Tag: physics

Questions Related to physics

A point source of light is kept in front of a convex mirror of radius of curvature $40 cm$. The image is formed at $10 cm$ behind the mirror. Calculate the object distance

physics reflection of light at curved surfaces the mirror formula derivation of formula for curved mirrors mirror formula and magnification

  1. 30

  2. 20

  3. 50

  4. 40

Show answer
Correct Option: B
Explanation:

Given: For a convex mirror, $R = -40 cm$.
$v = -10 cm$ (image is virtual).
From mirror formula, we have

$\displaystyle \frac {2}{R}=\frac {1}{u}+\frac {1}{v}$

$\displaystyle \frac {1}{u}=\frac {2}{R}-\frac {1}
{v}=\frac {2v-R}{vR}$

$\displaystyle u=\frac {vR}{2v-R}=\frac {(-10cm) \times (-40cm)}{2 \times (-10cm) - (-40cm)}$

$=+20cm$
Thus, object is placed $20 cm$ in front of the mirror.

Mirror formula is valid for:

physics reflection of light at curved surfaces the mirror formula derivation of formula for curved mirrors mirror formula and magnification

  1. Convex mirror

  2. Concave mirror

  3. Both A and B

  4. For lenses and mirrors

Show answer
Correct Option: C
Explanation:
Mirror formula is:
$\dfrac {1}{v} + \dfrac {1}{u} = \dfrac {1}{f}$
That is; $\dfrac {1}{\text {Object Distance}} + \dfrac {1}{\text {Image Distance}} = \dfrac {1}{\text {Focal length of the mirror}}$
This relationship is applicable for both concave and convex mirrors.

Mirror formula can also be written as:

physics reflection of light at curved surfaces the mirror formula derivation of formula for curved mirrors mirror formula and magnification

  1. $\dfrac {1}{2v} + \dfrac {1}{2u} = \dfrac {1}{2f}$

  2. $\dfrac {1}{v} + \dfrac {1}{u} = \dfrac {2}{R}$

  3. $\dfrac {1}{v} + \dfrac {2}{u} = \dfrac {1}{f}$

  4. $\dfrac {1}{2v} + \dfrac {4}{u} = \dfrac {3}{R}$

Show answer
Correct Option: B
Explanation:

Focal length, $f = \dfrac {R}{2}$, where $R$ = radius of curvature

So, $\dfrac {1}{v} + \dfrac {1}{u} = \dfrac {1}{f}$ can be written as
$\Rightarrow \dfrac {1}{v} + \dfrac {1}{u} = \dfrac {2}{R}$

Mark the incorrect statement regarding mirror formula.

physics reflection of light at curved surfaces the mirror formula derivation of formula for curved mirrors mirror formula and magnification

  1. Values of known and unknown parameters can be used with their proper signs

  2. Sign of unknown parameter comes of its own after calculation

  3. Mirror formula is applicable for both concave and convex mirrors

  4. All

Show answer
Correct Option: A
Explanation:

The values of known parameter should be used with their proper sign convention. No sign should be attached to the unknown parameter during calculation. Its sign will come of its own after calculation.

The relation among u, v and f for a mirror is

physics reflection of light at curved surfaces the mirror formula derivation of formula for curved mirrors mirror formula and magnification

  1. $f = uv/(u+v)$

  2. $v = fu/(u+f)$

  3. $u = fv/(f+v)$

  4. All of these

Show answer
Correct Option: A
Explanation:

According to mirror equation.

$\cfrac{1}{f} = \cfrac{1}{u}+ \cfrac{1}{v}$
$\cfrac{1}{f} = \cfrac{u+v}{uv}$
$f = \cfrac{uv}{u+v}$

Similarly,
$v = \cfrac{fu}{u-f}$
$u = \cfrac{fv}{v-f}$

The relation among $u,v$ and $f$ for a mirror is:

physics reflection of light at curved surfaces the mirror formula derivation of formula for curved mirrors mirror formula and magnification

  1. $f=uv(u+v)$

  2. $v=fu(u+f)$

  3. $u=fv(f+v)$

  4. None of these

Show answer
Correct Option: D
Explanation:

from mirror formula

              $\dfrac { 1 }{ v } +\dfrac { 1 }{ u } =\dfrac { 1 }{ f } $
              $\Rightarrow \quad \boxed { f=\dfrac { uv }{ u+v }  } $
and
          $1+\dfrac { v }{ u } =\dfrac { v }{ f } \Rightarrow \boxed { u=\dfrac { fv }{ f-v }  } $

A object moving at a speed of $5$ m/s towards a concave mirror of focal length $f=1$m is at a distance of $9$m. The average speed of the image is?

physics reflection of light at curved surfaces the mirror formula derivation of formula for curved mirrors mirror formula and magnification

  1. $\dfrac{1}{5}$ m/s

  2. $\dfrac{1}{10}$ m/s

  3. $\dfrac{4}{5}$ m/s

  4. $\dfrac{2}{5}$ m/s

Show answer
Correct Option: C

The size of E-coli bacteria is about :

physics electronic devices nanoscience and nanotechnology nanotechnology logic gates

  1. 500 nm

  2. 1000 nm

  3. 2000 nm

  4. 3000 nm

Show answer
Correct Option: C
Explanation:

Each E-coli bacterium measures 0.5 μm in width by 2 μm in length i.e, 2000 nm.
So option (b) is correct,

What is the $-$ve terminal of dry cell made up of?

physics electric current through conductors study about cells electric cell types of cells

  1. Carbon rod

  2. Zinc plate

  3. Magnesium plate

  4. Cadmium rod

Show answer
Correct Option: B
Explanation:
The dry cell consists of a zinc can, electrolyte paste, and a carbon rod in the center. A chemical reaction takes place between the zinc can and the paste. Electrons flow from the negative (zinc) terminal to the positive (carbon) terminal. Electrons are then pumped back to the negative terminal.

In secondary cells:

physics electric current through conductors study about cells electric cell types of cells

  1. Chemical changes can be reversed by heating electrodes

  2. Chemical changes can be reversed by passing electric current

  3. Current is produced by photo chemical reactions

  4. None of these

Show answer
Correct Option: B
Explanation:

Answer is B.

When a primary cell is used, chemical reactions in the battery use up the chemicals that generate the power; when they are gone, the battery stops producing electricity and is useless. In contrast, in a secondary cell, the reaction can be reversed by running a current into the cell with a battery charger to recharge it, regenerating the chemical reactants.