- 1(current)
Q. No.An electron falls through a distance of \(1.5\) cm in a uniform electric field of magnitude \(2\times10^4\) N/C [figure (a)]. The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance [figure (b)]. If \(t_e\) and \(t_p\) are the time of fall for electron and proton respectively, then:
1. \(t_e=t_p\)
2. \(t_e>t_p\)
3. \(t_e<t_p\)
4. none of these
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NEET MCQ Books for XIth & XIIth Physics, Chemistry & Biology
The mid-point of side AB is P while the circumcenter of ABC is O. Let the electric field at P be \(E_p\) and that at O be \(E_O.\)
Then, \(E_O:E_P=\)
1. \(\frac{2}{9}\)
2. \(\frac{4}{9}\)
3. \(\frac{9}{2}\)
4. \(\frac{9}{4}\)
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NEET MCQ Books for XIth & XIIth Physics, Chemistry & Biology
1. \(E_A>E_B>E_C\)
2. \(E_A=E_B=E_C\)
3. \(E_A=E_C>E_B\)
4. \(E_A=E_C<E_B\)
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NEET MCQ Books for XIth & XIIth Physics, Chemistry & BiologyNote: A, B, C are conductors. Other charges may be present in the vicinity.
1. I, III
2. II
3. I, II, III
4. none of I, II, III
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NEET MCQ Books for XIth & XIIth Physics, Chemistry & BiologyNow, identical point charges (same magnitude \(q\) each), are placed at the four corners of a square of side \(a.\) When one of the charges is removed, the electric field at the center becomes \(E_s.\) Then,
1. \(\frac{E_s}{2}=\frac{E_C}{3}\)
2. \(\frac{E_s}{3}=\frac{E_C}{2}\)
3. \(\frac{E_s}{\sqrt2}=\frac{E_C}{\sqrt3}\)
4. \(\frac{E_s}{\sqrt3}=\frac{E_C}{\sqrt2}\)
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NEET MCQ Books for XIth & XIIth Physics, Chemistry & Biology
| 1. | zero | 2. | \(4\dfrac{kq}{a^2}\) |
| 3. | \(2\dfrac{kq}{a^2}\) | 4. | \(2\sqrt2\dfrac{kq}{a^2}\) |
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NEET MCQ Books for XIth & XIIth Physics, Chemistry & Biology1. zero
2. \(\sqrt2\frac{kq}{a^2}\)
3. \(2\frac{kq}{a^2}\)
4. \(4\frac{kq}{a^2}\)
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NEET MCQ Books for XIth & XIIth Physics, Chemistry & Biology- 1(current)
Q. No.
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