datavorous/entrance-exam-dataset huggingface dataset:
A greenish yellow gas reacts with an alkali metal hydroxide to form a halate which can be used in fire works and safety matches. The gas and halate respectively are

Options:
A Br2,KBrO3
B Cl2,KClO3
C I2,NaIO3
D Cl2,NaClO3

The correct answer is:
Cl2,KClO3

Solution:
The gas is Cl2 and the halate used in fire works and safety matches is KClO3
3Cl2greenishyellowgas+6KOH→KClO3pot.chlorate(Halate)+5KCl+3H2O
As bromine is reddish liquid and I2 is violet solid so they are ignored here.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
A Grignard's reagent may be made by reacting magnesium with

Options:
A Methyl amine
B Diethyl ether
C Ethyl iodide
D Ethyl alcohol

The correct answer is:
Ethyl iodide

Solution:
$\underset{\text { Ethyl iodide }}{\mathrm{C}_2 \mathrm{H}_5 \mathrm{I}+\mathrm{Mg}} \xrightarrow{\text { Dry ether }} \underset{\text { Ethyl magnesiumiodide }}{\mathrm{C}_2 \mathrm{H}_5-M g-I}$

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
A group 13 element 'X' reacts with chlorine gas to produce a compound XCl3. XCl3 is electron deficient and easily reacts with NH3 to form Cl3X←NH3 adduct; however, XCl3 does not dimerize. X is:

Options:
A B
B Al
C In
D Ga

The correct answer is:
B

Solution:
BCl3

Boron react with chlorine gas to produce compound BCl3 which is electron deficient. It does not dimerize because of small boron atom. It react with ammonia to form aduct. In it ammonia act as lewis base and BCl3 act as lewis acid.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
A group 13 element 'X' reacts with chlorine gas to produce a compound XCl3. XCl3 is electron deficient and easily reacts with NH3 to form Cl3X←NH3 adduct; however, XCl3 does not dimerize. X is:

Options:
A B
B Al
C In
D Ga

The correct answer is:
B

Solution:
BCl3

Boron react with chlorine gas to produce compound BCl3 which is electron deficient. It does not dimerize because of small boron atom. It react with ammonia to form aduct. In it ammonia act as lewis base and BCl3 act as lewis acid.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2018 (16 Apr Online)
A group of students comprises of 5 boys and n girls. If the number of ways, in which a team of 3 students can randomly be selected from this group such that there is at least one boy and at least one girl in each team, is 1750, then n is equal to

Options:
A 24
B 27
C 25
D 28

The correct answer is:
25

Solution:
(5)	(n)

Boys	Girls

2	1

1	2

Here total number of ways
= 5 C 2 × n C 1 + 5 C 1 × n C 2 =1750
⇒10n+5 n( n−1 ) 2 =1750
⇒ n 2 +3n−700=0
⇒n=−28, 25
n cannot be negative
Hence n=25

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2019 (12 Apr Shift 2)
A guitar string of length 90 cm vibrates with a fundamental frequency of 120 Hz. The length of the string producing a fundamental of 180 Hz will be _____ cm

Options:

The correct answer is:
60

Solution:
The formula to calculate the fundamental frequency produced in a vibrating string is given by

f0=v2L ...1

For the first case, it can be written that

120=v2(90) ...2

Similarly, for the second case, it can be written that

180=v2(L) ...3

Divide equation (3) by equation (2) and solve to obtain the required length of the string.

180120=v2Lv290= 90L⇒L= 90×120180=60

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2023 (08 Apr Shift 2)
A gun fires a small bullet with kinetic energy $K$. Then kinetic energy of the gun while recoiling is

Options:
A $K$
B more than $K$
C less than $K$
D $\sqrt{K}$

The correct answer is:
less than $K$

Solution:
After firing, the momentum of gun and bullet is same. Therefore, by the relation,
$K=\frac{p^{2}}{2 m}$, we have
$K \propto \frac{1}{m} \quad$ (as $p$ is same)
Hence, as the mass of gun is greater, its kinetic energy will be less.

Topic: Physics, Work Power Energy, JEE Main
A gun (mass = M) fires a bullet (mass = m ) with speed Vr relative to the barrel of the gun which is inclined at an angle of 60Â° with horizontal. The gun is placed over a smooth horizontal surface. Find the recoil speed of the gun.

Options:
A V=12mVrm+M
B V=12mVrm-M
C V=12MVrm+M
D None of the above

The correct answer is:
V=12mVrm+M

Solution:
Using conservation of linear momentum in the horizontal direction.
mVrcos60°-V= MV
mVrcos 60om+M=V
V=12mVrm+M

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A gun of mass 10 kg fires 4 bullets per second. The mass of each bullet is 20 g and the velocity of the bullet, when it leaves the gun, is 300 m s-1. The force required to hold the gun while firing is

Options:
A 6 N
B 8 N
C 24 N
D 240 N

The correct answer is:
24 N

Solution:
Rate of change of momentum of the bullet in forward direction = Force required to hold the gun.
F=nmv=4Ã20Ã10-3Ã300=24 N

Topic: Physics, Laws of Motion, JEE Main
A+H2O→B+HCl
B+H2O→C+HCl
Compound A,B and C will be respectively

Options:
A PCl5,POCl3,H3PO3
B PCl5,POCl3,H3PO4
C SOCl2,POCl3,H3PO3
D PCl3,POCl3,H3PO4

The correct answer is:
PCl5,POCl3,H3PO4

Solution:
PCl5+H2O→POCl3+2HCl
POCl3+3H2O→H3PO4+3HCl

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
A + H 2 O → B + HCl B + H 2 O → C + HCl

Compound (A), (B) and (C) will be respectively :

Options:
A PCl5 , POCl3 , H3PO3
B PCl5 , POCl3 , H3PO4
C SOCl2, POCl3, H3PO3
D PCl3, POCl3, H3PO4

The correct answer is:
PCl5 , POCl3 , H3PO4

Solution:
PCl 5 + H 2 O → POCl 3 + 2 HCl A B POCl 3 + 3 H 2 O → H 3 PO 4 + 3 HCl B C

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
A He+ ion is in its first excited state. Its ionization energy is:

Options:
A 13.60 eV
B 48.36 eV
C 54.40 eV
D 6.04 eV

The correct answer is:
13.60 eV

Solution:
E=-13.6z2n2
=-13.62222=-13.60 eV
Total energy of He+ ion is in its first excited state is -13.60 eV . Hence, its ionization energy will be 13.60 eV

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2019 (09 Apr Shift 2)
A hall has a square floor of dimension 10 m×10 m (see the figure) and vertical walls. If the angle GPH between the diagonals AG and BH is cos-115, then the height of the hall (in meters) is:

Options:
A 52
B 53
C 510
D 5

The correct answer is:
52

Solution:
AG→=10i^+hj^+10k^

BH→=-10i^+hj^+10k^

cosθ=-100+h2+100200+h22

15=h2200+h2

200=4 h2

h2=50

⇒h=52

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2021 (26 Aug Shift 2)
A hall has the dimensions $10 \mathrm{~m} \times 12 \mathrm{~m} \times 14 \mathrm{~m}$. A fly starting at one corner ends up at a diametrically opposite corner. What is the magnitude of its displacement

Options:
A $1.7 \mathrm{~m}$
B $26 \mathrm{~m}$
C $36 \mathrm{~m}$
D $20 \mathrm{~m}$

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

Solution:
Diagonal of the hall $=\sqrt{l^2+b^2+h^2}$
$=\sqrt{10^2+12^2+14^2}$
$=\sqrt{100+144+196}$
$=\sqrt{400}=20 \mathrm{~m}$

Topic: Physics, Mathematics in Physics, JEE Main
A hanging block of mass m' prevents the smaller block of mass m from slipping over a movable triangular block of mass M. All the surfaces are frictionless and the strings and the pulleys are light. Value of mass m' in terms of m, M and Î¸ is

Options:
A m + M cot Î¸ - 1
B m - M cot θ + 1
C m - M cot θ - 2
D m + M cot θ + 2

The correct answer is:
m + M cot θ - 1

Solution:
Pulling force is m'g, while mass in motion is m+M+m'. So acceleration of the system will be given by
âa=m'gm+M+m'....(i)
Mass m will not slide over M if
a cosθ = g sinθ ....(ii)
From equation (i) in (ii), we get
m ′ g cos θ m + M + m ′ = g sin θ or m ′ = m + M cot θ - 1

Topic: Physics, Laws of Motion, JEE Main
A hanging block of mass m' prevents the smaller block of mass m from slipping over a movable triangular block of mass M . All the surfaces are frictionless and the strings and the pulleys are light. Value of mass m' in terms of m, M and Î¸ is

Options:
A m + M cot Î¸ - 1
B m - M cot θ + 1
C m - M cot θ - 2
D m + M cot θ + 2

The correct answer is:
m + M cot θ - 1

Solution:
Pulling force is m'g, while mass in motion is m+M+m'. So acceleration of the system will be given by
âa=m'gm+M+m'....(i)
Mass m will not slide over M if
a cosθ = g sinθ ....(ii)
From equation (i) in (ii), we get
m ′ g cos θ m + M + m ′ = g sin θ or m ′ = m + M cot θ - 1

Topic: Physics, Laws of Motion, JEE Main
A hanging mass M is connected to a four times bigger mass by using a string pulley arrangement, as shown in the figure. The bigger mass is placed on a horizontal ice-slab and being pulled by 2Mg force. In this situation, tension in the string is x5Mg for x=_____. Neglect mass of the string and friction of the block (bigger mass) with ice slab.
(Given g= acceleration due to gravity)

Options:

The correct answer is:
6

Solution:
Equation for 4M: 2Mg-T=4Ma ...1

Equation for M: T-Mg=Ma ...2

Adding both equation we get,

Mg=5Ma ⇒a=g5

Now, tension T=Mg+Ma=6Mg5

Therefore, x=6

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2022 (28 Jun Shift 1)
A harbour lies in the direction 60o south-west from a fort and at a distance of 30 km from it. A ship sets out from the harbour at noon and sails due east at 10 km/hour, then the time (in hours) in the evening at which the ship will be 70 km from the fort is

Options:

The correct answer is:
8

Solution:
Let A be the position of the harbour and O be of the fort. OA =30, OB=70. Let AB= x.

Applying cosine formula in ΔOAB, we get

cos60o=OA2+AB2−OB22OA.AB

⇒12=900+x2−49002×30×x

⇒x2−30x−4000=0

⇒x=80

AB=80 kms

Speed of ship =10 km/hr

∴ Time =8 hrs

Topic: Mathematics, Heights and Distances, JEE Main
A hard substance melts at high temperature and is an insulator in both solid and in molten state.
This solid is most likely to be a / an :

Options:
A Ionic solid
B Molecular solid
C Metallic solid
D Covalent solid

The correct answer is:
Covalent solid

Solution:
Covalent or network solid have very high melting point, and they are insulators in their solid and molten form.

Topic: Chemistry, Solid State, JEE Main, JEE Main 2021 (18 Mar Shift 2)
A health disorder that results from the deficiency of thyroxine in adults and characterized by (i) a low metabolic rate (ii) increase in body weight and (iii) tendency to retain water in tissues is :

Options:
A Cretinism
B Hypothyroidism
C Simple goiter
D Myxoedema

The correct answer is:
Myxoedema

Solution:
No Solution Available

Topic: Biology, Chemical Coordination and Integration, NEET, NEET 2009 (Mains)
A heat engine absorbs heat $Q_1$ at temperature $T_1$ and heat $Q_2$ at temperature $T_2$. Work done by the engine is $J\left(Q_1+Q_2\right)$. This data

Options:
A
violates $1^{\text {st }}$ law of thermodynamics

B
violates $1^{\text {st }}$ law of thermodynamics if $Q_1$ is $-v e$

C
violates $1^{\text {st }}$ law of thermodynamics if $Q_2$ is $-$ ve

D
does not violate $1^{\text {st }}$ law of thermodynamics

The correct answer is:
violates $1^{\text {st }}$ law of thermodynamics

Solution:
Some mechanical energy is always converted (lost) to other forms of energy.

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2002
A heat engine has an efficiency \( \eta \). Temperatures of source and sink are each decreased by \( 100 \mathrm{~K} \). Then, the efficiency of the engine.

Options:
A Increases
B Decreases
C Remains constant
D Becomes \( 1 \)

The correct answer is:
Increases

Solution:
η = 1 - T 2 T 1 = T 1 - T 2 T 1
where T1 and T2 are the temperatures of a source and sink respectively.
When T1 and T2 both are decreased by 100 K each, (T1 - T2) stays constant. T1 decreases.
∴ η increases.

Topic: Physics, Thermodynamics, JEE Main
A heat engine has an efficiency of 16. When the temperature of sink is reduced by 62°C, its efficiency get doubled. The temperature of the source is :

Options:
A 124°C
B 37°C
C 62°C
D 99°C

The correct answer is:
99°C

Solution:
η=1-TLTH ......(i)

2η=1-TL-62TH=1-TLTH+62 TH

⇒η=62 TH⇒16=62 TH⇒TH=6×62=372 K

In C∘⇒372-273=99°C

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2021 (25 Jul Shift 2)
A heat engine is involved with exchange of heat of 1915 J, –40 J, +125 J and –Q J, during one cycle achieving and efficiency of 50.0%. The value of Q is:

Options:
A 640 J
B 40 J
C 980 J
D 400 J

The correct answer is:
980 J

Solution:
Ist case when Q4 is negative η=WΣQ+=Q1+Q2+Q3+Q4Q1+Q3=0.5
⇒ 1915-40+125+Q41915+125=0.5
⇒ 1915-40+125+Q4=1020
⇒ Q4=1020-2000⇒Q4=-Q=-980J

Q=980J

IInd case when Q4 WΣQ+=Q1+Q2+Q3+Q4Q1+Q3+Q4=0.5 is positive
Q4=-1960

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2020 (02 Sep Shift 2)
A heat engine operates between a cold reservoir at temperature T2=400 K and a hot reservoir at temperature T1. It takes 300 J of heat from the hot reservoir and delivers 240 J of heat to the cold reservoir in a cycle. The minimum temperature of the hot reservoir has to be

Options:

The correct answer is:
500

Solution:
Recall the formula of efficiency of heat engine in terms of work done and heat supplied to the source, η=WQH=1-TLTH, here TL & TH, is the temperature of the sink and source⇒60300=1-400TH⇒TH=500 K.

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2021 (27 Aug Shift 2)
A heat engine operates with the cold reservoir at temperature 324 K. The minimum temperature of the hot reservoir, if the heat engine takes 300 J heat from the hot reservoir and delivers 180 J heat to the cold reservoir per cycle, is _____ K.

Options:

The correct answer is:
540

Solution:
Using the relation, Q2Q1=T2T1

Or 180300=324T1⇒T1=324×300180=540 K

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2022 (26 Jun Shift 2)
A heater coil is cut into two equal parts and only one part is now used in the heater. The heat generated will now be

Options:
A
doubled

B
four times

C
one fourth

D
halved

The correct answer is:
doubled

Solution:
$\mathrm{H}=\frac{\mathrm{V}^2 \Delta \mathrm{t}}{\mathrm{R}}$
$\mathrm{H}^{\prime}=\frac{\mathrm{V}^2}{\mathrm{R}^{\prime}} \Delta \mathrm{t} \quad$ Given $\mathrm{R}^{\prime}=\mathrm{R} / 2$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2005
A heater is designed to operate with a power of $1000 \mathrm{~W}$ in a $100 \mathrm{~V}$ line. It is connected in combination with a resistance of $10 \Omega$ and a resistance $R$, to a $100 \mathrm{~V}$ mains as shown in figure. For the heater to operate at $62.5 \mathrm{~W}$, the value of $\mathrm{R}$ should be _____$\Omega$.

Options:

The correct answer is:
5

Solution:
$\mathrm{R}_{\text {heater }}=\frac{\mathrm{V}^2}{\mathrm{P}}=\frac{(100)^2}{1000}=10 \Omega$
For heater $P=\frac{V^2}{R} \Rightarrow V=\sqrt{P R}$
$\begin{aligned} & \mathrm{V}=\sqrt{62.5 \times 10} \\ & \mathrm{~V}=25 \mathrm{v}\end{aligned}$
$\begin{aligned} & i_1=\frac{75}{10}=7.5 \mathrm{~A}, \mathrm{i}_{\mathrm{H}}=\frac{25}{10}=2.5 \mathrm{~A} . \\ & \mathrm{i}_{\mathrm{R}}=\mathrm{i}_1-\mathrm{i}_{\mathrm{H}}=5 \\ & \mathrm{~V}=\mathrm{IR} \\ & \mathrm{R}=\frac{25}{5}=5 \Omega\end{aligned}$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (08 Apr Shift 2)
A heater supplying constant power P watts is switched ON at time t=0 min to raise the temperature of a liquid kept in a calorimeter of negligible heat capacity. A student records the temperature of the liquid Tt at equal time intervals. A graph is plotted with Tt on the Y-axis versus t on the X-axis. Assume that there is no heat loss to the surroundings during heating. Then,

Options:
A the graph is a straight line parallel to the time axis
B the heat capacity of the liquid is inversely proportional to the slope of the graph
C if some heat were lost at a constant rate to the surroundings during heating, the graph would be a straight line but with a larger slope
D the internal energy of the liquid increases quadratically with time

The correct answer is:
the heat capacity of the liquid is inversely proportional to the slope of the graph

Solution:
Assuming no heat loss,

Heat gained by liquid in calorimeter = Heat supplied by heater

⇒mSTf-Ti=Pt

⇒Tf=PmSt+Ti

Comparing above equation with y=mx+c,

S!ope of line =PmS

or slope ∝1S

So, graph of T versus t is as shown below.

Topic: Physics, Thermal Properties of Matter, JEE Main
A heating curve has been plotted for a solid object as shown in the figure. If the mass of the object is 100 g, then latent heat of vapourisation for the material of the object is [Given, power supplied to the object is constant and equal to 1 kW]

Options:
A 1.5Ã106 J/kg
B 2.5Ã106 J/kg
C 3.5Ã106Â J/kg
D 9.0Ã106 J/kg

The correct answer is:
9.0Ã106 J/kg

Solution:
The total heat supplied to the block is

∫16002500pdt=mLv

1000×2500-1600=0.1Lv

Lv=9.0×106 J/kg

Topic: Physics, Thermal Properties of Matter, JEE Main
A heating element has a resistance of 100 Î© at room temperature. When it is connected to a supply of 220 V, a steady current of 2 A passes in it and temperature is 500 oC more than the room temperature. The temperature coefficient of resistance of the heating element is

Options:
A 5 ×10-4 oC-1
B 2 ×10-4 oC-1
C 1 ×10-4 oC-1
D 0.5 ×10-4 oC-1

The correct answer is:
2 ×10-4 oC-1

Solution:
Resistance of heating element after temperature increases by 500 oC=2202=110 Ω (from Ohm's law). The new resistance is related to the temperature coefficient of the heating element as 110=100 (1+α500)
α=10100 ×500⇒α=2 ×10-4 °C-1

Topic: Physics, Current Electricity, JEE Main, JEE Main 2018 (16 Apr Online)
A heating element has a resistance of 100 Î© at room temperature. When it is connected to a supply of 220 V, a steady current of 2 A passes in it and temperature is 500 oC more than the room temperature. The temperature coefficient of resistance of the heating element is

Options:
A 5 ×10-4 oC-1
B 2 ×10-4 oC-1
C 1 ×10-4 oC-1
D 0.5 ×10-4 oC-1

The correct answer is:
2 ×10-4 oC-1

Solution:
Resistance after temperature increases by 500 oC=2202=110 Î©
110=100 (1+Î±500)
Î±=10100 Ã500
Î±=2 Ã10-4 Â°C-1

Topic: Physics, Current Electricity, JEE Main
A heating element of mass $100 \mathrm{~g}$ and having specific heat of $1 \mathrm{~J} /\left(\mathrm{g}^{\circ} \mathrm{C}\right)$ is exposed to surrounding air at $27^{\circ} \mathrm{C}$. The element attains a steady state temperature of $127^{\circ} \mathrm{C}$, while absorbing 100W of electric power. If the power is switched Off, then approximate time taken by the element to cool down to $126^{\circ} \mathrm{C}$ will be (neglect radiation)

Options:
A $0.1 \mathrm{~s}$
B $1.0 \mathrm{~s}$
C $5.0 \mathrm{~s}$
D $10.0 \mathrm{~s}$

The correct answer is:
$1.0 \mathrm{~s}$

Solution:
Given, power absorbed by element, $P=100 \mathrm{~W}$
Mass of element, $m=100 \mathrm{~g}=0.1 \mathrm{~kg}$
Change in temperature,
$\Delta T=(127-126)^{\circ} \mathrm{C}=1{ }^{\circ} \mathrm{C}$
Specific heat of heating element,
$s=1 \mathrm{~J} /\left(\mathrm{g}^{\circ} \mathrm{C}\right)=1 \times 10^3 \mathrm{~J} /\left(\mathrm{kg}^{\circ} \mathrm{C}\right)$
$\therefore$ Heat rejected to cool from $127^{\circ} \mathrm{C}$ to $126^{\circ} \mathrm{C}$, $Q=m s \Delta T=0.1 \times 1 \times 10^3 \times 1=100 \mathrm{~J}$
$\therefore$ Time taken $=\frac{Q}{P}=\frac{100}{100}=1 \mathrm{~s}$

Topic: Physics, Thermal Properties of Matter, JEE Main
$A$ heating element of resistance $r$ is fitted inside an adiabatic cylinder which carries a frictionless piston of mass $m$ and cross section A as shown in diagram. The cylinder contains one mole diatomic current the element temperatures time $t$ as (a and $\beta$ are constants), while pressure remains constant. The atmospheric pressure above the piston is $P_{0}$. Then

Options:
A the rate of increase in internal energy is $\frac{5}{2} R(\alpha+\beta t)$
B the current flowing in the element is $\sqrt{\frac{5}{2 r} R(\alpha+\beta t)}$
C the piston moves upwards with constant acceleration
D the piston moves upwards with constant speed

The correct answers are:
the rate of increase in internal energy is $\frac{5}{2} R(\alpha+\beta t)$

Solution:
We know that Internal energy, $U=\frac{n f R T}{2}=\frac{5 R}{2}\left(\alpha t+\frac{1}{2} \beta t^{2}\right)$
Differentiate with respect to $t$
$$
\frac{d U}{d t}=\frac{5 R}{2}[\alpha+\beta t]
$$
But, $d Q=n C_{p} d T$
$$
\begin{array}{l}
\therefore \quad \frac{d Q}{d t}=n C_{p} \times \frac{d T}{d t} \Rightarrow i^{2} r=\frac{7}{2} R \times[\alpha+\beta t] \\
\quad i=\sqrt{\frac{7 R}{2 r}(\alpha+\beta t)} \\
p V=n R T
\end{array}
$$
or $\quad V=\frac{n R T}{p}=\frac{n R}{p}\left(\alpha t+\frac{1}{2} \beta t^{2}\right)$
$X=\frac{n R}{p A}\left(\alpha t+\frac{1}{2} \beta t^{2}\right)$
$$
v=\frac{n R}{p A}(\alpha+\beta t)
$$
and acceleration $=\frac{n R}{p A} \times \beta$
So, the rate of increase in internal energy is $\frac{5}{2} R(\alpha+\beta t)$ and the piston moves upwards with constant acceleration.

Topic: Physics, Thermodynamics, WBJEE, WBJEE 2014
A heavy ball of mass M is suspended from the ceiling of a car by a light string of mass m m≪M. When the car is at rest, the speed of transverse waves in the string is 60 ms-1. When the car has acceleration a, the wave-speed increases to 60.5 ms-1. The value of a, in terms of gravitational acceleration g, is closed to

Options:
A g10
B g20
C g5
D g30

The correct answer is:
g5

Solution:
For standing car, v=Tμ=Mgμ
For accelerating car,
υ′=(Mg)2+(Ma)2μ
⇒ 60.560=(Mg)2+(Ma)2Mg
⇒ 1+11202=(Mg)2+(Ma)2Mg
⇒ 1+2120=1+a2g2
≈2120=12a2g2
∴ a=g4120=g30≈g5

Topic: Physics, Gravitation, JEE Main, JEE Main 2019 (09 Jan Shift 1)
A heavy box is to be dragged along a rough horizontal floor. To do so, the person A pushes it at an angle 30° from the horizontal and requires a minimum force FA, while the person B pulls the box at an angle 60° from the horizontal and needs minimum force FB. If the coefficient of friction between the box and the floor is 35, the ratio FAFB is

Options:
A 32
B 23
C 3
D 53

The correct answer is:
23

Solution:
FAcos30°=μmg+FAsin30°

FBcos60°=μmg-FBsin60°

FAcos30°-μsin30°=μmg

FBcos60°+μsin60°=μmg

Divide the equation,

we get,

⇒FAFB=μmg/cos30°-μsin30°μmg/cos60°+μsin60°=23

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2014 (19 Apr Online)
A heavy box of mass $50 \mathrm{~kg}$ is moving on a horizontal surface. If co-efficient of kinetic friction between the box and horizontal surface is 0.3 then force of kinetic friction is :

Options:
A $1.47 \mathrm{~N}$
B $147 \mathrm{~N}$
C $14.7 \mathrm{~N}$
D $1470 \mathrm{~N}$

The correct answer is:
$147 \mathrm{~N}$

Solution:
$F_k=\mu_k N=0.3 \times 50 \times 9.8=147 N$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2024 (05 Apr Shift 2)
A heavy brass sphere is hung from a spring and it executes vertical vibrations with period T. The sphere is now immersed in a non-viscous liquid with a density (1/10)th that of brass. When set into vertical vibrations with the sphere remaining inside liquid all the time, the time period will be

Options:
A 910T
B 109T
C 910T
D Unchanged

The correct answer is:
Unchanged

Solution:
When the heavy brass sphere is hung from a spring and it executes vertical vibrations with time period T, time period of oscillation of spring is, T=2πmk

Now, the sphere is immersed in a non-viscous liquid with a density (1/10)th that of brass, since the time period doesn't depend upon the volume or area of sphere, it only depends upon mass of spring therefore time period will not be affected.

When set into vertical vibrations with the sphere remaining inside liquid all the time, the time period will remain unchanged.

Topic: Physics, Mechanical Properties of Fluids, AP EAMCET, AP EAMCET 2021 (19 Aug Shift 1)
A heavy, flexible uniform chain of length Ïr and mass Î»Ïr lies in a smooth semicircular tube AB of the radius r. Assuming a slight disturbance to start the chain in motion, find the velocity v with which it will emerge from the end B of the tube.

Options:
A 4gr⁡3π+π2
B 3gr2π+π5
C 2 g râ¡ 2 Ï + Ï 2
D 5gr4Ï+Ï3

The correct answer is:
2 g r⁡ 2 π + π 2

Solution:
Since friction absent, we can apply the law of conservation of energy. Centre of gravity of a semicircular arc is at a distance 2πr from the centre.
Initial potential energy=λπrg2rπ
Final potential energy=λπrg-πr2
When the chain is completely slipped off the tube, all the links of the chain have the same velocity v.

Kinetic energy of chain=12λπrv2
From COE,λπrg2πr=λπrg-πr2+12λπrv2
From which we find,v=2gr2π+π2

Topic: Physics, Work Power Energy, JEE Main