datavorous/entrance-exam-dataset huggingface dataset:
On the sides $\mathrm{AB}, \mathrm{BC}, \mathrm{CA}$ of a $\triangle \mathrm{ABC}, 3,4,5$ distinct points (excluding vertices $\mathrm{A}, \mathrm{B}, \mathrm{C}$ ) are respectively chosen. The number of triangles that can be constructed using these chosen points as vertices are :

Options:
A
210

B
205

C
215

D
220

The correct answer is:
205

Solution:
Required number of triangles
$$
={ }^{12} \mathrm{C}_3-\left({ }^3 \mathrm{C}_3+{ }^4 \mathrm{C}_3+{ }^5 \mathrm{C}_3\right)=205
$$

Topic: Mathematics, Properties of Triangles, JEE Main, JEE Main 2013 (23 Apr Online)
On the surface of polar stratospheric clouds, hydrolysis of chlorine nitrate gives A and B while its reaction with HCl produces B and C. A.B and C are, respectively

Options:
A HOCl,HNO3,Cl2
B Cl2,HNO3,HOCl
C HClO2,HNO2,HOCl
D HOCl,HNO2,Cl2O

The correct answer is:
HOCl,HNO3,Cl2

Solution:
In winter, special type of clouds called polar stratospheric clouds are formed over Antarctica.These polar stratospheric clouds provide surface on which chlorine nitrate formed gets hydrolysed to form hypochlorous acid.It also reacts with hydrogen chloride produced to give molecular chlorine

The reactions are:

ClONO2(g)+H2O(g)→HOCl(g)+HNO3(g)

ClONO2(g)+HCl(g)→Cl2(g)+HNO3(g)

Hence, A, B and C are HOCl,HNO3,Cl2

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2022 (27 Jun Shift 2)
On the treatment of the following compound with a strong acid, the most susceptible site for bond cleavage is:

Options:
A
$\mathrm{O} 2-\mathrm{C} 3$

B
O5-C6

C
$\mathrm{C} 4-\mathrm{O} 5$

D
$\mathrm{C} 1-\mathrm{O} 2$

The correct answer is:
O5-C6

Solution:
The lone pair of electrons on $\mathrm{O} 2$ is involved in resonance with $\mathrm{C}=\mathrm{C}$. Hence $\mathrm{O} 2$ will not be protonated.
The lone pair of electrons on O5 is not involved in resonance with $\mathrm{C}=\mathrm{C}$. Hence, O5 will be protonated. Chloride ion will then attack least substituted $\mathrm{C}$ atom (C6)

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
On treating a compound with warm dil. H2SO4, gas X is evolved which turns K2Cr2O7 paper acidified with dil. H2SO4 to a green compound Y. X and Y respectively are :

Options:
A X=SO2, Y=Cr2O3
B X=SO2, Y=Cr2SO43
C X=SO3, Y=Cr2SO43
D X=SO3, Y=Cr2O3

The correct answer is:
X=SO3, Y=Cr2SO43

Solution:
Sulfur dioxide gas turns acidified potassium dichromate (VI) solution from orange to green. Chromium is reduced from (VI) to (III) after reaction.

SO2+dil H2SO4⟶SO3g

SO3+K2Cr2O7→H2SO4dilCr2SO43

Hence, On treating a compound with warm dil. H2SO4, gas X is evolved which turns K2Cr2O7 paper acidified with dil. H2SO4 to a green compound Y. X and Y respectively are SO3 and Cr2(SO4)3.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2021 (26 Feb Shift 1)
On treatment of 100 mL of 0.1 M solution of CoCl3.6H2O with excess AgNO3; 1.2×1022 ions are precipitated. The complex is:

Options:
A CoH2O3Cl3.3H2O
B CoH2O6Cl3
C CoH2O5ClCl2. H2O
D CoH2O4Cl2Cl.2H2O

The correct answer is:
CoH2O5ClCl2. H2O

Solution:
Moles of CoCl3.6H2O=100×0.1×10-3=10-2mole

Moles of Cl- ions precipitated=1.2×10226×1023=2×10-2 mole

So, one mole of complex will release two moles of chloride ions so, formula of complex is

CoH2O5ClCl2. H2O

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2017 (02 Apr)
On treatment of 100 ml of 0.1 M solution of the complex CrCl3.6H2O with excess of AgNO3, 4.305 g of AgCl was obtained. The complex is

Options:
A [Cr(H2O)3 Cl3] . 3H2O
B [Cr(H2O)4Cl2] Cl.2H2O
C [Cr(H2O)5Cl)Cl2 .H2O
D [Cr(H2O)6]Cl3

The correct answer is:
[Cr(H2O)6]Cl3

Solution:
Mol of AgCl = 4 . 3 0 5 1 4 3 . 5 = 0 . 0 3 = mol of Cl- given by the complex.
Mol of the complex = 100 x 10-3 x 0.1 = 0.01 ;
Cr H 2 O 6 Cl 3 â¶ Cr H 2 O 6 3 + + 3 Cl - 0 . 0 1 mol 0 . 0 1 mol 0 . 0 3 mol

Topic: Chemistry, Coordination Compounds, JEE Main
On vigorous oxidation by permanganate solution
$$
\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{CH}-\mathrm{CH}_2-\mathrm{CHO} \text { gives }
$$

Options:
A

B

C

D

The correct answer is:

Solution:
Aldehydic group gets oxidised to carboxylic group
Double bond breaks and carbon gets oxidised to carboxylic group

Topic: Chemistry, Aldehydes and Ketones, JEE Main
On vigorous oxidation by permanganate solution
$$
\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{CH}-\mathrm{CH}_2-\mathrm{CHO} \text { gives }
$$

Options:
A

B

C

D

The correct answer is:

Solution:
Aldehydic group gets oxidised to carboxylic group
Double bond breaks and carbon gets oxidised to carboxylic group

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2002
On which of the following intervals is the function x100+sin x-1 decreasing?

Options:
A 0, π 2
B 0,1
C π 2 ,π
D None of these

The correct answer is:
None of these

Solution:
f x = x 100 +sinxâ1
â f â² x =100 x 99 +cosx
If 0<x< Ï 2 , then f â² x >0, , therefore f x is increasing on 0, Ï 2 .
If 0<x<1, then
100 x 99 >0 and cosx>0 [ âµx lies between 0 and 1 ]
â f â² x =100 x 99 +cosx>0
âf x is increasing on 0,1 .
If Ï 2 <x<Ï, then
100 x 99 >100â âµx>1â x 99 >1
⇒100 x 99 +cosx>0  ∵cosx≥−1⇒100 x 99 +cosx>99
â f â² x >0âf x is increasing in Ï 2 ,Ï .

Topic: Mathematics, Application of Derivatives, JEE Main
On which of the following lines lies the point of intersection of the line, x-42=y-52=z-31 and the plane, x+y+z=2?

Options:
A x-41=y-51=z-5-1
B x-11=y-32=z+4-5
C x-22=y-32=z+33
D x+33=4-y3=z+1-2

The correct answer is:
x-11=y-32=z+4-5

Solution:
A general point on the given line x-42=y-52=z-31=λ can we taken as 2λ+4, 2λ+5, λ+3
If it also lies on the given plane then
2λ+4+2λ+5+λ+3=2
⇒5λ+12=2
⇒λ=-2
⇒ Point of intersection is 0, 1, 1 , which lies on x-11=y-32=z+4-5

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2019 (10 Jan Shift 2)
On which of the following properties, does the coagulating power of an ion depend?

Options:
A Both the magnitude and sign of the charge on the ion.
B Size of the ion alone.
C The magnitude of the charge on the ion alone.
D The sign of charge on the ion alone.

The correct answer is:
Both the magnitude and sign of the charge on the ion.

Solution:
According to Hardy Schulze rule, the coagulating power of an ion depends on both the magnitude and sign of the charge of the ion.

Topic: Chemistry, Surface Chemistry, JEE Main
One advantage of cleistogamy is:

Options:
A It leads to greater genetic diversity.
B Seed dispersal is more efficient and widespread.
C Seed set is not dependent on pollinators.
D Each visit of a pollinator results in the transfer of hundreds of pollen grains.

The correct answer is:
Seed set is not dependent on pollinators.

Solution:
Cleistogamy is the phenomenon in which flower never opens up and remains closed in bud condition. In such flowers, only self-pollination can occur so assured sees formation is there and seed formation is not dependent on pollinators. Examples: Oxalis, Commelina.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
One ampere of current is passed for 9650 seconds through molten $\mathrm{AlCl}_3$. What is the weight in grams of $\mathrm{Al}$ deposited at cathode? (atomic weight of $\mathrm{Al}=27$ )

Options:
A $0.9$
B $9.0$
C $0.09$
D $90.0$

The correct answer is:
$0.9$

Solution:
$\begin{aligned} & \text { Gram equivalent weight of } \mathrm{Al}=\frac{27}{3}=9 \mathrm{~g} \\ & \qquad \begin{aligned} z & =\frac{E}{96500}=\frac{9}{96500} \\ w & =z . i . t=1 \times \frac{9}{96500} \times 9650 \\ & =\frac{9}{10}=0.9 \mathrm{~g}\end{aligned}\end{aligned}$

Topic: Chemistry, Electrochemistry, JEE Main
One bag contains 5 white balls and 3 black balls and a second bag contains 2 white balls and 4 black balls. One ball is drawn from the first bag and placed unseen in the second bag. What is the probability that a ball now drawn from the second bag is black?

Options:
A $\frac{15}{56}$
B $\frac{35}{56}$
C $\frac{37}{56} \quad$
D $\frac{25}{48}$

The correct answer is:
$\frac{35}{56}$

Solution:
Bag I. has 5 white $+3$ black balls. Bag II. has 2 white $+4$ black balls.
$\mathrm{P}(\text { Black })_{1 \text { st bag }}=\frac{3}{8} \& \mathrm{P}(\text { White })_{1 \text { st bag }}=\frac{5}{8}$
If one ball is drawn from bag I \& placed in bag II, bag II will have 7 balls.
If black ball is drawn, then; bag II contains. $2 \mathrm{~W}+5$ Black balls $=7$ balls $\mathrm{P}($ black ball from bag 1 and black ball from bag 2 )
$=\frac{3}{8} \times \frac{5}{7}=\frac{15}{56}$
If ball is white then bag II has $3 \mathrm{w}+4$ black balls $\mathrm{P}$ (white ball from bag 1 and black ball from bag 2)
$=\frac{5}{8} \times \frac{4}{7}=\frac{20}{56}$
$\Rightarrow$ Prob(blackball) $_{\text {bagII }}=\frac{15}{56}+\frac{20}{56}=\frac{35}{56}$

Topic: Mathematics, Probability, JEE Main
One can distinguish a leaflet from the leaf by the absence of

Options:
A midrib
B petiole
C axillary bud
D venation

The correct answer is:
axillary bud

Solution:
The leaf can be of two types, simple and compound. A simple leaf is the regular one which grows on the stem and branches. The compound leaf contains a small leaf-like part which is called the leaflet. It is like a leaf only but is not formed on the main plant stem as a leaf but develops on the branch of the leaf or the petiole. To distinguish between leaf and leaflet one should notice for the presence of axillary buds which are found in a leaf (formed at the junction of leaf and stem) but not in leaflets.

Topic: Biology, Morphology of Flowering Plants, NEET
One car with a mass of $400 \mathrm{~kg}$ is traveling east at $20 \frac{\mathrm{m}}{\mathrm{s}}$ and collides with a car of mass $800 \mathrm{~kg}$ traveling west at $15 \frac{\mathrm{m}}{\mathrm{s}}$. Assuming the collision is completely inelastic, what is velocity of the first car after the collision?

Options:
A $0.0 \frac{\mathrm{m}}{\mathrm{s}}$
B $1.0 \frac{\mathrm{m}}{\mathrm{s}}$ west
C $3.33 \frac{\mathrm{m}}{\mathrm{s}}$ west
D $1.0 \frac{\mathrm{m}}{\mathrm{s}}$ east

The correct answer is:
$3.33 \frac{\mathrm{m}}{\mathrm{s}}$ west

Solution:
Since the collision is completely inelastic, momentum is conserved but energy is not. Furthermore, the two cars stick to each other and travel as one. The equation for conservation of momentum is as follows:
$p_i=p_f$
There are two inital masses with different velocities and one final mass with a single velocity. Therefore, we can write:
$m_1 v_1+m_2 v_2=m v$
Rearranging for final velocity, we get:
$v=\frac{m_1 v_1+m_2 v_2}{m}$
At this point, we can denote which direction is positive and which is negative. Since the car traveling west has more momentum, we will consider west to be positive. Substituting our values into the equation, we get:
$\begin{aligned} & v=\frac{(400)(-20)+(800)(15)}{400+800} \\ & v=\frac{-800+12000}{1200}=\frac{4000}{1200}=3.33 \frac{\mathrm{m}}{\mathrm{s}}\end{aligned}$
Since this value is positive, the final answer is $3.33 \frac{\mathrm{m}}{\mathrm{s}}$ West.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
One card is drawn randomly from a pack of 52 cards, then the probability that it is a king or spade is

Options:
A $\frac{1}{26}$
B $\frac{3}{26}$
C $\frac{4}{13}$
D $\frac{3}{13}$

The correct answer is:
$\frac{4}{13}$

Solution:
Probability that it is queen $P(A)=\frac{4}{52}=\frac{1}{13}$
Probability that it is heart$P(B)=\frac{13}{52}=\frac{1}{4}$
Probability that it is queen and heart $P(A \cap B)=\frac{1}{52}$
$\therefore P($ Queen or heart $)=P(A)+P(B)-P(A \cap B)$
$=\frac{1}{13}+\frac{1}{4}-\frac{1}{52}$ $=\frac{4+13-1}{52}=\frac{16}{52}=\frac{4}{13}$

Topic: Mathematics, Probability, JEE Main
One card is missing in a pack of 52 playing cards. If two cards are drawn randomly from the remaining cards at a time and are found to be spades, then the probability that the missing card is not a spade is

Options:
A $\frac{3}{50}$
B $\frac{39}{50}$
C $\frac{39}{52}$
D $\frac{38}{52}$

The correct answer is:
$\frac{39}{50}$

Solution:
Let $\mathrm{E}_1=$ Event of choosing a spade card
$\mathrm{E}_2=$ Event of choosing a card which is not spade
Let A denote the lost card.
$\therefore \quad \mathrm{P}\left(\mathrm{E}_1\right)=\frac{13}{52}=\frac{1}{4}$
$\begin{aligned} & P\left(E_2\right)=\frac{39}{52}=\frac{3}{4} \\ & P\left(\frac{A}{E_1}\right)=\frac{{ }^{12} C_2}{{ }^{51} C_2}=\frac{22}{425} \\ & P\left(\frac{A}{E_2}\right)=\frac{{ }^{13} C_2}{{ }^{51} C_2}=\frac{26}{425}\end{aligned}$
By using Baye's theorem,
$P\left(\frac{E_2}{A}\right)=\frac{P\left(E_2\right) P\left(\frac{A}{E_2}\right)}{P\left(E_1\right) P\left(\frac{A}{E_1}\right)+P\left(E_2\right) P\left(\frac{A}{E_2}\right)}$
$\begin{aligned} & =\frac{\frac{3}{4} \times \frac{26}{425}}{\frac{1}{4} \times \frac{22}{425}+\frac{3}{4} \times \frac{26}{425}} \\ & =\frac{3 \times 26}{22+3 \times 26}=\frac{39}{50}\end{aligned}$

Topic: Mathematics, Probability, AP EAMCET, AP EAMCET 2023 (18 May Shift 1)
One conducting U tube can slide inside another as shown in figure, maintaining electrical contacts between the tubes. The magnetic field $B$ is perpendicular to the plane of the figure. if each tube moves towards the other at a constant speed V, then the emf induced in the circuit in terms of $B, \ell$ and $\mathrm{V}$ where $\ell$ is the width of each tube will be

Options:
A
$\mathrm{B}\ell \mathrm{V}$

B
$-\mathrm{B}\ell \mathrm{V}$

C
zero

D
$2 \mathrm{B}\ell \mathrm{V}$

The correct answer is:
$2 \mathrm{B}\ell \mathrm{V}$

Solution:
$\left|\frac{d \phi}{d t}\right|=2 B \ell v$

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2005
One conducting U tube can slide inside another as shown in the figure, maintaining electrical contacts between the tubes. The magnetic field B is perpendicular to the plane of the figure. If each tube moves towards the other at a constant speed v, then the emf induced in the circuit in terms of B , l and v where l is the width of each tube, will be

Options:
A zero
B 2Blv
C Blv
D -Blv

The correct answer is:
2Blv

Solution:
Induce electro motive force(emf) = Bvl
Relative velocity = v+v= 2v
â emf. = B.l (2v)

Topic: Physics, Electromagnetic Induction, NEET
One day in the morning, Ramesh filled up $1 / 3$ bucket of hot water from geyser, to take bath. Remaining $2 / 3$ was to be filled by cold water (at room temperature) to bring mixture to a comfortable temperature. Suddenly Ramesh had to attend to something which would take some times, say 5 10 min before he could take bath. Now, he had two options (i) fill the remaining bucket completely by cold water and then attend to the work, (ii) first attend to the work and fill the remaining bucket just before taking bath. Which option do you think would have kept water warmer? Explain.

Options:

None

Solution:
According to Newton's law of cooling which gives a consequence about rate of fall of temperature of a body with respect to the difference of temperature of body and surroundings.
Now considering the first option would have keep water warmer because according to Newton's law of cooling, the rate of loss of heat is directly proportional to the difference of temperature of surrounding and water is small. In the
first case the temperature difference is small so, less amount of loss of heat energy.
and again considering the second option the difference between water and surrounding is larger, so larger amount of heat energy lost.

Topic: Physics, Thermal Properties of Matter
One dice is thrown three times and the sum of the thrown numbers is 15. The probability for which number 4 appears in first throw

Options:
A $\frac{1}{18}$
B $\frac{1}{36}$
C $)^{\frac{1}{9}}$
D ${ }^{\frac{1}{3}}$

The correct answer is:
$\frac{1}{18}$

Solution:
We have to find the bounded probability to get sum 15 when 4 appears first. Let the event of getting sum 15 of three thrown number is $A$ and the event of apearing 4 is $B$. So we have to find $P\left(\frac{A}{B}\right)$ But $P\left(\frac{A}{B}\right)=\frac{n(A \cap B)}{n(B)}$
When $n(A \cap B)$ and $n(B)$ respectively denote the number of digits in $A \cap B$ and $B$.
Now $n(B)=36$, because first throw is of 4 . So another two throws stop by $6 \times 6=36$ types. Three dices have only two throws, which starts from 4 and give sum 15 i.e., $(4,5,6)$ and $(4,6,5)$.
So, $n(A \cap B)=2, n(B)=36$ $\therefore \quad P\left(\frac{A}{B}\right)=\frac{2}{36}=\frac{1}{18}$

Topic: Mathematics, Probability, JEE Main
One division on the main scale of a vernier callipers is equal to 1 mm. Its vernier scale has 20 equal divisions which match with 16 main scale division. The least count of the vernier calipers is

Options:
A 0.02 mm
B 0.2 mm
C 0.8 mm
D 0.08 mm

The correct answer is:
0.2 mm

Solution:
20 V.S.D=16 M.S.D=16 mm⇒1 V.S.D=45 mm

Least count = 1 M.S.D-1V.S.D=1-45=0.2 mm

Topic: Physics, Units and Dimensions, NEET
One end each of a resistance \( r \), capacitor \( C \) and resistance \( 2 r \) are connected together. The other ends are respectively connected to the positive terminals of batteries, \( P, Q, R \) having, respectively, emf's \( E, E \) and \( 2 E \). The negative terminals of the batteries are then connected together. In this circuit, with steady current the potential drop across the capacitor is,

Options:
A \( \frac{E}{3} \)
B \( \frac{E}{2} \)
C \( \frac{2 E}{3} \)
D \( E \)

The correct answer is:
\( \frac{E}{3} \)

Solution:
In the steady state, no current flows through the capacitor branch.
In the steady state, the middle branch having the capacitor acts as an open switch and hence, the effective emf of the circuit in the steady state is E+2E=3E and the current in the circuit is,
i=net emfnet resistance
=2E-Er+2r=E3r
So, potential drop across capacitor is,
V=ir=E3r×r=E3

Topic: Physics, Current Electricity, JEE Main
One end of a copper rod of uniform cross-section and of length 1.5 m is kept in contact with ice at 0Â°C and the other end with water at 100Â°C. If the whole system is insulated from its surroundings, then at what distance from the end of the rod which is in contact with the steam, the temperature should be maintained at 200Â°C, so that in steady-state the rate of melting of ice be equal to the rate of production of steam? [Lfusion=80 cal g-1, Lvaporization=540 cal g-1]

Options:
A 10.34 cm
B 16.48 cm
C 21.68 cm
D 27.87 cm

The correct answer is:
10.34 cm

Solution:
If the point is at a distance x from water at 100oC, heat conducted to ice in time t,
Q ice = KA 200 - 0 1.5 - x × t

So ice melted by this heat
m ice = Q ice L F = KA 80 200 - 0 1.5 - x × t
Similarly heat conducted by the rod to the water at 100oC in time t,
Q water = KA 200 - 100 x t
So steam formed by this heat
m steam = Q water L V = KA 200 - 100 540 × x t
According to given problem mice = msteam, i.e.,
200 80 1.5 - x = 100 540 × x or x = 6 58 m = 10.34 cm
i.e., 200oC temperature must be maintained at a distance 10.34 cm from water at 100oC.

Topic: Physics, Thermal Properties of Matter, JEE Main
One end of a light spring of natural length d and spring constant k is fixed on a rigid wall and the other is fixed to a smooth ring of mass m which can slide without friction in a vertical rod fixed at a distance d from the wall. Initially, the spring makes an angle of 37Â° with the horizontal as shown in the diagram. When the system is released from rest, find the speed of the ring when the spring becomes horizontal. (sin 37Â° = 3/5)

Options:
A d 3g 2d + k 16m
B d 2g 3d - k 21m
C d - 4g 2d + k 12m
D d - 8g 4d + k 28m

The correct answer is:
d 3g 2d + k 16m

Solution:
If lâ¡ is the stretched length of the spring
d lâ¡ = cos 37 o = 4 5 , i.e., lâ¡ = 5 4 d
So, the stretch y = lâ¡ - d = 5 4 d - d = d 4
and AB = h = lâ¡ sin 37 o = 5 4 d Ã 3 5 = 3 4 d
Now taking point B as a reference level and applying the law of conservation of Energy between A and B,
EA = EB
mgh + 1 2 ky 2 + 0 = 0 + 0 + 1 2 mv 2
as for B, h = 0 and y = 0
or 3 4 mgd + 1 2 k d 4 2 = 1 2 mv 2
as for A, h = 3 4 d and y = 1 4 d
or v = d 3g 2d + k 16m

Topic: Physics, Work Power Energy, JEE Main
One end of a long metallic wire of length L is tied to the ceiling. The other end is tied to a massless spring of spring constant k. A mass m hangs freely from the free end of the spring. The area of cross-section and Youngâs modulus of the wire are A and Y respectively. If the mass is slightly pulled down and released, it will oscillate with a time period T equal to

Options:
A 2π mk1/2

B 2ÏmYA+kLYAk
C 2πmYAkL12

D 2πmLYA12

The correct answer is:
2πmYA+kLYAk

Solution:
Equivalent force constant for a wire is given by k=YAL. Because in case of a wire, F=YAL∆L and in case of spring, F=k∆x. Comparing these two, we find k of wire =YAL

keq=k1k2k1+k2=YALkYAL+k=YAkYA+Lk
∴T=2πmkeq=2πmYA+LkYAk

Topic: Physics, Mechanical Properties of Solids, JEE Main
One end of a massless rope, which passes over a massless and frictionless pulley $P$ is tied to a hook $C$ while the other end is free. Maximum tension that the rope can bear is $360 \mathrm{~N}$. with what value of minimum safe acceleration (in $\mathrm{ms}^{-2}$ ) can a monkey of $60 \mathrm{~kg}$ move down on the rope

Options:
A 16
B 6
C 4
D 8

The correct answer is:
4

Solution:
If monkey move downward with acceleration a then its apparent weight decreases. In that condition
Tension in string $=m(g-a)$
This should not be exceed over breaking strength of the rope i.e.
$\begin{aligned}
& 360 \geq m(g-a) \Rightarrow 360 \geq 60(10-a) \\
& \Rightarrow a \geq 4 \mathrm{~m} / \mathrm{s}^2
\end{aligned}$

Topic: Physics, Laws of Motion, JEE Main
One end of a massless rope, which passes over a massless and frictionless pulley $P$ is tied to a hook $C$ while the other end is free. Maximum tension that the rope can bear is $360 \mathrm{~N}$. With what value of maximum safe acceleration (in $\mathrm{ms}^{-2}$ ) can a man of $60 \mathrm{~kg}$ climb on the rope?

Options:
A
16

B
6

C
4

D
8

The correct answer is:
4

Solution:
$\mathrm{T}-60 \mathrm{~g}=60 \mathrm{a} ; \mathrm{T}=3000 \mathrm{~N} ; \quad \therefore \mathrm{a}=4 \mathrm{~ms}^{-2}$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2002
One end of a massless spring of constant $100 \mathrm{~N} / \mathrm{m}$ and natural length 0.5 m is fixed and the other end is connected to a particle of mass 0.5 kg lying on a frictionless horizontal table. The spring remains horizontal. If the mass is made to rotate at angular velocity of $2 \mathrm{rad} / \mathrm{s}$, then elongation of spring is

Options:
A 0.1 m
B 10 cm
C 1 cm
D 0.01 cm

The correct answer is:
1 cm

Solution:
For the circular motion acceleration towards the centre is $\frac{v^2}{r}$. The horizontal force on the particle is due to the spring and $k l$, where $l$ is the elongation and $k$ is spring constant of spring.
$\begin{aligned} & k l=\frac{m v^2}{r} & =m \omega^2 r=m \omega^2\left(l_0+l\right) \\ \Rightarrow \quad & \left(k-m \omega^2\right) l & =m \omega^2 l_0 \\ \Rightarrow & l & =\frac{m \omega^2 l_0}{k-m \omega^2}\end{aligned}$
Putting the values $1=\frac{0.5 \times 4 \times 0.5}{100-0.5 \times 4}=\frac{1}{98} \mathrm{~m}=1 \mathrm{~cm}$

Topic: Physics, Motion In Two Dimensions, JIPMER, JIPMER 2015
One end of a massless spring of spring constant k and natural length l0 is fixed while the other end is connected to a small object of mass m lying on a frictionless table. The spring remains horizontal on the table. If the object is made to rotate at an angular velocity ω about an axis passing trough fixed end, then the elongation of the spring will be

Options:
A k-mω2l0mω2
B mω2l0k+mω2
C mω2l0k-mω2
D k+mω2l0mω2

The correct answer is:
mω2l0k-mω2

Solution:
The particle is moving in a horizontal circle, so it is accelerated towards the centre with magnitude v2r. The horizontal force on the particle is due to the spring and is given by kx, where x is the elongation and k is the spring constant.

kx=mv2r=mω2r=mω2l0+x

⇒k-mω2x=mω2l0

⇒x=mω2l0k-mω2

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2022 (27 Jun Shift 2)