wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A fixed pulley is driven by a 100 kgf mass falling at a rate of 8.0 m in 4.0 s. It lifts a load of 75.0 kgf. Calculate the efficiency of the pulley and the height to which the load is raised in 4.0 s.

A
0.65,4.0 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
0.85,8.0 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
0.75,4.0 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
0.75,8.0 m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is C 0.75,8.0 m
As given above effort (E)=100 kgf=10 kg
Load(L)=75 kgf=7.5 kg
M.A=LE=7.510=0.75
From the definition of velocity ratio, there is only single pulley. Hence, it will be dedl/n
where, n=number of pulley
de=distance moved by effort
dl = distance moved by load
So velocity of effort =84=2
Velocity of load=x4
Hence equating it with velocity ratio we find,
1 = 2x/4
x=8 m is the height at which load is raised
So, V.R=1
Efficiency, η = M.AV.R×100=0.751×100=75%

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Dearrangement
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon