datavorous/entrance-exam-dataset huggingface dataset:
A magnetic needle lying parallel to a magnetic field required W units of work to turn it through 60°. The torque required to maintain the needle in this position will be

Options:
A 3 W
B W
C 3W2
D 2W

The correct answer is:
3 W

Solution:
W=BM cos60°=MB2;MB=2W
τ=MB sinθ=2Wsin60°=2W32=W3

Topic: Physics, Magnetic Effects of Current, NEET
A magnetic needle lying parallel to a magnetic field requires $\mathrm{W}$ units of work to turn it through $60^{\circ}$. The torque needed to maintain the needle in this position will be

Options:
A
$\sqrt{3} \mathrm{~W}$

B
W

C
$\frac{\sqrt{3}}{2} \mathrm{~W}$

D
$2 \mathrm{~W}$

The correct answer is:
$\sqrt{3} \mathrm{~W}$

Solution:
$\tau=(\mathrm{H}) \tan 60^{\circ} \quad=\mathrm{W} \cdot \sqrt{3}$

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2003
A magnetic needle lying parallel to a magnetic field requires \(W\) units of work to turn it through \(60^{\circ}\). The torque required to maintain the needle in this position will be

Options:
A \(\sqrt{3} W\)
B \(W\)
C \(\frac{\sqrt{3}}{2} W\)
D \(2 \mathrm{~W}\)

The correct answer is:
\(\sqrt{3} W\)

Solution:
According to the question, work done required to rotate magnetic needle from \(\theta_1=0^{\circ}\) to \(\theta_2=60^{\circ}\).
\(W=M B\left(\cos \theta_1-\cos \theta_2\right)\)
\(\begin{aligned}
& =M B\left(\cos 0^{\circ}-\cos 60^{\circ}\right) \\
& =M B\left(1-\frac{1}{2}\right) \\
W & =\frac{M B}{2} \Rightarrow M B=2 W \quad \ldots (i) \\
\therefore \text {Torque, } \tau & =M B \sin 60^{\circ}=M B \frac{\sqrt{3}}{2} \\
& =2 W \times \frac{\sqrt{3}}{2} \quad \text { [from Eq. (i)] } \\
& =W \sqrt{3}=\sqrt{3} W
\end{aligned}\)

Topic: Physics, Magnetic Effects of Current, AP EAMCET, AP EAMCET 2020 (21 Sep Shift 2)
A magnetic needle lying parallel to the magnetic field requires $W$ units of work to turn it through an angle $45^{\circ} .$ The torque required to maintain the needle in this position will be

Options:
A $\sqrt{2} W$
B $\frac{1}{\sqrt{3} W}$
C $(\sqrt{2}-1) W$
D $\frac{W}{(\sqrt{2}-1)}$

The correct answer is:
$\frac{W}{(\sqrt{2}-1)}$

Solution:
Work done by magnet to turn from angle
$\begin{array}{l}
\begin{array}{l}
\theta_{1} \text { to } \theta_{2} \\
\mathrm{~W}=\mathrm{MB}\left(\cos \theta_{1}-\cos \theta_{2}\right) \\
=\mathrm{MB}\left(\cos 0^{\circ}-\cos 45^{\circ}\right)
\end{array} \\
=\mathrm{MB}\left(1-\frac{1}{\sqrt{2}}\right)=\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right) \mathrm{MB}
\end{array}$
Also torque acting on the magnet
$\begin{array}{l}
\tau=\mathrm{MB} \sin 45^{\circ}=\frac{\mathrm{MB}}{\sqrt{2}} \\
\mathrm{~W}=(\sqrt{2}-1) \cdot \tau \Rightarrow \tau=\frac{\mathrm{W}}{(\sqrt{2}-1)}
\end{array}$

Topic: Physics, Magnetic Properties of Matter, JEE Main
A magnetic needle lying parallel to the magnetic field requires $W$ units of work to turn it through an angle $45^{\circ} .$ The torque required to maintain the needle in this position will be

Options:
A $\sqrt{2} W$
B $\frac{1}{\sqrt{3} W}$
C $(\sqrt{2}-1) W$
D $\frac{W}{(\sqrt{2}-1)}$

The correct answer is:
$\frac{W}{(\sqrt{2}-1)}$

Solution:
Work done by magnet to turn from angle
$\begin{array}{l}
\begin{array}{l}
\theta_{1} \text { to } \theta_{2} \\
\mathrm{~W}=\mathrm{MB}\left(\cos \theta_{1}-\cos \theta_{2}\right) \\
=\mathrm{MB}\left(\cos 0^{\circ}-\cos 45^{\circ}\right)
\end{array} \\
=\mathrm{MB}\left(1-\frac{1}{\sqrt{2}}\right)=\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right) \mathrm{MB}
\end{array}$
Also torque acting on the magnet
$\begin{array}{l}
\tau=\mathrm{MB} \sin 45^{\circ}=\frac{\mathrm{MB}}{\sqrt{2}} \\
\mathrm{~W}=(\sqrt{2}-1) \cdot \tau \Rightarrow \tau=\frac{\mathrm{W}}{(\sqrt{2}-1)}
\end{array}$

Topic: Physics, Magnetic Properties of Matter, VITEEE, VITEEE 2015
A magnetic needle lying parallel to the magnetic field requires W units of work to turn it through an angle 45o . The torque required to maintain the needle in this position will be

Options:
A 2W
B 13W
C 2-1W
D W2-1

The correct answer is:
W2-1

Solution:
Work done by a magnet to turn from angle Î¸1Â toÂ Î¸2 is
W=Bcosâ¡Î¸1-cosÎ¸2
=MB1-12=2-12MB
Also torque acting on the magnet is given by
Ï=MBsinâ¡45o=MB2
So W=2-1⋅Ï
â´Â Â Â Ï=W2-1

Topic: Physics, Magnetic Effects of Current, NEET
A magnetic needle of magnetic moment 6.7×10-2 A m2 and moment of inertia 7.5×10-6 kg m2 is performing simple harmonic oscillations in a magnetic field of 0.01 T. Time taken for 10 complete oscillations is:

Options:
A 8.76 s
B 6.65 s
C 8.89 s
D 6.98 s

The correct answer is:
6.65 s

Solution:
Time period of oscillating magnetic needle is T=2πIMB
=2π7.5×10-66.7×10-2×0.01 sec
=0.6647 sec
Time for oscillations, t=10 T=6.647 sec⁡=6.65 sec

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2017 (02 Apr)
A magnetic needle of negligible breadth and thickness compared to its length, oscillates in a horizontal plane with a period T. The period of oscillation of each part obtained on breaking the magnet into n equal parts and perpendicular to the length is

Options:
A T
B T n
C nT
D 1n

The correct answer is:
T n

Solution:
T=2ÏIMB
on breaking n equal parts perpendicular to length
Iâ²=112 mn ln2 of each part
Iâ²=In3, Mâ²=Mn
Tâ²T=Iâ²Mâ²ÃMI=1n3 Ãn=1nâ Tâ²=Tn

Topic: Physics, Magnetic Properties of Matter, JEE Main
A magnetic needle suspended horizontally by an unspun silk fibre oscillates in the horizontal plane because of the restoring force originating mainly from

Options:
A the torsion of the silk fibre
B the horizontal component of earth's magnetic field
C the force of gravity
D all the above factors

The correct answer is:
the horizontal component of earth's magnetic field

Solution:
Earth behaves as a magnet which will interact with the magnetic needle and makes it oscillate.

Topic: Physics, Magnetic Properties of Matter, NEET
A magnetic needle suspended parallel to a magnetic field requires $\sqrt{3} \mathrm{~J}$ of work to turn it through $60^{\circ}$. The torque needed to maintain the needle in this position will be

Options:
A $2 \sqrt{3} \mathrm{~J}$
B $3 \mathrm{~J}$
C $\sqrt{3} \mathrm{~J}$
D $\frac{3}{2} \mathrm{~J}$

The correct answer is:
$3 \mathrm{~J}$

Solution:
In this case, work done
$$
\begin{aligned}
W & =M B\left(\cos \theta_1-\cos \theta_2\right) \\
& =M B\left(\cos 0^{\circ}-\cos 60^{\circ}\right) \\
& =M B\left(1-\frac{1}{2}\right)=\frac{M B}{2} \\
M B & =2 \sqrt{3} \mathrm{~J} \quad(\because \text { given } W=\sqrt{3} \mathrm{~J}) \\
\tau & =M B \sin 60^{\circ}=(2 \sqrt{3})\left(\frac{\sqrt{3}}{2}\right) \mathrm{J}=3 \mathrm{~J}
\end{aligned}
$$

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2012 (Mains)
A magnetised wire of moment M and length l is bent in the form of a semicircle of radius r. Then its new magnetic moment is

Options:
A 2Mπ
B 2M
C Mπ
D zero

The correct answer is:
2Mπ

Solution:
On bending a wire its pole strength remains unchanged whereas its magnetic length changes.

New magnetic moment,

M'=m×2r=m×2lπ=2Mπ

Topic: Physics, Magnetic Effects of Current, JEE Main
A magnetizing field of 500 A/m produces a magnetic flux of 4×10-5 Wb in an iron rod of cross-sectional area 0.4cm2 . The permeability of the rod in W/b/A-m is

Options:
A 4×10-6
B 1×10-3
C 2×10-4
D 3×10-5

The correct answer is:
2×10-4

Solution:
Given, magnetising field, H = 5000 A/m
Magnetic flux, ϕ=4×10-5Wb
Cross-sectional area, A = 0.4 cm2=4×10-5m2
∴ Permeability of rod,
μ=BH=ϕAH∴ϕ=BA⇒B=ϕA
=4×10-54×10-5×5000=2×10-4WbA-m

Topic: Physics, Magnetic Properties of Matter, JEE Main
A magnifying glass is to be used at the fixed object distance of $1 \mathrm{inch}$. If it is to produce an erect image magnified 5 times its focal length should be

Options:
A 0.2 inch
B 0.8 inch
C 1.25 inch
D 5 inch

The correct answer is:
1.25 inch

Solution:
$m=\frac{v}{u}=5 \Rightarrow v=5$ inch (Given $u=1 \mathrm{inch}$ )
Using sign convention $u=-1$ inch, $v=-5$ inch
$\therefore \frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{-5}-\frac{1}{-1} \Rightarrow f=1.25 \mathrm{inch}$

Topic: Physics, Ray Optics, JEE Main
A major component of Borsch reagent is obtained by reacting hydrazine hydrate with which of the following ?

Options:
A

B

C

D

The correct answer is:

Solution:
The major component of Borsch reagent is 2,4- dinitrophenyl hydrazine which can be obtained by reaction of 2,4-dinitrochloro benzene and hydrazine

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2013 (22 Apr Online)
A major percentage $(97 \%)$ of $\mathrm{O}_2$ is transported by RBCs in the blood. How does the remaining percentage (3\%) of $\mathrm{O}_2$ transported?

Options:

None

Solution:
About $97 \%$ of $\mathrm{O}_2$ is transported by RBCs in the blood. The remaining $3 \%$ of $\mathrm{O}_2$ is carried in a dissolved state through the plasma.

Topic: Biology, Breathing and Exchange of Gases
A man alternately tosses a coin and throw a dice, beginning with the coin. The probability that he gets a head in coin before he gets a 5 or 6 in dice, is

Options:
A 34
B 12
C 13
D 23

The correct answer is:
34

Solution:
A= coin show head
B= dice show 5 or 6
PA=12,PB=26=13
Hence, the required Probability =PA+PACPBCPA+PACPBC2PA+…
=12+12×23×12+12×232×12+…
=121-13=1223=34

Topic: Mathematics, Probability, JEE Main
A man carrying a monkey on his shoulder does cycling smoothly on a circular track of radius $9 \mathrm{~m}$ and completes 120 resolutions in 3 minutes. The magnitude of centripetal acceleration of monkey is $\left(\right.$ in $\mathrm{m} / \mathrm{s}^2$ ) :

Options:
A $57600 \pi^2 \mathrm{~ms}^{-2}$
B Zero
C $4 \pi^2 \mathrm{~ms}^{-2}$
D $16 \pi^2 \mathrm{~ms}^{-2}$

The correct answer is:
$16 \pi^2 \mathrm{~ms}^{-2}$

Solution:
$\begin{aligned} & \text { Given : } \mathrm{R}=9 \mathrm{~m}, \\ & 120 \text { revolution in } 3 \mathrm{~min} \\ & \omega=\frac{120 \mathrm{Rev} .}{3 \mathrm{~min} .}=\frac{120 \times 2 \pi \mathrm{rad}}{3 \times 60 \mathrm{sec}}=\frac{4 \pi}{3} \mathrm{rad} / \mathrm{s} \\ & \mathrm{a}_{\text {centripetal }}=\omega^2 \mathrm{R}=\left(\frac{4 \pi}{3}\right)^2 \times 9=16 \pi^2 \mathrm{~m} / \mathrm{s}^2\end{aligned}$

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2024 (05 Apr Shift 2)
A man crosses a river in a boat. If he crosses the river in minimum time, he takes 10min with a drift 120 m. If he crosses the river taking the shortest path, he takes 12.5min, find the width of the river.

Options:
A 215 m
B 150 m
C 100 m
D 200 m

The correct answer is:
200 m

Solution:
Let vr = velocity of river
vbr = velocity of river in still water and
w = width of river
Given, tmin = 10 min
or wvbr=10 ...(i)
Drift, in this case, will be,
x=vrt
∴ 120=10vr ...(ii)

The shortest path is taken when vb is along AB. In this case,
vb=vbr2-vr2
Now, 12.5=wvb=wvbr2-vr2 ...(iii)
Solving these three equations, we get
vbr = 20 m min-1,
vr = 12 m min-1
and w= 200 m

Topic: Physics, Motion In Two Dimensions, JEE Main
A man does a given amount of work in $10 \mathrm{sec}$. Another man does the same amount of work in $20 \mathrm{sec}$. The ratio of the output power of first man to the second man is

Options:
A 1
B $1 / 2$
C $2 / 1$
D None of these

The correct answer is:
$2 / 1$

Solution:
Power $=\frac{W}{t}$. If $W$ is constant then $P \propto \frac{1}{t}$
Le $\frac{P_1}{P_2}=\frac{t_2}{t_1}=\frac{20}{10}=\frac{2}{1}$

Topic: Physics, Work Power Energy, JEE Main
A man drops a ball downside from the roof of a tower of height 400 metres. At the same time another ball is thrown upside with a velocity 50 metre/sec. from the surface of the tower, then they will meet at which height from the surface of the tower

Options:
A 100 metres
B 320 metres
C 80 metres
D 240 metres

The correct answer is:
80 metres

Solution:
Let both balls meet at point $P$ after time $t$.
The distance travelled by ball $A, h_1=\frac{1}{2} g t^2$
The distance travelled by ball $B, l_2=u t-\frac{1}{2} g t^2$
$\begin{aligned} & h_1+h_2=400 \mathrm{~m} \Rightarrow u t=400, t=400 / 50=8 \mathrm{sec} \\ & \therefore h_1=320 \mathrm{~m} \text { and } h_2=80 \mathrm{~m}\end{aligned}$

Topic: Physics, Motion In One Dimension, JEE Main
A man fires a bullet of mass $200 \mathrm{~g}$ at a speed of $5 \mathrm{~m} / \mathrm{s}$. The gun is of one $\mathrm{kg}$ mass. by what velocity the gun rebounds backwards

Options:
A $0.1 \mathrm{~m} / \mathrm{s}$
B $10 \mathrm{~m} / \mathrm{s}$
C $1 \mathrm{~m} / \mathrm{s}$
D $0.01 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$1 \mathrm{~m} / \mathrm{s}$

Solution:
$v_G=\frac{m_B v_B}{m_G}=\frac{0.2 \times 5}{1}=1 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Laws of Motion, JEE Main
A man goes $10 m$ towards North, then $20 m$ towards east then displacement is

Options:
A 22.5m
B 25m
C 22.5m
D 30m

The correct answer is:
22.5m

Solution:
$\vec{r}=20 \hat{i}+10 \hat{j} \quad \therefore r=\sqrt{20^2+10^2}=22.5 \mathrm{~m}$

Topic: Physics, Motion In One Dimension, JEE Main
A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that his density remains same, the stress in the leg will change by a factor of:

Options:
A 181
B 9
C 19
D 81

The correct answer is:
9

Solution:
Stress =FA=MgA
=ρVolume .gA
=ρ93x3 g92x2
=9 times

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2017 (02 Apr)
A man has 7 friends. In how many ways he can invite one or more of them for a tea party

Options:
A $128$
B $256$
C $127$
D $130$

The correct answer is:
$127$

Solution:
Required number of ways $=2^7 - 1 = 128$.
$\{$Since the case that no friend be invited i.e., ${ }_7 C_0$ is excluded}.

Topic: Mathematics, Permutation Combination, JEE Main
A man has 10 friends. In how many ways he can invite one or more of them to a party

Options:
A $10!$
B $2^{10}$
C $10!-1$
D $2^{10}-1$

The correct answer is:
$2^{10}-1$

Solution:
Required number of ways $=2^{\wedge} 10-1$
(Since the case that no friend be invited i.e., ${ }_{10} C_0$ is excluded)

Topic: Mathematics, Permutation Combination, JEE Main
A man in a balloon throws a stone downwards with a speed of 5 m s-1 with respect to the balloon which is moving upwards with a constant acceleration of 5 m s-2. Find the velocity (in m s-1) of the stone relative to the man after 2 s. (Take g=10 m s-2)

Options:

The correct answer is:
35

Solution:
Relative velocity of stone = 5 m s-1
relative acceleration of stone = 10 + 5 = 15 m s-2
v = u + at = 5 + 15 Ã 2 = 35 m s-1
relative velocity after t = 2 s is 35 m s-1 â

Topic: Physics, Motion In One Dimension, JEE Main
A man in a car at location Q on a straight highway is moving with speed $\mathrm{v}$. He decides to reach a point $P$ in a field at a distance $d$ from highway (point $M$ ) as shown in the figure.Speed of the car in the field is half to that on the highway. What should be the distance RM, so that the time taken to reach $P$ is minimum?

Options:
A
$\frac{\mathrm{d}}{\sqrt{3}}$

B
$\frac{\mathrm{d}}{2}$

C
$\frac{\mathrm{d}}{\sqrt{2}}$

D
$\mathrm{d}$

The correct answer is:
$\frac{\mathrm{d}}{\sqrt{3}}$

Solution:
Let the car turn of the highway at a distance ' $x$ ' from the point $\mathrm{M}$. So, $\mathrm{RM}=x$ And if speed of car in field is $v$, then time taken by the car to cover the distance $Q R=$ $Q M-x$ on the highway,
$$
t_1=\frac{Q M-x}{2 v}
$$
Time taken to travel the distance ' $R P^{\prime}$ in the field
$$
t_2=\frac{\sqrt{d^2+x^2}}{v}
$$
Total time elapsed to move the car from $Q$ to $P$
$$
t=t_1+t_2=\frac{Q M-x}{2 v}+\frac{\sqrt{d^2+x^2}}{v}
$$
$$
\text { or } x=\frac{d}{\sqrt{2^2-1}}=\frac{d}{\sqrt{3}}
$$

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
A man is $180 \mathrm{~cm}$ tall and his eyes are $10 \mathrm{~cm}$ below the top of his head. In order to see his entire height right from toe to head, he uses a plane mirror kept at a distance of $1 \mathrm{~m}$ from him. The minimum length of the plane mirror required is

Options:
A $180 \mathrm{~cm}$
B $90 \mathrm{~cm}$
C $85 \mathrm{~cm}$
D $170 \mathrm{~cm}$

The correct answer is:
$90 \mathrm{~cm}$

Solution:
The minimum length of the plane mirror required for a man to see his entire height, from toe to head, is half his height. This is because the angle of incidence equals the angle of reflection in a plane mirror.
So, for a man who is $180 \mathrm{~cm}$ tall, the minimum length of the mirror required would be:
$180 / 2=90 \mathrm{~cm}$
This means that a mirror of at least $90 \mathrm{~cm}$ in length would be required for the man to see his entire height in the mirror. The distance of the man from the mirror and the position of his eyes do not affect the minimum length of the mirror required

Topic: Physics, Ray Optics, JEE Main
A man is admitted to a hospital and is found to be suffering from abnormally low body temperature, loss of appetite and extreme thirst. His brain scan would probably show a tumour in

Options:
A pons varoli.
B cerebellum.
C medulla oblongata.
D hypothalamus.

The correct answer is:
hypothalamus.

Solution:
The brain scan of the person will probably show a tumour in the hypothalamus. The hypothalamus forms an important part of the limbic system. It has centres which regulate body temperature and appetite. Thus, a hypothalamic tumour will result in low body temperature and loss of appetite. The hypothalamic nuclei are responsible for the secretion of the hormone vasopressin. This hormone is responsible for reabsorption of water and electrolytes from the renal tubules. This prevents excessive urination and dehydration. A hypothalamic tumour will affect the synthesis of vasopressin, which will lead to excessive urination, excessive thirst and dehydration.
A tumour in pons will lead to loss of coordination, as it contains fibres which connect different regions of the brain.
A tumour in cerebellum will lead to lack of balance, as the cerebellum is responsible for maintenance of balance.
A tumour in medulla oblongata can prove to be very fatal, as the medulla has the center which regulates the cardiorespiratory activities.

Topic: Biology, Neural Control and Coordination, NEET
A man is holding a chain of mass m and length l such that the lower end of the chain just touches a table. Now the man places half the length of the chain very slowly on the table while keeping the other half vertical. The work done by the man in this process is

Options:
A –mgl2
B –mgl4
C –3mgl8
D – mgl8

The correct answer is:
–3mgl8

Solution:
The work done by man is negative of magnitude of decrease in potential energy of chain.
ÎU=mg L2 â m2g L4=3 mg L8
â´W=â 3mgl8

Topic: Physics, Work Power Energy, JEE Main
A man is known to speak the truth 3 out of 4 times. He throws a die and reports that it is a six. The probability that it is actually a six, is

Options:
A ${ }^{\frac{3}{8}}$
B $\frac{1}{5}$
C $\frac{3}{4}$
D None of these

The correct answer is:
${ }^{\frac{3}{8}}$

Solution:
Let $E$ denote the event that a six occurs and $A$ the event that the man reports that it is a ' 6 ', we have
$P(E)=\frac{1}{6}, P\left(E^{\prime}\right)=\frac{5}{6}, P(A / E)=\frac{3}{4}$ and $P\left(A / E^{\prime}\right)=\frac{1}{4}$
$P(E / A)=\frac{P(E) \cdot P(A / E)}{P(E) \cdot P(A / E)+P\left(E^{\prime}\right) \cdot P\left(A / E^{\prime}\right)}$
$=\frac{\frac{1}{6} \times \frac{3}{4}}{\frac{1}{6} \times \frac{3}{4}+\frac{5}{6} \times \frac{1}{4}}=\frac{3}{8}$.

Topic: Mathematics, Probability, JEE Main
A man is known to speak the truth 3 out of 4 times. He throws a die and reports that it is a six. The probability that it is actually a six, is

Options:
A $\frac{3}{8}$
B $\frac{1}{5}$
C $\frac{3}{4}$
D None of these

The correct answer is:
$\frac{3}{8}$

Solution:
Let $E$ denote the event that a six occurs and $A$ the event that the man reports that it is a ' 6 ', we have
$P(E)=\frac{1}{6}, P\left(E^{\prime}\right)=\frac{5}{6}, P(A / E)=\frac{3}{4}$ and $P\left(A / E^{\prime}\right)=\frac{1}{4}$
From Baye's theorem,
$P(E / A)=$ $\frac{P(E) \cdot P(A / E)}{P(E) \cdot P(A / E)+P\left(E^{\prime}\right) \cdot P\left(A / E^{\prime}\right)}$
$=\frac{\frac{1}{6} \times \frac{3}{4}}{\frac{1}{6} \times \frac{3}{4}+\frac{5}{6} \times \frac{1}{4}}=\frac{3}{8}$.

Topic: Mathematics, Probability, JEE Main
A man is observing, from the top of a tower, a boat speeding towards the tower from a certain point A, with uniform speed. At that point, angle of depression of the boat with the man's eye is 30° (Ignore man's height). After sailing for 20 seconds, towards the base of the tower (which is at the level of water), the boat has reached a point B, where the angle of depression is 45°. Then the time taken (in seconds) by the boat from B to reach the base of the tower is :

Options:
A 10
B 10(3-1)
C 103
D 10(3+1)

The correct answer is:
10(3+1)

Solution:
Let speed of boat is u m/s and height of tower
is h meter & distance AB=x metre

∴x=hcot30°-hcot45°

⇒x=h(3-1)

∴u=x20=h(3-1)20 m/s

∴ Time taken to travel from B to C (Distance
=h meter)

=hu=hh(3-1)20=203-1=10(3+1)sec .

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2021 (25 Feb Shift 1)
A man is riding on a cycle with velocity $7.2 \mathrm{kn} / \mathrm{hr}$ up a hill having a slope 1 in 20 . The total mass of the man and cycle is $100 \mathrm{~kg}$. The power of the man is

Options:
A 200 W
B 175 W
C 125 W
D 98 W

The correct answer is:
98 W

Solution:
$v=7.2 \frac{\mathrm{km}}{\mathrm{h}}=7.2 \times \frac{5}{18}=2 \mathrm{~m} / \mathrm{s}$
Slope is given 1 in 20
$\therefore \quad \sin \theta=\frac{1}{20}$
When man and cycle moves up then component of weight opposes it motion is
$F=m g \sin \theta$
So power of the man $P=F \times v=m g \sin \theta \times v$
$=100 \times 9.8 \times\left(\frac{1}{20}\right) \times 2=98$ Watt

Topic: Physics, Work Power Energy, JEE Main
A man is standing on the platform and one train is approaching and another train is going away with the speed of 4 m s-1, frequency of sound produced by train is 240 Hz. What will be the number of beats heard by him per second? (velocity of sound = 300 m s-1)

Options:
A 12
B Zero
C 6
D 3

The correct answer is:
6

Solution:
The apparent frequency heard by the man is given by
napp=nv±vov±vs
On our case,vo=0
∴napp=nvv±vs
Given, v=300 m s-1, vs=4 m s-1, n=240 Hz
In the first case, the train is approaching the man, so frequency heard
n1=nvv±vs=240300300-4
=240×300296=243.24 Hz
In the second case, the train is going away from the man, so frequency heard
n2=nvv+vs=240 300300+4
=240×300304=236.84 Hz
Hence, the number of beats heard by the man per second
=n1-n2
=243.24-236.84=6.4≈6

Topic: Physics, Waves and Sound, JEE Main
A man is standing on top of a building $100 \mathrm{~m}$ high. He throws two balls vertically, one at $t=0$ and after a time interval (less than 2 seconds). The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is $+15 \mathrm{~m}$ at $t=2 \mathrm{~s}$. The gap is found to remain constant. Calculate the velocity with which the balls were thrown and the exact time interval between their throw.

Options:

None

Solution:
Let the speeds of the two balls ( 1 and 2 ) be $u_1$ and $u_2$ where if $u_1=2 u_2$ and $u_1=u_2=u$.
If $h_1$ and $h_2$ and the distance covered by the balls 1 and 2 , respectively, before coming to rest, then
$h_1=\frac{u_1^2}{2 g}=\frac{4 u^2}{2 g}$ and $h_2=\frac{u_2^2}{2 g}=\frac{u^2}{2 g}$
According to question,
$$
\begin{aligned}
&\left(h_1-h_2\right)=15 \mathrm{~m} \\
&\frac{4 u^2}{2 g}-\frac{u^2}{2 g}=15 \mathrm{~m} \text { or } \frac{3 u^2}{2 g}=15 \mathrm{~m} \\
&\text { or } u^2=\sqrt{5 \mathrm{~m} \times(2 \times 10)} \mathrm{m} / \mathrm{s}^2 \\
&u=10 \mathrm{~m} / \mathrm{s} \\
&
\end{aligned}
$$
So velocity of ball 1 and $2, u_1=20 \mathrm{~m} / \mathrm{s}$ and $u_2=10 \mathrm{~m} / \mathrm{s}$
As, $h_1=\frac{u_1^2}{2 g}=\frac{(20 \mathrm{~m})^2}{2 \times 10 \mathrm{~m} 15}=20 \mathrm{~m}$
$$
h_2=h_1-15 \mathrm{~m}=5 \mathrm{~m}
$$
For 1st ball, $v_1=u_t+g t_1$
$$
\begin{aligned}
&0=20-10 t_1 \\
&t_1=2 \mathrm{sec} .
\end{aligned}
$$
For second ball, $v_2=u_2+g t_2$
$$
\begin{aligned}
&0=10-10 t_2 \\
&t_2=1
\end{aligned}
$$
Exact time interval between the two balls
$$
=t_1-t_2=2 \mathrm{~s}-1 \mathrm{~s}=1 \text { second. }
$$

Topic: Physics, Motion In One Dimension