datavorous/entrance-exam-dataset huggingface dataset:
A block of mass \( m \) is on an inclined plane of angle \( \theta \). The coefficient of friction between the block and the plane is \( \mu \) and \( \tan \theta>\mu \). The block is held stationary by applying a force \( P \) parallel to the plane. The direction of force pointing up the plane is taken to be positive. As \( P \) is varied from \( P_{1}=m g(\sin \theta-\mu \cos \theta) \) to \( P_{2}=m g(\sin \theta+\mu \cos \theta) \), the frictional force \( f \) versus \( P \) graph will look like

Options:
A
B
C
D

The correct answer is:

Solution:
When
P=mg (sin⁡θ-μcos⁡θ)
f=μmgcos⁡θ (upwards)
when P=mgsin⁡θ
f=0
and when P=mg(sin⁡θ+μcos⁡θ)
f=μ mgcos⁡θ (downwards)
Hence, friction is first positive, then zero and then negative.

Topic: Physics, Laws of Motion, JEE Main
A block of mass m is placed at the lowest point of a smooth vertical track of radius R. In this position, the block is given a horizontal velocity u such that the block is just able to perform a complete vertical circular motion. The acceleration of block, when its velocity is vertical is

Options:
A g
B 3g
C g10
D 22g

The correct answer is:
g10

Solution:
The initial velocity of the block to undergo a complete vertical circular motion is u=5gR

Let us assume that the speed of the block when it is moving in the vertical direction is v, then using conservation of mechanical energy we get

12m5gR2=12mv2+mgR

⇒v2=3gR

The tangential and centripetal accelerations of the block are

at=g ↓ and ac=3g ←

anet=ac2+at2=g10

Topic: Physics, Laws of Motion, JEE Main
A block of mass m is placed at the top of a smooth wedge ABC. The wedge is rotated about an axis passing through C as shown in the figure. The minimum value of angular speed ω such that the block does not slip on the wedge is-

Options:
A ( gâsinÎ¸ l )secÎ¸
B ( g  l )cosθ
C ( g l cos θ )cosθ
D g sinθ l a

The correct answer is:
( g sinθ l )secθ

Solution:
mÏ2r cosâ¡Î¸=mg sinâ¡Î¸
m Ï 2 l cos 2 Î¸=mgâsinâÎ¸
Ï= gâsinâÎ¸ lâ cos 2 Î¸ = gâsinÎ¸ l secÎ¸

Topic: Physics, Laws of Motion, JEE Main
A block of mass M is placed on a smooth horizontal surface and it is pulled by a light spring as shown in the diagram. If the ends A and B of the spring are moving with 4 m s-1 and 2 m s-1 respectively and the rate at which spring energy is increasing is 20 J s-1. then what is the value of spring force (in N)?

Options:

The correct answer is:
10

Solution:
Let, distance moved by A and B in any time interval as measured from natural length of spring are xA and xB respectively. Let, l⁡ is the natural length of spring, then elongation in spring at any instant is
x=xA-xB-l⁡
U = kx 2 2
dU dt = P = k 2 × 2 x dx dt
⇒ 2 0 = kx v A - v B
⇒ kx = 2 0 4 - 2 = 1 0 N

Topic: Physics, Oscillations, JEE Main
A block of mass m is placed on a smooth inclined wedge ABC of inclination θ as shown in the figure. The wedge is given an acceleration a towards the right. The relation between a and θ for the block to remain stationary on the wedge is

Options:
A a=gcos⁡θ
B a=ssin⁡θ
C a=gcosce⁡θ
D a=gtan⁡θ

The correct answer is:
a=gtan⁡θ

Solution:
In non-inertial frame of wedge
Using pseudo force,
Free body diagram is given as

Normal reaction is balanced by mgcosθ

Where m is mass of the body,

g is the acceleration due to gravity

a is the acceleration produced by external force..
macos⁡θ=mgsin⁡θ

a=gsinθcosθ
a=gtan⁡θ

Topic: Physics, Laws of Motion, NEET, NEET 2018
A block of mass m is placed on a surface having vertical cross section given by y=x24. If coefficient of friction is 0.5, the maximum height above the ground at which block can be placed without slipping is:

Options:
A 14 m
B 12 m
C 16 m
D 13 m

The correct answer is:
14 m

Solution:
At equilibrium, tanθ=μ.

Now, tanθ is slope of the curve. Therefore,

tanθ=dydx=x2

Now,

x2=0.5 ⇒x=1

Required height

y=x24=14 m.

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2024 (30 Jan Shift 2)
A block of mass m is placed on a surface with a vertical cross section given by y=x36. If the coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is

Options:
A 16 m
B 23 m
C 13 m
D 12 m

The correct answer is:
16 m

Solution:
For block remain stationary.
μ=tanθ (θ is angle at that point)
∴tanθ=μ=12(slope)
y=x36
∴dydx=3x26=x22⇒slope=12⇒x=1
∴y=(1)36 =16 m

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2014 (06 Apr)
A block of mass m is placed on a vertical fixed circular track and then it is given velocity v along the track at position A on track. The coefficient of friction between the block and the track varies with the angle Î¸. If the block moves on track with constant speed then the coefficient of friction is

Options:
A Î¼=sinâ¡Î¸cotâ¡Î¸+v2Rg
B Î¼=sinâ¡Î¸cosâ¡Î¸+v2Rg
C μ=tanθcosθ+v2R g
D μ=cosθtanθ+v2R g

The correct answer is:
μ=sin⁡θcos⁡θ+v2Rg

Solution:
Since the block is moving with a constant speed, at any instant its tangential acceleration is zero. Hence,

f=mgsinθ

Also,

N-mgcosθ=mv2R

N=mv2R+mgcosθ

f=μN=μmv2R+μmgcosθ

mgsinθ=μmv2R+μmgcosθ

⇒μ=sinθcosθ+v2R g

Topic: Physics, Laws of Motion, JEE Main
A block of mass m is placed on the top of a 6 kg cart such that the time period of the system is 0.75 s assuming there is no slipping. If the cart is displaced by 50 mm from its equilibrium position and released, then the coefficient of static friction Î¼s between block and cart is just sufficient to prevent the block from sliding. The value of m and Î¼s respectively are (Take g=9.8 m/s2)

Options:
A 1.63kg; 0.251
B 2.55kg; 0.385
C 3.42kg; 0.632
D 4.28kg; 0.876

The correct answer is:
2.55kg; 0.385

Solution:
T = 2 Ï m + 6 600 ( T = 2 Ï m k )
or 0.75 = 2 Ï m + 6 600
â´ m = 0.75 2 Ã 600 2 Ï 2 - 6
= 2.55 kg
Maximum acceleration of SHM is,
amax = Ï2A (A = amplitude)
i.e., maximum force on mass 'm' is m Ï2 A which is being provided by the force of friction between the mass and the cart. Therefore,
Î¼smg â¥ mÏ2 A
or Î¼ s â¥ Ï 2 A g
or Î¼ s â¥ ( 2 Ï T ) 2 Â· A g
or Î¼ s â¥ ( 2 Ï 0.75 ) 2 ( 0.05 9.8 ) A = 50 mm
or Î¼s â¥ 0.358
Thus, the minimum value of Î¼s should be 0.358.

Topic: Physics, Oscillations, JEE Main
A block of mass $M$ is pulled along a horizontal frictionless surface by a rope of mass $\mathrm{m}$. If a force $\mathrm{P}$ is applied at the free end of the rope, the force exerted by the rope on the block is

Options:
A
$\frac{\mathrm{Pm}}{\mathrm{M}+\mathrm{m}}$

B
$\frac{\mathrm{Pm}}{\mathrm{M}-\mathrm{m}}$

C
$P$

D
$\frac{\mathrm{PM}}{\mathrm{M}+\mathrm{m}}$

The correct answer is:
$\frac{\mathrm{PM}}{\mathrm{M}+\mathrm{m}}$

Solution:
Since the displacement for both block and rope is same so, the acceleration must be same for both
$\Rightarrow \mathrm{p}=(\mathrm{m}+\mathrm{M}) \mathrm{a} \quad \Rightarrow \mathrm{a}=\frac{\mathrm{P}}{\mathrm{m}+\mathrm{M}}$
$\mathrm{T}=\mathrm{M} \cdot \mathrm{a}=\frac{\mathrm{PM}}{\mathrm{m}+\mathrm{M}}$

Topic: Physics, Laws of Motion, JEE Main
A block of mass $M$ is pulled along a horizontal frictionless surface by a rope of mass $\mathrm{m}$. If a force $\mathrm{P}$ is applied at the free end of the rope, the force exerted by the rope on the block is

Options:
A
$\frac{\mathrm{Pm}}{\mathrm{M}+\mathrm{m}}$

B
$\frac{\mathrm{Pm}}{\mathrm{M}-\mathrm{m}}$

C
$P$

D
$\frac{\mathrm{PM}}{\mathrm{M}+\mathrm{m}}$

The correct answer is:
$\frac{\mathrm{PM}}{\mathrm{M}+\mathrm{m}}$

Solution:
Since the displacement for both block and rope is same so, the acceleration must be same for both
$\Rightarrow \mathrm{p}=(\mathrm{m}+\mathrm{M}) \mathrm{a} \quad \Rightarrow \mathrm{a}=\frac{\mathrm{P}}{\mathrm{m}+\mathrm{M}}$
$\mathrm{T}=\mathrm{M} \cdot \mathrm{a}=\frac{\mathrm{PM}}{\mathrm{m}+\mathrm{M}}$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2003
A block of mass M is pulled along a smooth horizontal surface with a rope of mass m by force F. The acceleration of the block will be

Options:
A FM-m
B FM+m
C Fm
D FM

The correct answer is:
FM+m

Solution:
When a block of mass m is pulled along smooth horizontal surface with rope of mass m by force F, then (M + m) will be the total mass of the system.
∴F=M+ma
⇒a=FM+m

Topic: Physics, Motion In One Dimension, JEE Main
A block of mass m is pulled by a uniform chain of the same mass m tied to it by applying a force F at the other end of the chain. The tension at the point which is at a distance of a quarter of the length of the chain from the end on which the force is applied, will be

Options:
A 3F4
B 7F8m
C 6F7
D 4F5

The correct answer is:
7F8m

Solution:
a=Net external force in direction of motionTotal mass
a=Fm1+m2=F2m,
T=m1+34m2a=74m×F2m

T=7F8m

Topic: Physics, Laws of Motion, JEE Main
A block of mass m is resting on a smooth horizontal surface. One end of a uniform rope of mass m 3 is fixed to the block, which is pulled in the horizontal direction by applying force F at the other end. The tension in the middle of the rope is

Options:
A 86F
B 18F
C 15F
D 78F

The correct answer is:
78F

Solution:
Acceleration of the system
a=Fm+m3
â´ Tension in the middle of the rope
T=m+m6a
orÂ Â T=7m6⋅3F4m
=78F

Topic: Physics, Laws of Motion, JEE Main
A block of mass $\mathrm{m}$ is sliding down an inclined plane with constant speed. At a certain instant $\mathrm{t}_{0}$, its height above the ground is $\mathrm{h}$. The coefficient of kinetic friction between the block and the plane is $\mu$. If the block reaches the ground at a later instant $\mathrm{t}_{\mathrm{g}}$, then the energy dissipated by friction in the time interval $\left(\mathrm{t}_{\mathrm{g}}-\mathrm{t}_{0}\right)$ is-

Options:
A $\mu \mathrm{mgh}$
B $\mu \mathrm{mgh} / \sin \theta$
C $\mathrm{mgh}$ )
D $\mu \mathrm{mgh} / \cos \theta$

The correct answer is:
$\mathrm{mgh}$ )

Solution:
$\begin{array}{l}
\mathrm{W}_{\mathrm{f}}+\mathrm{W}_{\mathrm{mg}}=\Delta \mathrm{K} . \mathrm{E} .,(\Delta \mathrm{K} . \mathrm{E} \cdot=0) \\
\mathrm{W}_{\mathrm{f}}=-\mathrm{W}_{\mathrm{mg}} \\
\mathrm{W}_{\mathrm{f}}=-\mathrm{mgh} \\
\therefore \text { Energy dissipated }=\mathrm{mgh}
\end{array}$

Topic: Physics, Work Power Energy, JEE Main
A block of mass m is stationary with respect to the wedge of mass M moving with uniform speed v on horizontal surface. Work done by friction force on the block in t seconds is

Options:
A zero
B âmgvt2 sinâ¡2Î¸
C âÂ mgvt2
D âmgvt2 sinâ¡Î¸

The correct answer is:
âmgvt2 sinâ¡2Î¸

Solution:
Since block doesn't move with respect towedge, so friction force should balance the component of weight along the inclined plane.
f=mg sinâ¡Î¸
Component of displacement along the inclined plane in time t is given by vtcosÎ¸
Work done by f=â mg sinâ¡Î¸ cosâ¡Î¸ vt
= â mg sinâ¡2Î¸2v Ãt

Topic: Physics, Work Power Energy, JEE Main
A block of mass 'm', kept on a horizontal surface, is moved through a distance 's'
by applying a horizontal force (F) to it. What is the work done by the normal
reaction?

Options:
A $\frac{\mathrm{F}}{\mathrm{s}}$
B $\mathrm{Fs}$
C zero
D $\frac{\mathrm{S}}{\mathrm{F}}$

The correct answer is:
zero

Solution:
In the first case, the centre of gravity is at a height $=\frac{b}{2}$ $\therefore$ Work done $=\mathrm{mgh}=\mathrm{mg}\left(\frac{b}{2}\right)$
In the second case, the height of the centre of gravity $=\frac{L}{2}$
$\therefore$ Work done $=m g \frac{L}{2} \therefore L>b$
$\therefore$ Extra work done $\quad=m g\left(\frac{L-b}{2}\right)$

Topic: Physics, Work Power Energy, JEE Main
A block of mass m lying on a rough horizontal plane is acted upon by a horizontal force P and another force Q inclined at an angle θ to the vertical. The block will remain in equilibrium if the coefficient of friction between it and the surface is

Options:
A P+Qsin⁡θmg+Qcos⁡θ
B Pcos⁡θ+Qmg-Qsin⁡θ
C P+Qcos⁡θmg+Qsin⁡θ
D Psin⁡θ-Qmg-Qcos⁡θ

The correct answer is:
P+Qsin⁡θmg+Qcos⁡θ

Solution:
For equilibrium of the block
N=mg+Qcos⁡θ
Qsin⁡θ+P=μN
Qsin⁡θ+P=μ(mg+Qcos⁡θ)
μ=Qsin⁡θ+Pmg+Qcos⁡θ

Topic: Physics, Laws of Motion, NEET
A block of mass m, lying on a smooth horizontal surface, is attached to a spring (of negligible mass) of spring constant k. The other end of the spring is fixed, as shown in the figure. The block is initially at rest in its equilibrium position. If now the block is pulled with a constant force F, the maximum speed of the block is:

Options:
A Fmk
B 2Fmk
C πFmk
D Fπmk

The correct answer is:
Fmk

Solution:
When block attains maximum speed net force on m is zero. ∴F=kx
Or x=Fk
From work energy theorem
-12 k Fk2+FFk=12mVmax2
F22k=12mVmax2
Vmax=Fkm

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2019 (09 Jan Shift 1)
A block of mass M placed inside a box descends vertically with acceleration a. The block exerts a force equal to one-fourth of its weight on the floor of the box. The value of 'a' will be

Options:
A g
B 3g4
C g2
D g4

The correct answer is:
3g4

Solution:
The forces acting on the block are shown below.

Here, N is normal force and box is falling with acceleration a.

From figure, Mg-N=Ma

According to question, N=Mg4

Mg4=Mg-a

⇒a=3g4

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2022 (29 Jun Shift 2)
A block of mass m rests on a horizontal table with a co-efficient of static friction $\mu .$ What minimum force must be applied on the block to drag it on the table?

Options:
A $\frac{\mu}{\sqrt{1+\mu^{2}}} \mathrm{mg}$
B $\frac{\mu-1}{\mu+1} \mathrm{mg}$
C $\frac{\mu}{\sqrt{1-\mu^{2}}} \mathrm{mg}$
D $\mu \mathrm{mg}$

The correct answer is:
$\frac{\mu}{\sqrt{1+\mu^{2}}} \mathrm{mg}$

Solution:
Hint:

$F \sin \theta+N=m g \quad N=m g-F \sin \theta$
$F \cos \theta=\mu(m g-F \sin \theta) \quad F=\frac{\mu m g}{\cos \theta+\mu \sin \theta}$,
for $F_{\min }, \frac{d}{d \theta}[\cos \theta+\mu \sin \theta]=0$
$\therefore \tan \theta=\mu$
$\therefore F_{\min }=\frac{\mu \mathrm{mg}}{\sqrt{1+\mu^{2}}}$

Topic: Physics, Laws of Motion, JEE Main
A block of mass m sits on the top of a block of mass 2m which sits on a table. The coefficient of kinetic friction between all surface is Î¼ = 1 . A massless string is connected to each mass and wraps halfway around a massless pulley, as shown. Assume that you pull on the pulley with a force of 6mg. The acceleration of your hand is -

Options:
A g2
B 2g
C 5g4
D 4g5

The correct answer is:
5g4

Solution:
The free body diagrams are :
F1 is force of friction between blocks
F2 is force of friction between block and ground
Making FBD of pulley
Since pulley is light
T=F2=3mg
Let blocks move together with common acceleration a
a=6mg-F23m=6mg-3mg3m=g
then for upper block
T-F1=ma⇒3mg-F1=mg⇒F1=2mg
but F1 should be less than fmax=μmg

So, our assumption of no relative motion b/w blocks is incorrect that means there is relative motion
⇒ F1 is kinetic =μmg
am=3mg-μmgm2gtoward sright
a2m=3mg+μmg-3μmg2m=g2toward sright
acceleration of pulley = accelerated of hand
ap=am+a2m2=2g+g22=5g4toward sright

Topic: Physics, Laws of Motion, JEE Main
A block of mass m slides down the plane inclined at angle 30° with an acceleration g4 . The value of coefficient of kinetic friction will be :

Options:
A 23+12
B 123
C 32
D 23-12

The correct answer is:
123

Solution:
Applying Newton's second law along the direction perpendicular to inlined plane:

N-mgcos30∘=0 (As θ=30°)

(since the acceleration of block in the direction perpendicular to the inclined plane is 0)

∴N=mgcos30∘ .....(1)

Let μ be the coefficient of kinetic friction between the block and the inclined plane. Applying Newton's second law along inclined plane:

mgsin30∘-μN=ma=mg4 ....(2)

(since the acceleration along the inclined plane, a=g4)

∴mgsin30∘-μmgcos30∘=mg4 (from equation 1 and 2)

⇒mg2-32μmg=mg4

⇒μ=123

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2023 (29 Jan Shift 1)
A block of mass M slides on a frictionless surface with an initial speed of v0. On top of the block, there is a small block of mass m. The coefficient of friction between the blocks is Î¼. The system of blocks encounters an ideal spring with force constant k. The maximum value of k for which the blocks don't slip relative to each other is

Options:
A μgv02m+M2m
B μgv02m+M
C μgv02m+M2M
D μgv02M-m

The correct answer is:
μgv02m+M

Solution:
12M+mv02=12kx2

x=v0M+mk

Also for the upper block

fmax=μmg=mkm+Mx

μg=km+Mv0m+Mk

⇒km+M=μgv02

⇒k=μgv02m+M

Topic: Physics, Work Power Energy, JEE Main
A block of mass M with the semicircular track of radius R rests on a horizontal smooth surface. A cylinder of radius r slips on the track. If the cylinder is released from rest from the top, the distance moved by block when the cylinder reaches the bottom of the track is

Options:
A R-r
B mR-rM+m
C MM+m R- r
D MM-m r

The correct answer is:
mR-rM+m

Solution:
COM of system Block + cylinder remains at rest. If x is the distance moved by block when cylinder moves from top to bottom M + m x=m R-r
X=mR-rM+m

Topic: Physics, Center of Mass Momentum and Collision, NEET
A block of metal weighing 2 kg is resting on a frictionless plane (as shown in figure). It is struck by a jet releasing water at a rate of 1 kg s-1 and at a speed of 10 m s-1. Then, the initial acceleration of the block, in m s-2, will be

Options:
A 3
B 6
C 5
D 4

The correct answer is:
5

Solution:
Let velocity of water jet be v.

So, force applied is F=ΔpΔt=∆mvΔt

⇒ma=∆mvΔt

Acceleration of block, a=1×102

a=5 m s-2

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2022 (29 Jun Shift 1)
A block of the mass of 1 kg is moving on the x-axis. A force F acting on the block is shown. The velocity of the block at time t=2 s is -3 m s-1. What is the speed of the block at time t=4 s?

Options:
A 8 m s-1
B 2 m s-1
C 3 m s-1
D 5 m s-1

The correct answer is:
3 m s-1

Solution:
Given, the mass of block m=1 kg
The velocity of the block at time t=2  s is -3 m s-1
Impulse during t=2 s to t=4 s .
From the graph,
I=Area of ΔABC+Area of ΔCDE
=123-2×5+12(4-3)(-5)
=52-52=0
We know that,
Impulse = Change in momentum
0=mvf-mvi
⇒ mvf-mvi=0
⇒ mvf=mvi
vf=vi=-3 m s-1 or |3 m s-1|

Topic: Physics, Laws of Motion, JEE Main
A block of weight $W$ rests on a horizontal floor with coefficient of static friction $\mu$. It is desired to make the block move by applying minimum amount of force. The angle $\theta$ from the horizontal at which the force should be applied and magnitude of the force $F$ are respectively.

Options:
A
$\theta=\tan ^{-1}(\mu), F=\frac{\mu W}{\sqrt{1+\mu^2}}$

B
$\theta=\tan ^{-1}\left(\frac{1}{\mu}\right), F=\frac{\mu W}{\sqrt{1+\mu^2}}$

C
$\theta=0, F=\mu W$

D
$\theta=\tan ^{-1}\left(\frac{\mu}{1+\mu}\right), F=\frac{\mu W}{1+\mu}$

The correct answer is:
$\theta=\tan ^{-1}(\mu), F=\frac{\mu W}{\sqrt{1+\mu^2}}$

Solution:
Let the force $F$ is applied at an angle $\theta$ with the horizontal.
For horizontal equilibrium, $F \cos \theta=\mu R$
For vertical equilibrium, $R+F \sin \theta=\mathrm{mg}$ or, $R=m g-F \sin \theta$ Substituting this value of $R$ in eq. (i), we get
$$
\begin{aligned}
& F \cos \theta=\mu(\mathrm{mg}-F \sin \theta) \\
& =\mu \mathrm{mg}-\mu F \sin \theta \\
& \text { or, } F(\cos \theta+\mu \sin \theta)=\mu \mathrm{mg} \\
& \text { or, } \mathrm{F}=\frac{\mu \mathrm{mg}}{\cos \theta+\mu \sin \theta}
\end{aligned}
$$
For $F$ to be minimum, the denominator $(\cos \theta+\mu \sin \theta)$ should be maximum.
$$
\therefore \frac{d}{d \theta}(\cos \theta+\mu \sin \theta)=0
$$
or, $-\sin \theta+\mu \cos \theta=0$
or, $\tan \theta=\mu$
or, $\theta=\tan ^{-1}(\mu)$
Then, $\sin \theta=\frac{\mu}{\sqrt{1+\mu^2}}$ and
$$
\cos \theta=\frac{1}{\sqrt{1+\mu^2}}
$$
Hence, $F_{\text {min }}$
$$
=\frac{\mu w}{\frac{1}{\sqrt{1+\mu^2}}+\frac{\mu^2}{\sqrt{1+\mu^2}}}=\frac{\mu w}{\sqrt{1+\mu^2}}
$$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2012 (19 May Online)
A block of wood is floating in water at 0 Â°C. The temperature of water is slowly raised from 0 Â°C to 10 Â°C. With the rise in temperature, the volume of block V above water level will

Options:
A increase
B decrease
C first increase and then decrease
D first decrease and then increase

The correct answer is:
increase

Solution:
null

Topic: Physics, Thermal Properties of Matter, JEE Main
A block rests on a fixed wedge inclined at an angle $\theta$. The coefficient of friction between the block and plane is $\mu$. The maximum value of $\theta$ for the block to remain motionless on the wedge is

Options:
A $\mu=\tan \theta$
B $\mu=\sin \theta$
C $\mu=\cos \theta$
D $\mu=\cot \theta$

The correct answer is:
$\mu=\tan \theta$

Solution:
The given condition can be shown with a proper FBD of the block, as below
Let us suppose the mass of the block is $m$ and acceleration due to gravity is $g$.
Normal reaction, $N=m g \cos \theta$
Limiting friction, $(f r)_{\lim }=\mu_s N=\mu m g \cos \theta$
Net driving force, $(F)_{\text {net driving }}=m g \sin \theta$
If the block remain motionless on the wedge, then
$\begin{gathered}
(F)_{\text {net driving }} \leq(f r)_{\lim } \Rightarrow m g \sin \theta \leq \mu m g \cos \theta \\
\sin \theta \leq \mu \cos \theta \Rightarrow \tan \theta \leq \mu \Rightarrow \tan \theta_{\max }=\mu
\end{gathered}$

Topic: Physics, Laws of Motion, TS EAMCET, TS EAMCET 2020 (14 Sep Shift 1)
A block rests on a rough inclined plane making an angle of $30^{\circ}$ with the horizontal. The coefficient of static friction between the block and the plane is $0.8$. If the frictional force on the block is $10 \mathrm{~N}$, the mass of the block (in $\mathrm{kg}$ ) is (take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )

Options:
A
$2.0$

B
$4.0$

C
$1.6$

D
$2.5$

The correct answer is:
$2.0$

Solution:
$\mathrm{m}=2 \mathrm{~kg}$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2004
A block starts moving up a fixed inclined plane of inclination 60 ∘ with a velocity of 20 ms-1 and stops after 2 s. The approximate value of the coefficient of friction is [g=10 ms-2]

Options:
A 3
B 3.3
C 0.27
D 0.33

The correct answer is:
0.27

Solution:
Retardation of the block is

a=gsinθ+μcosθ=53+μ

v=u-at=0⇒a=ut⇒53+μ=202⇒μ=0.27

Topic: Physics, Laws of Motion, JEE Main
A block starts moving up an inclined plane of inclination 30° with an initial velocity of v0. It comes back to its initial position with velocity v02. The value of the coefficient of kinetic friction between the block and the inclined plane is close to 11000, The nearest integer to I is :

Options:

The correct answer is:
346

Solution:
A to B

a1=g sin 30°+μg cos 30°

=g2+μg32 ; g=10 m s2

v02-2a1s=0

s=v02a1 .....(i)

B to A

a2=g2-μ32g

V022=2a2s

s=V024a2 .....(ii)

From equation (i) and (ii)

V02a1=V024a2

⇒ a1=4a2

⇒ 5+53μ=45-53μ

⇒ 5+53μ ⇒25 3μ=15 ⇒ μ=35=0.346=3461000

So, 11000=3461000

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2020 (03 Sep Shift 2)
A boat carrying steel balls is floating on the surface of water in a tank. If the balls are thrown into the tank one by one, how will it affect the level of water

Options:
A It will remain unchanged
B It will rise
C It will fall
D First it will first rise and then fall

The correct answer is:
It will fall

Solution:
When the balls are floating on the boat, they displace their own MASS of water (Archimedes Principle), but when they are thrown into the tank, they displace their own VOLUME of water. Since the steel is denser than the water, less water is displaced when the ball is only displacing its volume, instead of its (greater) mass. Hence the water level falls.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A boat covers 24 km upstream and 36 km downstream in 6 hours while it covers 36 km upstream and 24 km downstream in 612 hours. The velocity (in kmph) of the current is

Options:

The correct answer is:
2

Solution:
Let rate of upstream = x kmph and rate of downstream = y kmph.
Then, 24x+36y=6â¦.(i) and 36x+24y=132â¦.(ii)
Solving (i) & (ii) y=12 km hr-1 and x=8 km hr-1
â´Speed upstream = 8 kmph, Speed downstream = 12 kmph.
Hence, rate of current =1212-8 kmph = 2 kmph.

Topic: Physics, Motion In One Dimension, JEE Main