Tag: physics

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

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

  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
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The magnetic force on a moving charge is F = q(v x B). If the charge moves parallel to the magnetic field, the angle between v and B is 0 degrees, so the cross product (and thus the force) is zero.

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

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?

  1. $4I$

  2. $2I$

  3. $I$

  4. $\dfrac{I}{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $\sin(x) = \sin(2\pi -x)$ 

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

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

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

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?

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

  2. $I$

  3. $2I$

  4. $4I$
Reveal answer Fill a bubble to check yourself
A Correct answer
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,

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

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

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

$ \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 :

  1. $3$

  2. $\sqrt3$

  3. $1/3$

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

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

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

  1. $I _1$

  2. $I _2$

  3. $I _3$

  4. $I _4$
Reveal answer Fill a bubble to check yourself
D Correct answer
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) $e^{-\frac {x^2}{2}}>e^{-x}$
$\therefore \int _{0}^{1}e^{-\frac {x^2}{2}}>\int _{0}^{1}e^{-x}$
$\therefore I _4>I _3>I _1>I _2$

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

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

  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$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $a=1$.



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

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

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
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 } $

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

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

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

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

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

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

  3. <span>$\dfrac { 2 }{ \pi } +\dfrac { 1 }{ { \pi }^{ 2 } }$</span>

  4. none of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
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] $ 

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

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

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