datavorous/entrance-exam-dataset huggingface dataset:
Two blocks A and B of mass m and 2m respectively are connected together by a light spring of stiffness K and natural length l. The system is lying on a smooth horizontal surface with the block A in contact with a fixed vertical wall as shown in the figure. The block B is pressed towards the wall by a distance x0 from the natural position of spring and then released. There is no friction anywhere.

If block B is released at time t=0 and A just loses contact with the wall at time t=Ît then the average acceleration of the centre of mass of the system during the time t=0 to t=Ît is:

Options:
A 29Kmx0Δt
B K2mx0Δt
C 2 Kmx0Δt
D  Kmx0Δt

The correct answer is:
29Kmx0Δt

Solution:
ucm=0  ;  vcm=2v03;<acm>=vcm-ucmΔT=x0Δt2K9m

Topic: Physics, Center of Mass Momentum and Collision, NEET
Two blocks A and B of masses 3m and m respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in figure. The magnitudes of acceleration of A and B immediately after the string is cut, are respectively

Options:
A g,g3
B g3, g
C g,Â g
D g3,g3

The correct answer is:
g3, g

Solution:
Before the string is cut, tension is acting in the string in downward position.
kx=T+3mg ......(i)
T=mg ......(ii)
⇒ kx=4mg
After the string is cut, T=0
a=kx-3mg3m
a=4mg-3mg3m
Acceleration of Ais a=g3↑ and acceleration of B is a=mgm=g↓.

Topic: Physics, Laws of Motion, NEET, NEET 2017
Two blocks A and B of masses m & 2m, respectively, are held at rest such that the spring is in natural length. Find out the accelerations of blocks A and B, respectively, just after release (pulley, string and spring are massless).

Options:
A g â , g â
B g 3 ↓ , g 3 ↑
C 0, 0
D g ↓ , 0

The correct answer is:
g â , g â

Solution:
In this case, spring force is zero initially F. B. D of A and B

Topic: Physics, Laws of Motion, NEET
Two blocks A and B of masses m and 2m, respectively, are connected by a massless spring of force constant k and are placed on a smooth horizontal plane. The spring is stretched by an amount x and then released. The relative velocity of the blocks when the spring comes to its natural length is

Options:
A 3k2mx
B 2k3mx
C 2kxm
D 3km2x

The correct answer is:
3k2mx

Solution:
Relative velocity will be v1+v2
Using conservation of linear momentum,
mv1=2mv2
â v1=2v2
Using conservation of energy
12kx2=12mv12+122mv22
12kx2=12m2v22+mv22
12kx2=2mv22+mv22
3mv22=kx22
v22=kx26m
v2= k6m.x
Relative velocity =3v2
=3k6m.x
= 3k2m.x

Topic: Physics, Center of Mass Momentum and Collision, NEET
Two blocks are in contact on a frictionless table. One has mass m and other 2m. A force F is applied on 2m as shown in figure. Next the same force F is applied from the right on m. In the two cases respectively, the ratio of the force of contact between the two blocks will be

Options:
A 2 : 1
B 1 : 3
C 1 : 2
D 3 : 1

The correct answer is:
1 : 2

Solution:
When force F is applied on 2m from left, contact force,
F1=mm+2mF=F3
When force F is applied on m from right, contact force
F2=2mm+2mF=2F3
∴ F1:F2=1 :2

Topic: Physics, Laws of Motion, NEET
Two blocks, each of mass m, are connected with springs, each of force constant K, as shown in the figure. The mass A is displaced to the left and mass B to the right by the same amount and released. Assuming the collision to be elastic, the time period of oscillation is

Options:
A 2πmK
B 2πm2K
C πmK
D πm2K

The correct answer is:
πmK

Solution:
When after coming back the blocks collide elastically, they interchange their velocities as the masses are equal. So, each mass completes half of the oscillation. So
T=122πmK=πmK

Topic: Physics, Oscillations, NEET
Two blocks each of mass m are joined together using an ideal spring of force constant K and natural length l0. The blocks are touching each other when the system is released from rest on a rough horizontal surface. Both the blocks come to rest simultaneously when the extension in the spring is l04. The coefficient of friction between each block and the surface (assuming it to be same between any of the blocks and the surface) is:

Options:
A Kl040mg
B Kl08mg
C 3Kl08mg
D 17Kl020mg

The correct answer is:
3Kl08mg

Solution:
Work energy theorem,
-μmgl0+l04=12Kl042-12Kl02
μ=3Kl08mg

Topic: Physics, Laws of Motion, NEET
Two blocks of mass 1 kg & 2 kg are hanged from a light pulley and resting on a horizontal surface. A time varying force F = 4t N is acting on pulley in the direction shown. Time after which block will break off the surface will be –

Options:
A 5
B 10
C 15
D 20

The correct answer is:
5

Solution:
1 kg block
N+F/2=10;
⇒F=2(10–N)
Block 1 will leave
surface when N = 0
⇒F=20⇒5sec⁡

Topic: Physics, Motion In Two Dimensions, NEET
Two blocks of mass m and M are connected by a massless thread which passes over a fixed massless pulley. The table under M is smooth. The force exerted by the pulley on the table is

Options:
A 2MmgM+m
B MmgM+m
C 2MmgM+m
D 3MmgM+m

The correct answer is:
2MmgM+m

Solution:
m+Ma=mg
a=mgM+m
T=MmgM+m
N=T2=Mmg2M+m

Topic: Physics, Laws of Motion, NEET
Two blocks of masses 2 kg and 1 kg respectively are tied to the ends of a string which passes over a light frictionless pulley. The masses are held at rest at the same horizontal level and then released. The distance traversed by centre of mass in 2 s is g=10 m s-2

Options:
A 1.42Â m
B 2.22 m
C 3.12Â m
D 3.33Â m

The correct answer is:
2.22 m

Solution:
Acceleration of each block is,
Here acceleration of center of mass is,
a=m2-m1m2-m1g
a=2-12+1gâ¡=103 m s-2
acm=m2a+m1-am2+m1
acm=2-12+1aâ¡=109 m s-2
Displacement of center of mass is,
SCM=12aCMt2=1210922
â SCM=2.22 m

Topic: Physics, Center of Mass Momentum and Collision, NEET
Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m s-1 to the heavier block in the direction of the lighter block. The velocity of the centre of mass is

Options:
A 30 m s-1
B 20 m s-1
C 10 m s-1
D 5 m s-1

The correct answer is:
10 m s-1

Solution:
vCM=m1v1+m2v2m1+m2=10Ã14+4Ã010+4=14014=10 m s-1

Topic: Physics, Center of Mass Momentum and Collision, NEET
Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m s-1 to the heavier block in the direction of the lighter block. The velocity of the centre of mass is

Options:
A 30 m s-1
B 20 m s-1
C 10 m s-1
D 5 m s-1

The correct answer is:
10 m s-1

Solution:
Hence, m1=10 kg, m2=4 kg
v1=14 m s-1, v2=0
vCM=m1v1+m2v2m1+m2
vCM=10Ã14+4Ã010+4=10 m s-1

Topic: Physics, Center of Mass Momentum and Collision, NEET
Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m s-1 to the heavier block in the direction of the lighter block. The velocity of the centre of mass is

Options:
A 30 m s-1
B 20 m s-1
C 10 m s-1
D 5 m s-1

The correct answer is:
10 m s-1

Solution:
v CM = m 1 v 1 + m 2 v 2 m 1 + m 2
10×14+4×010+4=10×1414=10 m s-1

Topic: Physics, Center of Mass Momentum and Collision, NEET
Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulsive force gives a velocity of 14 m s-1 to the heavier block in the direction of the lighter block. The velocity of the centre of mass of the system at that very moment is

Options:
A 30 m s-1
B 20 m s-1
C 10 m s-1
D 5 m s-1

The correct answer is:
10 m s-1

Solution:
At the time of applying the impulsive force block of 10 kg pushes the spring forward but 4 kg mass is at rest.
Hence,
vCM=m1v1+m2v2m1+m2=10Ã14+4Ã010+4=14014=10 m s-1

Topic: Physics, Center of Mass Momentum and Collision, NEET
Two blocks of masses 10âkg and 30âkg are placed along a vertical line. If 10 kg block is raised through a height of 7âcm, then the distance through which other mass should be moved to raise the center of mass of the system by 1âcm is

Options:
A 1 cmup
B 1 cmdown
C 2 cm down
D 2 cm up

The correct answer is:
1 cmdown

Solution:
x com = m 1 x 1 + m 2 x 2 m 1 + m 2
1=10×7+30x40
⇒x=−1 cm

Topic: Physics, Center of Mass Momentum and Collision, NEET
Two blocks of masses m1=4 kg and m2=2 kg are connected to the ends of a string which passes over a massless, frictionless pulley. The total downwards thrust on the pulley is nearly

Options:
A 27 N
B 54 N
C 0.8 N
D Zero

The correct answer is:
54 N

Solution:
T=2m1m2m1+m2g=2Ã4Ã2Ã104+2
=1606=26.6=27 N
Total downward thrust on the pulley
=2T=2Ã27=54 N

Topic: Physics, Laws of Motion, NEET
Two blocks of masses M=3 kg and m=2 kg, are in contact on a horizontal table. A constant horizontal force F=5 N is applied to block M as shown. There is a constant frictional force of 2 N between the table and the block m but no frictional force between the table and the first block M, then the acceleration of the two blocks is

Options:
A 0.4 m s-2â¡
B 0.6 m s-2â¡
C 0.8 m s-2â¡
D 1 m s-2â¡

The correct answer is:
0.6 m s-2â¡

Solution:
Net force = total mass Ã accelaration
a=F-fM+m
a=5-23+2
a=0.6 m s-2â¡

Topic: Physics, Laws of Motion, NEET
Two blocks of masses m and 2m are placed one over the other as shown in the figure. The coefficient of friction between m and 2m is Î¼ and between 2m and ground is Î¼3. If a horizontal force F is applied on the upper block and T is the tension developed in the string, then choose the incorrect alternative.

Options:
A If F=μ3mg,T=0
B If F=μmg,T=0
C If F=2μmg,T=μmg3
D If F=3μmg,T=0

The correct answer is:
If F=2μmg,T=μmg3

Solution:
The maximum frictional force on 1st block due to 2nd block is acting towards left as the accelerating force is acting towards the right. Its value is f1max=Î¼mg
Now for 2nd body, f1 acting towards right is responsible to move block towards the right, but f2 will act towards the left to hold the body in rest. Its value is
f2max=Î¼33 mg=Î¼mg
T+f2=f1
T+Î¼mgâ¡=Î¼mg
T=0
â Tension will be zero in all circumstances. Hence A, B and D all the three are correct.
Note- In the second block, we can very clearly see that T=0. It is independent of the force applied as 2nd block is free from any external force.

Topic: Physics, Laws of Motion, NEET
Two bodies A and B having masses in the ratio of 3:1 possess the same kinetic energies. The ratio of linear momenta of B to A is

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

The correct answer is:
1:3

Solution:
As 12mAvA2=12mBvB2
vAvB=mBmA;
PBPA=mBvBmAvA
=mBmAmAmB=mBmA=13

Topic: Physics, Work Power Energy, NEET
Two bodies A and B of definite shape are placed near one another. Electrostatic attraction is found between the bodies, then

Options:
A Both bodies must be positively charged
B Both bodies must be negatively charged
C Both bodies must be oppositely charged
D Body A may be neutral

The correct answer is:
Body A may be neutral

Solution:
Attraction may take place between like charged bodies, unlike charged bodies and charged body and neutral body.

Topic: Physics, Electrostatics, NEET
Two bodies A and B of equal mass are suspended from two separate massless springs of constant k1 and k2 respectively. If the bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of A to that of B is

Options:
A k1k2
B k2k1
C k1k2
D k2k1

The correct answer is:
k2k1

Solution:
vmax=ωA

ωAAA=ωBAB

⇒AAAB=ωBωA
ωBωA=k2m⁡÷k1m⁡=k2k1.
⇒ k2k1 is correct.

Topic: Physics, Oscillations, NEET
Two bodies $A$ and $B$ of same mass undergo completely inelastic one dimensional collision. The body $A$ moves with velocity $v_1$ while body $B$ is at rest before collision. The velocity of the system after collision is $v_2$. The ratio $v_1: v_2$ is

Options:
A 2:1
B 4:1
C 1:4
D 1:2

The correct answer is:
2:1

Solution:
It undergoes completely inelastic collision
Using conservation of linear momentum
Initial momentum = Final momentum
$\begin{aligned}
& \Rightarrow m v_1=m v_2+m v_2 \\
& \Rightarrow m v_1=2 m v_2 \\
& \Rightarrow \frac{v_1}{v_2}=\frac{2}{1}
\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2024
Two bodies $\mathrm{A}$ (of mass $1 \mathrm{~kg}$ ) and $\mathrm{B}$ (of mass $3 \mathrm{~kg}$ ) are dropped from heights of $16 \mathrm{~m}$ and $25 \mathrm{~m}$, respectively. The ratio of the time taken by them to reach the ground is:

Options:
A $4 / 5$
B $5 / 4$
C $12 / 5$
D $5 / 12$

The correct answer is:
$4 / 5$

Solution:
We know that time taken by a body from height $h$ to reach the ground is.
$$
\begin{aligned}
t & =\sqrt{\frac{2 h}{g}} \\
\therefore \quad \frac{t_{\mathrm{A}}}{t_{\mathrm{B}}} & =\frac{\frac{2 h_{\mathrm{A}}}{g}}{\frac{2 h_{\mathrm{B}}}{g}}=\sqrt{\frac{h_{\mathrm{A}}}{h_{\mathrm{B}}}} \\
& =\sqrt{\frac{16}{25}}=\frac{4}{5} .
\end{aligned}
$$

Topic: Physics, Motion In One Dimension, NEET, NEET 2006
Two bodies are in equilibrium when suspended in water from the arms of a balance. The mass of one body is 36 g and its density is 9 g cm-3â. If the mass of the other is 48 g, its density in g cm-3 is

Options:
A 4 3
B 3 2
C 3
D 5

The correct answer is:
3

Solution:
Apparent weight = V ρ - σ g = m ρ ρ - σ g
where m = mass of the body, ρ = density of the body and σ = density of water If two bodies are in equilibrium then their apparent weight must be equal.
∴ m 1 ρ 1 ρ 1 - σ g = m 2 ρ 2 ρ 2 - σ g
⇒ 3 6 9 9 - 1 = 4 8 ρ 2 ρ 2 - 1 g .
By solving we get ρ 2 = 3 .

Topic: Physics, Mechanical Properties of Fluids, NEET
Two bodies are in equilibrium when suspended in water from the arms of a balance. The mass of one body is 36 g and its density is 9 g / cm3â. If the mass of the other is 48 g, its density in g / cm3 is

Options:
A 4 3
B 3 2
C 3
D 5

The correct answer is:
3

Solution:
Apparent weight = V ρ - σ g = m ρ ρ - σ g
where m = mass of the body, ρ = density of the body and σ = density of water If two bodies are in equilibrium then their apparent weight must be equal.
∴ m 1 ρ 1 ρ 1 - σ g = m 2 ρ 2 ρ 2 - σ g
⇒ 3 6 9 9 - 1 = 4 8 ρ 2 ρ 2 - 1 g .
By solving we get ρ 2 = 3 .

Topic: Physics, Mechanical Properties of Fluids, NEET
Two bodies are projected at angles θ and 90o-θ to the horizontal with the same speed. The ratio of their times of flight is

Options:
A sin⁡θ:1
B cos⁡θ:1
C sin⁡θ:cos⁡θ
D cos⁡ θ:sin⁡θ

The correct answer is:
sin⁡θ:cos⁡θ

Solution:
Time of flight, T=2u sin⁡θg
For θ, T1=2u sin⁡θg ......(i)
For 90o-θ, T2=2u sin⁡90o-θg=2u cos⁡θg .....(ii)
Dividing Equation. (i) by Equation. (ii), we get
∴ T1T2=2u sin⁡θg×g2u cos⁡θ=sin⁡θcos⁡θ

Topic: Physics, Motion In Two Dimensions, NEET
Two bodies begin a free fall from rest from the same height 2 seconds apart. How long after the first body begins to fall, the two bodies will be 40 m apart? (Take g=10ms-2)

Options:
A 1 s
B 2 s
C 3 s
D 4 s

The correct answer is:
3 s

Solution:
Let after t time two bodies are 40m apart.
Then according to the problem
12gt2-12gt-22=40
or 12Ã102t-22=40
or 2t-2=4 or t=3 s

Topic: Physics, Motion In One Dimension, NEET
Two bodies collide as shown in the diagram. During the collision, they exert impulse of magnitude J on each other.

For what values of J (in N s) the 2 kg block will change its direction of velocity?

Options:
A J < 12

B J > 12

C J < 10

D J > 10

The correct answer is:
J > 12

Solution:
2 kg block will change its direction if v2 is negative.
or 6-J2<0

∴ J2>6
or J>12 N s

Topic: Physics, Center of Mass Momentum and Collision, NEET
Two bodies have their moments of inertia $l$ and $2 l$ respectively about their axis of rotation. If their kinetic energies of rotation are equal, their angular momenta will be in the ratio:

Options:
A $2: 1$
B $1: 2$
C $\sqrt{2}: 1$
D $1: \sqrt{2}$

The correct answer is:
$1: \sqrt{2}$

Solution:
K.E. rotation $K=\frac{1}{2} I \omega^2$
Angular momenta $L=100$
$\therefore \quad K=\frac{L^2}{2}$
Here $K_1+K_2, I_1=I_1, I_2=21$
$$
\therefore \quad \frac{L_1}{L_2}=\sqrt{\frac{2 I_1 K_1}{2 I_2 K_2}}
$$
$$
\begin{aligned}
& =\sqrt{\frac{I_1}{I_2}}=\sqrt{\frac{I}{2 I}} \\
& =\frac{1}{\sqrt{2}}
\end{aligned}
$$

Topic: Physics, Rotational Motion, NEET, NEET 2005
Two bodies of 6 kg and 4 kg masses have their velocity 5i^-2j^+10k^ and 10i^-2j^+5k^ respectively. Then the velocity of their centre of mass is

Options:
A 5i^+2j^-8k^
B 7i^+2j^-8k^
C 7i^-2j^+8k^
D 5i^-2j^+8k^

The correct answer is:
7i^-2j^+8k^

Solution:
Given, m1=6 kg, m2=4 kg
v1=5i^-2j^+10k^ , v2=10i^-2j^+5k^
The velocity of centre of mass is
v=m1v1+m2v2m1+m2
=65i^-2j^+10k^+410i^-2j^+5k^6+4
=70i^-20j^+80k^10=7i^-2j^+8k^

Topic: Physics, Center of Mass Momentum and Collision, NEET
Two bodies of 6 kg and 4 kg masses have their velocity (in m/s ) 5i^-2j^+10k^ and 10i^-2j^+5k^ respectively. Then the velocity of their centre of mass (in m/s ) is

Options:
A 5i^+2j^-8k^
B 7i^+2j^-8k^
C 7i^-2j^+8k^
D 5i^-2j^+8k^

The correct answer is:
7i^-2j^+8k^

Solution:
We know that
vcm=ΣmiviΣmi
v com = 6( 5 i ^ −2 j ^ +10 k ^ )+4( 10 i ^ −2 j ^ +5 k ^ ) 6+4
=7i^−2j^+8k^

Topic: Physics, Center of Mass Momentum and Collision, NEET
Two bodies of mass $1 \mathrm{~kg}$ and $3 \mathrm{~kg}$ have position vectors $\hat{i}+2 \hat{j}+\hat{k}$ and $-3 \hat{i}-2 \hat{j}+\hat{k}$, respectively. The centre of mass of this system has a position vector

Options:
A $-2 \hat{i}+2 \hat{k}$
B $-2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$
C $2 \hat{i}-\hat{j}-2 \hat{k}$
D $-\hat{i}+\hat{j}+\hat{k}$

The correct answer is:
$-2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$

Solution:
The position vector of centre of mass
$$
\begin{aligned}
\vec{r} & =\frac{m_1 \overrightarrow{r_1}+m_2 \vec{r}_2}{m_1+m_2} \\
& =\frac{1(\hat{i}+2 \hat{j}+\hat{k})+3(-3 \hat{i}-2 \hat{j}+\hat{k})}{1+3} \\
& =\frac{1}{4}(-8 \hat{i}-4 \hat{j}+4 \hat{k}) \\
& =-2 \hat{i}-\hat{j}+\hat{k}
\end{aligned}
$$
The centre of mass changes its position only under the translatory motion. There is no effect of rotatory motion on centre of mass of the body.

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2009 (Screening)
Two bodies of mass 4 kg and 6 kg are tied to the ends of a massless string. The string passes over a pulley which is frictionless (see figure). The acceleration of the system in terms of acceleration due to gravity g is:

Options:
A g2
B g5
C g10
D g

The correct answer is:
g5

Solution:
a=m2−m1gm1+m2

a=(6−4)g6+4=2g10

a=g5

Topic: Physics, Laws of Motion, NEET, NEET 2020 (Phase 1)
Two bodies of mass m and 9m are placed at a distance R. The gravitational potential on the line joining the bodies where the gravitational field equals zero, will be (G=gravitational constant):

Options:
A -20GmR
B -8GmR
C -12GmR
D -16GmR

The correct answer is:
-16GmR

Solution:
Let the gravitational field at point Q is zero. Therefore, we can write

-Gmx2+-G×9mR-x2=0⇒R-x2x2=9⇒x=R4

Now, net gravitational potential at point Q will be

Vnet=-Gmx+-G×9mR-x=-GmR4+-G×9m3R4=-16GmR

Topic: Physics, Gravitation, NEET, NEET 2023 (All India)
Two bodies of masses 2 kg and 8 kg are separated by a distance of 9 m. The point where the resultant gravitational field intensity is zero is at a distance of

Options:
A 4.5 m from each mass

B 6 m from 2 kg

C 6 m from 8 kg

D 2.5 m from 2 kg

The correct answer is:
6 m from 8 kg

Solution:
If x is the distance of point on the line joining the two masses from mass m2â² where gravitational field intensity is zero, then
Gmr-x2=Gm2x2 or 29-x2=8x2
or 19-x=2x
On solving, x=6 m

Topic: Physics, Gravitation, NEET
Two bodies of masses 10 kg and 100 kg are separated by a distance of 2 m. The gravitational potential at the mid-point of the line joining the two bodies is:

Options:
A -7.3Ã10-7 J/kg
B -7.3Ã10-8 J/kg
C -7.3Ã10-9 J/kg
D -7.3Ã10-6 J/kg

The correct answer is:
-7.3Ã10-9 J/kg

Solution:
As we know, gravitational potential
V=-GMR
Gravitational potential at mid point,
=-GÃ101-GÃ1001=-110 G
=- 110Ã6.67Ã10-11 J kg-1
=-7.3Ã10-9 Jâ kg â1

Topic: Physics, Gravitation, NEET
Two bodies of masses m1 and m2 are infinitely at rest at an infinite distance apart. They are then allowed to move towards each other under mutual gravitational attraction. The relative velocity of approach at a separation r between them is

Options:
A 2Gm1+m2r12
B 2Grm1m212
C 2Gm1-m2r12
D r2Gm1 m212

The correct answer is:
2Gm1+m2r12

Solution:
âµ Change in P.E = Change in K.E
12Î¼v2=Gm1m2r
Where, Î¼ is reduces mass
Î¼=m1m2m1+m2
â´12m1m2m1+m2 v2=Gm1m2r
v=2Grm1+m2

Topic: Physics, Work Power Energy, NEET
Two bodies of masses m1 and m2 are initially at rest placed infinite distance apart. They are then allowed to move towards each other under mutual gravitational attraction. Their relative velocity when they are r distance apart is

Options:
A 2G(m1+m2)r
B 2Gm1m2(m1+m2)r
C G(m1+m2)r
D Gm1m2m1+m2r

The correct answer is:
2G(m1+m2)r

Solution:
By conservation of momentum
m1v1-m2v2=0
by conservation of energy
12m1v12+12m2v22-Gm1m2r=0
12m1v12+12m2m1v1m22=Gm1m2r
v122m1+m12m2=Gm1m2r
v12=2Gm22m1+m2r
v1=m22Gm1+m2r
∴ v2=m12Gm1+m2r
Also, vrel=v1+v2
vrel=m2+m12Gm1+m2r
vrel=2Gm1+m2r

Topic: Physics, Gravitation, NEET
Two bodies of masses m and 4m are placed at a distance r. The gravitational potential at a point on the line joining them, where the gravitational field is zero, is

Options:
A -6Gmr
B -9Gmr
C Zero
D -4Gmr

The correct answer is:
-9Gmr

Solution:
Suppose, at point x, from mass m the gravitational field is zero
Gm x 2 = G 4 m r - x 2
⇒ r - x = 2 x
⇒ x = r 3
V=-Gm3r-Gm6r=-9Gmr

Topic: Physics, Gravitation, NEET
Two boys are standing at the ends $\mathrm{A}$ and $\mathrm{B}$ of a ground where $\mathrm{AB}=a$. The boy at $\mathrm{B}$ starts running in a direction perpendicular to $\mathrm{AB}$ with velocity $v_1$. The boy at A starts running simultaneously with velocity $v$ and catches the other in time $t$, where $t$ is:

Options:
A $\frac{a}{\sqrt{v^2+v_1^2}}$
B $\frac{a}{v+v_1}$
C $\frac{a}{v-v_1}$
D $\sqrt{\frac{a^2}{v^2-v_1^2}}$

The correct answer is:
$\sqrt{\frac{a^2}{v^2-v_1^2}}$

Solution:
If boy \(A\) catches boy \(B\) in time \(t\),
\(\begin{aligned}
& \Rightarrow(\mathrm{vt})^2=\left(\mathrm{v}_1 \mathrm{t}\right)^2+\mathrm{a}^2 \\
& \Rightarrow \mathrm{t}^2=\frac{\mathrm{a}^2}{\mathrm{v}^2-\mathrm{v}_1^2} \\
& \Rightarrow \mathrm{t}=\frac{\mathrm{a}}{\sqrt{\mathrm{v}^2-\mathrm{v}_1^2}}
\end{aligned}\)

Topic: Physics, Motion In Two Dimensions, NEET, NEET 2005
Two boys of masses 10 kg and 8 kg are moving along a vertical rope, the former climbing up with the acceleration of 2 m s-2 while the later coming down with a uniform velocity of 2 m s-1. Then tension in the rope at the fixed support is (Take g = 10 m s-2 )

Options:
A 200 N

B 120 N

C 180 N

D 160 N

The correct answer is:
200 N

Solution:
Since, m2 moves with constant velocity.
∴ f2=m2g
f2=8 ×10=80 N
Since, block m1 moves with acceleration
a=2 m s-2 in the upward direction.
∴ f1-m1g=m1a
∴ f1=m1g+m1a
=10 ×10+10×2
=120 N
For equilibrium of rope,
T=f1+f2
=120+80 N
=200 N

Topic: Physics, Laws of Motion, NEET
Two bulbs of equal power are connected in parallel and they totally consume 110 W at 220 V. The resistances of each bulb is

Options:
A 550Â Î©
B 440Â Î©
C 330Â Î©
D 880 Î©

The correct answer is:
880 Î©

Solution:
Bulbs are connected in parallel and they totally consume 110 W at 220 V
Net power, Pnet=P1+P2
P+P=110 W âµP1=P2=P
2P=110 WâP=55 W
We know that,
P=V2R
55=2202R
R=220255=220Ã22055=880 Î©

Topic: Physics, Current Electricity, NEET
Two bulbs when connected in parallel to a source take 100 W each. The total power consumed when they are connected in series with the same source is

Options:
A 25 W

B 50 W

C 100 W

D 200 W

The correct answer is:
50 W

Solution:
Let R be the resistance of each bulb.
We have P=V2R
or R=V2P=V2100
When the bulbs are connected in series, the voltage across each of them is . Hence, the total power consumed is
P′= V22R+ V22R=V22R
∴P′=P2=1002=50 W

Topic: Physics, Current Electricity, NEET