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Physics
Two protons are kept at a separation of 40 mathringA. Fn is the nuclear force and Fe, is the electrostatic force between them. Then
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Q. Two protons are kept at a separation of 40 $\mathring{A}$. $F_n$ is the nuclear force and $F_e$, is the electrostatic force between them. Then
KCET
KCET 2008
Nuclei
A
$F_n >> F_e $
26%
B
$F_n = F_e $
10%
C
$F_n << F_e $
61%
D
$F_n \approx F_e $
3%
Solution:
$F_{n}$ is stronger than $F_{e} . F_{n}$ operates at very short range inside the nucleus as little as $10^{-15} m$ As in the given case two protons are kept at a separation of $40 \,\mathring{A} F_{n} << F_{e}$