Fill in the blanks by suitable conversion of units:
(a) 1 kg m2s–2= ....g cm2 s–2
(b) 1 m =..... ly
(c) 3.0 m s–2=.... km h–2
(d) G = 6.67 × 10–11 N m2 (kg)–2=.... (cm)3s–2 g–1.
(a) 1 kg = 103 g
1 m2 = 104 cm2
1 kg m2 s–2 = 1 kg × 1 m2 × 1 s–2
=103 g × 104 cm2 × 1 s–2 = 107 g cm2 s–2
1 kg m2s–2 = 107 g cm2 s–2
(b) Distance = Speed × Time
Speed of light = 3 × 108 m/s
Time = 1 year = 365 days = 365 × 24 hours = 365 × 24 × 60 × 60 sec
Putting these values in above formula we get
1 light year distance = (3 × 108 m/s) × (365 × 24 × 60 × 60 s) = 9.46 × 1015 m 9.46 × 1015 m = 1 ly
So that 1 m = 1/ 9.46 × 1015 ly = 1.06 × 10-16 ly
(c) 1 hour = 3600 sec so that 1 sec = 1/3600 hour
1 km = 1000 m so that 1 m = 1/1000 km
3.0 m s–2 = 3.0 (1/1000 km)( 1/3600 hour)-2 = 3.0 × 10–3 km × ((1/3600)-2 h–2)
= 3.0 × 10–3 km × (3600)2 h–2 = 3.88 × 104 km h–2 3.0 m s–2
= 3.88 × 104 km h–2
(d) Given,
G= 6.67 × 10–11 N m2 (kg)–2
We know that
1 N = 1 kg m s–2
1 kg = 103 g
1 m = 100 cm = 102 cm
Putting above values,we get
6.67 × 10–11 N m2 kg–2 = 6.67 × 10–11 × (1 kg m s–2) (1 m2) (1Kg–2)
Solve and cancel out the units we get
⇒ 6.67 × 10–11 × (1 kg–1 × 1 m3 × 1 s–2)
Putting above values to convert Kg to g and m to cm
⇒ 6.67 × 10–11 × (103 g)-1 × (102 cm)3 × (1 s–2)
⇒ 6.67 × 10–11 × 10-3 g-1 × 106 cm3 × (1 s–2)
⇒ 6.67 × 10–8 cm3 s–2 g–1
G = 6.67 × 10–11 N m2 (kg)–2
= 6.67 × 10–8 (cm)3s–2 g–1.
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A physical quantity P is related to four observables a, b, c and d as follows :
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(d) a tumbling beaker that has slipped off the edge of a table. - Q:-
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1 Comment(s) on this Question
excellent explanation ever seen
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