datavorous/entrance-exam-dataset huggingface dataset:
A small pack of cards consists of 5 green cards, 4 blue cards and 3 black cards. The pack is shuffled through and first three cards are turned face up. The probability that there is exactly one card of each colour is

Options:
A 955
B 411
C 311
D 855

The correct answer is:
311

Solution:
Probability (One card of each colour) =C15.C14.C13C312=311

Topic: Mathematics, Probability, JEE Main
A small particle moves to position 5i^-2j^+k^ from its initial position 2i^+3j^-4k^ under the action of force 5i^+2j^+7k^ N. The value of work done will be ______ J.

Options:

The correct answer is:
40

Solution:
Work done by a constant force is given by W=ForceF→·Displacementr→, where, r→=r→2-r→1

Given here, r1→=5i^-2j^+k^ and r→2=2i^+3j^-4k^

So, W=F→·r→2-r→1

=(5i^+2j^+7k^)·(5i^-2j^+k^)-(2i^+3j^-4k^)

⇒W=40 J

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2023 (01 Feb Shift 1)
A small particle of mass 1 kg and charge 1 C enters perpendicularly in a triangular region of a uniform magnetic field of strength 2T as shown in the figure. What is the maximum velocity (in m s-1) of the particle with which it can enter so that it completes a half-circle in the region of the magnetic field?

Options:

The correct answer is:
3

Solution:
sinâ¡37Â°=R4-R
â35=R4-R
â 12 â 3R = 5R
â 8R =12
â R=128=32 m
V=qBRm=1Ã2Ã1.51=3 m s-1

Topic: Physics, Magnetic Effects of Current, JEE Main
A small particle of mass m and charge Q is dropped in uniform horizontal magnetic field B. The maximum vertical displacement of particle is given by h=nm2g2Q2B2 . Find the value of n.

Options:

The correct answer is:
4

Solution:
mgh=12mv2â v=2gh
At lowest position, velocity becomes horizontal.
â´ vx=v=2gh
Fsinâ¡Î¸=max
QvBsinâ¡Î¸=max
âQBvsinâ¡Î¸=max
â QBVy=max
â QBdydt=mdvxdt
â QBâ«0hdy=mâ«0vxdvx
â QBh=mvx
â QBh=m2gh
â h=2m2gQ2B2

Topic: Physics, Magnetic Effects of Current, JEE Main
A small particle of mass m and its retaining cord is spinning with angular velocity Ï on the horizontal surface of a smooth disc. As force F is slightly relaxed, r increases and Ï changes. Determine the rate of change of Ï with respect to r

Options:
A +ωr
B -ωr
C -2ωr
D +2ωr

The correct answer is:
-2ωr

Solution:
During the motion of the object, there is only one force tension which is acting towards a fixed point (centre of the table - disc). So that motion is under the effect to central force and thus, angular momentum will be conserved. At any radial distance, r is L=mωr2
dLdr=mr2dωdr+mω2r=0
So, dωdr= -2ωr

Topic: Physics, Rotational Motion, JEE Main
A small particle of mass $m$ is projected at an angle $\theta$ with the $\mathrm{x}$-axis with an initial velocity $\mathrm{v}_0$ in the $\mathrm{x}-\mathrm{y}$ plane as shown in the figure. At a time $t < \frac{v_0 \sin \theta}{g}$, the angular momentum of the particle is
where $\hat{\mathrm{i}}, \hat{\mathrm{j}}$ and $\hat{\mathrm{k}}$ are unit vectors along $\mathrm{x}, \mathrm{y}$ and $\mathrm{z}$-axis respectively.

Options:
A
$-\operatorname{mgv}_0 \mathrm{t}^2 \cos \theta \hat{\mathrm{j}}$

B
$\operatorname{mgv}_0 t \cos \theta \hat{\mathrm{k}}$

C
$-\frac{1}{2} m g v_0 t^2 \cos \theta \hat{k}$

D
$\frac{1}{2} m g v_0 t^2 \cos \theta \hat{i}$

The correct answer is:
$-\frac{1}{2} m g v_0 t^2 \cos \theta \hat{k}$

Solution:
$\begin{aligned} \overrightarrow{\mathrm{L}} & =m(\vec{r} \times \overrightarrow{\mathrm{v}}) \\ \overrightarrow{\mathrm{L}} & =m\left[\mathrm{v}_0 \cos \theta t \hat{\mathrm{i}}+\left(\mathrm{v}_0 \sin \theta \mathrm{t}-\frac{1}{2} g \mathrm{t}^2\right) \hat{\mathrm{j}}\right] \times\left[\mathrm{v}_0 \cos \theta \hat{\mathrm{i}}+\left(\mathrm{v}_0 \sin \theta-g \mathrm{~g}\right) \hat{\mathrm{j}}\right] \\ & =m \mathrm{v}_0 \cos \theta \mathrm{t}\left[-\frac{1}{2} g \mathrm{~g}\right] \hat{\mathrm{k}} \\ & =-\frac{1}{2} m g \mathrm{v}_0 \mathrm{t}^2 \cos \theta \hat{\mathrm{k}}\end{aligned}$

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2010
A small particle of mass m moves in such a way that its potential energy U=12mω2r2 where ω is constant and r is the distance of the particle from origin. Assuming Bohr’s quantization of momentum and circular orbit, the radius of nth orbit will be proportional to

Options:
A n
B 1n
C n2
D n

The correct answer is:
n

Solution:
The data given is

U=12mω2r2

From Bohr's quantization,

mvr=nh2π⇒v2=nh2πrm2 ...(i)

In an orbit the value of the kinetic energy is half of the potential energy,

K=mω2r24

Substituting the value of equation (i) in the kinetic energy,

K=12mn2h24π2m2r2=mω2r24⇒n2h22π2m2ω2=r4⇒r∝n

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2023 (06 Apr Shift 2)
A small planet is revolving around a very massive star in a circular orbit of radius r with a period of revolution T. If the gravitational force between the planet and the star is proportional to r-5/2, then T will be proportional to

Options:
A r3/2
B r5/3
C r7/4
D r3

The correct answer is:
r7/4

Solution:
Gravitational force provides the necessary centripetal force required by the given planet for circular motion. So
mv 2 r ∝ r - 5 / 2
⇒ mv 2 r = kr - 5 / 2 , k = constant
∴ v = kr - 3 / 2 m
T = 2 π r v = 2 π r × m k r - 3 / 2
i.e., T ∝ r 7 / 4 .

Topic: Physics, Gravitation, JEE Main
A small planet is revolving around a very massive star in a circular orbit of radius r with a period of revolution T. If the gravitational force between the planet and the star is proportional to r5/2, then T will be proportional to

Options:
A r3/2
B r5/3
C r7/4
D r3

The correct answer is:
r7/4

Solution:
According to the question, the gravitational force between the planet and the star is \(\mathrm{F} \propto \frac{1}{\mathrm{R}^{5 / 2}}\)
\(\therefore \mathrm{F}=\frac{\mathrm{GMm}}{\mathrm{R}^{5 / 2}}\)
Where M and M be masses of star and planet respectively.
For motion of planet in circular orbit,
\(\mathrm{mR} \omega^{2}=\frac{\mathrm{GMm}}{\mathrm{R}^{5 / 2}}\)
\(\mathrm{mR}\left(\frac{2 \pi}{\mathrm{T}}\right)^{2}=\frac{\mathrm{GMm}}{\mathrm{R}^{5 / 2}}\) \(\left(\therefore \omega=\frac{2 \pi}{\mathrm{T}}\right)\)
\(\frac{4 \pi^{2}}{\mathrm{~T}^{2}}=\frac{\mathrm{GM}}{\mathrm{R}^{7 / 2}} \Rightarrow \mathrm{T}^{2}=\frac{4 \pi^{2}}{\mathrm{GM}} \mathrm{R}^{7 / 2}\)
\(\mathrm{T}^{2} \propto \mathrm{R}^{7 / 2}\) or \(\mathrm{T}^{2} \propto \mathrm{R}^{7 / 4}\)

Topic: Physics, Gravitation, JEE Main
A small point mass carrying some positive charge on it, is released from the edge of a table. There is a uniform electric field in this region in the horizontal direction. Which of the following options then correctly describe the trajectory of the mass ? (Curves are drawn schematically and are not to scale)

Options:
A
B
C
D

The correct answer is:

Solution:
Net acceleration of particle is constant, initial velocity is zero therefore path is straight line.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2020 (02 Sep Shift 2)
A small ring of mass m is constrained to slide along a horizontal wire fixed between two rigid supports. The ring is connected to a particle of same mass by an ideal string & the whole system is released from rest as shown in the figure. If the coefficient of friction between ring A and wire is 3 5, the ring will start sliding when the connecting string will make an angle θ with the vertical, then θ will be (particle is free to move and ring can slide only)

Options:
A 30o
B 45o
C 60o
D None of these

The correct answer is:
45o

Solution:
Using energy conservation, we get

mgl cos θ = mv 2 2

⇒mv2l=2mgcosθ

As the mass is moving in a circular motion, applying circular motion equation,
T - mg cos θ = 2 mg cos θ
T = 3 mg cos θ

Tsinθ=μmg+Tcosθ
3 mg cos θ sin θ = μ mg + 3 mg cos 2 θ
θ = 4 5 ∘

Topic: Physics, Laws of Motion, JEE Main
A small ring of mass m is constrained to slide along a horizontal wire fixed between two rigid supports. The ring is connected to a particle of same mass by an ideal string & the whole system is released from rest as shown in the figure. If the coefficient of friction between ring A and wire is 3 5, the ring will start sliding when the connecting string will make an angle Î¸ with the vertical, then Î¸ will be (particle is free to move and ring can slide only)

Options:
A 30o
B 45o
C 60o
D None of these

The correct answer is:
45o

Solution:
Using energy conservation, we get

mgl cos θ = mv 2 2

⇒mv2l=2mgcosθ

As the mass is moving in a circular motion, applying circular motion equation,
T - mg cos θ = 2 mg cos θ
T = 3 mg cos θ

Tsinθ=μmg+Tcosθ
3 mg cos θ sin θ = μ mg + 3 mg cos 2 θ
θ = 4 5 ∘

Topic: Physics, Laws of Motion, JEE Main
A small satellite of mass m is revolving around the earth in a circular orbit of radius r 0 with speed v 0 . At a certain point in its orbit, the direction of motion of satellite is suddenly changed by angle Î¸ = cos -1 3 / 5 by turning its velocity vector in the same plane of motion, such that the speed remains constant. The satellite consequently goes into an elliptical orbit around earth. The ratio of speed at perigee to speed at apogee is

Options:
A 3
B 9
C 1/3
D 1/9

The correct answer is:
9

Solution:
Using conservation of angular momentum about O.
m V P r P = m V A r A = m V 0 r 0 cos Î¸
V A r A = V P r P = 3 v 0 r 0 5 ......(1)
using conservation of energy
1 2 m V A 2 + - GMm r A = - GMm r 0 + 1 2 m V 0 2
â 9 V 0 2 r 0 2 5 0 r A 2 - V 0 2 r 0 r A + V 0 2 2 = 0 Let r 0 r A = x This is a quadratic equation in r0 ra
â9x2-50x+25=0âx=5 or x=59

ârA=95ro â¦â¦.1
rp=ro5 â¦â¦.(2)
using (1)&(2)
â V P V A = r A r P = 9

Topic: Physics, Gravitation, JEE Main
A small signal voltage Vt=V0sin⁡ωt is applied across an ideal capacitor C:

Options:
A Current I(t), lags voltage V(t) by 90o
B Over a full cycle the capacitor C does not consume any energy from the voltage source
C Current I(t) is in phase with voltage V(t).
D Current I(t) leads voltage V(t) by 180o

The correct answer is:
Over a full cycle the capacitor C does not consume any energy from the voltage source

Solution:
In capacitor current leads the voltage by ϕ=π2 phase. Average power dissipated in AC circuit is given by

Pav=VrmsIrmscosϕ

In capacitor ϕ=π2 so cosπ2 is zero.

Topic: Physics, Alternating Current, JEE Main
A small smooth disc of mass m and radius r moving with an initial velocity v along the positive x-axis collided with a big disc of mass 2m and radius 2r which was initially at rest with its centre at origin as shown in the figure. If the coefficient of restitution is 0, then the velocity of the larger disc after the collision is

Options:
A 827v i^-2227v j^
B 827v i^-227v j^
C v3i^
D 2227v i^-827v j^

The correct answer is:
827v i^-2227v j^

Solution:
The larger disc will move along line of impact
As e=0, so velocity of larger disc
v ′ = mv cos θ m + 2 m = v cos θ 3
Velocity of larger disc
=v'cos θ i^-v'sin θ j^
=v3cos2θ i^-v3sinθ cos θ j^

( from the figure we can calculate cosθ=223, sinθ=13 )
=827v i^-2227v j^

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A small soap bubble of radius 4cm is trapped inside another bubble of radius 6cm without any contact. Let P2 be the pressure inside the inner bubble and P0, the pressure outside the outer bubble. Radius of another bubble with pressure difference P2-P0 between its inside and outside would be:

Options:
A

2.4 cm

B 12 cm
C 4.8 cm
D 6 cm

The correct answer is:

2.4 cm

Solution:
P2−P1=4Tr1      ....(1)P1−P0=4Tr2 .....(2)adding (1) & (2)P2−P0=4Tr1+4Tr2P2=P0+4Tr1+4Tr2P2=P0+4T6+4T4
for another bubble of radius r
P2= P0+4Tr
⇒ 1r=16+14
⇒ r=2.4 cm

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2018 (16 Apr Online)
A small speaker delivers 2 W of audio output. At what distance from the speaker will one detect 120 dB intensity sound? [Given reference intensity of sound as 10-12 W/m2 ]

Options:
A 40 cm
B 20 cm
C 10 cm
D 30 cm

The correct answer is:
40 cm

Solution:
Let the intensity of sound hear by observer is I
120=10log10⁡I10-12
12=log⁡I-log10⁡10-12
=log⁡I+12
log⁡I=0⇒I=1 wattm2
I=P4πr2
⇒1=24πr2⇒r=12π=0.4 m=40 cm

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2019 (12 Apr Shift 2)
A small sphere of mass $m$ is dropped from a great height. After it has fallen $100 \mathrm{~m}$, it has attained its terminal velocity and continues to fall at that speed. The work done by air friction against the sphere during the first $100 \mathrm{~m}$ of fall is

Options:
A Greater than the work done by air friction in the second $100 \mathrm{~m}$
B Less than the work done by air friction in the second $100 \mathrm{~m}$
C Equal to $100 \mathrm{mg}$
D Greater than $100 \mathrm{mg}$

The correct answer is:
Less than the work done by air friction in the second $100 \mathrm{~m}$

Solution:
In the first $100 \mathrm{~m}$ body starts from rest and its velocity goes on increasing and after $100 \mathrm{~m}$ it acquire maximum velocity (terminal velocity). Further, air friction i.e. viscous force which is proportional to velocity is low in the beginning and maximum at $v=v_r$.
Hence work done against air friction in the first $100 \mathrm{~m}$ is less than the work done in next $100 \mathrm{~m}$.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A small sphere of mass m suspended by a thread is first taken aside so that the thread forms a right angle with the vertical and then released. The total acceleration of the sphere when it makes an angle θ with the vertical is

Options:
A g1+3cos2θ
B g1+cos2θ
C g1+sin2θ
D g1+3sin2θ

The correct answer is:
g1+3cos2θ

Solution:
Between A and B, mgL cos Î¸ = 1 2 mv B 2
â´ v B 2 = 2 gL cos Î¸
Now, a r = v B 2 L = 2 g cos Î¸
and a t = g sin Î¸
â´ a = a t 2 + a r 2 = g 1 + 3 cos 2 Î¸

Topic: Physics, Laws of Motion, JEE Main
A small sphere of radius r falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains terminal velocity, is proportional to

Options:
A r5
B r2
C r3
D r4

The correct answer is:
r5

Solution:
Rate of heat produced = power of viscous force
terminal velocity V T = 2 r 2 ( ρ−σ )g 9η
⇒ V T ∝ r 2
dQ dt =6πηr  V T × V T
⇒ dQ dt ∝ r 5

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A small spherical ball (obeying Stoke's law for viscous force) is thrown up vertically with a speed 20 m s-1 and is received back by the thrower at the point of projection with a speed 10 m s-1. Neglecting the buoyant force on the ball, assuming the speed of the ball during its flight to be never equal to its terminal speed and taking the acceleration due to gravity g=10 m s-2, find the time of flight of the ball in seconds.

Options:

The correct answer is:
3

Solution:
mg→+kv→=mdv→dt

g→ dt+km v→ dt=d v→

Integrating both sides

∫0Tg→dt+km∫0Tv→dt=∫u→v→dv→

The total displacement of the ball in time T is

∫0Tv→dt = 0
g→T+0=v→-u→

gT=v--u=v+u

⇒T=v+ug=3 s

Topic: Physics, Laws of Motion, JEE Main
A small spherical ball (obeying Stoke's law for viscous force) is thrown up vertically with a speed 20 m s-1 and is received back by the thrower at the point of projection with a speed 10 m s-1. Neglecting the buoyant force on the ball, assuming the speed of the ball during its flight to be never equal to its terminal speed and taking the acceleration due to gravity g=10 m s-2, find the time of flight of the ball in seconds.

Options:

The correct answer is:
3

Solution:
mg→+kv→=mdv→dt

g→ dt+km v→ dt=d v→

Integrating both sides ∫0Tg→dt+km∫0Tv→dt=∫u→v→dv→
g→T+0=v→-u→

gT=v--u=v+u

⇒T=v+ug=3 s

Topic: Physics, Laws of Motion, JEE Main
A small spherical ball of radius 0.1 mm and density 104 kg m-3 falls freely under gravity through a distance h before entering a tank of water. If, after entering the water the velocity of ball does not change and it continue to fall with same constant velocity inside water, then the value of h will be ____ m. (Given g=10 m s-2, viscosity of water =1.0×10-5 N‐s m-2 ).

Options:

The correct answer is:
20

Solution:
Using equation of motion, v2=u2+2as

v2=0+2-g-h

⇒v2=2gh

Initial velocity of ball is v=2gh=20h.

If after entering the water this velocity does not change, then this value should be equal to the terminal velocity.

Then, 2gh=29r2ρ-σgη, here, ρ is density of ball, σ is density of water and η is viscosity.

⇒20h=290.1×10-32104-103101×10-5⇒20h=2910-810-1103×101×10-5

⇒20h=400

∴h=20 m

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2022 (29 Jun Shift 2)
A small spherical ball of radius r, falling through a viscous medium of negligible density has terminal velocity v. Another ball of the same mass but of radius 2r, falling through the same viscous medium will have terminal velocity:

Options:
A v2
B v4
C 4v
D 2v

The correct answer is:
v2

Solution:
Ignoring buoyancy force, we can write terminal velocity as

mg=6πηrv

⇒v∝1r (as mass is constant)

Now, vv'=r'r

As r'=2r

So v'=v2

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2024 (31 Jan Shift 2)
A small square loop of side a and one turn is placed inside a larger square loop of side b and one turn (b≫a). The two loops are coplanar with their centres coinciding. If a current I is passed in the square loop of side b, then the coefficient of mutual inductance between the two loops is :

Options:
A μ04π82a2 b
B μ04π82b2a
C μ04π82 b
D μ04π82a

The correct answer is:
μ04π82a2 b

Solution:
|
B=4×μ0i4πb22 ⇒B=22μ0iπb
⇒ϕ=MI ⇒BA=MI
⇒22μ0iπb·a2=Mi ⇒M=22μ0a2πb=μ04π82a2b

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2021 (31 Aug Shift 1)
A small square loop of wire of side $l$ is placed inside a large square loop of side $L(L>>l)$. If the loops are coplanar and their centres coincide, the mutual induction of the system is directly proportional to :

Options:
A $\frac{L}{l}$
B $\frac{l}{L}$
C $\frac{L^2}{l}$
D $\frac{l^2}{L}$

The correct answer is:
$\frac{l^2}{L}$

Solution:
Considering the larger loop to be made up of four rods each of length $L$, the field at the centre, i.e., at a distance $\left(\frac{L}{2}\right)$ from each rod, will be
$B=4 \times \frac{\mu_0}{4 \pi} \frac{l}{d}[\sin \alpha+\sin \beta]$
i.e., $\quad B=4 \times \frac{\mu_0}{4 \pi} \frac{I}{\left(\frac{L}{2}\right)} \times 2 \sin 45$
i.e., $\quad B_1=\frac{\mu_0}{4 \pi} \frac{8 \sqrt{2}}{L} I$
So, the flux with smaller loop
$\phi_2=B_1 S_2=\frac{\mu_0}{4 \pi} \frac{8 \sqrt{2}}{L} l^2 I$
and hence, $M=\frac{\phi_2}{I}=2 \sqrt{2} \frac{\mu_0}{\pi} \frac{l^2}{L}$
or $\quad M \propto \frac{l^2}{L}$

Topic: Physics, Electromagnetic Induction, JEE Main
A small steel ball bounces on a steel plate held horizontally. On each bounce the speed of the ball arriving at the plate is reduced by a factor
e (coefficient of restitution) in the rebound, so that $V_{upward}=eV_{downward}$
If the ball is initially dropped from a height of $0.4 \mathrm{m}$ above the plate and if 10 seconds later the bouncing ceases, the value of $e$ is

Options:
A $\sqrt{\frac{2}{7}}$
B $\frac{3}{4}$
C $\frac{13}{18}$
D $\frac{17}{18}$

The correct answers are:
$\frac{17}{18}$

Solution:
Given, steel ball bounces on a steel ple held horizontally, so $$
V_{\text {upward }}=e V_{\text {downward }}
$$
According to the relation,
$h_{n}=e^{2 n} h$
$\begin{aligned} \therefore \quad t &=\sqrt{\frac{2 h}{g}}+2 \sqrt{\frac{2 h e^{2}}{g}}+2 \sqrt{\frac{2 h e^{4}}{g}}+\ldots \\ &=\sqrt{\frac{2 h}{g}}\left[1+2 e+2 e^{2}+\ldots\right] \\ &=\sqrt{\frac{2 h}{g}}\left[\frac{1+e}{1-e}\right] \\ \therefore \quad t &=10 \mathrm{sec} \\ \therefore 10 &=\sqrt{\frac{2(0 \cdot 4)}{10}}\left(\frac{1+e}{1-e}\right) \end{aligned}$
Solving above relation $\therefore e=\frac{25 \sqrt{2}-1}{25 \sqrt{2}+1}=\frac{17}{18}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A small steel ball of radius $r$ is allowed to fall under gravity through a column of a viscous liquid of coefficient of viscosity $\eta$. After some time the velocity of the ball attains a constant value known as terminal velocity $v_T$. The terminal velocity depends on (i) the mass of the ball $m$, (ii) $\eta$, (iii) $r$ and (iv) acceleration due to gravity $g$. Which of the following relations is dimensionally correct?

Options:
A $v_T \propto \frac{m g}{\eta r}$
B $v_T \propto \frac{\eta r}{m g}$
C $v_T \propto \eta r m g$
D $v_T \propto \frac{m g r}{\eta}$

The correct answer is:
$v_T \propto \frac{m g}{\eta r}$

Solution:
By substituting dimension of each quantity in R.H.S. of option
(a) we get $\left[\frac{m g}{\eta r}\right]=\left[\frac{M \times L T^{-2}}{M {L}^{-1} T^{-1} \times L}\right]=\left[L T^{-1}\right]$.
This option gives the dimension of velocity.

Topic: Physics, Units and Dimensions, JEE Main
A small toy starts moving from the position of rest under a constant acceleration. If it travels a distance of 10 m in t s, the distance travelled by the toy in the next t s will be:

Options:
A 10 m
B 20 m
C 30 m
D 40 m

The correct answer is:
30 m

Solution:
Initial velocity of the particle is zero.

Applying equation of motion, we get 10=12×a×t2

Let distance covered in next t s is x.

Then, 10+x=12×a×2t2

⇒10+x=12at2×4

⇒10+x=10×4

⇒x=30 m

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2022 (29 Jun Shift 2)
A smaller cube with side b (depicted by dashed lines) is excised from a bigger uniform cube with side a as shown below such that both cubes have a common vertex $\mathrm{P}$. Let $\mathrm{X}=\mathrm{a} / \mathrm{b}$. If the centre of mass of the remaining solid is at the vertex $\mathrm{O}$ of smaller cube then $\mathrm{X}$ satisfies.

Options:
A $\mathrm{X}^{3}-\mathrm{X}^{2}-\mathrm{X}-1=0$
B $X^{2}-X-1=0$
C $X^{3}+X^{2}-X-1=0$
D $X^{3}-X^{2}-X+1=0$

The correct answer is:
$\mathrm{X}^{3}-\mathrm{X}^{2}-\mathrm{X}-1=0$

Solution:
Centre of mass of remaining cube $x$ coordinate $=b$
$$
\mathrm{X}_{\mathrm{CM}}=\frac{\rho \mathrm{a}^{3} \times \frac{\mathrm{a}}{2}-\rho \mathrm{b}^{3} \times \frac{\mathrm{b}}{2}}{\rho \mathrm{a}^{3}-\rho \mathrm{b}^{3}}
$$
We will consider removed mass as a negative mass
$$
\begin{array}{l}
b=\frac{\frac{\rho a^{4}}{2}-\frac{\rho b^{4}}{2}}{\rho a^{3}-\rho b^{3}} \\
a^{3} b-b^{4}=\frac{a^{4}}{2}-\frac{b^{4}}{2} \\
2 a^{3} b-2 b^{4}=a^{4}-b^{4} \\
\text { put a }=b x \Rightarrow 2 b^{4} x^{3} b-2 b^{4}=b^{4} x^{4}-b^{4} \\
2 x^{3}-1==x^{4} \\
2 x^{3}-2+1=x^{4} \\
2\left[x^{3}-1\right]=\left(x^{2}-1\right)\left(x^{2}+1\right) \\
2[x-1]\left[x^{2}+1+x\right]=[x-1][x+1]\left[x^{2}+1\right] \\
2 x^{2}+2+2 x=x^{3}+x+x^{2}+1 \\
x^{3}-x^{2}-x-1=0
\end{array}
$$

Topic: Physics, Center of Mass Momentum and Collision, KVPY, KVPY 2016 (SB/SX)
A smooth block is released at rest on a $45^{\circ}$ incline and then slides a distance d. The time taken to slide is $\mathrm{n}$ times as much to slide on rough incline than on a smooth incline. The coefficient of friction is

Options:
A
$\mu_{\mathrm{k}}=1-\frac{1}{\mathrm{n}^2}$

B
$\mu_{\mathrm{k}}=\sqrt{1-\frac{1}{\mathrm{n}^2}}$

C
$\mu_s=1-\frac{1}{n^2}$

D
$\mu_s=\sqrt{1-\frac{1}{n^2}}$

The correct answer is:
$\mu_{\mathrm{k}}=1-\frac{1}{\mathrm{n}^2}$

Solution:
$\mathrm{d}=\frac{1}{2} \frac{\mathrm{g}}{\sqrt{2}} \mathrm{t}_1^2$
$\mathrm{~d}=\frac{1}{2} \frac{\mathrm{g}}{\sqrt{2}}\left(1-\mu_{\mathrm{k}}\right) \mathrm{t}_2^2$
$\frac{\mathrm{t}_2^2}{\mathrm{t}_1^2}=\mathrm{n}^2=\frac{1}{1-\mu_{\mathrm{k}}}$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2005
A smooth rod OP is fixed vertically. A disc of mass m and radius R is rolling without slipping at the given instant on a rough horizontal surface as shown. Velocity of centre of mass C of disc is v0 . A rod AB of length L is connected with disc at A, R/2 distance vertically up from centre of disc with pin joint (about which it can rotate freely), other end of the rod is attached to a small smooth ring B, which can move freely over rod OP. At an instant rod AB makes an angle 60° with the vertical. The velocity of ring on the vertical rod at this instant is.

Options:
A 332 v0
B 32v0
C v0
D perpendicular to velocity of point Q ( Q and C are at same horizontal level)

The correct answer is:
332 v0

Solution:
V1 cos 60∘=3V02 cos 30∘⇒V1=332V0

Topic: Physics, Rotational Motion, JEE Main