datavorous/entrance-exam-dataset huggingface dataset:
A bird is perched on the top of a tree \( 20 \mathrm{~m} \) high and its elevation from a point on the ground is \( 45^{\circ} \). It flies off horizontally straight away from the observer and in one second the elevation of the bird is reduced to \( 30^{\circ} \). The speed of the bird is -

Options:
A \( 14.64 \mathrm{~m} / \mathrm{s} \)
B 17. \( 71 \mathrm{~m} / \mathrm{s} \)
C \( 12 \mathrm{~m} / \mathrm{s} \)
D None of these

The correct answer is:
\( 14.64 \mathrm{~m} / \mathrm{s} \)

Solution:
Let the bird be perched at B, the top of the tree BD and O be the observer. then, ∠BOD = 45° and BD = 20 m.
Now, the bird flying horizontally reaches M in 1 s.

Then, ∠MON = 30°, where, MN ⊥ ON

Now, BD = MN = 20 m

From triangle BOD.

tan 45°=BDOD=20OD

⇒OD = 20 m

Now, from ΔMON

tan30°=MNON

=2020+DN

⇒13=2020+DN

⇒ 20+DN=203

⇒DN=203-1

= 20 x 0.732

=14.64 m

Now,

Speed=distancetime

=BM1=DN1

= 14.64 m/s

Topic: Mathematics, Heights and Distances, JEE Main
A bird is sitting in a large closed cage which is placed on a spring balance. It records a weight of $25 \mathrm{~N}$. The bird (mass $\mathrm{m}=0.5 \mathrm{~kg}$ ) flies upward in the cage with an acceleration of $2 \mathrm{~m} / \mathrm{s}^2$. The spring balance will now record a weight of

Options:
A $24 \mathrm{~N}$
B $25 \mathrm{~N}$
C $26 \mathrm{~N}$
D $27 \mathrm{~N}$

The correct answer is:
$25 \mathrm{~N}$

Solution:
Since the cage is closed and we can treat bird, cage and the air as a closed (isolated) system. In this condition the force applied by the bird on cage is an internal force, due to this the reading of spring balance will not change.

Topic: Physics, Laws of Motion, JEE Main
A bird is sitting on the top of a vertical pole 20 m high and its elevation from a point O on the ground is 45°. It flies off horizontally straight away from the point O. After one second, the elevation of the bird from O is reduced to 30°. Then the speed (in m/s) of the bird is

Options:
A 2 0 2
B 2 0 3 - 1
C 4 0 2 - 1
D 4 0 3 - 2

The correct answer is:
2 0 3 - 1

Solution:
â
AD=20 m
AO=20tan45∘=201=20 m
CB=20 m
∴BO=20tan30∘=2013=203m

∴ AB=BO-AO

= 2 0 3 - 2 0

=203-1m
It took 1 sec to travel distance DC.
∴ Speed =203-1 m/sec.

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2014 (06 Apr)
A bird weighs 2 kg and is inside a closed cage of 1 kg. If it starts flying with constant velocity, then what is the weight of the bird and cage assembly

Options:
A 1.5 kg
B 2.5 kg
C 3 kg
D 4 kg

The correct answer is:
3 kg

Solution:
When the bird flies, it pushes air down to balance its weight. So the weight of the bird and closed cage assembly remains unchanged.

Topic: Physics, Laws of Motion, NEET
A black body at $227^{\circ} \mathrm{C}$ radiates heat at the rate of $7 \mathrm{cals} / \mathrm{cm}^2 \mathrm{~s}$. At a temperature of $727^{\circ} \mathrm{C}$, the rate of heat radiated in the same units will be :

Options:
A 80
B 60
C 50
D 112

The correct answer is:
112

Solution:
$\mathrm{P} \propto \mathrm{T}^4$
$$
\begin{aligned}
& \frac{\mathrm{P}_2}{\mathrm{P}_1}=\left(\frac{1000}{500}\right)^4 \\
& \mathrm{P}_2=16 \mathrm{P}_1=112
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter, JEE Main
A black body at $227^{\circ} \mathrm{C}$ radiates heat at the rate of $7 \mathrm{cals} / \mathrm{cm}^2 \mathrm{~s}$. At a temperature of $727^{\circ} \mathrm{C}$, the rate of heat radiated in the same units will be :

Options:
A 80
B 60
C 50
D 112

The correct answer is:
112

Solution:
$\mathrm{P} \propto \mathrm{T}^4$
$$
\begin{aligned}
& \frac{\mathrm{P}_2}{\mathrm{P}_1}=\left(\frac{1000}{500}\right)^4 \\
& \mathrm{P}_2=16 \mathrm{P}_1=112
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter, NEET, NEET 2009 (Mains)
A black body, at 200 K, is found to have maximum energy at a wavelength of 14 Î¼m. When its temperature is raised to 1000 K, the wavelength at which maximum energy is emitted is

Options:
A 14 Î¼m
B 70 μm
C 2.8 Î¼m
D 2.8 mm

The correct answer is:
2.8 Î¼m

Solution:
Î»m1T1=Î»m2T2âÎ»m2=Î»m1T1T2=14Ã2001000=2.8 Î¼m

Topic: Physics, Thermal Properties of Matter, JEE Main
A black body at a temperature of 227 Â°C radiates heat at the rate of 5 cal cm-2 s-1. At a temperature of 727 Â°C the rate of heat radiated per unit area in cal cm-2s-1 is

Options:
A 400
B 80
C 40
D 15

The correct answer is:
80

Solution:
According to Stefan's law radiant energy emitted by a perfectly black body per unit area per sec (ie, emissive power of black body) is directly proportional to the fourth power of its absolute temperature ie E∝T4
⇒ E1E2= T14T24
5E2= (273+227)4(273+727)4
E2=5×10005004
= 5×16=80 cal cm-2s-1

Topic: Physics, Thermodynamics, JEE Main
A black body emits radiation at the rate P when its temperature is T. At this temperature the wavelength at which the radiation has a maximum intensity is Î» 0. If at another temperature T' the power radiated is P' and wavelength at maximum intensity is Î» 0 / 2 then

Options:
A P'T'=32PT
B P'T'=16PT
C P'T'=8PT
D P'T'=4PT

The correct answer is:
P'T'=32PT

Solution:
For a black body, wavelength for maximum intensity :
λ∝1T and P∝T4 ⇒ P∝1λ4
P'=16P
∴ P'T'=32PT

Topic: Physics, Thermal Properties of Matter, JEE Main
A black body is at a temperature of 2880âK . The energy of radiation emitted by this body with wavelength between 499ânm and 500ânm is U 1, between 999ânm and 1000ânm , is U 2 and between 1499ânm and 1500ânm nm is U 3 . Wien's constant, b=2.88Ã 10 6 ânm-K, then,

Options:
A U 1 =0
B U 3 =0
C U 1 > U 2
D U 2 > U 1

The correct answer is:
U 2 > U 1

Solution:
Wien's displacement law is
Î» m T = b (b = Wien's constant)
Î» m = b T = 2.88 Ã 1 0 6 nm - K 2 8 8 0 K
Î» = 1 0 0 0 nm
Energy distribution with wavelength will be as follows :

From the graph it is clear that
U2 > U1 (in fact U2 maximum)

Topic: Physics, Thermal Properties of Matter, JEE Main
A black body is at a temperature of 5760 K. The energy of radiation emitted by the body at wavelength 250 nm is U1, at wavelength 500 nm is U2 and that at 1000 nm is U3. Wien's constant, b=2.88×106 nm K. Which of the following is correct?

Options:
A U1=0
B U3=0
C U1>U2
D \(\mathrm{U}_3 < \mathrm{U}_2 > \mathrm{U}_1\)

The correct answer is:
\(\mathrm{U}_3 < \mathrm{U}_2 > \mathrm{U}_1\)

Solution:
Given

Temperature =5760 K

Wien's constant b=2.88×106 nm K

According to Wien's displacement law, maximum amount of emitted radiation corresponding to λm=bT

λm=2.88×106 nm K5760K=500 nm

From the graph, U1<U2>U3

Topic: Physics, Thermal Properties of Matter, JEE Main
A black body of mass 34.38 g and surface area 19.2 cm2 is at an initial temperature of 400 K. It is allowed to cool inside an evacuated enclosure kept at constant temperature 300 K. The rate of cooling is 0.04 Â°C s-1. The specific heat of the body in J kg-1 K-1 is
(Stefan's constant Ï=5.73Ã10-8 W m-2 K-4)

Options:
A 2800
B 2100
C 1400
D 1200

The correct answer is:
1400

Solution:
According to Newtons's law of cooling
dÎ¸dt=ÏA(T4-T04)ms
â´ Specific heat sâ¡ =ÏA(T4-T04)mdÎ¸dt
Substituting the values
â´ s=5.73Ã10-819.2Ã10-4[44-34]Ã10834.38Ã10-3(4Ã10-2) =1400

Topic: Physics, Thermal Properties of Matter, JEE Main
A black body of mass 34.38 g and surface area 19.2 cm2 is at an initial temperature of 400 K. It is allowed to cool inside an evacuated enclosure kept at constant temperature of 300 K. The rate of cooling is 0.04 oC/s. The specific heat of body is

Options:
A 2800 J/kg-K
B 2100 J/kg-K
C 1400 J/kg-K
D 1200 J/kg-K

The correct answer is:
1400 J/kg-K

Solution:
From stefan's law,
mcdTdt=ÏT4-T04A
â c=ÏT4-T04 AmdTdt
=5.67Ã10-8 4004-3004Ã19.2Ã10-434.38Ã10-3Ã0.04
â c=1400 J/kg-K

Topic: Physics, Thermal Properties of Matter, JEE Main
A black body of temperature T is inside a chamber of temperature T0. Now the closed chamber is slightly opened to the sun such that the temperature of black body T and chamber T0 remains constant. Then,

Options:
A Black body will absorb more radiation
B Black body will absorb less radiation
C Black body will emit more energy
D Black body will emit energy equal to energy absorbed by it

The correct answer is:
Black body will emit energy equal to energy absorbed by it

Solution:
Since, the temperature of black body is constant, total heat absorbed = total heat radiated.

Topic: Physics, Thermal Properties of Matter, NEET
A black coloured solid sphere of radius R and mass M is inside a cavity with a vacuum inside. The walls of the cavity are maintained at temperature T0. The initial temperature of the sphere is 3T0. If the specific heat of the material of the sphere varies as αT3 per unit mass with the temperature T of the sphere, where α is a constant, then the time taken for the sphere to cool down to temperature 2T0 will be
(σ is Stefan Boltzmann constant)

Options:
A Mα16πR2σln32
B Mα16πR2σln163
C Mα4πR2σln32
D Mα4πR2σln163

The correct answer is:
Mα16πR2σln163

Solution:
dθdt=-σAeMsθ4-θ04
dθdt=-σAeMαθ3θ4-θ04

∫θ3θ4-θ04dθ=-σAeMα∫dt

Let θ4-θ04=x

4θ3dθ=dx

∫dx4x=-σAeMαt0t

14lnx=MσAeMαt0t

14lnθ4-θ042T03T0=-σAeMαt

14ln2T04-T043T04-T04=-σAeMαt

t=-αM4σAeln15T0480T04

=-αM4σAeln316

=-αM4σ4πR2eln316 where e=1

t=αM16πσR2ln163

Topic: Physics, Thermal Properties of Matter, JEE Main
A black coloured solid sphere of radius R and mass M is inside a cavity with a vacuum inside. The walls of the cavity are maintained at temperature T0. The initial temperature of the sphere is 3T0. If the specific heat of the material of the sphere varies as αT3 per unit mass with the temperature T of the sphere, where α is a constant, then the time taken for the sphere to cool down to temperature 2T0 will be
(σ is Stefan Boltzmann constant)

Options:
A Mα16πR2σln32
B Mα16πR2σln163
C Mα4πR2σln32
D Mα4πR2σln163

The correct answer is:
Mα16πR2σln163

Solution:
dθdt=-σAeMsθ4-θ04
dθdt=-σAeMαθ3θ4-θ04

∫θ3θ4-θ04dθ=-σAeMα∫dt

Let θ4-θ04=x

4θ3dθ=dx

∫dx4x=-σAeMαt0t

14lnx=MσAeMαt0t

14lnθ4-θ042T03T0=-σAeMαt

14ln2T04-T043T04-T04=-σAeMαt

t=-αM4σAeln15T0480T04

=-αM4σAeln316

=-αM4σ4πR2eln316 where e=1

t=αM16πσR2ln163

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2014 (19 Apr Online)
A black rectangular surface of area A emits energy E per second at 27 Â°C. If length and breadth are reduced to 13rd of initial value and temperature is raised to 327 Â°C, then energy emitted per second becomes

Options:
A 4E9
B 7E9
C 10E9
D 16E9

The correct answer is:
16E9

Solution:
E=eÏ.AT4-T04 and A=lb
When l and b changes to l3 and b3
AâA9
Eâ²E=Aâ²A327+273427+2734
â´ Eâ²E=196003004
â´ Eâ²=19Ã24ÃE
Eâ²=16E9

Topic: Physics, Thermal Properties of Matter, JEE Main
A black sulphide is formed by the action of $H_2 S$ on

Options:
A Cupric chloride
B Cadmium chloride
C Zinc chloride
D Sodium chloride

The correct answer is:
Cupric chloride

Solution:
$\mathrm{CuCl}_2+\mathrm{H}_2 \mathrm{~S} \rightarrow \underset{\text { (Black ppt.) }}{\mathrm{CuS}}+2 \mathrm{HCl}$

Topic: Chemistry, Practical Chemistry, JEE Main
A block A of mass 4 kg is placed on another block B of mass 5 kg, and the block B rests on a smooth horizontal table. If the minimum force that can be applied on A so that both the blocks move together is 12 N, the maximum force that can be applied on B for the blocks to move together will be :

Options:
A 0 N
B 25 N
C 48 N
D 27 N

The correct answer is:
0 N

Solution:
There is an error in this question. Given a smooth horizontal table, there would be no friction between the table and the lower (5 kg) block. As a result the "minimum" force on the upper (4kg) block for both blocks to move together would be 0 N.

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2014 (09 Apr Online)
A block A of mass m is in equilibrium after being suspended from the ceiling with the help of a spring of force constant k. The block B of mass m, strikes the block A with a speed v and sticks to it. The value of v for which the spring just attains its natural length is

Options:
A 6mg2k
B 3mg2k
C 2mg2k
D mg22k

The correct answer is:
6mg2k

Solution:
Let the velocity of the combined system just after the collision is v', then using conservation of linear momentum we get

mv=2mv′,
⇒v′=v2
Initial elongation of the spring is

x =mgk

Now, using the conservation of mechanical energy we get

122mv'2+12kx2=2mgx
mv24+m2g22k=2m2g2k
v=6mg2k

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A block $A$ with mass $100 \mathrm{~kg}$ is resting on another block $B$ of mass $200 \mathrm{~kg}$. As shown in figure a horizontal rope tied to a wall holds it. The coefficient of friction between $A$ and $B$ is 0.2 while coefficient of friction between $B$ and the ground is 0.3 . The minimum required force $F$ to start moving $B$ will be

Options:
A $900 \mathrm{~N}$
B $100 \mathrm{~N}$
C $1100 \mathrm{~N}$
D $1200 \mathrm{~N}$

The correct answer is:
$1100 \mathrm{~N}$

Solution:
$\begin{aligned}
& \quad F=f_{A B}+f_{B G} \\
& =\mu_{A B} m_a g+\mu_{B G}\left(m_A+m_B\right) g \\
& =0.2 \times 100 \times 10 \\
& +0.3(300) \times 10 \\
& =200+900=1100 \mathrm{~N}
\end{aligned}$

Topic: Physics, Laws of Motion, JEE Main
A block $B$ is pushed momentarily along a horizontal surface with an initial velocity $V$. If $\mu$ is the coefficient of sliding friction between B and the surface, block B will come to rest after a time.

Options:
A $\frac{g \mu}{V}$
B $\frac{g}{V}$
C $\frac{V}{g}$
D $\frac{V}{g \mu}$

The correct answer is:
$\frac{V}{g \mu}$

Solution:
By using $v=u+a t$
$\begin{aligned}
0 & =v-\mu g t \\
\Rightarrow \quad t & =\frac{V}{\mu g}
\end{aligned}$

Topic: Physics, Laws of Motion, JEE Main
A block $B$, lying on a table, weights $w$. The coefficient of static friction between the block and the table is $\mu$. Assume that, the cord between $B$ and the knot is horizontal. The maximum weight of the block $A$ for which the system will be stationary is

Options:
A $\frac{w \tan \theta}{\mu}$
B $\mu w \tan \theta$
C $\mu w \sqrt{1+\tan ^2 \theta}$
D $\mu w \sin \theta$

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

Solution:
Given that, weight of block $B=w$
Coefficient of friction between block and table $=\mu$
Let the maximum weight of block $A=w_A$
To keep the cord between block $B$ and knot horizontal, block must remains in equilibrium.
Now, from FBD of knot and block $B$, we get
$w_A=T \sin \theta$ ...(i)
and $\quad f=\mu w=T \cos \theta$ ...(ii)
By dividing Eq. (i) by Eq. (ii), we get
$\begin{aligned} \frac{w_A}{\mu w} & =\frac{T \sin \theta}{T \cos \theta} \\ w_A & =\mu w \tan \theta\end{aligned}$

Topic: Physics, Laws of Motion, AP EAMCET, AP EAMCET 2021 (23 Aug Shift 2)
A block having 12 g of a radioactive element is placed in a room. This half-life of the radioactive element is 15 years. After how many years there will be just 1.5 g of the element in the box?

Options:

The correct answer is:
45

Solution:
Given, T12=15 year
N0=12 g
N=1.5 g
∴ N=N0 e-λt
⇒ e-λt=NN0
⇒ -λt=ln⁡NN0
⇒ t=1λln⁡N0N=T120.693ln⁡N0N ∵ λ=0.693T12
=150.693×ln⁡121.5
=150.693ln(8)
=150.693×3×ln2
=15×3
=45 year

Topic: Physics, Nuclear Physics, JEE Main
A block having 12 g of an element is placed in a room. This element is a radioactive element with a half-life of 15 years. After how many years will there be just 1.5 g of the element in the box?

Options:
A 40 year
B 45 year
C 20 year
D 15 year

The correct answer is:
45 year

Solution:
Given, T12=15 year
N0=12 g
N=1.5 g
â´ N=N0 e-Î»t
â e-Î»t=NN0
â -Î»t=lnâ¡NN0
â t=1Î»lnâ¡N0N=T120.693lnâ¡N0N âµ Î»=0.693T12
=150.693Ãlnâ¡121.5
= 15 0.693 âân(8)
= 15 0.693 Ã3Ãâ n 2
=15Ã3
=45 year

Topic: Physics, Nuclear Physics, JEE Main
A block is being hung in the air from a wire which is stretched over a sonometer. The bridges of the sonometer are L cm apart when the wire is in unison with a tuning fork of frequency N. When the block is completely immersed in water, the length between the bridges is e cm for re-establishing unison. The specific gravity of the material of the block is

Options:
A L2+e2L2
B L2-e2L2
C L2L2-e2
D L2L2+e2

The correct answer is:
L2L2-e2

Solution:
When the block is being hung in air
N=12LMgÎ¼ ...(i)
Now, consider the case when the block is in water
T1+FB=Mg
âT1+ÏMÏg=Mg
â´T1=Mg1-ÏÏ
Where Ï: Density of block
Ï: Density of water
N=12eT1Î¼=12eMg1-ÏÏÎ¼
If Ïâ²: specific gravity of block then from Eqs. (i) and (ii)
eL=1-1Ïâ²
âe2L2=1-1Ïâ²
â1Ïâ²=1-e2L2
âÏâ²=L2L2-e2

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A block is dragged on a smooth horizontal plane with the help of a light rope which moves with a velocity v as shown in figure. The horizontal velocity of the block is

Options:
A v
B vsin⁡θ
C vsin⁡θ
D vcos⁡θ⁡

The correct answer is:
vsin⁡θ

Solution:
From geometry l2=x2+y2 but y is constant, hence differentiating, we have, 2ldldt=2xdxdt
But dldt=v. Hence horizontal velocity of block,
vx=dxdt
â lv=x.vx or vx=l.vx=vsinâ¡Î¸

Topic: Physics, Laws of Motion, JEE Main
A block is fastened to a horizontal spring. The block is pulled to a distance x=10 cm from its equilibrium position (at x=0) on a frictionless surface from rest. The energy of the block at x=5 cm is 0.25 J. The spring constant of the spring is ______ N m-1.

Options:

The correct answer is:
67

Solution:
At first the block is pulled to a distance x0=10 cm(extreme position).

Potential energy of block is Ui=12kx02 and kinetic energy os block, Ki=0.

Now, when the block is at distance 5 cm.

Potential energy is Uf=12kx022 and kinetic energy is Kf=0.25 J

Total energy of block will be conserved, so

12kx02+0=12kx024+0.25

⇒12kx0234=14

⇒12k3100=1⇒k=2003 N m-1

=67 N m-1

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2023 (01 Feb Shift 2)
A block is hanged from spring in a cage. Elongation in spring is x1=42 mm and x2=32 mm mm when cage moves up and down respectively with same acceleration. The expansion (in mm) in spring when the cage moves horizontally with the same acceleration.

Options:

The correct answer is:
5

Solution:
x1=mkg+a
x2=mkg-a
x3=mkg2+a2
∵ x12+x22=m2k22g2+a2=2x32

x3=x12+x222=2(16+9)2=5

Topic: Physics, Laws of Motion, JEE Main
A block is hanged from spring in a cage. Elongation in spring is x1 and x2 when the cage moves up and down respectively with the same acceleration. The expansion in spring when the cage move horizontally with the same acceleration -

Options:
A x1+x22
B x12-x222
C x12+x222
D x1x2

The correct answer is:
x12+x222

Solution:
Elongation in spring is given by x= m k ( g eff )
When the cage is moving up by acceleration a
x1=mk(g+a)
When the cage is moving down with acceleration a
x2=mkg-a
When the cage is moving with acceleration a horizontally
x3=mkg2+a2
m=mass of pendulum k=spring constant
using 12+22
∴x12+x22=m2k2.2(g2+a2)=2x32
⇒ x3=x12+x222

Topic: Physics, Oscillations, JEE Main
A block is held stationary at the position shown in the figure over the surface of a solid paraboloid. What should be the magnitude of the velocity, needed to be given to the block at this point such that it moves along the surface of the paraboloid without having any normal reaction anywhere? g=10 m s-2 r=2h, h=103 m.

Consider motion only along the plane of the paper.

Options:
A 20 m s-1
B 1023 m s-1
C 10 m s-1
D Cannot be calculated

The correct answer is:
10 m s-1

Solution:
Considering that the block is thrown horizontally with velocity u from the top point O.
-h=rtan⁡θ-12gr2u2cos2⁡θ
-h=gr22u2 ⇒ u2=gr22h
Velocity at point P
v2=u2+2gh
=gr22h+2gh=gr2+4h22h=g.3h=10×3×103=100
v=10 m s-1

Topic: Physics, Motion In Two Dimensions, JEE Main
A block is kept on a frictionless inclined surface with angle of inclination $\alpha$. The incline is given an acceleration a to keep the block stationary. Then a is equal to

Options:
A
$\mathrm{g} / \tan \alpha$

B
$\mathrm{g} \operatorname{cosec} \alpha$

C
g

D
$\mathrm{g} \operatorname{tan} \alpha$

The correct answer is:
$\mathrm{g} \operatorname{tan} \alpha$

Solution:
$m g \sin \alpha=m a \cos \alpha$
$\therefore \mathrm{a}=\mathrm{g} \tan \alpha$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2005
A block is kept on a smooth inclined plane of the angle of inclination 30Â° that moves with a constant acceleration so that the block does not slide relative to the inclined plane. Let F1 be the contact force between the block and the plane. Now the inclined plane stops and let F2 be the contact force between the two in this case. Then F1F2 is

Options:
A 1
B 43
C 2
D 32

The correct answer is:
43

Solution:
When block does not slide with respect to the inclined plane,
F1= normal reaction =mgcos⁡θ+masin⁡θ
∵a=gtan⁡θ
∴F1=mgcos⁡θ+mgtan⁡θsin⁡θ
=mgsec⁡θ=mgsec⁡30°=2mg3
When incline plane is stationary,
F2=mgcos⁡θ=mgcos⁡30°=32mg
∴F1F2=43

Topic: Physics, Laws of Motion, JEE Main
A block is lying at rest inside a bus. The maximum acceleration of the bus such that the block remain stationary (the static friction coefficient $=02$, acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

Options:
A $1 \mathrm{~ms}^{-2}$
B $0.5 \mathrm{~ms}^{-2}$
C $2 \mathrm{cms}^{-2}$
D $2 \mathrm{~ms}^{-2}$

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

Solution:
Due to acceleration of bus, block experiences a backward acceleration $a$
Block will move if $m a \geq f$
Where, $f=$ friction force
$\begin{aligned} & \Rightarrow \quad m a \geq \mu m g \Rightarrow a \geq \mu g=0.2 \times 10 \\ & \Rightarrow \quad a \geq 2 \mathrm{~ms}^{-2}\end{aligned}$
Hence, maximum acceleration of the bus such that the block remain stationary is $2 \mathrm{~m} / \mathrm{s}^2$.

Topic: Physics, Laws of Motion, JEE Main
A block is lying at rest inside a bus. The maximum acceleration of the bus such that the block remain stationary (the static friction coefficient $=02$, acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

Options:
A $1 \mathrm{~ms}^{-2}$
B $0.5 \mathrm{~ms}^{-2}$
C $2 \mathrm{cms}^{-2}$
D $2 \mathrm{~ms}^{-2}$

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

Solution:
Due to acceleration of bus, block experiences a backward acceleration $a$
Block will move if $m a \geq f$
Where, $f=$ friction force
$\begin{aligned} & \Rightarrow \quad m a \geq \mu m g \Rightarrow a \geq \mu g=0.2 \times 10 \\ & \Rightarrow \quad a \geq 2 \mathrm{~ms}^{-2}\end{aligned}$
Hence, maximum acceleration of the bus such that the block remain stationary is $2 \mathrm{~m} / \mathrm{s}^2$.

Topic: Physics, Laws of Motion, AP EAMCET, AP EAMCET 2022 (08 Jul Shift 2)
A block is placed on a rough horizontal plane. A time dependent horizontal force $\mathrm{F}=\mathrm{kt}$ acts on the block, where $\mathrm{k}$ is a positive constant. The acceleration - time graph of the block is :

Options:
A

B

C

D

The correct answer is:

Solution:
Graph (b) correctly depicts the acceleration-time graph of the block.

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2013 (25 Apr Online)
A block is placed on an inclined plane moving towards right horizontally with an acceleration a0=g. The length of the plane AC=1 m. Friction is absent everywhere. The time taken by the block to reach from C to A is: g=10 m/s2

Options:
A 1.2Â s
B 0.74 s
C 2.56Â s
D 0.42Â s

The correct answer is:
0.74 s

Solution:
Drawing free body diagram of block with respect to plane.
Acceleration of the block up the plane is
a=mg cos 30-mg sin 30m
=3-12g=3.66 m/s2
applying s =12at2
t=2sa=2Ã13.66=0.74 s

Topic: Physics, Laws of Motion, JEE Main
A block kept on a rough inclined plane, as shown in the figure, remains at rest upto a maximum force 2N down the inclined plane. The maximum external force up the inclined plane that does not move the block is 10N. The coefficient of static friction between the block and the plane is: [Take g=10 m/s2 ]

Options:
A 34
B 23
C 12
D 32

The correct answer is:
32

Solution:
2+mgsin⁡θ=μ mgcos⁡θ

mgsin⁡θ-μ mgcos⁡θ=2 ........1
mgsin⁡θ+ μ mgcos⁡θ=10 .......2

equation (1) devided by (2)
mg sin⁡θ-μcos⁡θmg sin⁡θ+ μcos⁡θ=-210=-15
1-μ35=-1+μ3
6=43 μ
μ=32

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2019 (12 Jan Shift 2)