Question

Hi guys...... please solve it................................​

Answer 1

Question :-

A point particle of mass M attached to one end of a mass-less rigid non-conducting rod of length I. Another point particle of the same mass is attached to the other end of the rod. The two particles carry charges +q and -q respectively. This arrangement is held in a region of a uniform electric field E such that the rod makes a small angle θ (say of about 5 degrees) with the field direction. Find an expression for the minimum time needed for the rod to become parallel to the field after it is set free.

Solution :-

The force acting at both the charged constitute a couple.

$\sf{Z=Force \times Perpendicular\ distance}$

$\sf{\hookrightarrow Z=qE \times ABsin\theta }\\\\\sf{\hookrightarrow Z== qElsin\theta}$

As θ is a small angle (taken 5°),

sinθ = 0 (approx. value)

$\sf{\hookrightarrow Z= qEl\theta}$

Restoring torque Z = $\sf{-PEl\theta}$

If α is the angular acceleration,

$\sf{\hookrightarrow Z=J\alpha }\\\\\sf{\hookrightarrow Z=\dfrac{Ml^2}{2}\alpha }\\\\\sf{\hookrightarrow \alpha =-W^2\theta}\\\\\sf{\hookrightarrow W=\dfrac{2qE}{Ml} }\\\\\sf{\hookrightarrow W=(\dfrac{2qE}{Ml} )^\frac{1}{2} }\:\:\:\: \cdots (i)$  

Time period of oscillation = $\sf{\dfrac{2\pi}{W}}$

$\sf{\hookrightarrow T=2\pi (\dfrac{MI}{2qE} )^\frac{1}{2} }\:\:\:\: \cdots [from (i)]$

Rotating in clock wise direction, minimum time taken by rod to align itself field is time taken to complete 1/4th oscillation

$\sf{\hookrightarrow t=\dfrac{T}{4} }\\\\\large\boxed{\boxed{\sf{\hookrightarrow t=\dfrac{\pi}{2}\times \dfrac{MI}{2qE}^\frac{1}{2} }}}\:\:\:\: ---\mathbf{ANSWER}$

Answer 2

Hie Mate ✌️

Question :-

A point particle of mass M attached to one end of a mass-less rigid non-conducting rod of length I. Another point particle of the same mass is attached to the other end of the rod. The two particles carry charges +q and -q respectively. This arrangement is held in a region of a uniform electric field E such that the rod makes a small angle θ (say of about 5 degrees) with the field direction. Find an expression for the minimum time needed for the rod to become parallel to the field after it is set free.

Solution :-

The force acting at both the charged constitute a couple.

\sf{Z=Force \times Perpendicular\ distance}Z=Force×Perpendicular distance

\begin{gathered}\sf{\hookrightarrow Z=qE \times ABsin\theta }\\\\\sf{\hookrightarrow Z== qElsin\theta}\end{gathered}

↪Z=qE×ABsinθ

↪Z==qElsinθ

As θ is a small angle (taken 5°),

sinθ = 0 (approx. value)

\sf{\hookrightarrow Z= qEl\theta}↪Z=qElθ

Restoring torque Z = \sf{-PEl\theta}−PElθ

If α is the angular acceleration,

\begin{gathered}\sf{\hookrightarrow Z=J\alpha }\\\\\sf{\hookrightarrow Z=\dfrac{Ml^2}{2}\alpha }\\\\\sf{\hookrightarrow \alpha =-W^2\theta}\\\\\sf{\hookrightarrow W=\dfrac{2qE}{Ml} }\\\\\sf{\hookrightarrow W=(\dfrac{2qE}{Ml} )^\frac{1}{2} }\:\:\:\: \cdots (i)\end{gathered}

↪Z=Jα

↪Z=

2

Ml

2

α

↪α=−W

2

θ

↪W=

Ml

2qE

↪W=(

Ml

2qE

)

2

1

⋯(i)

Time period of oscillation = \sf{\dfrac{2\pi}{W}}

W

2π

\sf{\hookrightarrow T=2\pi (\dfrac{MI}{2qE} )^\frac{1}{2} }\:\:\:\: \cdots [from (i)]↪T=2π(

2qE

MI

)

2

1

⋯[from(i)]

Rotating in clock wise direction, minimum time taken by rod to align itself field is time taken to complete 1/4th oscillation

\begin{gathered}\sf{\hookrightarrow t=\dfrac{T}{4} }\\\\\large\boxed{\boxed{\sf{\hookrightarrow t=\dfrac{\pi}{2}\times \dfrac{MI}{2qE}^\frac{1}{2} }}}\:\:\:\: ---\mathbf{ANSWER}\end{gathered}

↪t=

4

T

↪t=

2

π

×

2qE

MI

2

1

-StunningBabe27 ❤️