Tag: physics

Questions Related to physics

If E, M, J and G, respectively, denote energy, mass, angular momentum, and gravitational constant, then $EJ^2 / M^5G^2$ has the dimensions of

physics universe and space heliocentric model introduction to gravitation introduction to gravity

  1. Time

  2. Angle

  3. Mass

  4. Length

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Correct Option: B
Explanation:

$E=ML^2T^{-2}\ M+ML^0T^0\ G=M^{-1}L^3T^{-2}\ J=ML^2T^{-1}$

$EJ^2/M^5G^2=Ml^2T^{-2}\times M^2L^2T^{-1}$
$=M^5\times M^{-2}\times L^{-6}T^{-4}\ =M^0L^0T^0$

Choose the correct statement:

physics universe and space heliocentric model introduction to gravitation introduction to gravity

  1. The dimensional formula for $G$ is ${M}^{-1}{L}^{3}{T}^{-2}$

  2. $G$ is independent of medium

  3. $F=G\cfrac{{m} _{1}{m} _{2}}{{r}^{2}}$

  4. All of the above

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Correct Option: D

To double the orbital speed V and halve the angular velocity $\omega $, The centripetal acceleration of a revolving body:

physics universe and space heliocentric model introduction to gravitation introduction to gravity

  1. Is quadrupled

  2. Remains Unchanged

  3. Is halved

  4. Is doubled

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Correct Option: A

The force of attraction between two unit point masses separated by a unit distance is called

physics universe and space heliocentric model introduction to gravitation introduction to gravity

  1. Gravitational potential

  2. Acceleration due to gravity.

  3. Gravitational field

  4. Universal gravitational constant.

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Correct Option: D
Explanation:

Universal gravitational constant is the force of attraction between two bodies of unit mass and at a unit distance from each other.

In the relation F= $\dfrac{G M m}{r^{2}}$, the quantity G

physics universe and space heliocentric model introduction to gravitation introduction to gravity

  1. depends on the value of g at the place of observation.

  2. is used only when the earth is one of the two masses.

  3. is greatest at the surface of the earth.

  4. is universal constant in nature.

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Correct Option: D
Explanation:

G is the universal gravitational constant which remains constant irrespective of the place and time. G is the force of attraction between two bodies of unit mass and unit distance apart.

A body of mass $5\ kg$ is cut into two parts of masses (a) $\dfrac {m}{4}; \dfrac {3m}{4}$ (b) $\dfrac {m}{7}; \dfrac {5m}{7}$ (c) $\dfrac {m}{2}; \dfrac {m}{2}$ (d) $\dfrac {m}{5}; \dfrac {4m}{5}$. When these two pieces are kept apart by certain distance; In which case the gravitational force acting is maximum?

physics universe and space heliocentric model introduction to gravitation introduction to gravity

  1. In case a

  2. In case C

  3. In case d

  4. In case b

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Correct Option: B
Explanation:

Gravitation force between two masses is given by $F=\cfrac { G{ m } _{ 1 }{ m } _{ 2 } }{ { r }^{ 2 } } $

In case I:${ F } _{ 1 }=\cfrac { G\cfrac { m }{ 4 } .\cfrac { 3m }{ 4 }  }{ { r }^{ 2 } } =\cfrac { G{ m }^{ 2 } }{ { r }^{ 2 } } \times \cfrac { 1 }{ 16 } =\cfrac { G{ m }^{ 2 } }{ { r }^{ 2 } } \times 0.0625$
In case II: ${ F } _{ 2 }=\cfrac { G\cfrac { m }{ 7 } .\cfrac { 5m }{ 7 }  }{ { r }^{ 2 } } =\cfrac { G{ m }^{ 2 } }{ { r }^{ 2 } } \times \cfrac { 5 }{ 7 } =\cfrac { G{ m }^{ 2 } }{ { r }^{ 2 } } \times 0.714$
In case III: ${ F } _{ 3 }=\cfrac { G\cfrac { m }{ 2 } .\cfrac { m }{ 2 }  }{ { r }^{ 2 } } =\cfrac { G{ m }^{ 2 } }{ { r }^{ 2 } } .\cfrac { 1 }{ 4 } =\cfrac { G{ m }^{ 2 } }{ { r }^{ 2 } } \times 0.25$
${ F } _{ 2 }$ is maximum.

Magnetic induction due to a short bar magnet on its axial line is inversely proportional to cube of distance of the point.

physics magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

  1. True

  2. False

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Correct Option: A
Explanation:

Magnetic induction due to a short bar magnet on its axial line,

$B=\dfrac{\mu _0 M}{4\pi d^3}$
Magnetic induction due to a short bar magnet on its axial line is inversely proportional to cube of distance of the point.
$B\propto\dfrac{1}{d^3}$

The magnetic induction due to short bar magnet on its axial line at a distance 'd' is 'B'. What is the magnetic induction due to the same bar magnet on the same line at a distance $\displaystyle \frac{d}{4}?$

physics magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

  1. 16B

  2. 32B

  3. 64B

  4. 128B

Show answer
Correct Option: C
Explanation:

$B \displaystyle = \frac{\mu _0  2M}{4 \pi  d^3}$
At $\displaystyle \frac{d}{4} $ distance,
$B' \displaystyle = \frac{\mu _0 2M}{4 \pi (d/4)^3}$
$\displaystyle = \frac{\mu _0  2M}{4 \pi d^3} \times 64 = 64 B$

If r be the distance of a point on the axis of a bar magnet from its centre, the magnetic field at this point is proportional to :

physics magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

  1. (1/r)

  2. (1/r$^2$)

  3. (1/r$^3$)

  4. (1/r$^5$)

Show answer
Correct Option: C
Explanation:

For a short Bar Magnet, the magnetic induction at a point on the axix at a distance $r$ from centre is given by  the formula

$B = $   $(\dfrac{\mu _0}{4\pi} )\dfrac{2M}{r^3}$

$\Rightarrow$ $B= \dfrac{K}{r^3}$

$\Rightarrow$ $B\propto \dfrac{1}{r^3}$
Therefore, C is correct option.

A bar magnet of magnetic moment 'M' has a magnetic length '2d'. Find magnetic induction on its equatorial line at a distance $'\sqrt{13 d}'$.

physics magnetism the bar magnet magnetic field due to bar magnet intensity of magnetic field and torque on a bar magnet

  1. $\displaystyle \frac{\mu _0 M}{4 \pi (d^3)(17)^{\frac{3}{2}}}$

  2. $\displaystyle \frac{2\mu _0 M}{4 \pi (d^3)(17)^{\frac{3}{2}}}$

  3. $\displaystyle \frac{4\mu _0 M}{4 \pi (d^3)(17)^{\frac{3}{2}}}$

  4. $\displaystyle \frac{8\mu _0 M}{4 \pi (d^3)(17)^{\frac{3}{2}}}$

Show answer
Correct Option: A
Explanation:

$r = \sqrt{13} d ;  2l  = 2d$
$B _eq = \displaystyle \frac{\mu _0}{4 \pi} \times \frac{\mu}{(r^2 + 1^2)^{\frac{3}{2}}}$
$\displaystyle =\frac{\mu _0}{4\pi} \times \frac{M}{\left ((\sqrt{13}d)^2 + (2d)^2 \right )^{\frac{3}{2}}}$
$=\displaystyle \frac{\mu _0}{4\pi} \times \frac{M}{(17 d^2)^{\frac{3}{2}}} = \frac{\mu _0 M}{4 \pi (d^3)(17)^{\frac{3}{2}}}$