# KPK 11th Class Physics Chapter 2 Vectors and Equilibrium Short Questions Answers

KPK 11th Class Physics Chapter 2 Vectors and Equilibrium Short Questions with answers are combined for all 11th class(Intermediate/hssc) Level students.Here You can prepare all Physics Chapter 2 Vectors and Equilibrium short question in unique way and also attempt quiz related to this chapter. Just Click on Short Question and below Answer automatically shown. After each question you can give like/dislike to tell other students how its useful for each.

**Class/Subject: 11th Class ****Physics**

**Chapter Name: Vectors and Equilibrium**

**Board: All ****KPK Boards**

- Malakand Board 11th Class Physics Chapter 2 Vectors and Equilibrium short questions Answer
- Mardan Board 11th Class Physics Chapter 2 Vectors and Equilibrium short questions Answer
- Peshawar Board 11th Class Physics Chapter 2 Vectors and Equilibrium short questions Answer
- Swat Board 11th Class Physics Chapter 2 Vectors and Equilibrium short questions Answer
- Dera Ismail Khan Board 11th Class Physics Chapter 2 Vectors and Equilibrium short questions Answer
- Kohat Board 11th Class Physics Chapter 2 Vectors and Equilibrium short questions Answer
- Abbottabad Board 11th Class Physics Chapter 2 Vectors and Equilibrium short questions Answer
- Bannu Board 11th Class Physics Chapter 2 Vectors and Equilibrium short questions Answer

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## KPK 11th Class Physics Chapter 2 Vectors and Equilibrium Short Questions Answers

Consider two vector ‘A‘and ‘B‘such that the magnitude of ‘A‘ is greater than ‘B‘ as shown in the figure.

Now if the two vectors are acting along the same line and same direction, then their resultant is given by

R= A+B

If the two vectors are acting along the same line in opposite direction, then their resultant is given by,

R= A+–B

- R= A– B

Thus the range of the possible values of the resultant of the two vectors A and B varies from A+B to A–B.

The minimum number of unequal vectors to result into a null vector must be three. If we add three vectors of unequal magnitude in such a way that they forms the sides of a triangle, then their resultant must be zero.

In the given figure three vectors A.B and C are added according to head to tail rule and they form the sides of a triangle.

Now for getting their resultant, we will combine the tail of A with the head of C which already coincides each other. Thus we get a null vector or zero vector as a resultant. So we have.

R= A+B+C= 0

Let F_{1} and F_{2} represents the two forces, then according to the given conditions, we have,

F_{2}=2F_{1} ……………….(1)

And the resultant ‘R‘ is perpendicular to the smaller force ‘F1‘ as shown in the fig.

Now applying patha-gorem theorem to ∆abc, we get

(Hypotenuse)^{2}=(perpendicular)^{2}+(Base)^{2}

- (bc)
^{2}=(ac)^{2}+ (ab)^{2} - F
^{2}_{2}= R^{2}+ F^{2}_{1} - R
^{2}= F_{2}^{2}– F_{1}^{2}……………………………………..(2)

Putting eq (1) in eq(2), we get,

R^{2}=(2F_{1})^{2}–F_{1}^{2}

- R = √4F
_{1}^{2}–F_{1}^{2}= √(4-1)F_{1}^{2} - R= √3F
_{1}^{2} - R= √3 F
_{1}……………………………………..(3)

Now from the figure, we have, tan ac/ab

- Tan θ=R/F
_{1}………………………..(4)

Putting eq (3) in eq (4), we get,

Tan = (√3 F_{1} /F_{1} )

- Tan = √3 => =tan
^{-1}(3) - θ=60°

Thus the angle b./w the two force is 60°.

We know that objects with a high centre of gravity are likely to be uns-table. For this reason, the bases of the crones are made low and heavy, so that centre of gravity remains low.

The torque produced by a heavy load lifted by a crane is counterbalanced by its heavy body and large moment arm. Thus the cranes do not topple when they carry heavy loads.

We know that the torque dep-applied force. Mathematically, we have

T = r × F……………………….(1)

As we have given that the gravitational force acting on a satellite is directed towards the centre of the earth. As for central force, the moment arm is zero i.e. r= 0, so eq (1) => T=rF =0

- T=0 …………………………………..(2)
- eq 2shows that the torque produced by gravitational force acting an a satellite is zero.

T=r×F …………(1)

eq 1shows that to produce grater torque with a less

applied force, the value of moment arm ‘r’ must be taken as larger. Due to grater torque, we can turn the heavy trueks and buses easily. That is why heavy trucks and buses have large steering wheels to produce greater torque and make the drive more easy.

Let the point object is situated at origin ‘O’ as shown in the figure. Let F = 4N is acting along x= axis, F_{2} = 5N is acting along y-axis and F_{3} = 6N is acting along z-axis and as a result, the object is in equilibrium.

Now if 6N force is removed, then the resultant of 4N and 5N will be acting in xy-plane as shown in the figure.

Now applying pathagorem theorem to ∆OAB, we get

(OB)2 = (OA)2+(AB)2

- F
^{2}=F_{1}^{2}+F_{2}^{2} - F= √F
_{1}^{2}+F_{2}^{2} - F= √4
^{2}+5^{2}= √16+25 - F = √4
^{1}=> F = 6.4 N

Direction of F:

In ∆OAB, we have.

tanθ=(F2/F1)

- θ=tan
^{-1}(F_{2}/F_{1}) = tan^{-1}(5/4) - θ=51.3

Thus the resultant of F_{1} and F_{2}= F= 6.4N by making an angle of 51.3 with +ive x-axis in anti-clock wise direction.