Question

Try to answer atleast 3 Questions.
Irrelevant answers and copied answers will be reported.​

Answer 1

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2 .The maximum and minimum magnitudes of the resultant of two vectors of magnitudes P and Q are in the ratio 3:1 . Which of the following relations is true ?

3 .The maximum and minimum magnitude of the resultant of two given vectors are 17 units and 7 units respectively . If these two vectors are at right angles to each other , the magnitude of their resultant is

4 .Which pair of the following forces will never give resultant force of 2N ?

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2. Given is that : There are two vectors p and q and their maximum and minimum magnitudes are in the ratio 3:1

Now , There maximum magnitude would be p + q

Similarly ,

There minimum magnitude would be p - q ( letting p be greater than q )

We have , P + Q : P - Q = 3 : 1 ( Given ) - 1

From 1 we can assume that

P + Q = 3 - equation (2)

P - Q = 1 - equation (3)

Adding 2 and 3 we get ,

P + Q + P - Q = 3 + 1

Q gets cancelled ,

2P = 4

$\sf\implies\:P\:=\:\dfrac{4}{2}$

$\sf\implies\:P\:=\:2$ - equation (4)

$\sf\implies\:Q\:=\:1$ - equation (5)

From equation 4 and 5 we can conclude that P is twice Q that is P = 2 × 1

So hence Option A i.e P = 2Q is correct

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Let the two vectors be A and B . Also , let the magnitude of A be greater than that of B .

Their maximum and minimum magnitudes would be ,

$\rm | \overline{A}| + | \overline{B}| = 17\\ \\ \rm | \overline{A}| - | \overline{B}| = 7 \\ \\ \rm \: |A| = 12 \\ \\ |B| = 5$

It is given that both of the vectors are perpendicular to each other so we have ,

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Now by Pythagoras theorem we have the value of resultant vector as ,

$\implies\:\sqrt{12^2\:+\:5^2}$

$\implies\:\sqrt{144\:+\:25}$

$\implies\:\sqrt{169}$

$\large\boxed{\implies\:=\:13}$

Hence option D is correct i.e 13

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Let's try understanding the conditions why would the given options would be correct or not .

So as if the pair ( 1N , 1N ) are two forces acting in the same direction would give resultant force as 1N + 1N = 2N

Studying the second pair we have ( 3N , 1N ) this would give us a resultant 2N in case both of them are acting in opposite sense i.e 3N - 1N gives 2N

Observing the third pair we have ( 2N , 2N ) This pair would surely give resultant force as 2N if the angel formed between both of these is 60° ( forming an equilateral triangle ) .

Now comes the last pair that is ( 1N , 4N ) . Note that if the resultant of 1N and 4N is 2N then it doesn't form a triangle . Reason - A triangle has a property that the third side of a triangle is always greater than the sum of the other two sides . But here 1 + 4 > 2

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5 > 2. It fails the eligibility to form a triangle hence , the pair 1N , 4N won't give 2N as a resultant .

Therefore the correct option is D i.e 1N and 4N

Answer 2

2. Given,

$\longrightarrow|\vec{\sf{A}}+\vec{\sf{B}}|=|\vec{\sf{A}}|+|\vec{\sf{B}}|$

$\sf{\longrightarrow \sqrt{A^2+B^2+2AB\cos\theta}=A+B}$

Squaring both sides,

$\sf{\longrightarrow A^2+B^2+2AB\cos\theta=(A+B)^2}$

$\sf{\longrightarrow A^2+B^2+2AB\cos\theta=A^2+B^2+2AB}$

$\sf{\longrightarrow\cos\theta=1}$

$\sf{\longrightarrow\underline{\underline{\theta=0^o}}}$

Hence (c) is the answer.

3. The magnitude of resultant of two vectors $\vec{\sf{P}}$ and $\vec{\sf{Q}}$ are given by,

$\longrightarrow|\vec{\sf{P}}+\vec{\sf{Q}}|=\sf{\sqrt{P^2+Q^2+2PQ\cos\theta}}$

Condition for maximum value is that $\sf{\cos\theta=1}$ [$\cos\theta$ should be maximum].

$\longrightarrow\max|\vec{\sf{P}}+\vec{\sf{Q}}|=\sf{\sqrt{P^2+Q^2+2PQ}}$

$\longrightarrow\max|\vec{\sf{P}}+\vec{\sf{Q}}|=\sf{P+Q}$

Condition for minimum value is that $\sf{\cos\theta=-1}$ [$\cos\theta$ should be minimum].

$\longrightarrow\min|\vec{\sf{P}}+\vec{\sf{Q}}|=\sf{\sqrt{P^2+Q^2-2PQ}}$

$\longrightarrow\min|\vec{\sf{P}}+\vec{\sf{Q}}|=\sf{|P-Q|}$

Assume $\sf{P>Q.}$ In the question,

$\longrightarrow\dfrac{\max|\vec{\sf{P}}+\vec{\sf{Q}}|}{\min|\vec{\sf{P}}+\vec{\sf{Q}}|}=\sf{\dfrac{3}{1}}$

$\sf{\longrightarrow\dfrac{P+Q}{P-Q}=\dfrac{2+1}{2-1}}$

By rule of componendo and dividendo,

$\sf{\longrightarrow\dfrac{P}{Q}=\dfrac{2}{1}}$

$\sf{\longrightarrow\underline{\underline{P=2Q}}}$

Hence (a) is the answer.

4. Let the two vectors be $\vec{\sf{P}}$ and $\vec{\sf{Q}}.$ Let $\sf{P>Q.}$

The maximum magnitude of their resultant,

$\sf{\longrightarrow P+Q=17}$

The minimum magnitude of their resultant,

$\sf{\longrightarrow P-Q=7}$

Solving them we get,

$\sf{\longrightarrow P=12}$

$\sf{\longrightarrow Q=5}$

Then the magnitude of the resultant when both are right angled to each other,

$\sf{\longrightarrow R=\sqrt{P^2+Q^2+2PQ\cos90^o}}$

$\sf{\longrightarrow R=\sqrt{P^2+Q^2}}$

$\sf{\longrightarrow R=\sqrt{12^2+5^2}}$

$\sf{\longrightarrow\underline{\underline{R=13}}}$

Hence (d) is the answer.

5. The magnitude of resultant of two vectors $\vec{\sf{P}}$ and $\vec{\sf{Q}}$ are given by,

$\longrightarrow|\vec{\sf{P}}+\vec{\sf{Q}}|=\sf{\sqrt{P^2+Q^2+2PQ\cos\theta}}$

We know that the range of $\sf{\cos\theta}$ is,

$\sf{\longrightarrow\cos\theta\in[-1,\ 1]}$

Therefore,

$\longrightarrow\sf{\sqrt{P^2+Q^2+2PQ\cos\theta}\in\left[\sqrt{P^2+Q^2-2PQ},\ \sqrt{P^2+Q^2+2PQ}\right]}$

$\longrightarrow|\vec{\sf{P}}+\vec{\sf{Q}}|\in\sf{\left[|P-Q|,\ P+Q\right]}$

Thus,

$\longrightarrow|\vec{\sf{2}}+\vec{\sf{2}}|\in\sf{\left[0,\ 4\right]\ni2}$

$\longrightarrow|\vec{\sf{1}}+\vec{\sf{1}}|\in\sf{\left[0,\ 2\right]\ni2}$

$\longrightarrow|\vec{\sf{1}}+\vec{\sf{3}}|\in\sf{\left[2,\ 4\right]\ni2}$

$\longrightarrow|\vec{\sf{1}}+\vec{\sf{4}}|\in\sf{\left[3,\ 5\right]\not\ni2}$

We see the force of magnitude 1 N and 4 N cannot have a resultant force of magnitude 2 N.

Hence (d) is the answer.