MCQGeeks
0 : 0 : 1
CBSE
JEE
NTSE
NEET
English
UK Quiz
Quiz
Driving Test
Practice
Diagrams
Games
NEET
Physics NEET MCQ
Quiz 8
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Q.1
Two masses M and m are tied with a string and arranged as sjown. The velocity of block M when it loses the contact is
0%
a) 2√gh
0%
b) m√gh / (M+m)
0%
c)2m√gh /(m+m)
0%
d)2M√gh / (m+m)
Explanation
Velocity of m just when string becomes straight and tautu=√[2g(2h)] Now string exerts impules on M and m and are equal After that both the masses move with same speed as v change in momentum for m , m(u-v)=F(Δt) --(i) change in momentum of M , MV=F(Δt) --(ii) From (i) and (ii) m(u-v)=Mv v(M+m)=mu by substituting value of u we get Answer:(c)
Q.2
A uniform chain of length l and mass m is hanging vertically from its ends A and B which are close together. A t a given instant the end B is released. What is the tension at A when B has fallen a distance x [ x < l]?
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
When the end B has fallen by x, diagram is as shown. At this, point P has fallen through x/2 velocity of p vp=√[2g(x/2)]=√gx In one second, √gx length of chain will fall and loose all its velocity ∴ change in momentum per second=(µ√gx)(√gx)=µgx Where µ=mass per unit length=m/l Thus tension at A T=force due to weight of hanging part + force due to rate of change of momentum Answer: (a)
Q.3
A gun is mounted on a gun carriage movable on a smooth horizontal plane and the gun is elevated at an angle 45° to the horizon. A shell is fired and leave the gun inclines at an angle θ to horizon. IF the mass of gun and carriages is n times that of the shell, find the value of θ
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Let mass of shell=m Then mass of gun=mn velocity of shot with respect to gun=wrecoil velocity of gun=v Let inclination of gun be α=45° x-Componants of velocity of shell=wcosα y-component of velocity of shell=wsinα If shell leave the the gun at angle θ with respect to ground Since gun recoil along x-axis vertical complainant of shell will be same=wsinα But horizontal component with respect to ground will be=Wcosθ -v By conservation of momentum in horizontal direction mnv=m( wcosα-v) v=wcosα / ( n+1) substituting value of v in above equation we get Answer: (b)
Q.4
In head on elastic collision of two bodies of equal masses
0%
a) the velocities are interchanged
0%
b) the momentum are interchanged
0%
c)the faster body slows down and the slower body speeds up
0%
d)all are correct
Explanation
Answer: (d)
Q.5
A mass of 0.5 kg is just able to slide down the slope of an inclined rough surface when the angle of inclination is 60°. The necessary to pull the mass up the incline along the line of greatest slope is ( g=10 m/s2)
0%
a) 20.25 N
0%
b) 8.65 N
0%
c)100 N
0%
d)1 N
Explanation
Acceleration down the plane=g( sinθ-µcosθ)=0 so tanθ=µ √3=µ Minimum force necessary=mg(sinθ+µcosθ) Answer:(b)
Q.6
A 40 kg slab (B) rests on a smooth floor as shown in figure. A 10kg block (A) rests on the top of the slab. The coefficient of static friction between slab and block is 0.6 while the coefficient of kinetic friction is 0.The block (A) is acted upon by a horizontal force 100N. IF g=9.8 m/s2, the resulting acceleration of the slab (B) will be
0%
a) 0.98 m/s²
0%
b) 1.47 m/s²
0%
c) 1.52 m/s²
0%
d) 6.1 m/s²
Explanation
For a force of 100N on 10kg block, relative motion will take place ∴ the frictional between 10kg block and 40 kg block f=µ mg f=0.4×10×9.8 N The acceleration of slab of 40kg is a=(0.4×10×9.8) / 40=0.98 m/s2 Answer: (a)
Q.7
A uniform rope of length l lies on a table. If the coefficient of friction is µ, then the maximum length l1 of the hanging part of the rope which can overhang from the edge of the table without sliding down is
0%
a) l/µ
0%
b) l/(µ+1)
0%
c) µl/(µ+1)
0%
d) µl/ (l-1)
Explanation
To prevent sliding of rope friction due to rope on table = weight of rope hanging let m be the mass per unit length of rope. l1 be the length hanging m(l-l1)gµ = l1mg lµ - l1µ = l1 l1 ( 1+ µ) = lµ l1 = lµ / ( 1+ µ) Answer: (c)
Q.8
A body f mass 40 gm moving with constant velocity of 2cm/s on a horizontal frictionless table. The force on the table is
0%
a) 39200 dynes
0%
b) 160 dynes
0%
c) 80 dynes
0%
d) zero dynes
Explanation
N = mg Answer: (a)
Q.9
Water flows through a pipe bent at angle α to horizontal with a velocity v. What is the force exerted by water on the bend of the pipe of area of cross section S ?
0%
b) 2ρv2 Scosα
0%
c)2ρv2 Ssinα
0%
d)2ρv2 Scosα sinα
0%
a)2ρv² Ssin(α/2)
Explanation
mass of water per second=SvρLet water first move along the direction of positive x-axis Initial momentum P1=(Svρ)v iP1=(Sv2ρ) iFinal momentum of water since water is bend at angle α velocity will have two component mutually perpendicular one along positive x-axis and other along positive y axis v= vcosα i + v sinαj Thus final momentum P3=(Svρ) (cosα i + sinαj) Rate of change of momentum=force exerted by water on the bend of the pipe F=Sv2ρ (cosα i + sinαj) - Sv2ρi F=Sv2ρ [ (cosα - 1)i +sinαj] Answer: (a)
Q.10
A hammer head moving at speed of 10 m/s strikes a nail and drives it into a wooden block. The duration of impact is 0.005 seconds. IF the hammer head weighs 1.5 kg by how much distance the nail has penetrated into wood?
0%
a) 5 cm
0%
b) 2.5 cm
0%
c)1.0 cm
0%
d)1.5 cm
Explanation
When the hammer head (M) moving with 10m/s strikes the nail(m), the two move together with velocity v'. Mv=(M+m)v' as m<< M v'=v=10m/s Now duration of penetration=0.005 sec Deceleration of nail=v-u /Δt=10/0.0005=2000 distance penetrated by nail s=ut + ½ at2 s=10×0.005 - ½ ( 2000) (0.005)2=2.5 cmAnswer: (b)
Q.11
A spherical ball falls from a height h on a floor of coeffcient of restitution e. Then the height to which the ball rises after first inpact is
0%
a) e2h
0%
b) e4h
0%
c)e6h
0%
d)e8h
Explanation
velocit of sphere just before collision u=√(2gh) velocity of ball after collision v=e√(2gh)Height reached by the ball can be calculated as v2=u2 + 2as02=e2(2gh) + 2(-g)h1h1=e2h Answer:(a)
Q.12
A block of mass m moving with speed v collides with another block of mass 2m at rest. The lighter block comes to rest after collision. What is the value of coefficient of restitution?
0%
a) 1/2
0%
b) 1/3
0%
c) 3/4
0%
d) 1/4
Explanation
Suppose the second block moves at a speed of v' after collision By conservation of momentum mv=2mv' v'=v/2 velocity of separation=v/2 velocity of approach=v By definition e=velocity of separation / velocity approach=1/2 Answer: (a)
Q.13
How large must F be in the figure shown to give the 700 g block an acceleration of 30cm/s2? The coefficient of friction between all surface is 0.15
0%
a)4 N
0%
b) 2.18 N
0%
c)3.18 N
0%
d) 6 N
Explanation
From FBD of 200gm block Acceleration of block 0.3 ms2 ma=T - f f=µmg=0.15×0.2×9.8Thus 0.2×0.3=T - (0.15×0.2×9.8) T=0.354From FBD of 700gm blockF- T - f - f1f1=0.15×0.9×9.8F - 0.35 -(0.15×0.2×9.8) - (0.15×0.9×9.8) F=2.18NAnswer: (b)
Q.14
A sphere of mass m moving with a constant velocity u hits another stationary sphere of same mass. If e is the coefficient of restitution, the ratio of velocities of the two sphere after collision is
0%
a) (1-e)/(1+e)
0%
b) (1+e)/(1-e)
0%
c)(e+1) / (e-1)
0%
d)(e-1) / (e+1)
Explanation
By law of conservation of momentum mu=mv1 + mv2 u=v1 + v2 coefficient of restitution e Answer: (a)
Q.15
A ball approaches a moving wall of infinite mass with speed v along normal to the wall. The speed of the wall is u away from the ball and u < v. the speed of ball after an elastic collision is
0%
a) u + v away from the wall
0%
b) 2u+v away from the wall
0%
c)v-u towards the wall
0%
d)v - 2u away from the wall
Explanation
Let m1 be the mass of ball and velocity is u and m2 is mass of wall and velocity id v also m1 < < m2 By using following formula away from the wall Answer:(d)
Q.16
A neutron of mass m collides elastically with a nucleus of mass M which is at rest. If the initial kinetic energy of neutron is Ko, calculate the maximum kinetic energy that it can lose during the collision .
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
the maximum energy loss occurs in a head on collision. Let v be the vel;ocity of neutron before collision and V2 its velocity after collision and V1 the velocity of nucleus after collision. M=mass of nucleus m=mass of neutrons By principle of conservation of momentum MV1 + mV2=mV V1 - V2=VSolving we get Loss of kinetic kinetic energy E= Answer: (b)
Q.17
On slippery road with a coefficient of friction reduced to 0.2, the maximum speed at which a car can go round a curve of radius 100m is
0%
a)5 m/s
0%
b) 7 m/s
0%
c)14 m/s
0%
d)20 m/s
Explanation
v=√(µgr) v=√(0.2×100×9.8)=14 m/sAnswer: (c)
Q.18
A body is fired from point P and strikes at Q inside a smooth circular wall as shown in the figure. It rebounds to the point S ( diametrically opposite to P), then
0%
a) the coefficient of restitution is zero
0%
b) the coefficient of restitution is less than 1
0%
c)kentic energy is conserved in this collision
0%
d)the coefficient of restitution is (1/√3)
Explanation
In digram at Q angle of incidence=angle of reflection thus the collision is perfectly elastic. Kinetic energy remains constantAnswer: (c)
Q.19
A steel ball B of radius 2cm is initially at rest on a horizontal frictionless surface. It is struck head on by another steel ball A of radius 4cm, travelling with velocity of 0.81 m/s. the velocities of the two balls ( A and B) after the collision are
0%
a) 63 and 144 cm/s
0%
b) 31.5 and 72 cm/s
0%
c)21 and 48 cm/s
0%
d)42 and 96 cm/s
Explanation
mass of ball=Volume ×density and volume is proportional to cube of radius mB=k(2)3=k8 mA=k(4)3=k64 mA × 0.81=mAv1 + mBv2 64 × 0.81=64v1 + 8v1 8 × 0.81=8v1 + v2 --(i) and e=1 ∴ V2 - v1=0.81 --(ii) Solving v1=63 cm/s , v2=144 cm/s Answer:(a)
Q.20
A disk A of radius r moving on perfectly smooth surface at a speed v undergoes an elastic collision with an identical stationary disc B. Find the velocity of the disk B after collision if the impact parameter is d as shown in figure.
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
One of the disc is at rest before collision . After collision its velocity will be in the direction of the centre line at the moment of contact because in this direction in which the force acted on it from figure sinθ2=d/2r anad θ1 + θ2=π/2 since masses o both disk are equal, the triangle of momentum turns into triangle of velocities we have v1=vcosθ1=vsinθ2 v2=vcosθ2 Answer: (a)
Q.21
A particle of mass m moving with velocity v makes a head on elastic collision with another particle of same mass and initially at rest. The velocity of first particle after collision is
0%
a)v
0%
b) -v
0%
c)2v
0%
d)0
Explanation
In elastic collision, velocities of colliding bodies get interchanged if their masses are the same Answer: (d)
Q.22
A sphere A of mass 4kg is released from rest on a smooth hemispherical shell of radius 0.2m. the sphere A slides down and collides elastically with another sphere B of mass 1 kg placed on the bottom of shell. If the sphere B has to just reach top, the height h from where the sphere A should be released is
0%
a) 0.08 m
0%
b) 0.02 m
0%
c)0.18 m
0%
d)0.10 m
Explanation
Just before collision, velocity of A. v1=√(2gh) Just after collision, velocity of B The sphere is just able to reach the top from law of conservation of energy potential energy=kinetic energy Answer: (a)
Q.23
A disc of mass m moving on a horizontal surface with velocity v collides elastically identical stationary disc as shown in figure. The velocity of two discs after collision are
0%
a) v sinθ, vcosθ
0%
b) vsinθ, 0
0%
c)0, vcosθ
0%
d)vsinθ, (v/2)cosθ
Explanation
Let the velocities of two disc be v1 and v2conservation of momentum v=v1cosα + v2cosθ --(i)and v1sinα=v2sinθ --(ii) From (i) and (ii) we get equation (iii)Further K.E remains conserved equation (iv) from equation (iii) and (iv) v1=vsinθ and v2=vcosθ Answer:(a)
Q.24
Sand drops from a stationary hopper at the rate 5kg/s on to a conveyor belt moving with constant speed of 2m/s. What is the power delivered by the motor drawing the blet
0%
a) 10 watt
0%
b) 20 watt
0%
c) 30 watt
0%
d) 40 watt
Explanation
Mass of the belt is increasing at dm/dt rate Thus change in momentum per second=force=v (dm/dt ) force=2×5=10N power=force × velocity=10×2=20 watt Answer: (b)
Q.25
A body of mass m moving with velocity v collides head on which another body of mass 2m at rest. The ratio of K.E. of colliding body before and after collision is ( assuming collision is perfectly elastic)
0%
a)1:1
0%
b) 2:1
0%
c)4:1
0%
d)9:1
Explanation
mv=mv1 +2 mv2 V=v1+v2coefficient o restitution e=1 ∴ V2- v1=VSolving , 3v1=-v v1=v/3ratio of kinetic energy Answer: (d)
Q.26
A car is moving in circular horizontal track of radius 10m with constant speed of 10m/s. A plumb bob is suspended from roof by a light rod of length 1m. The angle made by the rod with the track is
0%
a) zero
0%
b) 30°
0%
c)45°
0%
d)60°
Explanation
The different forces acting on the boob are shown in figure. Resolving the forces along the length and perpendicular to the rod, we haveAnswer: (c)
Q.27
Block A and B of equal masses are arranged as shown in figure. The surface of A is smooth while B is rough and has a coefficient of friction 0.1 with surface. The block A moves with speed 10m/s and collides with B. The collision is perfectly elastic. Find the distance moved by B before it comes to rest.
0%
a) 25 m
0%
b) 100 m
0%
c)50 m
0%
d)75 m
Explanation
Since the collision is elastic and masses of colliding bodies are same, the velocities will be interchanged. Thus Block B will start to move at 10 m/s Retardation is due to frinctional force=µmg Retardation=µmg/m=µg=0.1×10=1 m/s2 Now v2=u2 +2as o=100 +2(-1)s s=100/2=50 m Answer:(c)
Q.28
Two persons standing on a floating boat run in succession along its length with a speed 4.2 m/s relative to the boat and dive off from the end. the mass of each man is 80kg and that of boat is 400 kg. If the boat was initially at rest, find the final velocity of the boat. Neglect friction.
0%
a) 0.6 m/s
0%
b) 0.7 m/s
0%
c) 0.1 m/s
0%
d) 1.3 m/s
Explanation
Mass of boat after first person dive off 480kg Let VMB : Velocity of man with respect to boat VBG = Velocity of boat with respect to ground VMG = Velocity of man with respect to ground According to law of conservation of momentum 80 VMG + 480 VBG = 0 80(VMB+ VBG)+ 480 VBG = 0 560VBG = -80 VMB Given VMB = 4.2 m/s VBG =- 0.6 m/s ( negative sign indicate boat is moving back) Thus momentum of boat = 0.6×480 =- 288 kg m/s Second person dive off, the mass of boat = 400 kg According to law of conservation of momentum 80 VMG + 400 VBG = -288 80(VMB+ VBG)+ 400 VBG = -288 480VBG = -80 VMB-288 Given VMB = 4.2 m/s VBG = -1.3 m/s ( Negative sign indicate velocity is opposite to velocity after first person dive off) Answer: (d)
Q.29
A ball hits the floor and rebounds after inelastic collision. In this case
0%
a)the momentum of ball just after collision is the same as that just before collision
0%
b) the mechanical energy of the ball remains the same in the collision
0%
c)total momentum of the ball and the earth is conserved
0%
d)the total energy of the ball and the earth is conserved
Explanation
Answer: (c)
Q.30
A particle is describing motion in a horizontal plane in contact with the smooth inside surface of fixed hemispherical bowl with its axis vertical. The radius of the bowl is 25cm and the radius of the orbit of motion of the particle is 15cm. Then the period of revolution in seconds of the particle is
0%
a) 2π/5
0%
b) 2π/7
0%
c)2π/3
0%
d)π/3
Explanation
From figure Nsinθ=mrω2 and Ncosθ=mg tanθ=ω2r / g and sinθ=15/25=3/5 ∴ tanθ=3/4 Answer: (b)
0 h : 0 m : 1 s
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)