Tag: forces - vectors and moments

Quizzes (8) Questions (83)

Questions Related to forces - vectors and moments

Multiple choice physics turning effects of forces centre of gravity forces - vectors and moments acceleration due to gravity

If a charge particle projected in a gravity-free room it does not deflect, 

  1. electric field and magnetic field must be zero

  2. both electric field and magnetic field may be present

  3. electric field will be zero and magnetic field may be zero

  4. electric field may be zero and magnetic field may be zero
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If a charged particle moves through a region without deflection, the net force must be zero. The Lorentz force is F = q(E + v x B). If E = 0 and B = 0, the force is zero. While other configurations exist (like E and B being parallel to v), the most fundamental condition for no deflection regardless of velocity is that both fields are zero.

Multiple choice physics turning effects of forces centre of gravity forces - vectors and moments acceleration due to gravity

If the earth shrinks such that its mass does not change but radius decreases to one quarter of its original value then one complete day will take:

  1. 96 h

  2. 48 h

  3. 6 h

  4. 1.5 h
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We know that angular momentum of spin $\displaystyle =I\omega $
Bythe conservation of angular momentum
$\displaystyle \frac { 2 }{ 5 } M{ R }^{ 2 }.\frac { 2\pi  }{ T } =\frac { 2 }{ 5 } M{ \left( \frac { R }{ 4 }  \right)  }^{ 2 }.\frac { 2\pi  }{ T' } $
$\displaystyle T'=\frac { T }{ 16 } =\frac { 24 }{ 16 } =1.5h$

Multiple choice physics turning effects of forces centre of gravity forces - vectors and moments acceleration due to gravity

Weight $W _ { m }$ of the body can be given as 

  1. $m g - m \frac { \left( v _ { e } + v \right) ^ { 2 } } { R }$

  2. $m g - m \frac { \left( v _ { e } - v \right) ^ { 2 } } { R }$

  3. $\frac { m } { R } \left[ v _ { e } ^ { 2 } - \left( v _ { e } + v \right) ^ { 2 } \right]$

  4. $m g + m \frac { \left( v _ { e } + v \right) ^ { 2 } } { R }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This relates to the effective weight of a body in a rotating frame or a non-inertial frame. The term m(v_e + v)^2 / R represents the centrifugal force component acting on the body, which reduces the effective gravitational weight.

Multiple choice physics turning effects of forces centre of gravity forces - vectors and moments acceleration due to gravity

Why are the passengers in the upper deck of a double-decker bus not allowed to stand?

  1. This ensures that the centre of gravity of the system may not rise up and the bus may not be toppled due to unstable equilibrium

  2. This ensures smaller centripetal force, thus helping the driver to negotiate the roundabouts properly

  3. If the passengers are in standing position, they may start oscillating due to jerks and there is a possibility of resonance, causing the bus to be toppled

  4. This is just for the safety reason
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When the passengers stand, the center of gravity rises. If it rises much it becomes unstable and may topple down.

Multiple choice physics forces - vectors and moments concept of force and its unit force - push or pull forces

Which of the following pairs of forces cannot be added to give a resultant force of $4N$?

  1. $2 N$ and $8 N$

  2. $2 N$ and $2 N$

  3. $2 N$ and $6 N$

  4. $2 N$ and $4 N$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Resultant force, 

$F=\sqrt{F _1^2+F _2^2+2F _1F _2cos\theta}$. . . . . . . . .(1)
Lets consider the option A,

$F _1=2N$
$F _2=8N$

For maximum resultant force, $cos\theta=1$
From equation (1),
$F _{max}=\sqrt{F _1^2+F _2^2+2F _1F _2}=\sqrt{(F _1+F _2)^2}$
$F _{max}=F _1+F _2$
$F _{max}=8+2=10N$
For minimum resultant force, $cos\theta=-1$
$F _{min}=\sqrt{F _1^2+F _2^2-2F1F _2}=\sqrt{(F _1-F _2)^2}$
$F _{min}=\sqrt{(2-8)^2}=6N$
So, the $4N$ does not lie within this range. Thus it is not possible to have it as resultant force.
The correct option is A.

Multiple choice physics forces - vectors and moments concept of force and its unit force - push or pull forces

Two forces 12 N and 5 N are acting perpendicular to each other. Then the net force acting is.

  1. 17 N

  2. 18 N

  3. 7 N

  4. zero

  5. 13 N
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

$ \vec F _{net} = \vec F _1 +\vec F _2$

$| \vec F _{net}| = \sqrt{F _1^2 +F _2^2 +2F _1F _2\cos\theta}$
$| \vec F _{net}| = \sqrt{12^2 +5^2 +2\times 12 \times 5\cos 90^0}$
$| \vec F _{net}| = \sqrt{12^2 +5^2 }$
$| \vec F _{net}| = 13N$
Therefore, E is correct option.

Multiple choice physics forces - vectors and moments concept of force and its unit force - push or pull forces

Two forces of $12 \mathrm { N } \text { and } 8 \mathrm { N }$ acts upon a body. The resultant force on the body has maximum value of

  1. $4 \mathrm { N }$

  2. $0 \mathrm { N }$

  3. $20 \mathrm { N }$

  4. $8 \mathrm { N }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given,

$|\vec F _1|=12N$
$|\vec F _2|=8N$

The resultant force on the body is 
$|\vec R|=\sqrt{F _1^2+F _2^2+2F _1F _2cos\theta}$

For maximum value, $cos\theta=1$
$|\vec R|=\sqrt{(12)^2+(8)^2+2\times 12\times 8\times 1}$

$|\vec R _{max}|=\sqrt{144+64+24\times 8}$

$|\vec R _{max}|=\sqrt{400}=20N$
The correct option is C.

Multiple choice physics forces - vectors and moments centre of gravity turning effects of forces acceleration due to gravity

Can the centre of gravity be situated outside the material of the body ?

  1. Yes

  2. No

  3. Can't say

  4. None
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Yes, it can. For example, in case of a ring, it is situated at the centre of that circle. But the material is only along the circumference. Hence centre of gravity is situated outside the material of the body.