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Standard XII

Physics

Rotational Inertia

A rod of weig...

Question

A rod of weight W is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at a distance d from each other. The centre of mass of the rod is at distance x from A. The normal reaction on A is :

W(d-x) /d

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Wx/d

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Wd/x

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W(d-x)/x

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Solution

The correct option is A
W(d-x) /d

For equilibrium

$N_{1} x = N_{2} (d - x)$ and $N_{1} + N_{2} = W$

$∴ N_{1} x = (W - N_{1}) (d - x)$

$N_{1} x + N_{1} (d - x) = W (d - x)$

$∴ N_{1} = \frac{W (d - x)}{d}$

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A rod of weight W is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at a distance d from each other. The centre of mass of the rod is at distance x from A. The normal reaction on A is :

The correct option is A W(d-x) /d For equilibrium N1x=N2(d−x) and N1+N2=W ∴N1x=(W−N1)(d−x) N1x+N1(d−x)=W(d−x) ∴N1=W(d−x)d

The correct option is A
W(d-x) /d

For equilibrium

$N_{1} x = N_{2} (d - x)$ and $N_{1} + N_{2} = W$

$∴ N_{1} x = (W - N_{1}) (d - x)$

$N_{1} x + N_{1} (d - x) = W (d - x)$

$∴ N_{1} = \frac{W (d - x)}{d}$