Tag: physics

Questions Related to physics

A charge particle is moving in the direction of a magnetic filed.The magnetic force acting on the particle:

physics magnetic effect of electric current fleming's left hand rule magnetic force magnetic force on a moving charge and current carrying wire

  1. is in the direction of its velocity

  2. is in the direction opposite to its velocity

  3. is perpendicular to its velocity

  4. is zero

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Correct Option: D

According to Fleming's left hand rule,whose direction is indicated by thumb?

physics magnetic effect of electric current fleming's left hand rule magnetic force magnetic force on a moving charge and current carrying wire

  1. $ Electric current O $

  2. $ Magnetic force O $

  3. $ Magnetic field O $

  4. $ None of these O $

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Correct Option: C

Consider the integral $I=\displaystyle\int^{\pi} _0 ln(\sin x)dx$.What is $\displaystyle\int^{\dfrac{\pi}{2}} _{0}$ ln $(\sin x)dx$ equal to?

physics definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

  1. $4I$

  2. $2I$

  3. $I$

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

Show answer
Correct Option: D
Explanation:

$I = \displaystyle \int _{0}^{\pi} {ln(\sin x)dx}$

 using property,
$I = \displaystyle \int _{0}^{\dfrac{\pi}{2}} {(ln(\sin (2\pi -x) +ln(\sin x)) dx}$

we know that $\sin(x) = \sin(2\pi -x)$ 

$I = 2\displaystyle \int _{0}^{\dfrac{\pi}{2}} {ln(\sin x)dx}$

$\dfrac{I}{2} = \displaystyle \int _{0}^{\dfrac{\pi}{2}} {ln(\sin x)dx}$

Consider the integral $I=\displaystyle\int^{\pi} _0 ln(\sin x)dx$.What is $\displaystyle\int^{\frac{\pi} {2}} _0 ln(\cos x)dx$ equal to?

physics definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

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

  2. $I$

  3. $2I$

  4. $4I$

Show answer
Correct Option: A
Explanation:

$I = \displaystyle \int _{0}^{\pi} {ln(\sin x)dx}$

 using property,
$I = \displaystyle \int _{0}^{\dfrac{\pi}{2}} {(ln(\sin (2\pi -x) +ln(\sin x)) dx}$

we know that $\sin(x) = \sin(2\pi -x)$ 
$I = 2\displaystyle \int _{0}^{\dfrac{\pi}{2}} {ln(\sin x)dx}$

$\dfrac{I}{2} = \displaystyle \int _{0}^{\dfrac{\pi}{2}} {ln(\sin x)dx}$
 by property,

$\dfrac{I}{2} = \displaystyle \int _{0}^{\dfrac{\pi}{2}} {ln(\sin (\dfrac{\pi}{2} - x))dx}$

$\dfrac{I}{2} = \displaystyle \int _{0}^{\dfrac{\pi}{2}} {ln(\cos x)dx}$

$ \int _{\sin x}^1 t^2 f(t) dt = 1 - \sin x \forall x \epsilon (0, \pi / 2 ) $ then $ f \left( \dfrac {1}{\sqrt3} \right) $ is :

physics definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

  1. $3$

  2. $\sqrt3$

  3. $1/3$

  4. None of these

Show answer
Correct Option: A
Explanation:

$ \int _{\sin x}^1 t^2 f(t) dt = 1 - \sin x \forall x \epsilon (0, \pi/2 ) $
Differentiating both sides we get 
$ \dfrac {d}{dx} (1) [ 1 \cdot f (1)] - \cos x ( \sin^2 x) f ( \sin x) = -\cos x  $
$ \Rightarrow f ( \sin x) = \dfrac {1}{ \sin^2 x} $
$ \therefore f \left( \dfrac {1}{\sqrt3} \right) = f \left( \sin \left( \sin^{-1} \dfrac {1}{\sqrt3} \right) \right) $
$ = \left[ \dfrac {1}{ \sin \left( \sin^{-1} \dfrac {1}{\sqrt3} \right)} \right]^2 = 3 $

Consider the integrals ${I _1} = \int _0^1 {{e^{ - x}}{{\cos }^2}xdx,} {I _2} = \int _0^1 {{e^{ - {x^2}}}{{\cos }^2}xdx,} {I _3} = \int _0^1 {{e^{ - x}}dx} $ and ${I _4} = \int _0^1 {{e^{ - (1/2){x^2}}}} dx$. The greatest of these integrals is

physics definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

  1. $I _1$

  2. $I _2$

  3. $I _3$

  4. $I _4$

Show answer
Correct Option: D
Explanation:

$I _1=\int _{0}^{4}e^{-x} cos^2x dx$
$I _2=\int _{0}^{1}e^{-x^2}cos^2x dx$
Both have $cos^2x$ so value get restricted more in (0, 1)
Now in (0, 1) $e^{-\frac {x^2}{2}}>e^{-x}$
$\therefore \int _{0}^{1}e^{-\frac {x^2}{2}}>\int _{0}^{1}e^{-x}$
$\therefore I _4>I _3>I _1>I _2$

Let $ f\left( a,b \right) =\int _{ a }^{ b }{ \left( { x }^{ 2 }-4x+3 \right) dx,\left( b>a \right)  }$ then

physics definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

  1. $ f\left( a,3 \right)$ is least when $a=1$

  2. $f\left( 4,b \right)$ is an increasing function $ \forall b\ge 4$

  3. $ f\left( 0,b \right)$ is least for $b=2$

  4. $ \min { \left{ f\left( a,b \right) \right} =-\dfrac { 4 }{ 3 } } \forall a,b\in R$

Show answer
Correct Option: A
Explanation:

First of all integrate the function.

$\displaystyle \int _{  }^{  }{ \left( { x }^{ 2 }-4x+3 \right) dx= } \left[ \frac { { x }^{ 3 } }{ 3 } -4\frac { { x }^{ 2 } }{ 2 } +3x \right] $

Apply the limits,
$\displaystyle f(a,b)={ \left[ \frac { { x }^{ 3 } }{ 3 } -4\frac { { x }^{ 2 } }{ 2 } +3x \right]  } _{ a }^{ b }$

Here $b=3$

Therefore,
$\displaystyle f(a,3)={ \left[ \frac { { x }^{ 3 } }{ 3 } -4\frac { { x }^{ 2 } }{ 2 } +3x \right]  } _{ a }^{ b }=\left[ \frac { 27 }{ 3 } -2\times 9+9 \right] -\left[ \frac { { a }^{ 3 } }{ 3 } -2{ a }^{ 2 }+3a \right] \ \displaystyle =2{ a }^{ 2 }-\frac { { a }^{ 3 } }{ 3 } -3a$

Hence,
Differentiate the above function to find the maximum or minimum.
$\displaystyle { f }^{ \prime  }\left( a,3 \right) =4a-{ a }^{ 2 }-3=0$

therefore $\displaystyle a=1,3$

Check whether the function is minimum or maximum.
$\displaystyle { f }^{ \prime \prime  }\left( a,3 \right) =4-{ 2a }=2$ which  is greater than $0$.
Hence the function $f(a,3)$ is minimum at $a=1$.



$\displaystyle \int _0^1 \dfrac{xe^x}{(x + 1)^2} dx =$

physics definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

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

  2. $\dfrac{e - 1}{2}$

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

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

Show answer
Correct Option: C

$\displaystyle\int _{ 0 }^{ 1 }{ \cfrac { \tan ^{ -1 }{ x }  }{ x }  } dx$ equals

physics definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

  1. $\displaystyle\int _{ 0 }^{ \pi /2 }{ \cfrac { x }{ \sin { x } } dx } $

  2. $\cfrac { 1 }{ 2 } \displaystyle\int _{ 0 }^{ \pi /2 }{ \cfrac { x }{ \sin { x } } dx } $

  3. $\displaystyle\int _{ 0 }^{ \pi /2 }{ \cfrac { \sin { x } }{ x } dx } $

  4. None of the above

Show answer
Correct Option: B
Explanation:

Let $I=\int _{ 0 }^{ 1 }{ \cfrac { \tan ^{ -1 }{ x }  }{ x }  } dx$


Put $\tan ^{ -1 }{ x } =\cfrac { z  }{ 2 } \Rightarrow x=\tan { \cfrac { z  }{ 2 }  } $

$\Rightarrow dx=\cfrac { 1 }{ 2 } \sec ^{ 2 } \cfrac { z  }{ 2 } dz$

$\therefore \quad I=\displaystyle\int _{ 0 }^{ \pi /2 }{ \cfrac { \cfrac { z }{ 2 } \left( \cfrac { 1 }{ 2 } \sec ^{ 2 } \cfrac { z  }{ 2 }  \right)  }{ \tan { \cfrac { z }{ 2 }  }  }  } dz$

$I=\cfrac { 1 }{ 2 } \displaystyle\int _{ 0 }^{ \pi /2 }{ \cfrac { z }{ \cfrac { \sin { \cfrac { z }{ 2 }  }  }{ \cos { \cfrac { z }{ 2 }  }  }  } .\cfrac { 1 }{ 2\cos ^{ 2 }{ \cfrac { z }{ 2 }  }  } dz } $

$=\cfrac { 1 }{ 2 } \int _{ 0 }^{ \pi /2 }{ \cfrac { z }{ 2\sin { \cfrac { z }{ 2 }  } \cos { \cfrac { z }{ 2 }  }  }  } dz$

$=\cfrac { 1 }{ 2 } \displaystyle\int _{ 0 }^{ \pi /2 }{ \cfrac { z }{ \sin { z }  }  } dz=\cfrac { 1 }{ 2 } \displaystyle\int _{ 0 }^{ \pi /2 }{ \cfrac { x }{ \sin { x }  }  } dx$

Evaluate $\displaystyle\int^{\frac{3}{2}} _{-1}|x\sin(\pi x)|dx$.

physics definite integral and it applications to areas fundamental theorem of calculus interpretation of define integral as an area fundamental theorem of integral calculus

  1. $\dfrac {3}{\pi} +\dfrac {1}{\pi^2}$

  2. $3\pi +\pi^2$

  3. $\dfrac { 2 }{ \pi } +\dfrac { 1 }{ { \pi }^{ 2 } }$

  4. none of the above

Show answer
Correct Option: A
Explanation:

$|x\sin(\pi x)|=\begin{cases}x\sin \pi x ,\,\,\,\,x\in(-1,1)\  -x\sin\pi x,x\in (1,\dfrac{3}{2})  \end{cases}$

$\displaystyle \int _{ 1 }^{ \frac { 3 }{ 2 }  }{ |x\sin(\pi x)| }dx=\int _{ -1 }^{ 1  }{ x\sin(\pi x)dx }+\int _{ 1 }^{ \frac { 3 }{ 2 }  }{ -x\sin(\pi x)dx }$

$=\displaystyle \left|\dfrac{-x\cos\pi x}{\pi}\right|^1 _{-1}-\int _{- 1 }^{ 1  }{ \left(\dfrac{-\cos\pi x}{\pi}dx\right) }-\left[\left|\dfrac{-x\cos (\pi x)}{\pi}\right|^{\dfrac{3}{2}} _1-\int _{ 1 }^{ \frac { 3 }{ 2 }  }{ \dfrac{-\cos x }{\pi}dx } \right] $ 

$=\dfrac{1}{\pi}-\left(\dfrac{-1}{\pi}\right)+0-\left[\dfrac{-1}{\pi}-(\dfrac{1}{\pi^2})\right]$

$=\dfrac{1}{\pi}+\dfrac{1}{\pi}+\dfrac{1}{\pi}+\dfrac{1}{\pi^2}$

$=\dfrac{3}{\pi}+\dfrac{1}{\pi^2}$