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Physics NEET MCQ
Quiz 7
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Q.1
From the top of twer of height 40 m a ball is projected upwards with a speed of 20m/s at an angle of elevation of 30°. Then the ratio of the total time taken by the ball to hit the ground to its time of flight ( time taken to come back to the same elevation ) is
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a) 2 : 1
0%
b) 3 : 1
0%
c)3 : 2
0%
d)4 : 1
Explanation
time taken to reach ball on ground h=-ut + ½ gt 2 40=-(20sin30) t + ½ ×10t2 5t2 - 10t -40=0 thus t=-2 or t=4 seconds time can not be negative thus t=4 sec time to come to same elevation T=2usinθ/g=(2×20×sin30 ) /10=2 sec ∴ t/T=4/2=2:1 Answer:(a)
Q.2
If the greatest range down an inclined plane is three times the greatest range up the plane, then the angle of inclination of the plane with the horizontal is
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a) 75°
0%
b) 45°
0%
c) 60°
0%
d) 30°
Explanation
Answer: (d)
Q.3
If two stones projected from the same point with same initial speed but at angle π/3 and π/6 respectively have their ranges R1 and R2 then
0%
a)R1=2R2
0%
c)R1=5R2
0%
d)R1=25R2
0%
b) R₁=R₂
Explanation
Answer: (b)
Q.4
The angle between the vector A and B is θ. The value of triple product A . B × A is
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a) A2B
0%
b) zero
0%
c)A2Bsinθ
0%
d)A2Bcosθ
Explanation
Let C=B × A then angle between C and B is 90 ∴ B.C=0Answer: (b)
Q.5
A projectile is projected in the upward direction making an angle of 60° with horizontal direction with a velocity of 147 m/sec. Then the time after which its inclination with the horizontal is 45° is
0%
a) 15 sec
0%
b) 10.98 sec
0%
c)5.49 sec
0%
d)2.745 sec
Explanation
we know that tanθ=Vy / Vxx componant of velocity do not change with time Answer:(c)
Q.6
The horizontal range of a projectile is 4√3 times its height, then the angle of projection is
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a) 90°
0%
b) 60°
0%
c) 45°
0%
d) 30°
Explanation
Answer: (d)
Q.7
A player kicks up a ball at an angle θ to horizontal. The horizontal range is maximum when θ equals:
0%
a)30°
0%
b) 45°
0%
c)60°
0%
d)90°
Explanation
Answer: (b)
Q.8
A body has an initial velocity of 3m/s and an acceleration of 1m/sec2 normal to the direction of initial velocity. Then the velocity of the body 4 sec after start is
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a) 7 m/sec along initial velocity
0%
b) 7 m/sec in the normal direction to initial velocity
0%
c)7m/sec midway between initial and normal direction
0%
d)5 m/sec at an angle of tan⁻¹ (4/3) with the direction of initial velocity
Explanation
Acceleration is perpendicular to initial velocity thus horizontal component of velocity will not change Vh=3 m/sec vertical componant of velocity initial velocity=0 Vv=at=(4) (1)=4 m/s Thus resultant velocity Answer: (d)
Q.9
An aeroplane is flying at height of 1960 m in horizontal direction with a velocity of 360 km/hr. When it is vertically above the point A on the ground, it drops a bomb . The bomb strikes at a point B on the ground. Then the time taken by the bomb to reach the ground is
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a) 20√2 sec
0%
b) 20 sec
0%
c)10√2
0%
d)10 sec
Explanation
Answer:(b)
Q.10
A bomb is fired from a canon with velocity V m/sec at an angle θ with the horizontal direction. At the highest point of its path it explodes into two pieces of equal masses. One of the piece retraces its path, then the speed of the other piece immediately after explosion is
0%
a) 3Vcosθ
0%
b) 2Vcosθ
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c) (3/2)Vcosθ
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d) (√3/2) Vcosθ
Explanation
Let the mass of the bomb be 2m and its velocity be V. Then the momentum of the bomb at the highest point of the path - 2mvcosθ At the highest point of the path, the bomb explodes into two equal fragments ( each of mass m). One of the fragments retraces its path, therefore its momentum is -mvcosθ. Let V'; be the velocity of other fragment , then form law of conservation of momentum 2mVcosθ=mVconsθ + mV' V'=3mVcosθ Answer: (a)
Q.11
An arrow is shot air. Its range is 200 m and its time of flight is 5sec. IF g=10m/sec2, then the horizontal component of velocity of the arrow is:
0%
a)12.5 m/sec
0%
b) 25 m/sec
0%
c)31.25 m /sec
0%
d)40 m/sec
Explanation
Range=Horizontal componant × time of fligfht 200=uscosΘ ×5 ucosθ=40 m/secAnswer: (d)
Q.12
Area of parallelogram formed by adjacent sides as the vectors A=3i + 2j and B=2j - 4k is
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a) (√244) /2
0%
b) √244
0%
c)√122
0%
d)(√122 )/ 2
Explanation
Answer: (b)
Q.13
If A.B=AB then the angle between A and B is
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a) 0°
0%
b) 45°
0%
c)90°
0%
d)180°
Explanation
Answer:(a)
Q.14
If A=4i- 2j + 6k and B=i -2j -3k, the angle which the A + B makes with x-axis is
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
A+B=(4i - 2j + 6k) + (i -2j -3k) A+B=5i - 4j + 3k|A+B|=√50 The nagle which A+B makes with x axis is Answer: (b)
Q.15
A particle is projected at an angle of 45° from a point lying 2m from the foot of a wall. It just touches the top of the wall and falls on the ground 4m from it. The height of the wall is
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a) 3/4 m
0%
b) 2/3 m
0%
c)4/3 m
0%
d)1/3 m
Explanation
Range of stone=6m Thus u2 /g=6 By using equation for trajectory Answer: (c)
Q.16
The equation of projectile is y=√3 x - gx2 / 2, then angle of projection is
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a) tanθ=1/√3
0%
b) tanθ=√3
0%
c)π/2
0%
d)zero
Explanation
Answer:(b)
Q.17
The equation of projectile is y=√3 x - gx2 / 2, the initial velocity is
0%
a) 4 m/s
0%
b) 2 m/s
0%
c) 1 m/s
0%
d) 5 m/s
Explanation
Answer: (b)
Q.18
i×(j×k) is
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a)i + j + k
0%
b) i - j + k
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c)zero vector
0%
d)unit vector
Explanation
Answer: (c)
Q.19
Many bullets are projected at equal angles with the horizontal in different directions with initial velocity v. The area in which these bullets are spread is
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a) πv² / g
0%
b) πv⁴/ g²
0%
c)π2v4 / g²
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d)π2v2 / g²
Explanation
Answer: (b)
Q.20
The magnitude of the vector product of two vectors is √3 times their scalar product. The angle between the vectors is
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a) π/2
0%
b) π/6
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c)π/3
0%
d)π/4
Explanation
Answer:(c)
Q.21
Which of the following statement is true
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a) Energy, force and weight are vector quantities
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b) Relative velocity, work and acceleration are vector quantities
0%
c) displacement, force and electric field are vector quantities
0%
d) Velocity, length and acceleration are vector quantities
Explanation
Answer: (c)
Q.22
The component of 9i. +17k along z-axis has magnitude
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a)zero
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b) 17
0%
c)9
0%
d)26
Explanation
Answer: (b)
Q.23
The magnitude of the resultant of two vectors is 12 units. If the magnitude of one vector is 20 units. what is the magnitude of the other vector? The resultant vector is at right angles to the "other vector"
0%
a) 12 units
0%
b) 16 units
0%
c)18 units
0%
d)28 unit
Explanation
Answer: (b)
Q.24
A ball thrown by one player is caught by another after 2 seconds. the maximum height attained by the ball is nearly.
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a) 2 m
0%
b) 5 m
0%
c)7 m
0%
d)9 m
Explanation
Answer:(b)
Q.25
If the range of projectile be R, then its K.E is maximum after covering ( from start) a distance equal to
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a) R/4
0%
b) R/2
0%
c) 3R/4
0%
d) R
Explanation
Answer: (d)
Q.26
A person is moving in a circle of radius r with a constant speed v. The change in velocity in moving from a to B is
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a)2vcos40
0%
b) 2vsin40
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c)2vcos20
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d)2vsin20
Explanation
Answer: (d)
Q.27
A blind person after walking 10 steps in one direction, each of length 80cm, turn randomly to by 90°. After walking a total of 40 steps, the maximum disparagement of the person from its starting point can be
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a) 32 m
0%
b) 16√2
0%
c)8√2
0%
d)0 m
Explanation
Maximum displacement=distance travelled=40 steps each of 80 m=32 m Answer: (a)
Q.28
Resultant of two forces acting at right angles to each other is 1414 dyne. The magnitude of each force is
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a)500 N
0%
b) 100 N
0%
c)1000 N
0%
d)none of the above
Explanation
force must be 1000 dyne , no option is given Answer:(d)
Q.29
At what angle the two forces A+B and A-B acts so that their resultant is √(3A2 + B2)
0%
a) π/6
0%
b) π/3
0%
c) π/2
0%
d) π
Explanation
Let P = A+B and Q = A-B Let θ be the angle between them R2 = P2 + Q2 + 2PQcosθ 3A2 + B2 = A2 +B2 +2AB + A2 +B2 -2AB + 2( A+B) (A-B)cosθ 3A2 +B2 = 2A2 + 2B2 + 2(A2 –B2) cosθ A2 – B2 = 2(A2 –B2) cosθ 1= 2cosθ θ = π/3 Answer: (b)
Q.30
A vector x, when added to two vectors A=3i -5j + 7k and B=2i + 4j - 3k gives a unit vector along y axis. The vector x is
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a)-5i + 2j - 4k
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b) 5i + 2j - 4k
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c)-5i - 2j + 4k
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d)5i - 2j - 4k
Explanation
Answer: (a)
0 h : 0 m : 1 s
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