Tag: physics

Quizzes (598) Questions (13441)

Questions Related to physics

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

A thin rod of length $\dfrac {f}{3}$ is placed along the optic axis of a concave mirror of focal length f such that its image which is real and elongated just touches the rod. The magnification is:

  1. $\dfrac {3}{4}$
  2. $\dfrac {1}{2}$
  3. $\dfrac {3}{2}$

  4. None of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Here, the object lies along the axis. The two ends of the object should be treated as two point objects and the difference between the corresponding image distances gives the length of the image. When one end of the image touches the rod, this end must be at 2 f. In this situation the other end of the rod can be towards the left or right of 2 f. Since the image of the rod is elongated, the other end of the rod must lie between f and 2 f, the image (when the object lies between f and 2f, the image is formed more far away behind 2 f).
So, object distance for closer end of the rod is 2f-f/3 and that of the farther end is 2 f. The difference between the corresponding image distance is found to be $\dfrac {f}{2},$ i.e. length of the image is $\dfrac {f}{2}$.
magnification$=\dfrac {\text {length of image}}{\text {length of object}}=\dfrac {f/2}{f/3}=\dfrac {3}{2}$
Note that the image is elongated and the only option which is greater than 1 is (c).

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

The focal length of a convex lens of refractive index $1.5$ is $f$ when it is places in air. When it is immersed in a liquid it behaves as a converging lens its focal length becomes $xf(x>1)$. The refractive index of the liquid

  1. $>3/2$
  2. $<(3/2)$ and $>1$
  3. $<3/2$

  4. all of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\dfrac { 1 }{ f } =\left( n-1 \right) \left( \dfrac { 1 }{ { R } _{ 1 } } +\dfrac { 1 }{ { R } _{ 2 } }  \right) $
$\Rightarrow \dfrac { 1 }{ f } =\left( \dfrac { 1.5 }{ 1 } -1 \right) \left( \dfrac { 1 }{ { R } _{ 1 } } +\dfrac { 1 }{ { R } _{ 2 } }  \right)$ when the lens is placed in air and 
$\dfrac { 1 }{ xf } =\left( \dfrac { 1.5 }{ y } -1 \right) \left( \dfrac { 1 }{ { R } _{ 1 } } +\dfrac { 1 }{ { R } _{ 2 } }  \right)$ when the lens is places in the liquid.
where $y=R.l.$ of the liquid
solving we get, $y=\dfrac {3}{2+1/x}$
Hence $(B)$ is correct.

Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

The nature of image of a candle flame located $40$cm from a concave spherical mirror is real, inverted and magnified four times. Then the radius of curvature of the mirror is:

  1. $32$ cm
  2. $64$ cm
  3. $48$ cm
  4. $80$ cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Distance of the Candle from the Concave Mirror (u) = $40 cm$.(negative)

Now, as per as the Question,

$Image Height = 4 × Object (or Candle) Height$

 $Image Height/Candle Height = 4$

 Magnification = $4$

[ Magnification = Image Height/Object Height]

Now, Magnification = $-v/u$

 $4 = -v/u$

 $-v = 4u$

 $v = -4u$

 $v = -4 \times 40$

 $v = -160 cm.$

Now, Image Distance(v) = - $160 cm.$

Using the Mirror's Formula,

 On Multiplying both sides by $160$ ,

We get,

⇒  $160/f = -1 - 4$

⇒ $160/f = -5$

⇒ $f = 160/-5$

⇒ $f = -32 cm$.

∴ Focal length of the Concave Mirror is 32 cm.

Now, For the Radius of the Curvature,

Using the Formula,

 $ Focal Length = Radius Of Curvature/2$

∴ $Radius of Curvature = Focal Length \times 2$

∴ $R = F \times 2$

∴ $R = 32 \times 2$

∴ $R = 64 cm.$

Hence, the Radius of the Curvature of the Concave mirror of Focal Length 3 cm is 64 cm.

Multiple choice physics mathematical methods multiplication of vectors products of vectors scalar product

Three vectors satisfy the relation $\displaystyle \overrightarrow { A } .\overrightarrow { B } =0$ and $\displaystyle \overrightarrow { A } .\overrightarrow { C } =0$, then $\displaystyle \overrightarrow { A } $ is parallel to:

  1. $\displaystyle \overrightarrow { C } $
  2. $\displaystyle \overrightarrow { B } $
  3. $\displaystyle \overrightarrow { B } \times \overrightarrow { C } $
  4. $\displaystyle \overrightarrow { B } .\overrightarrow { C } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using : $\vec{A} \times (\vec{B} \times \vec{C})  = \vec{B}  (\vec{A} . \vec{C})  - \vec{C} (\vec{A}.\vec{B})$

Given : $\vec{A}.\vec{C}  = 0$  and  $\vec{A}.\vec{B}  = 0$ 
$\therefore$         $\vec{A} \times (\vec{B} \times \vec{C})  = \vec{B}  (0)  - \vec{C} ( 0)   = 0$
Thus, $\vec{A}$ is parallel to  $(\vec{B} \times \vec{C})$.

Multiple choice physics mathematical methods multiplication of vectors products of vectors scalar product

Vectors $\bar { A }$, $\bar { B }$ and $\bar { C }$ are such that $ \bar { A } \bullet \bar { B } =0$ and $ \bar { A } \bullet \bar { C } =0$. Then the vector parallel to $\bar { A }$ is

  1. $\bar { A } \times \bar { B }$
  2. $\bar { A }+ \bar { B }$
  3. $\bar { B} \times \bar { C }$
  4. $\bar { B}$ and $\bar { B}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If A dot B = 0 and A dot C = 0, then A is perpendicular to both B and C. Therefore, A must be parallel to the cross product B x C.

Multiple choice physics mathematical methods multiplication of vectors products of vectors scalar product

$\vec {A}$ and $\vec {B}$ are vectors expressed as $\vec {A} =2\hat {i}+\hat {j}$ and $\vec {B} =\hat {i}-\hat {j}$. Unit vector perpendicular to $\vec {A}$ and $\vec {B}$ is

  1. $\dfrac{\hat {i}-\hat {j}+\hat {k}}{\sqrt{3}}$
  2. $\dfrac{\hat {i}+\hat {j}-\hat {k}}{\sqrt{3}}$
  3. $\dfrac{\hat {i}+\hat {j}+\hat {k}}{\sqrt{3}}$
  4. $\hat {k}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The unit vector perpendicular to A and B is (A x B) / |A x B|. A x B = (2i + j) x (i - j) = -2(i x j) + (j x i) = -2k - k = -3k. The unit vector is -k or k.

Multiple choice physics mathematical methods multiplication of vectors products of vectors scalar product

Two particles are simultaneously projected in opposite direction horizontally from a given point in space where gravity g is uniform.If $u _1 and u _2$ be their initial speeds, then the time t after which their velocities are mutually perpendicular is given by

  1. $\dfrac{\sqrt{u _1 u _2}}{g}$
  2. $\dfrac{\sqrt{u^2 _1 + u^2 _2}}{g}$
  3. $\dfrac{\sqrt{u _1(u _1 + u _2)}}{g}$
  4. $\dfrac{\sqrt{u _2(u _1 + u _2)}}{g}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

${v _1} = {u _1}\hat i - gt\hat j$

${v _2} =  - {u _2}\hat i - gt\hat j$

$for\,\,\,\,\,\,{v _1} \bot {v _2}$

${{\bar v} _1}.{{\bar v} _2} = 0$

$({u _1}\hat i - gt\hat j).( - {u _2}\hat i - gt\hat j) = 0$

$ - {u _1}{u _2} + {g^2}{t^2} = 0$

${g^2}{t^2} = {u _1}{u _2}$

$t = {{\sqrt {{u _1}{u _2}} } \over g}$

Multiple choice physics mathematical methods multiplication of vectors products of vectors scalar product

If the magnitude of two vectors are $8$ unit and $5$ and their scalar product is zero, the angle between the two vectors is

  1. Zero
  2. ${ 30 }^{ o }$
  3. ${ 60 }^{ o }$
  4. ${ 90 }^{ o }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The scalar product of two vectors is defined as A dot B = |A||B| cos(theta). If the scalar product is zero and the magnitudes are non-zero, then cos(theta) must be zero, which occurs at 90 degrees.