Answer is (a)
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Hello Friend !
Explanation is given below :
Consider a small length dx at adistance x on the rod from the axis of rotation
mass of liquid in dx legth = dm = M/L dx
centrifugal force is acting on dm mass
dF = dm x ω2
F = ∫df =∫dm x ω2
=M/L ω2∫x dx limit 0 to L
=MLω2/2
this force is balance by normal reaction on fluid by other end
N = MLω2/2
Hope this helps !
Explanation is given below :
Consider a small length dx at adistance x on the rod from the axis of rotation
mass of liquid in dx legth = dm = M/L dx
centrifugal force is acting on dm mass
dF = dm x ω2
F = ∫df =∫dm x ω2
=M/L ω2∫x dx limit 0 to L
=MLω2/2
this force is balance by normal reaction on fluid by other end
N = MLω2/2
Hope this helps !
AR17:
Thanks :-)
Answered by
4
suppose a small length dx at a distance x on the rod from the axis of rotation
mass of liquid in dx legth = dm = M/L dx
centrifugal force is acting on dm mass
dF = dm x w2
F = ∫df =∫dm x w2
=M/L w2∫x dx
limit 0 to L
=MLw2/2
this force is balance by normal reaction on fluid by other end
N = MLw2/2
mass of liquid in dx legth = dm = M/L dx
centrifugal force is acting on dm mass
dF = dm x w2
F = ∫df =∫dm x w2
=M/L w2∫x dx
limit 0 to L
=MLw2/2
this force is balance by normal reaction on fluid by other end
N = MLw2/2
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