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An L-shaped object, made of thin rods of uniform mass density, is suspended with a string as shown in the figure. If AB= BC, and the angle made by AB with downward vertical is 0, then:

  • Option 1)

         \tan \theta =\frac{1}{2\sqrt{3}}

      

  • Option 2)

    \tan \theta =\frac{1}{2}

  • Option 3)

    \tan \theta =\frac{2}{\sqrt{3}}

  • Option 4)

    \tan \theta =\frac{1}{3}

Answers (1)

best_answer

 

Torque -

\underset{\tau }{\rightarrow}= \underset{r}{\rightarrow}\times \underset{F}{\rightarrow}   

 

- wherein

This can be calculated by using either  \tau=r_{1}F\; or\; \tau=r\cdot F_{1}

r_{1} = perpendicular distance from origin to the line of force.

F_{1} = component of force perpendicular to line joining force.

mg C1X1 = mg sin \theta

mg C2X2 = (S1S2 - S2C2) mg = mg \frac{L}{2} cos  \theta - mg Lsin\theta

 

 

mg C1X1  = mg C2X2

\Rightarrow tan \theta = \frac{1}{3}

 

 


Option 1)

     \tan \theta =\frac{1}{2\sqrt{3}}

  

Option 2)

\tan \theta =\frac{1}{2}

Option 3)

\tan \theta =\frac{2}{\sqrt{3}}

Option 4)

\tan \theta =\frac{1}{3}

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