Question

An insulated ring having a charge q, = 2 x 10 C is uniformly distributed over it has radius 4m. Another
particle having charge q, 4 x 104C is released from rest along its axis at distance x = 3m from its
centre. Mass of both ring and the particle is 1kg each. Neglect gravitational effects. Ring is free to move.
7.
Maximum speed of particle will be :
(A) 4.4 m/s
(B) 3.8 m/s
(C) 5.2 m/s
(D) 6.1 m/s
8.
If ring is not free to move then maximum speed of particle will be
12
4
2
(A)
(B)
(C)
15
5
ms-1
ms-1
(D) None of these
ms-1​

Answer 1

Answer:

7.] (B) 3.8m/s is the correct answer.

8.] (A) $\frac{12}{\sqrt 5} m/s$ is the correct answer.

Explanation:

From law of conservation of energy : $(K.E. + P.E.)_{initial} =(K.E. + P.E.)_{final}$

7.] Initially, K.E. = 0

Potential energy = $\frac{Kq_1 q_2}{r}$

Finally, P.E. = 0

As mass of both ring and particle is same, Kinetic energy will be distributed equally among these,

K.E. = $\frac{1}{2} mv^2+\frac{1}{2} mv^2=\ mv^2$

Equating initial and final energies,

$\frac{Kq_1 q_2}{r}=mv^2$

$\implies v=\sqrt \frac{Kq_1 q_2}{mr}$

$v=\sqrt \frac{(9*10^{9})*( 2*10^{-5})* (4*10^{-4})}{1*\sqrt {4^2+3^2}}=3.8m/s$

8.] If the ring is fixed its, final K.E. = $\frac{1}{2} mv^2$

Equating initial and final energies,

$\frac{Kq_1 q_2}{r}=\frac{1}{2}mv^2$

$\implies v=\sqrt \frac{2Kq_1 q_2}{mr}$

$v=\sqrt \frac{2*(9*10^{9})*( 2*10^{-5})* (4*10^{-4})}{1*\sqrt {4^2+3^2}}=\sqrt \frac{144}{5}=\frac{12}{\sqrt 5} m/s$

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