MCQGeeks
0 : 0 : 1
CBSE
JEE
NTSE
NEET
English
UK Quiz
Quiz
Driving Test
Practice
Diagrams
Games
NEET
Physics NEET MCQ
Quiz 2
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
The vector A and B are such that|A + B |=| A - B | The angle between the two vectors is .. [ CBSE-PMT 2006]
0%
a) 60°
0%
b) 75°
0%
c)45°
0%
d)90°
Explanation
|A + B |2=| A - B |2 A2 + B2 + 2ABcosθ=A2 + B2 - 2ABcosθ 4ABcosθ=0 ∴ θ=90° So angle between vector A and B is 90°Answer: (d)
Q.2
The magnitude of vectors A, B, C are 3, 4 and 5 units respectively. If A + B=C then the angle between A and B is .. [ CBSE-PMT 1998]
0%
a) π /2
0%
b) cos-10.6
0%
c)tan-1 7/5
0%
d)π/4
Explanation
from given (A + B)2=C2A2 + B2 +2ABcosθ=C232 + 42 +2 (3)(4) cosθ=52∴ 2ABcosθ=0 cosθ=0thus angle between A and B is 90° Answer:(a)
Q.3
A × B|=AB √3, then the value of |A+B| is .. [ CBSE-PMT 2004]
0%
a) (A2 + B2 + AB√3) 1/2
0%
b) (A2 + B2 + AB) 1/2
0%
c) (A2 + B2 + AB/√3) 1/2
0%
d) A + B
Explanation
| A ×B|=ABsinθ A . B ABcosθ given |A × B|=AB √3ABsinθ=√3 ( ABcosθ)tanθ=√3 ∴ θ=60° ∴ | A + B|=(A2 + B2 + 2ABcos60) 1/2| A + B|=(A2 + B2 + AB) 1/2 Answer: (b)
Q.4
If the angle between the vectors A and B is θ, the value of the product (B ×A) . A is equal to ... [ CBSE-PMT 2005]
0%
a)BA2sinθ
0%
b) BA2cosθ
0%
c)BA2sinθcosθ
0%
d)zero
Explanation
B ×A=C Direction of vector C is perpendicular to both A And B Now C . A=CBCos90Thus (B ×A) . A=0Answer: (d)
Q.5
A bullet is fired from a gun with speed of 1000 m/s in order to fit a target 100 m away. At what height above the target should the gun be aimed? ( resistance of air is negligible and g=10 m/s2 [CBSE-PMT 1995]
0%
a) 5cm
0%
b) 10cm
0%
c)15cm
0%
d)20cm
Explanation
initial velocity of Bullet along vertical direction is zeroTime taken by bullet to cover horizontal distance t=Distance / speed of Bullett=100 / 1000=0.1secDuring this time, bullet will fall down due to gravitational accelerationheight 'h'=ut + ½ g t2h=½ 10(0.1) 2 h=0.05 m=5cmAnswer: (a)
Q.6
for angle of projection of a particle (45° - θ) and (45° + θ), the horizontal ranges described by the projectile are in the ratio [ CBSE-PMT 2006]
0%
a) 1 : 3
0%
b) 1 : 2
0%
c)2 : 1
0%
d)1 : 1
Explanation
(45° - θ) and (45° + θ) are complementaryWe know that angle of projection of two projectile makes complementary angles, their ranges are equal. In this case, the range will be same the ratio is 1:1 Answer:(d)
Q.7
A bus is moving on a straight road towards north with uniform speed of 50 km/h turns through 90° . If the speed remains un changes after turning the increase in the velocity of bus in the turning process is .. [ CBSE-PMT 1989]
0%
a) 70.7 km/h along south-west direction
0%
b) zero
0%
c) 50 km/h along west
0%
d) 70.7 km/h along north-west direction
Explanation
Let v1 due North Let V2 due west Angle between velocity is 90° Change in velocity| V2 - V1| change in velocity=( v22 + v12) 1/2 change in velocity=(502 + 502) 1/2Change in velocity=70.7Direction of this change in velocity is in South-West Answer: (a)
Q.8
A boat is sent across a river with velocity of 8 km/h. If the resultant velocity of boat is 10km/h, then the velocity of the river is .. [CBSE-PMT 1993]
0%
a)12.8 km/h
0%
b) 6 km/h
0%
c)8 km/h
0%
d)10 km/h
Explanation
Let Velocity of boat with respect to observer VBO=10 km/hVelocity of boat with respect to river VBR=8 km/hVelocity of River with respect to Observer VRO Now VBO=VBR + VROAngle between VBO and VRO is 90°thus V2BR=V2BO - V2RO82=102 - V2RO64=100 - V2ROV2RO=36VRO=6 km/hAnswer: (b)
Q.9
The maximum range of a gun of horizontal terrain is 16 km. If g=10 m/s2, then muzzle velocity of a shell must be.. [ CBSE-PMT 1990]
0%
a) 160 m/s
0%
b) 200√2 m/s
0%
c)400 m/s
0%
d)800 m/s
Explanation
Formula for maximum range=V2 / g 16000 (10)=V2∴ v=400 m/sAnswer: (c)
Q.10
A body constrained to move in y-direction , is subjected to force given byF=( -2i + 15j + 6k ) N. What is the work done by this force in moving the body through a distance of 10m along y-axis [ CBSE-PMT 1994]
0%
a)190 J
0%
b) 160 J
0%
c)150 J
0%
d)20 J
Explanation
Since displacement is along y-direction hence displacement vector S=10j work done=F . S Work done=( -2i + 15j + 6k )( 10j) Work Done=150J Answer:(c)
Q.11
Two particles A and B are connected by a rigid rod AB. The rod slides along perpendicular rails as shown here. The velocity of A to the left is 10 m/s. What is the velocity of B when angle α=60°? [ CBSE -PMT 1998]
0%
a) 5.8 m/s
0%
b) 9.8 m/s
0%
c) 10 m/s
0%
d) 17.3 m/s
Explanation
In one second angle become 60°. When the end A moves by 10m left, the end B moves up in one second=10×tan60=17.3 m/s Answer: (d)
Q.12
The result of (A × 0) will be equal to .. [ CBSE-PMT 1992]
0%
a)zero
0%
b) A
0%
c)zero vector
0%
d)unit vector
Explanation
When a vector is multiplied by a scalar result is vectorAnswer: (c)
Q.13
A particle moving with a velocity v=6i -4j+3k m/s under the influence of a constant force F=20i + 15j - 5kThe instantaneous power applied to the particle is .. [ CBSE-PMT 2000]
0%
a) 45 J/s
0%
b) 35 J/s
0%
c)25 J/s
0%
d)195 J/s
Explanation
power P=F . SP=(6i -4j+3k ) . ( 20i + 15j - 5k )P=45 J/sAnswer: (a)
Q.14
A boat which has a speed of 5 km/h in still water crosses a river of width 1 km along the shortest possible path in 15 minutes. the velocity of the river water in km/h is .. [ CBSE-PMT 2000]
0%
a) 3
0%
b) 4
0%
c)√21
0%
d)1
Explanation
Speed along the shortest path=distance / time Speed along shortest path=1/ (15/60)=4 km/hFrom geometry of figurevelocity of revere v2=52 - 42v=3 km/h Answer:(a)
Q.15
The angle between the two vectors A=3i+4j+5k and B=3i + 4j -5k will be.. [ CBSE-PMT 2001]
0%
a) zero
0%
b) 45°
0%
c) 90°
0%
d) 180°
Explanation
cos θ=A . B / (|A| |B|) Now A . B=(3i+4j+5k) . (3i+4j-5k) A . B=9+16-25=0 |A| and |B| are nonzero ∴ cosθ=0 ∴θ=90° Answer: (c)
Q.16
A body of 3kg moves in the XY plane under the action of force given by 6ti + 4tj . Assuming that the body is at rest at time t=0, the velocity of the body at t=3s is... [ CBSE - PMT 2002]
0%
a)6i + 6j
0%
b) 18i + 6j
0%
c)18i + 12j
0%
d)12i + 18j
Explanation
a=F / ma=(6ti + 4tj) /3a=2ti + (4/3)tjVelocity equation v=u +atv=0 + (2ti + (4/3)tj)tv=(2t2i + (4/3)t2j)v=2(3)2i + (4/3) (3)2j v=18i + 12jAnswer: (c)
Q.17
three forces acting on a body are shown in the figure. To have the resultant force only along the y direction, the magnitude of the minimum additional force needed is ... [ CBSE-PMT 2008]
0%
a) 0.5 N
0%
b) 1.5 N
0%
c)[(√3) / 4] N
0%
d)√3 N
Explanation
As shown in figure the component of 1N and 2 N forces along positive x-axis=1cos60+ 2sin30=1.5Ncomponent of 4 N force along negative x-axis is=4sin30=2NTherefore, if force of 0.5N is applied along x-axis, the resultant force along x-axis will become zero and the resultant force will be obtained only along y-axisAnswer: (a)
Q.18
Consider a F=4i - 3j. Another vector perpendicular to F is ..[MPPMT 1987]
0%
a) 4i + 3j
0%
b) 6i
0%
c)7k
0%
d)3i - 4j
Explanation
If vectors are perpendicular dot product is zero, Take dot product from the options, it is zero then that option is correctAnswer: (c)
Q.19
An aeroplane moves 400 m towards the north, 300 m towards west and then 1200m vertically upwards, then its displacement from the initial position is [ CBSE-PMT 1998]
0%
a) 1600 m
0%
b) 1800 m
0%
c)1500 m
0%
d)1300 m
Explanation
Here CD is perpendicular to the plane of page. Required distance DB From figure BD2=CB2 + CD2BD2=5002 + 12002BD=1300 m Answer:(d)
Q.20
for ordinary terrestrial experiments, the observer in an inertial frame in the following cases is .. [ AIIMS 2010]
0%
a) a child revolving in a giant wheel
0%
b) a driver in a sports car moving with constant speed of 200 km/h on a straight road
0%
c) the pilot of an aeroplane which is taking off
0%
d) a cyclist negotiating a sharp curve
Explanation
The car moving with a constant velocity has no acceleration. Hence, it is an inertial frame Answer: (b)
Q.21
Rain is falling vertically downwards with a velocity of 3km/h. A man walks in the rain with a velocity of 4km/h. The rain drop will fall on the man with a velocity... [ AIIMS 1997]
0%
a)5 km/h
0%
b) 4 km/h
0%
c)1 km/h
0%
d)3 km/h
Explanation
Let velocity of rain with respect to stationary observer be VRO Velocity of Man with respect to observer be VMOVelocity of rain with respect to man be VRMnow VRM=VRO + VOMbut VOM=- VMO∴ VRM=VRO - VMO Angle between VRO and VMO is 90°thus V2RM=V2RO + V2MOV2RM=9+16=25VRM=5Answer: (a)
Q.22
At the uppermost point of projectile, its velocity and acceleration are at an angle of ...[ AIIMS 2002]
0%
a) 180°
0%
b) 90°
0%
c)60°
0%
d)150°
Explanation
At uppermost point of projectile, velocity along vertical direction in zero, but have horizontal Velocity.Acceleration is gravitational and vertically down Thus angle is 90°Answer: (b)
Q.23
If R and H represent the horizontal range and the maximum height achieved by a projectile then which of the relation exists?
0%
a) H/R=4cotθ
0%
b) R/H=4cotθ
0%
c)H/R=4tanθ
0%
d)R/H=4tanθ
Explanation
Equation for range is : Equation for height By taking ratio of R/H we get Answer:(b)
Q.24
A projectile can have the same range R for two angle of projection. If t1 and t2 be the time of flights in the two cases, then the product of two time of flights is proportional to .. [ AIIMS 2008]
0%
a) 1/ R2
0%
b) R2
0%
c) R
0%
d) 1/R
Explanation
according to formula for same range different angles of projection Answer: (c)
Q.25
If A ×B=B×A , then angle between A and B is .. [ AIEEE 2004]
0%
a)π /2
0%
b) π/3
0%
c)π
0%
d)π/4
Explanation
given A ×B=B×AA ×B - B×A=0 Now - B×A=A×B ∴ A ×B +A ×B=0 ⇒ A ×B=0 ⇒ θ=0, π and 2π from given option θ=π Answer: (c)
Q.26
A ball whose kinetic energy is E, projected at an angle of 60° to the horizontal. The kinetic energy of the ball at the highest point of its flight will be .. [ AIEEE 2002]
0%
a) E
0%
b) E/ √2
0%
c)E/4
0%
d)zero
Explanation
let initial velocity be u at time t=0vertical component of velocity usinθ=0 at max heighthorizontal component will be ucosθ=u/2 which remains constant Now E ∝ u2Thus at max height E'=E/4Answer: (c)
Q.27
A boy playing on the roof of a 10 m height building throws a ball with speed of 10 m/s at an angle of 30° with the horizontal. How far from the throwing point will the ball be at height of 10m from the ground? [ AIEEE 2003][g=10 m/s2, sin30=1/2, cos30=√3 /2]
0%
a) 5.2 m
0%
b) 4.33 m
0%
c)2.6 m
0%
d)8.66 m
Explanation
As shown in figure we have to find the range of projectileFrom the equation for projectile u=10 m/s , θ=30° g=10 m/s2 Answer:(d)
Q.28
A particle is moving eastwards with a velocity of 5m/s. In 10 sec the velocity changes to 5m/s northwards. The average velocity in this time is ..[ AIEEE 2005]
0%
a) (1/2) ms-2 towards north
0%
b) (1/√2) ms-2 towards north-east
0%
c) (1/√2) ms-2 towards north-west
0%
d) zero
Explanation
From figure Final velocity=5j Initial velocity=5i Change in velocity Δv=5j - 5i acceleration=Δv/t=(5j - 5i) / 10 a=(1/2)j - (1/2)i |a|=[ (1/2)2 - (1/2)2]1/2 |a|=1 /√2 tanθ=-1, x co-ordinate is negative hence direction is north west Answer: (c)
Q.29
The coordinates of moving particle at any time 't' are given by x=αt3 and y=βtthe speed of the particle at time 't' is given by ... [ AIEE 2003]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
By taking first order derivative of both equations we get velocity along x-axis and y-axis vx=3αt2vy=3βt2Thus resultant velocity=Answer: (b)
Q.30
A projectile can have the same range 'R' for two angles of projection. If T1 and T2 to be time of flight in two cases, then the product of the two time of flights is directly proportional to .. [ AIEEE 2004]
0%
a) R
0%
b) 1/R
0%
c)1 /R2
0%
d)R2
Explanation
Equation for time of flight is=The angle of projection is different but range is same , then angle of projections are complimentary Let one angle be θ then other is (90 - θ) Taking product of T1 and T2 we get2sinθcosθ=sin2θformula of range is from above T1T2=2R/g Answer: (a)
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)