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Jee Main Advanced Physics Chapter Wise Mock Test
Quiz 2
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Q.1
A carpenter hits a nail with a hammer.which of the following is true for collision between nail and hammer?
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There is a force on hammer only
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There is force on hammer and nail both
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There is force on nail only
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Insufficient information
Explanation
Solution:
Newton's third law: whenever the hammer exerts a force on the nail (driving it in), the nail simultaneously exerts an equal and opposite force back on the hammer (which is why swinging a hammer hard enough against a nail can jar your hand) -- the force acts on both, not just one.
Q.2
Which one of the following equations is 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) M => is a Mass (M) F => is a force (ML/T2) p => is a momentum ( ML/T)
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$p=mv-Ft$
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$p^2=\frac {F}{M}$
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$v^2=2ax+ut$
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$v=at+ \frac {F}{m}$
Explanation
Solution:
mv has dimension [M][L/T] = [ML/T], exactly momentum's own dimension. Ft has dimension [ML/T²][T] = [ML/T], the same. So p=mv-Ft has matching dimensions on every term -- it's dimensionally consistent, even though it isn't a standard physics formula on its own.
Q.3
A plane traveling horizontally to the right at 100 m/s flies past a helicopter that is going straight up at 20 m/s. From the helicopter's perspective, the plane's direction and speed are?
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Right and down, 100 m/s
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Right and down, more than 100 m/s
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Right and down, less than 100 m/s
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none of these
Explanation
Solution:
Relative to the helicopter, the plane's velocity is the plane's velocity minus the helicopter's: (100, 0) - (0, 20) = (100, -20) in (horizontal, vertical) components -- still moving right, but now also appearing to move downward from the helicopter's point of view.
Its apparent speed is √(100²+20²) = √10400 ≈ 102 m/s, which is more than the plane's actual 100 m/s ground speed.
Q.4
which of these statement is not true?
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Momentum is product of mass and velocity
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If a net force acts on the systems,it momentum will change
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A heavy object will always have higher momentum than lighter object
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Impulse is the product of Force and time applied
Explanation
Solution:
Momentum being mass times velocity, changing when a net force acts, and impulse being force times time are all standard, correct definitions.
But momentum depends on BOTH mass and velocity together -- a light, fast-moving object can easily have more momentum than a heavy, slow one, so a heavier object doesn't automatically have more momentum.
Q.5
An open truck rolls along a friction-less track while it is raining. As it rolls, what happens to the speed of the truck as the rain collects in it? (assume that the rain falls vertically into the box) ?
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it increase
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it decrease
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remains same
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Insufficient information
Explanation
Solution:
Since the rain falls straight down, it carries no horizontal momentum of its own when it lands in the truck -- and with no friction, there's no external horizontal force on the truck+rain system either. So the system's total horizontal momentum stays fixed at its original value (mass × original speed) even as the total mass grows. Since momentum (mass × velocity) has to stay constant while mass increases, the truck's speed must actually decrease as it collects more rain -- the same fixed momentum has to be shared across a growing mass.
Q.6
Which of these is false?
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Momentum is conserved in elastic collision only not in inelastic collision
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When net force on the system is zero, Total momentum of the system remains conserved
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When two objects collides and completely bounce back with no deformation and no generation of heat,the collision is said to be elastic
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None of the above
Explanation
Solution:
Momentum is conserved in EVERY collision with no external force -- elastic or inelastic alike. What's special about an elastic collision is that kinetic energy is also conserved, not just momentum -- claiming momentum conservation is exclusive to elastic collisions mixes up these two separate conservation laws, which is why that's the false statement.
Q.7
A bomb that is stationary sitting at the origin on the xyz plane explodes into pieces of different sizes and shapes. ?
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After the explosion the momentum of all the pieces, exhaust and smoke add up to zero
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After the explosion the momentum of all the pieces, exhaust and smoke add up to non zero
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Total momentum is increased an energy is released by the explosion
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none of the above
Explanation
Solution:
Before the explosion, the bomb is stationary, so the total momentum of the system is zero. The explosion is entirely an internal force (pieces pushing off each other) -- with no external force involved, total momentum can't change, so it's still zero afterward, however the individual pieces end up moving.
Q.8
Momentum of the objects tells us about?
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object mass
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object velocity
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how difficult it to stop it
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object weight
Explanation
Solution:
Momentum measures an object's "quantity of motion" -- an object with more momentum requires a proportionally larger impulse (force applied over time) to bring it to rest, so momentum is really a measure of how hard an object is to stop, not simply its mass or its speed on their own.
Q.9
A 5 g rubber ball and a 5 g clay ball are thrown at a wall with equal speeds. The rubber ball bounces, the clay ball sticks. Which ball exerts a larger impulse on the wall?
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They exert equal impulses because they have equal momenta
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The clay ball exerts a larger impulse because it sticks
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The rubber ball exerts a larger impulse because it bounces.
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Neither exerts an impulse on the wall because the wall doesn't move
Explanation
Solution:
Impulse equals the change in momentum, and by Newton's third law, the impulse on the wall matches the impulse on the ball.
The clay ball sticks: its velocity goes from v to 0, a change of magnitude mv. The rubber ball bounces back: its velocity goes from v to roughly -v, a change of magnitude 2mv -- twice as large. So the bouncing rubber ball delivers the bigger impulse to the wall.
Q.10
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) M => is a Mass (M) F => is a force (ML/T2) p => is a momentum ( ML/T)
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p1i+p2i=p1f+p2f
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$F=\frac {d(mv)}{dt}$
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$m(dv)+v(dm)=F$
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none of these
Explanation
Solution:
p
1i
+p
2i
=p
1f
+p
2f
just equates momentum with momentum on both sides -- consistent. F=d(mv)/dt is force equated with (momentum)/(time) = [ML/T]/[T] = [ML/T²], matching force's own dimension -- consistent.
m(dv)+v(dm), however, has no /dt anywhere in it -- as written, both terms carry the dimension of momentum, [ML/T], not force's [ML/T²]. Setting that expression equal to F is missing a factor of 1/time compared to the (correct) product-rule form m(dv/dt)+v(dm/dt)=F.
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