भौतिक विज्ञान (Physics) NEET प्रश्न हिंदी में — उत्तर सहित MCQ अभ्यास

$Workdone=PEofthewater=mg\frac{h}{2}=4900J$

(b) $U=\frac{F^2}{2K}=\frac{T^2}{2K}$

(b)

 $$\begin{gathered} W=\frac{1}{2}\times F\times l=\frac{1}{2}mgl \\ =\frac{1}{2}\times 10\times 10\times 1\times {10}^{-3}=0.05 J \end{gathered}$$

(क) $Increase in energy =\frac{1}{2}\times 20\times 1\times {10}^{-3}=0.01J$

$$\begin{gathered} \left(2\right) \\ ∆\mathrm{P}=\frac{2\mathrm{T}}{\mathrm{r}} \\ \frac{∆{\mathrm{P}}_1}{∆{\mathrm{P}}_2}=\frac{{\mathrm{r}}_2}{{\mathrm{r}}_1}=\frac{2}{1} \end{gathered}$$

(a) $W=\frac{1}{2}Fl=\frac{1}{2}\times Mg\times l=\frac{1}{2}\times 5\times 10\times 3=75 J$

$$\begin{gathered} \mathrm{Force} \mathrm{on} \mathrm{element} \mathrm{of} \mathrm{width} \mathrm{dx} = \left(\frac{\mathrm{M}}{\mathrm{L}}\mathrm{dx}\right)\left(\mathrm{x}\right){\mathrm{\omega }}^2 \\ \mathrm{Total} \mathrm{force} \mathrm{F} = {\int }_0^{\mathrm{L}}\frac{\mathrm{M}}{\mathrm{L}}{\mathrm{x\omega }}^2 \mathrm{dx}= \frac{{\mathrm{ML\omega }}^2}{2} \end{gathered}$$

विकल्प A सही है।

(a)

 $$\begin{gathered} W=\frac{1}{2}Fl \therefore W\propto l \left(\text{F स्थिर है}\right) \\ \therefore \frac{W_1}{W_2}=\frac{l_1}{l_2}=\frac{l}{2l}=\frac{1}{2} \end{gathered}$$