$解：(1)电源电压U=2×1.5\ \text {V}=3\ \text {V}$
$当滑动变阻器连入阻值最大时，电压表示数最大，此时$
$I=\frac U{R_0+R_P}=\frac {3\ \text {V}}{5Ω+10Ω}=0.2\ \text {A}$
$最大示数U_P=IR_P=0.2\ \text {A}×10Ω=2\ \text {V}$
$(2)当U_P=1\ \text {V}时，U_0=3\ \text {V}-1\ \text {V}=2\ \text {V}$
$I_{ }=\frac {U_{0 }}{R_{0 }}=\frac {{ 2 }\ \text {V}}{{ 5 }Ω}={ 0.4 }\ \text {A}$
$R_{ P }=\frac {U_{ P }}{I_{ }}=\frac {{ 1 }\ \text {V}}{{ 0.4 }\ \text {A}}={ 2.5 }Ω$
$\Delta L=\frac {2.5Ω}{10Ω}×6\ \text {cm}=1.5\ \text {cm}$
$由图可知对应压力F=15\ \text {N}，即物重G=F=15\ \text {N}$
$m_{ }=\frac {G_{ }}g=\frac {{ 15 }\ \text {N}}{{ 10 }\ \text {N/kg}}={ 1.5 }\ \text {kg}$
$(3)偏大$
