datavorous/entrance-exam-dataset huggingface dataset:
A capacitor of capacitance C=1 μF is suddenly connected to a battery of 100 V through a resistance R=100 Ω. The time taken for the capacitor to be charged to get 50 V is:

(Take ln2=0.69)

Options:
A 1.44×10-4 s
B 3.33×10-4 s
C 0.69×10-4 s
D 0.30×10-4 s

The correct answer is:
0.69×10-4 s

Solution:
V=V01-e-tRC
50=1001-e-tRC
t=0.69×10-4 sec.

Topic: Physics, Capacitance, JEE Main, JEE Main 2021 (27 Jul Shift 1)
A capacitor of capacitance C and potential V has energy E. It is connected to another capacitor of capacitance 2C and potential 2V. Then the loss of energy is x3E, where x is _______.

Options:

The correct answer is:
2

Solution:
Given, E=12CV2

Required energy loss

=12C1C2C1+C2V1-V22=12C×2CC+2CV-2V2

=26CV2=23·E

∴x=2

Topic: Physics, Capacitance, JEE Main, JEE Main 2024 (30 Jan Shift 1)
A capacitor of capacitance C is charged to a potential V. The flux of the electric field through a closed surface enclosing the positive plate of the capacitor is:

Options:
A CVε0
B Zero
C 2CVε0
D CV2ε0

The correct answer is:
CVε0

Solution:
The charge enclosed is given by qinc=CV

Gauss' law states that the flux through any closed surface is equal to the charge enclosed by the surface.

Using Gauss' law the flux is, ϕ=qinc ε0=CVε0

Topic: Physics, Capacitance, JEE Main, JEE Main 2023 (11 Apr Shift 2)
A capacitor of capacitance C, is connected across an ac source of voltage V, given by V=V0sinωt. The displacement current between the plates of the capacitor would then be given by___

Options:
A Id=V0ωCsinωt
B Id=V0ωCcosωt
C Id=V0ωCcosωt
D Id=V0ωCsinωt

The correct answer is:
Id=V0ωCcosωt

Solution:
∵q=CV

And V=V0sinωt

Id=?

As we know,

∵Id=ε0dQdt

=ε0dEAdt∵ E=Vd

=ε0AdVddt

=ε0AdddtV0sinωt

Id=CV0ωcosωt

Topic: Physics, Electromagnetic Waves, JEE Main
A capacitor of capacity $C_1$ is charged to the potential of $V_0$. On disconnecting with the battery, it is connected with a capacitor of capacity $C_2$ as shown in the adjoining figure. The ratio of energies before and after the connection of switch $S$ will be

Options:
A $\left(C_1+C_2\right) / C_1$
B $C_1 /\left(C_1+C_2\right)$
C $\mathrm{C}_1 \mathrm{C}_2$
D $C_1 / C_2$

The correct answer is:
$\left(C_1+C_2\right) / C_1$

Solution:
$\begin{aligned} & \text { Energy }(C)=\frac{q^2}{2 C} . \text { q remains same so } U \propto \frac{1}{C} \\ & \Rightarrow \frac{U_{\text {Betore }}}{U_{\text {nner }}}=\frac{C_1+C_2}{C_1}\end{aligned}$

Topic: Physics, Capacitance, JEE Main
A capacitor of capacity $C$ has charge $Q$ and stored energy is $W$. If the charge is increased to $2 Q$, the stored energy will be

Options:
A $2 W$
B $w / 2$
C $4 W$
D $W / 4$

The correct answer is:
$4 W$

Solution:
We know that the energy stored in a capacitor is given by:
$\mathrm{W}=\frac{\mathrm{Q}^2}{2 \mathrm{C}}$
Now, if the charge $Q$ increases to 2Q, then the change in energy will be given by:
$\begin{aligned} & \mathrm{W}^{\prime}=\frac{(2 \mathrm{Q})^2}{2 \mathrm{C}} \\ & \Rightarrow \frac{4 \mathrm{Q}^2}{2 \mathrm{C}}=4 \mathrm{~W}\end{aligned}$

Topic: Physics, Capacitance, JEE Main
A capacitor with capacitance 5 μF is charged to 5 μC . If the plates are pulled apart to reduce the capacitance to 2 μF, how much work is done?

Options:
A 6.25×10–6 J
B 2.55×10–6 J
C 2.16×10–6 J
D 3.75×10–6 J

The correct answer is:
3.75×10–6 J

Solution:
Uinitial=Q22C1
Ufinal=Q22C2
Work done =Ufinal-Uinitial
=Q22C2-Q22C1
=5×10-622×2×10-6-5×10-622×5×10-6
=254-52×10-6
=154×10-6 =3.75×10-6 J

Topic: Physics, Capacitance, JEE Main, JEE Main 2019 (09 Apr Shift 1)
A capillary tube $(\mathrm{A})$ is dropped in water. Another identical tube $(\mathrm{B})$ is dipped in a soap water solution. Which of the following shows the relative nature of the liquid columns in the two tubes?

Options:
A

B

C

D

The correct answer is:

Solution:
Capillary rise $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\rho \mathrm{gr}}$. As soap solution has lower $\mathrm{T}, \mathrm{h}$ will be low.

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2008
A capillary tube is dipped in water up to a depth lâ¡ and the water rises to a height h<lâ¡ in the capillary tube. The lower end of the tube is closed now and the tube is taken out of water. When the lower end of the tube is opened now the length of the liquid column in the tube will be

Options:
A l⁡
B l⁡+h
C h
D 2h

The correct answer is:
2h

Solution:
The force of surface tension of water (due to top meniscus) can lift a water column of length h<l. So with the help of two surfaces (top and bottom meniscus), we will be able to hold a length 2h

Δ P = 2 × 2 T r = 4 T r
2 T r = ρ gh
and Δ P = ρ gh ′
∴ h ′ = 2 h

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A capillary tube is immersed vertically in water and the height of the water column is x⁡ . When this arrangement is taken into a mine of depth d, the height of the water column is y⁡ . If R is the radius of earth, the ratio x⁡ y⁡ is :

Options:
A R - d R + d
B R + d R - d
C 1 - 2 d R
D 1 - d R

The correct answer is:
1 - d R

Solution:
We know that
h = 2 T cos θ ρ gr
∴ x = 2 T cos θ ρ gr
Now, value of geff at depth 'd' is
g eff = g 1 - d R
∴ y = 2 T cos θ ρ g eff · r
∴ x y = g eff g = 1 - d R

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2014 (09 Apr Online)
A capillary tube made of glass of radius 0.15 mm is dipped vertically in a beaker filled with methylene iodide (surface tension = 0.05 N m-1, density = 667 kg m-3) which rises to height h in the tube. It is observed that the two tangents drawn from observed that the two tangents drawn from liquid-glass interfaces (from opp. sides of the capillary) make an angle of 60º with one another. Then h is close to (g=10 m s-2)

Options:
A 0.049 m
B 0.087 m
C 0.137 m
D 0.172 m

The correct answer is:
0.087 m

Solution:
h=2Tcosθρgr θ=30° P=667 kg m-3; T=1/20 N m-1; r=0.15×10-3

=2×120×32667×10×0.15×10-3=0.087m

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2020 (02 Sep Shift 2)
A capillary tube of radius r is lowered into a liquid of surface tension T and density Ï . The angle of contact between the solid and the free surface of the liquid is Î¸=0Â°. During the process in which the liquid rises in the capillary, the work done by surface tension is

Options:
A πT2ρg
B 4πT2ρg
C T2ρg
D 2T2ρg

The correct answer is:
4πT2ρg

Solution:
The liquid rise in capillary tube is :
h 0 = 2 T Ï gr
The work done by surface tension will be :
W = Fh 0 = 2 Ï Tr 2 T Ï gr = 4πT2ρg

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A car A is moving with speed 40 km h-1 along a straight line 30Â° north of east and another car B is moving with same speed along a straight line 30Â° south of east. The relative velocity of car A as observed from the car B is

Options:
A 40 km h-1 north-east
B 40 km h-1 south
C 40 km h-1 north
D 40 km h-1 south-east

The correct answer is:
40 km h-1 north

Solution:
V→A=40cos⁡30o i^+20j^
V→B=40cos⁡30o i^-20j^
∴V→AB=V→A-V→B=40j^

The relative velocity of car A as observed from car B is 40 km h-1 north

Topic: Physics, Motion In Two Dimensions, JEE Main
A car accelerates from rest at a constant rate α for some time after which it decelerates at a constant rate β to come to rest. If the total time elapsed is t seconds, the total distance travelled is:

Options:
A 4αβ(α+β)t2
B 2αβ(α+β)t2
C αβ2(α+β)t2
D αβ4(α+β)t2

The correct answer is:
αβ2(α+β)t2

Solution:
v0=αt1 and 0=v0-βt2⇒v0=βt2

t1+t2=t

v01α+1β=t

⇒v0=αβtα+β

Distance = area of v-t graph

=12×t×v0=12×t×αβtα+β=αβt22(α+β)

Topic: Physics, Motion In One Dimension, JEE Main
A car accelerates from rest at a constant rate α for some time after which it decelerates at a constant rate β to come to rest. If the total time elapsed is t seconds, the total distance travelled is:

Options:
A 4αβ(α+β)t2
B 2αβ(α+β)t2
C αβ2(α+β)t2
D αβ4(α+β)t2

The correct answer is:
αβ2(α+β)t2

Solution:
v0=αt1 and 0=v0-βt2⇒v0=βt2

t1+t2=t

v01α+1β=t

⇒v0=αβtα+β

Distance = area of v-t graph

=12×t×v0=12×t×αβtα+β=αβt22(α+β)

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2021 (17 Mar Shift 1)
A car accelerates from rest at a constant rate of 2 m s-2 for some time. Then, it retards at a constant rate of 4 m s-2 and comes to rest. If the total time for which it remains in motion is 3 s, Then the total distance travelled is

Options:
A 2 m
B 3 m
C 4 m
D 6 m

The correct answer is:
6 m

Solution:
Using v=u+at or v-u=at, we find that if aâ is doubled, t will be halved.
If t is the time for accelerations, then t2 is the time for retardation.
Now, t+t2=3

or 3t2=3

t=2 s
S=12Ã2Ã2Ã2+12Ã4Ã1Ã1=4+2 m=6 m

Topic: Physics, Motion In One Dimension, JEE Main
A car accelerates from rest of u m s-1. The energy spent in this process is E J. The energy required to accelerate the car from u m s-1 to 2u m s-1 is nE J. The value of n is _____.

Options:

The correct answer is:
3

Solution:
The kinetic energy of a moving particle is given by

E=12mu2 ...(1)

When the final velocity of the particle becomes twice the initial velocity, the kinetic energy of the particle can be written as

nE=12m(2u)2-u2= 12m3u2 ...2

Divide equation (2) by equation (1) and solve to calculate the value of n.

nEE=12m3u212mu2⇒n=3

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2023 (13 Apr Shift 2)
A car is fitted with a convex side-view mirror of focal length $20 \mathrm{~cm}$. A second car $2.8 \mathrm{~m}$ behind the first car is overtaking the first car at relative speed of $15 \mathrm{~m} / \mathrm{s}$. The speed of the image of the second car as seen in the mirror of the first one is :

Options:
A
$\frac{1}{15} \mathrm{~m} / \mathrm{s}$

B
$10 \mathrm{~m} / \mathrm{s}$

C
$15 \mathrm{~m} / \mathrm{s}$

D
$\frac{1}{10} \mathrm{~m} / \mathrm{s}$

The correct answer is:
$\frac{1}{15} \mathrm{~m} / \mathrm{s}$

Solution:
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$-\frac{1}{v^2} \frac{d v}{d t}-\frac{1}{u^2} \frac{d u}{d t}=0$
$\frac{d v}{d t}=-\frac{v^2}{u^2}\left(\frac{d u}{d t}\right)$
$f=20 \mathrm{~cm}$
$\frac{1}{u}+\frac{1}{-280}=\frac{1}{20}$
$\Rightarrow v=\frac{280}{15} \mathrm{~cm}$
$v_1=-\left(\frac{280}{15 \times 280}\right)^2 \times 15$
$=\frac{1}{15} \mathrm{~m} / \mathrm{s}$

Topic: Physics, Ray Optics, JEE Main, JEE Main 2011
A car is moving along the circle x2+y2=a2 in the anti-clockwise direction with a constant speed. The x-y plane is a rough horizontal stationary surface. When the car is at the point (a cosÎ¸, a sinÎ¸), the unit vector in the direction of the friction force acting on the car is

Options:
A cos θ i ^ + sin θ j ^
B cosθ i^-sinθ j^
C -cosθi^-sinθj^
D -cosθ i^+sinθ j^

The correct answer is:
-cosθi^-sinθj^

Solution:
The direction of the friction force should be towards the centre of the circle which is

-r^=-cosθi^-sinθj^

Topic: Physics, Laws of Motion, JEE Main
A car is moving in a circular horizontal track of radius 635m A plumb bob is suspended from the roof of the car by a light rigid rod. The angle made by the rod with the vertical is 60o. Then the car moves with a constant speed of _______ m/s.g=10 m/s2

Options:

The correct answer is:
6

Solution:
Tsinâ¡Î¸=mv2R
Tcosâ¡Î¸=mg
â´âtanâ¡Î¸=v2Rg
â´âv2=tanâ¡Î¸ÃRÃg
â´âv2=tanâ¡60oÃ635Ã10
â´âv2=3Ã635Ã10
â´âv2=36
â´âv=6 m/s

Topic: Physics, Laws of Motion, JEE Main
A car is moving on a circular path of radius 600 m such that the magnitudes of the tangential acceleration and centripetal acceleration are equal. The time taken by the car to complete first quarter of revolution, if it is moving with an initial speed of 54 km h-1 is t(1–e–π2) s. The value of t is ______.

Options:

The correct answer is:
40

Solution:
Given:

Initial speed =54×10003600=15 m s-1

at=ac

⇒vdvdx=v2R⇒∫15vdvv=1R∫0xdx

⇒v=15exR

⇒dxdt=15exR

Distance covered in first quarter of revolution =2πR4=πR2.

Therefore,

∫0πR2e-xRdx=15∫0t0dt

⇒t0=401-e-π2

Hence, t=40.

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2023 (29 Jan Shift 2)
A car is moving on a horizontal curved road with radius 50 m. The approximate maximum speed of car will be, if friction between tyres and road is 0.34. [Take g=10 m s-2]

Options:
A 3.4 m s-1
B 22.4 m s-1
C 13 m s-1
D 17 m s-1

The correct answer is:
13 m s-1

Solution:
For a car moving with uniform speed along a circular track, the centripetal acceleration is provided by the frictional force acting radially inwards. Therefore, applying Newton's second law along radial direction,

μmg=mv2r

⇒v=μrg

=0.34×50×10

≈13 m s-1

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2023 (29 Jan Shift 1)
A car is moving on a plane inclined at 30° to the horizontal with an acceleration of 10 m s-2 parallel to the plane upward. A bob is suspended by a string from the roof of the car. The angle in degrees which the string makes with the vertical is (Take g=10 m s-2)

Options:

The correct answer is:
30

Solution:
First, make the free body diagram, of an inclined plane with acceleration in an inclined plane moving in an upward direction.

With respect to the car, the F.B.D. is as follows at equilibrium, T is equal and opposite to the resultant of a1 Ã¢â¬â¹ and a2 Ã¢â¬â¹ As a1+a2 Ã¢â¬â¹ makes 30°with vertical; T and, the string is also making 30° with vertical.

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2021 (31 Aug Shift 1)
A car is moving towards a high cliff. The driver sounds a horn of frequency f. The reflected sound heard by the driver has frequency 2f. If v is the velocity of sound, then the velocity of the car, in the same velocity units, will be

Options:
A v2
B v3
C v4
D v2

The correct answer is:
v3

Solution:
1st the car is the source and at the cliff, one observes f'.
∴ f'=vv-vsf
2nd cliff is now source. It emits frequency f' and the observer is now the driver who observes f''.
∴ f''=v+v0vf' or 2f=v+v0v-vsf
⇒ 2v-2v0=v+v0⇒v=v0+2v0 as vs=v0
⇒ v0=v3

Topic: Physics, Waves and Sound, JEE Main
A car is moving towards an approaching bike. the bike is moving with a velocity 30 m s-1 . Frequency of horn sounded by the car is 100 Hz , while it is observed by bike rider as 120 Hz . The actual duration of horn is 6 s . Find the time interval for which the bike rider hears it. (Take velocity of sound in air as 330 m s-1)

Options:
A 5 s
B 6 s
C 7 s
D Cannot be calculated with given information

The correct answer is:
5 s

Solution:
Velocity of bike Vb=30 m s-1
Let the velocity of car be Vc

fobs = factualÃVso+VbVso-Vc
120 Hz =100 HzÃ 330+30 330â V c
âVc=30 m s-1

Distance between initial and final pulse of horn
=VsoÃÎt-VcÎt ( Ît is duration of emission of sound)
=( 330â30 )Ã6=1800âm

This wave train travels with 330âm s-1 and cross bike rider moving with velocity of 30âm s-1 towards it.

Time taken = length of wave train V so + V b = 1800 330+30 = 180 36 =5 s

Topic: Physics, Waves and Sound, JEE Main
A car is moving with a constant speed of 20 m s-1 in a circular horizontal track of radius 40 m. A bob is suspended from the roof of the car by a massless string. The angle made by the string with the vertical will be : (Take g=10 m s-2)

Options:
A π6
B π2
C π4
D π3

The correct answer is:
π4

Solution:
Let the frame of reference to be attached to the car with x-axis along horizontal direction and y-axis along vertical direction. Since the frame of reference is a rotating frame, a centrifugal force term should be added while applying Newton's second law.

Let θ be the angle made by the pendulum with the vertical.

Applying Newton's second law for horizontal direction:

Tsinθ -mv2R=0 (since the acceleration of the pendulum in the rotating frame is 0).

∴Tsinθ =mv2R ..........(1)

Applying Newton's second law for vertical direction:

Tcosθ-mg=0

⇒Tcosθ=mg ..........(2)

Dividing equation 1 by equation 2:

tanθ=v2Rg=2024010=1⇒θ=π4

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2023 (25 Jan Shift 1)
A car is moving with at a constant speed of $60 \mathrm{~km} \mathrm{~h}^{-1}$ on a straight road. Looking at the rear view mirror, the driver finds that the car following him is at a distance of $100 \mathrm{~m}$ and is approaching with a speed of 5 $\mathrm{kmh}^{-1}$.
In order to keep track of the car in the rear, the driver begins to glance alternatively at the rear and side mirror of his car after every $2 \mathrm{~s}$ till the other car overtakes. If the two cars were maintaining their speeds, which of the following statement (s) is/are correct?

Options:
A
The speed of the car in the rear is $65 \mathrm{~km} \mathrm{~h}^{-1}$

B
In the side mirror, the car in the rear would appear to approach with a speed of $5 \mathrm{kmh}^{-1}$ to the driver of the leading car

C
In the rear view mirror, the speed of the approaching car would appear to decrease as the distance between the cars decreases

D
In the side mirror, the speed of the approaching car would appear to increase as the distance between the cars decreases

The correct answer is:
In the side mirror, the speed of the approaching car would appear to increase as the distance between the cars decreases

Solution:
As we know that, the image formed by convex mirror does not depend on the relative position of object w.r.t. mirror.
So, when the car approaches in the rear side, initially it appear at rest as images is formed at focus. Hence the speed of the image of the car would appear to increase as the distance between the cars decreases.

Topic: Physics, Ray Optics
A car is moving with speed of 150 km h-1 and after applying the brake it will move 27 m before it stops. If the same car is moving with a speed of one third the reported speed then it will stop after travelling _____ m distance.

Options:

The correct answer is:
3

Solution:
In both cases, the final velocity is zero. Using equation of motion for constant acceleration

v2=u2+2as

⇒02-u2=-2a27 ...1

⇒02-u32=-2as' ...2

From 1 and 2,

2a27=92as'

⇒s'=3 m

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2022 (25 Jul Shift 1)
A car is negotiating a curved road of radius R . The road is banked at an angle θ. The coefficient of friction between the tyres of the car and the road is μs. The maximum safe velocity on this road is:

Options:
A gR2μs+tan⁡θ1-μstan⁡θ
B gRμs+tan⁡θ1-μstan⁡θ
C gRμs+tan⁡θ1-μstan⁡θ
D gR2μs+tanθ1-μstan⁡θ

The correct answer is:
gRμs+tan⁡θ1-μstan⁡θ

Solution:
N=mgcos⁡θ+mv2Rsin⁡θ
f max =μN ⇒ fmax= μ s mgcos⁡θ+ μ s mv2Rsin⁡θ
mg⁡sinθ+fmax=mv2R⁡cosθ
mgsin⁡θ+ μ s mgcos⁡θ+ μ s mv2Rsin⁡θ=mv2Rcos⁡θ
gsin⁡θ+ μ s gcos⁡θ=v2Rcos⁡θ- μ s sin⁡θ
gRtan⁡θ+ μ s 1- μ s tan⁡θ=v2

Topic: Physics, Laws of Motion, JEE Main
A car is standing 200 m behind a bus, which is also at rest. The two start moving at the same instant but with different forward accelerations. The bus has acceleration 2 m s-2 and the car has acceleration 4 m s-2 . The car will catch up with the bus after time :

Options:
A 120 s
B 15 s
C 110 s
D 102 s

The correct answer is:
102 s

Solution:
aCB=2 m sec-2
200=12×2t2
t=102 s

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2017 (09 Apr Online)
A car is travelling at a velocity 10 km h-1 on a straight road. The driver of the car throws a parcel with a velocity 102 km h-1 with respect to the car, when the car is passing by a man standing on the side of the road. If the parcel is to reach the man, the direction of throw makes the following angle with the direction of motion of the car

Options:
A 135Â°
B 45°
C tan-1â¡2
D tan-1â¡12

The correct answer is:
135°

Solution:
In the vector diagram, OAÂ¯ and OBÂ¯ represent the velocity vectors for the car and the parcel, respectively. VâR is the resultant velocity vector. Let Î¸ be the angle between Vâcar and VâParcel . Then,
cosâ¡(180Â°-Î¸)=VâcarVâparcel=10102=12
180Â°-Î¸=45Â°
â´Î¸=135Â°

Topic: Physics, Motion In Two Dimensions, JEE Main
A car moves from $\mathrm{X}$ to $\mathrm{Y}$ with a uniform speed $v_u$. The average speed for this round trip is :

Options:
A $\sqrt{v_u v_d}$
B $\frac{v_d v_u}{v_d+v_v}$
C $\frac{v_u+v_d}{2}$
D $\frac{2 v_d v_u}{v_d+v_u}$

The correct answer is:
$\frac{2 v_d v_u}{v_d+v_u}$

Solution:
Average speed
$\begin{aligned}
& =\frac{\text { Total distance }}{\text { Total time }} \\
& =\frac{2 d}{\frac{d}{V_u}+\frac{d}{V_d}}=\frac{2 V_d V_u}{V_d+V_u}
\end{aligned}$

Topic: Physics, Motion In One Dimension, JEE Main
A car moves in positive Y-direction with velocity v proportional to distance travelled y as vy∝yβ, where β is a positive constant. The car covers a distance L with average velocity v proportional to L as v∝L1/3. The constant β is given as

Options:
A 14
B 13
C 23
D 12

The correct answer is:
13

Solution:
Given that,

vy∝yβ ...1

v∝L1/3 ...2

On comparing 1 & 2,

β=13

Topic: Physics, Motion In One Dimension, JEE Main
A car moves uniformly along a horizontal sine curve y=asin(x/α), where a and α are certain constants. The coefficient of friction between the wheels and the road is equal to k. The maximum constant speed at which the car can travel without sliding is

Options:
A αkga
B -αkga
C kgαa
D -kgαa

The correct answer is:
αkga

Solution:
Step I : Determine the radius of curvature of given curve at point P (x, y).
âµ y = a sin x α ∴ d⁡ y d⁡ x = a α cos x α
∴ d⁡ 2 y d⁡ x 2 = - a α 2 sin x α
∴ the radius of curvature is R⁡ = 1 + d⁡ y d⁡ x 2 3 / 2 d⁡ 2 y d⁡ x 2
R⁡ = 1 + a 2 α 2 cos 2 x a 3 / 2 - a α 2 sin x α ...(i)
Step II: Relate the friction and centripetal acceleration. In this case, centripetal acceleration is provided by force of friction.
âµ f ≤ fL or f ≤ kmg
or mv 2 R⁡ ≤ kmg or v⁡ 2 R⁡ ≤ kg
∴ v⁡ max 2 R⁡ min = k g
∴ v⁡ max = k g R⁡ min
From equation (i), for the minimum value of the radius of curvature,
x α = π 2
∴ R⁡ min = α 2 a (from equation. (i))
∴ v⁡ max = k g α 2 a = α k g a

Topic: Physics, Laws of Motion, JEE Main
A car moving at a speed $v$ is stopped by a retarding force $F$ in a distance $s$. If the retarding force were $3 F$, then the car will be stopped in a distance

Options:
A $\frac{s}{3}$
B $\frac{s}{6}$
C $\frac{s}{9}$
D $\frac{s}{12}$

The correct answer is:
$\frac{s}{3}$

Solution:
Given that, initial speed of car, $v_i=v$
Final speed of car when it stops, $v_f=0$ Distance travelled $=s$
Retarding force $=F$
By Newton's law of motion,
$$
\begin{aligned}
F=m a & =m\left(\frac{v_f^2-v_i^2}{2 s}\right) \quad \ldots \text { (i) }\left[\begin{array}{l}
\text { using relation } \\
a=\frac{v_f^2-v_i^2}{2 s}
\end{array}\right] \\
& =\frac{m\left(-v^2\right)}{2 s} \\
F & =\frac{-m v^2}{2 s}
\end{aligned}
$$
When, retarding force, $F^{\prime}=3 F$
Let $s^{\prime}=$ New distance travelled.
Similarly, from Eq. (i) we can write
$$
3 F=\frac{-m v^2}{2 s^{\prime}}...(ii)
$$
Substituting value of $F$ from Eq. (i) into Eq. (ii), we get
$$
3\left(\frac{-m v^2}{2 s}\right)=\frac{-m v^2}{2 s^{\prime}}
$$
$$
\Rightarrow
$$
$$
s^{\prime}=s / 3
$$
Hence, the body will stop by travelling distance $\frac{s}{3}$, when retarding force is $3 F$.

Topic: Physics, Laws of Motion, JEE Main
A car moving with a speed of 50 km h-1 can be stopped by brakes in at least 6 m. If the same car is moving at a speed of 100 km h-1, then the minimum stopping distance (in m) is

Options:

The correct answer is:
24

Solution:
The stopping distance of a car is

s=u22a

Therefore, for the two situations given in the question, we can write

s1s2=u1u22=14

6s2=14âs2=24 m

Topic: Physics, Motion In One Dimension, JEE Main
A car, moving with a speed of $50 \mathrm{~km} / \mathrm{hr}$, can be stopped by brakes after at least $6 \mathrm{~m}$. If the same car is moving at a speed of $100 \mathrm{~km} / \mathrm{hr}$, the minimum stopping distance is

Options:
A
$12 \mathrm{~m}$

B
$18 \mathrm{~m}$

C
$24 \mathrm{~m}$

D
$6 \mathrm{~m}$

The correct answer is:
$24 \mathrm{~m}$

Solution:
Energy $=$ Work done by force $(\mathrm{F})$
$$
\Rightarrow \frac{1}{2} \mathrm{~m} .(50)^2=(\mathrm{F})(6) \quad \Rightarrow \mathrm{F}=\frac{2500 \mathrm{~m}}{2 \times 6}
$$
For $\mathrm{v}=100 \mathrm{~km} / \mathrm{hr} \frac{1}{2} \cdot \mathrm{m}(100)^2=(\mathrm{F})(\mathrm{S})$
$$
\begin{aligned}
& \Rightarrow \frac{1}{2} \mathrm{~m}(100)^2=\left(\frac{2500 \mathrm{~m}}{2 \times 6}\right) \mathrm{S} \\
& \Rightarrow \mathrm{S}=\frac{100 \times 100 \times 6 \times 2}{2500 \times 2}=24 \mathrm{~m}
\end{aligned}
$$

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2003