Class 11th

Tangents making angle $\frac{\pi }{4}$  with y = 3x + 5.

$tan\frac{\pi }{4}=\left|\frac{m-3}{1+3m}\right|\Rightarrow m=-2,\frac{1}{2}$  

So, these tangents are  . So ASB is a focal chord.

NLM at point Q

N – mg sin a =  $\frac{m{v}^2}{R}$  

$N=\frac{m{v}^2}{R}+mgsin\alpha$           - (1)

COE : between P & Q

$\frac{m{v}^2}{R}=2mgsin\alpha$               - (2)

Centripetal force = 2mg sin a

 N (Normal Reaction) = 2mg sin a + mg sin a

  = 3mg sin a

$\frac{Centripetalforce}{NormalReaction}=\frac{2mgsin\alpha }{3mgsin\alpha }=\frac{2}{3}$             

At maximum height velocity (v) = 0

So, momentum = mv = m * 0 = 0

$f\left(x\right)=x+x{\displaystyle \underset{0}{\overset{1}{\int }}f\left(t\right)dt}-{\displaystyle \underset{0}{\overset{1}{\int }}{t}_{0}f\left(t\right)dt}$

 Let $1+{\displaystyle \underset{0}{\overset{1}{\int }}f}\left(t\right)dt=\alpha$  

${\displaystyle \underset{0}{\overset{1}{\int }}tf\left(t\right)dt=\beta }$  

So, f(x) = ax - b

Now,  $\alpha ={\displaystyle \underset{0}{\overset{1}{\int }}f\left(t\right)dt+1}$  

$\alpha ={\displaystyle \underset{0}{\overset{1}{\int }}\left(at-\beta \right)dt+1}$  

$\beta ={\displaystyle \underset{0}{\overset{1}{\int }}t\cdot f\left(t\right)dt}$  

$\beta =\frac{4}{13},\alpha =\frac{18}{13}$  

f(x) = ax – b

$=\frac{18x-4}{13}$

option (D) satisfies

.B.D of person w.r.t. elevator

mg – N = ma

N = mg – ma

N decreases so, person experiences weight loss

$P=\frac{\alpha }{\beta }lo{g}_e\left(\frac{kt}{\beta x}\right)$

$\frac{kt}{\beta x}$ = Dimensionless

$\beta =\frac{kt}{x}=\frac{\left[M{L}^2{T}^{-2}{k}^{-1}\right]\left[k\right]}{\left[L\right]}$

$\frac{\alpha }{\beta }=$ dimensionless

a = dimensionless of b

a = MLT^-2

This refers to the highest limit of pressure which the body can tolerate in order to regain its original shape after removing the pressure. If the pressure exceeds the elastic limit, the body will never be able to regain its original shape agian. example: rubber stretching.

According to question, we can write

$d=d_0\left(1+\alpha \Delta T\right)\Rightarrow 6.241=6.230\left(1+1.4*10^{-5}*\Delta T\right)$

$\Rightarrow T=126.18+27=153.18°C$  

Let S = 1 $+\frac{5}{6}+\frac{12}{6^2}+\frac{22}{6^3}+....$  

$\underset{\_}{-\frac{S}{6}=\frac{1}{6}+\frac{5}{6^2}+....}$  

$\frac{5S}{6}=1+\frac{4}{6}+\frac{7}{6^2}+\frac{10}{6^3}+....$  

$\underset{\_}{-\frac{5S}{36}=\frac{1}{6}+\frac{4}{6^2}+\frac{7}{6^3}+.....}$  

$\frac{25S}{36}=1+\frac{3}{6}+\frac{3}{6^2}+\frac{3}{6^3}+......$  

$\frac{25S}{36}=1+\frac{3/6}{1-1/6}$  

$\frac{25S}{36}=1+\frac{3/5}{1}$  

$S=\frac{288}{125}$

According to question, we can write

10 = $\frac{1}{2}a{t}^2...............\left(1\right)and$

$10+x=\frac{1}{2}a{\left(2t\right)}^2................\left(2\right)$

$\Rightarrow X=\frac{1}{2}A\left[{\left(2t\right)}^2-t^2\right]=3\left(\frac{1}{2}a{t}^2\right)=30m$