Tag: physics

Quizzes (598) Questions (13441)

Questions Related to physics

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

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

  1. Time

  2. Angle

  3. Mass

  4. Length
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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

Choose the correct statement:

  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
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The dimensional formula for G is M^-1 L^3 T^-2, G is independent of the medium, and the formula F = G * m1 * m2 / r^2 is correct. All statements are true.

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

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

  1. Gravitational potential

  2. Acceleration due to gravity.

  3. Gravitational field

  4. Universal gravitational constant.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

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

  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.
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

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?

  1. In case a

  2. In case C

  3. In case d

  4. In case b
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
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.
$B\propto\dfrac{1}{d^3}$

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

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 $\displaystyle \frac{d}{4}?$

  1. 16B

  2. 32B

  3. 64B

  4. 128B
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

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

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 :

  1. (1/r)
  2. (1/r$^2$)
  3. (1/r$^3$)
  4. (1/r$^5$)
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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

  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. $\displaystyle \frac{8\mu _0 M}{4 \pi (d^3)(17)^{\frac{3}{2}}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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