datavorous/entrance-exam-dataset huggingface dataset:
A reversible heat engine converts one-fourth of the heat input into work. When the temperature of the sink is reduced by 52 K, its efficiency is doubled. The temperature in Kelvin of the source will be ___ .

Options:

The correct answer is:
208

Solution:
Efficiency of reversible engine is η=14=WQin=1-T2 T1

From above, T2 T1=34

When the temperature of the sink is reduced by 52 K then its efficiency is doubled.

Then, T2-52 T1=12

⇒T2 T1-52 T1=12⇒34-52 T1=12⇒52 T1=34-12=14⇒T1=52×4=208 K

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2021 (25 Feb Shift 2)
A reversible reaction is in equilibrium. If a factor is changed which affect it, then

Options:
A The speed of forward and backward reaction increases
B The speed of forward and backward reaction decreases
C Only the speed of that reaction increases which nullifies the factor causing increase of speed
D No difference

The correct answer is:
Only the speed of that reaction increases which nullifies the factor causing increase of speed

Solution:
Correct option is 3. Only the speed of that reaction increases which nullifies the factor causing increase of speed A reversible reaction is in equilibrium. If a factor is changed which affect it, then Only the speed of that reaction increases which nullifies the factor causing increase of speed
hence $\mathrm{3}$ is correct answer

Topic: Chemistry, Chemical Equilibrium, JEE Main
A rifle bullet loses $1 / 20^{\text {th }}$ of its velocity in passing through a plank. The least number of such planks required just to stop the bullet is

Options:
A 5
B 10
C 11
D 20

The correct answer is:
11

Solution:
$\mathrm{v}=\left(u-\frac{u}{20}\right)=\frac{19}{20} u$
from $v^2=u^2-2 a s$
$\Rightarrow\left(\frac{19}{20} u\right)^2=u^2-2 a s \Rightarrow \frac{400}{39}=\frac{u^2}{2 a s}$
Now if the $n$ planks are arranged just to stop the bullet then again from
$\begin{aligned}
& 0=u^2-2 a n s \\
& \Rightarrow n=\frac{u^2}{2 a s}=\frac{400}{39} \\
& \Rightarrow n=10.25
\end{aligned}$
As the planks are more than 10 so we can consider $n=11$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A rifle of $20 \mathrm{~kg}$ mass can fire 4 bullets per second. The mass of each bullet is $35 \times 10^{-3} \mathrm{~kg}$ and its final velocity is $400 \mathrm{~ms}^{-1}$. Then what force must be applied on the rifle so that it does not move backwards while firing the bullets ?

Options:
A $80 \mathrm{~N}$
B $28 \mathrm{~N}$
C $-112 \mathrm{~N}$
D $-56 \mathrm{~N}$

The correct answer is:
$-56 \mathrm{~N}$

Solution:
From conservation of momentum
$\begin{aligned} & M V+(4 m v)=0 \\ \Rightarrow \quad V & =-\frac{4 m v}{M}=-\frac{4 \times 35 \times 10^{-3} \times 400}{20} \\ = & -2.8 \mathrm{~ms}^{-1}\end{aligned}$
Force applied on the rifle,
$\begin{aligned} F & =\frac{M V}{t}=-\frac{20 \times 2.8}{1} \\ & =-56 \mathrm{~N}\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A rifle with a muzzle velocity of 100 m s-1 shoots a bullet at a small target 30 m away in the same horizontal line. How high above the target must the gun be aimed so that the bullet will hit the target?

Options:
A 0.45 m
B 0.60 m
C 0.75 m
D 0.90 m

The correct answer is:
0.45 m

Solution:
3 0 = u 2 sin2 Î¸ g â 3 0 Ã 1 0 = 1 0 0 2 sin 2 Î¸
sin2 Î¸ = 0 . 0 3 â 2 Î¸ = 0 . 0 3
Î¸ = 0 . 0 1 5 â tan Î¸ = 0 . 0 1 5
h 3 0 = 0 . 0 15 â h = 0 . 4 5 m

Topic: Physics, Motion In One Dimension, JEE Main
A right triangular plate ABC of mass m is free to rotate in the vertical plane about a fixed horizontal axis through A. It is supported by a string such that the side AB is horizontal. The reaction at the support A is

Options:
A mg3
B 2mg3
C mg2
D mg

The correct answer is:
2mg3

Solution:
For equilibrium of torque about A,
mgl3=TlorT=mg3
For verticle equilibrium of plate,
RA+T=mgâ´RA=2mg3

Topic: Physics, Rotational Motion, JEE Main
A rigid ball rolls without slipping on a surface shown below.
Which one of the following is the most likely representation of the distance traveled by the ball vs time graph?

Options:
A
B
C
D

The correct answer is:

Solution:
Ans. (D)

Topic: Physics, Motion In One Dimension, JEE Main
A rigid bar of mass 15 kg is supported symmetrically by three wires, each of length 2 m. The wires at the endpoints are made of copper and the middle one is made of steel. If the tension in each wire is the same, then the ratio of the diameter of copper wire to the diameter of steel wire is [Given, Ycopper=1.1Ã1011 N m-2 and Ysteel=1.9Ã1011 N m-2]

Options:
A 1911
B 1119
C 1119
D 1911

The correct answer is:
1911

Solution:
Young's modulus of copper (Y1) = 1.1×1011 N m-2

Young's modulus of steel (Y2) = 1.9×1011 N m-2

Let d1 and d2 be the diameters of copper and steel wires respectively.

Since the tension is the same in each wire, therefore each wire has the same extension. Also, as each wire is of the same length, the strain in each of them is the same.

Y=FA∆ll

Y1Y2=d2d12

d1d2=Y2Y1=1911

Topic: Physics, Mechanical Properties of Solids, JEE Main
A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume for this process is $\mathrm{TV}^{\mathrm{x}}=$ constant, then $\mathrm{x}$ is:

Options:
A $\frac{3}{5}$
B $\frac{2}{5}$
C $\frac{2}{3}$
D $\frac{5}{3}$

The correct answer is:
$\frac{2}{5}$

Solution:
Equation of adiabatic change is

$$

\mathrm{TV}^{\gamma-1}=\text { constant }

$$

Put $\gamma=\frac{7}{5}$, we get: $\gamma-1=\frac{7}{5}-1$

$\therefore \mathrm{x}=\frac{2}{5}$

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2019 (11 Jan Shift 1)
A rigid massless rod of length 3l has two masses attached at each end as shown in the figure. The rod is pivoted at point P on the horizontal axis. When released from the initial horizontal position, its instantaneous angular acceleration will be

Options:
A g2l
B 7g3l
C g3l
D g13l

The correct answer is:
g13l

Solution:
Both the weight will produce torques in opposite directions,
τnet=Iα (where I→moment if inertia about axis of rotation)
5mgl-2mg2l=2m2l2+5ml2α
α=g13l

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2019 (10 Jan Shift 2)
A rigid square loop of side ‘a’ and carrying current I2 is lying on a horizontal surface near a long current I1 carrying wire in the same plane as shown in figure. The net force on the loop due to the wire will be:

Options:
A Attractive and equal to μ0I1I23π
B Repulsive and equal to μ0I1I24π
C Repulsive and equal to μ0I1I22π
D Zero

The correct answer is:
Repulsive and equal to μ0I1I24π

Solution:
Force on segment 1-2 and 3-4 will get cancel out.
Force on segment 4-1,
F1=μ0I1I22πa×a toward Right
Force on segment 2-3:
F2=μ0I1I22π(2a)×a toward left
Net force on loop:
Fnet=F1-F2 toward Right
=μ0I1I22π-μ0I1I24π
=μ0I1I24π toward Right or repulsive

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2019 (09 Apr Shift 1)
A rigid square loop of side ' \( \mathrm{a} \) ' and carrying current \( \mathrm{I}_{2} \) is lying on a horizontal surface near a long current \( \mathrm{I}_{1} \) carrying wire in
the same plane as shown in figure. The net force on the loop due to the wire will be:

Options:
A Attractive and equal to \( \frac{\mu_{0} \mathrm{I}_{1} \mathrm{I}_{2}}{3 \pi} \)
B Repulsive and equal to \( \frac{\mu_{0} I_{1} I_{2}}{4 \pi} \)
C Repulsive and equal to \( \frac{\mu_{0} I_{1} I_{2}}{2 \pi} \)
D Zero

The correct answer is:
Repulsive and equal to \( \frac{\mu_{0} I_{1} I_{2}}{4 \pi} \)

Solution:
Force on segment 1-2 and 3-4 will get cancel out.
Force on segment 4-1,
F1=μ0I1I22πa×a toward Right
Force on segment 2-3:
F2=μ0I1I22π(2a)×a toward left
Net force on loop:
Fnet=F1-F2 toward Right
=μ0I1I22π-μ0I1I24π
=μ0I1I24π toward Right or repulsive

Topic: Physics, Magnetic Effects of Current, JEE Main
A rigid wire consists of a semicircular portion of radius R and two straight sections. The wire is partially immerged in a perpendicular magnetic field B=B0 j^ as shown in figure. The magnetic force on the wire if it has a current i is :

Options:
A -iBR j^
B 2iBR j^
C iBR j^
D -2iBR j^

The correct answer is:
-2iBR j^

Solution:
Direction of magnetic field is in +k^

Force acting on wire is given by, F→=il→×B→

As, l=2R, we get

⇒F→=-2iRB j^

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2024 (31 Jan Shift 1)
A ring and a solid sphere roll down the same inclined plane without slipping. They start from rest. The radii of both bodies are identical and the ratio of their kinetic energies is 7x, where x is ______.

Options:

The correct answer is:
7

Solution:
In pure rolling work done by friction is zero. Hence, potential energy is converted into kinetic energy. Since initially the ring and the sphere have same potential energy, finally they will have same kinetic energy too.
∴ Ratio of kinetic energies =1
⇒7x=1⇒x=7

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2024 (27 Jan Shift 2)
A ring and a solid sphere rotating about an axis passing through their centres have same radii of gyration. The axis of rotation is perpendicular to plane of ring. The ratio of radius of ring to that of sphere is 2x. The value of x is _____.

Options:

The correct answer is:
5

Solution:
The moment of inertia of the ring about an axis perpendicular to the plane of ring is Ir=mk2.

The moment of inertia of the sphere is

Is=25mR2s

It is given that

mk2=mR2r=25mRs2 ...(i)

From equation (i)

Rr=25Rs

From the question, it can be written that

RrRs=2x

⇒RrRs=25⇒x=5

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2023 (06 Apr Shift 2)
A ring is hung on a nail. It can oscillate, without slipping or sliding (i) in its plane with a time period T1 and (ii) back and forth in a direction perpendicular to its plane, with a period T2. The ratio T1T2 will be :

Options:
A 23
B 23
C 32
D 23

The correct answer is:
23

Solution:
The time period of oscillation of physical pendulum is given by

T=2πIMgl ...(1)

Where, I is the moment of inertia of ring about given axis and l is the distance of centre of mass of ring from the axis of rotation.

Moment of inertia of ring in its plane is given by

I1=Ic+ Md2=MR2+MR2=2MR2

Moment of inertia of ring perpendicular to its plane is given by

I2=Ic+ Md2=MR22+MR2=32MR2

Now the ratio of time period is given by

T1T2=I1I2

⇒T1T2=2MR232MR2

⇒T1T2=43=23

Topic: Physics, Oscillations, JEE Main, JEE Main 2020 (05 Sep Shift 2)
A ring is suspended from a point $S$ on its rim as shown in the figure. When displaced from equilibrium, it oscillates with time period of 1 second. The radius of the ring is (take $g=\pi^2$ )

Options:
A
$0.15 \mathrm{~m}$

B
$1.5 \mathrm{~m}$

C
$1.0 \mathrm{~m}$

D
$0.5 \mathrm{~m}$

The correct answer is:
$0.15 \mathrm{~m}$

Solution:
No Solution Available

Topic: Physics, Oscillations, JEE Main, JEE Main 2012 (19 May Online)
A ring of charge with radius 0.5 m has 0.002Ï m gap. If the ring carries a charge of +1 C, the electric field at the centre is

Options:
A 7.5Ã107 N C-1
B 7.2Ã107 N C-1
C 6.2Ã107 N C-1
D 6.5Ã107 N C-1

The correct answer is:
7.2Ã107 N C-1

Solution:
Approaximate charge on the element opposite to the gap is
Îq=Q2Ïr0.002Ï
=12Ï0.5Ã2Ï1000=2Ã10-3 C
E=9Ã109Ã2Ã10-30.52=7.2Ã107 N C-1

Topic: Physics, Electrostatics, JEE Main
A ring of mass 2Ï kg and of the radius 0.25 m is making 300 rpm about an axis through its centre perpendicular to its plane. The tension (in newtons) developed in the ring is:

Options:

The correct answer is:
250

Solution:
As 2 t sin θ 2 = dm ω 2 r (for small angle sin θ 2 → θ 2 )
but dm = m ℓ θ r
As ℓ = 2 π r
∴ T = m ω 2 r 2 π

Put m = 2 π kg , ω = 1 0 π rad s-1 and r=0.25 m
∴ T = 2 5 0 N

Topic: Physics, Laws of Motion, JEE Main
A ring of mass 4 kg is uniformly charged with a linear charge density Î»=4 C cm-1 and kept on a rough horizontal surface with a friction coefficient of Î¼=Ï4. A time-varying magnetic field B=B0t2 is applied in a circular region of radius a a<r perpendicular to the plane of the ring as shown in the figure. Find out the time when the ring just starts to rotate on the surface. (take a=5 cm, g=10 m s-2 and B0=125 SI units)

Options:

The correct answer is:
4

Solution:
Induced electric field,
E(2πr)=dφdt
E=πa22B0t2πr
Torque due to field about centre of ring,
τ1=(qE)r=(λ(2πr))2πa2B0t2πrr
The ring starts rotating when,
τ due to electric field =τ due to friction.
τ1=μmgr
Solving,
t=μmg2πa2B0λ=4 s .

Topic: Physics, Electromagnetic Induction, JEE Main
A ring of mass 5 kg sliding on a frictionless vertical rod connected by a block B of mass 10 kg by the help of a massless string. Then at equilibrium, the value of Î¸ (in degree) is

Options:

The correct answer is:
60

Solution:
For ring,
From force diagram of ring,
Tcosâ¡Î¸=m1g ......(i)
For block B.
mg=T ........ (ii)
Tcosâ¡Î¸=m1g
mgcosâ¡Î¸=m1g
â´ cosâ¡Î¸=m1m=510=12
â Î¸=60o

Topic: Physics, Laws of Motion, JEE Main
A ring of mass $M$ and radius $R$ is rotating about its axis with angular velocity $\omega$. Two identical bodies each of mass $m$ are now gently attached at the two ends of a diameter of the ring. Because of this, the kinetic energy loss will be :

Options:
A
$\frac{m(M+2 m)}{M} \omega^2 R^2$

B
$\frac{M m}{(M+m)} \omega^2 R^2$

C
$\frac{M m}{(M+2 m)} \omega^2 R^2$

D
$\frac{(M+m) M}{(M+2 m)} \omega^2 R^2$

The correct answer is:
$\frac{M m}{(M+2 m)} \omega^2 R^2$

Solution:
Kinetic energy $_{\text {(rotational) }} \mathrm{K}_{\mathrm{R}}=\frac{1}{2} \mathrm{I} \omega^2$
Kinetic energy $_{\text {(translational) }} \mathrm{K}_{\mathrm{T}}=\frac{1}{2} \mathrm{Mv}^2$
$\begin{aligned}
& (\mathrm{v}=\mathrm{R} \omega) \\
& \text { M.I. }_{\text {(initial) }} \mathrm{I}_{\text {ring }}=\mathrm{MR}^2 ; \omega_{\text {initial }}=\omega \\
& \text { M.I. } _{\text { (new) }}{\mathrm{I}^{\prime}}{ }_{\text {(system) }}=\mathrm{MR}^2+2 \mathrm{mR}^2 \\
& \omega_{(\text {system) }}^{\prime}=\frac{M \omega}{M+2 m} \\
&
\end{aligned}$
Solving we get loss in $\mathrm{K.E.}$
$=\frac{\mathrm{Mm}}{(\mathrm{M}+2 \mathrm{~m})} \omega^2 \mathrm{R}^2$

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2013 (25 Apr Online)
A ring of mass m free to slide on a fixed smooth horizontal rod is attached to a particle of mass M kg by a inextensible string of length l. Initially, both M and m are at rest and the string is vertical. A horizontal velocity v0 is imparted to the particle. The maximum height up to which block will rise w.r.t its initial position is (M = 2m)

Options:
A v022g
B v024g
C v026g
D v028g

The correct answer is:
v026g

Solution:
The reduced mass of the system is

mreduced=m×2mm+2m=2m3

∆K=122m3v02=2mgh

∴h=122m3v022mg=v026g

Topic: Physics, Work Power Energy, JEE Main
A ring of the radius R is placed in the y-z plane and it carries a current i. The conducting wires which are used to supply the current to the ring can be assumed to be very long and are placed as shown in the figure. The magnitude of the magnetic field at the centre of the ring A is

Options:
A Î¼0i4ÏR22Ï2-2Ï+1
B μ0i4πR32π2-2π+1
C μ0i3πR32π2+2π-1
D μ0i3πR32π2+2π+1

The correct answer is:
μ0i4πR22π2-2π+1

Solution:
â

Magnetic field due to wire along x- axis at A

Bâ1=Î¼04ÏiR-k^,

Magnetic field due to loop in yz- plane at A

Bâ2=Î¼04Ï2ÏiRi^

Magnetic field due to wire along z-axis at A
Bâ3=Î¼04ÏiR-i^
B â = B â 1 + B â 2 + B â 3
=Î¼04ÏiR2Ï-1i^+Î¼04ÏiRk^
B=Î¼04ÏiR2Ï-12+1 =Î¼04ÏiR22Ï2-2Ï+1

Topic: Physics, Magnetic Effects of Current, JEE Main
A ring of the radius r is suspended from a point on its circumference. If the ring is made to oscillate in the plane of the figure, then the angular frequency of these small oscillations is

Options:
A 2g3r
B g2r
C 5g4r
D 4g3r

The correct answer is:
g2r

Solution:
It is a physical pendulum, the time period of which is,
T = 2 Ï I mgl
Here, I = moment of inertia of the ring about the point of suspension
= mr2 + mr2
= 2mr2
âand l = distance of the point of suspension from the centre of gravity = r
â´ T = 2 Ï 2 mr 2 mgr = 2 Ï 2r g
â´ Angular frequency Ï= 2 Ï T
or Ï = g 2r

Topic: Physics, Oscillations, JEE Main
A ring-shaped tube contains two ideal gases with equal masses and relative molar masses M1=32 and M2=28. The gases are separated by one fixed partition and another movable stopper S which can move freely without friction inside the ring. What is the value of the angle α (in degree) at equilibrium?

Options:

The correct answer is:
192

Solution:
Pressure on both sides will be equal.
i.e., p1=p2
n 1 RT V 1 = n 2 RT V 2 n = m M
or m M 1 V 1 = m M 2 V 2 or V 2 V 1 = M 1 M 2 = 3 2 2 8 = 8 7
∴ α=360°8+7×8=192°

Topic: Physics, Thermodynamics, JEE Main
A river has a steady speed of v. A man swims upstream at a distance of d and swims back to the starting point in total time t. The man can swim at a speed of 2v in still water. If the time taken by the man in still water is t0 to complete the same length of swim, then tt0 is

Options:
A 12
B 32
C 34
D 43

The correct answer is:
43

Solution:
The total distance travelled by the man will be 2d. Therefore, the time taken by the man in still water will be,

t0=2d2v=dv

When the man is going downstream its speed will be equal to

v1=vm+vr=3v

Therefore, the time taken to go the distance d will be,

t1=dv1=d3v

When the man is going upstream its speed will be equal to

v2=vm-vr=v

Therefore, the time taken to go the distance d will be,

t2=dv2=dv

Therefore the total time taken is

t=t1+t2=4d3v⇒tt0=43

Topic: Physics, Motion In One Dimension, JEE Main
A river is flowing from $\mathrm{W}$ to $\mathrm{E}$ with a speed of $5 \mathrm{~m} / \mathrm{min}$. A man can swim in still water with a velocity $10 \mathrm{~m} / \mathrm{min}$. In which direction should the man swim so as to take the shortest possible path to go to the south

Options:
A $30^{\circ}$ with downstream
B $60^{\circ}$ with downstream
C $120^{\circ}$ with downstream
D South

The correct answer is:
$120^{\circ}$ with downstream

Solution:
For shortest possible path man should swim with an angle $(90+\theta)$ with downstream. From the fig,
$\begin{aligned} & \sin \theta=\frac{v_r}{v_m}=\frac{5}{10}=\frac{1}{2} \\ & \Rightarrow \therefore \theta=30^{\circ}\end{aligned}$
So angle with downstream $=90^{\circ}+30^{\circ}=120^{\circ}$

Topic: Physics, Motion In Two Dimensions, JEE Main
A road is banked at an angle of 30 â to the horizontal for negotiating a curve of radius 1 0 3 m . At what velocity will a car experience no friction while negotiating the curve? Take g=10â ms â2

Options:
A 54 km/hr
B 72 km/hr
C 36 km/hr
D 18 km/hr

The correct answer is:
36 km/hr

Solution:
For negotiating curve without friction tan Î¸ = V 2 rg
V = rg tan Î¸
= 10 m/s
= 36 km/hr

Topic: Physics, Laws of Motion, JEE Main
A rocket has an initial mass of $20 \times 10^3 \mathrm{~kg}$. If it is to blast off with an initial acceleration of $4 \mathrm{~ms}^{-2}$, the initial thrust needed is $\left(g \cong 10 \mathrm{~ms}^{-2}\right)$

Options:
A $6 \times 10^4 \mathrm{~N}$
B $28 \times 10^4 \mathrm{~N}$
C $20 \times 10^4 \mathrm{~N}$
D $12 \times 10^4 \mathrm{~N}$

The correct answer is:
$28 \times 10^4 \mathrm{~N}$

Solution:
$\begin{aligned} & \mathrm{F}=\mathrm{m}(\mathrm{g}+\mathrm{a}) \\ & =20 \times 10^3(10+4) \\ & =28 \times10^4 \mathrm{~N}\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A rocket has to be launched from earth in such a way that it never returns. If E is the minimum energy delivered by the rocket launcher, what should be the minimum energy that the launcher should have, if the same rocket is to be launched from the surface of the moon? Assume that the density of the earth and the moon are equal and that the earth's volume is 64 times the volume of the moon.

Options:
A E64
B E4
C E32
D E16

The correct answer is:
E16

Solution:
Mearth=64 Mmoon and Rearth=4 Rmoon
Given: E=G MearthmRearth
Now, for the moon,

E'=G MmmRm
=GMearth64 mRearth4=G Mearth m16 Rearth=E16

Topic: Physics, Gravitation, JEE Main, JEE Main 2019 (08 Apr Shift 2)
A rocket has to be launched from the Earth in such a way that it never returns. If E is the minimum energy delivered by the rocket launcher, the minimum energy that the launcher should have if the same rocket is to be launched from the surface of the moon is found to be EN. What is the value of N?
[Note: Assume ρm=ρe; Volume of Earth =64× Volume of the moon]

Options:

The correct answer is:
16

Solution:
The energy corresponding to escape speed is
E=12m2gR=12m2GMR2R
=12m2GMR=GMmR
Ve=64Vm
⇒43πRe3=43×π×Rm3×64
⇒Rc=4Rm
∵ ρc=ρm
MeVe=MmVm
⇒Me64Vm=MmVm
⇒Mc=64Mm
So EmEe=MmRm×ReMe
=164×4=116 …..∵Re=4Rm
Em=E16

Topic: Physics, Gravitation, JEE Main
A rocket is fired from the Earth towards the Sun. At what distance from the Earth's centre the gravitational force on the rockets is zero? Mass of the Sun =2Ã1030 kg and mass of the earth=6Ã1024 kg. Distance between earth and sun =15Ã1010 m

Options:
A 2.6Ã108 m
B 3.2×108 m
C 3.9×109 m
D 2.3×109 m

The correct answer is:
2.6Ã108 m

Solution:
Let P be a point at a distance r from the Earth's centre, where gravitational force due to Sun and Earth are equal and opposite and hence, the gravitational force on the rocket is zero.
GMSmx-r2=GMemr2âGMSx-r2=GMer2
2Ã1030x-r2=6Ã1024r2
x-r2r2=2Ã10306Ã1024 =13Ã106
â x-rr=1033âr=31735x
â r=31375Ã1.5Ã1011=3Ã15Ã100Ã1081735
r=2.594Ã108 m
r=2.6Ã108 m from Earth

Topic: Physics, Gravitation, JEE Main
A rocket is fired upwards. Its engine explodes fully in 12 s. The height reached by the rocket as calculated from its velocity-time graph is

Options:
A 79200 m
B 158400 m
C 18400 m
D 15400 m

The correct answer is:
79200 m

Solution:
The height reached by rocket is equal to the area under the velocity-time graph

h=12×132×1200

h= 79200 m

Topic: Physics, Motion In One Dimension, JEE Main
A rocket is fired vertically from the earth with an acceleration of 2g, where g is the gravitational acceleration. On an inclined plane inside the rocket, making an angle θ with the horizontal, a point object of mass m is kept. The minimum coefficient of friction μmin between the mass and the inclined surface such that the mass does not move is:

Options:
A tan⁡2θ
B tan⁡θ
C 3tan⁡θ
D 2tan⁡θ

The correct answer is:
tan⁡θ

Solution:
3mgsin⁡θ=μ3mgcos⁡θ
μ=tan⁡θ

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2016 (09 Apr Online)
A rocket is fired with a speed u=3gR from the earth surface. What will be its speed at interstellar space?

Options:
A zero
B 2gR
C 7gR
D 3gR

The correct answer is:
7gR

Solution:
From conservation of mechanical energy
K.E.+P.E.surface=K.E.+P.E.infinity
12m3gR2+-GMmR=12m v∞2+0
or v∞=7gR

Topic: Physics, Gravitation, JEE Main