A metal ball immersed in water weighs w1 at 0° C and w2 at 50° C . The coefficient of cubical expansion of metal is less than that of water. Then

Question

A metal ball immersed in water weighs w1 at 0° C and w2 at 50° C . The coefficient of cubical expansion of metal is less than that of water. Then (A) w1>w2 (B) w1 < w2 (C) w1=w2 (D) data is insufficient

Tardigrade Admin · Accepted Answer

As the coefficient of cubical expansion is more for alcohol then metal.
When we increase temperature, expansion in alcohol is more and hence density decreases,
The buoyant force

F = V ρ_{w} g

decreases as an increase in metal volume are less but the decrease in the density of water is more.
Hence

w_{1} < w_{2}

Q. A metal ball immersed in water weighs $w_{1}$ at $0^{\circ} C$ and $w_{2}$ at $5 0^{\circ} C$ . The coefficient of cubical expansion of metal is less than that of water. Then

$w_{1} > w_{2}$

$w_{1} < w_{2}$

$w_{1} = w_{2}$

data is insufficient

Q. A metal ball immersed in water weighs w1​ at 0∘C and w2​ at 50∘C . The coefficient of cubical expansion of metal is less than that of water. Then

w1​>w2​

w1​<w2​

w1​=w2​

data is insufficient

Q. A metal ball immersed in water weighs $w_{1}$ at $0^{\circ} C$ and $w_{2}$ at $5 0^{\circ} C$ . The coefficient of cubical expansion of metal is less than that of water. Then

$w_{1} > w_{2}$

$w_{1} < w_{2}$

$w_{1} = w_{2}$