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Work, Energy and Power Test - 2

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Work, Energy and Power Test - 2
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  • Question 1
    4 / -1

    A wheel is at rest. Its angular velocity increases uniformly and becomes 80 radian per second after 5 second. The total angular displacement is :

    Solution

    If the angular velocity increases uniformly, then the angular acceleration must be constant.

    Hence the formula,

    s= ut + 0.5 a t2

    Where.

    u= initial angular velocity

    a= angular acceleration (i.e = 80/5=16 rad/s^2)

    S= angular displacement

    t= time taken

    Now.

    s= 0 + 0.5× 16 * 52 = 200

    Hence angar displacement = 200 radians

  • Question 2
    4 / -1

    A wheel is subjected to uniform angular a cceleration about its axis. Initially its angular velocity is zero. In the first 2 sec, it rotates through an angle q1 ; in the next 2 sec, it rotates through an additional angle q2. The ratio of q2 / q1 is-

    Solution

    From the second equation of motion we get that
    Q1 = 0 + ½ a x 4
    Where a is constant angular acceleration
    Thus Q1 = 2a
    Similarly Q2 = 0 + ½ x a x 16
    Thus Q2 = 8a
    We get Q2 / Q1 = 4

  • Question 3
    4 / -1

    The linear and angular acceleration of a particle are 10 m/sec2 and 5 rad/sec2respectively it will be at a distance from the axis of rotation -

    Solution

    a = rα (where α is angular acceleration = 5 & a is linear acceleration = 10)

  • Question 4
    4 / -1

    A wheel having a diameter of 3 m starts from rest and accelerates uniformly to an angular velocity of 210 r.p.m. in 5 seconds. Angular acceleration of the wheel is -

    Solution

    https://cdn3.edurev.in/ApplicationImages/Temp/6100a654-ff6f-45d4-8ee8-e8e165f5da04_lg.gif
    α = 1.4 pi rad/s2 

  • Question 5
    4 / -1

    The second’s hand of a watch has length 6 cm. Speed of end point and magnitude of difference of velocities at two perpendicular positions will be

    Solution

  • Question 6
    4 / -1

    The ratio of angular speed of hours hand and seconds hand of a clock is-

    Solution

    The angular speed of second hand = 2/ 60rad/sec and angular speed of hour hand = 2/ 12 rad/hr
    = 2/ 12 x 60 x 60 rad/sec
    Hence the ratio is 1: 720

  • Question 7
    4 / -1

    The angular velocity of earth about its axis of rotation is-

    Solution

    Angular velocity = 2π/time period
    Time period of earth is one day means 24×60×60

  • Question 8
    4 / -1

    A stone of mass 0.5 kg tied with a string of length 1 metre is moving in a circular path with a speed of 4 m/sec. The tension acting on the string in Newton is-

    Solution

    https://cdn3.edurev.in/ApplicationImages/Temp/994aa29b-6e7d-4bbb-93a0-8baefa385818_lg.gif

  • Question 9
    4 / -1

    Two bodies A & B rotate about an axis, such that angle qA (in radians) covered by first body is proportional to square of time, & qB (in radians) covered by second body varies linearly. At t = 0, qA = qB = 0. If A completes its first revolution in sec. & B needs 4p sec. to complete half revolution then; angular velocity wA : wB at t = 5 sec. are in the ratio

  • Question 10
    4 / -1

    A car of mass 1000 kg moves on a circular track of radius 20 m. if the coefficient of friction is 0.64, what is the maximum velocity with which the car can be moved?

    Solution

    Using the formula Vmax = √urg taking coefficient of friction = (u) in this formula. Hence we have radius and also know gravitational constant. Solving the above question
    Vmax = √u*√r*√g
    u = 0.64
    g = 9.8
    r = 20
    Vmax = √9.8*√20*√0.64
    = √125.44 = 11.2m/s

  • Question 11
    4 / -1

    The earth, radius 6400 km, makes one revolution about its own axis in 24 hours. The centripetal acceleration of a point on its equator is nearly -

    Solution

    The time period of earth’s rotation is,
    T = 24 h = 86400 s
    Angular velocity, ω = 2π/T
    Radius of earth, r = 6400 km
    => 6.4 x 106 m
    Now, centripetal acceleration = ω2r
    => (2π/T)2r = 0.034 m/s2

  • Question 12
    4 / -1

    A stone of mass 0.1 kg tied to one end of a string 1.0 m long is revolved in a horizontal circle at the rate of 10/p revolution per second. Calculate the tension of the string ?

    Solution

    https://cdn3.edurev.in/ApplicationImages/Temp/510438_c87b1f89-df81-4a3d-bf68-fa2f67b411e2_lg.png

  • Question 13
    4 / -1

    A coin placed on a rotating turn table just slips if it is at a distance of 40 cm from the centre if the angular velocity of the turntable is doubled, it will just slip at a distance of

    Solution


  • Question 14
    4 / -1

    A particle moves in a circle of radius 25 cm at two revolutions per second. The acceleration of particle in m/s2 is -

    Solution

    The options should be changed to:

    1. π2
    2. 8π2
    3. 4π2
    4. 2π2 

    2 rev/sec means → ω = 2πn = 2π rad/sec
    Use the formula of Acceleration a = rω2, where r = 0.25 cm
    On substituting and calculating, we get
    Acceleration a = 4π2 m/s2

  • Question 15
    4 / -1

    The graphs below show angular velocity as a function of time. In which one is the magnitude of the angular acceleration constantly decreasing? 

    Solution

    As we know slope at any point on velocity time curve give us acceleration that means we have to choose the curve in which slope on curve is decreasing with time. The graph of angular velocity gives the angular acceleration. In option no 'a' we can see that the graph is constantly decreasing. So, angular acceleration will also constantly decrease.

  • Question 16
    4 / -1

    A particle moves along a circle of radius  m with constant tangential acceleration. If the velocity of the particle is 80 m/s at the end of the second revolution after motion has begun, the tangential acceleration is :

    Solution

  • Question 17
    4 / -1

     A particle moves with deaceleration along the circle of radius R so that at any moment of time its tangential and normal accelerations are equal in moduli. At the initial moment t = 0 the speed of the particle equals v0, then:

    (i) the speed of the particle as a function of the distance covered s will be

    (A) v = v0 e_s/R (B) v = v0es/R (C) v = v0e_R/s (D) v = v0eR/s

    (ii) the total acceleration of the particle as function of velocity and distance covered

    (A) a =  (B) a =  (C) a =  (D) a = 

    Solution

    (i) 

    (ii) 

  • Question 18
    4 / -1

    If angular velocity of a disc depends an angle rotated q as w = q2 + 2q, then its angular acceleration a at q = 1 rad is

    Solution

  • Question 19
    4 / -1

    A particle is going in a uniform helical and spiral path separately as shown in figure with constant speed.

     

    Solution

    The centripetal acceleration is inversely proportional to the radius of the circle. In case (A) the spiral path has same radius for every circle, so the particle will have a constant centripetal acceleration but in case (B) the radius of the circles goes on decreasing which results in decrease in centripetal acceleration of the particle.

  • Question 20
    4 / -1

    If the radii of circular paths of two particles of same masses are in the ratio of 1 : 2, then in order to have same centripetal force, their speeds should be in the ratio of :

    Solution

  • Question 21
    4 / -1

    A rigid body of mass m is moving in a circle of radius r with a constant speed v. The force on the body is  and is directed towards the centre. What is the work done by this force in moving the body over half the cirumference of the circle.

    Solution


    Force = mv2/r ( towards the centre )
    displacement = 2r
    Work done = FScosѲ
    here ,
    = mv2/r × s × cos 90
    = 0 answer.
     

  • Question 22
    4 / -1

    A body travels through a distance of 10 m on a straight line, under the influence of 5 N . If the work done by the force is 25J, the angle between the force and displacement is-

    Solution

    We Know that,
    Work = Force × displacement × cosine of angle between Force and displacement
    Given Work (W) = 25J
    Displacement (s) = 10 m
    Force (F) = 5 N
    Let angle between Force and displacement be θ
    so W=F × s × cos θ
    => cos θ = 1/2
    =>cos θ = cos 60
    Hence angle between Force and displacement is 60 degree

  • Question 23
    4 / -1

    A blocks is moved from rest through a distance of 4m along a straight line path. The mass of the blocks is 5 kg. and the force acting on it is 20 N . If the kinetic energy acquired by the block be 40J, at what angle to the path the force is acting-

    Solution

  • Question 24
    4 / -1

    The work done in pushing a block of mass 10 kg from bottom to the top of a frictionless inclined plane 5 m long and 3 m high is- (g = 9.8 m/sec2)

    Solution

    According to Work Energy Theorem
    W = -∆E
    So, W = mgh = 294 N

  • Question 25
    4 / -1

    A force F = (3x + 4) Newton (where x is in metres) acts on a particle which moves from a position (2m, 3m) to (3m, 0m). Then the work done is

    Solution

    https://cdn3.edurev.in/ApplicationImages/Temp/509081_9014cf1d-2fed-45b0-beee-12ce7caaa1d6_lg.png

  • Question 26
    4 / -1

    If a man increases his speed by 2 m/sec, his K.E. is doubled. The original speed of the man is-

    Solution

    Assume initial speed of man = u and kinetic energy = K
    Final speed of man = (u+2) then kinetic energy = 2K. Also,
    https://cdn3.edurev.in/ApplicationImages/Temp/509020_4b503235-1ba9-4a3c-9fe1-936981e48d1a_lg.png

  • Question 27
    4 / -1

    A block of mass m is suspended by a light thread from an elevator. The elevator is accelerating upward with uniform acceleration a. The work done by tension on the block during t seconds is :

                          

    Solution

    A block of mass m is suspended by a light thread from an elevator. The elevator is accelerating upward with uniform acceleration ‘a’. We know, Newton's law doesn't valid in non-inertial frame of reference. So, we have to add Pseudo force just opposite direction of it motion to make inertial frame of reference.
    So, tension in the thread = mg + ma → [Pseudo force]
    Tension = m(g + a)
    Now, we have to find displacement of elevator.
    Use formula, S = ut + 1/2 At2 
    Here A is net acceleration acting on the elevator.
    Here u = 0, A = a [ upward direction]
    So, S = 0 + 1/2 at2
    Hence, displacement is 1/2at2
    Now, work done = Tension x displacement = m(g + a) x 1/2 at2
    = 1/2 m(g + a)at2 

  • Question 28
    4 / -1

    In a typical power station, amid a series of energy transfers and conversions, about 70% of the energy input gets wasted in the form of thermal energy, the efficiency of such a power station is about

    Solution

    If 70% is wasted then 30% is used, so, the efficiency becomes 30/100 = 0.3

  • Question 29
    4 / -1

    A body of mass m accelerates uniformly from rest to a speed v0 in time t0. The work done on the body till any time t is

    Solution

    Acceleration is uniform
    vo = 0 + ato
    a = vo/to (equation 1)
    K.E. of the body at time t = ½ mv2 and v=at
    = ½ m(at)02 
    From equation 1
    = ½ mv2t2/to2 

  • Question 30
    4 / -1

    A light spring of length 20 cm and force constant 2 kg/cm is placed vertically on a table. A small block of mass 1 kg falls on it. The length h from the surface of the table at which the ball will have the maximum velocity is -

    Solution

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