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Jee Main Advanced Physics Chapter Wise Mock Test
Quiz 1
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Q.1
A car goes around a curve of radius r at speed v and experiences a centripetal acceleration a. If the car is to go around the same curve at a speed 4v, the required centripetal acceleration is?
0%
a
0%
4a
0%
16a
0%
2a
Explanation
Solution:
Centripetal acceleration is a = v²/r. Since a is proportional to the SQUARE of the speed, going 4 times faster around the same-radius curve multiplies the required acceleration by 4² = 16.
Q.2
A particle initially located at the origin has an acceleration of a=.3j m/s2 and an initial velocity of v=.5i m/s. Find the vector position and velocity at time t=2 sec?
0%
(.5i+6j),(i+.6j)
0%
(i+.1j),(.5i+6j)
0%
(i+.6j),(.5i+6j)
0%
None of these
Explanation
Solution:
Position: x = x
0
+ v
0
t + ½at² = 0 + (0.5i)(2) + ½(0.3j)(2²) = i + 0.6j.
Velocity: v = v
0
+ at = 0.5i + (0.3j)(2) = 0.5i + 0.6j.
Q.3
A small car had a head-on collision with bigger truck. The impulse exerted by the truck on car
0%
is less than impulse exerted by car on truck
0%
is equal to impulse exerted by car on truck
0%
is greater to impulse exerted by car on truck
0%
Insufficient information
Explanation
Solution:
By Newton's third law, whenever two objects collide, the force (and hence the impulse, force × the same shared contact time) that one exerts on the other is always equal in magnitude and opposite in direction to the force the second exerts back on the first -- true no matter how different their masses are.
Q.4
Two boxes P and Q, having mass m1 and m2 ( m1 > m2) are initially at rest on a horizontal friction-less surface. The same constant force F acts on each one for exactly 5 second. Which box has more momentum after the force acts ??
0%
P
0%
Q
0%
Both have same momentum
0%
insufficient information
Explanation
Solution:
By the impulse-momentum theorem, the change in momentum a box receives is exactly (force) × (time) = F × 5s -- which depends only on the shared force and shared time, not on mass at all. Since F and the duration are identical for both boxes and both start at rest, both end up with the same momentum (though, since p=mv, the lighter box Q ends up moving faster than the heavier box P).
Q.5
Robert (120 kg) and Bill (60 kg) are standing on slippery ice and push off each other. If Bill slides at 6 m/s, what speed does Robert have have?
0%
12 m/s
0%
6 m/s
0%
8 m/s
0%
3 m/s
Explanation
Solution:
Starting at rest, their total momentum is zero, so after pushing off, their momenta must be equal and opposite: m
Robert
v
Robert
= m
Bill
v
Bill
.
120 × v
Robert
= 60 × 6 = 360, so v
Robert
= 3 m/s.
Q.6
A man is standing on a weigh scale in an elevator. When the elevator is accelerating upward with constant acceleration a, the scale reads 867.0 N. When the elevator is accelerating downwards with the same constant acceleration a, the scale reads 604.5 N. Determine the magnitude of the acceleration a and the mass of the man?
0%
1.15 m/s2 ,75 kg
0%
1.25 m/s2,95 kg
0%
1.75 m/s2,85 kg
0%
1.75 m/s2,75 kg
Explanation
Solution:
Going up: the scale reading (normal force) exceeds the man's weight by the force needed to accelerate him upward: N = m(g+a) = 867.0 N.
Going down: the scale reads less than his weight: N' = m(g-a) = 604.5 N.
Adding the two equations: 2mg = 1471.5, so m = 1471.5 ÷ (2×9.8) ≈ 75 kg.
Subtracting them: 2ma = 867.0-604.5 = 262.5, so a = 262.5 ÷ (2×75) = 1.75 m/s².
Q.7
A canon fires ball at an angles 400.The ball would have landed at the same place if it were fired at any angle
0%
30 °
0%
50°
0%
90°
0%
None of these
Explanation
Solution:
Range depends on the launch angle through R = u²sin(2θ)/g, and sin(2θ) is the same for θ and its complement (90°-θ), since sin(2(90°-θ)) = sin(180°-2θ) = sin(2θ).
So firing at 40° gives the same range as firing at its complement, 90°-40° = 50°.
Q.8
Which one of the following equations is NOT dimensionally correct? Following things are given v => is a speed (L/T) a => is an acceleration (L/T2) x => is a distance (L) t => is a time (T)
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$x=at^2$
0%
$v=v_0+ax$
0%
$v^2=2ax$
0%
$v=at$
Explanation
Solution:
x = at²: [L/T²]×[T²] = [L], matching x's own dimension -- fine (the physics formula is missing the usual ½, but that doesn't affect dimensions).
v²=2ax: [L/T]² = [L/T²]×[L] = [L²/T²] on both sides -- fine. v=at: [L/T]=[L/T²]×[T]=[L/T] -- fine.
v=v
0
+ax, however, adds a speed [L/T] to the product ax = [L/T²]×[L] = [L²/T²] -- two DIFFERENT dimensions being added together, which is never valid. That's the dimensionally inconsistent one.
Q.9
Scalar product of two vector P and Q is zero.?
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P and Q are parallel
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either P = 0 or Q = 0
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P is perpendicular to Q
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either P = 0 or Q = 0 or P is perpendicular to Q
Explanation
Solution:
P·Q = |P||Q|cosθ. This is zero if either vector has zero magnitude (so there's nothing to the product regardless of angle), OR if neither is zero but cosθ=0, i.e. they're perpendicular. Both possibilities are covered by "either P=0 or Q=0 or P is perpendicular to Q."
Q.10
We knew that pushing heavy furniture across the carpet usually takes more force to get it moving than it takes to keep it moving. This is because
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the coefficient of static friction is greater than the coefficient of kinetic friction.
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It is due to law of inertia
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the coefficient of kinetic friction is greater than the coefficient of static friction
0%
None of these
Explanation
Solution:
Static friction resists the very first push that gets an object moving; kinetic friction resists it once it's already sliding. For most surfaces the coefficient of static friction is larger than the coefficient of kinetic friction, which is exactly why the initial push to start motion takes more force than sustaining it afterward.
Q.11
Ram drives from Rampur to hapur. Even without knowing anything about the roads between these two cities we can say with certainty that the magnitude of the Ram displacement from rampur when he reaches hapur is?
0%
always less than the total distance Ram traveled.
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always equal to the total distance ram traveled
0%
less than or equal to the total distance he traveled
0%
cannot be determined
Explanation
Solution:
Displacement is the straight-line distance between start and end points, while distance traveled adds up the actual path length -- which can only ever be the same as (for a perfectly straight, one-directional route) or more than (for any route with turns or curves) the displacement. So displacement is always less than or equal to distance traveled, and we can say this without knowing anything about the actual roads.
Q.12
Lets take the motion of the particle along x axis .Let the particle move in the positive direction, the slope of the velocity verses time graph will be negative when?
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velocity decreases with time
0%
velocity increases with time.
0%
the acceleration increases with time
0%
the acceleration decreases with time
Explanation
Solution:
The slope of a velocity-time graph, by definition, is the rate at which velocity is changing -- so a negative slope directly means the velocity is decreasing over time (whether or not the particle's direction or the sign of the acceleration itself is changing).
Q.13
A object is accelerated by the constant force F.Suddenly after some time, it is acted by a second force F with opposite direction to first force. With two force acting on it,the object ?
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it will come to halt in some time
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it comes to halt immediately
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continues with the same velocity which it has when second force start acting
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none of the above
Explanation
Solution:
Once the second force (equal in magnitude, opposite in direction to the first) starts acting, the two forces exactly cancel, leaving zero net force. By Newton's first law, zero net force means zero further acceleration -- the object simply keeps moving at whatever velocity it had at that moment, neither speeding up, slowing down, nor stopping.
Q.14
Two hot wheels toy cars A and B are released simultaneously on an inclined plane which is at an angle of 40° to the horizontal. A has a mass of m kg and the B a mass of 2 m kg. We assumed that friction and air drag are absent. which statement best describes the magnitudes of their accelerations after being released.?
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Both hot wheels cars accelerate at a rate of gsin50°
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Both hot wheels cars accelerate at a rate of g cos40°
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Car B acceleration is 8 times greater than the car A
0%
None of the above
Explanation
Solution:
On a frictionless incline, Newton's second law along the slope gives mg sinθ = ma, and the mass m cancels out of both sides -- so EVERY object, regardless of mass, accelerates down a frictionless incline at the same rate, g sinθ (here, g sin40°, not g sin50° or g cos40°). Since neither of the two specific formulas offered is actually g sin40°, and the accelerations of A and B are in fact equal (not 8× different), none of the given statements correctly describes what happens.
Q.15
A small car had a head-on collision with bigger truck. Which will have maximum change in velocity
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Truck
0%
car
0%
Both same
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Insufficient information
Explanation
Solution:
By Newton's third law, the car and the truck experience forces of equal magnitude for the same duration during the collision, so their momentum changes (impulses) are equal in size. Change in velocity is that impulse divided by mass -- since the car has much less mass than the truck, the same impulse produces a much bigger change in velocity for the car.
Q.16
A car accelerates on the road. Viewed from outside the car,what is the force that accelerates it?
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the wheel are pushing the car.
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Engine is pulling it forward
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Gravitational force
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The road pushes the car forward
Explanation
Solution:
The engine can only push against parts of the car itself, and internal forces can't accelerate the car as a whole -- something external has to. The tyres push backward against the road (via friction), and by Newton's third law the road pushes the car forward with an equal and opposite force -- that external push from the road is what actually accelerates the car.
Q.17
Two ball A and B are at the same height . Ball A is dropped while ball B is fired horizontally at same time. Which statement does not the describe the motion?
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The ball B has a larger net velocity when it hits the ground
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Both ball A and B hit the ground at the same time
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The ball B has greater vertical velocity than A when it hits the ground
0%
The ball A is having null horizontal velocity
Explanation
Solution:
Both balls start falling from the same height with zero vertical velocity (A is simply dropped, and B's launch velocity is purely horizontal, so it also starts with zero vertical component) -- since horizontal and vertical motion are independent, both experience identical vertical motion under gravity, hitting the ground at the same time with equal vertical velocities.
So the two balls reach the ground with the SAME vertical velocity, not different ones -- B only has an additional horizontal velocity on top of that, which is what makes its total (net) speed at impact larger, and why A alone has zero horizontal velocity throughout.
Q.18
An astronaut on planet A kicks a bowling ball and hurts his foot.A year later, the same astronaut kicks a bowling ball on the planet B with the same force. It is given gravitational pull of planet A is more than planet B?
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His foot will be hurt as same on planet B as planet A
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His foot will be hurt less on planet B as planet A
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His foot will be hurt greater on planet B as planet A
0%
insufficient information
Explanation
Solution:
The pain in the astronaut's foot comes from the reaction force the ball exerts back on it during the kick (Newton's third law) -- and that reaction is set by however hard the astronaut kicks, not by the local gravity. Since the same force was used both times, the foot takes the same reaction force, and hurts the same amount, on either planet.
Q.19
A person is attracted toward the earth by the force F, The force with which person attract the earth is ?
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slighting greater than F
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slighting lower than F
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F
0%
very very small
Explanation
Solution:
By Newton's third law, every force comes as an equal-and-opposite pair: if the Earth pulls the person down with force F, the person pulls the Earth back up with exactly the same magnitude of force, F -- just far too small to noticeably move something as massive as the Earth.
Q.20
A truck is traveling over the crest of a small semi-circular hill of Diameter D = 1600 m. How fast would it have to be traveling for it to leave the ground? Take g=10 m/s2
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85.46 m/s
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89.45 m/s
0%
90 m/s
0%
none of these
Explanation
Solution:
The truck leaves the ground exactly when gravity alone is just enough to supply the centripetal force needed to keep it on the circular crest, i.e. when the normal force drops to zero: mg = mv²/r, so v=√(gr).
With diameter 1600 m, the radius is r=800 m, so v=√(10×800)=√8000 ≈ 89.4 m/s.
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