datavorous/entrance-exam-dataset huggingface dataset:
A flat mirror M is arranged parallel to the wall W at a distance l from it. The light produced by a point source S kept on the wall is reflected by the mirror and produces a light spot on the wall. The mirror moves with velocity v towards the wall.

Options:
A The spot of light will move with the speed v on the wall
B The spot of light will move on the wall
C As the mirror comes closer the spot of light will become larger and shift away from the wall with speed larger than v
D The size of the light spot on the wall remains the same

The correct answer is:
The size of the light spot on the wall remains the same

Solution:
â
Since for refraction ∠i=∠r, so spot of light will not move on wall and will have same size of object

Topic: Physics, Ray Optics, NEET
A flat mirror revolves at a constant angular velocity making n=0.4 revolutions per second. With what velocity (in m s-1) will a light spot move along a spherical screen with a radius of 15 m, if the mirror is at the centre of curvature of the screen?

Options:
A 37.7
B 60.3
C 68.7
D 75.4

The correct answer is:
75.4

Solution:
∵ Angular velocity of the mirror =0.4 rev s-1
=0.4×2π=0.8π rad s-1
∴ Angular velocity of the reflected ray
=2×0.8π=1.6π rad s-1
Hence, the velocity of light spot over the screen
v=rω=15×1.6π=75.4 m s-1

Topic: Physics, Ray Optics, JEE Main
A flexible wire bent in the form of a circle is placed in a uniform magnetic field perpendicularly to the plane of the coil. The radius of the coil changes as shown in the diagram. The graph of the magnitude of induced emf in the coil is represented by

Options:
A

B

C

D

The correct answer is:

Solution:
In the r-t graph, it is clear that from a to b there is no change in radius and hence no change in area and magnetic flux. Same is the situation from c to d for b to c.
Now , e = d dt Ï
e = B d dt Ï r 2 = B Ï 2 r dr dt
Since r â t, â´ dr dt = constant
â´ e â r
This is represented in option (2).

Topic: Physics, Electromagnetic Induction, JEE Main
A fluid having a thermal coefficient of volume expansion Î³ is filled in a cylindrical vessel up to a height h0. The coefficient of linear thermal expansion of the material of the vessel is Î±. If the fluid is heated with the vessel, then find the level of liquid when the temperature increases by ÎÎ¸.

Options:
A h0(1+γΔθ)(1+2αΔθ)
B h0(1+γΔθ)(1+3αΔθ)
C h0(1+γΔθ)(1+αΔθ)
D h0(1+γΔθ)

The correct answer is:
h0(1+γΔθ)(1+2αΔθ)

Solution:
Abase=A01+βΔT=A0(1+2αΔθ )
Vfluid=V01+γΔT=A0h0(1+γΔθ)
Ah=A01+2αΔθh=A0h0(1+γΔθ)
h=h0(1+γΔθ)(1+2αΔθ)

Topic: Physics, Thermal Properties of Matter, JEE Main
A fluid is in streamline flow across a horizontal pipe of variable area of cross section. For this which of the following statements is correct?

Options:
A The velocity is maximum at the narrowed part of the pipe and pressure is a maximum at the widest part of the pipe.
B Velocity and pressure both are maximum at the narrowest part of the pipe.
C Velocity and pressure both are maximum at the widest part of the pipe
D The velocity is minimum at the narrowest part of the pipe and the pressure is minimum at the widest part of the pipe.

The correct answer is:
The velocity is maximum at the narrowed part of the pipe and pressure is a maximum at the widest part of the pipe.

Solution:
The velocity is maximum at the narrows part of the pipe and pressure is maximum at the widest part of the pipe.
(Using equation of continuity $\mathrm{AV}=$ constant.)
And Bernoulli's theorem
$$
P+\frac{1}{2} \rho V^2=\text { constant }
$$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A fly wheel is accelerated uniformly from rest and rotates through 5 rad in the first second. The angle rotated by the fly wheel in the next second, will be:

Options:
A 7.5 rad
B 15 rad
C 20 rad
D 30 rad

The correct answer is:
15 rad

Solution:
$\begin{aligned} & \theta_1=1 / 2 \alpha(2 \times 1-1)=5 \mathrm{rad} \\ & \Rightarrow a=10 \mathrm{rad} / \mathrm{sec}^2\end{aligned}$
So $\theta_2=12 \times \alpha(2 \times 2-1)=15 \mathrm{rad}$

Topic: Physics, Rotational Motion, JEE Main
A fly wheel is accelerated uniformly from rest and rotates through 5rad in the first second. The angle rotated by the fly wheel in the next second, will be :

Options:
A 7.5rad
B 15rad
C 20rad
D 30rad

The correct answer is:
15rad

Solution:
Here the angular acceleration is constant, so we have,

θ=ωt+12αt2.

When flywheel is rotated from rest in the first second, then θ1=12α12=5 rad

⇒α=10 rad s-2

The angle turned by flywheel in the next second is θ2=12α22=12104=20 rad s-2

Thus, the angle turned during next second is θ=20-5=15 rad

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2022 (24 Jun Shift 2)
A FM Broad cast transmitter, using modulating signal of frequency 20kHz has a deviation ratio of 10. The Bandwidth required for transmission is :

Options:
A 220kHz
B 180kHz
C 360kHz
D 440kHz

The correct answer is:
440kHz

Solution:
Given

FM broadcast

Modulating frequency =20 kHz=f

Deviation ratio = frequency deviation modulating frequency =∆ff

⇒ Frequency deviation, Δf=f×10

=20 kHz×10=200 kHz

⇒ Bandwidth =2f+Δf

=220+200 kHz

=440 kHz

Topic: Physics, Communication System, JEE Main, JEE Main 2022 (28 Jul Shift 2)
A focus of an ellipse is at the origin. The directrix is the line $x=4$ and the eccentricity is $1 / 2$. Then the length of the semi-major axis is

Options:
A
$\frac{8}{3}$

B
$\frac{2}{3}$

C
$\frac{4}{3}$

D
$\frac{5}{3}$

The correct answer is:
$\frac{8}{3}$

Solution:
Major axis is along $x$-axis.
$$
\begin{aligned}
& \frac{a}{e}-a e=4 \\
& a\left(2-\frac{1}{2}\right)=4 \\
& a=\frac{8}{3} .
\end{aligned}
$$

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2008
A force $\left(3 x^2+2 x-5\right) \mathrm{N}$ displaces a body from $x=2 \mathrm{~m}$ to $x=4 \mathrm{~m}$. Work done by this force is ________ $J$.

Options:

The correct answer is:
58

Solution:
$\begin{aligned} & W=\int_{x_1}^{x_2} F d x \\ & W=\int_2^4\left(3 x^2+2 x-5\right) d x \\ & W=\left[x^3+x^2-5 x\right]_2^4 \\ & W=[60-2] J=58 J\end{aligned}$

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2024 (09 Apr Shift 2)
A force $(5 \hat{\imath}-2 \hat{\jmath}+3 \hat{k}) \mathrm{N}$ acts on a body of mass $2 \mathrm{~kg}$ and displaces it from $(3 \hat{\imath}+2 \hat{\jmath}-\hat{k})$$\mathrm{m}$ to $(6 \hat{\imath}-\hat{\jmath}+4 \hat{k}) \mathrm{m} .$ The work done is

Options:
A 27 J
B 18 J
C $36 \mathrm{~J}$
D $9 \mathrm{~J}$

The correct answer is:
$36 \mathrm{~J}$

Solution:
No Solution

Topic: Physics, Work Power Energy, JEE Main
A force acts on a 2 kg object so that its position is given as a function of time as x=3t2+5. What is the work done by this force in first 5 seconds?

Options:
A 875 J
B 850 J
C 950 J
D 900 J

The correct answer is:
900 J

Solution:
Here, the position of the object is x=3t2+5, therefore the velocity of the object will be,

v=dxdt=d3t2+5dt

v=6t+0

From the work-energy theorem, all the work done by the force acting on it will be equal to the change in its kinetic energy, i.e.,

W=KEt=5 s-KEt=0 s
W=12×2×6×52-12×2×6×02
W=900 J.

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2019 (09 Jan Shift 2)
A force acts on a $30 \mathrm{gm}$ particle in such a way that the position of the particle as a function of time is given by $x=3 t-4 t^2+t^3$, where $x$ is in metres and $t$ is in seconds. The work done during the first 4 seconds is

Options:
A 5.28 J
B 450 mJ
C 490 mJ
D 530 mJ

The correct answer is:
5.28 J

Solution:
$\begin{aligned} & v=\frac{d x}{d t}=3-8 t+3 t^2 \\ & \therefore v_0=3 \mathrm{~m} / \mathrm{s} \text { and. } v_4=19 \mathrm{~m} / \mathrm{s}\end{aligned}$
$W=\frac{1}{2} m\left(v_4^2-v_0^2\right)$
(According to work energy theorem)
$=\frac{1}{2} \times 0.03 \times\left(19^2-3^2\right)=5.28 \mathrm{~J}$

Topic: Physics, Work Power Energy, JEE Main
A force applied by an engine on a train of mass $2.05 \times 10^6 \mathrm{~kg}$ changes its velocity from $5 \mathrm{~m} / \mathrm{s}$ to $25 \mathrm{~m} / \mathrm{s}$ in 5 minutes. The power of the engine is

Options:
A $1.025 \mathrm{MW}$
B $2.05 \mathrm{MW}$
C $5 \mathrm{MW}$
D $6 \mathrm{MW}$

The correct answer is:
$2.05 \mathrm{MW}$

Solution:
$\begin{aligned} m=2.05 & \times 10^6 \mathrm{~kg}, v_1=5 \mathrm{~m} / \mathrm{s} \\ v_2 & =25 \mathrm{~m} / \mathrm{s}, t=5 \mathrm{~min} \\ & =5 \times 60=300 \mathrm{sec} \\ \text { Power } & =\frac{W}{t}=\frac{\text { change in KE }}{t} \\ & =\frac{1}{2} \frac{m\left(v_2^2-v_1^2\right)}{300} \\ & =\frac{1}{2} \times \frac{2.05 \times 10^6\left[(25)^2-(5)^2\right]}{300} \\ & =\frac{1}{2} \times \frac{2.05 \times 10^6 \times 600}{300} \\ & =2.05 \times 10^6 \mathrm{~W} \\ & =2.05 \mathrm{MW}\end{aligned}$

Topic: Physics, Work Power Energy, JEE Main
A force F→=2+3xi^ acts on a particle in the x direction where F is in Newton and x is in meter. The work done by this force during a displacement from x = 0 to x=4 m is _____ J.

Options:

The correct answer is:
32

Solution:
The formula to calculate the work done by a variable force is given by

W=∫x1x2Fdx ...1

Substitute the values of the known parameters into equation (1) to calculate the required work done.

W=∫04 (2+3x)dx=2x+32x204=[8+24] J=32 J

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2023 (11 Apr Shift 1)
A force F→=4i^+3j^+4k^ is applied on an intersection point of x=2 plane and x-axis. The magnitude of torque of this force about a point 2,3,4 is ________.

(Round off to the Nearest Integer)

Options:

The correct answer is:
20

Solution:
τ→=r→×F→

r→=2i^-2i^+3j^+4k^=-3j^-4k^

and F→=4i^+3j^+4k^

τ→=r→×F→=i^j^k^0-3-4434

=i^-12+12-j^0+16+k^0+12

=-16i^+12k^

∴ |τ→|=162+122=20

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2021 (16 Mar Shift 2)
A force F=(5+3y2) acts on a particle in the y-direction, where F is newton and y is in meter. The work done by the force during a displacement from y=2 m to y=5 m is ______ J.

Options:

The correct answer is:
132

Solution:
If an object is moved by certain distance by the application of an external force along the direction of the force, then the force is said to perform some work.

Given, the force exerted on the object is

F=5+3y2

The formula to calculate the work done W by the force in moving the object from a position y1 to another position y2 can be written as

W= ∫y1y2Fdy..........................(1)

Substitute the values into equation (1) and solve to calculate the required work done by the force.

W=∫255+3y2dy=5y+3y3325=132

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2023 (01 Feb Shift 2)
A force $\vec{F}=(5 \hat{i}+3 \hat{j}+2 \hat{k}) N$ is applied over a particle which displaces it from its origin to the point $\vec{r}=(2 \hat{i}-\hat{j}) m$. The work done on the particle in joules is

Options:
A
$-7$

B
$+7$

C
$+10$

D
$+13$

The correct answer is:
$+7$

Solution:
Work done, W $=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{s}}$
Here $\vec{s}=\vec{r}_f-\vec{r}_i=(2 \hat{i}-\hat{j})$
$$
W=(5 \hat{i}+3 \hat{j}+2 \hat{k})(2 \hat{i}-\hat{j})=10-3=7 \mathrm{~J}
$$

Topic: Physics, Work Power Energy, JEE Main
A force $\vec{F}=(5 \hat{i}+3 \hat{j}+2 \hat{k}) N$ is applied over a particle which displaces it from its origin to the point $\vec{r}=(2 \hat{i}-\hat{j}) m$. The work done on the particle in joules is

Options:
A
$-7$

B
$+7$

C
$+10$

D
$+13$

The correct answer is:
$+7$

Solution:
Work done, W $=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{s}}$
Here $\vec{s}=\vec{r}_f-\vec{r}_i=(2 \hat{i}-\hat{j})$
$$
W=(5 \hat{i}+3 \hat{j}+2 \hat{k})(2 \hat{i}-\hat{j})=10-3=7 \mathrm{~J}
$$

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2004
A force $\boldsymbol{F}=(5 \hat{i}+3 \hat{j})$ mewton is applied over a particle which displaces it from its origin to the point $r=(2 \hat{i}-1 \hat{j})$ metres. The work done on the particle is

Options:
A -7 joules
B +13 joules
C +7 joules
D +11 joules

The correct answer is:
+7 joules

Solution:
$W=\vec{F} \cdot \hat{s}=(5 \hat{i}+3 \hat{j}) \cdot(2 \hat{i}-\hat{j})=10-3=7 \mathrm{~J}$

Topic: Physics, Work Power Energy, JEE Main
A force F→=(40i^+10j^) N acts on a body of mass 5 kg. If the body starts from rest, its position vector r→ at time t=10 s will be

Options:
A (100i^+400j^) m
B (100i^+100j^) m
C (400i^+100j^) m
D (400i^+400j^) m

The correct answer is:
(400i^+100j^) m

Solution:
dv→dt=a→=F→m=(8i^+2j^) m s-2

dr→dt=v→=(8ti^+2tj^) m s-1

r→=(i^+2j^)t22 m

At t=10 sec

r→=[(8i^+2j^)50] m

⇒r→=(400i^+100j^) m

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2021 (25 Jul Shift 2)
A force F is applied on a block of mass 3 kg which rests on a horizontal surface with a coefficient of friction 123. The maximum value of F for which the block doesn't move, is [Take g=10 ms-2]

Options:
A 20 N
B 10 N
C 15 N
D 25 N

The correct answer is:
20 N

Solution:
From acting on block are shown in adjoining figure
As the block does not move, hence
Fcos⁡60°=f=μN=μMg+Fsin⁡60°
∴F12=1233×10+F⋅32
On simplification, we get F=20 N

Topic: Physics, Laws of Motion, JEE Main
A force $F$ is applied on a square plate of length $L$. If the percentage error in the determination of $L$ is $3 \%$ and in $F$ is $4 \%$, then permissible error in the calculation of pressure is

Options:
A $13 \%$
B $10 \%$
C $7 \%$
D $12 \%$

The correct answer is:
$10 \%$

Solution:
Given,
Percentage error in length $=3 \%$
Percentage error in force $=4 \%$
We know that,
Pressure, $\quad P=\frac{F}{A}$
Area of square $=(L)^2$
$\begin{aligned}
P & =\frac{F}{L^2} \\
\frac{\Delta P}{P} & =\frac{\Delta F}{F}+\frac{2 \Delta L}{L} \\
\left(\frac{\Delta P}{P} \times 100\right) & =\left(\frac{\Delta F}{F} \times 100\right)+2\left(\frac{\Delta L}{L} \times 100\right) \\
& =4 \%+2(3 \%)=10 \%
\end{aligned}$

Topic: Physics, Mathematics in Physics, JEE Main
A force F is related to the position of a particle by the relation F = (10x2) N. The work done by the force when the particle moves from x = 2 m to x = 4 m is

Options:
A 5 6 3 J
B 560 J
C 5603 J
D 3 5 6 0 J

The correct answer is:
5603 J

Solution:
W=∫24Fdx=∫2410x2d⁡x=10x3324
=10343-23=103×56=5603 J

Topic: Physics, Work Power Energy, JEE Main
A force given by F â = f x i ^ + f y j ^ + f z k ^ acts on a particle which moves from a,b,c to d,e,f . The work done by the force F is : (Here A1,A2,A3 are magnitude of area bounded)

Options:
A A1+A2+A3
B A1-A2-A3
C -A1+A2-A3
D A1-A2+A3

The correct answer is:
A1-A2-A3

Solution:
w=â«abfxdx+â«befydy+â«cffzdz
Now â« a d f x dx is positive since f x > 0 and d > a
â« b e f y dy in negative since f y > 0 ; e < b
â« c f f z dz is negative f z < 0 ; f > c
â´ w=A1-A2-A3

Topic: Physics, Mathematics in Physics, JEE Main
A force is applied to a steel wire A, rigidly clamped at one end. As a result elongation in the wire is 0.2 mm. If same force is applied to another steel wire B of double the length and a diameter 2.4 times that of the wire A, the elongation in the wire B will be (wires having uniform circular cross sections)

Options:
A 6.06×10–2 mm
B 2.77×10–2 mm
C 3.0×10–2 mm
D 6.9×10–2 mm

The correct answer is:
6.9×10–2 mm

Solution:
As we know, Y=FA∆ll

⇒F=YAl∆l

For both cases force applied is the same, therefore

YA∆ll1=YA∆ll2

⇒∆l2∆l1=A1A2×l2l1

⇒∆l20.2=πR2π2.4R2×21

⇒∆l2=6.9×10-2 mm

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2023 (30 Jan Shift 2)
A force is represented by F=ax2+bt12, where x= distance and t= time. The dimensions of b2a are :

Options:
A [ML3 T –3]
B [MLT–2]
C [ML–1 T–1]
D [ML2 T–3]

The correct answer is:
[ML3 T –3]

Solution:
Given: F=ax2+bt12

Dimension of a=Fx2=MLT-2L2=M1L-1T-2

Dimension of b=Ft12=MLT-2T12=M1L1T-52

Required value, b2a=M2L2T-5M1L-1T-2=M1L3T-3

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2024 (31 Jan Shift 1)
A force of $10^3$ newton stretches the length of a hanging wire by 1 millimettre. The force required to stretch a wire of same material and length but having four times the diameter by 1 millimettee is

Options:
A $4 \times 10^3 N$
B $16 \times 10^3 \mathrm{~N}$
C $\frac{1}{4} \times 10^3 N$
D $\frac{1}{16} \times 10^3 \mathrm{~N}$

The correct answer is:
$16 \times 10^3 \mathrm{~N}$

Solution:
$F=r \times A \times \frac{l}{L}=F \propto r^2$ (Y, and $L$ are constant)
If diameter is made four times then force required will be 16 times. i.e. $16 \times 10^3 \mathrm{~N}$

Topic: Physics, Mechanical Properties of Solids, JEE Main
A force of $5 \mathrm{~N}$ acts on a particle along a direction making an angle of $60^{\circ}$ with vertical. Its vertical component be

Options:
A $10 \mathrm{~N}$
B $3 N$
C $4 \mathrm{~N}$
D $2.5 \mathrm{~N}$

The correct answer is:
$2.5 \mathrm{~N}$

Solution:
The component of force in vertical direction
$=F \cos \theta=F \cos 60^{\circ}=5 \times \frac{1}{2}=2.5 \mathrm{~N}$

Topic: Physics, Mathematics in Physics, JEE Main
A force of $40 \mathrm{~N}$ acts on a point $\mathrm{B}$ at the end of an L-shaped object, as shown in the figure. The angle $\theta$ that will produce maximum moment of the force about point $\mathrm{A}$ is given by:

Options:
A
$\tan \theta=\frac{1}{4}$

B
$\tan \theta=2$

C
$\tan \theta=\frac{1}{2}$

D
$\tan \theta=4$

The correct answer is:
$\tan \theta=\frac{1}{2}$

Solution:
To produce maximum moment of force line of action of force must be perpendicular to line $\mathrm{AB}$.
$\therefore \tan \theta=\frac{2}{4}=\frac{1}{2}$

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
A force of 40 N acts on a point B at the end of an L-shaped object as shown in the figure. The angle θ that will produce the maximum moment of the force about point A is given by:

Options:
A tan⁡θ=4
B tan⁡θ=14
C tan⁡θ=12
D tan⁡θ=2

The correct answer is:
tan⁡θ=12

Solution:
Moment of force will be maximum when the line of action of force is perpendicular to line AB, tan⁡θ=24=12.

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2018 (15 Apr)
A force of F=(5y+20)j^ N acts on a particle. The work done by this force when the particle is moved from y=0 m to y=10 m is ________J.

Options:

The correct answer is:
450

Solution:
F=(5y+20)j^

ω=∫Fdy=∫010(5y+20)dy

=5y22+20y010

=52×100+20×10

=250+200=450 J

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2021 (25 Jul Shift 2)
A force of $-F \hat{k}$ acts on $O$, the origin of the coordinate system. The torque about the point $(1,-1)$

Options:
A
$-F(\hat{i}-\hat{j})$

B
$F(\hat{i}-\hat{j})$

C
$-F(\hat{i}+\hat{j})$

D
$F(\hat{i}+\hat{j})$

The correct answer is:
$-F(\hat{i}+\hat{j})$

Solution:
$\vec{\tau}=(-\hat{i}+\hat{j}) \times(-F \hat{k})$
$=-F(\hat{i}+\hat{j})$

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2006
A force on an object of mass 100 g is 10i^+5j^ N. The position of that object at t=2 s is ai^+bj^ m after starting from rest. The value of ab will be _____ .

Options:

The correct answer is:
2

Solution:
a→=F→m=10i^+5j^0.1=100i^+50j^

Displacement covered in vector form can be written as, s→=u→t+12a→t2

⇒s→=12100i^+50j^22

, ⇒s→=200i^+100j^

Comparing it with, s→=ai^+bj^, we get a=200 & b=100

Therefore, ab=2

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2022 (25 Jun Shift 1)
A Fraunhofer diffraction pattern due to a single slit of width $0 \cdot 3 \mathrm{~mm}$ is obtained
on a screen placed at a distance of $3 \mathrm{~m}$ from the slit. The first minima lie at $5 \cdot 5 \mathrm{~mm}$
on either side of the central maximum on the screen. The wavelength of light used
is

Options:
A $6000 Å$
B $5500 Å$
C $4500 Å$
D $5000 Å$

The correct answer is:
$5500 Å$

Solution:
$a=0.3 \mathrm{~mm}=0.3 \times 10^{-3} \mathrm{~m}, \quad D=3 \mathrm{~m}$,
$x=5.5 \mathrm{~mm}=5.5 \times 10^{-3} \mathrm{~m}$
$x=\frac{\lambda D}{a}$
$\therefore \lambda=\frac{\mathrm{xa}}{\mathrm{D}}$
$\therefore \frac{\lambda=5.5 \times 10^{-3} \times 0.3 \times 10^{-3}}{3}=5.5 \times 10^{-7} \mathrm{~m}=5500 Å$

Topic: Physics, Wave Optics, JEE Main
A free body of mass $8 \mathrm{~kg}$ is travelling at 2 meter per second in a straight line. At a certain instant, the body splits into two equal parts due to internal explosion which releases 16 joules of energy. Neither part leaves the original line of motion finally

Options:
A Both parts continue to move in the same direction as that of the original body
B One part comes to rest and the other moves in the same direction as that of the original body
C One part comes to rest and the other moves in the direction opposite to that of the original body
D One part moves in the same direction and the other in the direction opposite to that of the original body

The correct answer is:
One part comes to rest and the other moves in the same direction as that of the original body

Solution:
As the body splits into two equal parts due to internal explosion therefore momentum of system remains conserved ie.
$8 \times 2=4 v_1+4 v_2 \Rightarrow v_1+v_2=4$
By the law of conservation of energy
Initial kinetic energy + Energy released due to explosion
= Final kinetic energy of the system
$\begin{aligned}
& \Rightarrow \frac{1}{2} \times 8 \times(2)^2+16=\frac{1}{2} 4 v_1^2+\frac{1}{2} 4 v_2^2 \\
& \Rightarrow v_1^2+v_2^2=16
\end{aligned}$
By solving eq. (i) and (ii) we get $v_1=4$ and $v_2=0$
i.e. one part comes to rest and other moves in the same direction as that of original body.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A free electron of 2.6 eV energy collides with a H+ ion. This results in the formation of a hydrogen atom in the first excited state and a photon is released. Find the frequency of the emitted photon. h=6.6×10-34 J s

Options:
A 9.0×1027 MHz
B 1.45×109 MHz
C 0.19×1015 MHz
D 1.45×1016 MHz

The correct answer is:
1.45×109 MHz

Solution:
By Energy conservation
K.E.e=T.E.H+ +EPhoton
2.6=-13.64+hf
hf=6eV
f=6×1.6×10-196.626×10-34
f=1.45×1015 Hz=1.45×109 MHz

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2021 (31 Aug Shift 2)
A free living nitrogen-fixing cyanobacterium which can also form symbiotic association with the water fern Azolla is:

Options:
A Tolypothrix
B Chlorella
C Nostoc
D Anabaena.

The correct answer is:
Anabaena.

Solution:
Azolla (water fern) and the blue-green alga Anabaena azollae maintain a symbiotic relationship where the alga provides nitrogen to the fern, and the fern provides a habitat for the alga.
Related Theory
Frankia show nitrogen fixation in both symbiotic association with actinorhizal plants, Casuarina and free-living aerobic conditions, while most Rhizobia fixes nitrogen in symbiotic form only.

Topic: Biology, Mineral Nutrition, NEET, NEET 2004
A free neutron decays into a proton, an electron and,

Options:
A a neutrino.
B an anti-neutrino.
C an \( \alpha \)-particle.
D a \( \beta \)-particle.

The correct answer is:
an anti-neutrino.

Solution:
An anti-neutrino is produced when β- is emitted.

n01 → p11 + e-10 +ν

Here, ν →anti-neutrino.
A free neutron (outside the nucleus) is unstable and it decays into a proton, an electron and an anti-neutrino with a mean lifetime of just under 15 min 881.5±1.5 s.
This is possible because the mass of the neutron is slightly greater than that of the proton. The free proton is stable.
The anti-neutrino is an anti-matter particle, the counterpart to a neutrino.

Topic: Physics, Nuclear Physics, JEE Main
A free neutron decays into a proton but a free proton does not decay into neutron. This is because

Options:
A neutron is an uncharged particle
B proton is a charged particle
C neutron is a composite particle made of a proton and an electron
D neutron has larger rest mass than proton

The correct answer is:
neutron has larger rest mass than proton

Solution:
Mass of the neutron is mn=1.67493×10-27 kg,

mass of the proton is mp=1.67262×10-27 kg.

Proton's rest mass is less than that of a neutron's rest mass. As particle can not decay to a state of greater mass, hence the proton does not decay to neutron outside nucleus whereas the neutron can decay into a proton though β-decay.

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2023 (31 Jan Shift 1)