Tag: physics

Questions Related to physics

If the mantissa of $\log 2125 =3.3275$, find the mantissa of $\log21.25$

  1. $1.3273$

  2. $2.3273$

  3. $0.3273$

  4. $32.2321$


Correct Option: A
Explanation:

Given that:

$\log2125=3.3273$
Now, $\log21.25=\log\cfrac{2125}{100}$
$=\log 2125-\log100$
$=3.3273-\log10^2$
$=3.3273-2$
$=1.3273$

The logarithm of $0.0625$ to the base $2$ is:

  1. $0.025$

  2. $0.25$

  3. $5$

  4. $-4$

  5. $-2$


Correct Option: D
Explanation:

$\log _{2}{\cfrac{625}{10000}}$

$=\log _{2}{\cfrac{1}{16}}$
$=\log _{2}{(16)^{-1}}$
$=\log _{2}{(2)^{-4}}$
$=-4\log _{2}{2}$
$= -4$

The number of zeros between the decimal point and first significant digit of ${\left(0.036\right)}^{16}$ where $log2=0.301$ and $log3=0.477$

  1. $21$

  2. $22$

  3. $23$

  4. $24$


Correct Option: C
Explanation:

$y=(0.036)^{16}$

$\Rightarrow \log (y)= 16 log(0.036)$
$\Rightarrow \log (y)= 16 \log(\cfrac{36}{1000})=16[\log 36- 3\log _{10}{10}]$
$\Rightarrow \log(y)= 16[\log (2^2\times 3^2)-3]=16[2(\log 2+\log 3)-3]$
$\Rightarrow \log(y)=16[2(0.301+0.477)-3]=-23.104<-23$
So, that number has $23$ zeros after this point.

Given $log _{10}2=a$ and $log _{10}3=b$, if $3x+2=25$, the value of x in terms of $a$ and $b$ is $x=(10^{k}+1)$. K=?

  1. $\dfrac{a-1}{b}$

  2. $a-b+1$

  3. $\dfrac{1+a}{b}$

  4. $\dfrac{b}{1-a}$


Correct Option: B
Explanation:

$\log _{10}{2}=a$

$\Rightarrow 2=(10^a)$
$\log _{10}{3}=b$
$\Rightarrow 3=(10^b)$
$3x+2=25$
$x=\cfrac{23}{3}=(2\times \cfrac{10}{3}+1)=10\cfrac{(10)^a}{(10)^b}+1$
$x=10^{(a+1-b)}+1$
$x=(10^{a-b+1}+1)$

Let $N=\dfrac{\log _{3}135}{\log _{15}3}-\dfrac{\log _{3}5}{\log{405}3}$, then $N$ is

  1. a natural number

  2. a prime number

  3. a rational number

  4. an integer


Correct Option: A,B,C,D

If $x=198!$ then value of the expression $\dfrac {1}{\log _{2}x}+\dfrac {3}{\log _{2}x}+...\dfrac {198}{\log _{2}x}$ equals ?

  1. $-1$

  2. $0$

  3. $1$

  4. $198$


Correct Option: C

The value of $\dfrac{log _2 24}{log _{96} 2}-\dfrac{log _2192}{log _{12}{2}}$ is

  1. $3$

  2. $0$

  3. $2$

  4. $1$


Correct Option: A
Explanation:
Consider
$\dfrac{log _2 24}{log _{96} 2}-\dfrac{log _2192}{log _{12}{2}}\\$
$=\dfrac{log24.log96-log192log12}{(log2)^2}$
$=\dfrac{log(2^3 \times 3)log(2^5\times 3)-log(2^6\times3)log(2^2\times3)}{(log2)^2}$
$=\dfrac{(3log2+log3)(5log2+log3)-(6log2+log3)(2log2+log3)}{(log2)^2}$
$=\dfrac{15(log2)^2-12(log2)^2}{(log2)^2}$
$=3\dfrac{(log2)}{log2}$
$=3$
Option A is the correct answer.

The greatest value of $(4\log _{10}{x}-\log _{2}{(0.0001)})$ for $0 < x < 1$ is

  1. $4$

  2. $-4$

  3. $8$

  4. $-8$


Correct Option: A

If $P$ is the number of natural numbers whose logarithm to the base $10$ have the characteristic $p$ and $Q$ is the number of natural numbers logarithm of whose reciprocals to the base $10$ have the characteristic $-q$, then find the value of $\log _{10}P-\log _{10}Q$.

  1. $p-q+1$

  2. $p+q-1$

  3. $p+q$

  4. $p-q$


Correct Option: A
Explanation:

Let $x$ and $y$ be the numbers whose logarithm to the base $10$ have the characteristic $p$ and $q$ respectively.
$10^{p}\leq x< 10^{p+1}\Rightarrow P=10^{p+1}-10^{p}\Rightarrow P=9\times 10^{p}$
Similarly, $10^{q-1}< y\leq 10^{q}$
$\Rightarrow $   $Q=10^{q}-10^{q-1}=10^{q-1}\left ( 10-1 \right )=9\times 10^{q-1}$
$\therefore $   $\log _{10}P-\log _{10}Q=\log _{10}\left ( P/Q \right )=\log _{10}10^{p-q+1}=p-q+1$

Find the number of positive integers which have the characteristic $3$, when the base of the logarithm is $7$.

  1. $2058$

  2. $1029$

  3. $1030$

  4. $2060$


Correct Option: A
Explanation:

Let there be $N$ integers whose characteristic is 3, when base of log is 7
Then, $\log _{ 7 }{ N } =x$ where $3\le x<4$
As $3\le x<4$
$3\le \log _{ 7 }{ N } <4\ \Rightarrow { 7 }^{ 3 }\le N<{ 7 }^{ 4 }\ \Rightarrow N={ 7 }^{ 4 }-{ 7 }^{ 3 }\ \Rightarrow N=2401-343=2058$