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2023～2024 学年第二学期高二年级期末学业诊断
物理参考答案及评分建议
一、单项选择题：本题包含10小题，每小题3分，共30分。
题号 1 2 3 4 5 6 7 8 9 10
选项 D B A C B A D C C A
二、多项选择题：本题包含5小题，每小题3分，共15分。
题号 11 12 13 14 15
选项 AD AC CD AC BD
三、实验题：共14分。
16.（6分）
（1） AC （2分）
（2）10.330（10.329、10.331）（1分） 660（2分）
（3）C（1分）
17.（8分）
（1）118（110~125） （2分）
（2）8×10-6 （2分）
（3）7×10-10 （2分）
（4）AC （2分）四、计算题：共41分。
18.（10分）
（1）从图中可知该简谐横波波长 ··········································（1分）
=
若波沿x轴正方向传播，0.2s内传播的最小距离为5m
最小波速为 ········································································（2分）
=
················································································（2分）
= /
（2）若波沿x轴负方向传播，0.2s经过 个波长，对应 个周期········（1分）
s···················································································（2分）
= .
T= s······················································································（2分）
19.（10分）
(1)如图，单色光转过θ=300
n= = ··············································································（1分）
为45°·······················································································（1分）
记此时光点位置为C，OO′=h，由几何关系知CO′=h························（2分）
(2)由临界条件可知，当玻璃砖转过 时，折射光线沿着玻璃砖边沿射向光屏，
用时最长光点为D点·····································································（1分）光点D到O′的距离x ＝h
DO′
x ＝ h···················································································（1分）
DO
v= = c················································································（1分）
在玻璃砖中传播时间t = = ······················································（1分）
1
从玻璃砖射出后到D点传播时间 t = = ·································（1分）
2
（ ）
t =t +t = ·····································································（1分）
总 1 2
+
20.（10分）
(1)初始状态时，左管内气体压强
） ··································································（1分）
=( −
设左管横截面积为S，右管横截面积为4S，管内气柱长度 ，体积 ，
= =
当左右两管内水银面相等时，管内气体压强 。设左管中水银面下降 ，
=
右管中水银面上升
············································································（1分）
+ =
·····················································································（1分）
=
末状态，管内气柱长度 ，体积
= =
由理想气体状态方程
···················································································（1分）
=
···············································································（1分）
= .
(2)末状态时左右液面相平，管内气体压强 ·····················（1分）
=
体积
=
根据波意耳定律 ·························································（1分）
=
················································································（1分）
= .
左边液面上升高度 （ ）
= − . = . 右边液面下降高度
= = .
右管需要加入的水银长度
= + +
····················································································（2分）
=
21.（11分）
（1）初始时设汽缸内气体压强为 ，大气压强P =
0
·········································································· （1分）
+ =
放待测物块m后汽缸内气体压强为
································································· （1分）
+ + =
由玻意耳定律得
·································································· （1分）
= ⋅ −
m=M·························································································（1分）
（2）若环境温度为T, 大气压强为
.
’=
··········································································（1分）
+ ’ =
由理想气体状态方程
··················································································（1分）
=
························································································（1分）
=
放上物块，活塞再次稳定时活塞到缸底的距离为 ，压强为
·································································（1分）
+ + ’ =
根据理想气体状态方程
··················································································（1分）
=
活塞再次稳定时下降的高度为
················································································（1分）
= −
····················································································（1分）
=
