Question of Exercise 2

Laws Of Motion of Class 11

A particle is projected with a speed of 100 m/s at angle θ = 60° with the horizontal at time
t = 0. At time t the velocity vector of the particle becomes perpendicular to the direction of velocity of projection. Its tangential acceleration at time t is


Option 1 10 m/s2

Option 2 m/s2

Option 3 5 m/s2

Option 4 zero

Frequently Asked Questions

A particle executes simple harmonic oscillation with an amplitude a

The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is

A: T/4

B: T/8

C: T/12

D: T/2





Equation of a SHM is given by X=Asin(ωt)


Final Answer: 

Option C is correct


The inertia of an object tends to cause the object

A: to increase its speed

B: to decrease its speed

C: to resist any change in its state of motion

D: to decelerate due to friction




The inertia is the physical property of a body of an object tends to cause the object to resist any change in its state of rest or motion. The inertia is directly dependent on the mass of the body. The object will stay at rest if it is at rest or will stay in motion if the object is already in motion.

Final Answer:

Option C is correct



Two waves are represented by the equations y1

y1 = asin ( ωt + kx + 0.57 )m and y2 = acos ( ωt + kx )m, where x is in metre and t in second. The phase difference between them is

A: 0.57 radian

B: 1.0 radian

C: 1.25 radian

D: 1.57 radian




The twwo waves can be written as 




So the phase difference is 




Final Answer:

The correct option is (B).


A Carnot engine has an efficiency of 1/6

When the temperature of the sink is reduced by 62ºC , its efficiency is doubled. The temperatures of the source and the sink are, respectively

A: 124ºC  ,62ºC

B: 37ºC  ,99ºC

C: 62ºC,124ºC

D: 99ºC  ,37ºC




The efficiency of the carnot’s engine can be given by



We can also write




Now Tsink=5/6×372K=310K=37ºC

Final answer:

The correct option is (D).⇒ ( Tsource=372K=99ºC

Now Tsink=5/6×372K=310K=37ºC ) 

The value of acceleration due to gravity

A: is same on equator and poles

B: is least on poles

C: is least on equator

D: increases from pole to equator




The acceleration can be given by the formula

g=G M/R2 where g= acceleration,G is the gravitional constant. M is the mass of the earth.R is the distance of the surface and the centre of earth.

The acceleration is less at the equator because the distance between the surface of the earth and the centre is more than the pole .

Final Answer:

The value of the acceleration due to gravity is least on equator.

So option (C ) is correct.



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