Motion in a Lane NEET प्रश्न हिंदी में — भौतिक विज्ञान (Physics) MCQ उत्तर सहित

(D)

$$\begin{gathered} \mathrm{Resultant} \mathrm{of} \overrightarrow{\mathrm{P}} \& \overrightarrow{\mathrm{Q}} \mathrm{is} \overrightarrow{\mathrm{R}} \\ {\mathrm{R}}^2= {\mathrm{P}}^2+{\mathrm{Q}}^2+2\mathrm{PQ} \mathrm{cos\theta } .....\left(\mathrm{i}\right) \\ \mathrm{If} \mathrm{Q} \mathrm{is} \mathrm{doubled}, \mathrm{resultant} \mathrm{is} 2\mathrm{R} \\ {\left(2\mathrm{R}\right)}^2= {\mathrm{P}}^2+{\left(2\mathrm{Q}\right)}^2+2\mathrm{P}\left(2\mathrm{Q}\right) \mathrm{cos\theta } ......\left(\mathrm{ii}\right) \\ \mathrm{From} \left(\mathrm{i}\right) \& \left(\mathrm{ii}\right) \\ 4\left({\mathrm{P}}^2+{\mathrm{Q}}^2+2\mathrm{PQ} \mathrm{cos\theta }\right)= {\mathrm{P}}^2+4{\mathrm{Q}}^2+4\mathrm{PQ} \mathrm{cos\theta } \\ 4{\mathrm{P}}^2+8\mathrm{PQ} \mathrm{cos\theta }={\mathrm{P}}^2+4\mathrm{QP} \mathrm{cos\theta } \\ 4\mathrm{Q} \mathrm{cos\theta }=-3\mathrm{P} \\ \mathrm{cos\theta }= \frac{-3\mathrm{P}}{4\mathrm{Q}} \end{gathered}$$

$$\begin{gathered} {\mathrm{v}}_{\mathrm{x}}=8\mathrm{t}-2 {\mathrm{v}}_{\mathrm{y}}=2 \\ \frac{\mathrm{dx}}{\mathrm{dt}}=8\mathrm{t}-2 \frac{\mathrm{dy}}{\mathrm{dt}}=2 \\ \int \mathrm{dx}=\int \left(8\mathrm{t}-2\right)\mathrm{dt} \int \mathrm{dy}=\int 2\mathrm{dt} \\ \mathrm{x}=4{\mathrm{t}}^2-2\mathrm{t}+\mathrm{c} \mathrm{y}=2\mathrm{t}+{\mathrm{c}}_1 \\ \mathrm{At} \mathrm{t}=2\mathrm{s}, \mathrm{x}=14 \mathrm{At} \mathrm{t}=2\mathrm{s}, \mathrm{y}=4 \\ 14=4{\left(2\right)}^2-2\left(2\right)+\mathrm{c} 4=2\left(2\right)+{\mathrm{c}}_1 \\ \mathrm{c}=2 {\mathrm{c}}_1=0 \\ \mathbf{x}\mathbf{=}\mathbf{4}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{t}\mathbf{+}\mathbf{2}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{y}\mathbf{=}\mathbf{2}\mathbf{t} \\ \mathrm{Putting} \mathrm{t}=\frac{\mathrm{y}}{2} \mathrm{in} \mathrm{x} \\ \mathrm{x}=4{\left(\frac{\mathrm{y}}{2}\right)}^2-2\left(\frac{\mathrm{y}}{2}\right)+2 \\ \mathrm{x}={\mathrm{y}}^2-\mathrm{y}+2 \end{gathered}$$

$$\begin{gathered} After time t_1: \\ Time to come back to height h after time t_1=t_2-t_1 \\ Time to reach highest point from height h=time to come back to height h from highest point=\frac{t_2-t_1}{2} \\ Time to reach highest point from ground=\frac{t_2-t_1}{2}+t_1=\frac{t_1+t_2}{2} \\ \Rightarrow \frac{t_1+t_2}{2}=\frac{u}{g} \\ \Rightarrow u=\frac{g{\left(t_1+t\right)}_2}{2} \end{gathered}$$

$$\begin{gathered} \overrightarrow{u}=u\cos \theta \hat{i}+u\sin \theta \hat{j} \\ After time t: \\ \overrightarrow{v}=u\cos \theta \hat{i}+\left(u\sin \theta -gt\right)\hat{j} \\ Change in velocity, ∆\overrightarrow{v}=\overrightarrow{v}-\overrightarrow{u}=u\cos \theta \hat{i}+\left(u\sin \theta -gt\right)\hat{j}-\left(u\cos \theta \hat{i}+u\sin \theta \hat{j}\right) \\ \Rightarrow ∆\overrightarrow{v}=\left(-gt\right)\hat{j} \\ \Rightarrow \left|∆\overrightarrow{v}\right|=\left(gt\right) \end{gathered}$$

$$\begin{gathered} Time of flight=\frac{2u_y}{g} \\ \frac{T_1}{T_2}=\frac{u_{y,1}}{u_{y,2}}=\frac{10}{10}=1 \end{gathered}$$

(c) दिया है, x=5t-2$t^2$

x-दिशा में कण का वेग,

   $v_x=\frac{dx}{dt}=\frac{d}{dt}\left(5t-2t^2\right)=5-4t$

त्वरण, $a_x=\frac{d{v}_x}{dt}=-4 m{s}^{-2}$

y-दिशा में कण का वेग,

$$\begin{gathered} v_y=\frac{dy}{dt}=10 \\ \therefore Acceleration a_y=\frac{d{v}_y}{dt}=0 \\ \therefore Net acceleration of the particle , \\ {\overrightarrow{a}}_{net}=a_x\hat{i}+a_y\hat{j}=\left(-4m{s}^2\right)\hat{i} \end{gathered}$$

अथवा   $a_{net}=-4 m{s}^{-2}$