Work and Energy Test

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Work and Energy Test - 25

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Weekly Quiz Competition

Question 1
1 / -0

How much energy in kilowatt hour is consumed in operating ten 50 watt bulbs for 10 hours per day in a month (30 days)

A
1500

B
5000

C
15

D
150

Solution

Energy consumed $$=10\times 50\times 10\times 30\times 3600 J$$
$$[1 Kwh=3600\times 1000J]$$
$$=\frac {10\times 50\times 10\times 30\times 3600}{3600\times 1000}kWh=150$$
Question 2
1 / -0

The hydroelectric plants do not produce electricity, if the water level in the dam is less than 34 m.

A
True

B
False

C
Ambiguous

D
Data insufficient
Question 3
1 / -0

A machine raises a load of $$750N$$ through a height of $$16m$$ in $$5s$$. Calculate work done by machine:

A
$$12000 kJ$$

B
$$12 kJ$$

C
$$1200 J$$

D
$$120 kJ$$

Solution

We know that $$Work = F\times s$$
Here, $$F = 750\ N$$ and $$s=16\ m$$.

$$W = 750\times16 = 12000\ J = 12\ kJ$$
Question 4
1 / -0

A pump is used to lift $$500\ kg$$ of water from a depth of $$80\ m$$ in $$10\ s$$.
(Take $$g=10\ ms^{-2}$$). Calculate the work done by the pump.

A
$$16 \times 10^5J$$

B
$$4\times 10^5J$$

C
$$4\times 10^8J$$

D
$$2\times 10^5J$$

Solution

Given,
Mass of water lifted, $$m=500\ kg$$
Displacement, $$d=80\ m$$
Time taken, $$t=10\ s$$

Force, $$F = m\times g$$
$$F = 500\times 10$$
$$F= 5000\ N$$

Work done, $$W= F\times d $$
$$W = 5000\times 80$$
$$W = 4\times 10^{5}\ J$$.
Question 5
1 / -0

A body when acted upon by a force of $$10\ kgf$$, gets displaced by $$0.5\ m$$. Calculate the work done by the force, when the displacement is in the direction of force.

A
$$5 J$$

B
$$500 J$$

C
$$0.5 J$$

D
$$50J$$

Solution

Given - $$F = 10\ kgf$$ and $$s = 0.5\ m$$ in the direction of force.

$$ 1 Kgf = 10 N $$
$$ \implies F = 10kgf = 10 \times 10N = 100N $$

$$ W = F \times s $$
$$ W = 100 \times 0.5 $$
$$ W = 50 J $$
Question 6
1 / -0

A boy carrying a box on his head is walking on a level road from one place to another. The work done by him against the gravitational force is zero. This statement is :

A
correct

B
incorrect

C
partly correct

D
insufficient data

Solution

Gravitational force on the box is acting in the downward direction. Since the boy walks horizontally, his vertical displacement is zero. Hence, when a boy carrying load on his head moves over a horizontal road, work done against gravitational force is zero.
Question 7
1 / -0

A man spends $$6.4 \ kJ$$ energy in displacing a body by $$64 \ m$$ in the direction in which he applies force, in $$2.5 \ s$$.
Calculate the force applied.

A
$$1N$$

B
$$10N$$

C
$$0.1N$$

D
$$100N$$

Solution

We know that $$Work = Force \times displacement $$.
Here, given $$ Energy = 6.4\ kJ = 6400\ J$$
$$ \implies Work =64000J$$ and $$ displacement = 64\ m$$.

$$Force = \dfrac{Work}{displacement} = 6400/64 = 100 N$$.
Question 8
1 / -0

Name the type of energy (kinetic energy $$K$$ or potential energy $$U$$) possessed in the following case.
The bob of a simple pendulum at its extreme position.

A
$$K$$

B
$$U$$

C
$$K$$ and $$U$$

D
No energy

Solution

Here simple pendulum is at a certain height and at extreme position it will not have any velocity. Hence , kinetic energy will be zero. Whereas $$ U = m\times g\times h$$.
As it is at certain height it has some value for $$h$$. Hence, it will have certain potential energy.
Question 9
1 / -0

A man raises a box of mass $$50 \ kg$$ to a height of $$2 \ m$$ in $$2 \ minutes$$, while another man raises the same box to the same height in $$5 \ minutes$$. What is the ratio of work done by them ?

A
$$1 : 1$$

B
$$2 : 1$$

C
$$1 : 2$$

D
$$4 : 1$$

Solution

$$Work = Fs\cos\theta$$
where, $$F$$ is the force applied, $$s$$ is the displacement, and $$\theta$$ is the angle between the force applied and displacement

Hence, work done is independent of time taken.

In the given cases, $$\theta = 0^{\circ}$$ as the force applied are in the same direction. Also, $$F=mg$$
So $$W=Fs=mgs$$

In both the cases, mass and displacement are the same.
$$W =50\times 10\times 2 = 1000 J$$

Hence work done is in the ratio of $$1000 : 1000 = 1:1$$.
Question 10
1 / -0

Before a rubber ball bounces off from the floor, the ball is in contact with the floor for a fraction of second. Which of the following statements are correct?

A
Conservation of energy is not valid during this period

B
Conservation of energy is valid during this period

C
Only the kinetic energy is conserved

D
Only the potential energy is conserved

Solution

$$\text{The total energy of the ball} =\text{Kinetic Energy}+\text{Potential Energy}$$
The kinetic energy or the potential energy changes as the ball falls down and hits the floor. But the total energy is always conserved.

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Re-Attempt Weekly Quiz Competition

Work and Energy Test - 25

Result

Work and Energy Test - 25

Score

-

Rank

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SHARING IS CARING

Question 1

1500

5000

15

150

Solution

Question 2

True

False

Ambiguous

Data insufficient

Question 3

$$12000 kJ$$

$$12 kJ$$

$$1200 J$$

$$120 kJ$$

Solution

Question 4

$$16 \times 10^5J$$

$$4\times 10^5J$$

$$4\times 10^8J$$

$$2\times 10^5J$$

Solution

Question 5

$$5 J$$

$$500 J$$

$$0.5 J$$

$$50J$$

Solution

Question 6

correct

incorrect

partly correct

insufficient data

Solution

Question 7

$$1N$$

$$10N$$

$$0.1N$$

$$100N$$

Solution

Question 8

$$K$$

$$U$$

$$K$$ and $$U$$

No energy

Solution

Question 9

$$1 : 1$$

$$2 : 1$$

$$1 : 2$$

$$4 : 1$$

Solution

Question 10

Conservation of energy is not valid during this period

Conservation of energy is valid during this period

Only the kinetic energy is conserved

Only the potential energy is conserved

Solution

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