A solid of mass 1 kg has 6 times 10^{24} ato | vedclass

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(B) Total number of atoms $N = 6 	imes 10^{24}$.Number of atoms from which one electron is removed is $n = 0.005 \% 	ext{ of } N$.$n = \frac{0.005}{

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$+48 \ C$

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(B) Total number of atoms $N = 6 	imes 10^{24}$.Number of atoms from which one electron is removed is $n = 0.005 \% 	ext{ of } N$.$n = \frac{0.005}{100} 	imes 6 	imes 10^{24} = 5 	imes 10^{-5} 	imes 6 	imes 10^{24} = 30 	imes 10^{19} = 3 	imes 10^{20}$.Since one electron is removed from each of these atoms,the total number of electrons removed is $3 	imes 10^{20}$.The charge gained by the solid is $q = n 	imes e$,where $e = 1.6 	imes 10^{-19} \ C$.$q = (3 	imes 10^{20}) 	imes (1.6 	imes 10^{-19}) = 4.8 	imes 10 = 48 \ C$.Since electrons are removed,the solid gains a positive charge of $+48 \ C$.

$A$ solid of mass $1 \ kg$ has $6 \times 10^{24}$ atoms. If one electron is removed from every one atom of $0.005 \%$ of the atoms,then the charge gained by the solid is

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