$解：(1)電源電壓U=2×1.5\ \text {V}=3\ \text {V}$
$當(dāng)滑動(dòng)變阻器連入阻值最大時(shí)，電壓表示數(shù)最大，此時(shí)$
$I=\frac U{R_0+R_P}=\frac {3\ \text {V}}{5Ω+10Ω}=0.2\ \text {A}$
$最大示數(shù)U_P=IR_P=0.2\ \text {A}×10Ω=2\ \text {V}$
$(2)當(dāng)U_P=1\ \text {V}時(shí)，U_0=3\ \text {V}-1\ \text {V}=2\ \text {V}$
$I_{ }=\frac {U_{0 }}{R_{0 }}=\frac {{ 2 }\ \text {V}}{{ 5 }Ω}={ 0.4 }\ \text {A}$
$R_{ P }=\frac {U_{ P }}{I_{ }}=\frac {{ 1 }\ \text {V}}{{ 0.4 }\ \text {A}}={ 2.5 }Ω$
$\Delta L=\frac {2.5Ω}{10Ω}×6\ \text {cm}=1.5\ \text {cm}$
$由圖可知對(duì)應(yīng)壓力F=15\ \text {N}，即物重G=F=15\ \text {N}$
$m_{ }=\frac {G_{ }}g=\frac {{ 15 }\ \text {N}}{{ 10 }\ \text {N/kg}}={ 1.5 }\ \text {kg}$
$(3)偏大$
