datavorous/entrance-exam-dataset huggingface dataset:
A boat is moving due east in a region where the earth's magnetic field is $5.0 \times 10^{-5} \mathrm{NA}^{-1} \mathrm{~m}^{-1}$ due north and horizontal. The boat carries a vertical aerial $2 \mathrm{~m}$ long. If the speed of the boat is $1.50 \mathrm{~ms}^{-1}$, the magnitude of the induced emf in the wire of aerial is :

Options:
A
$0.75 \mathrm{mV}$

B
$0.50 \mathrm{mV}$

C
$0.15 \mathrm{mV}$

D
$1 \mathrm{mV}$

The correct answer is:
$0.15 \mathrm{mV}$

Solution:
$\mathrm{E}=\mathrm{B}_{\mathrm{H}} \mathrm{VV}=0.15 \mathrm{mV}$

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2011
A boat is moving with velocity of $3 \hat{i}+4 \hat{j}$ in river and water is moving with a velocity of $-3 \hat{i}-4 \hat{j}$ with respect to ground. Relative velocity of boat with respect to water is :

Options:
A $-6 \hat{i}-8 \hat{j}$
B $6 \hat{i}+8 \hat{j}$
C $8 \hat{i}$
D $6 \hat{i}$

The correct answer is:
$6 \hat{i}+8 \hat{j}$

Solution:
The relative velocity of boat w.r.t. water
$=v_{\text {boat }}-v_{\text {water }}=(3 \hat{i}+4 \hat{j})-(-3 \hat{i}-4 \hat{j})=6 \hat{i}+8 \hat{j}$

Topic: Physics, Motion In Two Dimensions, JEE Main
A boat is sent across a river with a velocity of $8 \mathrm{~km} / \mathrm{hr}$. If the resultant velocity of boat is $10 \mathrm{~km} / \mathrm{hr}$, then velocity of the river is :

Options:
A 10 km/hr
B 8 km/hr
C 6 km/hr
D 4 km/hr

The correct answer is:
6 km/hr

Solution:
Given $\overrightarrow{A B}=\text { Velocity of boat }=8 \mathrm{~km} / \mathrm{hr}$
$\overrightarrow{A C}=\text { Resultant velocity of }$
$\text {boat }=10 \mathrm{~km} / \mathrm{hr}$
$\overrightarrow{B C}=\text { Velocity of river }$
$=\sqrt{A C^2-A B^2}$
$=\sqrt{(10)^2-(8)^2}=6 \mathrm{~km} / \mathrm{hr}$

Topic: Physics, Motion In Two Dimensions, JEE Main
A boat is sent across a river with a velocity of \( 8 \mathrm{~km} \mathrm{hr}^{-1} \). If the resultant velocity of the boat is \( 10 \mathrm{~km} \mathrm{hr}^{-1} \), then the velocity of the river is

Options:
A \( 12.8 \mathrm{~km} \mathrm{~h}^{-1} \)
B \( 6 \mathrm{~km} \mathrm{~h}^{-1} \)
C \( 8 \mathrm{~km} \mathrm{~h}^{-1} \)
D \( 10 \mathrm{~km} \mathrm{~h}^{-1} \)

The correct answer is:
\( 6 \mathrm{~km} \mathrm{~h}^{-1} \)

Solution:
Let the velocity of the river = v → R

∴|v→boat|=8 km h−1

v → bR = v → b − v → R
v → b = v → bR + v → R
∵ v → bR ⊥ v → R

vb=vbR2+vR2

vb2=vbR2+vR2

(10)2=(8)2+vR2

vR=6km h-1

Topic: Physics, Motion In Two Dimensions, JEE Main
A boat is travelling with a speed of 27 km h-1 due east. An observer is situated at 30 m south of the line of travel. The angular velocity of the boat relative to the observer in the position shown will be

Options:
A 0.125 rad s-1

B Zero
C 0.250 rad s-1

D 0.67 rad s-1

The correct answer is:
0.125 rad s-1

Solution:
v = 27 km h-1 = 7.5 m s-1

Let us assume the location of the observer as the origin
râ=30 tan Î¸i^+30j^
Vâ=drâdt ; 7.5=30 sec2 Î¸dÎ¸dti^
7.5=30Ï sec2Î¸ â 7.5=30Ï sec245Â°
7.530Ã2=Ï=7.560=0.125 rad s-1

Topic: Physics, Laws of Motion, JEE Main
A boat moves with a speed of $5 \mathrm{~km} / \mathrm{h}$ relative to water in a river flowing with a speed of $3 \mathrm{~km} / \mathrm{h}$ and having a width of 1 $\mathrm{km}$. The minimum time taken around a round trip is

Options:
A 5 min
B 60 min
C 20 min
D 30 min

The correct answer is:
30 min

Solution:
For the round trip he should cross perpendicular to the river $\therefore$ Time for trip to that side $=\frac{1 \mathrm{~km}}{4 \mathrm{~km} / \mathrm{hr}}=0.25 \mathrm{hr}$ To come back, again he take $0.25 \mathrm{hr}$ to cross the river.
Total time is $30 \mathrm{~min}$, he goes to the other bank and come back at the same point.

Topic: Physics, Motion In Two Dimensions, JEE Main
A boat which has a speed of 5 km h-1 in still water crosses a river of width 1 km along the shortest possible path in 15 minutes. The velocity of the river water in km h-1 is

Options:
A 1 km h-1

B 3 km h-1
C 4 km h-1
D 41 km h â1

The correct answer is:
3 km h-1

Solution:
Vertical displacement = 1 km

t=1560=14 h

∴   Vbcosθ=  1  (14)=4 km h−1

⇒cosθ= 4 V b = 4 5 ⇒sinθ= 3 5

By velocity triangle ABC, sinθ= V r V b

∴   VrVb=35⇒Vr5=35⇒Vr=3 km h−1

Topic: Physics, Motion In Two Dimensions, JEE Main
A bob of mass m attached to an inextensible string of length l⁡ is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed ω rad/s about the vertical. About the point of suspension :

Options:
A Angular momentum is conserved
B Angular momentum changes in magnitude but not in direction
C Angular momentum changes in direction but not in magnitude
D Angular momentum changes both in direction and magnitude

The correct answer is:
Angular momentum changes in direction but not in magnitude

Solution:
τ = mg × l sin θ . (Direction parallel to plane of rotation of particle)

as τ is perpendicular to L → , direction of L changes but magnitude remains same.

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2014 (06 Apr)
A bob of mass m is tied with a thread and is made to move in a circular path on a horizontal frictionless table about point O as shown in the figure. A hypothetical electric field exists in the radial direction along the surface of the table. In this condition, the bob is uncharged and the tension in the thread is T. If now the bob is given some charge, then

Options:
A Tension in thread must increase
B Tension in thread may increase or decrease
C Tension in thread will remain unchanged
D Tension in thread must decrease

The correct answer is:
Tension in thread may increase or decrease

Solution:
Tension may increase or decrease depending on the nature of charge given to sphere.

Topic: Physics, Electrostatics, NEET
A bob of mass m suspended by a thread of length ℓ undergoes simple harmonic oscillations with time period T. If the bob is immersed in a liquid that has density 14 times that of the bob and the length of the thread is increased by 13rd of the original length, then the time period of the simple harmonic oscillations will be:

Options:
A 34T
B 43T
C T
D 32T

The correct answer is:
43T

Solution:
T=2πℓg
geff=g-g4=3g4 ∴ T'=2π4ℓ/33g/4=2π16ℓ9g
T'=43 T

Topic: Physics, Oscillations, JEE Main, JEE Main 2021 (31 Aug Shift 2)
A body $A$ moves with a uniform acceleration a and zero initial velocity. Another body $B$, starts from the same point moves in the same direction with a constant velocity $v$. The two bodies meet after a time $t$. The value of $t$ is

Options:
A $\frac{2 v}{a}$
B $\frac{ v}{a}$
C $\frac{v}{2 a}$
D $\sqrt{\frac{v}{2 a}}$

The correct answer is:
$\frac{2 v}{a}$

Solution:
$\frac{1}{2} a t^2=v t \Rightarrow t=\frac{2 v}{a}$

Topic: Physics, Motion In One Dimension, JEE Main
A body A of mass m=0.1 kg has an initial velocity of 3i^ m s-1. It collides elastically with another body B of the same mass which has an initial velocity of 5j^ m s-1. After the collision, A moves with a velocity v→=4i^+j^ m s-1. The energy of B after the collision is written as x10 J. The value of x is

Options:

The correct answer is:
1

Solution:
Using the principle of conservation of linear momentum

0.1×3i^ + 0.1×5j^ = 0.1×4i^+j^ + 0.1×v→

v→=-i^ + j^ m s-1

v=v→=2 m s-1

KB=12×0.1×v2=110 J

x=1

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2020 (08 Jan Shift 1)
A body A of mass m=0.1 kg has an initial velocity of 3i^ m s-1. It collides elastically with another body B of the same mass which has an initial velocity of 5j^ m s-1. After the collision, A moves with a velocity vâ=4i^+j^. The energy of B after the collision is written as x10 J. The value of x is

Options:

The correct answer is:
1

Solution:
Using the principle of conservation of linear momentum

0.1×3i^ + 0.1×5j^ = 0.1×4i^+j^ + 0.1×v→

v→=-i^ + j^

KB=12×0.1×v2=110 J

x=1

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A body $A$ of mass $M$ while falling vertically downwards under gravity breaks into two parts; a body B of mass $1 / 3 \mathrm{M}$ and a body $\mathrm{C}$ of mass $2 / 3 \mathrm{M}$. The centre of mass of bodies $\mathrm{B}$ and $\mathrm{C}$ taken together shifts compared to that of body $\mathrm{A}$ towards

Options:
A
depends on height of breaking

B
does not shift

C
body C

D
body B

The correct answer is:
does not shift

Solution:
No horizontal external force is acting
$$
\begin{aligned}
& \therefore \mathrm{a}_{\mathrm{cm}}=0 \\
& \text { since } \mathrm{v}_{\mathrm{cm}}=0 \\
& \therefore \Delta \mathrm{x}_{\mathrm{cm}}=0
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2005
A body $A$ starts from rest with an acceleration ${ }^{a_1}$. After 2 seconds, another body $B$ starts from rest with an acceleration ${ }^{a_2}$. If they travel equal distances in the 5 th second, after the start of $A$, then the ratio $a_1: a_2$ is equal to

Options:
A 5:9
B 5:7
C 9:5
D 9:7

The correct answer is:
5:9

Solution:
According to problem
Distance travelled by body $A$ in $5^{\text {th }} \sec$ and distance travelled by body $B$ in $3^{\text {rd }}$ sec of its motion are equal.
$\begin{aligned} & 0+\frac{a_1}{2}(2 \times 5-1)=0+\frac{a_2}{2}[2 \times 3-1] \\ & 9 a_1=5 a_2 \Rightarrow \frac{a_1}{a_2}=\frac{5}{9}\end{aligned}$

Topic: Physics, Motion In One Dimension, JEE Main
A body, acted upon by a force of $50 N$ is displaced through a distance 10 meter in a direction making an angle of $60^{\circ}$ with the force. The work done by the force be

Options:
A $200 \mathrm{~J}$
B $1.00 \mathrm{~J}$
C 300
D $250 \mathrm{~J}$

The correct answer is:
$250 \mathrm{~J}$

Solution:
$W=\vec{F} \cdot \vec{S}=F S \cos \theta$
$=50 \times 10 \times \cos 60^{\circ}=50 \times 10 \times \frac{1}{2}=250 \mathrm{~J}$.

Topic: Physics, Mathematics in Physics, JEE Main
A body at a temperature of 27 Â°C emits heat energy at a rate of 0.3 W, then at 627 Â°C, it emits heat at a rate of

Options:
A 24.3 W

B 0.42 W

C 2.42 W

D 0.9 W

The correct answer is:
24.3 W

Solution:
u∝T4
u2u1=T2T14=9003004 (Temperature should be in kelvin)
⇒ u2u1=81
⇒ u2=81u1=24.3 W

Topic: Physics, Thermal Properties of Matter, JEE Main
A body at rest is moved along a horizontal straight line by a machine delivering a constant power. The distance moved by the body in time t is proportional to:

Options:
A t32
B t12
C t14
D t34

The correct answer is:
t32

Solution:
P= constant

12mv2=Pt

⇒v∝t

dxdt=Ct C= constant

by integration,

x=Ct12+112+1

x∝t3/2

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2021 (20 Jul Shift 2)
A body centre cubic lattice is made up of two different types of atoms A and B. Atom A occupies the body centre and B occupying the corner positions. One of the corners is left unoccupied per unit cell. Empirical formula of such a solid is

Options:
A AB
B A 2 B 2
C A 5 B 7
D A 8 B 7

The correct answer is:
A 8 B 7

Solution:
There is one A per unit cell
Number of B per unit cell =18Ã7=78
Empirical formula =â A 1 B 7 8
=  A 8 B 7

Topic: Chemistry, Solid State, JEE Main
A body centred cubic lattice is made up of hollow sphere of B. Sphere of solid A are present in hollow sphere of B. Radius of A is half of the radius of B. What is the ratio of total volume of sphere B unoccupied by A in unit cell and volume of unit cell?

Options:
A 2 9 π 3 6 4
B 7 Ï 3 6 4
C 1 9 π 3 6 4
D 2 π 3 6 4

The correct answer is:
7 π 3 6 4

Solution:
Effective number of atoms of B present in a unit cell = 2
Total volume of B unoccupied by A in a unit cell
= 2 Ã 4 3 R 3 - r 3 Ã Ï
= 7 3 Ï R 3 r = R 2
Volume of unit cell = a3
4 R 3 3 = 6 4 3 3 R 3 3 a = 4 R
Ratio of total volume of sphere B unoccupied by A in unit cell and volume of unit cell = 7 / 3 Ï R 3 6 4 3 3 R 3 = 7 Ï 3 6 4

Topic: Chemistry, Solid State, JEE Main
A body cools from 50 Â°C to 40 Â°C in 5 minutes. Its temperature comes down to 33.33 Â°C in the next 5 minutes. The temperature of the surroundings is

Options:
A 15 °C
B 20 Â°C
C 25 Â°C
D 10 °C

The correct answer is:
20 Â°C

Solution:
Let θ0 = temperature of surrounding,

Using Newton's law of cooling,
log θ2-θ0θ1-θ0=-Kt
log 40-θ050-θ0= -K×5 .....(i)
log 33.33-θ040-θ0 = -K×5 .....(ii)
From Eqs.(i) and (ii),
40-θ050-θ0 = 33.33-θ040-θ0
On solving, we get
θ0=19.95°C ≈20 °C

Topic: Physics, Thermal Properties of Matter, JEE Main
A body cools from 50 Â°C to 40 Â°C in 5 minutes. Its temperature comes down to 33.33 Â°C in the next 5 minutes. The temperature of the surroundings is

Options:
A 15 ℃
B 20 ℃
C 25 ℃
D 10 ℃

The correct answer is:
20 ℃

Solution:
Using Newton's law of cooling,
log Î¸2-Î¸0Î¸1-Î¸0=-Kt
log 40-Î¸050-Î¸0= -KÃ5 .....(i)
log 33.33-Î¸040-Î¸0 = -KÃ5 .....(ii)
From Eqs.(i) and (ii),
40-Î¸050-Î¸0 = 33.33-Î¸040-Î¸0
On solving, we get
Î¸0=19.95â â20â

Topic: Physics, Thermal Properties of Matter, JEE Main
A body cools from 60°C to 40°C in 6 minutes. If, temperature of surroundings is 10°C. Then, after the next 6 minutes, its temperature will be ______ °C.

Options:

The correct answer is:
28

Solution:
From Newton’s law of cooling

dTdt=kTavg-T0, here, T0 is surrounding temperature.

For case 1:

60-406=k60+402-10

206=k50-10 ...1

For case 2:

40-T6=k40+T2-10 ...2

From equation 1, we have k=112

Putting in equation 2

40-T6=11240+T2-10

⇒10+T2=80-2T⇒2T+T2=70

⇒T=1405=28°C

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2023 (29 Jan Shift 1)
A body cools from 80oC to 60o in 5 minutes. The temperature for the surrounding is 20oC. The time it takes to cool from 60oC to 40oC is

Options:
A 450 s
B 420 s
C 500 s
D 253 s

The correct answer is:
500 s

Solution:
The formula to calculate the rate of cooling of the object is given by

dTdt=KTf+Ti2-T0 ...1

For the first case, it can be written, using equation (1) that

80-605=K80+602-204=50K ...2

If t is the required time for the second case, from equation (1), it can be written that

60-40t=K60+402-2020t=30K ...3

Divide equation (2) by equation (3) and solve to calculate the required time.

420t=50K30K⇒t5=53⇒t=253 min=500 s

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2023 (12 Apr Shift 1)
A body cools from a temperature 3T to 2T in 10 minutes. The room temperature is T. Assume that Newton's law of cooling is applicable. The temperature of the body at the end of next 10 minutes will be

Options:
A 74T
B 32T
C 43T
D T

The correct answer is:
32T

Solution:
T1-T2∆t=KT1+T22-T0
3T-2T10=K2.5T-T
⇒T10=K1.5T
K=115
Now, 2T-x10=K2T+x2-T
Solving x=3T2

Topic: Physics, Thermal Properties of Matter, JEE Main
A body cools in 7 minutes from 60 Â°C to 40 Â°C. What will be its temperature after the next 7 minutes? The temperature of the surrounding is 10 Â°C.

Options:

The correct answer is:
28

Solution:
According to Newton's law of cooling,
- d Î¸ dt = K Î¸ - Î¸ 0
or â« 0 t dt = 1 K â« Î¸ 1 Î¸ 2 - d Î¸ Î¸ - Î¸ 0 or t = 1 K log e Î¸ 1 - Î¸ 0 Î¸ 2 - Î¸ 0
So here 7 = 1 K log e [ 60 - 10 40 - 10 ] and also 7 = 1 K log e 40 - 10 Î¸ - 10
â´ log [ 50 30 ] = log 30 Î¸ - 10 or 5 3 = 30 Î¸ - 10
i.e., 5θ-50=90 or θ=28°C

Topic: Physics, Thermal Properties of Matter, JEE Main
A body cools in 7 minutes from 60oC to 40oC. The temperature of the surrounding is 10oC. The temperature of the body after the next 7 minutes will be

Options:
A 30oC
B 32oC
C 34oC
D 28oC

The correct answer is:
28oC

Solution:
Newton's law of cooling states that

T1-T2t=KT1+T22-T0 ...(i)

where T0 is the temperature of the surrounding. From equation (i),

60-407×60=K60+402-10

⇒207×60=40K⇒K=207×60×40

Let the temperature of the body after the given time be x °C. By Newton's law of cooling,

40-x7×60=K40+x2-10⇒40-x7×60=114×60(20+x2)⇒160-4x=20+x⇒x=28 °C

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2023 (06 Apr Shift 2)
A body executes S.H.M. of period 20 seconds. Its velocity is 5 cm s-1, 2 seconds after it has passed the mean position. Find the amplitude of the bob

cos36Â°=0.809

Options:
A 21.45 cm

B 16.56 cm

C 19.67 cm

D 15.34 cm

The correct answer is:
19.67 cm

Solution:
Here, T=20 s
Also, when t=2 s, v=5 cm s-1
Now, v=Aωcos ωt=A×2πTcos2πTt
∴ 5 = A × 2 π 2 0 cos 2 π 2 0 × 2 or A π 1 0 cos π 5 = 5
or πA10cos 36°=5orπA10×0.8090=5
or A=19.67 cm

Topic: Physics, Oscillations, JEE Main
A body executes simple harmonic motion. The potential energy (P.E), the kinetic energy (K.E) and total energy (T.E) are measured as a function of displacement $x$. Which of the following statements is true?

Options:
A
K.E. is maximum when $x=0$

B
T.E is zero when $x=0$

C
K.E is maximum when $\mathrm{x}$ is maximum

D
P.E. is maximum when $x=0$

The correct answer is:
K.E. is maximum when $x=0$

Solution:
No Solution Available

Topic: Physics, Oscillations, JEE Main, JEE Main 2003
A body executes simple harmonic motion under the action of a force, F1 with a time period 45 s. If the force is changed to F2, it executes SHM with time period 35 s. If both the forces F1 and F2 act simultaneously in the same direction on the body, its time period (in seconds) is

Options:
A 1225
B 2425
C 3524
D 2512

The correct answer is:
1225

Solution:
F1=-k1x; F2=-k2x

a1=-k1mx; a2=-k2mx

Also, a1=-ω12x; a2=-ω22x

Now, resultant force F=F1+F2=-k1x-k2x

-kx=-k1x-k2x

k=k1+k2

mω2=mω12+mω22

ω2=ω12+ω22

2πT2=2πT12+2πT22

1T2=1T12+1T22

∴T=T1T2T12+T22

⇒T=45×35452+352=1225 s

Topic: Physics, Oscillations, JEE Main
A body falling from a height of $10 \mathrm{~m}$ rebounds from hard floor. If it loses $20 \%$ energy in the impact, then coefficient of restitution is

Options:
A 0.89
B 0.56
C 0.23
D 0.18

The correct answer is:
0.89

Solution:
As \(20 \%\) energy lost in collision therefore \(\mathrm{mgh}_2=80 \%\) of \(\mathrm{mgh}_1 \Rightarrow \frac{\mathrm{h}_2}{\mathrm{~h}_1}=0.8\)
Here \(h \propto v^2\)
and \(e=\frac{v_f}{v_i}\)
so in such cases:
\(\mathrm{e}=\sqrt{\frac{\mathrm{h}_2}{\mathrm{~h}_1}}=\sqrt{0.8}=0.89\)

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A body falling from rest under gravity passes a certain point $P$. It was at a distance of $400 \mathrm{~m}$ from $P, 4 \mathrm{~s}$ prior to passing through $P$. If $g=10 \mathrm{~m} / \mathrm{s}^2$, then the height above the point $\mathrm{P}$ from where the body began to fall is

Options:
A
720 m

B
900 m

C
320 m

D
680 m

The correct answer is:
720 m

Solution:
We have $h=\frac{1}{2}{g t^2}^2$ and $h+400=\frac{1}{2} g(t+4)^2$.
Subtracting we get $400=8 \mathrm{~g}+4 \mathrm{gt}$
$$
\begin{aligned}
& \Rightarrow \mathrm{t}=8 \mathrm{sec} \\
& \therefore \mathrm{h}=\frac{1}{2} \times 10 \times 64=320 \mathrm{~m} \\
& \therefore \text { Desired height }=320+400=720 \mathrm{~m} .
\end{aligned}
$$

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2006
A body falling under gravity covers two points A and B separated by 80 m in 2 s. The distance of upper point A from the starting point is _____m. Use g=10 m s-2

Options:

The correct answer is:
45

Solution:
From A→B, using the equation of motion, we have

-80=-v1t-12×10t2⇒-80=-2v1-12×10×22⇒-80=-2v1-20⇒-60=-2v1⇒v1=30 m s-1

From O to A, it follows that

v2=u2+2gS⇒302=0+2×(-10)(-S)⇒900=20 S⇒S=45 m

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2024 (27 Jan Shift 2)
A body floats in water with \(40 \%\) of its volume outside water. When the same body floats in an oil, \(60 \%\) of its volume remains outside oil. The relative density of oil is

Options:
A 0.9
B 1.0
C 1.2
D 1.5

The correct answer is:
1.5

Solution:
Hints: \(\mathrm{V} \sigma \mathrm{g}=0.6 \mathrm{~V} \sigma_1 \mathrm{~g}\) ....(1)
\(\mathrm{~V} \sigma \mathrm{g}=0.4 \mathrm{~V} \sigma_2 \mathrm{~g}\) ....(2)
\(\mathrm{V \sigma} \mathrm{g}=0.4 \mathrm{V \sigma}_2 \mathrm{~g}\)
Dividing (1) and (2) \(1=\frac{6}{4} \frac{\sigma_1}{\sigma_2} \therefore \frac{\sigma_2}{\sigma_1}=\frac{3}{2}\)

Topic: Physics, Mechanical Properties of Fluids, WBJEE, WBJEE 2011
A body floats in water with $40 \%$ of its volume outside water. When the same body floats in oil, $60 \%$ of its volume remains outside oil. The relative density of the oil is

Options:
A 0.9
B 1.2
C 1.5
D 1.8

The correct answer is:
1.5

Solution:
Hints : Fraction of immersed part $f=\frac{d}{\rho}$
Case-1,
$f=1-0.4=0.6$
$0.6=\frac{d}{1}$
$d=0.6$
Case-2,
$$
\begin{aligned}
& f=1-0.6=0.4 \\
& f=\frac{d}{\rho_{\text {oil }}} \\
& 0.4=\frac{0.6}{\rho_{\text {oil }}} \\
& \rho_{\text {oil }}=1.5
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Fluids, WBJEE, WBJEE 2010
A body hanging from a massless spring stretches it by 3 cm on earth's surface. At a place 800 km above the earth's surface, the same body will stretch the spring by (Radius of Earth =6400 km)

Options:
A 3427 cm
B 6427 cm
C 2764 cm
D 2734 cm

The correct answer is:
6427 cm

Solution:
Acceleration due to gravity,
g=GMr2
â´ gâ1r2
When the body is hanged on a spring
F=-kx=mg .....(i)
Acts on it, where x is extension in spring. Let 800 km above the Earth's surface, the stretch in the length of spring is xâ² and value of g is gâ² .
g=GMR2 and gâ²=GMR+h2
So, gâ²=g1+hR2 ....(ii)
Where, R= radius of Earth
From Equation (i), we get
kx=mg âxâg ......(iii)
Therefore, xâ²x=gâ²g
From Equations (ii) and (iii), we get
xâ²x=g1+hR2g=gg1+800Ã1036400Ã1032
=gg1+182
=gg982=gÃ64gÃ81=6481
xâ²=6481x x=3 cm
â xâ²=6481Ã3=6427
â xâ²=6427 cm

Topic: Physics, Gravitation, JEE Main
A body hanging from a spring stretches it by 1 cm at the earth's surface. How much will the same body stretch the spring at a place 1600 km above the earth's surface? (ans in cm) (Radius of earth =6400 km) If your answer is K, mark 100K as answer after rounding off to nearest integer.

Options:

The correct answer is:
64

Solution:
In equilibrium, weight of the body = stretching force
â´ At the earth's surface, mg'=kÃx'
âg'g=x'x=Re2(Re+h)2
=(6400)2(6400+1600)2=1625
âx'=1625Ãx=1625Ã1 cm=0.64 cm

Topic: Physics, Gravitation, JEE Main
A body hanging from a spring stretches it by 1 cm at the earth's surface. How much will the same body stretch the spring at a place 1600 km above the earth's surface? (ans in cm) (Radius of earth =6400 km )

Options:

The correct answer is:
0.64

Solution:
In equilibrium, weight of the body = stretching force
â´ At the earth's surface, mg'=kÃx'
âg'g=x'x=Re2(Re+h)2
=(6400)2(6400+1600)2=1625
âx'=1625Ãx=1625Ã1 cm=0.64 cm

Topic: Physics, Gravitation, JEE Main
A body having specific charge 8 μC g-1 is resting on a frictionless plane at a distance 10 cm from the wall (as shown in the figure). It starts moving towards the wall when a uniform electric field of 100 V m-1 is applied horizontally towards the wall. If the collision of the body with the wall is perfectly elastic, then the time period of the motion will be ____s.

Options:

The correct answer is:
1

Solution:
F=ma
qE=ma
a=qEm
Now d=12at2
t=2 da
t=2 dqEm

t=2×0.18×10-610-3×100=12
∴ Time period =2t=1 s
Answer =1.00

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2021 (20 Jul Shift 1)
A body initially at rest and sliding along a frictionless track from a height $\mathrm{h}$ (as shown in the figure) just completes a vertical circle of diameter $\mathrm{AB}=\mathrm{D}$. 'The height $h$ is equal to.

Options:
A $\frac{7}{5} \mathrm{D}$
B $\frac{3}{2} \mathrm{D}$
C $\frac{5}{4} \mathrm{D}$
D D

The correct answer is:
$\frac{5}{4} \mathrm{D}$

Solution:
The correct option is $\mathbf{3}: \frac{5}{4} \mathrm{D}$
Let total energy is conserved,
Total mechanical energy at start $=$ Total mechanical energy at ' $\mathrm{B}$ ' $\mathrm{mgh}=\frac{1}{2} \mathrm{mv}_{\mathrm{B}}^2+\mathrm{mgD}...(i)$
Minimum speed required to complete rotation is, $\mathrm{v}_{\mathrm{b}}=\sqrt{\mathrm{rg}}=\sqrt{\mathrm{Dg} / 2}$ Hence equation (1) becomes, $\mathrm{mgh}=\frac{1}{2} \mathrm{~m} \frac{\mathrm{D}_2}{2}+\mathrm{mgD}$
hence, $h=\frac{5 \mathrm{D}}{4}$

Topic: Physics, Work Power Energy, JEE Main