M.C.Q (1 Marks)

MCQ 511 Mark

Angle that the vector $\vec{\text{A}}=2\hat{\text{i}}+2\hat{\text{j}}$ makes with $y-$axis is:

A
$\tan^{-1}\Big(\frac{3}{2}\Big)$
✓
$\tan^{-1}\Big(\frac{2}{3}\Big)$
C
$\sin^{-1}\Big(\frac{2}{3}\Big)$
D
$\cos^{-1}\Big(\frac{3}{2}\Big)$

Answer

Correct option: B.

$\tan^{-1}\Big(\frac{2}{3}\Big)$

As $\vec{\text{A}}=2\hat{\text{i}}+2\hat{\text{j}},$
therefore $A_x = 2$ and $A_y = 3.$ If $\theta$ is the angle which $\vec{\text{A}}$ encloses with $y-$axis, then.
$\tan\theta=\frac{\text{A}_\text{x}}{\text{A}_\text{y}}=\frac{2}{3}$ or $\theta=\tan^{-1}\Big(\frac{2}{3}\Big)$

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MCQ 521 Mark

If a unit vector is represented by $0.5\hat{\text{i}}+0.8\hat{\text{j}}+\text{c}\hat{\text{k}},$ then the value of $'c\ '$ is:

A
$1$
✓
$\sqrt{0.11}$
C
$\sqrt{0.01}$
D
$\sqrt{0.39}$

Answer

Correct option: B.

$\sqrt{0.11}$

Here,$(0.5)^2 + (0.8)^2 + (c)^2 = 1$
or $\text{c}=\sqrt{0.11}$

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MCQ 531 Mark

A person moves $30m$ North, then $20m$ East then $30\sqrt{2}$ South$-$West. His displacement from the original position is:

A
$14m$ South$-$West.
B
$28m$ South.
✓
$10m$ West.
D
$15m$ East.

Answer

Correct option: C.

$10m$ West.

Resolving displacement $30\sqrt{2}\text{ m}$ south$-$west in two rectangular components:
we have $30\sqrt{2}\cos45^\circ=30\sqrt{2}\times\frac{1}{\sqrt{2}}=30\text{m}$ towards south
and $30\sqrt{2}\times\sin45^\circ=30\times\sqrt{2}\times\frac{1}{\sqrt{2}}=30\text{m}$ towards west.
The resultant of $30m$ north will neutralise the displacement of $30m$ south.
Hence, the effective displacement is the resultant of $30m$ west and $20m$ east $= 10m$ west.

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MCQ 541 Mark

The simple sum of two forces acting at a point is $16N$ and their sum is $8N$ and its direction is perpendicular to the smaller force, then the forces are:

✓
$6N$ and $10N$
B
$8N$ and $8N$
C
$4N$ and $12N$
D
$2N$ and $14N$

Answer

Correct option: A.

$6N$ and $10N$

Here $A + B = 16 ...(i)$
$\sqrt{\text{A}^2+\text{B}^2+2\text{AB}\cos\theta}=8\dots(\text{ii})$
and $\tan90^\circ=\frac{\text{B}\sin\theta}{\text{A}+\text{B}\cos\theta}$
or $\text{A}+\text{B}\cos\theta=\frac{\text{B}\sin\theta}{\tan90^\circ}=0$
or $\text{B}\cos\theta=-\text{A}$ or $\cos\theta=\frac{-\text{A}}{\text{B}}$
From $(ii), \text{A}^2+\text{B}^2+2\text{AB}\Big(\frac{-\text{A}}{\text{B}}\Big)=64$
or $B^2 - A^2 = 64$
Solving $(i)$ and $(iii),$ we get
$A = 6N$ and $B = 10N.$

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MCQ 551 Mark

For two vectors $A$ and $B, |\text{A} + \text{B}| = |\text{A} - \text{B}|$ is always true when

A
$|\text{A}|=|\text{B}|\neq0$
B
$\text{A}\bot\text{B}$
C
$|A|=|B|\neq 0$ and $A$ and $B$ are parallel or anti parallel
✓
$B$ and $C$

Answer

Correct option: D.

$B$ and $C$

According to the problem, $|\vec{\text{A}}+\vec{\text{B}}|=|\vec{\text{A}}-\vec{\text{B}}|$
$\Rightarrow\ \sqrt{|\vec{\text{A}}|^2+|\vec{\text{B}}|^2+2|\vec{\text{A}}||\vec{\text{B}}|\cos\theta}\\=\sqrt{|\vec{\text{A}}|^2+|\vec{\text{B}}|^2-2|\vec{\text{A}}||\vec{\text{B}}|\cos\theta}$
$\Rightarrow\ |\vec{\text{A}}|^2+|\vec{\text{B}}|^2+2|\vec{\text{A}}||\vec{\text{B}}|\cos\theta\\=|\vec{\text{A}}|^2+|\vec{\text{B}}|^2-2|\vec{\text{A}}||\vec{\text{B}}|\cos\theta$
$\Rightarrow\ 4|\vec{\text{A}}||\vec{\text{B}}|\cos\theta=0$
$\Rightarrow\ |\vec{\text{A}}||\vec{\text{B}}|\cos\theta=0$
$|\vec{\text{A}}|=0$ or $|\vec{\text{B}}|=0$ or $\cos\theta=0$
i.e. $\theta=90^\circ$
When $\theta=90^\circ,$ we can say that $\vec{\text{A}}\bot\vec{\text{B}}.$
Hence options $(B)$ and $(C)$ are correct.

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MCQ 561 Mark

A constant force is acting perpendicular to the velocity of a particle. For this situation which one is correct?

A
Velocity is constant.
✓
Acceleration is constant.
C
Momentum will be constant.
D
Particle will follow elliptical path.

Answer

Correct option: B.

Acceleration is constant.

When a constant force will be acting perpendicular to the velocity, the body will describe a circular path and its acceleration $($called centripetal acceleration$)$ will be constant.

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MCQ 571 Mark

If the resultant of three forces $\vec{\text{F}}_1=\text{p}\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}},\ \vec{\text{F}}_2=-5\hat{\text{i}}+2\hat{\text{k}},$ and $\vec{\text{F}}_3=6\hat{\text{i}}-\hat{\text{k}}$ acting on a particle has a magnitude equal to $5$ units, then the value of $p$ is:

A
$-6$
B
$-4$
✓
$3$
D
$4$

Answer

Correct option: C.

$3$

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Physics M.C.Q (1 Marks)