datavorous/entrance-exam-dataset huggingface dataset:
The figure shows a diagrammatic view of human respiratory system with labels $A, B, C$ and $D$. Select the option, which gives correct identification and main function and/or characteristic.

Options:
A A-trachea-long tube supported by complete cartilaginous rings for conducting inspired air
B B-pleural membrane-surround ribs on both sides to provide cushion against rubbing
C C-alveoli-thin walled vascular bag-like structures for exchange of gases
D D-lower end of lungs-diaphragm pulls it down during inspiration

The correct answer is:
C-alveoli-thin walled vascular bag-like structures for exchange of gases

Solution:
C-Alveoli are thin-walled vascular bag-like structures for exchange of gases. A-trachea or wind pipe is an air conducting tube through, which transport of gases takes place. B-pleural membrane is double layered, which reduces friction on the lung surface. D-diaphragm is involved in the inspiration and expiration process of breathing.

Topic: Biology, Breathing and Exchange of Gases, NEET, NEET 2013 (All India)
The figure shows a disc of radius 3R from which a circular hole of radius R is cut as shown in the figure. The distance of the centre of mass of the remaining object from the point O is

Options:
A R/4
B R/5
C R/3
D R/6

The correct answer is:
R/4

Solution:
The centre of mass of the remaining object is

xcm=m1x1-m2x2m1-m2=σA1x1-σA2x2σA1-σA2

where m1 and A1 are the mass and area of the total disc and m2 and A2 are the mass and area of the cut-out portion. σ is the mass per unit length of the disc.

A1=π(3R)2, A2=πR2

x1=0, x2=2R

⇒xcm=-R/4

Topic: Physics, Center of Mass Momentum and Collision, NEET
The figure shows a fixed incline plane on which a plank is being pulled upwards such that the centre C of the solid cylinder remains at rest with respect to ground. If no slipping takes place between the cylinder and the plank, then the acceleration of the plank is

Options:
A g 2 sin θ
B 2 g sin Î¸
C g sin θ
D 2 g sin θ

The correct answer is:
2 g sin θ

Solution:
Linear acceleration of cylinder is zero, i.e.,
mg sin Î¸=f ( here f = frictional force upwards )

Angular acceleration about the centre Î±=ÏI
or Î±=fR12mR2=2fmR
or Î±=2mg sinÎ¸mR=2g sinÎ¸R
For no slipping between cylinder and plank

a=RÎ±=2g sinÎ¸

Topic: Physics, Rotational Motion, JEE Main
The figure shows a logic circuit with two inputs $A$ and $B$ and the output $C$. The voltage wave forms across $A, B$ and $C$ are as given. The logic circuit gate is

Options:
A OR gate
B NOR gate
C AND gate
D NAND gate

The correct answer is:
OR gate

Solution:
From the given woveforms, the following truth table can be made
This truth table is required to 'OR' gate. So, logic circuit gate is OR gate.
Note The logic of the OR gate is if any of the input is one then output is one.

Topic: Physics, Semiconductors, JEE Main
The figure shows a long coaxial cable in which a current I flows down through the inner cylinder of radius a and the same current flows back up through the outer cylinder of radius 2a. The cylinders are insulated from each other and the current is uniformly distributed over the area of the cross-section in each cylinder. The strength of the magnetic field at a distance 3a2 from the axis of the cable is

Options:
A B=5μ0I36πa
B B=5μ0I12πa
C B=7μ0I12πa
D B=7μ0I36πa

The correct answer is:
B=7μ0I36πa

Solution:
Using ampere's law for a closed loop of radius r=3a2 around the axis of the cable,

∮B→.dl→=μ0I-Iπr2-πa24πa2-πa2

B×2πr=μ0I-Iπr2-πa24πa2-πa2

B×2π3a2=μ0I-Iπ9a24-πa23πa2

B×3πa=μ0I-5I12

B=7μ0I36πa

Topic: Physics, Magnetic Effects of Current, JEE Main
The figure shows a network of currents. Study the diagram and find the magnitude of current \( I \).

Options:
A \( 3 \mathrm{~A} \)
B \( 13 \mathrm{~A} \)
C \( 23 \mathrm{~A} \)
D \( -3 \mathrm{~A} \)

The correct answer is:
\( 23 \mathrm{~A} \)

Solution:
We have to use Kirchhoff's junction rule at each junction.

Currents in each branch are as shown in figure:

At junction A,

15 A=I1+8 A⇒I1=7 A

At junction B,

I1+3 A=I2⇒I2=10 A

At junction D,

8 A+5 A=I3⇒I3=13 A

At junction C,

I2+I3+I=0⇒I=-I2-I3

⇒I=-23 A

∴ I=23 A

∴ The magnitude of current I is 23 A.

Note: We can also use the fact that the net current going into the loop from outside is zero.

15+3+I+5=0⇒I=-23 A

Topic: Physics, Current Electricity, JEE Main
The figure shows a plot of photo current versus anode potential for a photo sensitive surface for three different radiations. Which one of the following is a correct statement ?

Options:
A Curves (b) and (c) represent incident radiations same frequencies having same intensity.
B Curves (a) and (b) represent incident radiations of different frequencies and different intensities
C Curves (a) and (b) represent incident radiations of same frequencies but of different intensities
D Curves (b) and (c) represent incident radiations of different frequencies and different intensities

The correct answer is:
Curves (a) and (b) represent incident radiations of same frequencies but of different intensities

Solution:
As a and $\mathrm{b}$ have same stopping potential and $\mathrm{c}$ has greater stopping potential, then $\mathrm{v}_{\mathrm{c}}>\mathrm{v}_{\mathrm{a}}=\mathrm{v}_{\mathrm{b}}$ as $\mathrm{b}$ and $\mathrm{c}$ have same saturation current and $a$ has lesser value.
So $\mathrm{I}_{\mathrm{a}} < \mathrm{I}_{\mathrm{b}}=\mathrm{I}_{\mathrm{c}}$

Topic: Physics, Dual Nature of Matter, JEE Main
The figure shows a plot of photo current versus anode potential for a photo sensitive surface for three different radiations. Which one of the following is a correct statement ?

Options:
A Curves (b) and (c) represent incident radiations same frequencies having same intensity.
B Curves (a) and (b) represent incident radiations of different frequencies and different intensities
C Curves (a) and (b) represent incident radiations of same frequencies but of different intensities
D Curves (b) and (c) represent incident radiations of different frequencies and different intensities

The correct answer is:
Curves (a) and (b) represent incident radiations of same frequencies but of different intensities

Solution:
As a and $\mathrm{b}$ have same stopping potential and $\mathrm{c}$ has greater stopping potential, then $\mathrm{v}_{\mathrm{c}}>\mathrm{v}_{\mathrm{a}}=\mathrm{v}_{\mathrm{b}}$ as $\mathrm{b}$ and $\mathrm{c}$ have same saturation current and $a$ has lesser value.
So $\mathrm{I}_{\mathrm{a}} < \mathrm{I}_{\mathrm{b}}=\mathrm{I}_{\mathrm{c}}$

Topic: Physics, Dual Nature of Matter, NEET, NEET 2009 (Mains)
The figure shows a portions of the graph $y=2 x-4 x^{3}$. The line $\mathrm{y}=\mathrm{c}$ is such that the areas of the regions marked I and II are equal. If $a, b$ are the $x$-coordinates of $A, B$ respectively, then $\mathrm{a}+\mathrm{b}$ equals

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

The correct answer is:
$\frac{2}{\sqrt{7}}$

Solution:

Topic: Mathematics, Area Under Curves, JEE Main
The figure shows a relationship between the sets $P$ and $Q$. Write this relation (i) in set builder form, (ii) roster form. What is its domain and range?

Options:

None

Solution:
It is obvious that the relation $R$ is "the difference between $x$ and $y$ is $2^n$.
In roster form, $R=\{(5,3),(6,4),(7,5)\}$.
In set builder form, $R=\{(x, y): x-y=2, x \in P, y \in Q\}$.
The domain of this relation is $\{5,6,7\}$.
The range of this relation is $\{3,4,5\}$.

Topic: Mathematics, Relations and Functions (Class 11)
The figure shows a section of the human ovary. Select the option which gives the correct identification of A, B, C.

(i) A is a group of primary follicles that gradually matures under the influence of LH and FSH.

(ii) C is the Graafian follicle which starts producing a hormone called progesterone.

(iii) C is a corpus luteum that secretes FSH.

(iv) B is the Graafian follicle which ruptures during mid of the cycle.

Options:
A i and ii are correct
B ii and iii are correct
C i and iii are correct
D i and iv are correct

The correct answer is:
i and iv are correct

Solution:
Corpus luteum, yellow hormone-secreting body in the female reproductive system that secretes progesterone which further maintains endometrium. Graafian follicle produce a hormone called estrogen by the follicular granulosa cells.
If fertilization does not occur, the corpus luteum involutes to form the corpus Albicans that is filled with fibrous tissue.

Topic: Biology, Human Reproduction, NEET
The figure shows a square loop L of side 5 cm which is connected to a network of resistances. The whole setup is moving towards the right with a constant speed of 1 cm s-1 . At some instant, a part of L is in a uniform magnetic field of 1T perpendicular to the plane of the loop. If the resistance of L is 1.7 Ω, the current in the loop at that instant will be close to:

Options:
A 115 μA
B 60 μA
C 150 μA
D 170 μA

The correct answer is:
170 μA

Solution:
If B is magnetic field, v is velocity of rod and l is the length of rod, then induced EMF in moving road is given by

ε=Bvl
=1T(1 cm s-1)(5 cm)
=1×1×10-2×5×10-2
=5×10-4  V
As given circuit forms balanced whetstone bridge, the current through 3 Ω is zero.
Equivalent resistance Req=4×24+2+1.7=3 Ω
Current in the circuit i=εReq=5×10-43
=167 μA≈170 μA

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2019 (12 Apr Shift 1)
The figure shows a square loop PQRS with an edge length a. The resistance of the wire PQR is r and that of PSR is 2r. The value of the magnetic field at the centre of the loop is found to be 2Î¼0ikÏa, then what is the value of k?

Options:

The correct answer is:
3

Solution:
The magnetic field due to a straight current-carrying wire is

B=μ0I4πdsinα+sinβ

BPQR=2×μ02i34πa2sin45°+sin45°

BPSR=2×μ0i34πa2sin45°+sin45°

Bnet=BPQR-BPSR

⇒Bnet=2μ0i3πa

⇒k=3

Topic: Physics, Magnetic Effects of Current, JEE Main
The figure shows a system of three concentric metal shells A, B and C with radii a, 2a and 3a respectively. Shell B is earthed and shell C is given a charge Q. Now if shell C is connected to shell A, then the final charge on the shell B is âQx11, find the value of x.

Options:

The correct answer is:
8

Solution:
From given conditions, VA=VCand  VB=0     ⇒VB=K(Q−q1)3a+Kq22a+Kq12a=0

⇒2Q+q1+3q2=0 ……(1)

Using VC=VA ; K(Q−q2)3a+Kq23a+Kq13a=Kq1a+K(Q−q1)3a+Kq22a  ⇒  q1=−q24 …….(2)

Using it in (1), q2=−811Q

Topic: Physics, Electrostatics, JEE Main
The figure shows a system of two concentric spheres of radii $r_1$ and $r_2$ and kept at temperatures $T_1$ and $T_2$ respectively. The radial rate of flow of heat in a substance between the two concentric sphere is proportional to

Options:
A
$\frac{r_2-r_1}{r_1 r_2}$

B
$\ln \left(\frac{r_2}{r_1}\right)$

C
$\frac{r_1 r_2}{r_2-r_1}$

D
$\ln \left(r_2-r_1\right)$

The correct answer is:
$\frac{r_1 r_2}{r_2-r_1}$

Solution:
$\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)=\left(\mathrm{T}_1-\mathrm{T}_2\right) \frac{4 \pi \mathrm{r}_1 \mathrm{r}_2 \mathrm{~K}}{\left(\mathrm{r}_2-\mathrm{r}_1\right)}$

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2005
The figure shows a system of two concentric spheres of radii r 1 and r 2 and kept at temperatures T 1 and T 2 , respectively. The radial rate of flow of heat in a substance between the two concentric spheres is proportional to

Options:
A r 1 r 2 r 2 - r 1
B r 2 - r 1
C r2-r1r1r2
D lnr2r1

The correct answer is:
r 1 r 2 r 2 - r 1

Solution:
For conduction from inner sphere to outer one,
dQ = - KA dT dr Ã time dt
or dQ dt = - K Ã 4 Ï r 2 dT dr
â´ Radial rate of flow dQdt=P=-4ÏKr2dTdr
â´ Pâ«r1r2drr2=-4ÏKâ«T1T2dT
or Pr1-r2r1r2=4ÏKT2-T1
or P=4ÏKT1-T2r1r2r2-r1
â´ dQdt is proportional to r 1 r 2 r 2 - r 1 .

Topic: Physics, Thermal Properties of Matter, JEE Main
The figure shows a Young's double slit experimental setup. It is observed that when a thin transparent sheet of thickness t and refractive index μ is put in front of one of the slits, the central maximum gets shifted by a distance equal to n fringe width. If the wavelength of light used is λ then t will be:

Options:
A Dλa(μ-1)
B nλ(μ-1)
C 2nλ(μ-1)
D 2Dλa(μ-1)

The correct answer is:
nλ(μ-1)

Solution:
Shift in central maxima =Daμ-1t
Fringe width =Dλa
It's given,
Daμ-1t=nDλa
t=nλμ-1

Topic: Physics, Wave Optics, JEE Main, JEE Main 2019 (09 Apr Shift 1)
The figure shows an electric line of force which curves along a circular arc. The magnitude of electric field intensity is the same at all points on this curve and is equal to E . If the potential at A is V, then the potential at B is

Options:
A V-ERθ
B V-2ERsin⁡θ2
C V+ERθ
D V+2ERsin⁡θ2

The correct answer is:
V-ERθ

Solution:
VA-VB= Work done by the electric field per unit charge.
Work done from A to B=ERÎ¸
â´VB=VA-ERÎ¸=V-ERÎ¸

Topic: Physics, Electrostatics, JEE Main
The figure shows an elliptical path ABCD of a planet around the sun S such that the area of triangle CSA is 14 the area of the ellipse. (see figure) with DB as the major axis, and CA as the minor axis. If t1 is the time taken for the planet to go over path ABC and t2 for path taken over CDA then:

Options:
A t1=4t2
B t1=2t2
C t1=3t2
D t1=t2

The correct answer is:
t1=3t2

Solution:
Using Kepler's law,
Area ABCDA=x.
Area SABCS=x4+x2=3x4.
Area SADCS=x2-x4=x4.

Since Arial velocity remains constant. Therefore, area swept will be proportional to time.
3x4x4=t1t2
⇒t1=3t2

Topic: Physics, Gravitation, JEE Main, JEE Main 2016 (09 Apr Online)
The figure shows an experimental plot for discharging of a capacitor in an $R-C$ circuit. The time constant $\tau$ of this circuit lies between:

Options:
A
$150 \mathrm{~sec}$ and $200 \mathrm{~sec}$

B
$0$ and $50 \mathrm{~sec}$

C
$50 \mathrm{~sec}$ and $100 \mathrm{~sec}$

D
$100 \mathrm{~sec}$ and $150 \mathrm{~sec}$

The correct answer is:
$100 \mathrm{~sec}$ and $150 \mathrm{~sec}$

Solution:
For discharging of an RC circuit,
$V=V_0 e^{-t / \tau}$
So, when
$V=\frac{V_0}{2}$
$\frac{\mathrm{V}_0}{2}=\mathrm{V}_0 \mathrm{e}^{-\mathrm{t} / \tau}$
$\ln \frac{1}{2}=-\frac{\mathrm{t}}{\tau} \Rightarrow \tau=\frac{\mathrm{t}}{\ln 2}$
From graph when $\mathrm{V}=\frac{\mathrm{V}_0}{2}, \mathrm{t}=100 \mathrm{~s} \quad \therefore \tau=\frac{100}{\ln 2}=144.3 \mathrm{~sec}$

Topic: Physics, Capacitance, JEE Main, JEE Main 2012 (Offline)
The figure shows an overhead view of a corridor with a plane mirror MN mounted at one end. A burglar B sneaks along the corridor directly towards the centre of the mirror. If d=3 m, how far from the mirror will the burglar be when the security guard S can first see him in the mirror?

Options:
A 3.0 m
B 4.5 m
C 6.0 m
D 1.5 m

The correct answer is:
1.5 m

Solution:
tan⁡θ=xMN2 (in the smaller triangle)
tan⁡θ=3 m3 m (in the bigger triangle)
∴ x1.5=33
∴ x=1.5 m

Topic: Physics, Ray Optics, JEE Main
The figure shows an RC circuit with a parallel plate capacitor. Before switching on the circuit, plate A of the capacitor has a charge -Q0 while plate B has no net charge. If at t=0, the switch is closed then after how much time (in seconds) will the net charge on plate A becomes zero?[Given: C=1 Î¼F, Q0=1 mC, Îµ=1000 V and R=2Ã106ln3 Î©]

Options:

The correct answer is:
2

Solution:
Let at any time t the charge that flows from the plate B to plate A is q and the instantaneous current is I.

2q-Q02C+IR-Îµ=0

âRdqdt=-2q+2ÎµC+Q02C

âdq2ÎµC+Q0-2q=dt2RC

Now for the charge on plate A to be zero q=Q0.

â«0Q0dq2ÎµC+Q0-2q=â«0tdt2RC

ât=RCln2ÎµC+Q02ÎµC-Q0=2 s

Topic: Physics, Current Electricity, JEE Main
The figure shows different human tissues labelled as $\mathrm{A}$ to $\mathrm{D}$. Which option gives the correct identification of the label, its location and one feature?

Options:
A D-Unicellular glandular epithelium, goblet cells, secrete saliva
B B-Squamous epithelium, walls of blood vessels, form a diffusion boundary
C A-Cuboidal epithelium, ducts of glands secretion and absorption
D C-Columnar epithelium, lining of stomach, secretion and absorption

The correct answer is:
C-Columnar epithelium, lining of stomach, secretion and absorption

Solution:
The given figure $\mathrm{A}$ is squamous epithelium, found in walls of blood vessels and air sacs of lungs and is involved in functions like forming a diffusion layer. B is cuboidal epithelium, found in ducts of glands and tubular parts of nephrons of kidney and its main functions are secretion and absorption. $\mathrm{C}$ is columnar epithelium, found in lining of stomach and intestine and helps in secretion and absorption. D is ciliated columnar epithelium, present in inner surface of hollow organs like bronchioles and Fallopian tubes and helps to move particles or mucus in a specific direction over the epithelium.

Topic: Biology, Structural Organisation in Animals, AIIMS, AIIMS 2017
The figure shows elliptical orbit of a planet $m$ about the sun $\mathrm{S}$. The shaded area SCD is twice the shaded are $\mathrm{SAB}$. It $t_1$ is the time for the planet to move from $C$ to $D$ and $t_2$ is the time to move from $A$ to $B$ then

Options:
A $\mathrm{t}_1=\mathrm{t}_2$
B $t_1>t_2$
C $t_1=4 t_2$
D $t_1=2 t_2$

The correct answer is:
$t_1=2 t_2$

Solution:
Kepler's $2^{\text {nd }}$ law
$$
\begin{aligned}
& \Rightarrow\left(\frac{\Delta \mathrm{A}}{\Delta \mathrm{t}}\right)_{\text {planet }}=\text { constant } \\
& \frac{\mathrm{A}_1}{\mathrm{t}_1}=\frac{A_2}{t_2} \\
& \Rightarrow \frac{2 \mathrm{~A}}{\mathrm{t}_1}=\frac{A}{t_2} \\
& \Rightarrow \mathrm{t}_1=2 \mathrm{t}_2
\end{aligned}
$$

Topic: Physics, Gravitation, NEET, NEET 2009 (Mains)
The figure shows four animals $(A),(B),(C),(D)$. Select the correct answer with respect to a common characteristics of two these animals.

Options:
A (A) and (D) respire mainly through body wall
B (B) and (C) show radial symmetry
C (A) and (B) have cnidoblasts for self-defense
D (C) and (D) have a true coelom

The correct answer is:
(C) and (D) have a true coelom

Solution:
From phylum-Arthropoda to phylum-Chordata, all are eucoelomate. In option (c), Octopus, a mollusc, is bilaterally symmetrical, triploblastic and coelomate animal. In option (D), scorpion, an arthropod, is bilaterally symmetrical, triploblastic, segmented and coelomate animal.

Topic: Biology, Animal Kingdom, NEET, NEET 2011 (Mains)
The figure shows the acceleration-time graph of a particle. Which of the following represents the corresponding velocity-time graph? ( consider initial velocity zero )

Options:
A
B
C
D

The correct answer is:

Solution:
Since acceleration is constant, therefore there is uniform increase in velocity. So, the v-t graph is a straight line slopping upward to the right. When acceleration becomes zero, velocity is constant. So v-t graph is a straight line parallel to the time-axis.

Topic: Physics, Motion In One Dimension, JEE Main
The figure shows the motion of a planet around the sun in an elliptical orbit with sun at the focus. The shaded areas $A$ and $B$ are also shown in the figure which can be assumed to be equal. If $t_1$ and $t_2$ represent the time for the planet to move from $a$ to $b$ and $d$ to $c$ respectively, then

Options:
A $t_1 \lt t_2$
B $t_1 \gt t_2$
C $t_1=t_2$
D $t_1 \leq t_2$

The correct answer is:
$t_1=t_2$

Solution:
A real velocity of the planet remains constant. If the areas $A$ and $B$ are equal then $t_1=t_2$.

Topic: Physics, Gravitation, JEE Main
The figure shows the position - time $(x-t)$ graph of one-dimensional motion of a body of mass $0.4 \mathrm{~kg}$. The magnitude of each impulse is

Options:
A
$0.4 \mathrm{Ns}$

B
$0.8 \mathrm{Ns}$

C
$1.6 \mathrm{Ns}$

D
$0.2 \mathrm{Ns}$

The correct answer is:
$0.8 \mathrm{Ns}$

Solution:
From the graph, it is a straight line so, uniform motion. Because of impulse direction of velocity changes as can be seen from the slope of the graph.
$$
\begin{aligned}
& \text { Initial velocity }=\frac{2}{2}=1 \mathrm{~m} / \mathrm{s} \\
& \text { Final velocity }=-\frac{2}{2}=-1 \mathrm{~m} / \mathrm{s} \\
& \qquad \bar{P}_{\mathrm{i}}=0.4 \mathrm{~N}-\mathrm{s} \\
& \qquad \overline{\mathrm{P}}_{\mathrm{i}}=-0.4 \mathrm{~N}-\mathrm{s} \\
& \overrightarrow{\mathrm{J}}=\overline{\mathrm{P}}_{\mathrm{i}}-\overline{\mathrm{P}}_{\mathrm{i}}=-0.4-0.4=-0.8 \mathrm{~N}-\mathrm{s}(\overrightarrow{\mathrm{J}}=\text { impulse }) \\
& |\overrightarrow{\mathrm{J}}|=0.8 \mathrm{~N}-\mathrm{s}
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
The figure shows the position - time $(x-t)$ graph of one-dimensional motion of a body of mass $0.4 \mathrm{~kg}$. The magnitude of each impulse is

Options:
A
$0.4 \mathrm{Ns}$

B
$0.8 \mathrm{Ns}$

C
$1.6 \mathrm{Ns}$

D
$0.2 \mathrm{Ns}$

The correct answer is:
$0.8 \mathrm{Ns}$

Solution:
From the graph, it is a straight line so, uniform motion. Because of impulse direction of velocity changes as can be seen from the slope of the graph.
$$
\begin{aligned}
& \text { Initial velocity }=\frac{2}{2}=1 \mathrm{~m} / \mathrm{s} \\
& \text { Final velocity }=-\frac{2}{2}=-1 \mathrm{~m} / \mathrm{s} \\
& \qquad \bar{P}_{\mathrm{i}}=0.4 \mathrm{~N}-\mathrm{s} \\
& \qquad \overline{\mathrm{P}}_{\mathrm{i}}=-0.4 \mathrm{~N}-\mathrm{s} \\
& \overrightarrow{\mathrm{J}}=\overline{\mathrm{P}}_{\mathrm{i}}-\overline{\mathrm{P}}_{\mathrm{i}}=-0.4-0.4=-0.8 \mathrm{~N}-\mathrm{s}(\overrightarrow{\mathrm{J}}=\text { impulse }) \\
& |\overrightarrow{\mathrm{J}}|=0.8 \mathrm{~N}-\mathrm{s}
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2010
The figure shows the position-time x-t graph of one-dimensional motion of a body of mass 0.4 kg. The magnitude of each impulse is

Options:
A 0.4 N s

B 0.8 N s

C 1.6 N s

D 0.2 N s

The correct answer is:
0.8 N s

Solution:
From the graph, it is a straight line so, uniform motion. Because of impulse direction of velocity changes as can be seen from the slope of the graph.
Initial velocity v1=22=1 m s-1
Final velocity v2=-22=-1 m s-1
pi=mv1=0.4 N s
pf=mv2=-0.4 N s
J=pf-pi=-0.4-0.4
=-0.8 N s (J=impulse)
∴ J=0.8 N s

Topic: Physics, Laws of Motion, JEE Main
The figure shows the v-t graph of a particle moving in a straight line. Starting from rest, the time after which the particle returns to its starting position is

â

Options:
A 36.2 s

B 40 s
C 25 s
D 42.6 s

The correct answer is:
36.2 s

Solution:
For a particle to return to origin. Net displacement is zero.

For t=0 to t=20 s,

a1=2020 =1 m s-2

For t=20 s to t s,

a2=0-2025-20=-4 m s-2
0 = 1 2 × 2 0 × 2 5 + 1 2 × - 4 × t - 2 5 2
(t - 25)2 = 125
t - 25 = ± 11.2

t >25 s
t = 36.2 s

Topic: Physics, Motion In One Dimension, JEE Main
The figure shows three circuits I, II and III which are connected to a 3 V battery. If the powers dissipated by the configurations I, II and III are P1, P2 and P3 respectively, then -

Options:
A P2>P1>P3
B P1>P2>P3
C P3>P2>P1
D P1>P3>P2

The correct answer is:
P2>P1>P3

Solution:
P=V2R
P∝1R
Req for I circuit by balanced Wheatstone bridge
R1=1 Ω

Req for circuit II by parallel combination
1R2=12+ 12+11

R2=12

Req for circuit III by parallel and series combination
R3=1+112+12
R3=2
so,P2>P1>P3.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2017 (09 Apr Online)
The figure shows two blocks of mass m each, connected by an ideal unstretched spring and then placed on a frictionless floor. If the blocks are given velocities v0 and 2v0 as shown, then the maximum extension in the spring is

Options:
A mv02k
B mv022k
C 9mv022k
D None of these

The correct answer is:
9mv022k

Solution:
2mv=m2v0-mv0

v=v02
ws=kf-ki

-k2x2-02=122mv2-12mv02+12m2v02
=mv022-5mv022
⇒ - kx22=-9mv024

x=92mv02k

Topic: Physics, Work Power Energy, JEE Main
The figure shows two diagrams of the same screw gauge. In the first case, nothing is kept in its jaw and in the second case a small ball is kept between the jaws for which the diameter is required to be measured. The number of circular divisions on the shown screw gauge is 50. It moves 0.5 mm on the main scale for one complete rotation and the main scale has 12 mm marks. The diameter of the ball is

Options:
A 2.25 mm
B 2.20 mm
C 1.20 mm
D 1.25 mm

The correct answer is:
1.20 mm

Solution:
Pitch = 0.5 mm
L.C = 0.5 50
= 0.01 mm
M.S.R = 2
Neck reading = 25
d = 2 Ã 0.5 + 25 Ã 0.01
= 1.25 mm
Considering zero error
Here zero error is
= + 5 Ã 0.01
= + 0.05 mm
so actual diameter
= d - 0.05
= 1.20 mm

Topic: Physics, Experimental Physics, NEET