goIIT: IIT JEE AIEEE CET WBJEE VITJEE BITSAT IIIT AICET Engineering Entrance Experts, Student Discussion Community, IIT JEE Preparation Online

$2As{H}_{3} \rightarrow 2As + 3{H}_{2}$

a 0 0 .......................P_o

a-2x 2x 3x .......................P_t

hence P_o = a = 0.9654

P_t = a +3x = 1.06

hence x = 0.0315

$k = \frac{2.303}{t} log(\frac{a}{a-x})$

substitute the values t = 5.5 hrs

a = 0.9654 , a-x = 0.9339

u get

k = 6 * 10^-3 h^-1

I am getting this answer pls nudge me if i am wrong......

see u get a rt angled triangle with slant height as the hypotenuse and radius and height as other sides....

hence

${l}^{2} = {r}^{2} + {h}^{2}$

$\Rightarrow {r}^{2} = {l}^{2} - {h}^{2}$ ..................................... 1.

now volume of cone is

$V = \frac{1}{3} \Pi {r}^{2} h$ ............................................. 2.

substitute 1. in eqn. 2.

u get

$V = \frac{1}{3} \Pi {l}^{2}h - \frac{1}{3}\Pi{h}^{3}$

now for vol. to be max.

$\frac{dV}{dh} = 0$ becoz l is a const. and only h is a variable..

u get

$h = \frac{l}{\sqrt{3}}$

substituting this value in 1st eqn for value of r

we get

$r = \frac{\sqrt{2} . l}{\sqrt{3}}$

from that rt angled triangle

let semivertical angle be $\theta$

then

$tan\theta = \frac{r}{h}\Rightarrow \theta = {tan}^{-1} (\sqrt{2})$

from the problem

eqn of tgt is

let the pt of contact be

$P({x}_{1} , {y}_{1})$

now eqn of tgt w .r.t the 2nd circle is

$x{x}_{1} + y{y}_{1} -4(x + {x}_{1}) + 3(y + {y}_{1}) + 20 = 0$

$\Rightarrow x ( {x}_{1} - 4) + y ({y}_{1} + 3) - 4{x}_{1} +3{y}_{1} + 20 = 0$

comparing both eqns

$\frac{{x}_{1} - 4}{1} = \frac{{y}_{1} + 3}{-2} = \frac{4{x}_{1} - 3{y}_{1} - 20}{5}$

you get

${x}_{1} = -3{y}_{1}$ and

$-2{x}_{1} + 5 = {y}_{1}$

solve and u will get

x₁ = 3

y₁ = -1

hence the pt of contact is (3,-1)

very funny dude...

good job....well done

let the third charge be q

for that charge to be in equilibrium, net force acting on particle should be zero

forces acting on ' q ' are due to individual charges in the problem

thus

${F}_{1} - {F}_{2} = 0$

F₁ = force due to 10^-6 charge

F₂ = force due to 2*10^-6 charge

thus

$\frac{2q}{{x}^{2}} = \frac{q}{{({10}^{-1}-x)}^{2}}$

hence

$x = \frac{2-\sqrt{2}}{10}$ m from 2*10^-6 C charge

x = 5.9 cm

let the point be P $({x}_{1}, {y}_{1})$

distance of the point from (0,3) is

${({{x}_{1}}^{2} + {{y}_{1}}^{2} - 6{y}_{1} + 9)}^{\frac{1}{2}}$

now this value should be minimum

thus

$f(y) = {({{y}_{1}}^{2} - 4{y}_{1} + 9)}^{\frac{1}{2}}$

because

${{x}_{1}}^{2} = 2{y}_{1}$

now differentiate f(y) w.r.t ' y '

u will get

${f}^{'}(y) = \frac{1}{2}{({{y}_{1}}^{2} - 4{y}_{1} + 9)}^{\frac{-1}{2}}(2{y}_{1} - 4)$

Equate f^'(y) = 0

y₁ = 2 but the other quadratic eqn has imaginary roots

hence

P is (2,2).

ques 3.) from q no 39 in hcverma vol.2 pg 122

$T = {[\frac{4{\Pi}^{2}m{R}^{3}}{kQq}]}^{\frac{1}{2}}$

substitute given values u get the answer as

$T = \frac{1}{5}\Pi$

i hope the answer is correct..

for subjective problems it is p.bahadur and my institute recommended OP.TANDON PHYSICAL CHEMISTRY which has good subject and lot of objective ques...i am presently doing R C MUKHERJEE which was easy as i did only chemical kinetics till now...why dont you ask the experts?

sinhx is a hyperbolic function

For the functions of the set

{ $x, sinx, tanx, {sin}^{-1}x, {tan}^{-1}x, sinhx, tanhx, {sinh}^{-1}x, {tanh}^{-1}x$ }

1. $\mathop{\lim}\limits_{{x} \to {0}} \frac{any function}{any function} = 1$

e.g.

1. $\mathop{\lim}\limits_{{x} \to {0}}\frac{sinx}{sinhx} =1$

--------------------------------------------------------------------------------

2. $\mathop{\lim}\limits_{{x} \to {0}} \frac{f(nx)}{g(mx)} = \frac{n}{m}$

e.g

1 . $\mathop{\lim}\limits_{{x} \to {0}} \frac{tan3x}{tan2x} = \frac{3}{2}$

---------------------------------------------------------------------------------

3. $\mathop{\lim}\limits_{{x} \to {0}} \frac{f(nx) \pm g(mx)}{h(px) \pm j(qx)} = \frac{n \pm m}{p \pm q}$

e.g.

1. $\mathop{\lim}\limits_{{x} \to {0}} \frac{sin3x + tan2x}{tanh4x - {sin}^{-1}2x} = \frac{3+2}{4-2} = \frac{5}{2}$

---------------------------------------------------------------------------------

4. $\mathop{\lim}\limits_{{x} \to {0}} \frac{{f}^{a}(mx) . {g}^{b}(nx)}{{h}^{c}(px) . {j}^{d}(qx)} = \frac{{m}^{a} . {n}^{b}}{{p}^{c} . {q}^{d}} if a+b = c+d$

e.g.

1. $\mathop{\lim}\limits_{{x} \to {0}} \frac{{sin}^{2}(2x) . {sinh}^{3}(x)}{tan3x. {tanh}^{4}(5x)} = \frac{{2}^{2} . {1}^{3}}{{3}^{1} . {5}^{4}} = \frac{8}{1875}$

---------------------------------------------------------------------------------

5. $\mathop{\lim}\limits_{{x} \to {0}} \frac{1 - cosax}{f(mx)} = 0$

e.g.

1. $\mathop{\lim}\limits_{{x} \to {0}} \frac{1 - cos3x}{tan4x} = 0$

---------------------------------------------------------------------------------

6. $\mathop{\lim}\limits_{{x} \to {0}} \frac{1 - {cos}^{p}ax}{f(mx).g(nx)} = \frac{p . {a}^{2}}{2mn}$

e.g.

1. $\mathop{\lim}\limits_{{x} \to {0}} \frac{1 - {cos}^{3}2x}{sinhx . tanh(4x)} = \frac{3 . {2}^{2}}{2.1.4} = \frac{3}{2}$

---------------------------------------------------------------------------------

7. $\mathop{\lim}\limits_{{x} \to {0}} \frac{cosax - cosbx}{f(mx) . g(nx)} = \frac{{b}^{2} - {a}^{2}}{2mn}$

e.g.

1. $\mathop{\lim}\limits_{{x} \to {0}} \frac{cosx - cos4x}{sinh(2x) . {tanh}^{-1}(3x)} = \frac{{4}^{2} - {1}^{2}}{2.2.3} = \frac{5}{4}$

---------------------------------------------------------------------------------

I hope these formulae will be useful in saving the time in competitive exams.

Although it is tough to remember some of them but few are very easy.

${F}_{AB} = \frac{2k{q}^{2}}{{d}^{2}}$

${F}_{BC} = \frac{2kq{q}_{1}}{{(d - x)}^{2}}$

${F}_{AC} = \frac{kq{q}_{1}}{{x}^{2}}$

if u take that charge q₁ is placed at a distance x from C

${F}_{AC} = {F}_{AB}$

${q}_{1} = (6 - 4\sqrt{2})q$

NOW

${F}_{BC} = {F}_{AB}$

$x = (\sqrt{2} - 1) d$

for the body to be in equilibrium

$F = mgsin\theta$

$\frac{k{q}^{2}}{{x}^{2}} = mgsin\theta$

on substituting the given values u will get

x = 0.271 m

i.e.

x = 27.1cm

i hope u got the b part

$F = -\frac{16kQqx}{{d}^{3}}$

for SHM

$a = -{\omega}^{2}.x$

hence

${\omega}^{2} = \frac{16kQq}{m{d}^{3}}$

we know that

$T = \frac{2\Pi}{\omega}$

hence u get

$T = {[\frac{m{\Pi}^{3}{\epsilon}_{o}{d}^{3}}{Qq}]}^{\frac{1}{2}}$