Tag: acceleration due to gravity

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


Correct Option: D
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.

Instrument used for determining specific gravity is

  1. Hydrometer

  2. Barometer

  3. Calorimeter

  4. None of these


Correct Option: A

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


Correct Option: A
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.

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


Correct Option: A

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


Correct Option: D
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' } $
$\displaystyle T'=\frac { T }{ 16 } =\frac { 24 }{ 16 } =1.5h$

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


Correct Option: A

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


Correct Option: A
Explanation:

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

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

  1. Yes

  2. No

  3. Can't say

  4. None


Correct Option: A
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.

Reference to ability of an object to return to its original position after it has been tilted slightly is termed as:

  1. Stability

  2. Equilibrium

  3. Centre of gravity

  4. Torque


Correct Option: A
Explanation:

stability of an object depends on the ability of an object to return back to its original equlibirum, when disturbed

If you place pivot at center of a meter rule, weight has no

  1. property

  2. Concern

  3. Turning effect

  4. Magnitude


Correct Option: C
Explanation:

The weight of an object is concentrated at the centre of gravity and hence a pivot at the centre of the metre scale does not record any changes in mass. Hence the scale does not turn around