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Standard XI

Physics

Floating

Two cylinders...

Question

Two cylinders of same cross-section and length $L$ but made of two different materials of densities $d_{1}$ and $d_{2}$ are cemented together to form a cylinder of length $2 L$ . The combination floats in a liquid of density $d$ with a length $\frac{L}{2}$ above the surface of the liquid. If $d_{1} < d_{2}$ then

d_{1} < \frac{3}{4} d

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\frac{d}{2} > d_{1}

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\frac{d}{4} > d_{1}

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d < d_{1}

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Solution

The correct option is A $d_{1} < \frac{3}{4} d$
From Archimedes principle ,
Net weight = upthrust
$(d_{1} + d_{2}) L A g = \frac{3}{2} L A d g$
$\Rightarrow (d_{1} + d_{2}) = \frac{3}{2} d$
$∵ d_{1} < d_{2}$ we can say that , $d_{1} < \frac{3}{4} d$
Hence, option (a) is the correct answer.

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Floating

Standard XI Physics

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Two cylinders of same cross-section and length L but made of two different materials of densities d1 and d2 are cemented together to form a cylinder of length 2L. The combination floats in a liquid of density d with a length L2 above the surface of the liquid. If d1<d2 then

The correct option is A d1<34dFrom Archimedes principle , Net weight = upthrust (d1+d2)LAg=32LAdg ⇒(d1+d2)=32d ∵d1<d2 we can say that , d1<34d Hence, option (a) is the correct answer.

The correct option is A $d_{1} < \frac{3}{4} d$
From Archimedes principle ,
Net weight = upthrust
$(d_{1} + d_{2}) L A g = \frac{3}{2} L A d g$
$\Rightarrow (d_{1} + d_{2}) = \frac{3}{2} d$
$∵ d_{1} < d_{2}$ we can say that , $d_{1} < \frac{3}{4} d$
Hence, option (a) is the correct answer.