datavorous/entrance-exam-dataset huggingface dataset:
A right isosceles triangle of side a has charges q, 3q and - q arranged on its vertices as shown in the figure. What is the electric potential at point P midway between the line connecting the + q and - q charges?

Options:
A 3qrε0a
B 3q2πε0a
C qπε0a
D 3q22πε0a

The correct answer is:
3q22πε0a

Solution:
Potential at point P due to charges - q, 3q, and + q is given by
Potential at P due to (+ q) charge + potential at P due to (- q) charge + potential at P due to (+3q) charge
i.e. VNet=VA+VB+VC
As given figure VC+VB=0, âµVC=Kq2a2 âµ VB=-Kq2a2
So, VNet=VA=3KqAP=3KQ2a2
âµ AP=a2-2a22=a2-2a24=2a24=2a2
VNet= 3qÃ24ÏÎµ02a
â 3q22ÏÎµ0a

Topic: Physics, Electrostatics, NEET
A right triangular plate ABC of mass m is free to rotate in the vertical plane about a fixed horizontal axis through A. It is supported by a string such that the side AB is horizontal. The reaction at the support A is

Options:
A mg3
B 2mg3
C mg2
D mg

The correct answer is:
2mg3

Solution:
For equilibrium of torque about A,
mgl3=TlorT=mg3
For verticle equilibrium of plate,
RA+T=mgâ´RA=2mg3

Topic: Physics, Rotational Motion, NEET
A rigid ball of mass, m strikes a rigid wall at 60o and gets reflected without any loss of speed as shown in the figure below. The value of impulse imparted by the wall on the ball will be

Options:
A mV
B 2mV
C mV2
D mV3

The correct answer is:
mV

Solution:
J→=P→2-P→1=2mVcosθ=mV.

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2016 (Phase 2)
A rigid body is made of three identical thin rods, each of length L fastened together in the form of letter H. The body is free to rotate about a horizontal axis that runs along the length of one of the legs of the H. The body is allowed to fall from rest from a position in which the plane of H is horizontal. What is the angular speed of the body when the plane of H is vertical ?

Options:
A gL
B 12 gL
C 32gL
D 2gL

The correct answer is:
32gL

Solution:
I=0+ML212+ML22+0+ML2=43ML2
( using parallel axis theorem )
âµ12IÏ2=mghcm
â´1243ML2Ï2=3MgL2
âÏ=32gL

Topic: Physics, Rotational Motion, NEET
A rigid circular loop of radius r and mass m lies in the x-y plane of a flat table and has a current i flowing in it. At this particular place, the Earth's magnetic field is Bâ=Bxi^+Bzk^ . The value of i, so that the loop starts tilting, is

Options:
A mgπBx2+B22
B mgπrBx
C mgπrBz
D mgπrBxBz

The correct answer is:
mgπrBx

Solution:
Mâ=iÏr2k^ ,
Bâ=Bxi^+Bzk^

To start tilting
Ïdue to mass=Ïmagnetic
mgr= MâÃ Bâ
mg r=iÏr2Bx
i=mgÏrBx

Topic: Physics, Magnetic Effects of Current, NEET
A rigid rod of length l leans against a vertical wall (y-axis), as shown in the figure. The other end of the rod is on the horizontal floor. Point A is pushed downwards with a constant velocity. Path of the centre of the rod is

Options:
A a straight line passing through origin
B a straight line not passing through origin
C a circle of radius l/2 and centre at origin
D a circle of radius l/2 but centre not at origin

The correct answer is:
a circle of radius l/2 and centre at origin

Solution:
Let l be the length of the rod and Î¸, the angle of rod with x-axis (horizontal) at some instant of time. Coordinates of the centre of rod at this instant of time are
x =l2cosâ¡Î¸ and y=l2sinâ¡Î¸
Squaring and adding both these equations, we get:
x2+y2=l24
This is an equation of a circle of radius l2 and centre at origin.

Topic: Physics, Laws of Motion, NEET
A ring consisting of two parts ADB and ACB of same conductivity K carries an amount of heat H. The ADB part is now replaced with another metal keeping the temperatures T1 and T2 constant. The heat carried increases to 2H. What should be the conductivity of the new ADB part? (Given ACBADB=3)

Options:
A 73K
B 2K
C 52K
D 3K

The correct answer is:
73K

Solution:
Let ADB=l and ACB=3l
HACBHADB=lADBlACB=13
HACB+HADB=H
HACB=H4
HADB=3H4=KAT2-T1l ............(i)
Now the ADB part is now replaced with another metal, and total heat is 2H.
HACB=H4
HACB+H′ADB=2H
In later case H′ADB=2H-HACB
H′ADB=2H-H4
H′ADB=7H4=K1AT2-T1l ............(ii)
Comparing (i) and (ii)
K1=7K3

Topic: Physics, Thermal Properties of Matter, NEET
A ring is made of a wire having a resistance $R_0=12 \Omega$. Find the points $A$ and $B$, as shown in the figure, at which a current carrying conductor should be connected so that the resistance $R$ of the sub circuit between these points is equal to $8 / 3 \Omega$

Options:
A $\frac{l_1}{l_2}=\frac{5}{8}$
B $\frac{l_1}{l_2}=\frac{1}{3}$
C $\frac{l_1}{l_2}=\frac{3}{8}$
D $\frac{l_1}{l_2}=\frac{1}{2}$

The correct answer is:
$\frac{l_1}{l_2}=\frac{1}{2}$

Solution:
We knows $R \propto l$
Here, $\quad R_1+R_2=12 \Omega$
and $\quad \frac{R_1 \times R_2}{R_1+R_2}=\frac{8}{3} \Omega$
$\Rightarrow \quad R_1 R_2=32 \Omega$
We get, $R_1=8$ and $R_2=4$
Again, $R_1=\frac{12 l_1}{l_1+l_2}$
and $R_2=\frac{12 l_2}{l_1+l_2}$
Hence, $\quad \frac{l_1}{l_2}=\frac{1}{2}$

Topic: Physics, Current Electricity, NEET, NEET 2012 (Screening)
A ring of charge with radius 0.5 m has 0.002Ï m gap. If the ring carries a charge of +1 C, the electric field at the centre is

Options:
A 7.5Ã107 N C-1
B 7.2Ã107 N C-1
C 6.2Ã107 N C-1
D 6.5Ã107 N C-1

The correct answer is:
7.2Ã107 N C-1

Solution:
Approaximate charge on the element opposite to the gap is
Îq=Q2Ïr0.002Ï
=12Ï0.5Ã2Ï1000=2Ã10-3 C
E=9Ã109Ã2Ã10-30.52=7.2Ã107 N C-1

Topic: Physics, Electrostatics, NEET
A ring of mass 5 kg sliding on a frictionless vertical rod connected by a block B of mass 10 kg by the help of a massless string. Then at the equilibrium of the system the value of Î¸ is

Options:
A 30°
B 60°
C 90°
D 0°

The correct answer is:
60°

Solution:
For ring,
From force diagram of ring,
Tcos⁡θ=m1g ......(i)
For block B
mg=T........ (ii)
Tcos⁡θ=m1g
mgcos⁡θ=m1g
∴ cos⁡θ=m1m=510=12
⇒ θ=60°

Topic: Physics, Laws of Motion, NEET
A ring of mass m and radius R has three particle attached to the ring as shown in the figure. The centre of the ring has a speed v0. The kinetic energy of the system is (slipping is absent).

Options:
A 6 m v 0 2
B 12 m v 0 2
C 4 m v 0 2
D 8 m v 0 2

The correct answer is:
6 m v 0 2

Solution:
KE = 1 2 m v 0 2 + 1 2 mR 2 v 0 R 2
= 1 2 3 m 2 v 0 2 + 1 2 m 2 v 0 2
= 6 m v 0 2

Topic: Physics, Rotational Motion, NEET
A ring of mass M is kept on a horizontal rough surface. A force F is applied tangentially at its rim as shown. The coefficient of friction between the ring and the surface is Î¼. Then

Options:
A Friction will act in the forward direction
B Friction will act in the backward direction
C Frictional force will not act
D Frictional force will be μ Mg

The correct answer is:
Frictional force will not act

Solution:
F - f = Ma
FR+fR=Iα
a = α R
Putting I = MR2 and solving we get f=0

Topic: Physics, Rotational Motion, NEET
A ring of radius R rolls without slipping on a rough horizontal surface with a constant velocity. The radius of curvature at the highest point of the path followed by any particle on the ring is

Options:
A 5 R
B 2 R
C 4 R
D none of these

The correct answer is:
4 R

Solution:
Radius of Curvature is given by
=velocity2Normal Acceleration

=2 v2v2R=4 R

Topic: Physics, Rotational Motion, NEET
A ring of radius R rolls without slipping on a rough horizontal surface with a constant velocity. The radius of curvature of the path followed by any particle of the ring at the highest point of its path will be :

Options:
A 5 R
B 2 R
C 4 R
D None of these

The correct answer is:
4 R

Solution:
Radius of Curvature
= velocity 2 Normal Acceleration

= 2 v 2 v 2 / R = 4 R

Topic: Physics, Rotational Motion, NEET
A ring shaped tube contains two ideal gases with equal masses and molar masses M 1 =32 and M 2 =28 . The gases are separated by one fixed partition and another movable stopper S which can move freely without friction inside the ring (at room temperature). The angle Î± is

Options:
A 182°
B 170°
C 192°
D 180°

The correct answer is:
192°

Solution:
PV = m M RT (for ideal gas)
â´ MV = mRT P
In the position of equilibrium of stopper S.
P 1 = P 2 , T 1 = T 2 , m 1 = m 2
â´âââMV= constant
M 1 V 1 = M 2 V 2
â A Ã 3 2 3 6 0 - Î± = 2 8 Î± Ã A
â´ Î± = 1 9 2 â

Topic: Physics, Thermodynamics, NEET
A ring starts to roll down the inclined plane of height h without slipping. The velocity when it reaches the ground is

Options:
A 10gh7
B 4gh7
C 4gh3
D gh

The correct answer is:
gh

Solution:
mgh=12mv2+12Iω2
=12 mv2+12mr2v2r2
gh=v2
v=gh

Topic: Physics, Rotational Motion, NEET
A river is flowing due east with a speed of 3 m/s. A swimmer can swim in still water at a speed of 4 m/s. If the swimmer starts swimming due north, what will be his resultant velocity?

Options:
A 3 m/s
B 4 m/s
C 5 m/s
D 7 m/s

The correct answer is:
5 m/s

Solution:
Correct Option is : (C)
5 m/s

Topic: Chemistry, Motion In Two Dimensions, NEET
A river with an inflow of domestic sewage rich in organic waste may result in:

Options:
A Drying of the river very soon due to algal bloom.
B Increased population of aquatic food web organisms.
C An increased production of fish due to biodegradable nutrients.
D Death of fish due to lack of oxygen.

The correct answer is:
Death of fish due to lack of oxygen.

Solution:
A river with an inflow of domestic sewage rich in organic waste will reduce the dissolved oxygen (DO) and may result in death of fish due to lack of oxygen.

Topic: Biology, Microbes in Human Welfare, NEET, NEET 2016 (Phase 1)
A road of 10 m width has a radius of curvature 50 m. Its outer edge is raised above the inner edge by a distance of 1.5 m. The road is most suited for vehicles moving with a velocity of

Options:
A 8.5 m s-1
B 6.5 m s-1
C 5.5 m s-1
D none of these

The correct answer is:
8.5 m s-1

Solution:
tanÎ¸=v2rg
v=rg tanÎ¸
=rg hb (where h is the elevation of the other end of the road and b is the width of the road)
=1.5501010
=8.5 m s-1

Topic: Physics, Laws of Motion, NEET
A rocket is fired from the earth towards the sun. At what distance from the earth’s centre is the gravitational force on the rocket zero ? Mass of the sun = 2×10^30 kg, mass of the earth = 6×10^24 kg. Neglect the effect of other planets etc. (orbital radius = 1.5 × 10^11 m)

Options:
A 2.59 ×10^8
B 3.59 ×10^8
C 2.11 ×10^8
D 3.11 ×10^8

The correct answer is:
2.59 ×10^8

Solution:
Correct Option is : (A)
2.59 ×10^8

Topic: Chemistry, Gravitation, NEET
A rocket is fired from the Earth towards the Sun. At what distance from the Earth's centre the gravitational force on the rockets is zero? Mass of the Sun =2Ã1030 kg and mass of the earth=6Ã1024 kg. Distance between earth and sun =15Ã1010 m

Options:
A 2.6Ã108 m
B 3.2×108 m
C 3.9×109 m
D 2.3×109 m

The correct answer is:
2.6Ã108 m

Solution:
Let P be a point at a distance r from the Earth's centre, where gravitational force due to Sun and Earth are equal and opposite and hence, the gravitational force on the rocket is zero.
GMSmx-r2=GMemr2âGMSx-r2=GMer2
2Ã1030x-r2=6Ã1024r2
x-r2r2=2Ã10306Ã1024 =13Ã106
â x-rr=1033âr=31735x
â r=31375Ã1.5Ã1011=3Ã15Ã100Ã1081735
r=2.594Ã108 m
r=2.6Ã108 m from Earth

Topic: Physics, Gravitation, NEET
A rocket is fired with a speed u=3gR from the earth surface. What will be its speed at interstellar space ?

Options:
A zero
B 2gR
C 7gR
D 3gR

The correct answer is:
7gR

Solution:
From conservation of mechanical energy
K.E.+P.E.surface=K.E.+P.E.infinity
12m3gR2+-GMmR=12m v∞2+0
or v∞=7gR

Topic: Physics, Gravitation, NEET
A rocket of mass 4000 kg is set for vertical firing. How much fuel must be consumed per second so that the rocket may have initial upwards acceleration of magnitude 19.6 ms2. [Exhaust gases speed = 980 ms]

Options:
A 240 kg s-1
B 60 kg s-1
C 120 kg s-1
D None

The correct answer is:
120 kg s-1

Solution:
ma = u dm dt - mg
â u dm dt = m a + g
â dm dt = 4 0 00 Ã 1 9 . 6 + 9 . 8 9 8 0
dm dt = 1 2 0 kg/s

Topic: Physics, Center of Mass Momentum and Collision, NEET
A rod AB of length L and mass M is free to move on a frictionless horizontal surface. It is moving with a velocity v, as shown in figure. End B of rod AB strikes the end of the wall. Assuming elastic impact, the angular velocity of the rod AB, just after impact, is

Options:
A v2L
B 3vL
C 3v2L
D vL

The correct answer is:
3vL

Solution:
From the conservation of angular momentum,

âMvcmL2+ML212Ï=MvL2âvcm+ÏL6=v
From the conservation of linear momentum,

â ÏL2-vcm=vâÏL2+L6=2v
â´ Ï=3vL

Topic: Physics, Rotational Motion, NEET
A rod of length is $3 \mathrm{~m}$ and its mass acting per unit length is directly proportional to distance $x$ from one of its end, then its centre of gravity from that end will be at:

Options:
A $1.5 \mathrm{~m}$
B $2 \mathrm{~m}$
C $2.5 \mathrm{~m}$
D $3.0 \mathrm{~m}$

The correct answer is:
$2 \mathrm{~m}$

Solution:
Here $r=k x$ where
$k=$ constant mass of small element of $d x$ length is
$\begin{aligned}
& d m=k x \cdot d x \\
& \mathrm{X}_{\mathrm{CM}}=\frac{\int x d m}{\int d m}=\frac{\int_0^3 x(x d x)}{\int_0^3 x d x} \\
&=\frac{\left[\frac{x^3}{3}\right]_0^3}{\left[\frac{x^2}{2}\right]_0^3}=\frac{\frac{27}{3}}{\frac{9}{2}}=2
\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2002
A rod of length l and radius r is joined to a rod of length l2 and radius of same material. The free end of small rod is fixed to a rigid base and the free end of larger rod is given a twist of Î¸0, the twist angle at the joint will be

Options:
A Î¸4
B Î¸2
C 5 Î¸6
D 8 Î¸9

The correct answer is:
8 Î¸9

Solution:
Ï=CÎ¸
=ÏÉ³r4Î¸2L=constant
â¹ Î¸-Î¸02=Î¸016
â¹ Î¸0=89Î¸

Topic: Physics, Mechanical Properties of Solids, NEET
A rod of length l rotates with a small but uniform angular velocity ω about its perpendicular bisector. A uniform magnetic field B exists parallel to the axis of rotation. The potential difference between the two ends of the rod is

Options:
A zero
B 12ωBl2
C ωBl2
D 2ωBl2

The correct answer is:
zero

Solution:
Induced emf is generated between centre and end, and at two end. Potential will be same, hence, no potential difference is produced.

Topic: Physics, Electromagnetic Induction, NEET
A rod of mass m and length l is hinged at one of its end A as shown in figure. A force F is applied at a distance x from A. The acceleration of the centre of mass a varies with x as -

Options:
A
B
C
D

The correct answer is:

Solution:
The rod will rotate about point A with angular acceleration:
α=τI=Fxml23=3Fxml2
∴a =l2 α=32Fxml
or a ∝ x
i.e., a-x graph is a straight line passing through the origin.

Topic: Physics, Rotational Motion, NEET
A rod of mass M kg and length L metre is bent in the form of an equilateral triangle. The moment of inertia of triangle about a vertical axis to perpendicular to the plane of triangle and passing through the centre (in units of kg m2 ) is

Options:
A ML212
B ML254
C ML2162
D ML2108

The correct answer is:
ML254

Solution:
Moment of inertia of any one side about its perpendicular bisector is
Icm=12M3L32
Moment of inertia of this rod about about O
I=Icm+Md2
I=112M3L32+M3L3232
I=ML2162
Moment of Inertia of a triangle =3I=ML254

Topic: Physics, Rotational Motion, NEET
A rod of weight W is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at a distance d from each other. The centre of mass of the rod is at distance x from A . the normal reaction on A is:

Options:
A Wxd
B Wdx
C Wd-xx
D Wd-xd

The correct answer is:
Wd-xd

Solution:
Equation for force balance
NA+NB=W ...(i)
NB=W-NA
Equation for torque balance about centre of mass of rod.
NAx=NBd-x
NAx=W-NAd-x
NAx=Wd-Wx-NAd+NAx
NAd=Wd-x
NA=Wd-xd

Topic: Physics, Laws of Motion, NEET, NEET 2015 (Phase 1)
A rod P of length 1 m, is hinged at one end A and there is a ring attached to the other end. Another long rod Q is hinged at B and it passes through the ring. The rod P is rotated about an axis which is perpendicular to the plane in which both rods are present and the variations between the angles Î¸ and Ï are plotted as shown.The distance between the hinges A and B is

Options:
A 3 m
B 1 m
C 2âm
D 2 2 m

The correct answer is:
2 m

Solution:
From the graph it can be seen that the max. value Ï is at Î¸ = 6 0 â . Ï is max. when the rod Q is tangent on the circle on which the ring attached to P moves.
from the fig. d=1 sec 60
=2 m

Topic: Physics, Mathematics in Physics, NEET
A rod PQ of mass M and length L is hinged at end P. The rod is kept horizontal by a massless string tied to point Q as shown in the figure.

When string is cut, the initial angular acceleration of the rod is

Options:
A 3g2L
B gL
C 2gL
D 2g2L

The correct answer is:
3g2L

Solution:
Torque on the rod = Moment of weight of the rod about P .
τ=MgL2 ..... (i)
Moment of inertia of the rod about P
I=ML33 ..... (ii)
As, τ=Iα
From Eqs. (i) and (ii), we get
MgL2=ML23α
⇒α=3g2L

Topic: Physics, Rotational Motion, NEET
A rod, pivoted at O, is resting on a block of height h. If the block moves with a constant velocity v, then the angular velocity of the rod, at the instant shown, is

Options:
A vcos2θh
B vtanθh
C vsec2θh
D vcosθh

The correct answer is:
vcos2θh

Solution:
tanθ=xh⇒x=htanθ
∴ dxdt=hsec2θdθdt
⇒ dθdt=vhcos2θ

Topic: Physics, Laws of Motion, NEET
A $\operatorname{rod} P Q$ of mass $M$ and length $L$ is hinged at end $P$. The rod is kepts horizontal by a massless string tied to point $Q$ as shown in figure. When string is cut, the initial angular acceleration of the rod is

Options:
A $\frac{3 g}{2 L}$
B $\frac{G}{L}$
C $\frac{2 g}{L}$
D $\frac{2 g}{3 L}$

The correct answer is:
$\frac{3 g}{2 L}$

Solution:
Torque on the rad = Mament of weight of the rod about $\rho$
$\tau=m g \frac{1}{2}$
$\because$ Moment of inertia of rod about
$P=\frac{M L^2}{3}$
As $\tau=1 \alpha$
From Eqs. (i) and (ii), we get
$\begin{aligned}
& M g \frac{L}{2}=\frac{M L^2}{3} \alpha \\
\therefore \alpha & =\frac{3 g}{2 L}
\end{aligned}$

Topic: Physics, Rotational Motion, NEET, NEET 2013 (All India)
A roller coaster is designed such that riders experience "weightlessness" as they go round the top of a hill whose radius of curvature is $20 \mathrm{~m}$. The speed of the car at the top of the hill is between

Options:
A $14 \mathrm{~m} / \mathrm{s}$ and $15 \mathrm{~m} / \mathrm{s}$
B $15 \mathrm{~m} / \mathrm{s}$ and $16 \mathrm{~m} / \mathrm{s}$
C $16 \mathrm{~m} / \mathrm{s}$ and $17 \mathrm{~m} / \mathrm{s}$
D $13 \mathrm{~m} / \mathrm{s}$ and $14 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$14 \mathrm{~m} / \mathrm{s}$ and $15 \mathrm{~m} / \mathrm{s}$

Solution:
Balancing the forces, we get
$M g-N=M \frac{v^2}{R}$
For weightlessness, $N=0$
$\therefore \frac{M v^2}{R}=M g$
where $R$ is the radius of curvature and $v$ is the speed of car.
Therefore, $v=\sqrt{R g}$
Putting the values, $R=20 \mathrm{~m}, g=10.0 \mathrm{~m} / \mathrm{s}^2$
So, $y=\sqrt{20 \times 10.0}=14.14 \mathrm{~m} / \mathrm{s}^2$
Thus, the speed of the car at the top of the hill is between $14 \mathrm{~m} / \mathrm{s}$ and $15 \mathrm{~m} / \mathrm{s}$.
Note: The roller coaster is a popular amusement ride developed for amusement parks and modern theme parks.

Topic: Physics, Work Power Energy, NEET, NEET 2008 (Screening)
A room (cubical) is made of mirrors. An insect is moving along the diagonal on the floor such that the component of velocity of direct image of insect on two adjacent wall mirrors along those mirrors is 10 cm s-1 . The velocity of image of the insect in the ceiling mirror is

Options:
A 10 cm s-1
B 20 cm s-1
C 102 cm s-1
D 102 cm s-1

The correct answer is:
102 cm s-1

Solution:
vcosâ¡45o=10
â´ v=102 cm s-1
In the ceiling mirror the original velocity will be seen.

Topic: Physics, Motion In Two Dimensions, NEET
A root was described as adventitious root because it

Options:
A Arose from plumule
B Was used variously for storage of food
C Was swollen
D Was growing in marshy place

The correct answer is:
Arose from plumule

Solution:
Roots developing from any part of the plant, expect radicle, are called adventituous roots.

Topic: Biology, Morphology of Flowering Plants, NEET
A rope is wound around a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N?

Options:
A 25 m s-2
B 0.25 rad s-2
C 25 rad s-2
D 5 m s-2

The correct answer is:
25 rad s-2

Solution:
The torque about centre is τ=I α, here I is the moment of inertia and α is angular acceleration.

The formula of moment of inertia is MR2.
F×R=MR2α
30×0.4=3×0.42α
12=3×0.16α
400=16α
Thus, angular acceleration is α=25 rad/s2

Topic: Physics, Rotational Motion, NEET, NEET 2017
A rope thrown over a pulley has a ladder with a man of mass m on one of its ends and a counterbalancing mass M on its other end. The man climbs with a velocity Vr relative to the ladder. Ignoring the masses of the pulley and the rope, as well as the friction on the pulley axis, the velocity of the centre of mass of this system, is :

Options:
A mMVr
B m2MVr
C MmVr
D 2MmVr

The correct answer is:
m2MVr

Solution:
The masses of load, ladder and man are M, M-m and m respectively.
Their velocities are V (upward), -V and Vr-V respectively
∴Vcm=ΣmiViΣmi
=Mv+M-m-v+mVr-v2M
=m2MVr

Topic: Physics, Center of Mass Momentum and Collision, NEET
A rotating wheel changes angular speed from 1800 rpm to 3000 rpm in 20 s. What is the angular acceleration assuming it to be uniform?

Options:
A 60π rad s-2
B 90π rad s-2
C 2π rad s-2
D 40π rad s-2

The correct answer is:
2π rad s-2

Solution:
We know that,
ω=2πn
∴ω1=2πn1
Where n1=1800 rpm
n2=3000 rpm
Δt=20 s
ω1=2π×180060
=2π×30=60π
Similarly, ω2=2πn2=2π×300060
=2π×50
=100π
If the angular velocity of a rotting wheel about on axis changes by change in angular velocity in a time interval Δt , then the angular acceleration of rotating wheel about that axis is
α=Change in angular velocityTime interval
α=ω2-ω1Δt
=100π-60π20
=40π20
=2π rad/s2

Topic: Physics, Laws of Motion, NEET
A rubber cord of density d, Young's modulus Y and length L is suspended vertically. If the cord extends by a length 0.5L under its own weight, then L is

Options:
A Y2dg
B Ydg
C 2Ydg
D dg2Y

The correct answer is:
Ydg

Solution:
âµ Young's modulus Y=StressStrain
Y=FAâLL
ââl=FLAY
extennsion in dx at âxâ
de=x d A gdxAY
total extension e= â«0Lx d Ag dxAY
=dgY L22
0.5 L=dgL2Y 2 âL=Ydg

Topic: Physics, Mechanical Properties of Solids, NEET
A running man has half the kinetic energy of that of a boy of half of his mass. The man speeds up by 1 m/s so as to have same K.E. as that of the boy. The original speed of the man will be

Options:
A 2Â m/s
B 2-1Â m/s
C 1(2-1) m/s
D 12Â m/s

The correct answer is:
1(2-1) m/s

Solution:
Let m= mass of boy, M= mass of man v= velocity of boy, V= velocity of man
1 2 M V 2 = 1 2 [ 1 2 m v 2 ] … (i)
1 2 M ( V+1 ) 2 =1[ 1 2 m v 2 ] …(ii)
Putting m= M 2 and solving V= 1 2 −1

Topic: Physics, Work Power Energy, NEET
A running man has the same kinetic energy as that of a boy of half his mass. The man speed up by 2 m s-1 and the boy changes his speed by x m s-1, so that the kinetic energies of the boy and the man are again equal. Then x in m s-1 is

Options:
A 42
B 22
C 2
D 2

The correct answer is:
22

Solution:
Let vM be velocity of man,vBof boy, then kinetic energy according to question ,
ie K=12MvM2=12.M2.vB2
Or vM2=VB22
Or 2vM=vB
When the man speeds up 2 m s-1and the boy changes his speed by x m s-1.Then ,
12MvM+22=12⋅M2⋅vB+x2
Or vM+22=vB+x22
2vM+22=2vM+x2 ∵vB=2vm
Or 2vM+2=2vM+x
Or x= 22

Topic: Physics, Work Power Energy, NEET
AsF5 reacts with XeF4 to form an adduct. The shapes of cation and anion in the adduct are respectively.

Options:
A square planar, octahedral
B T-shaped, octahedral
C square pyramidal, octahedral
D square planar, trigonal bipyramidal

The correct answer is:
T-shaped, octahedral

Solution:
AsF5 reacts with XeF4 to form an adduct. The shapes of cation and anion in the adduct are T-shaped, octahedral respectively.
AsF5+XeF4→[XeF3]+T-shaped+[AsF6]-octahedral

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET