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A body of mass 1g and carrying a charge ...

A body of mass 1g and carrying a charge `10^(-8)`C passes from two points P and Q. P and Q are at electric potentials 600 V and 0 V respectively. The velocity of the body at Q is 20 `cms^(-1)`. It velocity in `ms^(-1)`at P is

A

`sqrt(0.028)`

B

`sqrt(0.056)`

C

`sqrt(0.56)`

D

`sqrt(5.6)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step by step, we will use the principle of conservation of energy. The total mechanical energy (kinetic energy + potential energy) remains constant when a charged body moves in an electric field. ### Step 1: Understand the Given Data - Mass of the body, \( m = 1 \, \text{g} = 1 \times 10^{-3} \, \text{kg} \) - Charge of the body, \( q = 10^{-8} \, \text{C} \) - Electric potential at point P, \( V_P = 600 \, \text{V} \) - Electric potential at point Q, \( V_Q = 0 \, \text{V} \) - Velocity at point Q, \( V_2 = 20 \, \text{cm/s} = 20 \times 10^{-2} \, \text{m/s} = 0.2 \, \text{m/s} \) ### Step 2: Write the Energy Conservation Equation The conservation of energy states that the gain in kinetic energy is equal to the loss in potential energy. Therefore, we can write: \[ \text{Gain in K.E.} = \text{Loss in P.E.} \] This can be expressed mathematically as: \[ \frac{1}{2} m V_1^2 - \frac{1}{2} m V_2^2 = q V_P - q V_Q \] ### Step 3: Substitute Known Values Substituting the known values into the equation: \[ \frac{1}{2} (1 \times 10^{-3}) V_1^2 - \frac{1}{2} (1 \times 10^{-3}) (0.2)^2 = (10^{-8})(600) - (10^{-8})(0) \] ### Step 4: Simplify the Equation This simplifies to: \[ \frac{1}{2} (1 \times 10^{-3}) V_1^2 - \frac{1}{2} (1 \times 10^{-3}) (0.04) = 6 \times 10^{-6} \] Now, simplifying further: \[ \frac{1}{2} (1 \times 10^{-3}) V_1^2 - 2 \times 10^{-5} = 6 \times 10^{-6} \] ### Step 5: Rearranging the Equation Rearranging gives: \[ \frac{1}{2} (1 \times 10^{-3}) V_1^2 = 6 \times 10^{-6} + 2 \times 10^{-5} \] Calculating the right side: \[ 6 \times 10^{-6} + 2 \times 10^{-5} = 2.6 \times 10^{-5} \] So we have: \[ \frac{1}{2} (1 \times 10^{-3}) V_1^2 = 2.6 \times 10^{-5} \] ### Step 6: Solve for \( V_1^2 \) Multiplying both sides by 2: \[ (1 \times 10^{-3}) V_1^2 = 5.2 \times 10^{-5} \] Now, divide by \( 1 \times 10^{-3} \): \[ V_1^2 = \frac{5.2 \times 10^{-5}}{1 \times 10^{-3}} = 5.2 \times 10^{-2} \] ### Step 7: Take the Square Root Taking the square root gives: \[ V_1 = \sqrt{5.2 \times 10^{-2}} \approx 0.228 \, \text{m/s} \] ### Final Answer The velocity of the body at point P is approximately \( 0.228 \, \text{m/s} \). ---
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