datavorous/entrance-exam-dataset huggingface dataset:
A bomb of $12 \mathrm{~kg}$ explodes into two pieces of masses $4 \mathrm{~kg}$ and $8 \mathrm{~kg}$. The velocity of $8 \mathrm{~kg}$ mass is 6 $\mathrm{m} / \mathrm{sec}$. The kinetic energy of the other mass is

Options:
A $48 \mathrm{~J}$
B $32 \mathrm{~J}$
C $24 \mathrm{~J}$
D $288 \mathrm{~J}$

The correct answer is:
$288 \mathrm{~J}$

Solution:
As the initial momentum of bomb was zero, therefore after explosion two parts should possess numerically equal momentum
i.e. $m_A v_A=m_B v_B \Rightarrow 4 \times v_A=8 \times 6 \Rightarrow v_A=12 \mathrm{~m} / \mathrm{s}$
$\therefore$ Kinetic energy of other mass $A_1=\frac{1}{2} m_A v_A^2$
\(K=\frac{1}{2} 4 \times 144=288 \mathrm{~J}\)

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A bomb of mass $16 \mathrm{~kg}$ at rest explodes into two pieces of masses of $4 \mathrm{~kg}$ and $12 \mathrm{~kg}$. The velocity of the $12 \mathrm{~kg}$ mass is $4 \mathrm{~ms}^{-1}$. The kinetic energy of the other mass is

Options:
A
$96 \mathrm{~J}$

B
$144 \mathrm{~J}$

C
$288 \mathrm{~J}$

D
$192 \mathrm{~J}$

The correct answer is:
$288 \mathrm{~J}$

Solution:
$m_1 v_1=m_2 v_2$
$\mathrm{KE}=\frac{1}{2} \mathrm{~m}_2 \mathrm{v}_2^2=\frac{1}{2} \times 4 \times 144=288 \mathrm{~J}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2006
A bomb of mass 30 kg at rest explodes into two pieces of masses 18 kg and 12 kg. The velocity of 18 kg mass is $6 \mathrm{~ms}^{-1}$. The kinetic energy of the other mass is

Options:
A 256 J
B 486 J
C 524 J
D 324 J

The correct answer is:
486 J

Solution:
The linear momentum of exploding part will remain conserved.
Applying conservation of linear momentum,
we write,
$m_1 u_1=m_2 u_2$
Here,
$\begin{gathered}m_1=18 \mathrm{~kg}, m_2=12 \mathrm{~kg} \\ u_1=6 \mathrm{~ms}^{-1}, u_2=?\end{gathered}$
$\therefore 18 \times 6=12 u_2$
$\Rightarrow u_2=\frac{18 \times 6}{12} = 9 \mathrm{~ms}^{-1}$
Thus, kinetic energy of 12 kg mass
$\begin{aligned} K_2 & =\frac{1}{2} m_1 u_2^2 \\ & =\frac{1}{2} \times 12 \times(9)^2 \\ & =6 \times 81 \\ & =486 \mathrm{~J}\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, JIPMER, JIPMER 2007
A bomber plane moves horizontally with a speed of 500 m s-1 and a bomb released from it, strikes the ground in 10 s. The angle at which it strikes the ground will be (g=10ms-2)

Options:
A tan-115
B tan-1 115
C tan-1(1)
D tan-1(5)

The correct answer is:
tan-115

Solution:
Horizontal velocity remains some v x = 500 ms - 1
vertical velocity v y = 0 + 10 t
v y = 100 ms - 1

tan θ = 100 500
θ = tan - 1 1 5

Topic: Physics, Motion In Two Dimensions, NEET
$\mathrm{A} \sigma$ bonded molecule $\mathrm{MX}_{3}$ is T-shaped. The number of non bonded pair of electrons is

Options:
A 0
B 2
C 1
D can be predicted only if atomic number is known

The correct answer is:
2

Solution:
For T-shape geometry the molecule must have 3 bonded pair and 2 lone pair of electrons.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
A book of weight 20 N is pressed between two hands and each hand exerts a force of 40 N. If the book just starts to slide down. What is the value of the coefficient of friction?

Options:

The correct answer is:
0.25

Solution:
Block is in horizontal equilibrium. In the vertical direction, it is in limiting equilibrium.
μs=FsR=mgR=2040=0.5
mg=2fr1=2μF
μ=mg2F=202×40=0.25

Topic: Physics, Laws of Motion, JEE Main
A bottle has an opening of radius a and length b. A cork of length b and radius a+âa where âaâªa, is compressed to fit into the opening completely (see figure). If the bulk modulus of cork is B and the coefficient of friction between the bottle and cork is Î¼, then the force needed to push the cork into the bottle is

Options:
A πμBba
B 2πμBb∆a
C πμBb∆a
D 4πμBb∆a

The correct answer is:
4πμBb∆a

Solution:
Pressure=NA=N2πab
⇒ stress=B×strain
N2πab=B2πa∆a×b πa2b
⇒N=B4πb△a
f=μN=μ4πbBΔa

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2016 (10 Apr Online)
A bottle has an opening of radius a and length b. A cork of length b and radius a+âa where âaâªa, is compressed to fit into the opening completely (see figure). If the bulk modulus of cork is B and the coefficient of friction between the bottle and cork is Î¼, then the force needed to push the cork into the bottle is

Options:
A πμBba
B 2πμBb∆a
C πμBb∆a
D 4πμBb∆a

The correct answer is:
4πμBb∆a

Solution:
Pressure=NA=N2πab
⇒ stress=B×strain
N2πab=B2πa∆a×b πa2b
⇒N=B4πb△a
f=μN=μ4πbBΔa

Topic: Physics, Mechanical Properties of Solids, JEE Main
A bottle has an opening of radius a and length b. A cork of length b and radius a+âa where âaâªa, is compressed to fit into the opening completely (see figure). If the bulk modulus of cork is B and the coefficient of friction between the bottle and cork is Î¼, then the force needed to push the cork into the bottle is

Options:
A πμBba
B 2πμBb∆a
C πμBb∆a
D 4πμBb∆a

The correct answer is:
4πμBb∆a

Solution:
Pressure=NA=N2πab
⇒ stress=B×strain
N2πab=B2πa∆a×b πa2b
⇒N=B
f=μN=μ4πbBΔa

Topic: Physics, Mechanical Properties of Solids, JEE Main
A bowl filled with very hot soup cools from 98°C to $86^{\circ} \mathrm{C}$ in 2 minutes when the room temperature is $22^{\circ} \mathrm{C}$. How long it will take to cool from $75^{\circ} \mathrm{C}$ to $69^{\circ} \mathrm{C}$ ?

Options:
A

2 minute

B 1.4 minute
C 0.5 minute
D 1 minute

The correct answer is:
1.4 minute

Solution:
From Newton's law of cooling, we have

ΔQΔt=-KTavg-T0

Where, Q=ms∆T, m is the mass, s is the specific heat and ∆T is the temperature change.

Case 1:

∆t=2 minutes, T0=22°C

ms×122=-K98+862-22

6=-Kms70 ...1

Case 2:

ms×6Δt=-K75+692-22

6Δt=-Kms(50) ...2

Dividing (2) by (1), we get

⇒6Δt(6)=5070

⇒Δt=75=1.4 min

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2023 (25 Jan Shift 1)
A box A contains 2 white, 3 red and 2 black balls. Another box B contains 4 white, 2 red and 3 black balls. If two balls are drawn at random, without replacement from a randomly, selected box and one ball turns out to be white while the other ball turns out to be red, then the probability that both balls are drawn from box B is :

Options:
A 78
B 916Â
C 716
D 932Â

The correct answer is:
716

Solution:
Probability that box A is selected PA=12

Probability that box B is selected PB=12

Let E be the event that one ball is white while the other is red.

∴ PE=PA·PE/A+PB·PE/B

=122·37C2+4·29C2=12621+836=1227+29=1663

Thus, PB/E=PB PE/BPE=191663=716

Topic: Mathematics, Probability, JEE Main, JEE Main 2018 (15 Apr)
A box ' $A$ ' contanis 2 white, 3 red and 2 black balls. Another box ' $B^{\prime}$ contains 4 white, 2 red and 3 black balls. If two balls are drawn at random, without replacement, from a randomly selected box and one ball turns out to be white while the other ball turns out to be red, then the probability that both balls are drawn from box ' $B^{\prime}$ is

Options:
A
$\frac{7}{16}$

B
$\frac{9}{32}$

C
$\frac{7}{8}$

D
$\frac{9}{16}$

The correct answer is:
$\frac{7}{16}$

Solution:
Probability of drawing a white ball and then a red ball from bag $B$ is given by
$$
\frac{{ }^4 C_1 \times{ }^2 C_1}{{ }^9 C_2}=\frac{2}{9}
$$
Probability of drawing a white ball and then a red ball from bag $A$ is given by
$$
\frac{{ }^2 C_1 \times{ }^3 C_1}{{ }^7 C_2}=\frac{2}{7}
$$
Hence, the probability of drawing a white ball and then a red ball from bag $B$
$$
=\frac{\frac{2}{9}}{\frac{2}{7}+\frac{2}{9}}=\frac{2 \times 7}{18+14}=\frac{7}{16}
$$

Topic: Mathematics, Probability, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
A box contains 1 black and 1 white ball. A ball is drawn randomly and replaced in the box with an additional ball of the same colour, then a second ball is drawn randomly from the box containing 3 balls. The probability that the first drawn ball was white given that at least one of the two balls drawn was white is

Options:
A 12
B 34
C 45
D 511

The correct answer is:
34

Solution:
B→ first ball drawn is white
A→ At least one of two balls drawn are white
PA=WW+WB+BW
=12×23+12×13+12×13=46=23
=PBA=PB∩APA=12×23+12×1323=1223=34

Topic: Mathematics, Probability, JEE Main
A box contains 3 coins B1,B2,B3 and the probability of getting heads on the coins are 12,14,18 respectively. If one of the coins is selected at random and tossed for 3 times and exactly 3 heads appeared, then the probability that it was coin B1 is

Options:
A 973
B 1073
C 3673
D 6473

The correct answer is:
6473

Solution:
Aâ selected coin got exactly 3 heads
B1â coin B1 is selected
B2â coin B2 is selected
B3â coin B3 is selected
PB1=PB2=PB3=13
Using Bayes Theorem,
P B 1 A = P Aâ© B 1 P A = P B 1 âP A B 1 P B 1 P A B 1 +P B 2 P A B 2 +P B 3 P A B 3
=13Ã12313Ã123+13Ã143+13183=6464+8+1=6473

Topic: Mathematics, Probability, JEE Main
A box contains 6 distinct dolls. From this box, 3 dolls are randomly selected one by one with replacement. What is the probability of selecting 3 distinct dolls?

Options:
A $5 / 54$
B $12 / 25$
C $1 / 20$
D $5 / 9$

The correct answer is:
$5 / 9$

Solution:
Required probability $=\frac{{ }^{6} \mathrm{C}_{1} \times{ }^{5} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{1}}{{ }^{6} \mathrm{C}_{1} \times{ }^{6} \mathrm{C}_{1} \times{ }^{6} \mathrm{C}_{1}}$
$=\frac{6 \times 5 \times 4}{6 \times 6 \times 6}=\frac{5}{9}$

Topic: Mathematics, Probability, JEE Main
A box contains 6 nails and 10 nuts. Half of the nails and half of the nuts are rusted. If one item is chosen at random, what is the probability that it is rusted or is a nail

Options:
A $\frac{3}{16}$
B $\frac{5}{16}$
C $\frac{11}{16}$
D $\frac{14}{16}$

The correct answer is:
$\frac{11}{16}$

Solution:
Total rusted items $=3+5=8$; unrusted nails $=3$
$\therefore$ Required probability $=\frac{3+8}{6+10}=\frac{11}{16}$.

Topic: Mathematics, Probability, JEE Main
A box contains 6 pens, 2 of which are defective. Two pens are taken randomly from the box. If random variable x: Number of defective pens obtained, then standard deviation of x=

Options:
A ± 43√5
B 83
C 1645
D 435

The correct answer is:
435

Solution:
x : no. of defective pens
Two pens are taken from box
â´ x can taken values 0, 1, 2
px=0= 4C2 6C2=4Ã36Ã5=25=615
p x=1= 2C1Ã 4C1 6C2=2Ã4Ã2Ã16Ã5=815
px=2= 2C2 6C2=1Ã2Ã16Ã5=115

x	P	xipi	xi2pi

0	615	0	0

1	815	815	815

2	115	215	415

Ex=1015
=23
Ex2=1215
=45
Standard deviation =Ex2-Ex2
Standard deviation =45-232
=45-49
=4Ã445
=43â5

Topic: Mathematics, Probability, JEE Main
A box contains 9 slips bearing numbers -3, -2, -1, 0, 1, 2, 3, 4 and 5. An experiment consists of drawing a slip from this box and replacing it back in the box after noting the number. This experiment is repeated 9 times. These 9 numbers are now chosen as elements of a 3Ã3 matrix, then the probability that the matrix is skew symmetric is

Options:
A 196
B 34399
C 199
D 197

The correct answer is:
34399

Solution:
Total cases =99
All diagonal elements are zero and aij=-aji
So favourable cases =73
Hence, required probability =7399

Topic: Mathematics, Probability, JEE Main
A box contains 10 identical electronic components of which 4 are defective. If 3 components are selected at random from the box in succession, without replacing the units already drawn, what is the probability that two of the selected components are defective?

Options:
A $1 / 5$
B $5 / 24$
C $3 / 10$
D $1 / 40$

The correct answer is:
$3 / 10$

Solution:
Total number of selecting 3 components out of $10=$ ${ }^{10} \mathrm{C}_{3}$. Out of 3 selected components two defective pieces can be selected in ${ }^{4} \mathrm{C}_{2}$ ways and one non-defective piece will be selected in ${ }^{6} \mathrm{C}_{1}$ ways, hence.
Required probability $=\frac{{ }^{6} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{2}}{{ }^{10} \mathrm{C}_{3}}=\frac{6 \times 6 \times 6}{10 \times 9 \times 8}=\frac{3}{10}$

Topic: Mathematics, Probability, JEE Main
A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn, one-by-one, with replacement, then the variance of the number of green balls drawn is:

Options:
A 125
B 6
C 4
D 625

The correct answer is:
125

Solution:
For Bernoulli's trials

Variance =n.p.q
Given, number of trials n=10
Probability of drawing a green ball p=1525=35 and q=25
Hence, Variance =10×35×25=125

Topic: Mathematics, Statistics, JEE Main, JEE Main 2017 (02 Apr)
A box contains 30 toys of same size in which 10 toys are white and all the remaining toys are blue. A toy is drawn at random from the box and it is replaced in the box after noting down its colour. If 5 toys are drawn in this way, then the probability of getting atmost 2 white toys is

Options:
A $\left(\frac{6}{9}\right)^2$
B $\left(\frac{8}{9}\right)^2$
C $\left(\frac{7}{9}\right)^2$
D $\left(\frac{2}{3}\right)^5$

The correct answer is:
$\left(\frac{8}{9}\right)^2$

Solution:
Required probability
$=$ No white ball + one white ball + two white ball
$$
\begin{aligned}
& =\left(\frac{2}{3}\right)^5+{ }^5 C_1\left(\frac{2}{3}\right)^4\left(\frac{1}{3}\right)^1+{ }^5 C_2\left(\frac{2}{3}\right)^3\left(\frac{1}{3}\right)^2 \\
& =\left(\frac{1}{3}\right)^5\left[2^5+5 \cdot 2^4+10 \cdot 2^3\right] \\
& =\left(\frac{1}{3}\right)^5[32+80+80]=\frac{192}{3^5}=\frac{64 \times 3}{3^5} \\
& =\frac{64}{3^4}=\frac{8 \times 8}{3^2 \times 3^2}=\left(\frac{8}{9}\right)^2
\end{aligned}
$$

Topic: Mathematics, Probability, AP EAMCET, AP EAMCET 2019 (21 Apr Shift 1)
A box contains tickets numbered 1 to N. n tickets are drawn from the box with replacement. The probability that the largest number on the tickets is k, is

Options:
A k N n
B k - 1 N n
C 0
D None of these

The correct answer is:
None of these

Solution:
Number of ways of selecting n tickets from N tickets numbered from 1 to N with replacement =Nn
Now, the largest number on selected tickets must be k. Hence, favorable cases =kn−(k−1)n
Therefore, required probability =kn−(k−1)nNn

Topic: Mathematics, Probability, JEE Main
A box contains two white, three black and four red balls. The number of ways of selecting 3 balls from the box with atleast one black ball is

Options:
A 64
B 74
C 54
D 84

The correct answer is:
64

Solution:
Number of ways of selecting 3 balls out of 9 balls without any restriction is 9C3.
Number of ways of selecting 3 balls such that no black ball is selected is 6C3.

Number of ways such that atleast one black ball is selected is C 9 3 â C 6 3 =84â20=64

Topic: Mathematics, Permutation Combination, JEE Main
A box contains x red balls and 10 black balls. 3 balls are drawn one by one without replacement. If the probability of choosing 3 red balls is equal to the probability of choosing 2 red and 1 black ball, then the possible value of x can be

Options:
A 1
B 32
C 53
D 40

The correct answer is:
32

Solution:
Probability of choosing 3 red balls =xC310+xC3
Probability of choosing 2 red balls and one black ball =xC2 10C110+xC3
So, xC310+xC3=xC2 10C110+xC3⇒xC3=xC2 10C1
⇒xC3xC2=10⇒x-3+13=10⇒x=32

Topic: Mathematics, Probability, JEE Main
A box filled with water has a small hole on its side near the bottom. It is dropped from the top of a tower. As it falls, a camera attached on the side of the box records the shape of the water stream coming out of the hole. The resulting video will show

Options:
A the water coming down forming a parabolic stream

B the water going up forming a parabolic stream

C the water coming out in a straight line
D no water coming out

The correct answer is:
no water coming out

Solution:
When box with hole is in free fall, both water and box cover equal distance downwards in equal time. Hence, no water comes out of hole in free fall of box.

Topic: Physics, Motion In One Dimension, JEE Main
A box has ten chits numbered $0,1,2,3, \ldots$, 9. First, one chit is drawn at random and kept aside. From the remaining, a second chit is drawn at random. What is the probability that the second chit drawn is " 9 "?

Options:
A $\frac{1}{10}$
B $\frac{1}{9}$
C $\frac{1}{90}$
D None of the above

The correct answer is:
$\frac{1}{90}$

Solution:
First chit can be drawn in 10 ways. Second chit can be drawn in 9 ways. Total number of events, $\mathrm{n}(\mathrm{S})=10 \times 9=90$. Number of events of drawing chit numbered $9=\mathrm{n}(\mathrm{E})=1$
Probability $=\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{1}{90}$

Topic: Mathematics, Probability, NDA, NDA 2018 (Phase 1)
A box is lying on an inclined plane what is the coefficient of static friction if the box starts sliding when an angle of inclination is $60^{\circ}$

Options:
A 1.173
B 1.732
C 2.732
D 1.677

The correct answer is:
1.732

Solution:
$\mu=\tan ($ Angle of repose $)=\tan 60^{\circ}=1.732$

Topic: Physics, Laws of Motion, JEE Main
A box is moved along a straight line by a machine delivering constant power. The distance moved by the body in time \(\mathrm{t}\) is proportional to

Options:
A \(\mathrm{t}^{\frac{1}{2}}\)
B \(t^{\frac{3}{4}}\)
C \(\mathrm{t}^{\frac{3}{2}}\)
D \(\mathrm{t}^2\)

The correct answer is:
\(\mathrm{t}^{\frac{3}{2}}\)

Solution:
\(\begin{aligned}
& \text {Hints: } \mathrm{P}=\mathrm{Fv}=\mathrm{m} \cdot \frac{\mathrm{dv}}{\mathrm{dt}} \cdot \mathrm{v} \\
& \int \mathrm{vdv}=\int \mathrm{P} / \mathrm{mdt} ; \quad \frac{\mathrm{v}^2}{2}=\frac{\mathrm{Pt}}{\mathrm{m}} \\
& V=\sqrt{\frac{2 p}{m}} \mathrm{t}^{\frac{1}{2}} ; \frac{\mathrm{dx}}{\mathrm{dt}}=\sqrt{\frac{2 \mathrm{p}}{\mathrm{m}}} \mathrm{t}^{\frac{1}{2}} \\
& \int d x=\sqrt{\frac{2 p}{m}} \int t^{\frac{1}{2}} d t ; \quad x=\sqrt{\frac{2 p}{m}} \frac{t^{\frac{3}{2}}}{\frac{3}{2}}=\frac{2}{3} \sqrt{\frac{2 p}{m}} \mathrm{t}^{\frac{3}{2}} \\
& \mathrm{x} \alpha \mathrm{t}^{\frac{3}{2}} \\
&
\end{aligned}\)

Topic: Physics, Work Power Energy, JEE Main
A box is put on a scale which is adjusted to read zero , when the box is empty. A stream of pebbles is then poured into the base from a height h above its bottom at a rate of n pebbles/s. Each pebble has a mass m. If the pebbles collide with the box such that they immediately come to rest after collision, then the scale reading at time t after the pebbles begin to fill the box is [ neglect piling up of pebbles]

Options:
A mnt
B mn 2gh+ gt
C mngt
D zero

The correct answer is:
mn 2gh+ gt

Solution:
Scale reading will be the sum of weight and force due to their impulse
Scale reading = v ∆m∆t + mnt g
= 2ghmn + mng t
= mn 2gh+ gt

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A box of mass $2 \mathrm{kg}$ is placed on the roof of a car. The box would remain stationary until the car attains a maximum acceleration. Coefficient of static friction between the box and the roof of the car is 0.2 and $g=10 \mathrm{ms}^{-2}$ This maximum acceleration of the car, for the box to remain stationary, is

Options:
A $8 \mathrm{ms}^{-2}$
B $6 \mathrm{ms}^{-2}$
C $4 \mathrm{ms}^{-2}$
D $2 \mathrm{ms}^{-2}$

The correct answer is:
$2 \mathrm{ms}^{-2}$

Solution:
Given, $m=2 \mathrm{kg}, \mu=0.2$ and $g=10 \mathrm{ms}^{-2}$
Here, $\quad m a=\mu m g$
$$
\begin{array}{l}
a=\mu g \\
a=0.2 \times 10 \\
a=2 \mathrm{m} / \mathrm{s}^{2}
\end{array}
$$

Topic: Physics, Laws of Motion, WBJEE, WBJEE 2012
A box weighs 196N on a spring balance at the north pole. Its weight recorded on the same balance if it is shifted to the equator is close to (Take g=10ms-2 at the north pole and the radius of the earth =6400km ):

Options:
A 195.66N
B 194.32N
C 194.66N
D 195.32N

The correct answer is:
195.32N

Solution:
At pole, weight mg=196
m=19.6kg
At equator, weight =mg-mω2R
=196-19.62π24×36002×6400×103
=195.32N

Topic: Physics, Gravitation, JEE Main, JEE Main 2020 (07 Jan Shift 2)
A box, when hung from a spring balance shows a reading of $50 \mathrm{~kg}$. If the same box is hung from the same spring balance inside an evacuated chamber, the reading on the scale will be

Options:
A $50 \mathrm{~kg}$ because the mass of the box remains unchanged
B $50 \mathrm{~kg}$ because the effect of the absence of the atmosphere will be indentical on the box and the spring balance
C Less than $50 \mathrm{~kg}$ because the weight of the column of air on the box will be absent
D More than $50 \mathrm{~kg}$ because the atmospheric buoyancy force will be absent

The correct answer is:
More than $50 \mathrm{~kg}$ because the atmospheric buoyancy force will be absent

Solution:
Sol. No Bnoycncy force in vaccum

Topic: Physics, Laws of Motion, KVPY, KVPY 2014 (SA)
A boy and a man carry a uniform rod of length L horizontally in such a way that the boy gets 14th of the load. If the boy is at one end of the rod, the distance of the man from the other end is

Options:
A L3
B L4
C 2L3
D 3L4

The correct answer is:
L3

Solution:
Let normal reaction on the rod due to boy is N1 and due to man is N2
Given, N1=W4
∴ N2=W-W4=3W4
Let x be the distance of the man from the other end.

For rotational equilibrium net torque must be zero.
⇒N1×L2=N2×L2-x
⇒ W4×L2=3W4×L2-x
⇒ x=L3

Topic: Physics, Rotational Motion, JEE Main
A boy can throw a stone up to a maximum height of $10 \mathrm{~m}$. The maximum horizontal distance that the boy can throw the same stone up to will be

Options:
A
$20 \sqrt{2} \mathrm{~m}$

B
$10 \mathrm{~m}$

C
$10 \sqrt{2} \mathrm{~m}$

D
$20 \mathrm{~m}$

The correct answer is:
$20 \mathrm{~m}$

Solution:
maximum vertical height $=\frac{u^2}{2 g}=10 \mathrm{~m}$
Horizontal range of a projectile $=\frac{u^2 \sin 2 \theta}{\mathrm{g}}$
Range is maximum when $\theta=45^{\circ}$
Maximum horizontal range $=\frac{\mathrm{u}^2}{\mathrm{~g}}$
Hence maximum horizontal distance $=20 \mathrm{~m}$.

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2012 (Offline)
A boy is rolling a 0.5 kg ball on the frictionless floor with the speed of 20 m s-1. The ball gets deflected by an obstacle on the way. After deflection it moves with 5% of its initial kinetic energy. What is the speed of the ball now?

Options:
A 19.0 m s-1
B 4.4 m s-1
C 14.41 m s-1
D 1.00 m s-1

The correct answer is:
4.4 m s-1

Solution:
Given, m=0.5 kg and u=20 m s-1

Initial kinetic energy ki=12mu2

=12×0.5×20×20=100 J

After deflection it moves with 5% of ki

∴kf=5100×ki⇒5100×100

⇒kf=5 J

Now, let the final speed be 'v' m s-1, then:

kf=5=12mv2

⇒v2=20

⇒v=20=4.47 m s-1

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2021 (17 Mar Shift 1)
A boy is trying to start a fire by focusing Sunlight on a piece of paper using an equiconvex lens of focal length $10 \mathrm{~cm}$. The diameter of the Sun is $1.39 \times 10^9 \mathrm{~m}$ and its mean distance from the earth is $1.5 \times 10^{11} \mathrm{~m}$. What is the diameter of the Sun's image on the paper?

Options:
A $12.4 \times 10^{-4} \mathrm{~m}$
B $9.2 \times 10^{-4} \mathrm{~m}$
C $6.5 \times 10^{-4} \mathrm{~m}$
D $6.5 \times 10^{-5} \mathrm{~m}$

The correct answer is:
$9.2 \times 10^{-4} \mathrm{~m}$

Solution:
Here, $\frac{D / 2}{u}=\frac{d / 2}{f}$
$$
\begin{aligned}
\Rightarrow \quad d & =\frac{D f}{u} \\
& =\frac{1.39 \times 10^8 \times 10 \times 10^{-2}}{1.5 \times 10^{11}}=9.2 \times 10^{-4} \mathrm{~m}
\end{aligned}
$$

Topic: Physics, Ray Optics, NEET, NEET 2008 (Mains)
A boy needs to select five courses from 12 available courses, out of which 5 courses are language courses. If he can choose at most two language courses, then the number of ways he can choose five courses is

Options:

The correct answer is:
546

Solution:
Case I -

When 0 language course is selected from 5 particular language courses, then total number of ways

=C57=21

Case II -

When 1 language course is selected from 5 particular language courses, then total number of ways

=C47C15=175

Case III -

When 2 language course is selected from 5 particular language courses, then total number of ways

=C37C25=350

So, required ways=546

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2023 (24 Jan Shift 1)
A boy of mass 4 kg is standing on a piece of wood having mass 5 kg. If the coefficient of friction between the wood and the floor is 0.5 , the maximum force that the boy can exert on the rope so that the piece of wood does not move from its place is _______ N. (Round off to the Nearest Integer)

[Take g=10 m s-2

Options:

The correct answer is:
30

Solution:
N+T=90

T=μN=0.5(90-T)

1.5T=45

T=30 N

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2021 (17 Mar Shift 2)