Numerical Problem No. 2.11 (Vectors and Equilibrium) Physics HSSC - I (Class -11) (Punjab Curriculum and Text Board, Lahore) With Verified Answers

Vectors $\vec{A}$ $\vec {A}$ , $\vec{b}$ $\vec {b}$ and $\vec{C}$ $\vec {C}$ are 4 units north, 3 units west and 8 units east, respectively. Describe carefully (a) $\vec{A}$ $\vec {A}$ x $\vec{B}$ $\vec {B}$ (b) $\vec{A}$ $\vec {A}$ x $\vec{C}$ $\vec {C}$ (c) $\vec{B}$ $\vec {B}$ x $\vec{C}$ $\vec {C}$

[Ans: (a)12 units vertically up (b) 32 units vertically down (c) Zero]

Given:

$\vec{A}$ $\vec {A}$ = 4 units North
$\vec{B}$ $\vec {B}$ = 3 units West
$\vec{C}$ $\vec {C}$ = 8 units East

So, keeping in view the given directions of the vectors
Angle 𝜽 between $\vec{A}$ $\vec {A}$ and $\vec{B}$ $\vec {B}$ = 90⁰
Angle 𝜽 between $\vec{A}$ $\vec {A}$ and $\vec{C}$ $\vec {C}$ = 90⁰
Angle 𝜽 between $\vec{B}$ $\vec {B}$ and $\vec{C}$ $\vec {C}$ = 180⁰

To Find:

(a)   $\vec{A}$ $\vec {A}$ ｘ $\vec{B}$ $\vec {B}$
(b)   $\vec{A}$ $\vec {A}$ ｘ $\vec{C}$ $\vec {C}$
(c)   $\vec{B}$ $\vec {B}$ ｘ $\vec{C}$ $\vec {C}$

Solution:

(a) $\vec{A}$ $\vec {A}$ ｘ $\vec{B}$ $\vec {B}$

Here we have

$\vec{A}$ $\vec {A}$ ｘ $\vec{B}$ $\vec {B}$ = | $\vec{A}$ $\vec {A}$ | | $\vec{B}$ $\vec {B}$ | Sin Ө $\hat{n}$ $\hat {n}$

by putting values

$\vec{A}$ $\vec {A}$ ｘ $\vec{B}$ $\vec {B}$ = 4 ｘ 3 ｘ Sin 90⁰ $\hat{n}$ $\hat {n}$

[ Sin 90⁰ = 1 ]

$\vec{A}$ $\vec {A}$ ｘ $\vec{B}$ $\vec {B}$ = 4 ｘ 3 ｘ 1 ｘ 1

$\vec{A}$ $\vec {A}$ ｘ $\vec{B}$ $\vec {B}$ = 12 units ------Ans

According to the right-hand rules the direction will be vertically upwards.

(b) $\vec{A}$ $\vec {A}$ ｘ $\vec{C}$ $\vec {C}$

Here we have

$\vec{A}$ $\vec {A}$ ｘ $\vec{C}$ $\vec {C}$ = | $\vec{A}$ $\vec {A}$ | | $\vec{C}$ $\vec {C}$ | Sin Ө $\hat{n}$ $\hat {n}$

by putting values

$\vec{A}$ $\vec {A}$ ｘ $\vec{C}$ $\vec {C}$ = 4 ｘ 8 ｘ Sin 90⁰ $\hat{n}$ $\hat {n}$

[ Sin 90⁰ = 1 ]

$\vec{A}$ $\vec {A}$ ｘ $\vec{C}$ $\vec {C}$ = 4 ｘ 8 ｘ 1ｘ 1

$\vec{A}$ $\vec {A}$ ｘ $\vec{C}$ $\vec {C}$ = 32 units ------Ans

By right-hand rules, the direction will be vertically downwards.

(c) $\vec{B}$ $\vec {B}$ ｘ $\vec{C}$ $\vec {C}$

Here we have

$\vec{B}$ $\vec {B}$ ｘ $\vec{C}$ $\vec {C}$ = | $\vec{A}$ $\vec {A}$ | | $\vec{C}$ $\vec {C}$ | Sin Ө $\hat{n}$ $\hat {n}$

by putting values

$\vec{B}$ $\vec {B}$ ｘ $\vec{C}$ $\vec {C}$ = 3 ｘ 8 ｘ Sin 90⁰ $\hat{n}$ $\hat {n}$

[ Sin 180⁰ = 0 ]

$\vec{B}$ $\vec {B}$ ｘ $\vec{C}$ $\vec {C}$ = 3 ｘ 8 ｘ 0 ｘ 1

$\vec{B}$ $\vec {B}$ ｘ $\vec{C}$ $\vec {C}$ = 0 units -------Ans

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Numerical Problem 2.11
⇑

List of 11th Physics Numerical Problems of Chapter - 2

⇑

Numerical Problem 2.10 ⇚ 11th Physics Main Content List ⇛ Numerical Problem 2.12

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Numerical Problem No. 2.11 (Vectors and Equilibrium) Physics HSSC - I (Class -11) (Punjab Curriculum and Text Board, Lahore) With Verified Answers

[Ans: (a)12 units vertically up (b) 32 units vertically down (c) Zero]

Given:

To Find:

(a)   $\vec{A}$ $\vec {A}$ ｘ $\vec{B}$ $\vec {B}$
(b)   $\vec{A}$ $\vec {A}$ ｘ $\vec{C}$ $\vec {C}$
(c)   $\vec{B}$ $\vec {B}$ ｘ $\vec{C}$ $\vec {C}$

Solution:

(a) $\vec{A}$ $\vec {A}$ ｘ $\vec{B}$ $\vec {B}$

Here we have

$\vec{A}$ $\vec {A}$ ｘ $\vec{B}$ $\vec {B}$ = | $\vec{A}$ $\vec {A}$ | | $\vec{B}$ $\vec {B}$ | Sin Ө $\hat{n}$ $\hat {n}$

by putting values

$\vec{A}$ $\vec {A}$ ｘ $\vec{B}$ $\vec {B}$ = 4 ｘ 3 ｘ Sin 90⁰ $\hat{n}$ $\hat {n}$

[ Sin 90⁰ = 1 ]

$\vec{A}$ $\vec {A}$ ｘ $\vec{B}$ $\vec {B}$ = 4 ｘ 3 ｘ 1 ｘ 1

$\vec{A}$ $\vec {A}$ ｘ $\vec{B}$ $\vec {B}$ = 12 units ------Ans

(b) $\vec{A}$ $\vec {A}$ ｘ $\vec{C}$ $\vec {C}$

Here we have

$\vec{A}$ $\vec {A}$ ｘ $\vec{C}$ $\vec {C}$ = | $\vec{A}$ $\vec {A}$ | | $\vec{C}$ $\vec {C}$ | Sin Ө $\hat{n}$ $\hat {n}$

by putting values

$\vec{A}$ $\vec {A}$ ｘ $\vec{C}$ $\vec {C}$ = 4 ｘ 8 ｘ Sin 90⁰ $\hat{n}$ $\hat {n}$

[ Sin 90⁰ = 1 ]

$\vec{A}$ $\vec {A}$ ｘ $\vec{C}$ $\vec {C}$ = 4 ｘ 8 ｘ 1ｘ 1

$\vec{A}$ $\vec {A}$ ｘ $\vec{C}$ $\vec {C}$ = 32 units ------Ans

By right-hand rules, the direction will be vertically downwards.

(c) $\vec{B}$ $\vec {B}$ ｘ $\vec{C}$ $\vec {C}$

Here we have

$\vec{B}$ $\vec {B}$ ｘ $\vec{C}$ $\vec {C}$ = | $\vec{A}$ $\vec {A}$ | | $\vec{C}$ $\vec {C}$ | Sin Ө $\hat{n}$ $\hat {n}$

by putting values

$\vec{B}$ $\vec {B}$ ｘ $\vec{C}$ $\vec {C}$ = 3 ｘ 8 ｘ Sin 90⁰ $\hat{n}$ $\hat {n}$

[ Sin 180⁰ = 0 ]

$\vec{B}$ $\vec {B}$ ｘ $\vec{C}$ $\vec {C}$ = 3 ｘ 8 ｘ 0 ｘ 1

$\vec{B}$ $\vec {B}$ ｘ $\vec{C}$ $\vec {C}$ = 0 units -------Ans

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Numerical Problem No. 2.11 (Vectors and Equilibrium) Physics HSSC - I (Class -11) (Punjab Curriculum and Text Board, Lahore) With Verified Answers

Vectors →AA→\vec {A}, →bb→\vec {b} and →CC→\vec {C} are 4 units north, 3 units west and 8 units east, respectively. Describe carefully (a) →AA→\vec {A} x →BB→\vec {B} (b) →AA→\vec {A} x →CC→\vec {C} (c) →BB→\vec {B} x →CC→\vec {C}

[Ans: (a)12 units vertically up (b) 32 units vertically down (c) Zero]

Given:

To Find:

(a) →AA→\vec {A} ｘ →BB→\vec {B}(b) →AA→\vec {A} ｘ →CC→\vec {C}(c) →BB→\vec {B} ｘ →CC→\vec {C}

Solution:

(a) →AA→\vec {A} ｘ →BB→\vec {B}

Here we have

→AA→\vec {A} ｘ →BB→\vec {B} = |→AA→\vec {A}| |→BB→\vec {B}| Sin Ө ˆnn^\hat {n}

by putting values

→AA→\vec {A} ｘ →BB→\vec {B} = 4 ｘ 3 ｘ Sin 90⁰ ˆnn^\hat {n}

[ Sin 90⁰ = 1 ]

→AA→\vec {A} ｘ →BB→\vec {B} = 4 ｘ 3 ｘ 1 ｘ 1

→AA→\vec {A} ｘ →BB→\vec {B} = 12 units ------Ans

(b) →AA→\vec {A} ｘ →CC→\vec {C}

Here we have

→AA→\vec {A} ｘ →CC→\vec {C} = |→AA→\vec {A}| |→CC→\vec {C}| Sin Ө ˆnn^\hat {n}

by putting values

→AA→\vec {A} ｘ →CC→\vec {C} = 4 ｘ 8 ｘ Sin 90⁰ ˆnn^\hat {n}

[ Sin 90⁰ = 1 ]

→AA→\vec {A} ｘ →CC→\vec {C} = 4 ｘ 8 ｘ 1ｘ 1

→AA→\vec {A} ｘ →CC→\vec {C} = 32 units ------Ans

By right-hand rules, the direction will be vertically downwards.

(c) →BB→\vec {B} ｘ →CC→\vec {C}

Here we have

→BB→\vec {B} ｘ →CC→\vec {C} = |→AA→\vec {A}| |→CC→\vec {C}| Sin Ө ˆnn^\hat {n}

by putting values

→BB→\vec {B} ｘ →CC→\vec {C} = 3 ｘ 8 ｘ Sin 90⁰ ˆnn^\hat {n}

[ Sin 180⁰ = 0 ]

→BB→\vec {B} ｘ →CC→\vec {C} = 3 ｘ 8 ｘ 0 ｘ 1

→BB→\vec {B} ｘ →CC→\vec {C} = 0 units -------Ans

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(a) $\vec{A}$ $\vec {A}$ ｘ $\vec{B}$ $\vec {B}$
(b) $\vec{A}$ $\vec {A}$ ｘ $\vec{C}$ $\vec {C}$
(c) $\vec{B}$ $\vec {B}$ ｘ $\vec{C}$ $\vec {C}$

(a) $\vec{A}$ $\vec {A}$ ｘ $\vec{B}$ $\vec {B}$

$\vec{A}$ $\vec {A}$ ｘ $\vec{B}$ $\vec {B}$ = | $\vec{A}$ $\vec {A}$ | | $\vec{B}$ $\vec {B}$ | Sin Ө $\hat{n}$ $\hat {n}$

$\vec{A}$ $\vec {A}$ ｘ $\vec{B}$ $\vec {B}$ = 4 ｘ 3 ｘ Sin 90⁰ $\hat{n}$ $\hat {n}$

$\vec{A}$ $\vec {A}$ ｘ $\vec{B}$ $\vec {B}$ = 4 ｘ 3 ｘ 1 ｘ 1

$\vec{A}$ $\vec {A}$ ｘ $\vec{B}$ $\vec {B}$ = 12 units ------Ans

(b) $\vec{A}$ $\vec {A}$ ｘ $\vec{C}$ $\vec {C}$

$\vec{A}$ $\vec {A}$ ｘ $\vec{C}$ $\vec {C}$ = | $\vec{A}$ $\vec {A}$ | | $\vec{C}$ $\vec {C}$ | Sin Ө $\hat{n}$ $\hat {n}$

$\vec{A}$ $\vec {A}$ ｘ $\vec{C}$ $\vec {C}$ = 4 ｘ 8 ｘ Sin 90⁰ $\hat{n}$ $\hat {n}$

$\vec{A}$ $\vec {A}$ ｘ $\vec{C}$ $\vec {C}$ = 4 ｘ 8 ｘ 1ｘ 1

$\vec{A}$ $\vec {A}$ ｘ $\vec{C}$ $\vec {C}$ = 32 units ------Ans

(c) $\vec{B}$ $\vec {B}$ ｘ $\vec{C}$ $\vec {C}$

$\vec{B}$ $\vec {B}$ ｘ $\vec{C}$ $\vec {C}$ = | $\vec{A}$ $\vec {A}$ | | $\vec{C}$ $\vec {C}$ | Sin Ө $\hat{n}$ $\hat {n}$

$\vec{B}$ $\vec {B}$ ｘ $\vec{C}$ $\vec {C}$ = 3 ｘ 8 ｘ Sin 90⁰ $\hat{n}$ $\hat {n}$

$\vec{B}$ $\vec {B}$ ｘ $\vec{C}$ $\vec {C}$ = 3 ｘ 8 ｘ 0 ｘ 1

$\vec{B}$ $\vec {B}$ ｘ $\vec{C}$ $\vec {C}$ = 0 units -------Ans