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（659.9、660.1）（2分）
（3）C（1分）
17.（8分）
（1）118（113~123） （2分）
（2）8×10-6 （2分）
（3）7×10-10 （2分）
（4）AC （2分）
{#{QQABDYCAogCAAoBAAAgCAQ2YCECQkBCACQgGAFAMsAIAQRFABAA=}#}
四、计算题：共 41分。
18.（10分）
（1）波沿 x轴正方向传播，t = 0.2s内传播的最小距离为 x = 5m，最小波速为
=
··························································································（3分）
= / ··················································································（2分）
（2）从图中可知波长 = ，波沿 x轴负方向传播有最大周期，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 = = ··············································································（2分）
为 45°·······················································································（1分）
记此时光点位置为 C，OO′ = h，由几何关系知 CO′ = h························（1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）
{#{QQABDYCAogCAAoBAAAgCAQ2YCECQkBCACQgGAFAMsAIAQRFABAA=}#}
光点 D到 O′的距离 xDO′＝h
xDO＝ h···················································································（1分）
v = = c················································································（1分）
t = 在玻璃砖中传播时间 1 =
······················································（1分）
从玻璃砖射出后到 D点传播时间 t 2= = ·································（1分）
t （ + ）总= t1+ t2= ·····································································（1分）
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 = （ . ） = .
{#{QQABDYCAogCAAoBAAAgCAQ2YCECQkBCACQgGAFAMsAIAQRFABAA=}#}
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 P0=
+ = ·········································································· （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
=
·································································· （2分）
m = M·························································································（1分）
（2 6）若环境温度为 T , . 大气压强为 ’ = ，不放物块活塞与缸底距离为 5
+ ’ = ··········································································（1分）
由理想气体状态方程
=
··················································································（2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）
{#{QQABDYCAogCAAoBAAAgCAQ2YCECQkBCACQgGAFAMsAIAQRFABAA=}#}4.蟾蜍在池塘边平静的水面上鸣叫，某时形成如图所示的水波。若蟾蜍的鸣叫频率不变，
2023~2024学年第二学期高二年级期末学业诊断
下列选项正确的是
A.岸边的人接收到鸣叫的声波是横波
物理试卷
B.水波从浅水区传入深水区，频率变小
C.水面上的落叶遇到水波后做受迫振动
(考试时间：下午4：10—5：40)
说明：本试卷为闭卷笔答，答题时间90分钟，满分100分。
D.水波遇到大石头比遇到小石头更容易发生衍射现象
题号
二
三
四
总分
5.关于衍射图样，下列选项正确的是
得分
一、单项选择题：本题共10小题，每小题3分，共30分。请将正确选项前字母标号填入下表内
甲
相应位置。
A.甲图是白光的单缝衍射条纹
题号
1
2
3
4
6
7
9
10
B.乙图是单色光单缝衍射产生的图样
答案
C.丙图是法国科学家泊松在实验中首次观察到的泊松亮斑
1.下列选项正确的是
D.丁图是英国物理学家沃森和克里克做实验获得的一系列DNA纤维的电子射线衍射图样
A.可利用激光亮度高的特点来进行精确测距
6.固定在振动片上的金属丝周期性触动水面可形成水波。振动片在水面上沿x轴匀速移动
B.激光与自然光相比，自然光更容易发生干涉现象
C.观看立体电影时，观众戴的眼镜是一对透振方向互相平行的偏振片
时，拍得某时刻波的图样如图所示。下列选项正确的是
D.通过转动加装在相机镜头前的偏振滤光片，可使照片中水下和玻璃后的景象清晰
A.波源向x轴负方向运动
2.下列选项正确的是
B.A位置处波速大于B位置处波速
A.液晶和晶体都有各向异性的特点
C.A位置处波速小于B位置处波速
B.水黾能停在水面上与液体的表面张力有关
崇
C.分子间的相互作用力随着分子间距离的增大，一定先减小后增大
D.A位置处接收到的波的频率小于B位置处接收到的波的频率
D.理想气体分子间距较大，分子力表现为引力，分子势能随分子间距的增大而增大
7.如图所示为冰箱工作原理示意图。制冷剂在蒸发器中汽化吸收冰箱内的热量，经过冷凝器
3.体操彩带由短杆和一定长度的彩带组成。运动员周期性地上下抖动短杆，彩带上形成简谐
时液化，放出热量到冰箱外。下列选项正确的是
箱体
干燥过滤器
波，某一时刻彩带的形状如图所示。彩带重力不计，下列选项正确的是
冷凝器
A.热量可以自发地从冰箱内传到冰箱外
A.图中彩带上的质点a正在向下运动
蒸发器
B.冰箱的工作原理违背热力学第一定律
B.彩带上的质点b向右移动，把能量传递给质点a
短杆
C.短杆与彩带的接触点在竖直方向做匀速运动
C.冰箱的工作原理违背热力学第二定律
压缩机
形带
D.由于短杆在竖直方向振动，所以该简谐波为纵波
D.冰箱的制冷系统能够不断地从冰箱内向冰箱外传递热量，但同时消耗了电能
高二物理第1页（共8页）
高二物理第2页（共8页）
部
