Work Energy And Power NEET प्रश्न हिंदी में — भौतिक विज्ञान (Physics) MCQ उत्तर सहित

$ Spring constant K = { F \over x} , U = { 1 \over 2} kx^2 $

$E_1 = E and E_2 = 1.44E $ $ Than P \alpha \sqrt E $

$ m_1 v_1 + m_2 v_2 =0 and ,than K.E = {1 \over 2} m_2 v_2^2 $

$$\begin{gathered} If vector \overrightarrow{A } is \perp to \overrightarrow{B, } \\ \overrightarrow{B }.\overrightarrow{A }=0 \\ Checking all options \\ a) (\hat{i} - \hat{j} -\hat{k}). (\hat{i} + \hat{j} +\hat{k}) = 1-1-1 =-1 (Not perpendicular\left) \\ b\right) (\hat{i} - \hat{j} -\hat{k}). (\widehat{-i} + \hat{j} +\hat{k}) = -1-1-1 =-3 (Not perpendicular) \\ c) (\hat{i} - \hat{j} -\hat{k}). (\hat{i} + \hat{j} -\hat{k}) = -1-1+1 =1 (Not perpendicular) \\ \mathit{S}\mathit{o}\mathbf{ }\mathbf{,}\mathbf{ }\mathit{n}\mathit{o}\mathbf{ }\mathit{o}\mathit{p}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{ }\mathit{m}\mathit{a}\mathit{t}\mathit{c}\mathit{h}\mathit{e}\mathit{s}\mathbf{.}\mathbf{ }\mathit{A}\mathit{n}\mathit{s}\mathit{w}\mathit{e}\mathit{r}\mathbf{ }\mathit{i}\mathit{s}\mathbf{ }\mathit{o}\mathit{p}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{ }\mathit{D} \end{gathered}$$

(A)

$$\begin{gathered} \mathrm{Net} \mathrm{force}= {\overrightarrow{\mathrm{F}}}_1+{\overrightarrow{\mathrm{F}}}_2 \\ \left({\overrightarrow{\mathrm{F}}}_{\mathrm{net}}\right)= \left(2\hat{\mathrm{i}}-3\hat{\mathrm{j}}+3\hat{\mathrm{k}}\right)+\left(\hat{\mathrm{i}}+\hat{\mathrm{j}}-2\hat{\mathrm{k}}\right) \\ = 3\hat{\mathrm{i}}-2\hat{\mathrm{j}}+\hat{\mathrm{k}} \\ \mathrm{Displacement} ∆\overrightarrow{\mathrm{r}}= {\overrightarrow{\mathrm{r}}}_2-{\overrightarrow{\mathrm{r}}}_1 \\ = 6\hat{\mathrm{i}}+8\hat{\mathrm{j}}+7\hat{\mathrm{k}} \\ \mathrm{Work} \mathrm{done}= \mathrm{Force}. \mathrm{Displacement} \\ = {\overrightarrow{\mathrm{F}}}_{\mathrm{net}}.∆\overrightarrow{\mathrm{r}} \\ = \left(3\hat{\mathrm{i}}-2\hat{\mathrm{j}}+\hat{\mathrm{k}}\right).\left(6\hat{\mathrm{i}}+8\hat{\mathrm{j}}+7\hat{\mathrm{k}}\right) \\ = 18-16+7 \\ = 9 \mathrm{J} \end{gathered}$$

(D)

$$\begin{gathered} R=\sqrt{{\left(P\right)}^2+{\left(Q\right)}^2+2PQ\cos \theta } \\ When \theta =\pi , \cos \theta = -1 \\ then R is minimum \end{gathered}$$

वेग

चूँकि Fv= P= नियत

अर्थात्  $m\frac{dv}{dt}v=P$
  $\left[As F= \frac{mdv}{dt}\right]$
अथवा  $\int vdv= \int \frac{P}{m}dt$

दोनों ओर समाकलन करने पर हमें प्राप्त होता है $\frac{v^2}{2}=\frac{P}{m}t+C_1$

चूँकि प्रारंभ में पिंड विरामावस्था में है अर्थात् t= 0 पर v= 0, अतः $C_1= 0 \therefore v={\left(\frac{2Pt}{m}\right)}^{1/2}$

स्थिति

उपरोक्त व्यंजक से $v={\left(\frac{2Pt}{m}\right)}^{1/2}$

अथवा $\frac{ds}{dt}={\left(\frac{2Pt}{m}\right)}^{1/2}$  $\left[As v= \frac{ds}{dt}\right]$

अर्थात्  $\int ds= \int {\left(\frac{2Pt}{m}\right)}^{1/2}dt$

दोनों ओर समाकलन करने पर हमें प्राप्त होता है $s={\left(\frac{2P}{m}\right)}^{1/2}.\frac{2}{3}{t}^{3/2}+C_2$

अब चूँकि t= 0 पर, s= 0, अतः $C_2=0 \therefore s={\left(\frac{8P}{9m}\right)}^{1/2}{t}^{3/2}$

$F=\frac{mdv}{dt} \therefore F dt= mdv \Rightarrow v= \frac{F}{m}t$

अब $P= F\times v= F\times \frac{F}{m}t =\frac{F^{2}t}{m}$

यदि बल तथा द्रव्यमान नियत हैं तो $P \propto t$

$$\begin{gathered} \mathrm{dW}=\overrightarrow{\mathrm{F}}.\overrightarrow{\mathrm{dr}} \\ =\left(\mathrm{y}\hat{\mathrm{i}}+{\mathrm{x}}^2\hat{\mathrm{j}}\right).\left(\mathrm{dx}\hat{\mathrm{i}}+\mathrm{dy}\hat{\mathrm{j}}\right) \\ =\mathrm{ydx}+{\mathrm{x}}^2\mathrm{dy} \\ \mathrm{From} \mathrm{y}=\mathrm{x} , \mathrm{putting} \mathrm{y} \mathrm{in} \mathrm{terms} \mathrm{of} \mathrm{x} \mathrm{and} \mathrm{x} \mathrm{in} \mathrm{terms} \mathrm{of} \mathrm{y} \mathrm{in} \mathrm{above} \mathrm{relations} \\ \mathrm{dW}=\mathrm{xdx}+{\mathrm{y}}^2\mathrm{dy} \\ \mathrm{Integrating} \mathrm{both} \mathrm{sides} - \\ \mathrm{W}=\underset{0}{\overset{1}{\int }}\mathrm{xdx} + \underset{0}{\overset{1}{\int }}{\mathrm{y}}^2\mathrm{dy} \\ ={\left[\frac{{\mathrm{x}}^2}{2}\right]}_0^1 + {\left[\frac{{\mathrm{y}}^3}{3}\right]}_0^1 \\ =\frac{1}{2} + \frac{1}{3} =\frac{5}{6}\mathrm{J} \end{gathered}$$