One gram of mass is equal to:

5 \times 10^{10} e r g

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9 \times 10^{20} e r g

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7 \times 10^{5} e r g

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11 \times 10^{12} e r g

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Solution

The correct option is B $9 \times 10^{20} e r g$
$E = m c^{2}$
$E = 1 g \times (3 \times 10^{10} c m / s)^{2}$
$= 9 \times 10^{20} e r g$

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a

$\frac{5 + 9 + 13 + . . . . . . . u p t o n . t e r m s}{I f 7 + 9 + 11 + . . . . . u p t o (n + 1) t e r m s} = \frac{17}{16}$ then n is equal to

(- \frac{8}{9} \times \frac{1}{- 5}) + (- \frac{8}{9} \times - \frac{- 7}{11}) = - \frac{8}{9} \times (\frac{1}{- 5} + \frac{- 7}{11})

(10)^{9} + 2 (11)^{1} (10)^{8} + 3 (11)^{2} (10)^{7} + . . . + 10 (11)^{9} = k (10)^{9}

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