A conductor has been given a charge -3 × 10-7 by transferring electrons. Mass increase (in kg ) of the conductor and the number of electrons added to the conductor are respectively

Question

A conductor has been given a charge -3 × 10-7 by transferring electrons. Mass increase (in kg ) of the conductor and the number of electrons added to the conductor are respectively (A)  2 × 10-16 and 2 × 1031  (B)  5 × 10-31 and 5 × 1019  (C)  3 × 10-19 and 9 × 1016  (D)  2 × 10-18 and 2 × 1012

Tardigrade Admin · Accepted Answer

Number of electrons added to the conductor n=(q/e) =(-3×10-7/1.6×10-19)=1.8×1012 Mass increase of the conductor =1.8×1012×9.1×10-31 =20×10-19=2×10-18

Q. A conductor has been given a charge $- 3 \times 1 0^{- 7}$ by transferring electrons. Mass increase (in $k g$ ) of the conductor and the number of electrons added to the conductor are respectively

$2 \times 1 0^{- 16}$ and $2 \times 1 0^{31}$

$5 \times 1 0^{- 31}$ and $5 \times 1 0^{19}$

$3 \times 1 0^{- 19}$ and $9 \times 1 0^{16}$

$2 \times 1 0^{- 18}$ and $2 \times 1 0^{12}$

Number of electrons added to the conductor
$n = \frac{q}{e}$
$= \frac{- 3 \times 1 0 ^{- 7}}{1.6 \times 1 0 ^{- 19}} = 1.8 \times 1 0^{12}$
Mass increase of the conductor
$= 1.8 \times 1 0^{12} \times 9.1 \times 1 0^{- 31}$
$= 20 \times 1 0^{- 19} = 2 \times 1 0^{- 18}$

Q. A conductor has been given a charge −3×10−7 by transferring electrons. Mass increase (in kg ) of the conductor and the number of electrons added to the conductor are respectively

2×10−16 and 2×1031

5×10−31 and 5×1019

3×10−19 and 9×1016

2×10−18 and 2×1012