Work and Energy Test

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Work and Energy Test - 47

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Weekly Quiz Competition

Question 1
1 / -0

An elevator is run by the cables at constant speed. The total work done by the elevator is:

A
Negative

B
Positive

C
Zero

D
Positive or negative depending on the direction of motion

Solution

Work done is defined as $$W = Fd \cos\theta$$
where, $$F$$ is the force applied, $$d$$ is the displacement and $$\theta$$ is the angle between force applied and displacement.

Since, the elevator is moving at a constant speed, the work done is only due to the force of gravity and the tension force exerted on the elevator by the cable which pulls it up. Since, the force of tension and displacement are in the same direction, the angle $$\theta = 0$$ and hence the work done is Positive. Hence, option (B) is correct.
Question 2
1 / -0

Kinetic energy of the liquid per unit mass is

A
$$\frac { 1 } { 2 } m v ^ { 2 }$$

B
$$\frac { 1 } { 2 } v ^ { 2 }$$

C
$$\frac { 1 } { 2 } m ^ { 2 } v$$

D
$$m v ^ { 2 }$$

Solution

Let $$m$$ be the mass of liquid and $$v$$ be the velocity of liquid
Hence ,
Kinetic energy of liquid $$,K=\dfrac12 mv^2 $$

$$\therefore $$ Kinetic energy of liquid per unit mass $$= \dfrac Km = \dfrac12 v^2$$
Question 3
1 / -0

An energy of $$4 kJ$$ causes a displacement of $$64 m$$ in $$2.5 s$$. The power delivered is

A
$$16 W$$

B
$$160 W$$

C
$$1600 W$$

D
$$16000 W$$

Solution

Given that:

Energy, $$E=4\ kJ=4000\ J$$
Time taken, $$t=2.5\ sec$$

Power delivered , $$P=\dfrac Et=\dfrac{4000 \ J}{2.5 \ sec}= 1600\ W$$
Option C is correct.
Question 4
1 / -0

Work done in lifting a body is calculated by

A
Mass of the body $$ \times $$ vertical distance moved

B
Force acting on a body $$ \times $$ vertical distance moved

C
Weight acting on the body $$ \times $$ vertical distance moved

D
None of the above.

Solution

Given W= work done

and mass = m

and vertical height= H

and gravitational force= g

W= mgh
Question 5
1 / -0

A stone is projected vertically up to reach maximum height $$h$$. The ratio of its kinetic to potential energies at a height $$\cfrac{4h}{5}$$ will be

A
$$5:4$$

B
$$4:5$$

C
$$1:4$$

D
$$4:1$$

Solution
Question 6
1 / -0

Two masses $$10 \ gm$$ and $$40 \ gm$$ are moving with kinetic energies in the ratio $$9:25$$. The ratio of their linear momentum is:

A
$$5:6$$

B
$$3:10$$

C
$$6:5$$

D
$$10:3$$

Solution
Question 7
1 / -0

When the mass of body is halved and velocity is doubled, then the kinetic energy of the body

A
remains same

B
is doubled

C
is 4 times

D
is $$\frac{1}{4}$$ th

Solution

$$K.E = \dfrac{1}{2}mv^2$$

$$New\ K.E = \dfrac{1}{2} (\dfrac{m}{2})\ (2v)^2$$ = $$2 \times \dfrac{1}{2}mv^2$$ = $$2 \times K.E$$

Hence, kinetic energy of body is doubled.
Question 8
1 / -0

Mass of $$B$$ is four times that of $$A . B$$ moves with a velocity half that of $$A$$. Then, $$B$$ has

A
kinetic energy equal to that of $$A$$

B
half the kinetic energy of $$A$$

C
twice the kinetic energy of $$A$$

D
kinetic energy one-fourth of $$A$$

Solution

According to question given,
$$M_B = 4 \times M_A $$ and $$ V_B = \dfrac{V_A}{2} $$

$$K.E_A = \dfrac{1}{2}M_AV_A^2 $$ .........(1)

$$K.E_B = \dfrac{1}{2}M_BV_B^2 $$ = $$\dfrac{1}{2}\ (4 \times M_A) \left(\dfrac{V_A}{2}\right)^2=\dfrac{1}{2}M_AV_A^2 $$........(2)

From above equation
$$K.E_A=K.E_B$$
Question 9
1 / -0

A body of mass $$15\ kg$$ moving with a velocity of $$10\ ms^{-1}$$ is bought to rest. The work done by the brake is

A
$$-250\ J$$

B
$$-500\ J$$

C
$$-750\ J$$

D
$$-1000\ J$$

Solution

$$W = change\ in\ kinetic\ energy =\dfrac{1}{2}m(v^2-u^2) $$

$$W = \dfrac{1}{2} \times 15 \times (0^2-10^2)$$

$$W = - 750J$$

Hence, option $$C$$ is correct answer.
Question 10
1 / -0

If the unit of force and length each be increased by four times, then the unit of energy is increased by

A
$$16$$ times

B
$$8$$ times

C
$$2$$ times

D
$$4$$ times

Solution

Work = Force $$ \times $$ Displacement ( length)
$$\Rightarrow W=FS$$
Now,
$$F'=4F$$
$$S' = 4s$$
So energy,
$$W' = 4F \times 4S$$
$$\Rightarrow W' = 16F S$$
$$\Rightarrow W' = 16W$$
If the unit of force and length be increased by four times then the unit of energy will increase by $$ 16 $$ times.

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Work and Energy Test - 47

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Work and Energy Test - 47

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SHARING IS CARING

Question 1

Negative

Positive

Zero

Positive or negative depending on the direction of motion

Solution

Question 2

$$\frac { 1 } { 2 } m v ^ { 2 }$$

$$\frac { 1 } { 2 } v ^ { 2 }$$

$$\frac { 1 } { 2 } m ^ { 2 } v$$

$$m v ^ { 2 }$$

Solution

Question 3

$$16 W$$

$$160 W$$

$$1600 W$$

$$16000 W$$

Solution

Question 4

Mass of the body $$ \times $$ vertical distance moved

Force acting on a body $$ \times $$ vertical distance moved

Weight acting on the body $$ \times $$ vertical distance moved

None of the above.

Solution

Question 5

$$5:4$$

$$4:5$$

$$1:4$$

$$4:1$$

Solution

Question 6

$$5:6$$

$$3:10$$

$$6:5$$

$$10:3$$

Solution

Question 7

remains same

is doubled

is 4 times

is $$\frac{1}{4}$$ th

Solution

Question 8

kinetic energy equal to that of $$A$$

half the kinetic energy of $$A$$

twice the kinetic energy of $$A$$

kinetic energy one-fourth of $$A$$

Solution

Question 9

$$-250\ J$$

$$-500\ J$$

$$-750\ J$$

$$-1000\ J$$

Solution

Question 10

$$16$$ times

$$8$$ times

$$2$$ times

$$4$$ times

Solution

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