An object having 10 kg mass weighs 9.81 kg on a spring balance. The value of 'g' at this place is
A. 10 m/sec2
B. 9.81 m/sec2
C. 9.75 m/sec2
D. 9 m/sec
Answer: Option A
Solution(By Examveda Team)
Standard weight = mass × acceleration = 10 × 9.81 = 98.1 NActual weight = 9.81 × g = 98.1 N
$${\text{g}} = \frac{{98.1}}{{9.81}} = 10\,{\text{m}}/{\sec ^2}$$
This Question Belongs to Mechanical Engineering >> Hydraulics And Fluid Mechanics In ME
Join The Discussion
Comments ( 4 )
Related Questions on Hydraulics and Fluid Mechanics in ME
A flow in which the quantity of liquid flowing per second is constant, is called __________ flow.
A. Steady
B. Streamline
C. Turbulent
D. Unsteady
A hydraulic ram is a device used to
A. Store the energy of water
B. Increase the pressure of water
C. To lift water from deep wells
D. To lift small quantity of water to a greater height when a large quantity of water is available at a smaller height
A. The weight of the body
B. More than the weight of the body
C. Less than the weight of the body
D. Weight of the fluid displaced by the body
10* 9.81 = 9.81* gravity. So that g=10 m/sec2
Ans should be 0.981. The solution is wrong
0.981
Hello ...how u solved this problem.my result is 0.981...w=m*g ...