Tag: exponentiation

Questions Related to exponentiation

The value of $\displaystyle\frac{2^{m+3}\times3^{2m-n}\times5^{m+n+3}6^{n+1}}{6^{m+1}\times10^{n+3}\times15^m}$ is equal to

maths repeated multiplication exponentiation index notation and products of prime factors powers

  1. 0

  2. 1

  3. $2^m$

  4. none of these

Show answer
Correct Option: B
Explanation:

Numerator $=2^{m+3}\cdot3^{2m-n}\cdot5^{m+n+3}\cdot2^{n+1}\cdot3^{n+1}$
$=2^{m+n+4}\cdot3^{2m+1}\cdot5^{m+n+3}$ (i)
Denominator $=2^{m+1}\cdot3^{m+1}\cdot2^{n+3}\cdot5^{n+3}\cdot3^m\cdot5^m$
$=2^{m+n+4}\cdot3^{2m+1}\cdot5^{m+n+3}$ (ii)
Given expression $=1$

What is the unit digit in ${({6374}^{1793}\times {625}^{317}\times{341}^{491})}$?

maths repeated multiplication exponentiation index notation and products of prime factors powers

  1. $0$

  2. $2$

  3. $3$

  4. $5$

Show answer
Correct Option: A
Explanation:

Unit digit in ${6374}^{1993}=$ Unit digit in ${(4)}^{1793}$
=Unit digit in $[{({4}^{2})}^{896}\times 4]$
=Unit digit in $(6\times 4)=4$
Unit digit in ${(625)}^{317}=$ Unit digit in ${(5)}^{317}=5$
Unit digit in ${(341)}^{491}=$ Unit digit in ${(1)}^{491}=1$
Required digit$=$ Unit digit in $(4\times 5\times1)=0$

The number of values of $x\ \epsilon \ [0,5]$ at which $f(x)=|x-\dfrac{1}{4}|+|x-2|+\tan{x}$ is not differentiable are

maths repeated multiplication exponentiation index notation and products of prime factors powers

  1. $0$

  2. $2$

  3. $3$

  4. $4$

Show answer
Correct Option: C

The greatest of the number 
$1,2^{1/2},3^{1/3},4^{1/4},5^{1/5}, 6^{1/6}, and \ 7^{1/7}$ is

maths repeated multiplication exponentiation index notation and products of prime factors powers

  1. $2^1/2$

  2. $3^1/3$

  3. $7^1/7$

  4. $4^1/4$

Show answer
Correct Option: B

${ 5 }^{ n }\left( n\in N \right) $ ends with ......

maths repeated multiplication exponentiation index notation and products of prime factors powers

  1. 4

  2. 0

  3. 5

  4. 2

Show answer
Correct Option: C
Explanation:
 $n$  $1$  $2$  $3$  $4$  $5$
 ${5}^{n}$  $5$  $25$  $125$  $625$  $3125$
 Ending number  $5$  $5$  $5$  $5$  $5$

Thus, the number ends with $5$ for $n\in N$ and $n$ is odd or even.

Find the value of $\displaystyle\frac{5^0+5^{-1}}{5^0-5^{-1}}-\left(\frac{8}{27}\right)^{\displaystyle\frac{1}{3}}-\left(\frac{36}{25}\right)^{-\displaystyle\frac{1}{3}}$

maths repeated multiplication exponentiation index notation and products of prime factors powers

  1. 0

  2. $\displaystyle\frac{1}{2}$

  3. 1

  4. 2

Show answer
Correct Option: A
Explanation:

$\displaystyle\frac{\displaystyle1+\frac{1}{5}}{\displaystyle1-\frac{1}{5}}-\left[\left(\frac{2}{3}\right)^3\right]^{\displaystyle\frac{1}{3}}-\left[\left(\frac{6}{5}\right)^2\right]^{\displaystyle-\frac{1}{2}}=\frac{\displaystyle\frac{6}{5}}{\displaystyle\frac{4}{5}}-\left(\frac{2}{3}\right)^1-\left(\frac{6}{5}\right)^{-1}=\frac{6}{4}-\frac{2}{3}-\frac{5}{6}=\frac{18-8-10}{12}=0$

Arrange the following lengths in the increasing order of their magnitudes :
1 meter, 1 centimeter, 1 kilometer, 1 millimeter .

physics dynamics - explaining motion exponentiation index notation and products of prime factors powers

  1. 1 meter < 1 centimeter < 1 kilometer < 1 millimeter

  2. 1 millimeter < 1 meter < 1 centimeter < 1 kilometer

  3. 1 millimeter < 1 centimeter < 1 meter < 1 kilometer

  4. none of these

Show answer
Correct Option: C
Explanation:

1 kilometer = 10$^3$ meter

1 centimeter = 10$^{-2}$ meter
1 millimeter = 10$^{-3}$ meter

Ascending order of length :
1 millimeter < 1 centimeter < 1meter < 1 kilometer

The wrong unit conversion among the following is

physics dynamics - explaining motion exponentiation index notation and products of prime factors powers

  1. 1 angstrom = $10^{-10} m$

  2. 1 fermi = $10^{-15}$ m

  3. 1 light year = $9.46 \times 10^{15} m$

  4. 1 astronomical unit = $1.496 \times 10^{-11} m$

Show answer
Correct Option: D
Explanation:
$A)1 $Angstrom: $1$ Ångström is equal to$ 0.1 $nanometers. Expressed in the SI base unit meters $1$ Ångström is equal to $1 \times 10^{-10}$ meters. $1 Å = 0.0000000001 m.$

$B)$Definition of fermi unit: $≡ 1\times 10^{−15} m.$

$C)$The light-year is a unit of length used to express astronomical distances and measures about $9.46$ trillion kilometres $(9.46 \times 10^{12} km)$ or $5.88$ trillion miles $(5.88 \times 10^{12} mi).$

$D)$The total radiation power received from the sun on a unit area perpendicular to the sun rays at the mean earth-sun distance, termed an astronomical unit, is called the solar constant (SC), where $1$ astronomical unit $= 1 AU = 1.496 \times 10^{11} m.$

So $D$ is wrong unit conversion 

A calorie is a unit of heat energy and its value is 4.18 J where $1 J = 1 kg m^2 s^{-2}$. Suppose we use a new system of units in which unit of mass equals $\alpha$ kg, the unit of length equals $\beta$ m and the unit of the time is $\gamma$ sec. Then the value of a calorie in the new system of units is then

physics dynamics - explaining motion exponentiation index notation and products of prime factors powers

  1. 4.18 $\displaystyle \frac{\gamma^2}{\alpha \beta^2}$

  2. 4.18 $\displaystyle \frac{\alpha \beta^2}{\gamma^2}$

  3. 4.18 $\displaystyle \frac{\gamma^2}{\alpha}$

  4. 4.18 $\displaystyle \frac{\beta^2}{\alpha \gamma^2}$

Show answer
Correct Option: A
Explanation:

$1 J = (1 kg) ( 1 m)^2 (1 sec)^{-2}$
$1 x = (\alpha kg) (\beta m)^2 (\gamma sec)^{-2}$
$\displaystyle \therefore \frac{1 J}{1x} = \left( \frac{1}{\alpha} \right) \left( \frac{1}{\beta} \right)^2 (\gamma)^2 = \frac{\gamma^2}{\alpha \beta^2} $
$\displaystyle \therefore 1 J = \frac{\gamma^2}{\alpha \beta^2}$ or $\displaystyle 1 cal = 4.18 \displaystyle \frac{\gamma^2}{\alpha \beta^2}$

1 fermi is equal to 

physics dynamics - explaining motion exponentiation index notation and products of prime factors powers

  1. $\displaystyle { 10 }^{ -15 }$m

  2. $\displaystyle { 10 }^{ 15 }$m

  3. $\displaystyle { 10 }^{ -12 }$m

  4. $\displaystyle { 10 }^{ 12 }$m

Show answer
Correct Option: A
Explanation:
Fermi is used to express length.
$1 \ $ fermi $= \ 10^{-15} \ m$