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MCQ 11 Mark
Suppose the magnitude of Nuclear force between two protons varies with the distance between them as shown in figure. Estimate the ratio "Nuclear force/Coulomb force" for:

  • A
    $x = 8fm$
     
  • B
    $x = 4fm$
     
  • C
    $x = 2fm$
     
  • D
    $x = 1fm (1fm = 10^{-15} m).$
Answer
First let us calculate the coulomb force between 2 protons for distance:
  1. $x = 8fm$
$ \mathrm{F}=\mathrm{K} \mathrm{q}^2 \mathrm{r}^2=9 \times 109 \times(1.6 \times 10-19)^2(8 \times 10-15) $
$ =3.6 \mathrm{NFN}$
$ =0.05 \mathrm{NFNFC}=0.053 .6=0.0138 \mathrm{~N}$
  1. $x = 4fm$
$FC = 9 × 109 × (1.6 × 10 - 19)^2(4 × 10 - 15)$
$= 23.04 × 10 - 29(4 × 10 - 15) = 14.4NF$
$= 1NFNFC = 114.4 = 0.0694N$
  1. $x = 2fm$
$FC = 9 × 109 × (1.6 × 10 - 19)^2(2 × 10 - 15)$
$= 57.6 NFN = 10 NFNFC = 1057.6 =0.173$
  1. $x = 1fm$
$FC = 9 × 109 × (1.6 × 10 - 19)^2(1 × 10 - 15)^2$
$= 230.4 NFN = 1000 NFNFC = 1000230.4 = 4.34$
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1 Marks Question - Physics STD 12 Science Questions - Vidyadip