Tag: physics

Quizzes (598) Questions (13441)

Questions Related to physics

Multiple choice physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

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

  1. $log _{a3}$
  2. $log _{a1}$
  3. $log _{a2}$
  4. $log _{a5}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

Multiple choice physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

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

  1. $\displaystyle \frac{1}{100}$ or $100$
  2. $\pm$ 2
  3. 10 or $\displaystyle \frac{1}{10}$

  4. 100
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

Multiple choice physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

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

  1. x = 4

  2. x = + 1 or -1
  3. x = $\pm$ 2

  4. x = 0
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

Multiple choice physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

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

  1. no natural solution

  2. two rational solutions

  3. no prime solution

  4. one irrational solution
Reveal answer Fill a bubble to check yourself
B,C,D Correct answer
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} $

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

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

  1. $\frac { f }{ 3 }$
  2. $\frac { 2f }{ 3 }$
  3. $1f$
  4. $2f$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Magnification m = -v/u = 1/3 for a convex mirror. Using the mirror formula 1/v + 1/u = 1/f, we substitute v = -u/3. This gives -3/u + 1/u = 1/f, leading to -2/u = 1/f, so u = -2f. The distance is 2f.

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

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:

  1. $1.25cm$
  2. $2.5 cm$
  3. $2 cm$
  4. $0.75cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice physics reflection of light in spherical mirrors sign convention for mirrors sign convention for spherical mirrors sign convention for the measurement of distances

The object distance $u$ for a concave mirror:

  1. must be positive

  2. must be negative

  3. must not be negative

  4. may be negative
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

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).

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

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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

  1. focal length

  2. reciprocal of the focal length

  3. radius of curvature

  4. reciprocal of the radius of curvature
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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