From a metallic charged body a current is drawn. The rate of increase of current at an instant is equal to the charge on the body at that instant. If the initial charge on the body is Q

Options

A. the minimum time it will take for the charge to become zero is `(pi)/(2)sec`
B. the minimum time it will take for the charge to become zero is 2 sec
C. The value of the current when the charge on the body is `Q//2` is `Q sqrt((3)/(2))`
D. The value of the current when the charge on the body is `Q//2` is `(Q)/(sqrt(3))`

Explanation

To solve the problem step by step, we need to analyze the situation involving the metallic charged body, the current drawn from it, and the relationship between charge and current.

### Step 1: Understand the relationship between charge and current
From the problem, we know that the current \( I \) drawn from the charged body is related to the rate of change of charge \( Q \) on the body. The current can be expressed as:
\[ I = -\frac{dQ}{dt} \]
This negative sign indicates that the charge \( Q \) is decreasing as current is drawn from the body.

### Step 2: Establish the rate of increase of current
The problem states that the rate of increase of current at an instant is equal to the charge on the body at that instant. Mathematically, this can be expressed as:
\[ \frac{dI}{dt} = Q \]

### Step 3: Substitute the expression for current
Now, we can substitute the expression for current \( I \) into the equation for \( \frac{dI}{dt} \):
\[ \frac{dI}{dt} = \frac{d}{dt}\left(-\frac{dQ}{dt}\right) = -\frac{d^2Q}{dt^2} \]
So, we have:
\[ -\frac{d^2Q}{dt^2} = Q \]
This can be rearranged to form a second-order differential equation:
\[ \frac{d^2Q}{dt^2} + Q = 0 \]

### Step 4: Solve the differential equation
The equation \( \frac{d^2Q}{dt^2} + Q = 0 \) is characteristic of simple harmonic motion. The general solution can be expressed as:
\[ Q(t) = Q_0 \cos(\omega t) \]
where \( Q_0 \) is the maximum charge and \( \omega \) is the angular frequency.

### Step 5: Find the second derivative of charge
To find the current, we need to differentiate \( Q(t) \):
\[ \frac{dQ}{dt} = -Q_0 \omega \sin(\omega t) \]
And the second derivative is:
\[ \frac{d^2Q}{dt^2} = -Q_0 \omega^2 \cos(\omega t) \]

### Step 6: Substitute back into the differential equation
Substituting \( \frac{d^2Q}{dt^2} \) back into the differential equation gives:
\[ -Q_0 \omega^2 \cos(\omega t) + Q_0 \cos(\omega t) = 0 \]
This simplifies to:
\[ Q_0(\omega^2 - 1) \cos(\omega t) = 0 \]
Since \( Q_0 \neq 0 \), we have:
\[ \omega^2 = 1 \implies \omega = 1 \]

### Step 7: Determine the time period
The time period \( T \) of the oscillation is given by:
\[ T = \frac{2\pi}{\omega} = \frac{2\pi}{1} = 2\pi \text{ seconds} \]

### Step 8: Find the time when charge becomes zero
To find the time when the charge becomes zero:
\[ Q(t) = Q_0 \cos(\omega t) = 0 \]
This occurs when:
\[ \cos(\omega t) = 0 \implies \omega t = \frac{\pi}{2} \implies t = \frac{\pi}{2} \text{ seconds} \]

### Conclusion
The minimum time it will take for the charge to become zero is:
\[ t = \frac{\pi}{2} \text{ seconds} \]