logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

A uniform rod of length $2.0 m$ specific $0.5$ and mass$2 \;kg$ is hinged at one end to bottom of tank of water (Specific gravity =1.0) filled upto a height of $1m$. as shown when $\theta \neq 0$. the force exerted by hinge on the rod is $(g=10 \;m/s^2)$

\[\begin {array} {1 1} (a)\;10.2\;N\;upward \\ (b)\;4.2\;N\; downward \\ (c)\;8.3\;N\;downward \\ (d)\;6.2\;N\;upward \end {array}\]

 

Can you answer this question?
 
 

1 Answer

0 votes
We consider the portion of rod inside water and the upthrust on this portion (F) and the weight of the rod (W) acting down wards.
Length of rod inside water $= 1.0\; \sec \theta $
$\qquad= \sec \theta$
upthrust (F)=Volume of water displaced $\times $ density of water $\times$ g
Volume of rod inside water:
$\qquad= \large\frac{(mass/unit \;length) \times length}{density}$
$\qquad= \large\frac{\Large\frac{2}{2} \times \sec \theta}{500}$
[Specific gravity=0.5]
$F= \large\frac{2}{2}$$ \sec \theta \times \large\frac{1}{500}$$ \times 1000 \times 10$
$F= 20\;sec \theta$
Weight of rod $(W)=2 \times 10=20\;N$
For rotational equilibrium net torque about 0 should be zero.
$F \large\frac{\sec \theta}{2}$$ \sin \theta$$=W(1.0 \sin \theta)$
$\large\frac{20}{2}$$ \sec^2 \theta=20$
$\theta=45^{\circ}$
$F= 20 \sec 45^{\circ}=20 \sqrt 2 \;N$
For vertical equilibrium of rod force exerted by the hinge will be
$(20 \sqrt {2} -20)=8.28\;N\; down ward$
Hence c is the correct answer.
answered Oct 8, 2013 by meena.p
edited Mar 21, 2014 by balaji.thirumalai
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...