Class 11th

Slope of $(x-t)$  graph, gives velocity

$\frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}=\frac{\mathrm{t}\mathrm{a}\mathrm{n}?{\theta }_1}{\mathrm{t}\mathrm{a}\mathrm{n}?{\theta }_2}=\frac{\mathrm{t}\mathrm{a}\mathrm{n}?{30}^{?}}{\mathrm{t}\mathrm{a}\mathrm{n}?{45}^{?}}=\frac{1}{\sqrt[]{3}}$

Yes. Angular Velocity is a vector quantity i.e. it has directions. The direction of rotation can influence the sign of angular velocity's value. If the movement of the object is anticlockwise, the sign will be positive. Similarly, if the movement of the object is clockwise, the sign will be negative.

Even if the torque is same, the rotation of a body can be slow or fast depending on the distribution of mass along the surface area. If the mass of a body is more on the axis area, it's speed of rotation will be faster as compared to the body whose mass is accumulated away from the axis due to which the body will resist the angular acceleration.

In simple words, a push or pull applied on a body is known as a force. Force is a concept used in terms of linear motion. On the other hand, the force which causes a body to spin around an axis instead of moving from one point to aother is called torque. Torque is a theory which comes into account while studying rotational motion and is considered as the rotational analogue of force.

Initially speed is zero, then increases & after some time it becomes constant.

$\text{?}\mathrm{}\stackrel{?}{\mathrm{E}}\cdot \mathrm{d}\stackrel{?}{\mathrm{s}}=0\Rightarrow {?}_{\text{net}}={?}_{\text{in}}-{?}_{\text{out}}=0$

$\Rightarrow {?}_{\text{in}}={?}_{\text{out}}$

$\stackrel{?}{\tau }=\stackrel{?}{\mathrm{P}}*\stackrel{?}{\mathrm{E}}$

$\left|\stackrel{?}{\tau }\right|=\mathrm{P}\mathrm{E}\mathrm{s}\mathrm{i}\mathrm{n}?\theta \mathrm{}\Rightarrow 4=\mathrm{q}*2\mathrm{a}*\mathrm{E}\mathrm{s}\mathrm{i}\mathrm{n}?{30}^{?}$

$\mathrm{q}=\frac{4}{\left(2*{10}^{-2}\right)*2*{10}^5*\left(\frac{1}{2}\right)}$

$=2*{10}^{-3}\mathrm{C}$

$=2\mathrm{m}\mathrm{C}$

$\mathrm{T}\mathrm{i}=-{50}^{?}\mathrm{C}$

$\begin{array}{rr}& =223\mathrm{K}\\ & {\mathrm{V}}_{\mathrm{r}\mathrm{m}\mathrm{s}}\propto \sqrt[]{\mathrm{T}}\end{array}$

As ${\mathrm{v}}_{\mathrm{r}\mathrm{m}\mathrm{s}}$  increased by 3 times

So ${\left({\mathrm{v}}_{\mathrm{r}\mathrm{m}\mathrm{s}}\right)}_{\mathrm{f}}=4{\left({\mathrm{v}}_{\mathrm{r}\mathrm{m}\mathrm{s}}\right)}_{\text{initial}}$

${\mathrm{T}}_{\mathrm{f}}=16{\mathrm{T}}_{\mathrm{i}}$

$=16*223$

$=3568\mathrm{K}$ ${\mathrm{T}}_{\mathrm{f}}=(3568-273{)}^{?}\mathrm{C}$

$={3295}^{?}\mathrm{C}$

15. (c) Fundamental harmonic frequency open pipe

$=\frac{\mathrm{V}}{2\mathrm{L}}={\mathrm{v}}_1$

(say)

Fundamental harmonic frequency of closed pipe $=\frac{\mathrm{V}}{4\mathrm{L}}$

$={\mathrm{v}}_2$ (say)

$\Rightarrow \frac{{\mathrm{v}}_1}{{\mathrm{v}}_2}=\frac{\frac{\mathrm{V}}{2\mathrm{L}}}{\frac{\mathrm{V}}{4\mathrm{L}}}\mathrm{}2:1$

$\eta =1-\frac{T_{\text{sin}k}}{T_{\text{source}}}=0.5$

$\begin{array}{rr}& \frac{{\mathrm{T}}_{\text{sin}}}{{\mathrm{T}}_{\text{source}}}=0.5\\ & \Rightarrow {\mathrm{T}}_{\text{sin}}=\frac{1}{2}*(273+327)\\ & =\frac{1}{2}*600\\ & =300\mathrm{K}\\ & ={27}^{?}\mathrm{C}\end{array}$