Gujarati
6.System of Particles and Rotational Motion
normal

एक " $V "$-आकार के दृढ पिंड में दो समरूप एकसमान भुजाएं हैं। भुजाओं के बीच क्या कोण होगा जिससे कि इस पिंड को एक भुजा से लटकाने पर दूसरी भुजा पूर्णतया क्षैतिज हो जायेगी?

A

$\cos ^{-1}(1 / 3)$

B

$\cos ^{-1}(1 / 2)$

C

$\cos ^{-1}(1 / 4)$

D

$\cos ^{-1}(1 / 6)$

(KVPY-2016)

Solution

(a)

Let length of each of rod is $l$ and angle between them is $\theta$.

Let the lower rod is horizontal and upper rod makes $\theta$ angle with horizontal. Weights of rods acts vertically downwards from their centres $A$ and $B$ as shown in above figure.

Now, perpendicular distance of weight acting through point $A$ from point $D$ is $C D=l \cos \theta-\frac{l}{2} \cos \theta$

$C D=\frac{l}{2} \cos \theta$

and perpendicular distance of weight acting through $B$ from point $D$ is

$B D=\frac{l}{2}-l \cos \theta=\frac{l}{2}(1-2 \cos \theta)$

At equilibrium torque of these two weights about $D$ must balance each other,

i.e. $m g \times \frac{l}{2} \cos \theta=m g \times \frac{l}{2}(1-2 \cos \theta)$

$\Rightarrow \frac{3}{2} \cos \theta=\frac{1}{2}$

$\Rightarrow \cos \theta=\frac{1}{3}$

or $\theta=\cos ^{-1}\left(\frac{1}{3}\right)$

Standard 11
Physics

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