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Class 12th Application of Derivatives

x and y are the sides of two squares such that y = x – $x^{2}$ . Find the rate of change of the area of the second square with respect to the area of the first square.

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t a r e a o f t h e f i r s t s q u a r e A_{1} = x^{2} \\ a n d a r e a o f t h e s e c o n d s q u a r e A_{2} = y^{2} \\ N o w A_{1} = x^{2} a n d A_{2} = y^{2} = {(x - x^{2})}^{2} \\ D i f f e r e n t i a t i n g b o t h A_{1} a n d A_{2} w . r . t . t, w e g e t \\ \frac{d A_{1}}{d t} = 2 x . \frac{d x}{d t} a n d \frac{d A_{2}}{d t} = 2 (x - x^{2}) . (1 - 2 x) . \frac{d x}{d t} \\ ∴ \frac{d A_{2}}{d A_{1}} = \frac{\frac{d A_{2}}{d t}}{\frac{d A_{1}}{d t}} = \frac{2 (x - x^{2}) . (1 - 2 x) . \frac{d x}{d t}}{2 x . \frac{d x}{d t}} \\ = \frac{x (1 - x) (1 - 2 x)}{x} = (1 - x) (1 - 2 x) \\ = 1 - 2 x - x + 2 x^{2} = 2 x^{2} - 3 x + 1 \\ H e n c e, t h e r a t e o f c h a n g e o f a r e a o f t h e s e c o n d s q u a r e w i t h r e s p e c t t o f i r s t i s 2 x^{2} - 3 x + 1 . \end{array}$

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New answer posted

a year ago

Class 12th Application of Derivatives

The volume of a cube increases at a constant rate. Prove that the increase in its surface area varies inversely as the length of the side.

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Question as classified in NCERT Exemplar

Sol.

$\begin{array}{l} L e t x b e t h e l e n g t h o f t h e c u b e \\ ∴ V o l u m e o f t h e c u b e V = x^{3} \dots (i) \\ G i v e n t h a t \frac{d V}{d t} = K \\ D i f f e r e n t i a t i n g e q (i) w . r . t . t, w e g e t \\ \frac{d V}{d t} = 3 x^{2} . \frac{d x}{d t} = K (constant) \\ ∴ \frac{d x}{d t} = \frac{K}{3 x^{2}} \\ N o w s u r f a c e ? ? a r e a o f t h e c u b e, S = 6 x^{2} \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . t, w e g e t \\ \frac{d S}{d t} = 6.2 . x . \frac{d x}{d t} = 12 x . \frac{K}{3 x^{2}} \\ \Rightarrow \frac{d S}{d t} = \frac{4 K}{x} \Rightarrow \frac{d S}{d t} \propto \frac{1}{x} (4 K = constant) \\ H e n c e, s u r f a c e ? ? a r e a o f t h e c u b e v a r i e s i n v e r s e l y a s t h e l e n g t h o f t h e s i d e . \end{array}$

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New answer posted

a year ago

Class 12th Application of Derivatives

A swimming pool is to be drained for cleaning. If L represents the number of liters of water in the pool t seconds after the pool has been plugged off to drain and L = 200(10 – t)², how fast is the water running out at the end of 5 seconds? What is the average rate at which the water flows out during the first 5 seconds?

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Question as classified in NCERT Exemplar

Sol.

$\begin{array}{l} G i v e n t h a t L = 200 {(10 - t)}^{2} \\ w h e r e L r e p r e s e n t s t h e n u m b e r o f l i t r e s o f w a t e r i n t h e p o o l . \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t ., t, w e g e t \\ \frac{d L}{d t} = 200 * 2 (10 - t) (- 1) = - 400 (10 - t) \\ B u t t h e r a t e a t w h i c h t h e w a t e r i s r u n n i n g o u t \\ = - \frac{d L}{d t} = 400 (10 - t) \dots (i) \\ R a t e a t w h i c h t h e w a t e r i s r u n n i n g a f t e r 5 seconds \\ = 400 (10 - 5) = 2000 L / s (f i n a l r a t e) \\ F o r i n i t i a l r a t e p u t t = 0 \\ = 400 (10 - 0) = 4000 L / s \\ T h e a v e r a g e r a t e a t w h i c h t h e w a t e r i s r u n n i n g o u t \\ = \frac{I n i t i a l r a t e + F i n a l r a t e}{2} = \frac{4000 + 2000}{2} = \frac{6000}{2} = 3000 L / s \\ H e n c e, t h e r e q u i r e d r a t e = 3000 L / s . \end{array}$

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New answer posted

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Class 12th Application of Derivatives

Find the approximate volume of metal in a hollow spherical shell whose internal and external radii are 3 cm and 3.0005 cm, respectively.

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} I n t e r n a l r a d i u s r = 3 c m \\ E x t e r n a l r a d i u s R = r + Δ r = 3.0005 c m \\ ∴ Δ r = 3.0005 - 3 = 0.0005 c m \\ L e t y = r^{3} \Rightarrow y + Δ y = {(r + Δ r)}^{3} = R^{3} = {(3.0005)}^{3} \dots (i) \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t ., r, w e g e t \\ \frac{d y}{d x} = 3 r^{2} \\ ∴ Δ y = (\frac{d y}{d r}) . Δ r = 3 r^{2} * 0.0005 \\ = 3 * {(3)}^{2} * 0.0005 = 27 * 0.0005 = 0.0135 \\ ∴ {(3.0005)}^{3} = y + Δ y [F r o m e q n (i)] \\ = {(3)}^{3} + 0.0135 = 27 + 0.0135 = 27.0135 \\ V o l u m e o f t h e s h e l l = \frac{4}{3} π [R^{3} - r^{3}] \\ = \frac{4}{3} π [27.0135 - 27] = \frac{4}{3} π * 0.0135 \\ = 4 π * 0.0045 = 4 * 3.14 * 0.0045 = 0.018 π c m^{3} \\ H e n c e, a p p r o x i m a t e V o l u m e o f t h e m e t a l i n t h e s h e l l i s 0.018 π c m^{3} . \end{array}$

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New answer posted

a year ago

Class 12th Dual Nature of Radiation and Matter

Consider a 20 W bulb emitting light of wavelength 5000 Å and shining on a metal surface kept at a distance 2 m. Assume that the metal surface has work function of 2 eV and that each atom on the metal surface can be treated as a circular disk of radius 1.5 Å.
(i) Estimate number of photons emitted by the bulb per second. [Assume no other losses]
(ii) Will there be photoelectric emission?

(iii) How much time would be required by the atomic disk to receive energy equal to work function (2 eV)?
(iv) How many photons would atomic disk receive within time duration calculated in (iii) above?
(v) Can you explain how photoelectric effect was

...more

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Pallavi Pathak

Contributor-Level 10

Explanation-number of photon emitted per second n= $\frac{p}{\frac{h c}{?}} = \frac{p ?}{h c} = \frac{20 * 5000 * 10^{- 10}}{6.62 * 10^{- 34} * 3 * 10^{8}}$

$= 5 * 10^{19} s^{- 1}$

(ii) E=hc $/ ?$ = $\frac{6.62 * 10^{- 34} * 3 * 10^{8}}{5000 * 10^{- 10} * 1.6 * 10^{- 19}} = 2.48 e V$ this enegy is greater than 2 so emission is possible

(iii) work function $?$ = $\frac{p}{4 ? d^{2}} * ? r^{2} ? t$ = $?_{o}$

$? t$ = $\frac{4 ? d^{2}}{p r^{2}}$ = $\frac{4 * 2 * 16 * 1.6 * 10^{- 19} * 2^{2}}{20 * {(1.5 * 10^{- 10})}^{- 2}} = 28.4 s$

(iv) N= $\frac{n ? r^{2}}{4 ? d^{2}} * ? t$

= $\frac{5 * 10^{19} * {(1.5 * 10^{- 10})}^{2} * 28.4}{4 * ({2)}^{2}}$ =2

(v) as the time of emission is 11.04s so photoelectric is not spontaneous.

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New answer posted

a year ago

Class 12th Application of Derivatives

Find the approximate value of (1.999) ?.

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} {(1.999)}^{5} = {(2 - 0.001)}^{5} \\ L e t x = 2 a n d Δ x = - 0.001 \\ L e t y = x^{5} \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t ., x, w e g e t \\ \frac{d y}{d x} = 5 x^{4} = 5 {(2)}^{4} = 80 \\ N o w Δ y = (\frac{d y}{d x}) . Δ x = 80 . (- 0.001) = - 0.080 \\ ∴ {(1.999)}^{5} = y + Δ y \\ = x^{5} - 0.080 = {(2)}^{5} - 0.080 = 32 - 0.080 = 31.92 \\ H e n c e, a p p r o x i m a t e v a l u e o f {(1.999)}^{5} i s 31.92 . \end{array}$

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New answer posted

a year ago

Class 12th Application of Derivatives

Find an angle θ, 0 < < $\frac{π}{2}$ , which increases twice as fast as its sine.

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} A s p e r t h e g i v e n c o n d i t i o n, \\ \frac{d θ}{d t} = 2 \frac{d}{d t} (s i n θ) \\ \Rightarrow \frac{d θ}{d t} = 2 c o s θ \frac{d θ}{d t} \Rightarrow 1 = 2 c o s θ \\ ∴ c o s θ = \frac{1}{2} \Rightarrow c o s θ = c o s \frac{π}{3} \Rightarrow θ = \frac{π}{3} \\ H e n c e, t h e r e q u i r e d a n g l e i s \frac{π}{3} . \end{array}$

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New answer posted

a year ago

Class 12th Dual Nature of Radiation and Matter

A particle A with a mass m_A is moving with a velocity v and hits a particle B (mass m_B) at rest (one dimensional motion). Find the change in the de-Broglie wavelength of the particle A. Treat the collision as elastic.

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Pallavi Pathak

Contributor-Level 10

Explanation- according to law of conservation of momentum

S0 m_Av+m_b0=m_Av₁+m_Bv₂

So m_A(v-v₁)= m_Bv₂

according to law of conservation of kinetic energy

1/2m_Av²=1/2m_Av₁²+1/2m_Bv₂²

So m_A(v²-v₁²)= m_Bv₂²

From above eqn we can say that v+v₁=v₂ or v=v₂-v₁

So v₁= $\frac{m_{A} {- m}_{B}}{m_{A} {+ m}_{B}}$ v and v₂= $\frac{2 m_{A}}{m_{A} {+ m}_{B}}$ v

_{$?$ initial}=h/m_Av

_{$?$ final=}h/m_Av₁= $\frac{h (m_{A} {+ m}_{B})}{m_{a (m_{A} {- m}_{B}) v}}$

$d ?$ = $?$ _final- $?$ _initial= $\frac{h}{m_{A} v} {\frac{m_{A} {+ m}_{B}}{m_{A} {- m}_{B}} - 1}$

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New answer posted

a year ago

Class 12th Application of Derivatives

If the area of a circle increases at a uniform rate, then prove that the perimeter varies inversely as the radius.

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e k n o w t h a t \\ A r e a o f c i r c l e, A = π r^{2}, w h e r e r = r a d i u s o f t h e c i r c l e a n d p e r i m e t e r = 2 π r \\ A s p e r Q u e s t i o n, \frac{d A}{d t} = Κ, w h e r e K = constant \\ \Rightarrow \frac{d}{d t} (π r^{2}) = K \\ \Rightarrow π . 2 r . \frac{d r}{d t} = K \\ ∴ \frac{d r}{d t} = \frac{Κ}{4 π r^{2}} \dots (1) \\ N o w P e r i m e t e r c = 2 π r \\ D i f f e r e n t i a t i n g b o t h s i d e s ? w . r . t . t, w e g e t \\ \Rightarrow \frac{d c}{d t} = \frac{d}{d t} (2 π r) \\ \Rightarrow \frac{d c}{d t} = 2 π . \frac{d r}{d t} \\ \Rightarrow \frac{d c}{d t} = 2 π . \frac{K}{2 π r} = \frac{K}{r} [F r o m e q n (1)] \\ \Rightarrow \frac{d c}{d t} \propto \frac{1}{r} \\ H e n c e, t h e p e r i m e t e r o f t h e c i r c l e varies inversely a s the r a d i u s o f t h e c i r c l e . \end{array}$

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New answer posted

a year ago

Class 12th NCERT

A student performs an experiment on photoelectric effect, using two materials A and B. A plot of V_stopversus v is given in figure. Which has higher work function?

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Pallavi Pathak

Contributor-Level 10

Explanation -Given threshold frequency of A is given by v_0A= 5 $* 10^{14}$ hz

V_OB= 10 $*$ 10¹⁴hz

$? = h v_{0}$

$\frac{?_{O A}}{?_{O B}} = \frac{5 * 10^{14}}{10 * 10^{14}} 1$

$?_{O A}$ < $?_{O B}$

$(i i)$ for metal A slope=h/e= $\frac{2}{(10 - 5) 10^{14}}$

$h = \frac{2 e}{5 * 10^{14}} = \frac{2 * 1.6 * 10^{- 19}}{5 * 10^{14}}$ = 6.4 $* 10^{- 34}$ js

$f o r m e t a l B$ slope=h/e= $\frac{2.5}{(15 - 10) 10^{14}}$ = 8 $* 10^{- 34}$ js

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