Đề tham khảo Tốt nghiệp THPT Vật Lí 2026-2027 (Đề 26)

$m{}_{1}c{}_{1}(t{}_{1}-t)=m{}_{2}(λ+c{}_{2}t)\Rightarrow m{}_{1}\cdot 4000\cdot (85-40)=m{}_{2}(334\cdot 10{}^{3}+4200\cdot 40)\Rightarrow m{}_{2}=\frac{90}{251}m{}_{1}$

$\frac{C{}_{1}-C{}_{2}}{C{}_{1}}=1-\frac{C{}_{2}}{C{}_{1}}=1-\frac{\frac{m{}_{cp}}{m{}_{1}+m{}_{2}}}{\frac{m{}_{cp}}{m{}_{1}}}=1-\frac{m{}_{1}}{m{}_{1}+m{}_{2}}=1-\frac{1}{1+\frac{90}{251}}\approx 0{,}264=26{,}4\%\text{.}\$Đáp án đúng B

Độ bền vững của hạt nhân càng cao khi năng lượng liên kết riêng của hạt nhân càng lớn.

Đáp án đúng D

$n=\frac{m}{M}=\frac{0{,}5\cdot 10{}^{3}}{235}=\frac{100}{47}\text{m}\text{o}\text{l}$

$N=nN{}_{A}=\frac{100}{47}\cdot 6{,}02\cdot 10{}^{23}\approx 1{,}281.10{}^{24}$

$Q=NΔE=1{,}281\cdot 10{}^{24}\cdot 200\cdot 1{,}6\cdot 10{}^{-13}\approx 4{,}1\cdot 10{}^{13}J$

$A=HQ=0{,}2.4{,}1.10{}^{13}=8{,}2.10{}^{12}J$

$t=\frac{Q}{P}=\frac{8{,}2.10{}^{12}}{160.10{}^{3}}=5125.10{}^{4}s\approx 593$ ngày. Đáp án đúng B

$\{\begin{matrix}N{}_{X}=N{}_{0}\cdot 2{}^{\frac{-t}{T}} \\ N{}_{Y}=N{}_{0Y}+N{}_{0}\cdot (1-2{}^{\frac{-t}{T}})\Rightarrow \{\begin{matrix}0{,}75=2{}^{\frac{-t{}_{1}}{T}} \\ 0{,}5N{}_{0}=N{}_{0Y}+N{}_{0}\cdot (1-2{}^{\frac{-t}{T}})\end{matrix}\end{matrix}$

$\Rightarrow 0{,}5N{}_{0}=N{}_{0Y}+N{}_{0}\cdot (1-0{,}75)\Rightarrow N{}_{0Y}=0{,}25N{}_{0}$

$\frac{N{}_{Y}}{N{}_{X}}=\frac{0{,}25N{}_{0}+N{}_{0}\cdot (1-2{}^{\frac{-t}{T}})}{N{}_{0}\cdot 2{}^{\frac{-t}{T}}}\Rightarrow 1=\frac{0{,}25+(1-2{}^{\frac{-t{}_{2}}{T}})}{2{}^{\frac{-t{}_{2}}{T}}}\Rightarrow 2{}^{\frac{-t{}_{2}}{T}}=0{,}625\overset{t{}_{2}=6{,}78}{\rightarrow }T\approx 10\ \text{n}\text{g}\text{à}\text{y}$

Đáp án đúng B

$m{}_{n}=VD=1{,}5.1=1{,}5\text{k}\text{g}$

$Q=(m{}_{n}c{}_{n}+m{}_{nh}c{}_{nh})Δt+0{,}3m{}_{n}L=(1{,}5\cdot 4190+0{,}6\cdot 880)\cdot (100-20)+0{,}3\cdot 1{,}5\cdot 2{,}26\cdot 10{}^{6}=1562040J$

$A=\frac{Q}{H}=\frac{1562040}{0{,}75}=2082720\text{J}$

$P=\frac{A}{t}=\frac{2082720}{40.60}=867{,}8\text{W}\Rightarrow$ a)Đúng; b)Sai; c)Sai; d)Đúng

$p{}_{1}=p{}_{0}+\frac{m{}_{gh}g}{S}=10{}^{5}+\frac{5\cdot 10}{50\cdot 10{}^{-4}}=1{,}1\cdot 10{}^{5}\text{P}\text{a}\Rightarrow$ a)Sai

$p{}_{2}=p{}_{0}+\frac{(m{}_{hs}+m{}_{gh})g}{S}=10{}^{5}+\frac{(55+5)\cdot 10}{50\cdot 10{}^{-4}}=2{,}2\cdot 10{}^{5}Pa\Rightarrow$ b)Đúng

Quá trình ghế hạ xuống, khí trong xilanh nhận công $\Rightarrow$ c) Sai

$pV=$ const $\Rightarrow p{}_{1}\cdot Sl=p{}_{2}\cdot S(l-8)\Rightarrow 1{,}1\cdot 10{}^{5}.l=2{,}2\cdot 10{}^{5}.(l-8)\Rightarrow l=16\text{c}\text{m}\Rightarrow$ d) Đúng

a) Đúng

$\frac{p{}_{1}V{}_{1}}{T{}_{1}}=nR\Rightarrow \frac{8{,}31\cdot 10{}^{5}\cdot V{}_{1}}{27+273}=8{,}31\Rightarrow V{}_{1}=3\cdot 10{}^{-3}\text{m}{}^{3}$

$T{}_{1}=aV{}_{1}^{2}\Rightarrow 27+273=a\cdot (3\cdot 10{}^{-3}){}^{2}\Rightarrow a=\frac{10}{3}\cdot 10{}^{7}(\text{K}/\text{m}{}^{6})\Rightarrow$b) Sai

$\frac{p{}_{2}V{}_{2}}{T{}_{2}}=nR\Rightarrow \frac{13{,}85\cdot 10{}^{5}\cdot V{}_{2}}{\frac{10}{3}\cdot 10{}^{7}\cdot V{}_{2}^{2}}=8{,}31\Rightarrow V{}_{2}=5\cdot 10{}^{-3}\text{m}{}^{3}\Rightarrow T{}_{2}=833{,}3\text{K}\Rightarrow$c)Đúng

Từ 1 đến 2 có $nR=\frac{pV}{T}=\frac{pV}{aV{}^{2}}=\frac{p}{aV}\Rightarrow p=nRaV\Rightarrow \text{p}$ theo V là hàm bậc nhất

$A{}_{12}^{′}=\frac{1}{2}(p{}_{1}+p{}_{2})(V{}_{2}-V{}_{1})=\frac{1}{2}(8{,}31\cdot 10{}^{5}+13{,}85\cdot 10{}^{5})(5-3)\cdot 10{}^{-3}=2216J\Rightarrow$d)Đúng

a) Đúng

b) Đúng. Sau 1T thì một nửa số hạt Na hiện có bị phân rã.

c) Sai. $n{}_{0}=\frac{m}{M}=\frac{0{,}248\cdot 10{}^{-3}}{25}=9{,}92\cdot 10{}^{-6}\text{m}\text{o}\text{l}$

$N{}_{0}=n{}_{0}N{}_{A}=9{,}92\cdot 10{}^{-6}\cdot 6{,}02\cdot 10{}^{23}\approx 5{,}972\cdot 10{}^{18}$

$H=λN=\frac{ln2}{T}\cdot N{}_{0}\cdot 2{}^{\frac{-t}{T}}=\frac{ln2}{62}\cdot 5{,}972\cdot 10{}^{18}\cdot 2{}^{\frac{-10.60}{62}}\approx 8{,}15\cdot 10{}^{15}\text{B}\text{q}$

d) Đúng. $\frac{H}{H{}_{0}}=2{}^{\frac{-t}{T}}=2{}^{\frac{-186}{62}}=0{,}125=12{,}5\%$

$T=5\widehat{o}=5.0{,}004=0{,}02s\Rightarrow f=\frac{1}{T}=50\text{H}\text{z}$

Đáp án: 50

$I{}_{0}=5\widehat{o}=5A\Rightarrow I=\frac{I{}_{0}}{\sqrt{2}}\approx 3{,}5A$

Đáp án: 3,5

$S=πr{}^{2}+2πrh=π\cdot 4{}^{2}+2π\cdot 4\cdot 10=96π(\text{c}\text{m}{}^{2})$

$V=πr{}^{2}h=π\cdot 4{}^{2}\cdot 10=160π(\text{c}\text{m}{}^{3})$

$m=VD=160π\cdot 10{}^{-6}\cdot 1000=0{,}16π(\text{k}\text{g})$

$-mc\cdot dt{}_{t}=\frac{kS(t{}_{t}-t{}_{p})}{d}\cdot dτ\Rightarrow -\frac{1}{t{}_{t}-t{}_{p}}\cdot dt{}_{t}=\frac{kS}{mcd}\cdot dτ\Rightarrow -{\int }_{100}^{4}\frac{1}{t{}_{t}-25}\cdot dt{}_{t}={\int }_{0}^{2}\frac{96π\cdot 10{}^{-4}}{0{,}16π\cdot 4180\cdot 5\cdot 10{}^{-3}}\cdot dτ$

Casio shift solve $\Rightarrow t{}_{t}\approx 99{,}5706{}^{^{\circ}}\text{C}$

$Q=mcΔT=0{,}16π\cdot 4180.(100-99{,}5706)\approx 902J$

Đáp án: 902

$-mc\cdot dt{}_{t}=\frac{kS(t{}_{t}-t{}_{p})}{d}\cdot dτ\Rightarrow -\frac{1}{t{}_{t}-t{}_{p}}\cdot dt{}_{t}=\frac{kS}{mcd}\cdot dτ\Rightarrow -{\int }_{100}^{t}\frac{1}{t{}_{t}-25}\cdot dt{}_{t}={\int }_{0}^{30}\frac{96π\cdot 10{}^{-4}}{0{,}16π\cdot 4180\cdot 5\cdot 10{}^{-3}}\cdot dτ$

Casio shift solve $\Rightarrow t{}_{t}\approx 93{,}8{}^{^{\circ}}\text{C}\Rightarrow Δt{}_{t}=100-93{,}8=6{,}2{}^{^{\circ}}\text{C}$

Đáp án: 6,2

$A=210$

Đáp án: 210

$n{}_{0}=\frac{m{}_{0}}{M}=\frac{100}{210}=\frac{10}{21}mol$

$N{}_{0}=n{}_{0}N{}_{A}=\frac{10}{21}\cdot 6{,}02\cdot 10{}^{23}\approx 2{,}87\cdot 10{}^{23}$

$N=N{}_{0}\cdot 2{}^{\frac{-t}{T}}=2{,}87\cdot 10{}^{23}\cdot 2{}^{\frac{-69}{138}}\approx 20\cdot 10{}^{22}$

Đáp án: 20