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Physics NEET MCQ
Motion In Two And Three Dimensions Mcq
Quiz 4
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Q.1
A particle moves in plane with constant acceleration in a direction different from the initial velocity. The path of the particle is a
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a)straight line
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b) arc of a circle
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c)parabola
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d)ellipse
Explanation
Answer: (c)
Q.2
The vector sum of N coplanar forces each of magnitude F, when each force is making an angle (2π / N) with the preceding one is
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a) NF
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b)NF /2
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c)F/2
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d)zero
Explanation
Since angle between 1 st and nth vector is 2π vectors are symmetrically distributed hence resultant will be zeroAnswer: (d)
Q.3
If A=Acosθi - Asinθj be any vector. Another vector B , which is normal to A can be expressed as [ NCRT 1984]
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a)Bcosθi - Bsinθj
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b) Bcosθi + Bsinθj
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c)Bsinθi - Bcosθj
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d)Bsinθi + Bcosθj
Explanation
Dot product of perpendicular vector is zero.. Out of given option , option (d) is correct Answer:(d)
Q.4
A grasshopper can jump a maximum horizontal distance of 20 cm. If it spends negligible time on the ground, what is its speed of travel along the road
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a) √2
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b) 1 m/s
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c) 1/√2 m/s
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d) can not be calculated
Explanation
Maximum horizontal rage of the grass-hopper, thus angle of projection θ=45° 20=u2 / g u2=20×980 u=140 cm/sec The horizontal component of velocity=u cosθ horizontal component of velocity=140 cos45=140×(1/√2) horizontal component of velocity=100 cm/sec=1 m/sec ∴ speed of travel along the road=1 m/sec Answer: (b)
Q.5
An object is thrown along the direction inclined at an angle of 45° with horizontal direction. The horizontal range of the particle is .. [ MPPMT 1985]
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a)vertical height
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b) twice the vesicle height
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c)thrice the vertical height
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d)four times the vertical height
Explanation
We know that H/R=(1/4) tanθ given θ=45 ∴ H/R=1/4 R=4HAnswer: (d)
Q.6
The height y and the distance x along the horizontal plane of projectile on a certain planet ( with no surrounding atmosphere) are given byy=8t - 5t2 meter and x=6t meter, where t is in seconds. the angle with which the projectile is projected is
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a) tan-1(3/4)
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b) tan-1(4/3)
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c)sin-1(3/4)
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d)not obtainable form given data
Explanation
vx is component of velocity along x axis at time to=0 dx/dt=vx=6 vy is component of velocity along y axis at time to=0 dy/dt=vy=8-10t for t=0 vy=8 tanθ=vy / vxtanθ=8/6=4/3θ=tan -1 (4/3)Answer: (b)
Q.7
A gun fires two bullets at 60° and 30° with horizontal. The bullets strikes at same horizontal distance. The ratio of maximum height for the two bullets is in the ratio..
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a) 2:1
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b) 3:1
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c)4:1
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d)1:1
Explanation
We know that H=(R/4) tanθ Thus R=4H / tanθ Given Range is same 4H1/ tan60=4H2 / tan30H1/H2=tan60/tan30 H1/H2=3/1 Answer:(b)
Q.8
A hose pipe lying on the ground shoots a stream of water upward at an angle of 60° to the horizontal. The speed of the water is 20m/s as it leaves the hose. It will strike a wall 10m away at height of
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a) 10.5 m
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b) 12.32 m
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c) 10 m
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d) 20 m
Explanation
Horizontal displacement is 10 m Horizontal component of water=20cos60=10 m/s Time taken to travel 10 m, t=10/10=1 sec Vertical component of velocity=20sin60=10√3 displacement along vertical h=ut - ½gt2 h=10√3(1) - ½ (10)(1)=17.32- 5=12.32 m Answer: (b)
Q.9
A boy aims a gun at a bird from a point, at horizontal distance of 100 m. if gun can impart the velocity of 500 m/s to the bullet, what height above the bird must he aim his gun in order to hit it? ( take g=10 m/sec2
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a)100 cm
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b) 50 cm
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c)40 cm
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d)20 cm
Explanation
Horizontal distance traveled by bullet in time t=d/v=100/500=1/5 secIn time t=1/5 sec bullet will travel a vertically down ward distance of h=ut +½g(t)2h=½g(t)2 h=½ ×10×(1/5)2=0.2 m h=20 cm Thus hunter should aim the gun 20cm above the targetAnswer: (d)
Q.10
In the arrangement shown in figure, the ends of P and Q of an un stretchable string move downwards with speed u. Pulley are fixed. Mass M moves upwards with speed of [CPMT 1990]
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a) 2ucosθ
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b) u/cosθ
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c)2u/cosθ
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d)ucosθ
Explanation
Let MP = h , MO = y and PO = x. h2 = y2 + x2 Taking derivative of above equation, note PO is constant h/y = 1/cosθ and dh/dt = u given , dy/dt is velocity of mass =v ∴v = u/cosθ Answer: (b)
Q.11
A bus is moving on the straight road towards north with a uniform speed of 50km/he then it turns left through 90°. If the speed remains unchanged after turning, the increase in the velocity of bus in turning process is [ CBSE 1990]
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a) 70.7 km/he along south-west direction
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b) zero
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c)50km/he along west
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d)70.7 km/h along north-west direction
Explanation
To find the change in velocity change initial velocity of bus in south direction as shown in figureSince angle between both the velocity is 90° resultant change in velocity is ΔV=50√2=70.7 km/hr and from figure direction is south-west direction Answer:(a)
Q.12
The angle between the vector A and B is θ. The value of the triple product A⋅B ×A is
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a) A2B
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b) zero
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c) A2Bsinθ
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d) A2Bsinθ
Explanation
A⋅B ×A can be written as B⋅A ×A As A ×A=0 answer is zero Answer: (b)
Q.13
A projectile is projected upward direction making an angle of 60° with horizontal direction with velocity 147m/sec. Then, the time after which inclination with the horizontal is 45° is
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a)15 sec
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b) 10.98 sec
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c)5.49 sec
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d)2.745 sec
Explanation
Horizontal component Initial velocity u=147 m/s Initial horizontal component u=147cos60Let after some time velocity of projectile be and angle made by horizontal component is 45°Thus final horizontal component is v'=Vcos45We know that horizontal velocity don't change thus 147cos60=Vcos45V=147cos60 /cos45V=147/√2 Vertical component Initial vertical component=usin60since due to gravitational acceleration velocity decreasesThus final velocity=usin60- gtSince V is the velocity when angle is 45° then vertical component=VsinθThus usin60- gt=Vsin45By substituting valuesAnswer: (c)
Q.14
From the top of tower of height 40m ball is projected upwards with speed of 20m/s at an angle of elevation of 30°. Then the ratio of total time taken by the ball to hit the ground to its time of flight( time take to come back to the same elevation) is
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a) 2:1
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b) 3:1
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c)3:2
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d)4:1
Explanation
Vertical component of velocity=usinθLet total time taken by ball to hit ground be tNow h=-(usinθ)t + ½gt240=-(20sin30)t + ½ (10)×t2∴ 5t2 - 10t - 40=0 t2 - 2t - 8=0 t=-2 and t=4 since time can't be negative t=4 sec Time to come back to same elevation is T=(2u sinθ)/ g T=(2×20×sin30)/10=2sec T=2 sec ∴ t/T=4/2=2:1Answer: (a)
Q.15
A body has an initial velocity of 3m/s and acceleration of 1m/s2, normal to the direction of initial velocity. Then the velocity of the body 4 seconds after start is
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a) 7 m/s along initial velocity
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b) 7 m/s in the normal direction to initial velocity
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c)7 m/s midway between initial and normal direction
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d)5 m/sec at an angle of tan-1(4/3) with the direction of initial velocity
Explanation
Initial velocity is in horizontal direction Vx=3 m/sInitial velocity along vertical=0 Acceleration is vertical, Final Velocity vertical=Vy=at=1×4=4 m/sThus magnitude of velocity at end of 4 sec=[ Vx2 +Vy2]1/2velocity=5 m/s direction is tanθ=Vy / Vxtanθ=4/3direction=tan-1(4/3) Answer:(d)
Q.16
A bomb is fired from a canon with velocity V m/sec at an angle θ with the horizontal direction. At highest point of its path it explodes into two pieces of equal masses. One of the pieces retraces its path, then the speed of the other piece immediately after explosion is
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a) 3Vcosθ
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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 velocity V At maximum height velocity of bomb u'=Vcosθ ∴ momentum at the time of explosion=2mVcosθ Given that one part retrace its path after explosion thus momentum of that part=-mcosθ Now let v' be the velocity of other part , momenum=mv' According to law of conservation of momentum Momentum before explosion=momentum after explosion 2mVcosθ=-mVcosθ + mv' v'=3vcosθ Answer: (a)
Q.17
The equation of projectile is y=√3 x -gx2/2 angle of projection is
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a)tanθ=1/ √3
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b) tanθ=√3
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c)π/2
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d)zero
Explanation
Comparing given equation with standard equation for path of projectile tanθ=√3Answer: (b)
Q.18
A blind person after walking 10 steps in one direction, each of length 80cm, turns randomly to left or right by 90°. After waking of total of 40 steps, the maximum displacement of the person from its starting point can be
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a) 32 m
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b) 16√2 m
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c)8√2 m
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d)0 m
Explanation
Maximum displacement=distance Distance=40×80=3200 cm=32mAnswer: (a)
Q.19
A boat is moving with velocity 3i+4j with respect to ground. The water in the river is moving with velocity -3i-4j with respect to ground. The relative velocity of the boat with respect to water is [ EAMCET 1991]
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a)-6i-8j
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b)3i+3j
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c)6i+8j
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d)3i+8j
Explanation
Let Vbg be the velocity of boat with respect to groundVbg=3i+4jVrg be the velocity of water in river with respect to groundVrg=-3i-4jVbr be the velocity of boat with respect to riverNow Vbg=Vbr+Vrg Vbr=Vbg - Vrg Vbr=3i+4j - (-3i-4j)Vbr=6i+8j Answer:(c)
Q.20
Two particles start to move with velocity u and v, along x-axis and y-axis respectively, from the origin. The acceleration of the first particle with respect to the second is
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a) √(u2+v2)
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b) zero
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c) u+v
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d) none of these
Explanation
Answer: (d)
Q.21
Two like parallel forces of 4 N and 6N acts on a body at a separation of 4m. The resultant will be ..[ UPSC 1994]
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a)2.4 m from 4 N
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b) 2.4m from 6N
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c)2.0 m from 6N
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d)1. m from 4N
Explanation
Let the resultant force at a distance of x from 4N forceThus 4x=6(4-x) ∴ x=2.4 mAnswer: (a)
Q.22
If two non-zero vectors A and B obey the relation A+B=-B , the angle between them is .. [ M.N.R. 1994]
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a) 120°
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b) 90°
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c)60°
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d)0°
Explanation
A+B=-B A=-2BThus Angle is zeroAnswer: (d)
Q.23
The friction of the air causes a vertical retardation of 10% in acceleration due to gravity. The maximum height will decreased by
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a) 11%
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b) 10%
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c)9%
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d)8%
Explanation
Air is giving retardation 10% of gravitational acceleration Thus effective gravitational retardation g'=1.1g Now Max height H ∝ (1/g) Thus H2 / H1=g/g'H2 / H1=g/1.1g=0.909(H2 / H1)×100=0.909×100=91%Thus decrease in height=9% Answer:(c)
Q.24
Given that A + B=C and that C is ⊥ to A. Further if |A|=|C| , then what is the angle between A and B [ CPMT 1998]
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a) π/4 radian
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b) π/2 radian
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c) 3π/4 radian
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d) π radian
Explanation
The vectors must form a right Δ in which A⊥C and |A|=|C| ∴ The other two angles must be 45° each Answer: (a)
Q.25
Component of 3i+4j perpendicular to i+j and in the same plane as (3i+4j) is
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a)(1/2) (j-i)
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b) (3/2) (j-i)
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c)(5/2) (j-i)
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d)(7/2) (j-i)
Explanation
Let A=3i+4j and B=i+j Vector perpendicular to B is B'=i - j Now A⋅B'=|B'|(component of A along B') -1=√2 ×(component of A along B') (component of A along B')=-1 /√2 Thus component of A along B’=(component of A along B')(unit vector B') component of A along B’=(-1 /√2)[i - j)] Thus component of A along B’ =-½ (i - j)=½ (j - i) Answer: (a)
Q.26
If A=iAcosΘ -jAsinΘ be any vector. Another vector B, which is normal to B can be expressed as
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a) BcosΘi+BsinΘj
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b) BcosΘi-BsinΘj
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c)BsinΘi+BcosΘj
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d)BsinΘi - BcosΘj
Explanation
B is perpendicular to A if A⋅B=0. Apply this condition to all the four expressions for B in the given question, the expression (d) will satisfy the above conditionAnswer: (d)
Q.27
If A=B + C and the magnitude of A, B and C are 5, 4 and 3 units respectively. The angle between A and C is ...
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a) cos-1(3/5)
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b) cos-1(4/5)
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c)π/2
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d)sin-1(3/5)
Explanation
A=B + C so B=A - C B2=A2 + C2 - 2ACcosθ42=52 -2×5×3 cosθcosθ=3/5 Answer:(a)
Q.28
During projectile motion quantities that remain unchanged are [ C.P.M.T 1994]
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a) force and vertical velocity
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b) acceleration and horizontal velocity
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c) kinetic energy and acceleration
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d) acceleration and momentum
Explanation
Answer: (b)
Q.29
The simple sum of two forces acting at a point is 16N and there sum is 8N and its direction is perpendicular to the smaller force, then the forces are.. [ CPMT 1997]
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a)6 newton and 10 newton
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b) 8 newton and 8 newton
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c)4 newton and 12 newton
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d)2 newton and 14 newton
Explanation
Let A be the smaller force and B be bigger forceGiven that resultant force R is perpendicular to A thus vector B is hypotenuse of triangle Thus B2=R2 + A2R2=B2 - A2R2=(B+A) (B-A) 82=(16)(B-A) 64 /16=(B-A) ∴ B-A=4 only option (a) satisfy above conditionAnswer: (a)
Q.30
A stuntman plans to run across a roof top and then horizontally off it to land on the roof of the next building. The roof of the next building is 4.9 meter below the first one and 6.2 meter away from it. What should be his minimum roof top speed in m/s, so that he can successfully make the jump [A.M.U.P.M.T 1997]
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a) 3.1
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b) 4.0
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c)4.9
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d)6.2
Explanation
Vertical velocity is zero thus time taken by the stuntman to travel a distance of 4.9 m is 4.9=½ 9.8 t2 t=1Now stunt man has to travel distance of 6.2 m in one sec thus his speed must be 6.2m/sAnswer: (d)
0 h : 0 m : 1 s
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