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Physics NEET MCQ
Quiz 1
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Q.1
A given object takes n times as much times to slide down a 45° rough incline as its takes to slide down a perfectly smooth 45° incline. The coefficient f kinetic friction between the objects and incline is given by.
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$ 1 - \frac {1}{n^2}$
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$\frac { 1}{1- n^2}$
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$ \sqrt {1 - \frac {1}{n^2}}$
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$\sqrt {\frac { 1}{1- n^2}}$
Explanation
Solution: On a smooth incline: a_smooth = g sinθ. On a rough incline: a_rough = g(sinθ − μcosθ). For the same distance, time ∝ 1/√a, so n = t_rough/t_smooth = √(a_smooth/a_rough) = √[sinθ / (sinθ − μcosθ)]. At θ = 45°, sinθ = cosθ, so this simplifies to n = √[1/(1−μ)] → n² = 1/(1−μ) → 1−μ = 1/n² → μ = 1 − 1/n².
Q.2
According to Newton’s third law all forces come in action-reaction pairs which
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may act on the same object.
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always act in the same direction.
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may be at right angles to each other.
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always act on different objects.
Explanation
Newton's third law action-reaction pairs are always equal in magnitude, opposite in direction, and — crucially — always act on two DIFFERENT objects (never both on the same body, which is why they never cancel each other out for a single object).
Q.3
The tension in cable supporting an elevator is equal to the weight of the elevator’s elevator may be (more than one correct)
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going up with increasing speed
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going down with increasing speed
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going up with uniform speed
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going down with uniform speed.correct
Explanation
This question notes "(more than one correct)", and indeed both cases are correct: if the elevator moves at CONSTANT (uniform) speed — whether going up or going down — its acceleration is zero, so by Newton's second law the net force on it is zero, meaning cable tension exactly equals its weight. It's only when the speed is changing (increasing or decreasing) that tension differs from weight.
Q.4
When train stops, the passenger move forward. It is due to
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inertia of passenger
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inertia of train
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gravitational pull by earth
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None of the above
Explanation
By inertia, a passenger's body tends to keep moving at the speed the train was travelling even after the train itself decelerates and stops — so relative to the now-stopped train, the passenger's body continues forward.
Q.5
By means of rope ,a body of weight W is moved vertically upward with constant acceleration a.Find the tensile force in the rope
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$W(1+ \frac {a}{g})$
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$W(1- \frac {a}{g})$
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W
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$\frac {Wa}{g}$
Explanation
Using Newton's second law (mass = W/g since weight W = mg): T − W = (W/g) × a → T = W + (W/g)a = W(1 + a/g).
Q.6
A rockets works on the principle of conservation of
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Linear momentum
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Mass
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Energy
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angular momentum
Explanation
A rocket expels exhaust gases backward at high speed; by conservation of linear momentum, the rocket itself gains an equal and opposite forward momentum — this is the principle a rocket works on.
Q.7
A smooth wedge A is fitted in a chamber hanging from a fixed ceiling near the earth’s surface. A block B placed at the top of the wedge takes time T to slide down the length of the wedge and the cable supporting the chamber is broken ath the same instant, the block will be
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take a time longer than T to slide down the wedge
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take a time shorter than T to slide down the wedge
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jump off the wedge
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remain at the top of the wedge
Explanation
The instant the cable breaks, the entire chamber — wedge, block, and all — enters free fall. In free fall, everything inside accelerates downward together at g, which is equivalent to a state of effective weightlessness inside the chamber: there is no longer any net force pushing the block down the incline relative to the wedge. So the block simply falls together with the wedge, without sliding — it stays at the same spot on the wedge.
Q.8
A IITJEE text book of mass M rests flat on a horizontal table of mass m placed on the ground. Let RX->Y be the constant force exerted by the body x on body Y.Which of the following is true for the forces Rground->table and Rtable->ground (More than one correct)
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It is action and reaction pair
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have equal magnitudecorrect
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Opposite directioncorrect
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Have resultant zerocorrect
Explanation
The ground pushing up on the table, and the table pressing down on the ground, are a genuine Newton's third law action-reaction pair — so all of the following hold true simultaneously: they are an action-reaction pair, they have equal magnitude, they point in opposite directions, and (added together as vectors) their resultant is zero.
Q.9
Assertion and Reason (a) If both assertion and reason are true and reason is the correct explanation of the assertion. (b) If both assertion and reason are true but reason is not correct explanation of the assertion. (c) If assertion is true, but reason is false. (d) If both assertion and reason are false. (e) If reason is true but assertion is false. Assertion: On a rainy day, it is difficult to drive a car or bus at high speed. Reason: The value of coefficient of friction is lowered on wetting surface.
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a
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b
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c
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d
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e
Explanation
On a wet road, water reduces the coefficient of friction between the tyres and the road surface, cutting down on the grip/traction available for braking and steering — which is exactly why high-speed driving becomes risky and difficult in rain. The Reason directly explains the Assertion.
Q.10
The coefficient of static and kinetic friction between a body and the surface are .75 and .50 respectively. A force is applied to the body to make it just slide with a constant acceleration which is
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g/4
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g/2
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3g/4
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g
Explanation
The applied force needed to just start the body sliding must overcome the maximum static friction: F = μs × mg = 0.75mg. Once sliding begins, that same force F continues to act, but now it's opposed by the (smaller) kinetic friction: μk × mg = 0.50mg. Net force = F − kinetic friction = 0.75mg − 0.50mg = 0.25mg = mg/4. Acceleration = net force / m = g/4.
Q.11
(a) If both assertion and reason are true and reason is thecorrect explanation of the assertion. (b) If both assertion and reason are true but reason is not correct explanation of the assertion. (c) If assertion is true, but reason is false. (d) If both assertion and reason are false. (e) If reason is true but assertion is false. Assertion: Linear momentum of a body changes even when it is moving uniformly in a circle. Reason: In uniform circular motion velocity remains constant.
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a
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b
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c
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d
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e
Explanation
The Assertion is true — even though the speed stays constant in uniform circular motion, the direction of the velocity is continuously changing, so the momentum vector (mass × velocity) keeps changing too. But the Reason is false: velocity is a vector, and in uniform circular motion only its magnitude (speed) stays constant — its direction is always changing, so the velocity itself is NOT constant.
Q.12
A skydiver jumps from a high-flying plane. When she reaches her terminal velocity
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the force of gravity on her is zero.
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her acceleration has reached magnitude g.
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the drag force of the air on her has become zero.
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the net force on her is zero.
Explanation
At terminal velocity, the skydiver's velocity is constant, so her acceleration is zero — by Newton's second law, that means the net force acting on her is zero (gravity pulling down is exactly balanced by air drag pushing up, not that either one individually vanishes).
Q.13
A block of mass m is placed on a smooth wedge of inclination $\theta$. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block will be (g is acceleration due to gravity)
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$mgcos \theta$
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mg
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$mg sin \theta$
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$\frac {mg}{cos \theta}$
Explanation
On the smooth (frictionless) wedge, the only forces on the block are gravity (mg, downward) and the normal force N (perpendicular to the incline). For the block to stay in place on the incline while the whole system accelerates, the VERTICAL forces alone must balance (since friction, which could otherwise help vertically, is absent): N cosθ = mg → N = mg/cosθ.
Q.14
A hockey player is moving northward and suddenly turns westward with the same speed to avoid an opponent. The force that acts on the player is
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frictional force along westward.
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muscle force along southward.
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frictional force along south-west
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muscle force along south-west.
Explanation
The player's velocity changes from v northward to v westward — the CHANGE in velocity (and hence the required external force) points diagonally between "reverse of north" and "west", i.e. towards the south-west. Since the player is running on the ground, this external force comes from the friction between his feet and the ground (a runner changes direction only through this ground-reaction friction force, since muscular force alone is internal to the body and can't change the body's total momentum without an external push from the ground).
Q.15
A uniform chain of length L is lying on the horizontal surface of a table. If the coefficient of friction between the chain and the tabletop is μ what is the maximum length of the chain that can hang over the edge of the table without disturbing the rest of the chain on table
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$\frac {L}{(1+\mu)}$
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$\frac {\mu L}{(1+\mu)}$
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$\frac {L}{(1-\mu)}$
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$\frac {\mu L}{(1-\mu)}$
Explanation
Let the hanging length be x (out of total length L), with weight per unit length w = W_total/L. The hanging part's weight = (x/L)W_total pulls the chain down; the friction available from the part still on the table = μ × (weight of that part) = μ(L−x)/L × W_total. At the point where the chain is just about to slide, these balance: μ(L−x) = x → μL − μx = x → μL = x(1+μ) → x = μL/(1+μ).
Q.16
If the normal force is doubled, the coefficient of friction, is
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doubled
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tripled
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halved
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not changed
Explanation
The coefficient of friction μ = (friction force)/(normal force) is a ratio characteristic of the two surfaces in contact. While the actual friction FORCE does scale up if the normal force doubles, the RATIO μ itself stays the same — it doesn't change with normal force.
Q.17
A flat car of weight W roll without resistance along on a horizontal track .Initially the car together with weight w is moving to the right with speed v.What increment of the velocity car will obtain if man runs with speed relative to the floor of the car and jumps off at the left?
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$\frac {wu}{w+W}$
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$\frac {Wu}{W+w}$
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$\frac {(W+w)u}{w}$
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$\frac {(W+w)u}{2w}$
Explanation
Using conservation of momentum (masses W/g and w/g, frictionless track). If the car's new velocity is V, and the man (jumping off with speed u relative to the car) then has ground velocity (V − u): (W/g)v + (w/g)v = (W/g)V + (w/g)(V − u) (W+w)v = (W+w)V − wu V = v + wu/(W+w) So the car's velocity INCREASES by wu/(W+w).
Q.18
A car is moving with uniform velocity on a rough horizontal road. Therefore, according to Newton's first law of motion
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No force is being applied by its engine
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An acceleration is being produced in the car
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The kinetic energy of the car is increasing
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force is surely being applied by its engine
Explanation
The road is rough, meaning friction opposes the car's motion. For the car to keep moving at a constant (uniform) velocity despite this opposing friction, Newton's first law requires the NET force to still be zero — which means the engine must be supplying a forward force exactly equal and opposite to the friction force. So the engine is definitely applying force, even though the car's net force (and hence its velocity) doesn't change.
Q.19
A coin is dropped in a lift. It takes time a to reach the floor, when lift is stationary. It takes time b, when lift is moving up with constant acceleration. Then,
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a > b
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b > a
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a=b
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none of the above
Explanation
Time to fall a given height is t = √(2h/g_eff). When the lift accelerates upward, the effective gravity felt inside it increases to (g + a), making objects fall "faster" relative to the lift floor — so it takes LESS time to reach the floor when the lift is accelerating upward than when it's stationary. So a (stationary case) is greater than b (accelerating upward case): a > b.
Q.20
A reference frame attached to the earth (more than one correct)
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is an inertial frame by definition
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Cannot be inertial frame as earth is resolving around the sun
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is an inertial frame because Newton's law are applicable in this frame
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Cannot be inertial frame because earth is rotating about its own axiscorrect
Explanation
This question notes "(more than one correct)", and indeed a reference frame attached to the Earth is NOT truly inertial for two separate reasons: the Earth revolves around the Sun (introducing an orbital/centripetal acceleration), AND the Earth rotates about its own axis (introducing its own centripetal acceleration due to that spin) — either effect alone is already enough to make the Earth's frame only approximately, not exactly, inertial.
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