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Physics NEET MCQ
Quiz 5
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Q.1
Two like parallel forces P and 3P are 40cm apart. If the directions of P is reversed, then their resultant shifts through a distance .. [ Roorkee 1998]
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a) 30 cm
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b) 40 cm
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c)50 cm
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d)60 cm
Explanation
Case I when forces are parallel let resultant force be at distance x from force Pp(x)=3P(40-x)x=30 ( 30cm from P)Now force P is reversed let the resultant force be at distance y from force P -p(y)=3P(40-y) y=60 cm resultant shift through distance of (60-30)=30cm Answer: (a)
Q.2
Two balls are projected making an angle of 30° and 45° respectively with the horizontal. if both have same velocity at the highest points of their path, then the ratio of their velocities of projection is [ C.S.E.P 2001]
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a) √3 : √2
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b) √2 : 1
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c) √2 : √ 3
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d) √3 : 2
Explanation
let u be the velocity of first ball horizontal component=ucos30 let v be the velocity of second ball horizontal component=vcos45 given ucos30=vcos45 thus u/v=cos45/cos30 Answer: (c)
Q.3
If a stone is to hit at a point which is at a distance d away and at a height h above the point from where the stone starts, then what is the value of initial speed u if stone is launched at an angle θ [ Hariyana C.E.T 2002]
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a)
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b)
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c)
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d)
Explanation
equation for path of projectile substituting y=h in above equation and solving for u we getAnswer: (b)
Q.4
Let F be the force acting on a particle having position vector r and T be the torque of this force about the origin . Then.. [ AIEEE2003]
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a) r⋅T=0 and F⋅T ≠ 0
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b) r⋅T ≠ 0 and F ⋅ T=0
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c)r⋅T ≠ 0 and F⋅T ≠ 0
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d)r⋅T=0 and F⋅T=0
Explanation
Torque is perpendicular to plane containing r and F thus dot product of r and T is zero and dot product of F and T is zero. option (d) is correctAnswer: (d)
Q.5
Three forces start acting simultaneously on a particle moving with velocity. These forces are represented in magnitude and direction by three sides of triangle as shown in figure. The particle will move with velocity
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a) less than v
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b) greater than v
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c)|v| in the direction of the largest force
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d)v remains unchanged
Explanation
Since three forces acting on the particle are represented by the three sides of a triangle taken in one order, their resultant is zero. So the particle velocity remains unchanged Answer:(d)
Q.6
A block is dragged on a smooth plane with the help of a rope which moves with velocity v as shown in figure. The horizontal velocity of the block is
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a) v
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b) vsinθ
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c) v/sinθ
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d) v/cosθ
Explanation
As shown in adjacent figure. h2 = x2+y2 Note y is constant. By taking derivative of above equation we get Answer: (c)
Q.7
A body is projected at time t=0 from a certain point on a planet surface with a certain velocity at a certain angle with the planet,s surface ( assumed horizontal ). The horizontal and vertical displacement x and y (in meters) respectively vary with time t (in second) as x=10t√3 and y=10t-tWhat is the magnitude and direction of the velocity with which the body is projected?
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a)20m/s at an angle of 30° with the horizontal
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b)20m/s at an angle of 60° with the horizontal
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c)10m/s at an angle of 30° with the horizontal
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d)10m/s at an angle of 60° with the horizontal
Explanation
we know that the position co-ordinates x and y are given by x=(vocosθ)t comparing above equation with given equation x=10√3 (t) we get vocosθ=10√3 --(1) y=vosinθt - ½ gt2 comparing given equation y=10t - t2 with standard equation we get vosinθ=10 --(2) by taking ratio of equation 2 to equation 1 we get tanθ=1/√3 θ=30° Substituting value of θ in equation (1) we get vocos30=10√3 vo (&radic3 /2)=10√3 vo=20 m/s Answer: (a)
Q.8
The area of the parallelogram formed from the vectorsA=i -2j + 3k and B=3i - 2j + k as adjacent sides is
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a) 4√3
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b) 4√6
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c)8√3
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d)8√6
Explanation
We know that |A×B|=Area of parallelogram Answer: (b)
Q.9
The angle between A and B is θ. R=A + B makes an angle θ/2 with A. Which of the following is true?
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a) A=2B
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b) 2A=B
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c) AB=1
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d) none of the above
Explanation
Angle made by the resultant vector with A given by None of the above options , β comes out to θ/2 Answer: (d)
Q.10
A particle is projected from a point O with velocity u in a direction making an angle α upward with the horizontal. At P, it is moving at right angle to its initial direction of projection, its velocity at P is ..
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a)u tanα
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b) ucotα
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c)ucosecα
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d)usecα
Explanation
As shown in figure component of u along x-axis is ucosα At point P it changes its direction by 90° let velocity be v Then component of v along x-axis is vcos(90-α) = vsinα Thus ucosα = vsinα V = ucotα Answer: (b)
Q.11
The velocity of a particle P moving freely under gravity is 4.9 m/s, the direction being 30° with the downward normal
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a) its acceleration normal to the direction of motion at P=9.8 m/s2
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b) the radius of curvature of P of the parabolic trajectory of particle is 4.9m
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c)the particle has no acceleration normal to the direction of motion
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d)the radius of curvature at P of the path depends upon the initial velocity of projection
Explanation
As shown in figure centripetal acceleration ac=gcos60 ac=9.8( 1/2)=4.9 m/sec2Centripetal acceleration is given by V2 /r V2 /r=4.9 (4.9)2 / r=4.9r=4.9 mAnswer: (b)
Q.12
A particle is projected up an inclined as shown in figure. For maximum range over the inclined plane the value of θ should be
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a) 45°
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b) 15°
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c)30°
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d)60°
Explanation
Angle made by u with horizontal α for maximum range α=(45 + β/2) Here β is angle made by the inclined plane with horizontal In given problem α=θ +30 and β=30 (45 + 30/2)=θ +30 θ=30° Answer:(c)
Q.13
A particle located at x=0 at time t=0, starts moving along with the positive x-direction with velocity 'v' that varies as v=α√x. The displacement of the particle varies with time as [ AIEEE 2006]
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a)t2
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b) t
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c)√t
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d)t3
Explanation
Answer: (a)
Q.14
A particle is projected at 60° to the horizontal with a kinetic energy K. The kinetic energy at the highest point is [ AIEEE 2007]
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a) K/2
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b) K
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c)zero
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d)K/4
Explanation
Horizontal velocity v'=vcos60=v/2Thus kinetic energy at top=K/4Answer: (d)
Q.15
The velocity of a particle is v=vo +gt + ftIf its position is x=0 at t=0, then its displacement after unit time (t=1) is [ AIEEE 2007]
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a) vo + g/2 + f
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b) vo + 2g + 3f
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c)vo + g/2 + f/3
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d)vo + g + f
Explanation
Answer:(c)
Q.16
A particle has an initial velocity of 3i + 4j and an acceleration of 0.4i+0.3j . Its speed after 10s is [ AIEEE 2009]
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a) 7√2 units
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b) 7 units
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c) 8.5 units
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d) 10 units
Explanation
Given u=3i + 4j, a=0.4i+0.3j, t=10 s v=u + at v=3i + 4j + (0.4i+0.3j)×10=7i + 7j |v|=7√2 units Answer: (a)
Q.17
If the magnitudes of A, B and C are 12, 5 and 13 units respectively and A + B=C, then the angle between A and B is
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a)0
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b) π
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c)π/2
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d)π/4
Explanation
Answer: (c)
Q.18
If A=B + C and the magnitude of A, B and C are 5,4,3 units, the angle between A and C is .. [ CBSE 1990]
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c)π/2
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a) cos⁻¹ (3/5)
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b) cos⁻¹(4/5)
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d)sin⁻¹(3/4)
Explanation
Answer: (a)
Q.19
Out of the following sets of forces, the resultant of which cannot be zero [ CPMT 1985]
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a) 10,10,10
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b) 10,20,40
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c)10,10,20
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d)10,20,20
Explanation
Answer:(b)
Q.20
Out of following pairs, the resultant of which cannot be 4 Newton [ CPMT 1985]
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a) 2N and 2N
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b) 2N and 6 N
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c) 2N and 4N
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d) 2N and 8N
Explanation
Answer: (d)
Q.21
A person moves 30 m North, then 20m East then 30√2 m South-West. His displacement from the original position is [ CPMT 1989]
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a)15 m East
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b) 28 m South
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c)10 m West
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d)14 m South-West
Explanation
Answer: (c)
Q.22
The length of seconds hand of watch is 1cm. the change in velocity of its tip in 15 seconds is [ MPPMT 1987]
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a) zero
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b) π/ (30√2) cm/sec
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c)(π/30) cm/sec
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d)2π/(30√2) cm/sec
Explanation
Answer: (d)
Q.23
Consider a F=4 i - 3 j. Another vector perpendicular to F is ..[ MPPMT 1987]
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a) 4i+3j
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b) 6i
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c)7k
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d)3i - 4j
Explanation
Answer:(c)
Q.24
Consider a F1=4 i + 5 k and F2=3 j - 4 k. The magnitude of the scalar product of these vector is [ MPPMT 1987]
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a) 20
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b) 23
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c) 5√(33)
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d) 26
Explanation
Answer: (a)
Q.25
A river is flowing from west to east at a speed of 5m/min. A man on the south bank of the river, capable of swimming at 10 m/min in still water, wants to swim across the river in shortest time. He should swim in a direction
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a)due north
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b) 30° east of north
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c)30° west of north
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d)60° east of north
Explanation
Answer: (a)
Q.26
A particle is moving eastwards with velocity of 5m/sec. In 10 sec the velocity changes to 5m/sec north-wards. the average acceleration in this time is [ CPMT 1998]
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a) zero
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b) (1/√2) m/sec2 towards north-west
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c)30° west of north
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d)60° east of north
Explanation
Answer: (b)
Q.27
The magnitude of a given vectors with end points (4, - 4, 0) and (-2, - 2, 0) must be
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a) 6
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b) 5√2
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c)4
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d)2√10
Explanation
Answer:(d)
Q.28
Choose the wrong statement
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a) Three vectors of different magnitudes may be combined to give zero resultant
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b) Two vectors of different magnitudes can be combined to give a zero resultant
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c) The product of a scalar and vector is a vector quantity
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d) All above statements are wrong statement
Explanation
Answer: (b)
Q.29
A boat is sent across a river with velocity of 8km/h. If the resultant velocity of the boat is 10km/h, the river is flowing with velocity of [ CBSE 1994]
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a)12.8 km/hr
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b) 6 km/hr
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c)8 km/hr
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d)10 km/hr
Explanation
Answer: (b)
Q.30
The angle between vectors ( A×B) and (B×A) is
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a) zero
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b) π
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c)π/4
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d)π/2
Explanation
Answer: (b)
0 h : 0 m : 1 s
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