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# Question Particle A is moving in a horizontal plane with constant velocity V as shown. Another particle B is moving in a circle of radius 1 m with same speed V. At the moment when A is diametrically opposite to B, find the radius of curvature (in m) of B as seen by A at this moment.

#### 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

Solution:

Explanation:

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

Solution:

Explanation:

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

Solution:

Explanation:

The twwo waves can be written as

y1=asin(ωt+kx+0.57)

y1=acos(ωt+kx)

y1=asin(ωt+kx+0.57+π/2)

So the phase difference is

ϕ=π/2-0.57

⇒ϕ=1.57-0.57

⇒ϕ=1.00

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

Solution:

Explanation:

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

1/6=1-Tsink/Tsource

⇒Tsink/Tsource=5/6

We can also write

2/6=1-Tsink-62/Tsource

⇒6/2Tsource=1/6

⇒Tsource=372K=99ºC

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

Solution:

Explanation:

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.