Вес полого алюминиевого шарика в воздухе P0 = 0,405 Н, в воде P1 = 0,205 Н, а в керосине...

Дано:
$P_0=0,405H$
$P_1=0,205H$
$P_2=0,245H$
$g=10 \frac{H}{kg}$
$p_0=2.7 \frac{g}{cm^3}$
$p_1=1 \frac{g}{cm^3}$
$p_2=0,8 \frac{g}{cm^3}$
$-----$
$V_t-?$
$V_p-?$

Решение:
$P_1=P_0-F_A_1$
$P_2=P_0-F_A_2$
$P_0=F_t=m_tg;$ $F_A=pV_tg$
$P_1=m_tg-p_1V_tg=p_tV_tg-p_1V_tg=V_tg(p_t-p_1)$
$V_t= \frac{P_1}{g(p_t-p_1)}$
$P_2=m_tg-p_2V_tg=p_tV_tg-p_2V_tg=V_tg(p_t-p_2)$
$V_t= \frac{P_2}{g(p_t-p_2)}$
$\frac{P_1}{g(p_t-p_1)}= \frac{P_2}{g(p_t-p_2)}$
$P_1g(p_t-p_2)=P_2g(p_t-p_1)$
$P_1(p_t-p_2)=P_2(p_t-p_1)$
$P_1p_t-P_1p_2=P_2p_t-P_2p_1$
$P_1p_t-P_2p_t=P_1p_2-P_2p_1$
$p_t(P_1-P_2)=P_1p_2-P_2p_1$
$p_t= \frac{P_1p_2-P_2p_1}{P_1-P_2} = \frac{0,205H*800 \frac{kg}{m^3}-0,245H*1000 \frac{kg}{m^3} }{0,205H-0,245H}= 2025 \frac{kg}{m^3}$
$m_t= \frac{P_0}{g}$
$m_t= \frac{0,405H}{10 \frac{H}{kg} }=0,0405kg$
$V_t= \frac{m_t}{p_t}$
$V_t= \frac{0,0405kg}{2025 \frac{kg}{m^3} }= 0,00002m^3=20cm^3$
$V_p=V_t-V_a_l$
$V_a_l= \frac{m_t}{p_a_l}$
$V_a_l= \frac{0,0405kg}{2700 \frac{kg}{m^3} }=0,000015m^3=15cm^3$
$V_p=20cm^3-15cm^3=5cm^3$

Ответ: $V_t=20cm^3;$ $V_p=5cm^3$