Lakshya Education MCQs

Question: A car moves on a circular path of radius 10 metre. It complete 5 revolution in 5 minute. What is its average speed?
Options:
A.10 π m / minute
B.20 π m / minute
C.30 π m / minute
D.5 π m / minute
Answer: Option B
: B

The linear speed is, v = r ω ;ω = θt Substituting, we get, v=20 π m / minute

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Question 1.  

A particle moves along a circular path over a horizontal xy coordinate system, at constant speed. At time t1 = 5.00 s, it is at point (5.00 m, 6.00 m) with velocity (3π m/s)^j  and acceleration in the positive x direction. At time t2 = 10.0 s, it has velocity (-3π m/s)^i and acceleration in the positive y direction. What are the (a) x and (b) y coordinates of the center of the circular path if t2 - t1 is less than one period?
  1.    (15 m, 6 m)
  2.    (-5 m, 6)
  3.    (15 m, 0)
  4.    (-5 m, 0)
Answer: Option A
: A

We know the body is undergoing uniform circular motion,

only possible path is
As acceleration should always be directed toward the centre To calculate the position of the centre, we need the radius, to calculate radius,v=ωr we can use the relation,as we can calculate ω
Δθ=3π2,Δt=5s
=3π2×5=3π10S1
usingv=ωr
r=vω
|v|=3πm/sgiven
r=3π3π10=30π3π=10m
ifr=10m, coordinates of the centre (15 m,6m)
Question 2.  

Assume that the earth goes round the sun in a circular orbit with a constant speed of 30 km/s. Then,
  1.    The average velocity of the earth from 1st Jan, 90 to 30th June, 90 is zero
  2.    The average acceleration during the above period is 60 km/s2
  3.    The average speed from 1st Jan, 90 to 31st Dec, 90 is zero
  4.    The instantaneous acceleration of the earth points towards the sun.
Answer: Option D
: D


(a) we can see there is a displacement from A to B . average velocity cannot be zero
(b) aavg=v2v1t=30^i(30^i)6months=606×30×3600km/s2 which is already not 60 km/s2
(c) average speed is not zero as total distance covered is not zero
(d) The instantaneous acceleration will point to the sun as the above described motion is uniform circular motion with sun at the centre of the circular path.
Question 3. Consider two masses m1 and m2 are moving in circles of radii r1 and r2 respectively. Their speeds are such that they complete circular motion in the same time t. The ratio of their untripetal acceleration is,
  1.    m1 : m2
  2.    r1 : r2
  3.    1 : 1
  4.    m1 r1   : m2r2
Answer: Option B
: B

Centripetal acceleration, a = w2 r = (2πT)2. r a1a2 = r1r2
Question 4.  

A circular disc is rotating with constant angular velocity ω about an axis that passes through the centre. A particle 'P' is kept at a distance of 2m from the centre and another particle 'Q' at a distance of 3m from the centre.

Which of these will have higher centripetal acceleration?
  1.    P
  2.    Q
  3.    both have equal
  4.    can't say until value of ω is given
Answer: Option B
: B

If a particle goes in a circle with constant ω then it has a centripetal acceleration given by ω2R since ω is constant for both particles. So the acceleration depends on R RQ>RP
ω2RQ>ω2RP So particle q will have higher centripetal acceleration.
Question 5. Two cars having masses m1 and m2 move in circles of radii r1 and r2 respectively. If they complete the circle in equal time, the ratio of their angular speeds ω1ω2 is _______
  1.    m1m2
  2.    r1r2
  3.    m1r1m2r2
  4.    1
Answer: Option D
: D

ω1=θt If time taken to complete a revolution is t. Then ω1=2πt,ω2=2πt t=2πω1=2πω2 we get ω1=ω2orω1ω2=1
Question 6.  

At t1 = 2.00 s, the acceleration of a particle in counter clockwise circular motion is  It moves at constant speed. At time t2 = 5.00 s, the particle's acceleration is  What is the radius of the path taken by the particle if t2 - t1 is less than one period?                                                     
  1.    40π2
  2.    360π2
  3.    30π2
  4.    60π2
Answer: Option A
: A

Att1=2s,a=(53ms2)^i+5^j
Att2=5s,a=(5ms2)^i(53ms2)^i+^j
Given the body is moving in anti-clockwise direction

Since,the speed is constant, net acceleration will always be directed toward the centre and since it is

This will be the only path possible using this, we can calculate ω as we can find the change in angle


tanθ1=153=13
tanθ2=335=3
θ1=30
θ2=60

We can see that angle between the vectors is 90 which means particle would have covered an angle of 36090=270=3π/2radin3s
ω=ΔθΔt=3π23=π2rad/s
a=ω2r
r=aω2=(53)2+52(π2)2=100π24
=4×10π2=40π2



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