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Work, Energy and Power Test - 14

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Work, Energy and Power Test - 14

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Question 1
1 / -0

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

A
800 Red

B
400 Red

C
200 Red

D
100 Red

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
1 / -0

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-

A
2

B
1

C
4

D
3

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
1 / -0

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 -

A
50 m

B
1/2 m

C
1 m

D
2 m

Solution

a = rα (where α is angular acceleration = 5 & a is linear acceleration = 10)
Question 4
1 / -0

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 -

A
1.4p

B
3.3p

C
2.2p

D
1.1p

Solution

α = 1.4 pi rad/s²
Question 5
1 / -0

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

A
6.28 and 0 mm/s

B
8.88 and 4.44 mm/s

C
8.88 and 6.28 mm/s

D
6.28 and 8.88 mm/s

Solution
Question 6
1 / -0

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

A
1 : 1

B
1 : 60

C
1 : 720

D
3600 : 1

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
1 / -0

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

A
2p / ( 60 × 60 × 24 ) rad / sec

B
2p / ( 60 × 60 ) rad / sec

C
2p / 60 rad / sec

D
2p / ( 365 × 24 × 60 × 60 ) rad / sec

Solution

Angular velocity = 2π/time period
Time period of earth is one day means 24×60×60
Question 8
1 / -0

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-

A
2

B
8

C
0.2

D
0.8

Solution
Question 9
1 / -0

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

A
4 : 1

B
80 : 1

C
20 : 1

D
20 : 4
Question 10
1 / -0

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?

A
1.12 m/s

B
11.2 m/s

C
m/s

D
m/s

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
1 / -0

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 -

A
340

B
3.4

C
34

D
0.34

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
1 / -0

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 ?

A
30 N

B
40 N

C
50 N

D
60 N

Solution
Question 13
1 / -0

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

A
10 cm

B
20 cm

C
40 cm

D
80 cm

Solution
Question 14
1 / -0

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

A
p²

B
8p²

C
4p²

D
2p²
Solution
The options should be changed to:
π²
8π²
4π²
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
1 / -0

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

A

B

C

D

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
1 / -0

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 :

A
160 p m/s²

B
40 p m/s²

C
40 m/s²

D
640 p m/s²

Solution
Question 17
1 / -0

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

A
The velocity of the particle is constant in both cases

B
The acceleration of the particle is constant in both cases

C
The acceleration of the particle is constant in a and decreasing in b cases

D
The magnitude of acceleration is decreasing continuously in both the cases

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 18
1 / -0

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 :

A
1 : 4

B
4 : 1

C
1 :

D
: 1

Solution
Question 19
1 / -0

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.

A

B
Zero

C

D

Solution

Force = mv2/r ( towards the centre )
displacement = 2r
Work done = FScosѲ
here ,
= mv2/r × s × cos 90
= 0 answer.
Question 20
1 / -0

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-

A
0º

B
30º

C
60º

D
90º

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 21
1 / -0

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-

A
30º

B
60º

C
45º

D
none of the above

Solution
Question 22
1 / -0

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)

A
392 J

B
294 J

C
98 J

D
0.98 J

Solution

According to Work Energy Theorem
W = -∆E
So, W = mgh = 294 N
Question 23
1 / -0

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

A
- 7.5J

B
-12J

C
- 4.5 J

D
+4.5 J

Solution
Question 24
1 / -0

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

A
(2 +) m/s

B
(2 + 2) m/s

C
4 m/s

D
(1 +) m/s

Solution

Assume initial speed of man = u and kinetic energy = K
Final speed of man = (u+2) then kinetic energy = 2K. Also,
Question 25
1 / -0

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 :

A

B

C

D
0

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 26
1 / -0

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

A
0.3

B
0.4

C
0.7

D
1

Solution

If 70% is wasted then 30% is used, so, the efficiency becomes 30/100 = 0.3
Question 27
1 / -0

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

A

B

C

D

Solution

Acceleration is uniform
v_o = 0 + at_o
a = v_o/t_o (equation 1)
K.E. of the body at time t = ½ mv2 and v=at
= ½ m(at)₀²
From equation 1
= ½ mv²t²/t_o2
Question 28
1 / -0

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 -

A
20 cm

B
15 cm

C
10 cm

D
5 cm

Solution

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