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Questions Related to physics

Express in logarithmic form and find x: $\displaystyle 10^{x} = 0.001$ (i.e base 10)

physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

  1. $3$

  2. $-3$

  3. $2$

  4. $-2$

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Correct Option: B
Explanation:

$10^{-3}=0.001$
Taking log on both sides
$-3log10= log0.001$
$log _{10}0.001= -3$.....(since $\dfrac{loga}{logb}=log _ba$)

State whether true or false:
$\displaystyle \log F = \log : G + \log : m _1 + \log : m _2 - 2 \log : d$ gives $\displaystyle F = G \frac {m _2m _1}{d^2}$.

physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

  1. True

  2. False

Show answer
Correct Option: A
Explanation:
$\displaystyle \log F = \log  \: G + \log  \: m _1 + \log  \: m _2 - 2 \: \log  \: d$

$\therefore \log F= \log G+\log m _1+\log m _2-\log d^2$....($\log a^b=b\log a$)

$\therefore \log F= \log \cfrac{Gm _1m _2}{d^2}$.....($\log a.b=\log a+\log b, \log \cfrac{a}{b}=\log a-\log b$)

$\therefore F=\cfrac{Gm _1m _2}{d^2}$

State whether true or false:
$\displaystyle \log : V = 2 \log : 2 - \log : 3 + \log : \pi + 3 \log : r$ gives $\displaystyle V = \frac {4}{3} \pi r^3$

physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

  1. True

  2. False

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Correct Option: A
Explanation:

$\displaystyle \log : V = 2 : \log : 2 - : \log : 3 + : \log : \pi + 3 : \log : r$
$\therefore \log V = \log 2^2-\log 3+\log \pi+ \log r$....($\log a^b=b \log a$)
$\therefore \log V= \log \cfrac{4}{3}\pi r^3$.....($\log a.b = \log a + \log b, \log \cfrac {a}{b} = \log a - \log b$)
$\therefore V=\cfrac{4}{3}\pi r^3$

The logarithm form of $10^{-3} = 0.001$ is $\log _{10} 0.001 = -m$, then value of $m$ is 

physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

  1. $-1$

  2. $-2$

  3. $3$

  4. $-4$

Show answer
Correct Option: C
Explanation:
The logarithm of the $10^{-3}=\frac{1}{1000}=0.0001$
So $log _{10}0.0001=-3$
$log _{10}0.0001=-m$
Then m=3

The value of $\displaystyle \log _{10}0.001 $ is equal to

physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

  1. $-3$

  2. $3$

  3. $-2$

  4. $2$

Show answer
Correct Option: A
Explanation:

$ \ \log _{ 10 }{ 0.001 } =x\ 0.001 = { 10 }^{ x }\ \dfrac { 1 }{ 1000 } = { 10 }^{ -3 }\ x=-3  $

The value of $\log _{0.5}16$ is equal to

physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

  1. $-4$

  2. $-1$

  3. $-2$

  4. $0$

Show answer
Correct Option: A
Explanation:

$ \log _{ 0.5 }{ 16 } = x $


$ 16 = {0.5 }^{ x }$

$ { 2 }^{ 4 } = { 2 }^{ -1x }$

$ x = -4\ \ 
$

The logarithm of $0.001$ to the base $10$ is equal to

physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

  1. $5$

  2. $-1$

  3. $6$

  4. $-3$

Show answer
Correct Option: D
Explanation:

$0.001$ to the base $10$
So, as per logarithm,
$\log _{ 10 }{ 0.001 } =\log _{ 10 }{ { 10 }^{ -3 } } $
$\log _{ 10 }{ 0.001 } =-3\log _{ 10 }{ { 10 } } $
$\log _{ 10 }{ 0.001 } =-3\times 1$
$\log _{ 10 }{ 0.001 } =-3$

$\log V = 2 \log 2 - \log 3 + \log \pi + 3 \log r$ can be expressed as

physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

  1. $V = \dfrac{4}{3} \pi r^{3}$

  2. $ V = \dfrac{2}{3} \pi r^{3}$

  3. $ V = \dfrac{4}{3} \pi r$

  4. $ V = \dfrac{2}{3} \pi r$

Show answer
Correct Option: A
Explanation:
$ \log { V }  = 2\log { 2 } -\log { 3 } +\log { \pi  }  + 3\log { r }$
$ \log V = \log ({ 2 }^{ 2 }\times \pi \times { r }^{ 3 }) - \log { 3 }$
$ \log V = \log \dfrac { 4\pi { r }^{ 3 } }{ 3 } $
Removing log
$V = \dfrac {4}{3} \pi r^3$

Which of the following is true for $\log _25$?

physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

  1. An integer

  2. A rational number

  3. An irrational number

  4. A whole number

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Correct Option: C
Explanation:

Let us assume that $\log _{ 2 }{ 5 } =\frac { p }{ q } $ , where $p,q$ are integers

We have $5=2^{\frac{p}{q}}$
$\Rightarrow 5^{q}=2^{p}$
This suggest that $5$ and $2$ are not mutually prime , But $2$ and $5$ are mutually prime
Therefore $\log _{ 2 }{ 5 } $ is an irrational number

If $\log _{10}(x - 10) = 1$, then value of $x$ is

physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

  1. $10$

  2. $13$

  3. $20$

  4. $26$

Show answer
Correct Option: C
Explanation:

$ \log _{ 10 }{ (x-10) } =1$


$ \log _{ 10 }{ (x-10) } =\log _{ 10 }{ 10 }$

$x-10=10$

$  = 20 $