Tag: physics

Questions Related to physics

If $log 2 = 0.3010 $ and $3 = 0.4771$, the value of $log _5 512$ is

  1. 2.870

  2. 2.967

  3. 3.876

  4. 3.912


Correct Option: C
Explanation:

$log _5 512 = \dfrac{log 512}{log 5}$
$=\dfrac{log 2^9}{log (10/2)}$
$=\dfrac{9 log  2}{log 10 - log 2}$
$=\dfrac{9 \times 0.3010}{1- 0.3010}$
$= \dfrac{2.709}{0.699}$
$=\dfrac{2709}{699}$
$= 3.876$

If $log 2 = 0.30103$, the number of digits in $2^{64}$ i

  1. 18

  2. 19

  3. 20

  4. 21


Correct Option: C
Explanation:

$log (2^{64}) = 64 \times log 2$
$= (64 \times 0.30103)$
$= 19.2592$
Its characteristic is 19.
Hence, then number of digits in $2^{64}$ is 20.

If $log _x \left( \dfrac{9}{16} \right) = - \dfrac{1}{2}$, then x is equal to

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

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

  3. $\dfrac{81}{256}$

  4. $\dfrac{256}{81}$


Correct Option: D
Explanation:

$log _x \left( \dfrac{9}{16} \right ) = - \dfrac{1}{2}$
$\Rightarrow x^{-1/2} = \dfrac{9}{16}$
$\Rightarrow \dfrac{1}{\sqrt x} = \dfrac{9}{16}$
$\Rightarrow \sqrt x = \dfrac{16}{9}$
$\Rightarrow x = \left( \dfrac{16}{9} \right)^2$
$\Rightarrow x = \dfrac{256}{81}$

What is the value of $\dfrac {1}{2}\log _{10} 25 - 2 \log _{10} 3 +\log _{10} 18$?

  1. $2$

  2. $3$

  3. $1$

  4. $0$


Correct Option: C
Explanation:

The value of $\dfrac {1}{2}\log _{10} 25 - 2 \log _{10} 3 +\log _{10} 18$ is
$= \log _{10}(25)^{1/2} - \log _{10} (3)^{2} + \log _{10}18$
$= \log _{10}5 - \log _{10}9 + \log _{10}18$
$= \log _{10} \left (\dfrac {5}{9}\times 18\right ) $

$= \log _{10} 10 $    ....Using the identity $\log _aa=1$
$= 1$

The value of $log _2$ 16 is

  1. $\dfrac{1}{8}$

  2. 4

  3. 8

  4. 16


Correct Option: B
Explanation:

Let $log _2      16 = n$
Then, $2^n = 16 = 2^4    \Rightarrow n = 4$
$\therefore log _2  16 = 4$

The logarithmic form of ${5}^{2}=25$ is

  1. $\log _{ 5 }{ 2 } =25$

  2. $\log _{ 2 }{ 5 } =25$

  3. $\log _{ 5 }{ 25 } =2$

  4. $\log _{ 25 }{ 5 } =2$


Correct Option: C
Explanation:

$5^2=25$

Taking log with base $5$ both sides, we get
$\log _55^2=\log _525$
$\Rightarrow \log _525=2\log _55$
$\Rightarrow \log _525=2$     $(\log _aa=1)$
Hence, C is the correct option.

The exponential form of $\log _{ 2 }{ 16 } =4$ is

  1. ${2}^{4}=16$

  2. ${4}^{2}=16$

  3. ${2}^{16}=4$

  4. ${4}^{16}=2$


Correct Option: A
Explanation:

Exponential form of $\log _2 16=4$

Taking antilog both sides, we get
$\Rightarrow 16=2^4$
Hence, A is the correct option.

If mantissa of logarithm of 719.3 to the base 10 is 0.8569 , then mantissa of logarithm  of 71.93 is

  1. 0.8569

  2. $\overline 1 .8569$

  3. 1.8569

  4. 0.1431


Correct Option: A
Explanation:

Mantissa of logarithm of $719.3$ to base $10$ is $0.8569$

Then, mantissa of logarithm of $71.93$ is also $0.8569$
As, $\log _{10}{(719.3)}=2+(0.8569)$(mantissa)
So, $\log _{10}{(71.93)}=1+(0.8596)$ (mantissa)

If $2\log y -\log x -3=0$, express $x$ in terms of $y.$

  1. $x=\dfrac{y^2}{e^3}$

  2. $x=\dfrac{y^2}{e^2}$

  3. $x^2=\dfrac{y^2}{e^3}$

  4. $x=\dfrac{y^3}{e^3}$


Correct Option: A
Explanation:
$\Rightarrow$$\log { { y }^{ 2 } } -\log { x } -\log { { e }^{ 3 } } =0$.......$\log e=1$

$\Rightarrow$$ \log { x } =\log { \left (\cfrac { { y }^{ 2 } }{ { e }^{ 3 } } \right ) } $

$\Rightarrow$$ x=\cfrac { { y }^{ 2 } }{ { e }^{ 3 } } $

If $2\log y -\log x-3=0$ express $x$ in terms of $y.$

  1. $x^2=1000y$

  2. $x^2= \dfrac{y^2}{e^3}$

  3. $y^2= \dfrac{x}{1000}$

  4. $y^2= 1000x$


Correct Option: B
Explanation:
Given: $2\log y -\log x-3=0$

$\log { { y }^{ 2 } } -\log { x } -3\log { { e }=0 } $.......$(\log e=1)$

$\log { { y }^{ 2 } } -\log { x } -\log { { e }^{ 3 }=0 } $

$ \log { x } =\log { { y }^{ 2 } } -\log { { e }^{ 3 } } =\log { \left (\cfrac { { y }^{ 2 } }{ { e }^{ 3 } } \right ) } $

$ x=\cfrac { { y }^{ 2 } }{ { e }^{ 3 } } $