 v} & : & \frac{3 x}{5 v} & : & \frac{2 x}{4 v} \\ & 10 & : & 18 & : & 15 & \end{array}$

Topic: Mathematics, Basic of Mathematics, KVPY, KVPY 2011 (SA)
A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them are ladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in this party is:

Options:
A 485
B 468
C 469
D 484

The correct answer is:
485

Solution:
Here, m - men and L - ladies
Total=9×16+1+9×36+16
=144+1+324+16
=485

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2017 (02 Apr)
A manometer connected to a close tap reads 3.5Ã105 N m-2. When the valve is opened, the reading of manometer falls to 3.0Ã105 N m-2, then velocity of flow of water is

Options:
A 100 m s-1
B 10 m s-1
C 1 m s-1
D 1010 m s-1

The correct answer is:
10 m s-1

Solution:
Bernoulli's theorem for unit mass of liquid
PÏ+12v2 = constant
As the liquid starts flowing , it pressure energy decreases
12v2=P1-P2Ï
â¹ 12v2=3.5Ã105-3Ã105103
â¹ v2=2Ã0.5Ã105103
â¹v2=100
â¹v=10 m s-1

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A Mapping from IN to IN is defined as follows :
$$
\begin{aligned}
& \mathrm{f}: \mathrm{IN} \rightarrow \mathrm{IN} \\
& \mathrm{f}(\mathrm{n})=(\mathrm{n}+5)^2, \mathrm{n} \in \mathrm{IN}
\end{aligned}
$$
(IN is the set of natural numbers). Then

Options:
A $\mathrm{f}$ is not one-to-one
B $\mathrm{f}$ is onto
C $\mathrm{f}$ is both one-to-one and onto
D $\mathrm{f}$ is one-to-one but not onto

The correct answer is:
$\mathrm{f}$ is one-to-one but not onto

Solution:
$$
\begin{aligned}
& \text { Hints: } \mathrm{f}: \mathrm{IN} \rightarrow \mathrm{IN} ; \mathrm{f}(\mathrm{n})=(\mathrm{n}+5)^2 \\
& \left(\mathrm{n}_1+5\right)^2=\left(\mathrm{n}_2+5\right)^2 \\
& \Rightarrow\left(\mathrm{n}_1-\mathrm{n}_2\right)\left(\mathrm{n}_1+\mathrm{n}_2+10\right)=0 \\
& \Rightarrow \mathrm{n}_1=\mathrm{n}_2 \rightarrow \text { one-to-one }
\end{aligned}
$$
There does not exist $n \in$ IN such that $(n+5)^2=1$
Hence $\mathrm{f}$ is not onto

Topic: Mathematics, Functions, JEE Main
A marble block of mass $2 \mathrm{~kg}$ lying on ice when given a velocity of $6 \mathrm{~m} / \mathrm{s}$ is stopped by friction in $10 \mathrm{~s}$. Then the coefficient of friction is

Options:
A
$0.02$

B
$0.03$

C
$0.04$

D
None of these

The correct answer is:
None of these

Solution:
$a=\mu g=\frac{6}{10}$ [ using $\mathrm{v}=\mathrm{u}+\mathrm{at}$ ]
$\Rightarrow \mu=\frac{6}{10 \times g}=\frac{6}{10 \times 10}=0.06$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2003
A marble of mass $0.1 \mathrm{~kg}$ falls off a bed with a height of $1.2 \mathrm{~m}$. What is the impulse on the marble as it hits the ground?
$g=10 \frac{m}{s^2}$

Options:
A 1.22 N-s
B 0.84 N-s
C 0.49 N-s
D 1.74 N-s

The correct answer is:
0.49 N-s

Solution:
Impulse can be written as either of two popular expressions:
$I=F \Delta t=m \Delta v$
From the problem statement, we can determine the velocity of the marble as it hits the floor, allowing us to use the latter expression. To determining the velocity of the marble, we can use the equation for conservation of energy:
$E=U_i+K_i=U_f+K_f$
Assuming the final height is zero, we can eliminate initial kinetic energy and final potential energy. Therefore, we can write:
$m g h_i=\frac{1}{2} m v_f^2$
Canceling out mass and rearranging for final velocity, we get:
$v_f=\sqrt{2 g h_i}$
We know these variables, allowing us to solve for the velocity:
$v_f=\sqrt{2\left(10 \frac{m}{s^2}\right)(1.2 m)}=4.9 \frac{\mathrm{m}}{\mathrm{s}}$
Plugging this value into the expression for impulse, we get:
$I=(0.1 \mathrm{~kg})\left(4.9 \frac{\mathrm{m}}{\mathrm{s}}\right)=0.49 \mathrm{~N} \cdot \mathrm{s}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A mass $1 \mathrm{~kg}$ is suspended by a thread. It is
(i) lifted up with an acceleration $4.9 \mathrm{~m} / \mathrm{s}^2$
(ii) lowered with an acceleration $4.9 \mathrm{~m} / \mathrm{s}^2$
The ratio of the tensions is

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

The correct answer is:
$3: 1$

Solution:
$\begin{aligned} & T_1=m(g+a)=1 \times\left(g+\frac{g}{2}\right)=\frac{3 g}{2} \\ & T_2=m(g-a)=1 \times\left(g-\frac{g}{2}\right)=\frac{g}{2} \quad \therefore \quad \frac{T_1}{T_2}=\frac{3}{1}\end{aligned}$

Topic: Physics, Laws of Motion, JEE Main