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Q1. Select the pair of quantities which are introduced by Newton’s first law of motion
Answer: Force and Inertia
Q2. A cyclist taking a curve leans inwards. This is due to ………….
Answer: Provide the necessary centripetal force
Answer: 3.2 m/s2
Hint:
Q4. The force of friction acts.....
Answer: Tangential to the surface of contact
Q5. When a constant force acts on a body it moves with .....
Answer:Uniform acceleration
Q6. Walking is related to.....
Answer: Newton’s third law of motion
Q7. The net force acting on a car of mass ‘M’ kg moving with a constant velocity ‘v’ m/s on a rough road is ......
Answer: Zero
Hint : Constant velocity means zero acceleration
Q8. A force of 10 N acts on a body for 0.3 seconds. The impulse is…..
Answer: 3 Ns
Hint: Impulse = Force x time
Q9. What is the unit of "coefficient of static friction" ?
Answer: No unit
Q10. A lift of mass (m) moving up with acceleration (a)
. Then tension on the rope of the lift is....
Answer: T = mg + ma
Dear Student, Read question and answers
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| Sameena C S |
| GHSS |
| ELAMKUNNAPUZHA |
Q1. Select the pair of quantities which are introduced by Newton’s first law of motion
Answer: Force and Inertia
Q2. A cyclist taking a curve leans inwards. This is due to ………….
Answer: Provide the necessary centripetal force
Q3. A body is
moving on a circular path of 10m radius. At any instant of time ,
its speed is 5m/s and the speed is increasing at a rate of 2 m/s2
. The magnitude of net acceleration at any instant is
Answer: 3.2 m/s2
Hint:
Here
r = 10m , v = 5m/s , at
=2 m/s2
, ar
=
v2/r
= 5x5/10 = 2.5 m/s2
The
net acceleration is
a = (ar2+at2)1/2 = ( 2.52+ 22)1/2 = (10.25)1/2 = 3.2 m/s2
a = (ar2+at2)1/2 = ( 2.52+ 22)1/2 = (10.25)1/2 = 3.2 m/s2
Q4. The force of friction acts.....
Answer: Tangential to the surface of contact
Q5. When a constant force acts on a body it moves with .....
Answer:Uniform acceleration
Q6. Walking is related to.....
Answer: Newton’s third law of motion
Q7. The net force acting on a car of mass ‘M’ kg moving with a constant velocity ‘v’ m/s on a rough road is ......
Answer: Zero
Hint : Constant velocity means zero acceleration
Q8. A force of 10 N acts on a body for 0.3 seconds. The impulse is…..
Answer: 3 Ns
Hint: Impulse = Force x time
Q9. What is the unit of "coefficient of static friction" ?
Answer: No unit
Q10. A lift of mass (m) moving up with acceleration (a)
. Then tension on the rope of the lift is....
Answer: T = mg + ma



I dont understand the 3rd question
ReplyDeletea(r) is radial and a(t) is tangential acceleration
Deletea = (ar2+at2)1/2 = is the magnitude resultant linear resultant acceleration, since a(r) and a(t) are acting perpendicular to each other.. more doubt?..please ask
No hint for tangential acceleration in the question
ReplyDeleteI also don't understand the 3rd question
ReplyDeleteI also didn't understood the 3rd question
ReplyDeleteI dont understand the 10 th question
ReplyDeleteI dont understand the 10 th question
ReplyDelete