A uniform rod weighs $10 k g$ and with a load $5 k g$ attached to one end, it balances on a knife edge at $2 m$ from that end. The length of rod is :

10.34 m

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8.56 m

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6.21 m

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4.35 m

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Solution

The correct option is D $4.35 m$
From the figure.

Net moment on P is zero.

(Take clockwise direction positive.)

$10 (L - x) \frac{(L - x)}{2} - 5 x - 10 x \frac{x}{2} = 0$

$5 {(L - 2)}^{2} = 30$ (Putting, $x = 2$ )

$\Rightarrow L = 2 + \sqrt{6} = 4.449 = 4.45 m$

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A uniform rod weighs 10kg and with a load 5kg attached to one end, it balances on a knife edge at 2m from that end. The length of rod is :

The correct option is D 4.35 mFrom the figure. Net moment on P is zero. (Take clockwise direction positive.) 10(L−x)(L−x)2−5x−10xx2=0 5(L−2)2=30 (Putting,x=2) ⇒L=2+√6=4.449=4.45m

A uniform rod weighs $10 k g$ and with a load $5 k g$ attached to one end, it balances on a knife edge at $2 m$ from that end. The length of rod is :

The correct option is D $4.35 m$
From the figure.

Net moment on P is zero.

(Take clockwise direction positive.)

$10 (L - x) \frac{(L - x)}{2} - 5 x - 10 x \frac{x}{2} = 0$

$5 {(L - 2)}^{2} = 30$ (Putting, $x = 2$ )

$\Rightarrow L = 2 + \sqrt{6} = 4.449 = 4.45 m$