Question

A metal ball immersed in alcohol weighs $W_1$ at $0^{\circ }$C and $W_2$ at ${59}^{\circ }$C. The coefficient of cubical expansion of metal is less than that of alcohol. If the density of the metal is large compared to that of alcohol, then

• A. $W_1$ > $W_2$
• B. $W_1$ = $W_2$
• C. $W_1$ < $W_2$
• D. $W_2=\frac{W_1}{2}$

Answer 1

The correct option is B $W_1$ = $W_2$

Let ${}^{\prime }{v}_0^{\prime }$ be the volume of the metal at $0^{\circ }$C and $v_t$ its volume at $t^{\circ }$C. At temperature ‘t’ the up thrust is
Where ${}^{\prime }{p}_t^{\prime }$ is the density of alcohol at temperature ‘t’. Now $v_t=v_0(1+rt)$
Where $v_0$ = volume of alcohol displaced at temperature $t=0^{\circ }$C.
Now the density of alcohol at temperature ‘t’ is
$p_t=\frac{p_0}{1+rt}$
where $p_0$ is the density of alcohol at $t=0^{\circ }$C
Hence, $U_t=v_0(1+rt)\frac{p_{0}g}{(1+rt)}=v_0{\rho }_{0}g=U_0$
Where ${}^{\prime }{U}_0^{\prime }$ is the up thrust at $0^{\circ }$C, Since the up thrust is independent of temperature, $w_1=w_2$.