Force and Pressure Test

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Force and Pressure Test - 40

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Weekly Quiz Competition

Question 1
1 / -0

With the increase in the area of contact of an object the pressure

A
increases

B
decreases

C
is not affected

D
none of these

Solution

The pressure is the force per unit area. So, $$P=F/A$$
Thus the pressure P is inversely proportional to contact area A.
So when area increases , the pressure will be decreasing.
Question 2
1 / -0

The resultant of two opposite parallel forces $$12\ N$$ and $$7\ N$$ is _____ N

A
$$12$$

B
$$5$$

C
$$19$$

D
$$7$$

Solution

$$\vec{F}_{resultant}=\vec{F_{1}}+\vec{F_{2}}$$
$$\left | \vec{F_{1}}+\vec{F_{2}} \right |=\sqrt{F_{1}^{2}+F_{2}^{2}+2F_{1}F_{2}cos\theta }$$
$$\theta =\pi $$
$$\left | \vec{F_{1}}+\vec{F_{2}} \right |=\sqrt{144+49-168}$$
$$=5N$$
Question 3
1 / -0

Which of the following is a correct statement regarding the pressure due to fluids?

A
Force acting on the wall due to pressure is always perpendicular to it because of collision between the molecules takes place perpendicular to it.

B
Force acting on the wall due to pressure is always perpendicular to it because of collision between the molecules takes place oblique to it.

C
Force acting on the wall due to pressure is always oblique to it because of collision between the molecules takes place perpendicular to it.

D
Force acting on the wall due to pressure is always perpendicular to it because of collision between the molecules takes place at any direction to it but its normal reaction on the wall is always perpendicular to it.

Solution

The net force acting on a wall of fluid containing vessel is always normal to it because effective force is due to collisions of molecules to the wall of the vessel.
Question 4
1 / -0

In the superhit film 'Raja Hindustani', Amir Khan greets his beloved by shaking hand. What kind of force do they exert?

A
Nuclear

B
Gravitational

C
Weak

D
Electromagnetic

Solution

Shaking hands will involve electromagnetic forces because their hands interaction is kind of friction between their hands and Friction force is a type of Electromagnetic force.
Therefore option $$D$$ is correct
Question 5
1 / -0

The two femurs each of the cross-sectional area 10 cm$$^2$$ support the upper part of a human body of mass 40 kg. The average pressure sustained by the femurs is then (Takes g= 10 m s$$^{-2}$$)

A
2 x 10$$^3$$ N m$$^{-2}$$

B
2 x 10$$^4$$ N m$$^{-2}$$

C
2 x 10$$^5$$ N m$$^{-2}$$

D
2 x 10$$^6$$ N m$$^{-2}$$.

Solution

Given that,
$$Area=10 \ cm^{2}$$
$$Mass=40 \ kg$$
$$g=10 \ m/s^{2}$$

Total cross-sectional area of the femurs is,
$$A = 2 \times 10 \ cm^2$$

$$ = 2 \times 10\times 10^{-4} m^2 $$

$$ = 20 \times 10^{-4} m^2$$

Force acting on them is

$$F= mg =40\,kg \times 10\,m\,s^{-2} =400\,N$$

$$\therefore$$ Average pressure sustained by them is
$$\displaystyle P=\dfrac{F}{A}= \dfrac {400\,N}{20 \times 10^{-4}\,m^2}$$

$$ = 2 \times 10^5\,N\,m^{-2}$$
Question 6
1 / -0

Atmospheric pressure exerted on earth is due to the.

A
Gravitational pull

B
Revolution of earth

C
Rotation of earth

D
Uneven heating of earth

Solution

The atmospheric pressure exerted on earth is due to gravitational pull of the earth and is about 14.7 psi.
Hence, option A is correct.
Question 7
1 / -0

A 50 kg girl wearing heel shoes balances on a single heel. The heel is circular with a diameter 1 cm. The pressure exerted by the heel on the horizontal floor is then (Take g = 10 m s$$^{-2}$$)

A
6.4 x 10$$^4$$ Pa.

B
6.4 x 10$$^5$$ pa

C
6.4 x 10$$^6$$ Pa

D
6.4 x 10$$^7$$ Pa

Solution

Here,$$m =50 kg, $$

$$D = 1 cm = 10^{-2} m$$,

$$g= 10 m s^{-2}$$

$$\therefore$$ Pressure exerted by the heel on the horizontal floor is
$$Pressure=\dfrac{Force}{Area}$$ Force here is WEIGHT and area is area of circular portion of heel.

$$\displaystyle P=\frac{F}{A} = \frac{mg}{\pi \left(\dfrac{D}{2}\right)^{2}}=\frac{4 mg}{\pi D^2}$$

$$\displaystyle = \frac{4\times 50 kg \times 10\,m\,s^{-2}}{3.14\times(10^{-2}\,m)^2}=6.4 \times 10^6\,pa$$
Question 8
1 / -0

A force acts for 10s on a body of mass $${10^{ - 2}}{\text{kg}}$$ initially at rest. Then after the force ceases to act. The body traverses 0.5m in next 5s. The magnitude of the force is

A
$${10^{ - 1}}N$$

B
$${10^{ - 2}}N$$

C
$${10^{ - 3}}N$$

D
$${10^{ - 4}}N$$

Solution
Question 9
1 / -0

Force acts on an object may change

A
Direction

B
Speed

C
Shape

D
All the above

Solution

Force acting on an object may change its direction, shape and speed. To change the direction of any moving object, force component should exist perpendicular to direction of motion. if applied force is in the direction of motion then it will change only its speed.
Force can also change the shape of an object exp: Rubber, deformable-body.
Question 10
1 / -0

Which of the following pairs of forces will never give resultant force of $$2N$$

A
$$2N$$ and $$2N$$

B
$$1N$$ and $$1N$$

C
$$1N$$ and $$3N$$

D
None of these

Solution

Because if these two forces Act then it cancelled out because they have the same value. Either it acts in same or opposite direction

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Re-Attempt Weekly Quiz Competition

Force and Pressure Test - 40

Result

Force and Pressure Test - 40

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SHARING IS CARING

Question 1

increases

decreases

is not affected

none of these

Solution

Question 2

$$12$$

$$5$$

$$19$$

$$7$$

Solution

Question 3

Force acting on the wall due to pressure is always perpendicular to it because of collision between the molecules takes place perpendicular to it.

Force acting on the wall due to pressure is always perpendicular to it because of collision between the molecules takes place oblique to it.

Force acting on the wall due to pressure is always oblique to it because of collision between the molecules takes place perpendicular to it.

Force acting on the wall due to pressure is always perpendicular to it because of collision between the molecules takes place at any direction to it but its normal reaction on the wall is always perpendicular to it.

Solution

Question 4

Nuclear

Gravitational

Weak

Electromagnetic

Solution

Question 5

2 x 10$$^3$$ N m$$^{-2}$$

2 x 10$$^4$$ N m$$^{-2}$$

2 x 10$$^5$$ N m$$^{-2}$$

2 x 10$$^6$$ N m$$^{-2}$$.

Solution

Question 6

Gravitational pull

Revolution of earth

Rotation of earth

Uneven heating of earth

Solution

Question 7

6.4 x 10$$^4$$ Pa.

6.4 x 10$$^5$$ pa

6.4 x 10$$^6$$ Pa

6.4 x 10$$^7$$ Pa

Solution

Question 8

$${10^{ - 1}}N$$

$${10^{ - 2}}N$$

$${10^{ - 3}}N$$

$${10^{ - 4}}N$$

Solution

Question 9

Direction

Speed

Shape

All the above

Solution

Question 10

$$2N$$ and $$2N$$

$$1N$$ and $$1N$$

$$1N$$ and $$3N$$

None of these

Solution

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