If the acceleration due to gravity is $10 m^{- 2}$ and the units of length and times are changes in kilometer and hour respectively. The numerical value of the acceleration is

360000

No worries! We‘ve got your back. Try BYJU‘S free classes today!

72000

No worries! We‘ve got your back. Try BYJU‘S free classes today!

36000

No worries! We‘ve got your back. Try BYJU‘S free classes today!

129600

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $129600$
$g = 10 \frac{M}{S^{2}} = 10 \times \frac{M}{S . S}$
$M = 10^{- 3} k m$
$S = \frac{1}{3600} h$
$∴$ $g = \frac{10 \times 10^{- 3}}{\frac{1}{3600} h \times \frac{1}{3600} h}$
$= 36 \times 36 \times 10^{2} k m / h^{2}$
$= 1296 k m / h^{2}$

Suggest Corrections

Similar questions

Q. The value of acceleration due to gravity is

980 c m s^{- 2}

. What will be its value if the unit of length is kilometer and that of time is minute?

Q. If unit of length and time is doubled, the numerical value of

g

(acceleration due to gravity) will be

Q. The percentage errors in the measurement of length and time period of a simple pendulum are 1% and 2% respectively. Then, the maximum error in the measurement of acceleration due to gravity is:

The percentage errors in the measurements of the length of a simple pendulum and its time period are 3% and 5% respectively. The maximum error in the value of the acceleration due to gravity obtained from these measurements is

Q. Let us assume that the acceleration due to gravity be

10 m / s^{2}

. If acceleration due to gravity (

g

), the velocity (

v

) acquired by a body after falling from rest for

5

seconds and the momentum (

p

) acquired by a mass of

1

kg in falling freely from rest for

10

seconds are taken as the fundamental units, then the value of the unit of time in terms of second?

Join BYJU'S Learning Program

If the acceleration due to gravity is 10m−2 and the units of length and times are changes in kilometer and hour respectively. The numerical value of the acceleration is

The correct option is D 129600g=10MS2=10×MS.SM=10−3kmS=13600h∴ g=10×10−313600h×13600h =36×36×102km/h2 =1296km/h2

If the acceleration due to gravity is $10 m^{- 2}$ and the units of length and times are changes in kilometer and hour respectively. The numerical value of the acceleration is

The correct option is D $129600$
$g = 10 \frac{M}{S^{2}} = 10 \times \frac{M}{S . S}$
$M = 10^{- 3} k m$
$S = \frac{1}{3600} h$
$∴$ $g = \frac{10 \times 10^{- 3}}{\frac{1}{3600} h \times \frac{1}{3600} h}$
$= 36 \times 36 \times 10^{2} k m / h^{2}$
$= 1296 k m / h^{2}$