131. (2019). Stol ustidagi ikkita bir xil silindrsimon idishlarga suv quyilgan. Birinchi idishdagi suv sathining balandligi 50 cm , ikkinchi idishdagisi \( h_{2}\left(h_{2}>50 \mathrm{~cm}\right) \). Idishlarni ingichka nay bilan tutashtirganimizda og'irlik kuchi 100 J ish bajardi. Agar idish asosining yuzasi \( 1 \mathrm{~m}^{2} \) bo' \( 1 \mathrm{sa}, h_{2} \mathrm{ni}(\mathrm{cm}) \) aniqlang.p.pw \( =1 \mathrm{~g} / \mathrm{cm}^{3}, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2} \). \( \begin{array}{llll}\text { A) } 70 & \text { B) } 90 & \text { C) } 80 & \text { D) } 60\end{array} \)

Berilgan masalani yechib chiqamiz: **Ma'lumotlar:** - Birinchi idishdagi suv balandligi, \( h_1 = 50 \) cm. - Ikkinchi idishdagi suv balandligi, \( h_2 > 50 \) cm. - Idishlarning maydoni, \( S = 1 \) m². - Suv zichligi, \( \rho = 1 \) g/cm³ \( = 1000 \) kg/m³. - Gravitatsiya o'zgarishi, \( g = 10 \) m/s². - Qilingan ish, \( W = 100 \) J. **Yechish jarayoni:** 1. **Potensial energiyaning o'zgarishi:** Ikki idishdagi suvlar bog'langanidan so'ng, suvlar teng balandlikka ega bo'ladilar. Tenge balandlik, \( h = \frac{h_1 + h_2}{2} \). 2. **Potensial energiya formulasi:** \[ U = \frac{1}{2} \rho g S h^2 \] Dastlabki energiya: \[ U_{\text{initial}} = \frac{1}{2} \rho g S h_1^2 + \frac{1}{2} \rho g S h_2^2 \] Yakuniy energiya: \[ U_{\text{final}} = \rho g S \left( \frac{h_1 + h_2}{2} \right)^2 \] 3. **Ish va energiya o'rtasidagi munosabat:** \[ W = U_{\text{initial}} - U_{\text{final}} = 100 \text{ J} \] Hisoblashni davom ettirsak: \[ (h_2 - h_1)^2 = \frac{400}{\rho g S} \] \[ (h_2 - 50)^2 = \frac{400}{1000 \times 10 \times 1} = 0.04 \text{ m}² \] \[ h_2 - 50 = \sqrt{0.04} \times 100 \text{ cm/m} = 20 \text{ cm} \] \[ h_2 = 50 + 20 = 70 \text{ cm} \] **Javob:** **A) 70**