The value of 120Joule per second on a system that has 10^1 gram ,10^2 centimetre and 10^3 second as base unit is

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Q. The value of gravitational constant

G

in Meter-Kilogram-Second system is

6.67 \times 10^{- 11} N - m^{2} k g^{- 2}

. What will be its value in centimetre gram second system.

What should be the momentum (in gram centimetre per second) of a particle if its de-Broglie wavelength is 1 angstrom and the value of his $6.6252 \times 10^{- 27}$ erg second ?

Q. If 125 x centimetre per second = (x+7) km/hour then the value of x is

32.689 × $10^{1}$ = ?

32.689 ÷ $10^{2}$ = ?

32.689 × $10^{3}$ = ?

Write the numeral whose expanded form is given below:
(i) $6 \times 10^{4} + 3 \times 10^{3} + 0 \times 10^{2} + 7 \times 10^{1} + 8 \times 10^{0}$
(ii) $9 \times 10^{6} + 7 \times 10^{5} + 0 \times 10^{4} + 3 \times 10^{3} + 4 \times 10^{2} + 6 \times 10^{1} + 2 \times 10^{0}$
(iii) $8 \times 10^{5} + 6 \times 10^{4} + 4 \times 10^{3} + 2 \times 10^{2} + 9 \times 10^{1} + 6 \times 10^{0}$

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