First slide

Physical quantities and units

Question

If the acceleration due to gravity is $10 m s^{- 2}$ and the units of length and time are changed in kilometre and hour respectively, the numerical value of acceleration is

Moderate

A

360000
B

72000
C

36000
D

129600

Solution

As, $n_{2} = n_{1} {(\frac{L_{1}}{L_{2}})}^{1} {(\frac{T_{1}}{T_{2}})}^{- 2} = 10 {(\frac{m}{k m})}^{1} {(\frac{\sec}{h r})}^{- 2}$
(Here, $n_{1} a n d n_{2}$ are numerical values)
$n_{2} = 10 {(\frac{m}{10^{3} m})}^{1} {(\frac{\sec}{3600 \sec})}^{- 2} = 129600$

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