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

$$\begin{gathered} The unit vector=\frac{\overrightarrow{A}}{\left|\overrightarrow{A}\right|} \\ =\frac{5\hat{i}-12\hat{j}}{\sqrt{{\left(5\right)}^2+{\left(12\right)}^2}} \\ =\frac{1}{13}(5\hat{i}-12\hat{j}) \end{gathered}$$

सदिश द्वारा y-अक्ष के साथ बनाया गया कोण tan^-1(y/x) द्वारा दिया जाता है। यहाँ, सदिश A = 2i + 3j में x = 2 और y = 3 है। इसलिए, इसके द्वारा y-अक्ष के साथ बनाया गया कोण tan^-1(3/2) है।

$$\begin{gathered} Initial position (x_1, y_1) = (1,-2) \\ Final position (x_2, y_2) = (4,2) \\ Displacement = Final position -Initial position \\ =( x_2 -x_1, y_2 -y_1) \\ =(3, 4) \\ = 3 \hat{i} + 4 \hat{j} \\ Direction of velocity is same as diplacement \\ So unit vector along displacement = \frac{Vector}{Magnitude} \\ =\frac{3 \hat{i} + 4 \hat{j}}{\sqrt{3^2+4^2}}=\frac{3 \hat{i} + 4 \hat{j}}{5} \\ Velocity = Magnitude of velocity \times Unit Vector along velocity \\ =10 \times \left(\frac{3 \hat{i} + 4 \hat{j}}{5}\right) \\ =6 \hat{i} + 8 \hat{j} \end{gathered}$$

(C)

$$\begin{gathered} v = \frac{dx}{dt}== a_1{t}^{1-1} + 2a_2{t}^{2-1}=a_1 + 2a_{2}t \\ a = \frac{dv}{dt} = 2a_2{t}^{1-1}=2a_2 \end{gathered}$$

$$\begin{gathered} \left(D\right) \\ For \overrightarrow{P} \perp \stackrel{ }{ \overrightarrow{Q}} , \\ Dot product \overrightarrow{P}.\overrightarrow{Q} =0 \\ (2\hat{i} +bj +2\hat{k}). (\hat{i} +j +\hat{k}) = 0 \\ 2 +b +2 = 0 \\ b =-4 \end{gathered}$$

(B)$$\begin{gathered} (4\hat{i} - 3\hat{j}).7\widehat{k } = 0 \\ So \widehat{k } is the required vector \end{gathered}$$

(A)

$$\begin{gathered} F = \frac{dp}{dt} \\ =4 N \\ So force is cons\tan t always \end{gathered}$$

$$\begin{gathered} Given, y=a+bt+c{t}^2-d{t}^4 \\ \Rightarrow v=\frac{dv}{dt}=b+2ct-4d{t}^3 \\ \Rightarrow a=\frac{dv}{dt}=2c-12d{t}^2 \\ at t=0, v=b and a=2c \end{gathered}$$

$F = \frac{dP}{dt}$

x को t के सापेक्ष अवकलित करने पर ,

$v = \frac{dx}{dt}$

$v = 2at - 3b{t}^2$

$a = 2a - 6bt$

पर $$\begin{gathered} t = \frac{a}{3b} \\ a= 0 \end{gathered}$$