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

A uniform rod AB,12 m long and weighing 24 kg, is supported at end B by a flexible light string and a lead mass (of very small size) of 12 kg is attached at end A. The rod floats in water with half of its length submerged. For the situation described, choose the correct statement(s). [Take g=10 m/s2, density of water =1000 kg/m3]


A
The tension in the string is 40 N.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
The tension in the string is 120 N.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
The volume of the rod is 6.4×102 m3.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
The point of action of the buoyant force is C (centre of mass of rod).
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C The volume of the rod is 6.4×102 m3.
The length CP=OC=l2
Bouyant force or upthrust (U) will act at the mid point(M) of the length of the rod submerged in water.
i.e option (d) is wrong.
OM=MC=l4


From the FBD of rod, applying equilibrium condition in vertical direction,
T+U=W1+W2=360 N ...(i)
Upthrust will be given by
U=Vρg, where V= volume of immersed rod, i.e half of total volume.
U=V2ρwg=(V/2)(103)(10)
U=0.5×104V ...(ii)
For rotational equilibrium, balancing the moments of the forces on the rod about point M:
Net anticlockwise moment=Net clockwise moment
120(l4cosθ)+T(3l4cosθ)=240(l4cosθ)
T=(6030)×43
T=40 N
i.e Option (a) is correct and (b) is wrong.
Now from eqs. (i) and (ii) we get,
U=360T=36040=320 N
320=0.5V×104
V=6.4×102 m3
where V is the volume of rod.
Option (c) is also correct.

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon