The vector $V_{b}$ representing the velocity of the motorboat and the vector $v_{c}$ representing the water current are oriented as per problem.
Using the parallelogram method of addition, the magnitude of resultant R can be obtained using the law of cosine:
$R = \sqrt{v_{b}^{2} + v_{c}^{2} + 2v_{b}v_{c}\cos 120^{\circ}}$
$R = \sqrt{25^{2} + 10^{2} + 2 \times 25 \times 10(-1/2)} \approx 22 \text{ km/h}$
$|\Delta\vec{a}| = 2a \sin\left(\frac{d\theta}{2}\right) = a d\theta$ (since $d\theta$ is small, $\sin \frac{d\theta}{2} \approx \frac{d\theta}{2}$)
$\Delta|\vec{a}| = a - a = 0$
Among the given units, kilogram is the fundamental or base unit, whereas others (joule, watt, newton) are derived units.
Angular frequency = $2\pi \times \text{Frequency} = [T^{-1}] = [M^{0}L^{0}T^{-1}]$
Comparing with $[M^{a}L^{b}T^{c}]$, we get $a=0, b=0, c=-1$.
Let regular interval = $t$. When the 4th drop begins to fall, the 1st drop has travelled for time $3t$.
For first drop: $18 = \frac{1}{2}g(3t)^{2} = \frac{1}{2} \times 10 \times 9t^{2} \Rightarrow t = \sqrt{\frac{4}{10}}$ s.
For 2nd drop ($x_2$): $x_{2} = \frac{1}{2}g(2t)^{2} = \frac{1}{2} \times 10 \times 4 \times \frac{4}{10} = 8 \text{ m}$.
For 3rd drop ($x_3$): $x_{3} = \frac{1}{2}g(t)^{2} = \frac{1}{2} \times 10 \times \frac{4}{10} = 2 \text{ m}$.
Average velocity is displacement over time, whereas average speed is total distance over time. Since |displacement| $\le$ distance, |Average velocity| $\le$ average speed. Hence, stating they are strictly equal is not correct.
Magnitude of cross product of two vectors is $|\vec{A} \times \vec{B}| = AB \sin\theta$. Since the maximum value of $\sin\theta$ is 1, the cross product can be equal to AB or less than AB, but never greater than AB.
For 18 equal magnitude co-planar vectors forming a closed polygon to maintain equilibrium, the angle between two adjacent vectors is $\frac{360^{\circ}}{18} = 20^{\circ}$.
Given $\vec{a}+\vec{b}+\vec{c}+\vec{d}=0$. The statements (b), (c) and (d) are conceptually correct following vector addition polygon properties.
Density $\rho = \frac{m}{l\pi r^{2}}$
Max percentage error: $\frac{\Delta\rho}{\rho} \times 100 = \left(\frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l}\right) \times 100$
$= \left(\frac{0.003}{0.3} + 2 \times \frac{0.005}{1} + \frac{0.06}{12}\right) \times 100 = (0.01 + 0.01 + 0.005) \times 100 = 2.5\%$
$\frac{M^{6}G^{2}}{EJ^{2}} = \frac{M^{6} \cdot [M^{-1}L^{3}T^{-2}]^{2}}{[ML^{2}T^{-2}] \cdot [ML^{2}T^{-1}]^{2}} = \frac{M^{6} \cdot M^{-2}L^{6}T^{-4}}{ML^{2}T^{-2} \cdot M^{2}L^{4}T^{-2}} = \frac{M^{4}L^{6}T^{-4}}{M^{3}L^{6}T^{-4}} = [M] = \text{mass}$
Resonant frequency of LC circuit is given by $\nu = \frac{1}{2\pi\sqrt{LC}}$ or $\sqrt{LC} = \frac{T}{2\pi}$.
Therefore, $\sqrt{LC}$ has the dimension of time $[T]$.
Let $M = k f^{x}g^{y}p^{z}$. Equating dimensions: $[M] = [MLT^{-2}]^{x}[LT^{-2}]^{y}[ML^{-1}T^{-2}]^{z}$.
For simplicity: $f/g = \text{mass}$. Therefore dimensional formula is $[f^{1}g^{-1}p^{0}]$.
Time constant of LR circuit is $t = \frac{L}{R}$.
Frequency $= \frac{1}{\text{time}} = \frac{R}{L}$.
For a body crossing the same height at $t_1$ and $t_2$, time of flight $T = t_1 + t_2$.
$T = \frac{2u}{g} \Rightarrow u = \frac{g(t_1 + t_2)}{2} = \frac{10 \times (4+6)}{2} = 50 \text{ m/s}$.
Vertical motion: $h = \frac{1}{2}gt^{2} = \frac{1}{2} \times 9.8 \times (3)^{2} = 4.9 \times 9 = 44.1 \text{ m}$. (Closest is 44.2m if $g=9.81$)
Measurement of a physical quantity is essentially a process of comparing an unknown magnitude with a standard known quantity using an instrument.
$[F] = [MLT^{-2}]$
$1000 \text{ N} = [x \text{ kg}]^{1} [1000 \text{ m}]^{1} [100 \text{ s}]^{-2}$
$1000 = x \times 1000 \times \frac{1}{10000} \Rightarrow x = 10,000 \text{ kg}$.
Displacement $y = a\sin\omega t$.
Velocity $v_{y} = \frac{dy}{dt} = a\omega\cos\omega t$.
Average velocity $\vec{v}_{av} = \frac{\vec{r}_{2} - \vec{r}_{1}}{t_{2} - t_{1}}$
$\vec{v}_{av} = \frac{(13\hat{i} + 14\hat{j}) - (2\hat{i} + 3\hat{j})}{5 - 0} = \frac{11\hat{i} + 11\hat{j}}{5} = \frac{11}{5}(\hat{i} + \hat{j})$.
The component of acceleration in the direction of velocity changes the speed. This is given by the projection of $\vec{a}$ onto $\vec{v}$, which is $(\vec{a} \cdot \hat{v})\hat{v}$.
Speed at max height $= u\cos\theta$.
Speed at half max height: $v^2 = (u\cos\theta)^2 + (u\sin\theta)^2/2$
Given $u\cos\theta = \sqrt{\frac{2}{5}} \sqrt{u^2\cos^2\theta + \frac{u^2\sin^2\theta}{2}}$. Solving gives $\theta = 60^{\circ}$.
They cannot meet each other because the necessary relative velocity conditions to intercept are not satisfied. They will be closest at time $t = \frac{La}{a^{2}+b^{2}}$.
At the highest point, the instantaneous velocity is purely horizontal, and acceleration (due to gravity) acts vertically downward. Therefore, the angle between them is $90^{\circ}$, not $180^{\circ}$. Assertion is incorrect, Reason is correct.
$\omega_{\text{hour hand}} = \frac{2\pi}{12} \text{ rad/hr}$
$\omega_{\text{earth}} = \frac{2\pi}{24} \text{ rad/hr}$
Ratio $\frac{\omega_{\text{hour}}}{\omega_{\text{earth}}} = \frac{2\pi/12}{2\pi/24} = \frac{24}{12} = \frac{2}{1}$.
Average acceleration $\bar{a} = \frac{\Delta V_{1} + \Delta V_{2}}{t_{1} + t_{2}} = \frac{a_{1}t_{1} + a_{2}t_{2}}{t_{1} + t_{2}}$.
$t = \sqrt{\frac{x-a}{b}} \Rightarrow t^{2} = \frac{x-a}{b} \Rightarrow x = a + bt^{2}$
Velocity $v = \frac{dx}{dt} = 2bt$
Acceleration $a = \frac{dv}{dt} = 2b$, which is a constant, hence uniformly accelerated motion.
$\frac{1}{2}gt^{2} - \frac{1}{2}g(t-T)^{2} = L$
$\frac{1}{2}g(t^{2} - t^{2} - T^{2} + 2tT) = L \Rightarrow (2t-T)T = \frac{2L}{g}$
$2t = \frac{2L}{gT} + T \Rightarrow t = \frac{L}{gT} + \frac{T}{2}$.
Average acceleration $\bar{a} = \frac{v_{2}-v_{1}}{t_{2}-t_{1}}$
$v_{1} = \tan 45^{\circ} = 1$
$v_{2} = \tan 135^{\circ} = -1$
$\bar{a} = \frac{-1 - 1}{2 - 1} = -2 \text{ m/s}^{2}$.
Acceleration of P normal to the plane is $g\cos 60^{\circ} = \frac{g}{2}$.
P will hit the plane after time T: $0 = VT - \frac{1}{2}(\frac{g}{2})T^{2} \Rightarrow T = \frac{4V}{g}$.
Given $T = 4 \text{ s} \Rightarrow 4 = \frac{4V}{10} \Rightarrow V = 10 \text{ m/s}$.
From triangle law of vector addition, $\vec{c} - \vec{a} = \vec{b} - \vec{c} \Rightarrow 2\vec{c} = \vec{a} + \vec{b}$.
Resultant of equal forces $R = 2x\cos(\frac{\theta}{2})$.
Given $R^{2} = 3x^{2} \Rightarrow (2x\cos\frac{\theta}{2})^{2} = 3x^{2}$
$4\cos^{2}\frac{\theta}{2} = 3 \Rightarrow \cos\frac{\theta}{2} = \frac{\sqrt{3}}{2} \Rightarrow \frac{\theta}{2} = 30^{\circ} \Rightarrow \theta = 60^{\circ}$.
$\vec{A}+\vec{B}+\vec{C}=0 \Rightarrow \vec{C}=-(\vec{A}+\vec{B})$. Since A and B are equal and separated by $120^{\circ}$, their resultant is equal to A and bisects the angle ($60^{\circ}$ from A). Since $\vec{C}$ is opposite to the resultant, the angle between $\vec{A}$ and $\vec{C}$ is $60^{\circ}$.
$\frac{C^{2}}{A^{2}} > 1 + \frac{B^{2}}{A^{2}} \Rightarrow C^{2} > A^{2} + B^{2}$
$C^2 = A^2 + B^2 + 2AB\cos\theta > A^2 + B^2$
$\Rightarrow 2AB\cos\theta > 0 \Rightarrow \cos\theta > 0 \Rightarrow \theta < 90^{\circ}$.
"Velocity of light in vacuum = velocity of light in medium" is numerically incorrect (light is slower in a medium) but dimensionally correct (both are speeds). Thus, the student's statement is numerically incorrect but dimensionally correct. The question options might be interpreted uniquely based on context, but Option (4) represents "Both (1) and (3)".
$V^{2} = 108 - 9x^{2}$
Differentiating w.r.t x: $2V\frac{dV}{dx} = -18x$
Since $a = V\frac{dV}{dx}$, $2a = -18x \Rightarrow a = -9x \text{ m/s}^{2}$.
Vertical fall $h = 10 \text{ cm} = 0.1 \text{ m}$. Time taken $t = \sqrt{\frac{2h}{g}}$.
Horizontal velocity $v = \frac{R}{t} = R\sqrt{\frac{g}{2h}} = 100\sqrt{\frac{10}{2 \times 0.1}} = 100 \times \sqrt{50} \approx 700 \text{ m/s}$.
By symmetry of upward and downward motion under gravity, distance covered in $n^{\text{th}}$ second during ascent equals distance in $(2T - n + 1)^{\text{th}}$ second during descent, where T is ascent time. $T = 5$s. 1st sec $\leftrightarrow$ 10th sec, 2nd sec $\leftrightarrow$ 9th sec, etc. Option (1) is correct.
By Galilean transformation, $x' = x - vt$. Differentiating once: $v' = v_x - v$. Differentiating twice: $\frac{d^{2}x'}{dt'^{2}} = \frac{d^{2}x}{dt^{2}}$, meaning acceleration is invariant in inertial frames.
$\vec{u} = v_{0}\hat{j}$. $\vec{v}(t) = at\hat{i} + (v_{0} - gt)\hat{j}$.
For $\vec{v}$ to rotate by $90^{\circ}$, it must become horizontal (i.e., y-component becomes 0).
$v_{0} - gt = 0 \Rightarrow t = \frac{v_{0}}{g}$.
Horizontal distance $x = \frac{1}{2}at^{2} = \frac{1}{2}a\left(\frac{v_{0}}{g}\right)^{2} = \frac{av_{0}^{2}}{2g^{2}}$.
Let the first stone fall for time $t$. $h = \frac{1}{2}gt^{2}$.
The second stone falls for time $t-1$. $h - 20 = \frac{1}{2}g(t-1)^{2}$.
Subtracting the two: $20 = \frac{1}{2}g [t^{2} - (t-1)^{2}] = 5(2t-1)$.
$2t - 1 = 4 \Rightarrow 2t = 5 \Rightarrow t = 2.5 \text{ s}$.
$h = \frac{1}{2} \times 10 \times (2.5)^{2} = 5 \times 6.25 = 31.25 \text{ m}$.
$a = \frac{dv}{dt} = \alpha - \beta v$
$\int_{0}^{v} \frac{dv}{\alpha - \beta v} = \int_{0}^{t} dt$
$\frac{\ln(\alpha - \beta v)}{-\beta}\Big|_{0}^{v} = t \Rightarrow \ln\left(\frac{\alpha - \beta v}{\alpha}\right) = -\beta t \Rightarrow \alpha - \beta v = \alpha e^{-\beta t}$
$v = \frac{\alpha}{\beta}(1 - e^{-\beta t})$.
From the given acceleration-time graph:
For $0 < t \le 1$: $a=2$, so $v = 2t$, $S_1 = \frac{1}{2}(2)(1^2) = 1 \text{ m}$. Velocity at $t=1$ is $2 \text{ m/s}$.
For $1 < t \le 2$: $a=0$, constant velocity $v=2 \text{ m/s}$. $S_2 = 2 \times 1 = 2 \text{ m}$.
For $2 < t \le 3$: $a=-2$, retarding to 0. $S_3 = ut + \frac{1}{2}at^2 = 2(1) - \frac{1}{2}(2)(1^2) = 1 \text{ m}$.
Total distance $= 1 + 2 + 1 = 4 \text{ m}$.
Velocity after 30s: $v(30) = 0 + 10 \times 30 = 300 \text{ m/s}$.
After fuel finishes, it undergoes free fall under gravity. Time to reach max height from this point $= \frac{v}{g} = \frac{300}{10} = 30 \text{ s}$.
Total time from firing $= 30 + 30 = 60 \text{ s}$.
Initial velocity of stone $u = 12 \text{ m/s}$ (downwards).
$S = ut + \frac{1}{2}gt^{2} = 12(10) + \frac{1}{2}(9.8)(10)^{2} = 120 + 490 = 610 \text{ m}$.