Tag: turning effects of forces

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Questions Related to turning effects of forces

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

Two planets have radii $r _1$ and $r _2$ and their densities are $\rho _1$ and $\rho _2$, respectively. The ratio of acceleration due to gravity on them will be.

  1. $r _1\rho _1:r _2\rho _2$

  2. $r _1\rho^2 _1:r _2\rho^2 _2$

  3. $r^2 _1\rho _1:r^2 _2\rho _2$

  4. $r _1\rho _2:r _2\rho _1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\dfrac { { g } _{ 1 } }{ { g } _{ 2 } } =\dfrac { { GM } _{ 1 } }{ { r } _{ 1 }^{ 2 } } \div \dfrac { { GM } _{ 2 } }{ { r } _{ 2 }^{ 2 } } $
$=\dfrac { { Gp } _{ 1 }\dfrac { 4 }{ 3 } \pi { r } _{ 1 }^{ 3 } }{ { r } _{ 1 }^{ 2 } } \div \dfrac { { Gp } _{ 2 }\dfrac { 4 }{ 3 } \pi { r } _{ 2 }^{ 3 } }{ { r } _{ 2 }^{ 2 } } $
$=\dfrac { { p } _{ 1 }{ r } _{ 1 } }{ { p } _{ 2 }{ r } _{ 2 } } $
$=\dfrac { { r } _{ 1 }{ p } _{ 1 } }{ { r } _{ 2 }{ p } _{ 2 } } $
$={ r } _{ 1 }{ p } _{ 1 }:{ r } _{ 2 }{ p } _{ 2 }$
Multiple choice physics turning effects of forces centre of gravity forces - vectors and moments acceleration due to gravity

An object after deforming 

  1. may have different centre of mass.

  2. have same centre  of mass

  3. doesn't change its centre of gravity 

  4. none of the above 
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Centre of mass is a point on a physical body which behaves as if the whole mass is concentrated over there. Whenever a force acts on a body it appears to be acting on the centre of mass. 

The centre of mass depends upon how the particles of the body are scattered around it. It is closer towards the heavier regions of the body as compared to the lighter regions.  Whenever an object is deformed, its heavier regions may change their locations. As a result, the centre of mass may change its location

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

Racing car can corner rapidly without turning due to

  1. having high centre of gravity

  2. having lower centre of gravity

  3. equal mass distribution of the car

  4. none of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Racing car can corner rapidly without turning due to low centre of gravity. The lower the centre of gravity is more stable is the object The higher it is more likely  the object is to topple over if it is pushed.Racing cars have really low centre of gravity so that they can corner rapidly withoit turning.

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

A man can jump 1.5m high on the earth. Calculate the approximate height he might be able to jump on a planet whose density is one-quarter that of the earth and whose radius is one-third of the earth's radius.

  1. $1.5\ m$

  2. $15\ m$

  3. $18\ m$

  4. $28\ m$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The height a person can jump is inversely proportional to the acceleration due to gravity g. g = G * M / R^2. Since M = density * volume = rho * (4/3) * pi * R^3, g is proportional to rho * R. If rho' = rho/4 and R' = R/3, then g' = g * (1/4) * (1/3) = g/12. The jump height h' = h * (g/g') = 1.5 * 12 = 18 m.

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

Which among the following is the major cause of nutation in Earth's axis:

  1. Earth's Oblate spheroid shape

  2. The difference between Land and Sea hemispheres

  3. Oceanic Currents

  4. TIdal Forces
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
In case of earth the principal sources of tidal forces are the sun and moon while continuously change location relative to each other and cause nutation in earth's axis, So option (D) correct.
Multiple choice physics turning effects of forces centre of gravity forces - vectors and moments acceleration due to gravity

The mean radius of the earth is $R$, its angular speed about its own axis $\omega$ and the acceleration due to gravity at the earth surface is $g$. The cube of radius of orbit of 'geostationary satellite' will be

  1. $(R^{2}g/ \omega)$

  2. $(R^{2}\omega/ g)$

  3. $(Rg/ \omega^{2})$

  4. $(R^{2}g/ \omega^{2})$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$mr\omega^{2} = \dfrac {GMm}{r^{2}}$
$\Rightarrow r\omega^{2} = \dfrac {GM}{r^{2}}$
$\Rightarrow r^{3} = \dfrac {GM}{\omega^{2}} = \dfrac {GM}{R^{2}} . \dfrac {R^{2}}{\omega^{2}}$
$\Rightarrow r^{3} = g \dfrac {R^{2}}{\omega^{2}}$
Hence (D) is correct.

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

For a ball falling in a iiquid with constant velocity, ratio of resistance force due to the liquid to that due to gravity is 

  1. $\frac{{2{a^2}\rho g}}{{9xv}}$

  2. $\dfrac { 2 a ^ { 2 } \rho g } { 9 \eta ^ { 2 } }$

  3. $\dfrac { 2 a ^ { 2 } ( \rho - \sigma ) g } { 9 \eta }$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Force due to liquid (stokes law):

$f = 6\pi xrv$
Force due to gravity:
$\begin{array}{l} f=mg \ =\rho .\dfrac { { 4\pi { r^{ 3 } } } }{ 3 } .g \end{array}$
Ratio =
$\ \begin{array}{l} \dfrac { { 4\rho .\pi { r^{ 2 } }.g } }{ { 3.6\pi xr.v } }  \ =\dfrac { { 2{ a^{ 2 } }\rho g } }{ { 9xv } }  \end{array}$
$\therefore \,\,\,\,\,\dfrac{{2{a^2}\rho g}}{{9xv}}$
so,
Option $A$ is correct answer.

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.