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Multiple choice physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

The radius of the earth is $6.37 \times 10^6 m$ and its mass is $5.975 \times 10^{24} kg$. Find the earth's average density to appropriate significant figures.

  1. $5 \times 10^3 kg m^{-3}$

  2. $5.52 \times 10^3 kg m^{-3}$

  3. $2 \times 10^3 kg m^{-3}$

  4. $5.52 \times 10^3 kg m^{-4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Density = Mass / Volume. Volume = (4/3) * pi * r^3. Density = 5.975e24 / ((4/3) * pi * (6.37e6)^3) = 5518 kg/m^3, which is 5.52e3 kg/m^3.

Multiple choice physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

The number of significant figures in $5.69 \times 10^{15}$kg is

  1. $1$

  2. $2$

  3. $3$

  4. $18$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

No. of significant figures in $\underbrace { 5.69 } \times { 10 }^{ 15 }$ Kg is $3$.

                                           Digits that cannot meaning to measurement and measurement resolution.

Multiple choice physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

A vernier callipers has its main scale of 10 cm equally divided into 200 equal parts. Its vernier scale of 25 divisions coincides with 12 mm on the main scale. The least count of the instrument is - 

  1. 0.020 cm

  2. 0.002 cm

  3. 0.010 cm

  4. 0.001 cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$10\ cm$ of Main Scale $=200$ div of Main scale 

$\Rightarrow IMSD=0.05\ cm$
given $25$ vernier scale division coincide with $12\ mm$ on main scale
$\Rightarrow 25VSD=12\ mm$ on main scale
$=\dfrac{12}{0.5}=24\ MSD$
$\Rightarrow 25VSD=24MSD$
$\Rightarrow 1VSD=\dfrac{24}{25}MSD=0.048\ cm$
Least Count $\Rightarrow 1MSD-1VSD$
$\Rightarrow (0.050-0.048)cm$
$\Rightarrow 0.002\ cm$

Multiple choice physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

Calculate focal length of a spherical mirror from the following observations.

Object distance, u=(50.1±0.5)u=(50.1±0.5) cm
Image distance, v=(20.1±0.2) cm

  1. $\displaystyle \left ( 14.3\pm 0.4\right )$ cm

  2. $\displaystyle \left ( 14.3\pm 0.2\right )$ cm

  3. $\displaystyle \left ( 12.3\pm 0.4\right )$ cm

  4. $\displaystyle \left ( 12.3\pm 0.2\right )$ cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let focal length  of the mirror be $f$  cm.

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

$\dfrac{1}{20.1} + \dfrac{1}{50.1} =  \dfrac{1}{f}$            $\implies  f =  14.3$ cm

As, $f = \dfrac{uv}{u+v}$
$\therefore$ Taking log and differentiating, we get $    \Delta f = f \bigg( \dfrac{\Delta u}{u}  + \dfrac{\Delta v}{v}  + \dfrac{\Delta u + \Delta v}{u+ v} \bigg)$
  $    \Delta f = 14.3 \bigg( \dfrac{0.5}{50.1}  + \dfrac{ 0.2}{20.1}  + \dfrac{0.5 +0.2}{50.1+20.1} \bigg)  = 0.4$  cm

Thus focal length of the mirror $ = (14.3 \pm 0.4)$  cm

Multiple choice physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

In an experiment the following observations were recorded 
$L=2.890$ metre
$\mathrm{M}=3.0$ kg
$\ell=0.87$ cm
$\mathrm{D}=0.041$ cm
$\mathrm{g}=981\mathrm{c}\mathrm{m}/\sec^{2}$
and the formula used for calculatlon of Young's modulus ($\mathrm{Y}$) is $\displaystyle \mathrm{Y}=\frac{\mathrm{M}\mathrm{g}\mathrm{L}}{\pi \mathrm{r}^{2}l}$
What is the maximum possible error expressed in percentage?

  1. 6.36%

  2. 4.8%

  3. 3%

  4. 1%
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$Y=\dfrac{4MgL}{\pi D^2 l}$

Hence, $log(Y)=log4+log(M)+log(g)+log(L)-2log(D)-log(l)$
$\dfrac{\Delta Y}{Y}=\dfrac{\Delta M}{M}+\dfrac{\Delta g}{g}+\dfrac{\Delta L}{L}+\dfrac{2\Delta D}{D}+\dfrac{\Delta l}{l}$
For maximum possible error,
$\Delta M=0.05kg$
$\Delta g=0.5cm/sec^2$
$\Delta L=0.0005metre$
$\Delta D=0.0005cm$
$\Delta l=0.005cm$
Putting the values in $\dfrac{\Delta Y}{Y}\times 100$
Hence maximum possible error is 4.8%

Multiple choice physics refraction of light the nature of light speed of light and optical density introduction to light

Blue light of wavelength $480\ nanometers$, is most strongly reflected off a thin film of oil on a glass slide, when viewed near normal incidence. Assuming that the index of refraction of the oil is $1.2$ and that of the glass is $1.6$, what is the minimum thickness of the oil film (other than zero)?

  1. $150\ nm$

  2. $200\ nm$

  3. $300\ nm$

  4. $400\ nm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For constructive interference in a thin film (n_oil=1.2, n_glass=1.6, n_air=1.0), there is a phase shift at both surfaces. The condition for constructive interference is 2nt = m * lambda. For minimum thickness, m=1: 2 * 1.2 * t = 480 nm. t = 480 / 2.4 = 200 nm.

Multiple choice physics refraction of light the nature of light speed of light and optical density introduction to light

When light is refracted into a denser medium,

  1. its wavelength and frequency both increase

  2. its wavelength increases but frequency remains unchanged

  3. its wavelength decreases but frequency remains unchanged

  4. its wavelength and frequency both decrease
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$ _1\mu _2 = \dfrac{v _1}{v _2} = \dfrac{v _{rarer}}{v _{denser}} \Rightarrow v _1 > v _2$


$\Rightarrow v _1 > v _2$


$\Rightarrow \lambda _1 f _1 > \lambda _2 f _2$
Where $f$ is the frequency

when light goes from any rarer to any denser medium frequency remain unchanged.

$\therefore f _1 = f _2 = f$

$\Rightarrow \lambda _1 > \lambda _2$

hence wavelength decreases but frequency remains unchanged. 

Option C is correct.

Multiple choice physics refraction of light the nature of light speed of light and optical density introduction to light

Mark the correct statements (s) for a very thin soap film (RI = 1.33) whose thickness is much less then wavelength of visible light.

  1. Film appears dark.

  2. Film appears shiny.

  3. Film on glass (RI = 1.5) appears dark.

  4. Film on glass (RI = 1.5) appears shiny.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a very thin film (t << lambda), the path difference is negligible. The phase change occurs at the reflection from the glass interface (n=1.5 > n=1.33), leading to destructive interference, making it appear dark.

Multiple choice physics refraction of light the nature of light speed of light and optical density introduction to light

When a ray of light enters glass from water, it bends :

  1. Away from the normal due to increase in the speed of light

  2. Away from the normal due to decrease in the speed of light

  3. Towards the normal due to decrease in the speed of light.

  4. Towards the normal due to increase in the speed of light.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

When a ray of light enters from glass to water, it bends towards normal due to increase in speed of light. 

As we know,

          $\mu=\dfrac{c}{v}$

         $\mu$ for water is less than that of glass, as glass acts as a denser medium. $\therefore$ the speed of light in glass will be less than that in water.

Multiple choice physics refraction of light the nature of light speed of light and optical density introduction to light

A convex lens forms a real image with magnification $m _1$ on a screen. Now, the screen is moved by a distance x and the object is also moved so as to obtain a real image with magnification $m _2 (> m _1) $ on the screen. Then, the focal length of the lens is

  1. $\left [ \dfrac{m _1}{m _2} \right ]x$

  2. $\left [ \dfrac{m _2}{m _1} \right ]x$

  3. $x(m _2-m _1)$

  4. $ \dfrac{x}{(m _2-m _1)} $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using the lens formula and magnification m = f / (f + u), one can derive the relationship between the shift in screen position and focal length. The correct relation is f = x / (m2 - m1).