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Questions Related to physics

The value of $7 log _a \displaystyle \frac{16}{15} + 5 log _a \frac{25}{24} + 3 log _a \frac{81}{80}$ is

physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

  1. $log _{a3}$

  2. $log _{a1}$

  3. $log _{a2}$

  4. $log _{a5}$

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Correct Option: C
Explanation:

We have,

$7log _a\dfrac{16}{15}+5log _a\dfrac{25}{24}+3log _a\dfrac{81}{80}$
$\Rightarrow log _a(\dfrac{16}{15})^7+log _a(\dfrac{25}{24})^5+log _a(\dfrac{81}{80})^3$
$\Rightarrow log _a(\dfrac{16}{15})^7\times (\dfrac{25}{24})^5\times (\dfrac{81}{80})^3$
$\Rightarrow log _a\dfrac{16^3\times 16^4}{5^7\times 3^7}\times \dfrac{5^5\times 5^5}{8^5\times 3^5}\times \dfrac{3^6\times 3^6}{16^3\times 5^3}$
$\Rightarrow log _a\dfrac{16^4}{1\times 1}\times \dfrac{1}{8^5\times 1}\times \dfrac{1}{1}$
$\Rightarrow log _a\dfrac{2^4\times 8^4}{8^5}$
$\Rightarrow log _a\dfrac{16}{8}$
$\Rightarrow log _a2$

Hence, this is the answer.

If $log _{10} x - log _{10} \sqrt x = \displaystyle \frac{2}{log _{10} x}$, then value of x is

physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

  1. $\displaystyle \frac{1}{100}$ or $100$

  2. $\pm$ 2

  3. 10 or $\displaystyle \frac{1}{10}$

  4. 100

Show answer
Correct Option: A
Explanation:

Consider the given equation.

$log _{10}x-log _{10}\sqrt x=\dfrac{2}{log _{10}x}$
$log _{10}\dfrac{x}{\sqrt x}=\dfrac{2}{log _{10}x}$
$log _{10}\sqrt x=\dfrac{2}{log _{10}x}$
$\dfrac{1}{2}log _{10}\ x=\dfrac{2}{log _{10}x}$
$\dfrac{1}{2}(log _{10}\ x)^2=2$
$(log _{10}\ x)^2=4$
$log _{10}\ x=\pm 2$
$x=10^{\pm2}$
$x=100\ or\ \dfrac{1}{100}$

Hence, this is the answer.

If $\displaystyle \frac{log _2 (9 - 2^x)}{3 - x} = 1$, then value of x is

physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

  1. x = 4

  2. x = + 1 or -1

  3. x = $\pm$ 2

  4. x = 0

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Correct Option: D
Explanation:

We have,

$\dfrac{log _2(9-2^x)}{3-x}=1$
$log _2(9-2^x)=3-x$
$9-2^x=2^{3-x}$                $ .......... (1)$

From option $(D)$
$9-2^0=2^{3-0}$
$9-1=2^3$
$8=8$

Hence, $x=0$ is the root of this equation.

Hence, only option $D$ is correct.

The value of $\log _{ \frac{1}{2} }{ 4 } $ is

physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

  1. $-2$

  2. $0$

  3. $\dfrac{1}{2}$

  4. $2$

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Correct Option: A
Explanation:

Let $\log _{\frac{1}{2}}4 = x $

Then, $(\dfrac{1}{2})^x = 4 $

$Or, (\dfrac{1}{2})^x =2^2$

$Or, 2^{-x} = 2^2$

$x= -2$

The equation  ${ \left( \log _{ 10 }{ x+2 }  \right)  }^{ 3 }+{ \left( \log _{ 10 }{ x-1 }  \right)  }^{ 3 }={ \left( 2\log _{ 10 }{ x+1 }  \right)  }^{ 3 }$ has

physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

  1. no natural solution

  2. two rational solutions

  3. no prime solution

  4. one irrational solution

Show answer
Correct Option: B,C,D
Explanation:

Let $ \log _{ 10 }{ x+2 } =a$ and $ \log _{ 10 }{ x-1 } =b$
$\therefore a+b=2\log _{ 10 }{ x+1 } $ (from the question)
Thus, the given equation(in the question) reduces to ${a}^{3}+{b}^{3}={(a+b)}^{3}$
$\Rightarrow 3ab(a+b)=0$
$\Rightarrow a=0$ or $b=0$ or $a+b=0$
$\Rightarrow \log _{ 10 }{ x+2 }=0$  or $\log _{ 10 }{ x-1 }=0$ or $2\log _{ 10 }{ x } +1=0$
$\Rightarrow x={10}^{-2}$  or  $x=10$ or  $x={ 10 }^{ -\frac { 1 }{ 2 }  }$
Hence  $x=\left{ \dfrac { 1 }{ 100 },10 ,\dfrac { 1 }{ \sqrt { 10 }  }  \right} $

Magnification produced by a convex mirror is $\frac { 1 }{ 3 }$, then distance of the object from mirror is

physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

  1. $\frac { f }{ 3 }$

  2. $\frac { 2f }{ 3 }$

  3. $1f$

  4. $2f$

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Correct Option: D

A convex lens of focal length 30 cm forms an image of height 2 cm for an object situated at infinity. If a concave lens of focal length 20 cm is placed coaxially at a distance of 26 cm in front of convex lens. then size of final image would be:

physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

  1. $1.25cm$

  2. $2.5 cm$

  3. $2 cm$

  4. $0.75cm$

Show answer
Correct Option: B

The object distance $u$ for a concave mirror:

physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

  1. must be positive

  2. must be negative

  3. must not be negative

  4. may be negative

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Correct Option: D
Explanation:

Positive and negative sign depend on the assumption of sign conversion.
either side we can consider positive or negative.
Hence Option D.

The linear magnification for a mirror is the ratio of the size of the image to the size of the object, and is denoted by m. Then m is equal to (symbols have their usual meanings).

physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

  1. $\displaystyle \frac { uf }{ u-f } $

  2. $\displaystyle \frac { uf }{ u+f } $

  3. $\displaystyle \frac { f }{ u-f } $

  4. None of these

Show answer
Correct Option: C
Explanation:

we now,$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$
multiplying by u in above eq.
$\dfrac{u}{f}=\dfrac{u}{v}+\dfrac{u}{u}$
$\dfrac{u}{f}=\dfrac{u}{v}+1$
$\dfrac{u}{f}-1=\dfrac{u}{v}$
$\dfrac{u}{v}=\dfrac{u-f}{f}$
$\dfrac{v}{u}=\dfrac{f}{u-f}  ,  As, m=\dfrac{v}{u}$
$m=\dfrac{f}{u-f}$
hence,option C is correct.

The sum of the reciprocals of object distance and image distance is equal to the __________ of a mirror.

physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

  1. focal length

  2. reciprocal of the focal length

  3. radius of curvature

  4. reciprocal of the radius of curvature

Show answer
Correct Option: B
Explanation:

The sum of the reciprocals of object distance and image distance is equal to the reciprocal of the focal length of a mirror.