Class 11th

Inertia, as the First Law says, is the resistance to any change in motion. The law states that every object at rest stays at rest and one in motion stays in motion unless acted on by a force. For example, in a bus that starts or stops suddenly, your body resists the change, making you feel pushed backward or forward.

$x_1=5sin\left(2\pi t+\frac{\pi }{4}\right)$

$x_2=5\sqrt[]{2}\left(sin2\pi t+cos2\pi t\right)$

$=10sin\left(2\pi t+\frac{\pi }{4}\right)$

$x_{2,max}=2.x_{1,max}$

$\left[\lambda *3*g\right]*\frac{3}{2}-\left[\lambda *1*g\right]*\frac{1}{2}=K$         

=> $\lambda \Rightarrow \frac{mass}{length}$

=> $\lambda g\left(\frac{9}{2}-\frac{1}{2}\right)=K$

$\frac{3}{3}*10*4=K$

K = 40 J

  $p_1-p_2=\frac{4T}{r_1}.......\left(1\right)$

$p_2-p_0=\frac{4T}{r_2}.......\left(2\right)$

(1) + (2), p₁ – p₀ = 4T $\left(\frac{1}{r_1}+\frac{1}{r_2}\right)$

$\Rightarrow \frac{4T}{r_{eq}}=4T\left(\frac{1}{r_1}+\frac{1}{r_2}\right)$

$r_{eg}=\frac{r_1{r}_2}{r_1+r_2}=\frac{3*6}{3+6}=2cm$

$Let,E\propto m^a{g}^b{T}^c$

So, $\left[E\right]={\left[m\right]}^a{\left[g\right]}^b{\left[T\right]}^c$  

$\Rightarrow M{L}^2{T}^{-2}=M^a{\left(L{T}^{-2}\right)}^b{T}^c$  

Thus, E = k mg² T²

$\frac{\Delta E}{E}=2.\frac{\Delta g}{g}+2.\frac{\Delta T}{T}$   

Percentage error in E = 2 * 4 + 2 * 3 = 14%

Kindly go through the solution 

 $\frac{F_{max}}{1+2}=0.5*10\Rightarrow F_{max}=15N$

= $l_{xx\text{'}}=\left[l_{yy\text{'}}+M{\left(\frac{5r}{2}\right)}^2\right]+\left[l_{ZZ\text{'}}+M{\left(\frac{13r}{2}\right)}^2\right]$

$M\left(\frac{36r^2}{12}\right)+\frac{M*25r^2}{4}=\frac{M{r}^2}{2}+M*\frac{169M{r}^2}{4}$

= 52 Mr²

ms_x (16 – 10) = ms_y (20 – 16)

  $\Rightarrow \frac{s_x}{s_y}=\frac{2}{3}$             

$m{s}_y\left(26-20\right)=m{s}_z\left(30-26\right)$               

  $\Rightarrow \frac{s_y}{s_z}=\frac{2}{3}$             

$\frac{s_x}{s_z}=\frac{s_x}{s_y}*\frac{s_y}{s_z}=\frac{2}{3}*\frac{2}{3}=\frac{4}{9}$              

  $m{s}_x\left(T-10\right)=m{s}_z\left(30-T\right)$            

->4(T – 10) = 9(30 – T)

->T = 23.84°C

y₁ = y₂

=> $35t-\frac{1}{2}g{t}^2=35\left(t-3\right)-\frac{1}{2}g{\left(t-3\right)}^2$

=> 35 * 3 = $\frac{1}{2}g\left(t+t-3\right)*3$

105 = $\frac{10}{2}*3*\left(2t-3\right)$

t = 5

$y_1=35*5-\frac{1}{2}*10*5^2$  

= 50 m

Tension in wire, $T=\frac{2*3*5*10}{3+5}=\frac{75}{2}N$

$\frac{T}{\pi r_{min}^2}=\frac{24}{\pi }*10^2$

$\Rightarrow r_{min}^2=\frac{75}{2*24}*10^{-2}=\frac{25}{16}*10^{-2}$

$\Rightarrow r_{min}=\frac{5}{4}*10^{-1}m=12.5cm$