All the curves in Column 1 are parabolic and those in Column 2 and Column 3 are straight lines.

‘x’, ‘v’ and ‘a’ denote position, velocity and acceleration of a particle and consider positive in the direction of increasing ‘x’.

In which one of the following options the particle starts with positive velocity and crosses the origin with negative velocity?

(I) (iii) (P)

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(II) (ii) (P)

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(III) (iv) (R)

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(IV) (i) (R)

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Solution

The correct option is D
(IV) (i) (R)

Concept Based

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Similar questions

All the curves in Column 1 are parabolic and those in Column 2 and Column 3 are straight lines.

‘x’, ‘v’ and ‘a’ denote position, velocity and acceleration of a particle and consider positive in the direction of increasing ‘x’.

Which one of the following options correctly represents a particle starting with positive position, positive velocity and uniform positive acceleration?

All the curves in Column 1 are parabolic and those in Column 2 and Column 3 are straight lines.

‘x’, ‘v’ and ‘a’ denote position, velocity and acceleration of a particle and consider positive in the direction of increasing ‘x’.

Which of the following options is the only INCORRECT combination?

A charged particle is projected with velocity $\to v = u^i$ from the origin, in a region having electric field $\to E$ and magnetic field $\to B$ both. Value of $\to B$ and $\to E$ are given in column 1 and column 2 respectively and path of the particle is given in column 3.

$\begin{matrix} Column 1 & Column 2 & Column 3 (Magnetic field) & (Electric field) & (Path of particle) (i) & \to B = 0 & (i) & \to E = 0 & (P) & Circular path (i i) & \to B = B_{0}^i + B_{0}^j & (i i) & \to E = E_{0}^i & (Q) & Helical path (i i i) & \to B = B_{0}^j + B_{0}^k & (i i i) & \to E = B_{0} u (^j -^k) & (R) & Cycloid (i v) & \to B = B_{0}^k & (i v) & \to E = E_{0}^k & (S) & Straight line \end{matrix}$

In which of the following combination particle moves in uniform circular motion?

A charged particle (electron or proton) is introduced at the origin $(x = 0, y = 0, x = 0)$ with a given initial velocity $\to v$ . A uniform electric field $\to E$ and a uniform magnetic field $\to B$ exist everywhere. The velocity $\to v$ , electric field $\to E$ and magnetic field $\to B$ are given in column $1, 2$ and $3,$ respectively. The quantities $E_{0}, B_{0}$ are positive in magnitude.

Column 1	Column 2	Column 3
(I) Electron with $\to v = 2 \frac{E_{0}}{B_{0}}^x$	(i) $\to E = E_{0}^z$	(P) $\to B = - B_{0}^x$
(II) Electron with $\to v = \frac{E_{0}}{B_{0}}^y$	(ii) $\to E = - E_{0}^y$	(Q) $\to B = - B_{0}^x$
(III) Electron with $\to v = 0$	(iii) $\to E = - E_{0}^x$	(R) $\to B = B_{0}^y$
(IV) Electron with $\to v = 2 \frac{E_{0}}{B_{0}}^x$	(iv) $\to E = E_{0}^x$	(S) $\to B = B_{0}^z$

In which case would the particle move in a straight line along the negative direction of

y

-axis (i.e, move along

-^y) ?

Which of the following combination is correct?

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The correct option is D (IV) (i) (R) Concept Based

The correct option is D
(IV) (i) (R)

Concept Based