**CBSE Previous Year Question Papers Class 12** will help student to understand the weightage of marks carried by every chapter. Based on that student can study for the exam, by preparing the important topic which has more weightage. Students should practice these question papers to examine themselves and check whether they are prepared perfectly for the board exam or not. They should solve the questions in the given timing, which will help to maintain their timings during the final examination.

CBSE Previous Year Question Papers Class 12 are available here for all the important subjects like Maths, English, Physics, Chemistry, Biology, Hindi, Commerce and Humanities. All these previous year question papers are available in PDF file by SelfStudys and can be downloaded simply. To make the students of class 12 prepare well for the board examination, we are giving here the last 10 year’s question papers. By solving these previous year question papers students will get an idea of the paper pattern, like the number of questions asked from every chapter and the marking pattern as well.

Question 1 :

Explain the following : (a) Sky appears blue. (b) The Sun appears reddish at (i) sunset, (ii) sunrise.

(1 Marks)
Answer / Solution :

**(a) Sky appears blues :**Light from the sun reaches the atmosphere that is comprised of the tiny particles of the atmosphere. These act as a prism and cause the different component; to scatter. As blue light travels in shorter and smaller waves in comparison to the other colours of spectrum. It is scattered the most , causing the sky to appear bluish.

**(b} The Sun appears reddish at (i) sunset, (ii) sunrise: **The molecules of the atmosphere and other particles that are smaller than the longest wavelength of visible light are more effective in scattering light of shorter wavelengths than light of longer wavelengths. The amount of scattering is inversely proportional to the fourth power of the wavelength, (Rayleigh Effect) Light from the Sun near the horizon passes through a greater distance in the .Earth's atmosphere than does the light received when the Sun is overhead. The correspondingly greater scattering of short wavelengths accounts for the reddish appearance of the Sun at rising and at setting.

Question 2 :

Under which conditions can a rainbow be observed? Distinguish between a primary and a secondary rainbow.

(2 Marks)
Answer / Solution :

The rainbow is an example of the dispersion of sunlight by the water drops in the atmosphere. This is a phenomenon due to combined effect of dispersion, refraction and reflection of Sunlight by spherical water droplets of rain.

The conditions for observing a rainbow are that the Sun should be shining in one part of the sky (say near western horizontal while it is raining in the opposite part of the sky (say eastern horizon).

**Difference between Primary and Secondary Rainbow**

**Primary Rainbow :**Three steps process (Reflection-Reflection and Refraction).**| Secondary Rainbow :**Four Step process (Refraction - Reflection and Refraction).**Primary Rainbow :**Appearance intensity better than Secondary.**| Secondary Rainbow :**Appearance intensity lesser than Primary.**Primary Rainbow :**Single reflection occurs.**| Secondary Rainbow :**Double reflection occurs.**Primary Rainbow :**Two Degree range occurs.**| Secondary Rainbow :**Three Degree range.

Question 3 :

A proton is accelerated through a potential difference V, subjected to a uniform magnetic field acting normal to the velocity of the proton. If the potential difference is doubled, how will the radius of the circular path described by the proton in the magnetic field change ?

(2 Marks)
Answer / Solution :

Given, proton accelerated through potential difference V, the direction of magnetic field is normal to velocity of proton.

As we know 1/2 m_{p}v^{2} = eV [during acceleration of proton, P.R will converted to kinetic energy]

v = √(2eV/m_{p})

When V is doubled, V' = 2V

v' = √(2e2V/m_{p}) = √2v

qvB = m_{p}v/r

r = m_{p}v/qB

r' = m_{p}/qB v'

r' = m_{p}/qB √2v

r' = √2r

Question 4 :

Draw a labelled diagram of cyclotron. Explain its working principle. Show that cyclotron frequency is independent of the speed and radius of the orbit.

(4 Marks)
Answer / Solution :

**What is Cyclotron :** Cyclotron is a device by which the positively charged particles like protons, deutrons, etc. can be accelerated.

**Principle of Cyclotron ** : Cyclotron works on the principle that a positively charged particle can be acclerated by making it to cross the Same electric field repeatedly with the help of a mimetic field.

**Construction :** The construction of a simple cyclotron is shown in figure above.

It consist of two-semi cylindrical boxes D_{1} and D_{2}, which are called Dees. They are enclosed in an evacuated chamber. Chamber is kept between the pules of a powerful magnet so that uniform magnetic field acts perpendicular to the plane of the dees. An alternating voltage is applied in the gap between the two dees by the help of a high quency oscillator. The electric field is zero inside the dees.

**Working and theory : **At a certain instant, let D_{1} be positive and D_{2} be negative. A proton from an ion source will be accelerated towards D_{1}, it describes a semi-circular path with a constant speed and is acted upon only by the magnetic field. The radius of the circular path is given by.

qvB =mv^{2}/r

From the above equation we get,

r = mv/qvB

The period of revolution is given by,

T = 2πr/v = 2π/v • mv/qB

T = 2πm/qB

The frequency of revolution is given by,

f ≈ 1/T = qB/2πm

From the above equation it follows that frequency ,f is independent of both v and r and is called cyclotron frequency. Also if we make the frequency of applied a.c. equal tot then every time the proton reaches the gap between the dees, the direction of electric field is reversed and proton receives a push and finally it gains very high kinetic energy. The proton follows a spiral path and finally gets directed towards the farget and comes out from it.

Question 5 :

A capacitor made of two parallel plates, each of the area is A and separation d is charged by an external d.c. (direct current) source. Show that during charging, the displacement current inside the capacitor is the same as the current charging the capacitor.

(2 Marks)
Answer / Solution :

From Ampere's Law, ∮ BdI = μ_{0}i(t). Let the **case 1**, where a point P is considered outside the capacitor charging. From the Ampere's Law magnetic field at Point P will be :

**CASE 1**

B•(2πr) – μ_{0} i(t) ⇒ B = μ_{0} i(t) / 2πr

Now, take another case 2, where shape of the surface under consideration covers capacitor's plate as we consider there is no current through capacitor then this value of B will be zero.

**CASE 2**

Hence, there is a contradiction. Therefore, this Ampere's law was modified with addition of displacement current inside the capacitor.

φ_{E} = | E | A = 1/_{ }• Q/A • A = Q/ε_{0}

dφ_{E}/dt = 1/ε_{0} • dQ/dt

ε_{0 }dφ_{E}/dt = dQ/dt = i_{d}, where i_{d} is the displacement current.

During charging of capacitor, outside the capacitor, i_{c} (conduction current) flows and inside i_{d} (displacement current) flows.

i = i_{c} + i_{d} = ic + ε_{0}dφ_{E}/dt

outside the capacitor, i_{d} = 0, hence, i = i_{c} and inside the capacitor, i_{c} = 0, hence i = i_{d}. Thus, i_{c} = i_{d} as capacitor gets charged.

Question 6 :
(4 Marks)

- Derive, with the help of a diagram, the expression for the magnetic field inside a very long solenoid having n tams per unit length carrying a current I.
- How is a toroid different from a solenoid?

Answer / Solution :

(a) **Magnetic field inside the solenoid**

∮B • dl = µ_{0} I_{0}= µ_{0} naI ............................(i)

∮B • dl = ∮_{∞} B•dl + ∮_{OR} B•dl + ∮_{0} B•dl + ∮_{0} B•dl

∮B • dl = B∮ dl cos 0 + ∮ B•dl cos 90 + 0 dl cos 0 + ∮ B•dl cos 90

B ∮ dl = B a .............................(ii)

From equation (i) and (ii),

B = µ_{0} naI

µ_{0} I_{0 }– µ_{0}naI (where, n = Number of turns per unit length | I = Current passing through)

(b) Torpid is a form in which a conductor is wound around a circular body. In this case we get magnetic field inside the corn but poles are absent because circular body don't have ends. Zooid is used in toroidal inductor, toroidal transformer.

Solenoid is a form in which conductor is wound around a cylindrical body with limb. In this case magnetic field creates two poles N and S. Solenoids have some flux leakage. This Is used in relay, motors, cicctro-magnetes.

**Solenoid Coil**

**Toroidal Coil**

Question 7 :

What is the value of **| A |**, If A is a square matrix satisfying** A' A = I**.

Answer / Solution :

⇒ Given that **A'A = I**

Then, | A' A | = | I |

⇒ | A' | | A | = | I |

⇒ | A | | A | = | I | [∵ | A' | = | A| ]

⇒ | A | ^{2} = 1 [∵ | I | = 1 ]

⇒ | A | = ±1 [∵ | I | = 1 ]

Question 8 :

Find the direction cosines (or directional cosines) of a line which makes equal angles with the coordinate axes.

(1 Marks)
Answer / Solution :

Let the direction cosines of the line make and angle α with each of the coordinate axex and direction cosines be *l, m, *and* n*. Therefore, ** l = cos α, m = cos α and n = cos α**.

l^{2} + m^{2} + n^{2} = 1 ⇒ *cos*^{2} α + *cos*^{2} α + *cos*^{2} α = 1

⇒ 3 *cos*^{2} α = 1

⇒ *cos*^{2} α = 1/3

⇒ *cos* α = ±1/√3

Therefore, the direction cosine are **(±1/√3, ±1/√3, ±1/√3) **

Question 9 :

Find the differential equation of the family of the curves y = Ae^{2x} + Be^{–2x}, where A and B are arbitrary constants.

Answer / Solution :

Given differential equation is y = Ae^{2x} + Be^{–2x}. ___ (i)

Now differentiating this equation w.r.t. x, we get

dy/dx = 2Ae^{2x} – 2Be^{–2x} ____(ii)

Again differentiating equation w.r.t. x, we get

d^{2}y/dx^{2} = 4Ae^{2x} + 4Be^{–2x}

d^{2}y/dx^{2} = 4 (Ae^{2x} + Be^{–2x}) = 4y

Therefore, required differential equation is d^{2}y/dx^{2} – y = 0.

Question 10 :

if P(not A) = 0.7, P(B) = 0.7 and P(B/A) = 0.5, then find the value of P(A/B).

(2 Marks)
Answer / Solution :

Given

P(not A) = 0.7 ⇒ P(A) = 1 – 0.7 = 0.3

P(B) = 0.7

P(B/A) = 0.5

We know that **P(B/A) = P(A ∩ B) / P(A)**, therefore

0.5 = P(A ∩ B)/0.3

⇒ P(A ∩ B) = 0.5 x 0.3 = 0.15

Now, Also we know **P(A/B) = P(A ∩ B) / P(B)**, therefore

P(A/B) = 0.15/0.7 = 15/7

Question 11 :

A coin is tossed 5 times. What is the probability of getting **3 heads, **and **at most 3 heads?**

Answer / Solution :

Here, A coin is tossed 5 times, therefore n = 5, p = 1/2 and q = 1/2,

We know that **p(x) = ^{n}C_{x} p^{x} q^{n – x},** therefore p(x) =

Now for 3 heads, x = 3, then the probability of getting 3 heads is

P(3) = ^{5}C_{3} (1/2)^{3} (1/2)^{5 – 3} = ^{5}C_{3} (1/2)^{3} (1/2)^{2} = ^{5}C_{3} (1/2)^{5}

^{n}C_{r} = n! / r! (n – r)! ⇒ ^{5}C_{3} = 5! / 2! (5 – 2)! = 5x4x3! / 2! x 3! = 10

Hence P(3) = 10 (1/2)^{5} = 5/16

Probability of getting at most 3 heads is P(x ≤ 3) = P(0) + P(1) + P(2) + P(3), then

P(0) = ^{5}C_{0} (1/2)^{0} (1/2)^{5 – 0 }= ^{5}C_{5} (1/2)^{5 }= 1/32

P(1) = ^{5}C_{1} (1/2)^{1} (1/2)^{5 – 1 }= ^{5}C_{1} (1/2)^{5 }= 5/32

P(2) = ^{5}C_{2} (1/2)^{2} (1/2)^{5 – 2 }= ^{5}C_{2} (1/2)^{5 }= 5/16

P(3) = ^{5}C_{3} (1/2)^{3} (1/2)^{5 – 3 }= ^{5}C_{3} (1/2)^{5 }= 5/16

Therefore, P(x ≤ 3) = 1/32 + 5/32 + 5/16 + 5/16 = 26/32 = 13/16

Question 12 :

Check whether the relation R defined on the set **A = {1, 2, 3, 4, 5, 6} **as **R = {(a, b) : b = a + 1}** is a reflexive, symmetric or transitive.

Answer / Solution :

Here **R = {(a, b) : b = a + 1}, **therefore R = {(a, a + 1) : a, a + 1 ∈ (1, 2, 3, 4, 5, 6) } ⇒ R = {(1, 2),(2, 3),(3, 4),(4, 5),(5, 6)}

- R is not reflexive as (a, a) ∉ R ∀ a
- R is not symmetric as (1, 2) ∈ R but (2, 1) ∉ R
- R is not transitive as (1, 2) ∈ R, (2, 3) ∈ R but (1, 3) ∈ R

Question 13 :

Let f : N → Y be a function defined as f(x) = 4x + 3, where Y = { y ∈ N : y = 4x + 3, for some x ∈ N}. Show that f is invertible. Find its inverse.

(4 Marks)
Answer / Solution :

First, consider any one arbitrary element of Y. By the definition of y, y = 4x + 3, for some value of x in the domain N. This shows that

4x = y – 3 ⇒ x = (y – 3)/4

Now, define g : Y → N by g(y) =& (y – 3)/4

gof (x) = g ( f(x) ) = g(4x + 3) = (4x + 3 – 3)/4 = 4x/4 = x

fog (y) = f ( g(y) ) = g( (y – 3)/4) ) = 4( y – 3)/4 + 3 = y – 3 + 3 = y

This shows that **gof = I _{N} and fog = I_{Y} **which implies that f is

Question 14 :

Find the value of sin (cos^{–1} 4/5 + tan^{–1} 2/3).

Answer / Solution :

sin (cos^{–1} 4/5 + tan^{–1} 2/3)

= sin (tan^{–1} 3/4 + tan^{–1} 2/3) [ ∵ cos^{–1} 4/5 = tan^{–1} 3/4) ]

We know that **tan ^{–1} x + tan^{–1} y = tan^{–1} (x + y)/(1 **–

= sin(tan^{–1} (3/4 + 2/3)/(1 – 3/4x2/3)

= sin(tan^{–1} ( (9+8)/12 / (1 – 6/12) )

= sin(tan^{–1} ( 17/12 / 6/12) )

= sin(tan^{–1} (17/6) )

Now, **tan ^{–1} 17/6 = sin^{–1} 17/√325**

= sin(sin^{–1} (17/√325) )

= 17/√325

Question 15 :

Prove that **dy/dx = –1/(x + 1) ^{2}**, if

Answer / Solution :

Given that x√(1 + y) + y√(1 + x) = 0

x√(1 + y) =–y√(1 + x)

After Squaring both the sides, we get

⇒ x^{2}(1 + y) = y^{2}(1 + x)

⇒ x^{2} + x^{2}y = y^{2} + y^{2}x = 0

⇒ x^{2} + x^{2}y – y^{2} – y^{2}x = 0

⇒ x^{2} – y^{2} + x^{2}y – y^{2}x = 0

⇒ (x + y)(x – y) + xy (x – y) = 0

⇒ (x – y)[x + y + xy] = 0

⇒ x + y + xy = 0 [∵ x ≠ y ]

⇒ y(1 + x) + x = 0

⇒ y(1 + x) = –x

⇒ y(1 + x) = – x/(1 + x)

On differentiating above equation with respect to x, we get

dy/dx = (1 + x)d/dx (–x) – (–x) d/dx (1 + x) / (1 + x)^{2}

dy/dx = (1 + x)(–1) + x(0 + 1) / (1 + x)^{2}

dy/dx = –1 – x + x / (1 + x)^{2}

dy/dx = –1/(1 + x)^{2}

Question 16 :

If (cos x)^{y} = (sin y)^{x}, find dy/dx.

Answer / Solution :

Given (cos x)^{y} = (sin y)^{x}, taking log both the sides, we get

log (cos x)^{y} = log (sin y)^{x }⇒ y log (cos x) = x log (sin y)

Differentiating with respect to x, we get

d/dx [ y log (cos x)] = d/dx [ x log (sin y)]

y d/dx log (cos x) + log (cos x) d/dx (y) = x d/dx log (sin y) + log (sin y) d/dx (x)

y • 1/cos x (– sinx) + log (cos x) dy/dx = x 1/sin y • cos y dy/dx + log (sin y)

– y tan x + log (cos x) dy/dx = x cot y dy/dx + log (sin y)

log (cos x) dy/dx – x cot y dy/dx = log (sin y) + y tan x

dy/dx [log (cos x) dy/dx – x cot y ] = log (sin y) + y tan x

**dy/dx = [log (sin y) + y tan x] / [log (cos x) dy/dx – x cot y]**

Question 17 :

Find the equation of the normal curve x^{2} = 4y which passes through the point (–1, 4).

Answer / Solution :

Suppose the normal at point P(x_{1}, y_{1}) on the given equation of parabola x^{2} = 4y passes through (–1, 4). Since the point (x_{1}, y_{1}) lies on the parabola x^{2} = 4y

⇒ x_{1}^{2} = 4y_{1} ...............(i)

The equation of the curve is x^{2} = 4y

Differentiating with respect to x, we get

2x = 4 dy/dx

dy/dx = 2x/4 = x/2

dy/dx at point P(x_{1}, y_{1}) = x_{1}/2

m (Slope) = dy/dx = x_{1}/2

Now, the equation (x_{1}, y_{1}) of normal at P(x_{1}, y_{1}) is given by **y – y _{1} = –1/m (x – x_{1})**

y – y_{1} = –2/x_{1} (x – x_{1}) ..........(ii)

This is passing through the given point (–1, 4). So putting the value as x = –1 and y = 4 in the above equation, we get the result

4 – y_{1} = –2/x_{1} (–1 – x_{1})

x_{1}(4 – y_{1}) = 2(1 + x_{1})

4x_{1} – x_{1}y_{1} = 2 + 2x_{1}

x_{1}y_{1} = 4x_{1 }– 2x_{1} – 2

x_{1}y_{1} = 2x_{1} – 2

y_{1} = (2x_{1} – 2)/x_{1} ................(iii)

Now, in next step we elemination y_{1} from equation (i) x_{1}^{2} = 4 y_{1}, we get

x_{1}^{2} = 4(2x_{1} – 2)/x_{1}

x_{1}^{3} = 8x_{1} – 8

⇒ x_{1 }= 2

Substitute x = 1 in equation (iii), we get y_{1} = 1. Putting these values of x_{1} and y_{1} in equation (ii), we get

y – 1 = –1 (x – 2)

y – 1 = –x + 2

y – 1 + x – 2 = 0

x + y – 3 = 0

Which is the required equation of normal to the given curve.

Question 18 :

Solve the differential equation : **xdy/dx = y − x tan (y/x)**.

Answer / Solution :

Given **xdy/dx = y – x tan (y/x)**

Let **y = vx**, (**v = y/x**) differentiate w.r.to x, we get

**dy/dx = v + x dv/dx**

Substitute dy/dx in the given differentiation, we get

⇒ x(v + xdv/dx) = vx – x tan (y/x)

⇒ xv + x^{2} dv/dx = vx – x tan v

⇒ x^{2} dv/dx = – x tan v

⇒ x dv/dx = –tan v

⇒ dv/tan v = –dx/x

⇒ cot v dv = –dx/x

⇒ ∫cot v dv = –∫dx/x

⇒ log sin v = –log x + log c

⇒ log sin v = log c/x

⇒ sin v = c/x

⇒ sin (y/x) = c/x

**⇒ x sin (y/x) = c**

Question 19 :

If the lines (x – 1)/–3 = (y – 2)/2λ = (z – 3)/2 and (x – 1)/3λ = (y – 1)/2 = (z – 6)/–5 are perpendicular, find the value of λ. Hence find whether the lines are intersecting or not.

(4 Marks)
Answer / Solution :

The equation of the given lines are (x – 1)/–3 = (y – 2)/2λ = (z – 3)/2 and (x – 1)/3λ = (y – 1)/2 = (z – 6)/–5

if the lines are perpendicular then it satisft the condition a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0

⇒ –3 × 3λ + 3λ × 2 + 2 × (–5) = 0

⇒ –9λ + 4λ –10 = 0

⇒ –5λ = 10

⇒ λ = –2

Now, the lines are **(x – 1)/–3 = (y – 2)/–4 = (z – 3)/2 and (x – 1)/–6 = (y – 1)/2 = (z – 6)/–5**. The coordinates of any point on first line are given by

⇒ (x – 1)/–3 = (y – 2)/–4 = (z – 3)/2 = α

⇒ (x – 1) = –3α ⇒ x = –3α + 1

⇒ (y – 2) = –4α ⇒ y = –4α + 2

⇒ (z – 3) = 2α ⇒ z = 2α + 3

Therefore, the coordinates of any point on first line are **(–3α + 1, –4α + 2, 2α + 3)**.

The coordinates of any point on second line are given by

⇒ (x – 1)/–6 = (y – 1)/2 = (z – 6)/–5 = β

⇒ (x – 1) = –6β ⇒ x = –6β + 1

⇒ (y – 1) = 2β ⇒ y = 2β + 1

⇒ (z – 6) = –5β ⇒ z = –5β + 6

Therefore, the coordinates of any point on second line are **(–6**β** + 1, 2**β** + 1, –5**β** + 6)**. If the linses intersect then they have a common point, So, for some value of **α and β**

**–3α + 1 = –6β + 1 **

**⇒ –3α = –6β **

**⇒ α = 2β **

and

**–4α + 2 = 2β + 1 **

**⇒ –4α + 1 = 2β**.

On solving we get **α = 1/5 and β = 1/10**

The values of **α and β** do not satisfy the third equation. Hence, lines do not intersect each other.

Question 20 :

There are three coins. One is a two-headed coin another is a biased coin that comes up heads 75% of the time and the third is an unbiased coin. One of the three coins is chosen at random and tossed. If it shows heads, what is the probability that It is the two-headed coin?

(6 Marks)
Answer / Solution :

Given, there are three coins. Let, E_{1} = Coin is two headed, E_{2} = Biased coin, E_{3} = Unbiased coin and A = Shows only head. Here P(E_{1}) = P(E_{2}) = P(E_{3}) = 1/3. Then

P(A/E_{1}) = 1

P(A/E_{2}) = 75/100 = 3/4 (This is Given value)

P(A/E_{3}) = 1/2

Now, the probability of two headed coin is

**P(E _{1}/A) = P(E_{1}) P(A/E_{1}) / [P(E_{1}) × P(A/E_{1}) + P(E_{2}) × P(A/E_{2}) + P(E_{3}) × P(A/E_{3})]**

= (1/3 × 1) / [1/3 × 1 + 1/3 × 3/4 + 1/3 × 1/2]

= 1/[1/3 + 1/4 + 1/6]

= 1/(9/4)

= 4/9

Question 21 :

A company produces two types of goods, A and B, that require gold and silver. Each unit of type A requries 3g of silver and 1 g of gold while that of type B requires 1 g of silver and 2 g of gold. The company can use at the most of 9 g of silver and 8 of gold. If each unit of type A brings a profit of Rs. 40 and that of type B Rs. 50. Find the number of units of each type that the company should produce to maximize profit.

Formulate the above LPP and solve it graphically and also find the maximum profit.

(6 Marks)
Answer / Solution :

There am two types of goods, A and B and let units of type A be x and units of type B be y.

A | B | |

Gold | 1 | 2 |

Silver | 3 | 1 |

Profit | 40 | 50 |

Then, Total Profit of goods is **P = 40x + 50y, x ≥ 0, y ≥ 0**

Hence, the mathematical formulation of the problem is as follows :

**Maximize P = 40x + 50y**

Subject to the constraints

x + 2y ≤ 8,

3x + y ≤ 9

x ≥ 0, y ≥ 0

To solve this LPP, we draw the lines **x + 2y = 8, 3x + y = 9, x = 0 and y = 0.**

x | 0 | 8 |

y | 4 | 0 |

and

x | 0 | 3 |

y | 9 | 0 |

Plotting these points on the graph

The shaded region is the required feasible region.

Corner Points |
Maximize P = 40x + 50y |

A (0, 4) | 0 + 4 × 50 = 200 |

B (2, 3) | 2 × 40 + 3 × 50 = 80 + 150 = 230 |

C (3, 0) | 3 × 40 + 0 = 120 |

O (0, 0) | 0 + 0 = 0 |

Clearly, **P is Maximum at Point B (2, 3) and the maximum profit is Rs. 230.**

Question 22 :

A line passing through the point with position vector 2 −+ 4 and is in the direction of the vector + − 2. Find the equation of the line in cartesian form.

(1 Marks)
Answer / Solution :

The line passing through the point with position (2, −1, 4) and has a direction ratio proportional to (1, 1, −2). Then cartesian equation of a line is

= (x − x_{1})/ a = (y − y_{1})/b = (z − z_{1})/c

**= (x − 2)/1 = (y + 1)/1 = (z − 4)/−2**

Question 23 :

Find the volume of a cuboid whose edges are given by −3 + 7 − 5, −5 + 7 − 3 and 7 − 5 − 3.

(2 Marks)
Answer / Solution :

We know that if a, b and c are the edges of a cuboid, then volume of a coboid is given by • (×). Here

= −3 + 7 − 5

= −5 + 7 − 3

= 7 − 5 − 3

• (×) = −3 (−21 − 15) −7 (15 + 21) + 5 (25 − 49)

= −3 × (−36) − 7 × 36 + 5 × (−24)

= 108 − 252 − 120

= −264 cubit units

Question 24 :

If || = 2, || = 7 and × = 3 + 2 + 6. find the angle between and .

(2 Marks)
Answer / Solution :

Given that || = 2, || = 7 and × = 3 + 2 + 6.then the angle between and is given by sin θ = | × | / | | ||

| × | = √(3^{2} + 2^{2} + 6^{2}) = √(9+4+36) = √49 = 7

sin θ = 7/2×7 = 7/14 = 1/2 = sin π/6 ⇒ θ = π/6

Question 25 :

Find the general solution of the differential equation dy/dx = e^{x + y}.

Answer / Solution :

Given dy/dx = e^{x + y}

⇒ dy/dx = e^{x }e^{y}

⇒ dy/e^{y} = e^{x }dx

On Integrating both the sides, we get

⇒ ∫dy/e^{y} = ∫e^{x }dx

⇒ ∫e^{−y}dy = ∫e^{x }dx

⇒ e^{−y} = e^{x} + c, which is the required solution.

Question 26 :

If **(a + bx) e ^{y/x} = x**, then prove that

Answer / Solution :

We have

(a + bx) e^{y/x} = x

⇒ e^{y/x} = x/(a + bx)

⇒ y/x = log [ x/(a + bx) ]

⇒ y/x = log x − log (a + bx)

Now, differentiating with respect to x, we get

⇒ [x dy/dx − y] / x^{2} = 1/x − 1/(a + bx) × b

⇒ [x dy/dx − y] (1/x^{2}) = 1/x − b/(a + bx)

⇒ x dy/dx − y = x^{2}[1/x − b/(a + bx)]

**⇒ x dy/dx − y = ax/(a + bx) ..........(i)**

Again. differentiating both the sides w.r.to x, we get

x d^{2}y/dx^{2} + dy/dx − dy/dx = [ (a + bx) a − ax × b ] / (a + bx)^{2}

⇒ x d^{2}y/dx^{2} = [ a^{2} + abx − abx ] / (a + bx)^{2}

⇒ x d^{2}y/dx^{2} = a^{2}/(a + bx)^{2}

On multiplying both the sides by x^{2}, we get

⇒ x^{3} d^{2}y/dx^{2} = a^{2}x^{2}/(a + bx)^{2}

⇒ x^{3} d^{2}y/dx^{2} = [ax / (a + bx) ]^{2} ..........(ii)

From equation (i) and (ii), we get

⇒ **x ^{3} d^{2}y/dx^{2} = [ax/(a + bx)]^{2}**

Question 27 :

The volume of a cube is increasing at the rate of 8 cm^{3}/s. How fast is the surface area increasing when the length of its edge is 12 centimeter.

Answer / Solution :

let x be the length of side, V be the Volume and S be the surface area of cube. Then V = x^{3} and S = 6 x^{2}, where x is a function of time t. Now, dV/dt = 8cm^{3}/s (it is given), therefore

8 = dV/dt = d/dt (x^{3}) = 3x^{2} dx/dt

⇒ 8 = 3x^{2} dx/dt

⇒ 8/3x^{2} = dx/dt .......(i)

Now, dS/dt = d/dt (6x^{2})

⇒ dS/dt = 12x dx/dt

⇒ dS/dt = 12x (8/3x^{2}) .... (from equation i)

⇒ dS/dt = 32/x

Hence, when x = 12 cm, then

dS/dt = 32/12 = 8/3 cm^{2}/s

Question 28 :

Find the point on the curve y^{2} = 4x, which is nearest to the point (2, −8).

Answer / Solution :

Given curve is y^{2} = 4x and say point P(x, y) is on the curve which is nearest to the given point (2, −8). Because y^{2} = 4x ⇒ x = y^{2}/a. Therefore, a point P(x, y) will be P(y^{2}/4, y).

Now, the distance between both the points A and P is given by this distance formula √ [(x_{1 }− x_{2})^{2} + (y_{1 }− y_{2})^{2}] i.e.

⇒ AP = √ [(x_{ }− 2)^{2} + (y_{ }+ 8)^{2}]

⇒ AP = √ [(y^{2}/4 − 2)^{2} + (y_{ }+ 8)^{2}]

⇒ AP = √ [(y^{4}/16 − 2 × y^{2}/4 × 2_{ }+ 4) + (y^{2}_{ }+ 2 × y × 8 + 64)]

⇒ AP = √ [(y^{4}/16 − y^{2}_{ }+ 4) + (y^{2}_{ }+ 16y + 64)]

⇒ AP = √ [y^{4}/16 − y^{2}_{ }+ 4 + y^{2}_{ }+ 16y + 64)]

⇒ AP = √ [y^{4}/16 + 16y + 68)]

Let Z = AP^{2} = y^{4}/16 + 16y + 68

Now differentiating Z with respect to y, we get

⇒ dZ/dy = 1/16 × 4y^{3} + 16 = y^{3}/4 + 16

For maximum of minimum value of Z, we have dZ/dy = 0 i.e.

⇒ y^{3}/4 + 16 = 0

⇒ y^{3} + 64 = 0

⇒ (y + 4) (y^{2} − 4y + 16) = 0

(∵ y^{2} − 4y + 16 = 0 gives the imaginary value of y)

⇒ y = −4

Now, again differentiating with respect to y, we get

⇒ d^{2}Z/dy^{2} = d/dy(y^{3}/4 + 16)

⇒ d^{2}Z/dy^{2} = 3y^{2}/4

At y = −4, d^{2}Z/dy^{2} = 3(−4)^{2}/4 = 3 × 16/4 = 3 × 4 = 12 > 0. Thus, Z is maximum when y = −4. Substituting y = −4 in equation of given curve y^{2} = 4x.

x = (−4)^{2}/4 = 16/4 = 4

Hence, the **point P(4, −4) on the curve y ^{2} = 4x is nearest to the point (12, −8).**

Question 29 :

A factory manufactures two types of screws A and B each type requiring the use of two machines, an automatic and a hand operated it takes 4 minutes on the automatic and 6 minutes on the hand operated machines to manufacture a packet of screws A while it takes 6 minutes on the automatic and 3 minutes on the hand operated machine to manufacture a packet of screw B. Each machine is available for at most 4 hours on a day. The manufacturer can sell a packet of screw A at a profit of 70 paise and screw B at a profit of Rupee 1. Assuming that he can sell all the screws he manufactures, how many packets of each type should be the factory owner produce in a day in order to maximize his profit. Also formulate the above LPP and solve it graphically and the maximum profit.

(6 Marks)
Answer / Solution :

Let the number of packets of screw a manufactured in a day be x and the screw B be y. Therefore,x ≥ 0, and y ≥ 0.

List of Items | Number | Machine A | Machine B | Profit |

Screw A | x | 4 minutes | 6 minutes | Rs. 0.7 |

Screw B | y | 6 minutes | 3 minutes | Rs 1 |

Maximum time available | 4 hrs = 240 min | 4 hrs = 240 mi |

Then the constants are

⇒ 4x + 6y ≤ 240 or 2x + 3y ≤ 120

⇒ 6x + 3y ≤ 240 or 2x + y ≤ 80

and total profit is Z = 0.7 x + y.

So the LPP (Linear Programming Problem) will be **Max Z = 0.7 x + y**

Subject to the constraints : **2x + 3y ≤ 120, 2x + y ≤ 80 and x ≥ 0, y ≥ 0**

Now we solve the table for the equation **2x + 3y ≤ 120, 2x + y ≤ 80. **i, e.

**For 2x + 3y = 120**

x | 0 | 0 |

60 |

**For 2x + y ≤ 80 **

X + 3 y less than equal to 120 2x + Y less than is equal to 80 and x y greater Now 2 X + 3 Y less than is equal to 120 and 2 X + Y less than is equal to 180 draught a Floating the points on the graph we get feasible region oabc is shown in the shaded region

Question 30 :

Find the magnitude of each of the two vectors and having the same magnitude such that the angle between them is 60° and their scalar product is 9/2.

(1 Marks)
Answer / Solution :

Let and are the two such vectors. Now • = || || cos θ .......(i)

It is given that || = ||, θ = 60° and • = 9/2

⇒ • = || || cos θ

⇒ 9/2 = ||^{2} 60°

⇒ 9/2 = ||^{2} 1/2

⇒ ||^{2 }= 9

⇒ ||^{ }= 3

⇒ ||^{ }= || = 3

Question 31 :

The total cost C(x) associated with the production of x units of an item is given by C(x) = 0.005x^{3} − 0.02x^{2} + 30x + 5000. Find the marginal cost when 3 units are produced, where by marginal cost we mean the instantaneous rate of change of total cost at any level of output.

Answer / Solution :

Cost function is given by C(x) = 0.005x^{3} − 0.02x^{2} + 30x + 5000

Marginal Cost (MC) = d/dx [C(x)]

= d/dx (0.005x^{3} − 0.02x^{2} + 30x + 5000)

= 0.005(3x^{2}) − 0.02(2x) + 30

= 0.015x^{2} − 0.04x + 30

When x = 3,

MC = 0.015 × 3^{2} − 0.04 × 3 + 30

= 0.135 − 0.12 + 30

= 30.015

Question 32 :

Find the differential equation representing the family of curves **y = ae ^{bx + 5}**, where a and b are the arbitrary constants.

Answer / Solution :

Given curve is **y = ae ^{bx + 5}**

Differentiating equation with respect to x, we get

⇒ dy/dx = ae^{bx + 5} d/dx (bx + 5)

⇒ dy/dx = ae^{bx + 5} b

⇒ dy/dx = by ......(i)

Differentiating again with respect to x, we get

⇒ d^{2}y/dx^{2} = b dy/dx

⇒ d^{2}y/dx^{2} = dy/dx (dy/dx • 1/y) [∵ **b = dy/dx • 1/y **... from (i)]

⇒ y d^{2}y/dx^{2} = (dy/dx)^{2}

Question 33 :

A black and a red die are rolled together. Find the conditional probability of obtaining the sum 8. Given that the red die resulted in a number less than 4.

(2 Marks)
Answer / Solution :

The sample space has 36 outcomes.

Let A be the event that the sum of observations is 8 :

**A = {(2, 6), (3, 5), (5, 3), (4,4), (6, 2)} **

∴ n(A) = 5, P(A) = 5/36

Let B be event that observation on red die is less than 4

**B = {(1, 1), (2, 1), (3, 1), (4,1), (1, 1), , (6, 1), , (1, 2), , (2, 2), , (3, 2), , (4, 2), (5, 2), , (6, 2), (1, 3), (2, 3), (3, 3), (4, 3), (5, 3), (6, 3)}**

∴ n(B) = 18 ⇒ P(B) = 18/36 = 1/2

Clearly A ∩ B = {(5, 3), (6, 2)}

∴ n(A ∩ B) = 2 P(A ∩ B) = 2/36 = 1/18

Now P(A/B) = P(A ∩ B) /P(B) = (1/18)/(1/2) = 2/18 = 1/9

Question 34 :

Find the equation of the tangent and the normal to the curve 16x^{2} + 9y^{2} = 145 at the point (x_{1}, y_{1}) where x_{1} = 2 and y_{1} > 0.

Answer / Solution :

Given curve is 16x^{2} + 9y^{2} = 145 .......(i)

Since the point (x_{1}, y_{1}) lies on the given curve therefore 16x_{1}^{2} + 9y_{1}^{2} = 145.

Here, given x_{1} = 2,

⇒ 16(2)^{2} + 9y_{1}^{2} = 145

⇒ 64 + 9y_{1}^{2} = 145

⇒ 9y_{1}^{2} = 145 − 64 = 81

⇒ y_{1}^{2} = 9

⇒ y_{1} = 3 [∵ y1 > 0]

Therefore, **Point of contact is (2, 3)**

Differentiating equation (i) with respect to x, which will give the slope of the tangent i.e.

32x + 18y dy/dx = 0

⇒ dy/dx = −32x/18y

Now, slope of the tangent at the point (2, 3) is

m = (dy/dx)_{(2 ,3)} = −32(2)/18(3) = −64/54 = −32/27

Therefore, the equation of the tangent is y − 3 = m(x − 2)

⇒ y − 3 = −32/27 (x − 2)

⇒ 27(y − 3) = −32(x − 2)

⇒ 27y − 81 = −32x + 64

⇒ 32x + 27y = 64 + 81

⇒ 32x + 27y = 145

Now, the slope of the normal = −1/Slope of tangent = −1/(−32/27) = 27/32

Equation of the normal is

⇒ y − 3 = 27/32 (x − 2)

⇒ 32(y − 3) = 27(x − 2)

⇒ 32y − 96 = 27x − 54

⇒ 27x − 32y = 54 − 96

⇒ 27x − 32y = − 42

**Equation of the tanget is 32x + 27y = 145 and normal is 27x − 32y = − 42**

Question 35 :

Find the intervals in the function f(x) = x^{4}/4 − x^{3} −5x^{2} + 24 x + 12 is (a) strictly increasing and (b) strictly decreasing.

Answer / Solution :

Given function is **f(x) = x ^{4}/4 − x^{3} −5x^{2} + 24 x + 12.**

Differentiate w.r.to x, we get

⇒ f '(x) = 4x^{3}/4 − 3x^{2} − 5(2x) + 24

⇒ f '(x) = x^{3} − 3x^{2} − 10x + 24

For critical points, put f '(x) = 0. Therefore x^{3} − 3x^{2} − 10x + 24 = 0

⇒ (x − 2) (x^{2} − x − 12) = 0

⇒ (x − 2) (x − 4) (x + 3) = 0

⇒ x = 2, 4, −3

Therefore, we have the intervals (−∞, 3), (−3, 2), (2, 4) and (4, ∞).

Since, f '(x) > 0 in (−3, 2) U (4, ∞). Therefore f(x) in strictly increasing in interval (−3, 2) U (4, ∞) and f '(x) < 0 in (−∞, −3) U (2, 4). Therefore f(x) is strictly decreading in the interval (−∞, −3) U (2, 4).

Question 36 :

Using Integration, find the area of the triangular region whose sides have the equation y = 2x + 1, y = 3x + 1 and x = 4.

(6 Marks)
Answer / Solution :

The given equations of the sides of triangle is y = 2x + 1 ...(i), y = 3x + 1 ...(ii), and x = 4 ...(iii). Preapre the table of this equations and draw the graph shown below.

Now, from the graph, equation y = 2x + 1 meets x and y axes at (−1/2, 0) and (0, 1). By joining these points we obtain the graph of x + 2y = 2. Similarly graph of other equations are drawn and points are found. On solving all the given three equations in pairs, we obtain intersection point. This method we obtain the coordinates of the vertices of Triange ABC are A(0, 1), B(4, 13) and C(4, 9).

Then area of the ∆ABC = Area (OLBAO) + Area (OLCAO)

= ∫_{0}^{4} (3x + 1)dx − ∫_{0}^{4} (2x + 1)dx

= ∫_{0}^{4} (3x + 1 − 2x − 1)dx

= ∫_{0}^{4} x dx

= [x^{2}/2]_{0}^{4}

=1/2 [4^{2} − 0^{2}]

=1/2 × 16

= 8 square units

Question 37 :

Find the differential equation representing the family of curves y = ae^{2x} + 5, where a is an arbitary constant.

Answer / Solution :

Given that y = ae^{2x} + 5. On differentiating w.r.t x, we get

dy/dx = ae^{2x }• (2) [Using Chaing Rule]

⇒ dy/dx = 2ae^{2x}

Again differentiating w.r.t x, we get

d^{2}y/dx^{2} = 2[ae^{2x }• (2)]

⇒ d^{2}y/dx^{2} = 2[dy/dx]

⇒ d^{2}y/dx^{2} = 2 dy/dx

⇒ d^{2}y/dx^{2} − 2 dy/dx = 0

Question 38 :

If cos √(3x), then find dy/dx.

(1 Marks)
Answer / Solution :

Given y = cos √(3x). On differentiating w.r.t x, we get

dy/dx = −sin √(3x) × √3 × 1/2√x

dy/dx = −√3 sin√(3x) / 2√x

Question 39 :

If x = ae^{t} (sin t + cos t) and y = ae^{t} (sin t − cos t), then prove that dy/dx = (x + y)/(x − y).

Answer / Solution :

Given

x = ae^{t} (sin t + cos t)

y = ae^{t} (sin t − cos t)

On differentiating above equation with respect to t, we get

x = ae^{t} (sin t + cos t)

⇒ dx/dt = a d/dt [e^{t} (sin t + cos t)]

= a [e^{t} d/dt(sin t + cos t) + (sin t + cos t) d/dt e^{t}]

= a [e^{t} (cos t − sin t) + (sin t + cos t) e^{t}]

= a [e^{t} cos t − e^{t} sin t + e^{t} sin t + e^{t} cos t]

**dx/dt = a [2e ^{t} cos t] = 2ae^{t} cos t ...(i)**

y = ae^{t} (sin t − cos t)

⇒ dy/dt = a d/dt [e^{t} (sin t − cos t)]

= a [e^{t} d/dt(sin t − cos t) + (sin t − cos t) d/dt e^{t}]

= a [e^{t} (cos t + sin t) + (sin t − cos t) e^{t}]

= a [e^{t} cos t + e^{t} sin t + e^{t} sin t − e^{t} cos t]

**dy/dt = a [2e ^{t} sin t] = 2ae^{t} sin t ...(ii)**

On dividing equation (ii) by equation (i), we get

dy/dx = dy/dt / dx/dt

dy/dx = 2ae^{t} sin t / 2ae^{t} cos t

dy/dx = sin t /cos t

**dy/dx = tan t**

Also we solve (x + y)/(x − y) i.e.

x + y = ae^{t} (sin t + cos t) + ae^{t} (sin t − cos t)

= ae^{t} [sin t + cos t + sin t − cos t]

**x + y = 2 ae ^{t} sin t**

x − y = ae^{t} (sin t + cos t) − ae^{t} (sin t − cos t)

= ae^{t} [sin t + cos t − sin t + cos t]

**x + y = 2 ae ^{t} cos t**

Now **(x + y)/(x − y) = 2 ae ^{t} sin t/2 ae^{t} cos t**

⇒ (x + y)/(x − y) = sin t/cos t

⇒ **(x + y)/(x − y) = sin t/cos t = tant **

Hence Proved

**dy/dx = (x + y)/(x − y)**

Question 40 :

A bag contains 5 red and 4 black balls, a second bag contains 3 red and 6 black balls. One of the two bags is selected at random and two balls are drawn at random (without replacement) both of which are found to be red. Find the probability that the balls are drawn from the second bag.

(6 Marks)
Answer / Solution :

Let E_{1} be the event of choosing the First bag and E_{2} be the event of choosing the Second bag and A be the event of drawing a red ball. Then, **P(E _{1}) = P(E_{2}) = 1/2**.

Now,

P(A/E_{1}) = 5/9 × 4/8 = 20/72

P(A/E_{2}) = 3/9 × 2/8 = 6/72

Now the probability of drawing a ball from Secong Bag, if it is given that it is red is P(A/E_{2}). Use Baye's theorem of probability, we have

P(E_{2}/A) = P(E_{2}) • P(A/E_{2}) / [ P(E_{1}) × P(A/E_{1}) + P(E_{2}) × P(A/E_{2}) ]

= 1/2 × 6/72 / [1/2 × 20/72 + 1/2 × 6/72]

= 6/(20 + 6)

= 6/26

**P(E _{2}/A) = 3/13**

Question 41 :

An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quantity of water. Show that the cost of material will be least when depth of the tank is half of its width. If the cost is to be borne by nearer settled lower income families for whom water will be provided, what kind of value is hidden this equation?

(4 Marks)
Answer / Solution :

Let the length breadth and height of the open tank be x, x and y units respectively. Then

Volume (V) = x^{2}y ....(i)

Total Surface Area (S) = x^{2 }+ 4xy .....(ii)

S = x^{2 }+ 4xy .....(ii)

S = x^{2 }+ 4x × V/x^{2} [Using Equation (i) y = V/x^{2}]

S = x^{2 }+ 4V/x

Differentiating w.r.to x, we get

dS/dx = d/dx [x^{2 }+ 4V/x^{2}]

dS/dx = d/dx (x^{2}) + 4V d/dx (1/x^{2})

dS/dx = 2x − 4V/x^{2}

For critical points, put dS/dx = 0

⇒ 2x − 4V/x^{2} = 0

⇒ 2x^{3} − 4V = 0

⇒ 2x^{3} = 4V

⇒ 2x^{3} = 4 x^{2}y

⇒ x = 2y .... (iii)

Now, Again differentiating w.r.to x, we get

d^{2}S/dx^{2} = d/dx (2x − 4V/x^{2})

= 2 − 4 × (−2V/x^{3})

= 2 + 8V/x^{3}

= 2 + 8V/(2y)^{3} [Using equation (iii)]

= 2 + 8V/8y^{3}

= 2 + V/y^{3} > 0

**Area is minimum, therefore the cost is minimum when x = 2y that means depth of the tank is half of the width.**

Question 42 :

Find the particular solution of the differential equation **e ^{x} tan y dx + (2 − e^{x}) sec^{2} y dy = 0**, given that y = π/4 when x = 0.

Answer / Solution :

Given differential equation is **e ^{x} tan y dx + (2 − e^{x}) sec^{2} y dy = 0**

⇒ e^{x} tan y dx = − (2 − e^{x}) sec^{2} y dy

⇒ e^{x} tan y dx = (e^{x }− 2) sec^{2} y dy

⇒ e^{x}/(e^{x }− 2) dx = tan y sec^{2} y dy

Integrating both the sides, we get

⇒ ∫ e^{x}/(e^{x }− 2) dx = ∫ tan y sec^{2} y dy

⇒ log |e^{x }− 2| = log tan y + log C

⇒ **e ^{x }− 2 = C tan y ... (i)**

Put y = π/4, when x = 0

e^{0 }− 2 = C tan π/4

1^{ }− 2 = C

C = −1

Therefore, from equation (i),

e^{x }− 2 = −1 tan y

tan y = 2 − e^{x}

y = tan^{−1}(2 − e^{x}) is the required solution of differential equation.

Question 43 :

Find the particular solution of the differential equation **dy/dx + 2y tan x = sin x**, given that y = 0 when x = π/3.

Answer / Solution :

Given differential equation is **dy/dx + 2y tan x = sin x**. It is a linear differential equation of the form **dy/dx + Py = Q**.

When we compare both, we get P = 2 tan x and Q = sin x. Now we find Integrating factor, with the formula

Integrating Factor (IF) = e ^{∫ Pdx }

= e ^{∫ 2 tan x dx}

= e^{2 ∫ tan x dx}

= e^{2 log sec x dx}

Integrating Factor = sec^{2 }x

Now solution of the equation (i) is given by

y • IF = ∫ Q • IF dx + C

⇒ y sec^{2 }x = ∫ sin x • sec^{2} x dx + C

⇒ y sec^{2 }x = ∫ sec x • tan x dx + C [ ∵ ∫ sec x • tan x dx = sec x + C]

⇒ y sec^{2 }x = sec x + C

Putting y = 0 when x = π/3

0 × sec^{2 }π/3 = sec π/3 + C

0 = 2 + C

C = −2

Hence the pariicular solution is **y sec ^{2}x = sec x **−

Question 44 :

Suppose a girl throws a die. If she gets 1 or 2, she tosses a coin three times and notes the number of tails. If she gets 3, 4, 5 or 6, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one tail, what is the probability that she threw 3, 4, 5 or 6 with the dice?

(4 Marks)
Answer / Solution :

Let E_{1} be the event that girl gets 1 or 2 on the role and E_{2} be the event that girl gets 3, 4, 5 or six on the role of a die. Therefore, P(E1) = 2/6 = 1/3 and P(E2) = 4/6 = 2/3.

If she tossed coin only once and exactly one tail shows, then P(A/E_{1}) = 3/8

**P(E _{1}/A) = P(E_{2}) P(A/(E_{2}) / [P(E_{1}) P(A/E_{1}) + P(E_{2}) P(A/E_{2})]**

= 2/3 × 1/2 / [1/3 × 3/8 + 2/3 × 1/2]

= 1/3 / [1/8 + 1/3]

= 1/3 / 11/24

**P(E _{1}/A) = 8/11**

Question 45 :

Two numbers are selected at random without replacement from the first five positive integers. Let X denote the larger of the two numbers obtained. Find the mean and variance of X.

(4 Marks)
Answer / Solution :

First five positive integers are 1 2 3 4 and 5. We select two positive integers in 5 times 4 equal to 20 ways. Out of these two numbers are selected at random. Let X denote larger of the two selected numbers. Then X can have the values 2 3 4 or 5.

P(X = 2) = P(Larger Number is 2) = {(1, 2), (2, 1)}

⇒ P(X = 2) = 2/20

Similarly,

⇒ P(X = 3) = 4/20

⇒ P(X = 4) = 6/20

⇒ P(X = 5) = 8/20

Thus, the probability distribution of X is

X | 2 | 3 | 4 | 5 |

P(X) | 2/20 | 4/20 | 6/20 | 8/20 |

Therefore,

Mean = E(X) = ∑_{i=1}^{n }x_{i} P(x_{i})

= 2 × 2/20 + 3 × 4/20 + 4 × 6/20 + 5 × 8/20

= 4/20 + 12/20 + 24/20 + 40/20

= (4 + 12 + 24 + 40)/20 = 80/20 = 4

E(X^{2}) = ∑_{i=1}^{n }x_{i}^{2} P(x_{i})

= 2^{2} × 2/20 + 3^{2} × 4/20 + 4^{2} × 6/20 + 5^{2} × 8/20

= 4 × 2/20 + 9 × 4/20 + 16 × 6/20 + 25 × 8/20

= 8/20 + 36/20 + 96/20 + 200/20

= (8 + 36 + 96 + 200)/20 = 340/20 = 34/2 = 17

Variance = E(X^{2}) − [E(X)]^{2} = 17 - (4)^{2} = 17 - 16 = 1.

**Therefore, mean and variance are 4 and 1 respectively.**

Question 46 :

Show that the height of a cylinder of maximum volume that can be inscribed in a sphere of radius R is 2R/√3. Also find the maximum volume.

(6 Marks)
Answer / Solution :

Let x be the diameter of the base of a cylinder and let h be the height of the cylinder.

In Triangle ABC, we have BC^{2} + AB^{2} = AC^{2}

⇒ h^{2} + x^{2} = (2R)^{2}

⇒ x^{2} = 4R^{2} − h^{2} ... (i)

Now, the formula to find the volume of a cylinder is given by V = πr^{2}h i.e.

⇒ V = π (x/2)^{2 }× h

⇒ V = π × x^{2}/4 × h

⇒ V = π × (4R^{2} − h^{2})/4 × h

⇒ V = 4πR^{2}h/4 − πh^{3}/4

⇒ V = πR^{2}h − πh^{3}/4 ....(ii)

On differentiating equation (ii) with respect to h, we get

dV/dh = πR^{2} d/dh h − π/4 d/dh (h^{3})

⇒ dV/dh = πR^{2} − 3πh^{2}/4 ....(iii)

⇒ 0 = πR^{2} − 3πh^{2}/4 [Because dV/dh = 0]

⇒ 3h^{2}/4 = R^{2}

**⇒ h = 2R/√3**

Again differentiating equation (iii) with respect to h, we get

d^{2}V/dh^{2} = d/dh (πR^{2}) − 3π/4 d/dh(h^{2})

⇒ d^{2}V/dh^{2} = 0 − 3π/4 (2h)

⇒ d^{2}V/dh^{2} = −3πh/2

At h = 2R/√3, we have

⇒ d^{2}V/dh^{2} = −3π/2 × 2R/√3

⇒ d^{2}V/dh^{2} = −√3πR < 0

Hence, h = 2R/√3 is a point of maxima. So V is maximum when h = 2R/√3. Therefore, the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is 2R/√3.

Also from equation (i) we have

x^{2} = 4R^{2} − h^{2}

⇒ x^{2} = 4R^{2} − (2R/√3)^{2}

⇒ x^{2} = 4R^{2} − 4R^{2}/3 = 8R^{2}/3

Now, maximum volume of a cylinder is π × (x/2)^{2 }× h

V = π/4 × x^{2 }× h

V = π/4 8R^{2}/3^{ }× 2R/√3

V = 4πR^{3}/3√3

Question 47 :

Solve the differential equation : **dy/dx = **− **[(x + y cos x) / (1 + sin x)]**

Answer / Solution :

dy/dx = −[(x + y cos x) / (1 + sin x)]

dy/dx = −x/(1 + sin x) − y cosx /(1 + sin x)

dy/dx + y cosx /(1 + sin x) = −x/(1 + sin x)

Compare this equation with dy/dx + Py = Q. we have

**P = cosx /(1 + sin x) and **

**Q = −x/(1 + sin x)**

Now Intergrating factor is IF = e ^{∫ P dx}

= e ^{∫ cosx /(1 + sin x) dx}

= e ^{log (1 + sin x)}

IF = (1 + sin x)

Then,

y × (1 + sin x) = ∫ −1/(1 + sin x) × (1 + sin x) dx + C

⇒ y × (1 + sin x) = y (1 + sin x) = − ∫xdx + C

⇒ **y × (1 + sin x) = y (1 + sin x) = −x ^{2}/2 + C**

Question 48 :

Find dy/dx, if (x^{2} + y^{2})^{2} = xy,

Answer / Solution :

Given (x^{2} + y^{2})^{2} = xy

Differentiating with respect to x, we get

d/dx (x^{2} + y^{2})^{2} = d/dx (xy)

⇒ 2(x^{2} + y^{2}) d/dx (x^{2} + y^{2}) = x d/dx (y)+ y d/dx (x)

⇒ 2(x^{2} + y^{2}) (2x + 2y dy/dx) = x dy/dx + y

⇒ 4(x^{2} + y^{2})(x + y dy/dx) = x dy/dx + y

⇒ 4x(x^{2} + y^{2}) + 4y(x^{2} + y^{2}) dy/dx = x dy/dx + y

⇒ 4y(x^{2} + y^{2}) dy/dx − x dy/dx = y − 4x(x^{2} + y^{2})

⇒ (4x^{2}y + 4y^{3 }− x) dy/dx = y − 4x^{3} + 4xy^{2}

⇒ **dy/dx = (y − 4x ^{3} + 4xy^{2}) / (4x^{2}y + 4y^{3 }− x)**

Question 49 :

If x = a (2θ − sin 2θ) and y = a (1 − cos 2θ), find the value of dy/dx when θ = π/3.

(4 Marks)
Answer / Solution :

Given that **x = a (2θ − sin 2θ) and y = a (1 − cos 2θ)**, differentiating with respect to **θ**, we get

x = a (2θ − sin 2θ)

dx/dθ = a(2 − 2 cos 2θ) = 2a(1 − cos 2θ)

dy/dθ = a (2 sin 2θ) = 2a sin 2θ

Now dy/dx = dy/dθ / dx/dθ

= 2a sin 2θ / 2a(1 − cos 2θ)

= sin 2θ / (1 − cos 2θ)

= 2 sin θ cos θ / 2 sin^{2} θ

= cos θ / sin θ

**dy/dx = cot θ**

Now, **[dy/dx] _{θ=π/3} = cot π/3 = 1/√3**

Question 50 :

If y = sin (sinx), then prove that **d ^{2}y/dx^{2} + tan x dy/x + y cos^{2} x = 0.**

Answer / Solution :

Given, y = sin(sin x). Differentiating with respect to x we get

**dy/dx = cos (sin x) • cos x **...(i)

Again differentiating with respect to x we get

d^{2}y/dx^{2} = d/dx (cos (sin x) • cos x)

= cos (sin x) d/dx (cos x) + cos x d/dx (cos (sin x))

= cos (sin x) (− sinx) + cos x (− sin (sin x) • cos x)

= cos (sin x) (− sinx) + cos x (− sin (sin x) • cos x)

**d ^{2}y/dx^{2 }= − sinx cos (sin x) − cos^{2} x sin (sin x)** ...(i)

Now, find LHS → d^{2}y/dx^{2} + tan x dy/x + y cos^{2} x

= − sinx cos (sin x) − cos^{2} x sin (sin x) + tan x cos (sin x) • cos x + y cos^{2} x

= − sinx cos (sin x) − cos^{2} x sin (sin x) + tan x cos (sin x) • cos x + sin(sin x) cos^{2} x

= − sinx cos (sin x) − cos^{2} x sin (sin x) + sin x cos (sin x) + cos^{2} x sin(sin x)

= 0

Question 51 :

Using the Integration, find the area of the region in the first quadrant enclosed by the X-axis, the line y = x and the circle x^{2} + y^{2} = 32.

Answer / Solution :

×Given curve is **x ^{2} + y^{2} = 32 ⇒ x^{2} + y^{2} = (4√2)^{2}**. It is a cricle with center is (0, 0) and radius is 4√2. Also given line is y = x. Solving both the given equations, we get

x^{2} + y^{2} = 32 ⇒ x^{2} + x^{2} = 32 ⇒ 2x^{2} = 16 ⇒ x^{2} = 16 ⇒ x = ± 4

Therefore, the Point of Intersection is (4, 4).

Now, **Required Area = Area OMA**

= Area OMP + Area MPA

= ∫_{0}^{4} x dx + ∫_{0}^{4√2} √ [ (4√2)^{2} − x^{2}] dx

= [x^{2}/2]_{0}^{4 }+ [ x/2 ( (4√2)^{2} − x^{2}) ) + (4√2)^{2}/2 sin^{−1} (x/4√2)]_{0}^{4}

= 16/2 + [{4√2/2 ( (4√2)^{2} − (4√2)^{2}) + 32/2 sin^{−1} 1 } − {4/2 ( (4√2)^{2} − (4)^{2}) + 32/2 sin^{−1} 1/√2 }]

= 8 + (2√2 (0) + 16 × π/2) − (2 × 4 + 16 × π/4)

= 8 + 8π − 8 − 4π

= 4π square units

Question 52 :

Find the distance of the points (−1, −5, −10) from the point of intersection of the line = (2 − + 2) + λ(3 + 4 + 2) and the plane ( − + ) = 5.

(6 Marks)
Answer / Solution :

Equation of the line is = (2 − + 2) + λ(3 + 4 + 2) ....(i)

Coordinates of any point on this line are (2 + 3λ) + (−1 + 4λ) + (2 + 2λ)

Given equation of the plane is ( − + ) = 5 .....(ii)

Since the point on line lies on the plane ( − + ) = 5, Therefore,

[(2 + 3λ) + (−1 + 4λ) + (2 + 2λ) ] ( − + ) = 5

⇒ (2 + 3λ) − (−1 + 4λ) + (2 + 2λ) = 5

⇒ 2 + 1 + 2 + 3λ − 4λ + 2λ = 5

⇒ 5 + λ = 5

⇒ λ = 0

So substitute the value λ = 0 in equation (i), we have

= (2 − + 2) + 0 (3 + 4 + 2)

⇒ = (2 − + 2) .... (iii)

Let point of intersection be (x, y, z). Therefore (x + y + z) ......(iv)

Comparing equations (iii) and (iv) we get x = 2, y = −1 and z = 2. Thererfore, point of intersection is (2, −1, 2) and (−1, −5, −10) is

= √ [(−1 −2)^{2} + (−5 + 1)^{2} + (−10 − 2)^{2}]

= √ [(−3)^{2} + (−4)^{2} + (−12)^{2}]

= √ [9 + 16 + 144]

= √169

= 13 units

Question 53 :

Let A = {x ∈ Z : 0 ≤ x ≤ 12}. Show that R { (a, b) : a, b ∈ A, |a − b| is divisible by 4} is an equialence relation. Find the set of all elements related to 1. Also write the equivalence class [2].

(6 Marks)
Answer / Solution :

Given R { (a, b) : a, b ∈ A, |a − b| is divisible by 4}

**⇒ Reflexivity : **

for any a ∈ A

|a − a| = 0, which is divisible by 4

(a, a) ∈ R

So R is reflexive.

**⇒ Symmertry : **

for any (a, b) ∈ R

⇒ |a − b| is divisible by 4

⇒ |b − a| is divisible by 4

Because, |a − b| = |b − a|

⇒ (b, a) ∈ R

So R is symmetric.

**⇒ Transitive : **

Let (a, b) ∈ R and (b, c) ∈ R

⇒ |a − b| is divisible by 4

⇒ |a − b| = 4k

⇒ a − b = ±4k, where k ∈ Z

Also

⇒ |b − c| is divisible by 4

⇒ |b − c| = 4m

⇒ b − c = ±4q, where q ∈ Z

Adding both the equations, we get

a − b + b − c = ±4 (k + q)

⇒ a − c = ±4(k + q)

⇒ |a − c| is divisible by 4

⇒ (a, c) ∈ R

So R is transitive.

⇒ __R is reflexive, symmetric and transitive. Therefore, R is an equivalence relation.__

Now again, let x be an element of R such that (x, 1) ∈ R. Then |x − 1| is divisible by 4. i.e.

x − 1 = 0, 4, 8, 16 ...

x = 1, 5, 9 (since x ≤ 12)

Therefore, set of all the elements of A which are related to 1 are {1,5, 9}.

Equivalance class of 2 i.e.

[2] = {(a, 2) : a ∈ A. |a − 2| is divisible by 4}

⇒ |a − 2| = 4k (where k is a whole number, k ≤ 3)

⇒ a = 2, 6, 10

Therefore. equivalence class [2] is [2, 6, 10]

Question 54 :

If θ is the angle between two vectors − 2 + 3 and 3 − 2 + , find sin θ.

(2 Marks)
Answer / Solution :

Let = − 2 + 3 and = 3 − 2 +

We know that, • = | | | | cos θ

⇒ cos θ = • / | | | |

= ( − 2 + 3) • (3 − 2 + ) / | − 2 + 3 | | 3 − 2 + |

= 1(3) + (−2)(−2) + 3(1) /√(1+4+9) √(9+4+1)

= 3 + 4 + 3 / √(14) √(14)

= 10 /14 = 5/7

Now sin θ = √(1 − sin^{2} θ)

= √ [1 − (5/7)^{2}]

= √ [1 − 25/49]

= √ [24/49]

**sin θ = 2√6 / 7**

Question 55 :

Differentiate **tan ^{−1} [(1 + cos x)/sin x] **with respect to x.

Answer / Solution :

Let y = tan^{−1} [(1 + cos x)/sin x]

= tan^{−1}(2cos^{2} x/2 / 2sinx/2 cos x2)

= tan^{−1}(cos x/2 / sinx/2)

= tan^{−1}(cot x/2)

= tan^{−1}(tan (π/2 − x/2))

y = π/2 − x/2

Now, differentiating with respect to x, we get

dy/dx = d/dx (π/2) − d/dx (x/2) = 0 − 1/2

**dy/dx = −1/2**

Question 56 :

Find : ∫ (sin^{2} x − cos^{2} x)/sin x cos x dx.

Answer / Solution :

∫ (sin^{2} x − cos^{2} x)/sin x cos x dx

= − ∫ (cos^{2} x − sin^{2} x)/sin x cos x dx

= − ∫ cos 2x / sin 2x dx

= − ∫ cot 2x dx

= − log sin 2x + C

Question 57 :

Solve the integration : ∫ (sin^{3} x + cos^{3} x)/ sin^{2} x cos^{2 }x dx.

Answer / Solution :

∫ (sin^{3} x + cos^{3} x)/ sin^{2} x cos^{2 }x dx

= ∫ sin^{3} x / sin^{2} x cos^{2 }x dx + ∫ cos^{3} x/sin^{2} x cos^{2 }x dx

= ∫ sin x / cos^{2 }x dx + ∫ cos x/sin^{2} x dx

= ∫ tan x sec x dx + ∫ cot x cosec x dx

= sec x − cosec x + C

[ ∫ tan x sec x dx = sec x and ∫ cot x cosec x dx = − cosec x ]

Question 58 :

Using method of integration, find the area of the triangle whose vertices are (1, 0), (2, 2) and (3, 1).

(6 Marks)
Answer / Solution :

Let A(1, 0), B(2, 2) and C(3, 1) be the vertices of a triangle.

**Area of Triangle ABC = Area of Triangle ABD + Area of Trapezium BDEC − Area of Triangle AEC**

Now, first we find the equation of side AB using the formula y − y_{1} = m (x − x_{1}) where m = (y_{2} − y_{1}) / (x_{2} − x_{1})

m = (2 − 0) / (2 − 1) = 2/1 = 1

⇒ y − 0 = 2 (x − 1)

**⇒ y = 2(x − 1)**

Similarly, Equation of line BC

m = (1 − 2) / (3 − 2) = −1/1 = −1

⇒ y − 2 = −1 (x − 2)

⇒ y − 2 = −(x − 2)

**⇒ y = 4 − x**

Similarly, Equation of line AC

m = (1 − 0) / (3 − 1) = 1/2

⇒ y − 0 = 1/2 − (x − 1)

**⇒ y = 1/2 (x − 1)**

Hence, area of a Triangle ABC

= ∫_{0}^{2} 2(x − 1) dx − ∫_{1}^{3} 1/2 (x − 1) dx + ∫_{2}^{3} (4 − x) dx

= 2[x^{2}/2 − x]_{0}^{2} − 1/2 [x^{2}/2 − x]_{1}^{3} + [4x − x^{2}/2]_{2}^{3}

= 2[ (2^{2}/2 − 2) − (0^{2}/2 − 2)] − 1/2 [(3^{2}/2 − 3) − (1^{2}/2 − 1)] + [(4 × 3 − 3^{2}/2) − (4 × 2 − 2^{2}/2)]

= 2[ 0− (− 2)] − 1/2 [(9/2 − 3) − (1/2 − 1)] + [(12 − 9/2) − (6)]

= 3/2 square units.

Question 59 :

Using the method of integration, find the area of the region enclosed between two circles x^{2} + y^{2} = 4 and (x − 2)^{2} + y^{2} = 4.

Answer / Solution :

Given equation of circles are

x^{2} + y^{2} = 4 ...... (i)

(x − 2)^{2} + y^{2} = 4 ......(ii)

Equation x^{2} + y^{2} = 4 is a circle with center O at the origin (0, 0) and radius is 2 units and equation (x − 2)^{2} + y^{2} = 4 is a ciecle with center is (2, 0) and radius is 2 units.

On Solving both the equations, we get

(x − 2)^{2} + y^{2} = x^{2} + y^{2}

⇒ x^{2} − 4x + 4 + y^{2}= x^{2} + y^{2}

⇒ x = 1 which gives y = ±√3

Thus, the points of intersection of the given circle are **A(1, √3) and A' (1, −√3).**

Therefore, required area of the enclosed region in the figure is OACA'O between these two circles.

= 2[Area of Region ODCAO]

= 2[Area of Region ODAO + Area of Region OCAD]

= 2[∫_{0}^{1} ydx + ∫_{1}^{2} ydx ]

= 2[∫_{0}^{1} √(4 − (x − 2)^{2 }) dx + ∫_{1}^{2} √(4 − x^{2}) dx ]

= 2[1/2 (x − 2) √ (4 − (x − 2)^{2 }) + 1/2 × 4 sin^{−1} x ( (x − 2 )/ 2) ]_{0}^{1 }+ 2 [ 1/2 x √ (4 − x^{2}) + 1/2 × 4 sin^{−1} x/2]_{1}^{2}

= [(x − 2) √(4 − (x − 2)^{2 }) + 4 sin^{−1} x ((x − 2 )/ 2) ]_{0}^{1 }+ [ x √(4 − x^{2}) + 4 sin^{−1} x/2]_{1}^{2}

= [−√3 + 4 sin^{−1} (−1/2) − 4 sin^{−1} (−1)] + [ 4 sin^{−1} 1 − √3 −4 sin^{−1} 1/2]

= [(−√3 + 4 × π/6) + 4 × π/2] + [ 4 × π/2 −√3 −4 × π/6]

= 8π/3 − 2√3

**Area of the enclosed figure OACA'O = 8π/3 − 2√3**

Question 60 :

Differentiate e^{√(3x)}, with respect to x.

Answer / Solution :

y = e^{√(3x)}

On differentiating with respect to x, we get

dy/dx = d/dx (e^{√(3x)})

= e^{√(3x) }d/dx √(3x)

= e^{√(3x) }√3 d/dx √x

**= √3//2 × e ^{√(3x)}/√x**

Question 61 :

If (a + bx) e^{y/x} = x, then prove that x^{3} d^{2}y/dx^{2} = (x dy/dx − y)^{2}

Answer / Solution :

Given, (a + bx) e^{y/x} = x

⇒ e^{y/x} = x/(a + bx)

⇒ y/x = log [ x/ (a + bx)]

⇒ y/x = log [ x/ (a + bx)]

⇒ y/x = log x − log (a + bx)

Differertiating with respect to x, we get

⇒ (x dy/dx − y)/x^{2} = 1/x − b/(a + bx)

⇒ x dy/dx − y = x^{2} (1/x − b/(a + bx))

⇒ x dy/dx − y = x − bx^{2}/(a + bx)

⇒ x dy/dx − y = ax/(a + bx) ...(i)

Again differentiating with respect to x, we get

⇒ d/dx (x dy/dx − y) = d/dx (ax/(a + bx))

⇒ x d^{2}y/dx^{2} + dy/dx − dy/dx = [(a + bx)a − ax × b] / (a+ bx)^{2}

⇒ x d^{2}y/dx^{2} = [(a^{2} + abx − abx] / (a+ bx)^{2}

⇒ x d^{2}y/dx^{2} = a^{2}/(a+ bx)^{2}

On multiplying x^{2} both the sides, we get

⇒ x^{3} d^{2}y/dx^{2} = a^{2}x^{2}/(a+ bx)^{2}

⇒ **x ^{3} d^{2}y/dx^{2} = [ax/(a+ bx) ]^{2}**

Question 62 :

Solve for x : tan^{−1} (x + 1) + tan^{−1} (x − 1) = tan^{−1}(8/31).

Answer / Solution :

Given, tan^{−1} (x + 1) + tan^{−1} (x − 1) = tan^{−1}(8/31)

We know that **tan ^{−1 }x + tan^{−1 }y = tan^{−1 }(x+y / 1 − xy)**.

Therefore,

tan^{−1} (x + 1) + tan^{−1} (x − 1) = tan^{−1} [(x + 1) + (x − 1) / (1 − (x + 1)(x − 1))

= tan^{−1} [2x / (1 − (x^{2} − 1))

= tan^{−1} [2x / (2 − x^{2}))

Now, tan^{−1} [2x / (2 − x^{2})) = tan^{−1} (8/31)

⇒ 2x/(2 − x^{2}) = 8/31

⇒ 62x = 8(2 − x^{2})

⇒ 31x = 8 − 4x^{2}

⇒ 4x^{2} + 31x − 8 = 0

⇒ 4x(x + 8) − 1(x + 8) = 0

⇒ (4x − 1) (x + 8) = 0

⇒ x = 1/4, −8

Question 63 :

The magnetic susceptibility χ of a magnesium at 300 K is 1.2 × 10^{5}. At what temperature will its magnetic susceptibility became 1.44 × 10^{5 }?

Answer / Solution :

Given χ_{mg} at 300 K = 1.2 × 10^{5},^{ }χ'_{mg} at t temperature = 1.44 × 10^{5 }, t = ?

From curies law, χ ∝ 1/T

χ'_{mg }/ χ_{mg} = 300/T

⇒ 300/T = 1.44 × 10^{5}_{ }/1.2 × 10^{5}

⇒ T = 300 ×1.2_{ }/1.44

⇒ T = 250 K

Question 64 :

How are electronagnetic waves produced by accelerating charges?

(1 Marks)
Answer / Solution :

An oscillting electric field in space, produces an oscillating magnetic field, which is turn, in a source of oscillting electric field, and so on.... The oscillting electric and magnetic fields thus regenerate each other.

Question 65 :

Draw the ray diagram of an astronomical telescope showing image formation in the normal adjustment position. Write the expression for its magnifying power.

Also Draw a labeled ray diagram to show image formation by a compound microscope and write the expression for its resolving power.

(2 Marks)
Answer / Solution :

The magnifying power m is the ratio of the angle β subtended at the eye by the final image to the angle α which the object subtends at the lens or the eye. Hence

**m = angle β/ / angle α**

m = h/f_{e} • f_{0}/h = f_{0}/f_{e}

The resolving power of microscope is the reciprocal of the maximum distance. Therefore, we have

**R.P = 1/d _{min} = 2n sin β / 1.22 λ**

Question 66 :

A capacitor (C) and register (R) are connected in the series with an AC source of voltage of frequency 50 Hz. The potential difference across C and R are respectively 120 volt and 90 volt, and the current in the circuit is 3 ampere. Calculate

(i) the impedance of the circuit

(ii) the value of the inductance which when connected in series with C and R will make the power factor of the circuit unity.

(3 Marks)
Answer / Solution :

Given that V_{C} = 120, V_{R} = 90 V, I = 3A and we find Z = ?

(i) From Kirchhoff's law,

V = V_{C} + V_{R}

= 230 V

I = V/Z ⇒ Z = V/I = 230/3 = 76.67 Ω

(ii)

cos φ = R/Z = 1

R = Z = R + j (ωL − 1/ωC)

Hence L = 1/C

Question 67 :

The figure shows a series LCR circuit connected to a variable frequency 230 volt source.

i) Determine the source frequency which derives the circuit in resonance.

ii) Calculate the impedance of the circuit and amplitude of current at resonance.

iii) Shows that potential drop across LC combination is zero at resonating frequency.

(3 Marks)
Answer / Solution :

i) Source frequency will be same as resonance frequency of LC circuit.

f_{R} = 1 / 2π√LC

= 1 / 2π √(400 × 10^{−6})

= 1 / 2π × 2 × 10^{−2}

= 100 / 4π = 7.957 Hz [Because ω = 2πf]

ω_{r} = 50 Hz

(ii) Impedance of a circuit

z_{i} = RωL + 1/jωC

= 40 + 50j × 5 + 1/ ( j × 50 × 80 × 10^{−6})

= 40 + 250j + 10^{3}/ 4j

= 40 + 250j − 250j (at resonance)

z_{i} = 40Ω

Now, Amplitude of a current is

I = 230/z_{i} = 230/40 = 5.75 A

iii) As at the resonance frequency impedance of combination of L and C is 0. Hence, voltage drop across LC combination is zero at resonating frequency.

Question 68 :

Prove that the magnetic moment of the electron revolving around a nucleus in an orbit of radius r with orbital speed V is equal to evr/2. Hence using Bohr's postulate of quantization of angular momentum, deduce the expression for the magnetic movement of hydrogen atom in the ground state.

(3 Marks)
Answer / Solution :

µ = − (e/2m) × L where, negative sign indicates µ direction is opposite to L. As Bohr's atomic model, L = mvr. Therefore,

µ = − (e/2m) × mvr

**µ = evr/2**

But from Bohr's second postulate,

m_{e}v_{r} = nh/2π × h/2π (for the value of n = 1)

**v _{r} = nh/2π m_{e}**

Hence, the magnetic movement is

m = e/2 × h/2π m_{e} (Here n = 1)

Therefore, **m = eh/4πm _{e}**

Question 69 :

Why it is difficult to detect the presence of an anti-neutrino during β-decay? Define the term decay constant of a radioactive nucleus and derived the expression for its mean life in terms of the decay constant.

(3 Marks)
Answer / Solution :

The symbols and V present anti-neutrino and neutrino respectively during β-decay both the neutral particles with very little or no mass. These particles are emitted from the nucleus along with the electron or positron during the decay process. Neutrons interact very weakly with matter. They can even penetrate the earth without being absorbed. It is for this reason that they are deduction is extremely difficult and their presence went unnoticed for a long.

**Decay constant **

Decay constant of a radioactive element is the reciprocal of time during which the number of atoms left in the sample reduces to 1/e times the number of atoms in the original sample .

**Derivation of mean life**

Let us consider N_{0} the total number of radioactive atoms present initially. After some time t, total number of atoms present undecayed be N, Further dt time dN be the number of atoms distinguished. So the life of dN atoms ranges lies between t + dt and dt. Since dt is very small time, so the most appropriate life of dN atom is t. So the total life of N atom is equal to t • dN.

Sum of edges of all atoms = ∫_{0}^{N0 }t dN .... (i)

N = N_{0} e^{−λt}

dN = N_{0} (−λt)e^{−λt }dt

Now, substituting the values of dN and changing the limits, we get the final result

= ∫_{∞}^{0 }N_{0} (−λt)e^{−λt }dt

= N_{0} λ ∫_{∞}^{0 }t e^{−λt }dt

= Sum of life of all atoms

Now, Mean Life = τ = N_{0} λ ∫_{∞}^{0 }t e^{−λt }dt / N_{0}

τ = λ ∫_{∞}^{0 }t e^{−λt }dt

τ = λ × 1/λ^{2}

**τ = 1/λ**

This expression gives the relationship relation between mean life and a decay constant. Hence, mean life is reciprocal of every decay constant.

Question 70 :

a) State two distinguishing features of a nuclear force and b) Draw a plot showing the variation of potential energy of a pair of nucleons as a function of their separations. Mark the region on the graph where the force is attractive and repulsive.

(3 Marks)
Answer / Solution :

a) Some features of nucleus force are

- It is independent of the charges protons and neutrons i.e. charge independent.
- Nuclear forces are very strong binding forces i.e. attractive forces.
- It depends on the spins of the nucleones.

b) Plot showing variation of potential energy of a pair of nucleons as a function of separation mark attractive and repulsive region. In the following figure, the x-axis shows separations between pair of nucleons and the y-axis shows the variation of potential energy with respect to the separation in 10^{−15} m.

Question 71 :

Explain with the help of diagram, how many polarized light can be produced by scattering of light from the sun? Two polaroids P_{1} and P_{2} are placed with their pass axis perpendicular to each other. Unpolarized light of intensity I is incident on P_{1}. A third polarized P_{3} is kept between P_{1} and P_{2} such that its pass axis makes an angle of 45 degree with that of P_{1}. Calculate the intensity of light transmitted P_{1}, P_{2} and P_{3}.

Answer / Solution :

Molecules behave like a dipole radiators and scatter no energy along the dipole axis by this way plane polarized light can be produced during scattering of light.

Intensity of a light after passing through the polarized P_{1}(I),

**⇒ I _{1} = I_{0}/2**

Intensity of light after passing through P_{3},

**⇒ I _{2 }= I_{1} cos^{2} 45° = I_{1} /2 = I/4**

Intensity of light passing through P_{2}

**⇒ I _{3 }= I_{2} cos^{2} 45° = I/4 × I/2 = I/8**

Question 72 :

a) Why cannot the phenomena of interference be observed by Illuminating to pin holes with two sodium lamps?

b) Two monochromatic waves having displacements y_{1} = a cos ωt and y_{2} = a cos (ωt + φ) from two coherent sources interfere to produce and interference pattern. Derive the expression for the resultant intensity and obtain the conditions for constructive and destructive interference.

c) Two wavelengths of sodium light of 590 nm and 596 nm are used in turn to study the diffraction taking place at a single slit of aperture 2 × 10^{−6}m. if the distance between the slit and the screen is 1.5 M, calculate the separation between the positions of the second maxima of diffraction pattern obtained in the two cases.

Answer / Solution :
**1. Consecutive interference **

**⇒ Destructive interference**

**Solution a) **

Phenomenon of interference can't be observed by iluminating two pin holes with two sodium lamps because these sources are not coherent sources. That means it is they are not in same phase.

**Solution b) **

Consider two monochromatic coherent sources A and B with waves y_{1} = a cos ωt and y_{2} = a cos (ωt + φ) respectively.

From the principle of superposition y = y_{1} + y_{2}

= a sin ωt + b sin (ωt + φ)

= a sin ωt + b sin ωt cos φ + b cos ωt sin φ

= (a + b cos φ) sin ωt + b cos ωt sin φ

let a + b cos φ = A cos ς

b sin φ = A sin ς

⇒ y = A sin ωt cos ς + A cos ωt sin ς

⇒ y = A sin (ωt + ς)

A = √(a^{2} + b^{2} + 2ab cos φ)

tan ς = b sin φ / (a + b cos φ)

For I - Maxima, I ∝ A^{2 }and for A to be maximum cos φ = 1 i.e.

cos φ = cos 2nπ, n = 1, 2, 3, 4, ...

φ = cos 2nπ

and the differene is Δx = nλ, A_{max} = a + b and I → I_{max} = k(a + b)^{2}

For I - Minima cos φ = −1, Δφ = (2n + 1)π,

Path difference is Δx = (2n + 1)λ/2, A_{min} = a − b, I → I_{min} = k(a − b)^{2}

**Solution c) **

θ = (n + 1/2) λ/a, where a = 2 × 10^{−6}

θ_{1} = λ/2a = (590 × 10^{−9}) / (4 × 10^{−6}) = 147.5 × 10^{−3}

θ'_{1} = λ'/2a = (596 × 10^{−9}) / (4 × 10^{−6}) = 149 × 10^{−3}

θ_{2 }−_{ }θ_{1 }= 1.5 × 10^{−3}

λ_{1} = 596 nm, λ_{2} = 590 nm, a = 2 × 10^{−3}, D = 1.5 m, y = 3λD/2a

y_{1} − y_{2} = 3D/2a (λ_{1} − λ_{2})

= (3 × 1.5) / (2 × 2 × 10^{6}) × [596 − 590] × 10^{−9} m

= 4.5 / 4 × 10^{6} × 6 × 10^{−9} m

= 6.78 nm

Question 73 :

a) State the working principle of a meter Bridge used to measure an unknown resistance.

b) Give reasons

(i) Why the connections between the registers in a meter bridge are made of thick copper strips.

(ii) Why is it generally preferred to obtain the balance length near the midpoint of the bridge wire.

(iii) Calculate the potential difference across the 4 Ohm resistor in the given electrical circuit using kirchhoff's law.

(6 Marks)

Answer / Solution :

**Solution a: **

Meter bridge is the practical apparatus working which works on principle of Wheat-Stone bridge. It is used to measure unknown resistance experentally.

Hence, as per Wheat Stone Bridge balance condition

R_{1}/R_{2} = Resistance of wire AD / Resistance of wire DC

where, D is the point of balance.

R_{1}/R_{2} = ρl_{1} / ρl_{2} = l_{1 }/ l_{2} = l_{1}/(100 − l_{1}) (bacause l_{1 }+_{ }l_{2 }= 100 cm

ρ = Resistance per unit length of wire

R_{1} = R_{2} l_{1 }/ (100 − l_{1})

**Solution b) **

(ii) It is preferred to obtain the balance length near the midpoint of the bridge wire because it increased the sensitivity of meter bridges.

(ii) Connection between registers are made of thick copper is strips so that it will have maximum resistance and location of point of balance D will be more accurate which results in correct measurements of unknown resistance.

**Solution c)**

From KCl (kirchhoff's current law) at point D

Current flowing through 4 Ohm resistance is i = i_{1} + i_{2}

kirchhoff's current law in loop DEFCD is ∑V = 0 i.e

−4(i_{1} + i_{2}) − i_{2} + 6 = 0

−4i_{1} − 5i_{2} + 6 = 0 (Voltage drop is negative and gain his positive)

**4i _{1} + 5i_{2} = 6 ...(i)**

kirchhoff's current law in loop AE FBA is ∑V = 0 i.e.

−4(i_{1} + i_{2}) − 2i_{1} + 8 = 0

−6i_{1} − 4i_{2} + 8 = 0

**6i _{1} + 4i_{2} = 8 ...(2)**

After solving both the equations, we get i_{1} = 8/7 and i_{2} = 2/7

Question 74 :

a) Derive an expression foe the induced emf (electro magnetic field) developed when a coil of n turns, and area of cross-section A, is rotated at a constant angular speed ω in a uniform magnetic field B.

b) A wheel with hundred metallic spokes each 0.5 meter long is rotated with a speed of 120 rev/min in a plane normal to the horizontal component of the Earth's magnetic field. If the resultant magnetic field at that place is 4 × 10^{−4} T and the angle of dip at the place is 30 degree, find the emf induced between the axle and the rim of the wheel.

Answer / Solution :

**Solution a): **

As the armature coil is rotated in the magnetic field, angle θ between the field and normal to the coil changes continuously. Therefore, magnetic flux linked with the coil changes. An emf is induced in the coil. According to Fleming's right-hand rule current induced in AB is from A to B, and it is from C to D in CD in the external circuit current flows from B_{2} to B_{1}.

**To to calculate the magnitude of EMF induced : **

Suppose A is area of each turn of the coil, N is the Number of turns in the coil, B is the trength of magnetic field and θ is the angle which normal to the coil makes.

Therefore, magnetic flux linked with the coil in this position.

φ = N (•) = NBA cos θ = NBA cos ωt ...(i) where ω is the angular velocity of the coil as the coil.

As the coil rotates, angle θ changes. Therefore magnetic flux φ linked of with the coil changes and therefore the emf is induced in the coil. At this instant t, if e is the emf induced in the coil, then

e = − d/dt (θ) = − d/dt (NBA cos ωt)

= − NBA d/dt (cos ωt)

= − NBA (−sin ωt) ω

Therefore, e = NAB ω sin ωt

e = ε_{0} sin ωt (Here ε_{0} = NAB ω)

**Solution b) **

We have,

Number of spokes = N = 100

Length of each spoke = L = 0.5 m

Magnetic field = B = 0.4 × 10^{−4} = 4 × 10^{−5} T

Frequency f = 120 rpm = 2 rps

Induced EMF between axle and rim his given by

e = N × B × l^{2} × π × f

= 100 × 4 × 10^{−5 }× (0.5)^{2} × 3.14 × 2

**e = 6.28 × 10 ^{−3} V **

Question 75 :

a) Describe briefly, with the help of a circuit diagram, the method of measuring the internal resistance of a cell.

b) Give reasons why potentiometer is preferred over a voltmeter for the measurement of EMF of a cell?

c) In the potentiometer circuit given below, calculate the balancing length l. Give reason, whether the circuit will work, if the driver cell of EMF 5 volt is replaced with a cell of 2 volt keeping all other factors constant.

(5 Marks)
Answer / Solution :

Then K_{2} is open, balance is obtained at l_{1} = AN_{1}, ε = φl_{1}, when k_{2} is closed V = φl_{2.}

**(Circuit for determining Internal Resistance of a Cell) **

ε/V = l_{1 }/ l_{2}

But ε = I (r + R) and V = IR

ε/V = r + R / R = l_{1 }/ l_{2}

**r = R (l _{1 }/ l_{2 }− 1)**

where, r is the internal resistance of the cell.

**Solution b) **

Potentiometer is preferred over voltmeter for measurement of EMF of cell because a voltmeter draws some current from the cell where potentiometer draws no current. Therefore the potentiometer measures the actual EMF of the cell whereas voltmeter measures the terminal voltage.

**Solution c) **

For current through a AB

I = V/R = 5/500 = 10 mA

10 mA current passes through AB. Thus, voltage drop through AB = V_{AB }= 10 × 10^{−3} × 50 = 500 mV

Voltage drop per unit length = 500mV / 10 = 50 mV/m.

Balancing point is at 300 mV. Hence l will be 6 m.

When 5V is replaced with 2V, then

I_{AB} = 2/500 = 4mA

V_{AB} = 4 mA × 50 = 200 mV

Therefore, balancing will not possible as it needs to cater 300 mV.

Question 76 :

A triangular prism of refracting angle 60 degree is made of a transparent material of refractive index 2/√3. A ray of light is incident normally on the face KL as shown in the figure. Trace the path of the ray as it passes through the prism and calculate the angle of emergence and angle of deviation.

(4 Marks)
Answer / Solution :

From diagram, it is clear that incidence angle at face KM is 60 degree. Therefore,

sin C = 1/µ = 1/ 2/√3 = √3/2

Hence, critical angle is also 60 degree. Therefore incident light ray will not emerge from KM face due to total internal reflection at the face. So, it will move along face KM. Angle of emergence = 90 degree therefore, Angle of deviation = 30 degree.

Question 77 :
Answer / Solution :

How is the equation for Ampere's circuital law modified in the presence of displacement current? Explain them?

(2 Marks)From Ampere's Law, ∮ BdI = μ_{0}i(t). Let the **case 1**, where a point P is considered outside the capacitor charging. From the Ampere's Law magnetic field at Point P will be :

**CASE 1**

B•(2πr) – μ_{0} i(t) ⇒ B = μ_{0} i(t) / 2πr

Now, take another case 2, where shape of the surface under consideration covers capacitor's plate as we consider there is no current through capacitor then this value of B will be zero.

**CASE 2**

Hence, there is a contradiction. Therefore, this Ampere's law was modified with addition of displacement current inside the capacitor.

φ_{E} = | E | A = 1/_{ }• Q/A • A = Q/ε_{0}

dφ_{E}/dt = 1/ε_{0} • dQ/dt

ε_{0 }dφ_{E}/dt = dQ/dt = i_{d}, where i_{d} is the displacement current.

During charging of capacitor, outside the capacitor, i_{c} (conduction current) flows and inside i_{d} (displacement current) flows.

i = i_{c} + i_{d} = ic + ε_{0}dφ_{E}/dt

outside the capacitor, i_{d} = 0, hence, i = i_{c} and inside the capacitor, i_{c} = 0, hence i = i_{d}. Thus, i_{c} = i_{d} as capacitor gets charged.

Question 78 :

a) How is the stability of hydrogen atom in Bohr model explained by de−Broglie's hypothesis?

b) A hydrogen atom initially in the ground state absorbs a photon which excites it to n = 4. When it gets de-excited, find the maximum number of lines which are emitted by the atom. Identify the series to which these lines belong. Which of them has the shortest wavelength?

Answer / Solution :

**Solution a) **

From Bohr's model, an atom has a number of stable orbits in which an electron can reside without the emission of radiant energy. Each orbit correspondent to a certain energy level. Electron revolves in his circular orbit is given by

The motion of an electron in circular orbits is restricted in such a manner that its angular momentum is an integral multiple of h/2π.

Thus, L = mvr = nh/2π

E_{n} = −13.6/r^{2 }• z^{2}eV

Z = 1 for H_{2} atom

E_{n} = −13.6/r^{2 }• eV

From de-broglie hypothesis λ = h/p = h/mv and from Bohr model

⇒ nλ = 2πr

⇒ nh/2π = mvr = L

**Solution b)**

n_{i }= 1, n_{f} = 4 and Possible transitions are: ⇒ 4 → 3, 4 → 2, 4 → 1, ⇒ 3 → 2, 3 → 1 and ⇒ 2 → 1. Also six lines are possible.

**Has a smallest Wavelength :**4 → 1**Paschen Series :**4 → 3**Balmer Series :**4 → 2, and 3 → 2**Lyman Series :**4 → 1, 3 → 1 and 2 → 1

Question 79 :

A hundred µF parallel plate capacitor having plate separation of 4 mm is charged by 200 volt DC. The source is now disconnected. When the distance between the plates is doubled and dielectric slab of thickness 4 mm and dielectric constant 5 is introduced between the plates, how will

- its capacitance
- the electric field between the plates and
- energy density of the capacitor get affected ?

Justify your answer in each case.

(3 Marks)
Answer / Solution :

Given c = 100 µF, d = 4 × 10^{-3 }m, V = 250 V , k = 5 and

Q = CV = 200 × 100 × 10^{−6} = 2 × 10^{−2} coulomb

As dielectric of 4 mm is inserted between the plates of capacitor and the spacing between the plates is doubled then it will both acts as following see in figure A and figure B.

Here, C' will be capacitance with dielectric of 4 mm and 8 mm separation between the plates.

C' = KC = 5 × 100 × 10^{−6 }= 0.5 × 10^{−3} F

(i) Equivalent Capacitance is get by adding 1/C and 1/C' i.e.

1/C_{eq} = 1/C +1/C' = 1 / 100 × 10^{−3} + 1 / 0.5 × 10^{−3}

= 10 × 10^{3} + 2 × 10^{3}

= 12 × 10^{3}

**C _{eq} = 1/12 × 10^{−3} = 8.33 µF**

(ii) Electric field inside dielectric will be

E' = E/K = 50 / (5 × 4 × 10^{−3}) = 1000 V/m

and electric field inside capacitor but out of dielectric area will be E = 50×10^{3} V/m

(iii) Energy density of the capacitor is given by U = Q^{2}/2C

⇒ U' + U = Q^{2}/2C + Q^{2}/2C'

= Q^{2}/2 [1/C + 1/C']

= (2 × 10^{−2} × 2 × 10^{−2})/2 × (12 × 10^{3})

= 2 × 10^{1 }× 12 = 2.4 J

Question 80 :

A Ray of light incident on the face AB of an isosceles triangular prism makes an angle of incidence and (i) the deviates by angle β as shown in the figure. Show that in the position of minimum deviation ∠β = ∠α. Also find out the condition when the refracted ray QR suffers total internal reflection.

(3 Marks)
Answer / Solution :

(i) Condition of minimum deviation is **A = 180° − 2α and µ = sin i / sin(90° − β) **

when r_{1} = r_{2} = r > critical angle

r_{1} + r_{2} = 180° − 2α

2r = 180° − 2α [Because r_{1} = r_{2}]

r = 90° − α

β = 90° − r_{1 }= 90° − (90° − α)_{ }

β = α

__Condition when QR have total internal reflection__

∠QRC ≥ critical angle for the prism

∠180° − β ≥ critical angle or ∠180° − α ≥ critical angle

Because 180° − α = ∠BAC. Therefore, ∠BCA ≥ critical angle

Question 81 :

What is the reason to operate photodiodes in Reverse biases? A ph photo diode has fabricated from a semiconductor with a band gap of range of 2.5 to 2.8 eV. Calculate the range of wavelengths of the radiation which can be detected by the photodiode.

(3 Marks)
Answer / Solution :

Photodiodes are reverse biased for working in photo conductive mode. These reduces the response time because the additional reverse bias increases the width of the depletion layer, which decreases the junction capacitance.

The reverse bias also increases the dark current without much change in the photo current.

Given, W = 2.5 − 2.28 eV

E = hc/λ

λ_{1} = hc/E_{1}

= 6.63 × 10^{−34} × 3 × 10^{8} / 2.5 × 1.6 × 10^{−19}

= 3 × 6.63 × 10^{−7} / (−2.5) × 1.6

**λ _{1} = 497.25 nm**

λ_{2} = hc/E_{2}

= 6.63 × 10^{−34} × 3 × 10^{8} / 2.8 × 1.6 × 10^{−19}

**λ _{2 }= 443.97 nm**

Question 82 :

How are electromagnetic waves produced by oscillating charges? What is the source of the energy associated with the em waves.

(2 Marks)
Answer / Solution :

Oscillating charges are responsible for generation of periodically varying electric field in the space. The oscillating charges generate varying electric current which in turn is responsible for the generation of periodical varying magnetic field. This way the electromagnetic waves are generated.

Question 83 :

State with the help of ray diagram the working principle of optical fibers. Write one important use of optical fibers.

(2 Marks)
Answer / Solution :

Optical fibers works on principle of total internal reflection. When angle of incidence is greater than critical angle then incident rays are totally reflected back in same media. When θ_{i} > θ_{c}, Total internal reflection occurs and if θ_{i} < θ_{c},_{ }refraction occurs. **Application Optical fibers are used for communication due to very high bandwidth of media**.

Question 84 :

The wavelength of light light from the spectral emission line of sodium is 5 90 nm. Find the kinetic energy at which the electron would have the same de-Broglie wavelength.

(2 Marks)
Answer / Solution :

Given, λ = 590 nm = 590 × 10^{−9}, KE = ?

For some de-Broglie wavelength, (λ) = h/mv where h is a planck constant

h/m_{e} × 590 × 10^{−9} = v

KE = 1/2 m_{e }v^{2}

= 1/2 m_{e }h^{2}/m_{e}^{2} × (590 × 10^{−9})^{2}

**KE = 6.91 × 10 ^{−17} J**

Question 85 :

Draw the equipotential surfaces due to an electric dipole. Also derive an expression for the electric field if at a point on its perpendicular bisector.

(3 Marks)

Answer / Solution :

The magnitudes of the current electric field due to the two charges +q and −q given by

E_{+q} = q/4πε_{0} × 1/(r^{2} + a^{2})

E_{−q} = q/4πε_{0} × 1/(r^{2} + a^{2})

⇒ E_{+q} = E_{−q}

The directions of E_{+q} and E_{−q }are shown in the given above figure. The components normal to the dipole axis cancel away. The components along the dipole axis add up. Therefore, Total electric field is

E = −(E_{+q} + E_{−q}) cos θ

E = − 2qa / 4πε_{0} × (r^{2} + a^{2})^{3/2}

In this expression negative sign shows that the field is opposite to . Ar large distance (r >>a), this reduces to

E = − 2qa / 4πε_{0} × r^{3}

= q × 2a

Therefore, E = −/ 4πε_{0 }(r >>a)

Question 86 :

State the underlying principle of a cyclotron. Explain its working with the help of systematic diagram. Obtain the expression for cyclotron frequency.

(3 Marks)
Answer / Solution :

**Principle of a cyclotron : **

It is based on a principle that a positive ion can acquire sufficiently charged energy from a comparatively smaller alternating potential difference by making it to cross the same electric field again and again by making use of a strong magnetic field.

**Working of a cyclotron :**

The positive ions are produced from the source at the center are accelerated by a dee which is at negative potential at that moment. Due to the presence of perpendicular magnetic fields the ion will move in a circular path inside the dees. The magnetic field and the frequency of AC Source are so to that as the ions comes out of a dee. It changes it polarity and the ion is further accelerated and moves with higher velocity along a circular path of greater radius. This phenomena is continued till the ion reaches at the periphery of the dees where an deflecting plates deflects the accekerated ion on the target to be bombarded.

**Expression of cyclotron frequency:**

Suppose a position ion with charge q moving with a velocity v, then qvb = mv^{2}\r.

⇒ r = mv/qb

Therefore angular velocity ω is ω = v/r = qB/m

The time taken by iron in describing a semicircle is t = π/ω = πm/qB

This is a semi periodic time T/2 = t = πm/qB. i.e. T = 2πm/qB, time period of revolution. Therefore, frequency of revolution is **f = 1/T = qB/2πm**. This frequency is called the cyclotron frequency.

Question 87 :

Defined the terms (i) Threshold frequency and (ii) stopping potential in photoelectric effect.

(2 Marks)
Answer / Solution :

**Threshold frequency : **The minimum frequency of incident light which is just capable of ejecting electrons from a metal is called the threshold frequency. It is denoted by v_{0}.

**Stopping potential :** The minimum retarding potential applied to anode of a photoelectric tube which is just capable of stopping photoelectric current is called the stopping potential. It is donated by V_{0 }or V_{s}.

Question 88 :

Distinguish between unpolarized and linearly polarized light.

(1 Marks)
Answer / Solution :

**Unpolarized light :**

The light having way vibration of electric field vector in all possible directions are perpendicular to direction of wave propagation the light is known as unpolarized light.

**Linearly polarized light :**

The light having vibrations of electric field vector in only one direction of perpendicular to the direction of a propagation of light is as plane or linearly polarized light.

Question 89 :

Explain briefly how Rutherford scattering of alpha particle by a target nucleus can provide information on the size of the nucleus. Also show that density of nucleus is independent of its mass number A.

(3 Marks)
Answer / Solution :

In Rutherford's scattering experiment of an alpha particle, it was observed that the fast and heavy alpha particles could be deflected through 180 degree. But only very small number of particles that is 1 in about 8,000 alpha particles are deflected through 180 degree that too from center only.

So by this it was assumed that the size of central part. That is nucleus is about 1/10,000 th of the size of the atom and whole positive charge is concentrated in it.

Consider an atom whose mass number is A and R be the radius of the nucleus. If we neglect the mass of orbital electrons, than mass of the nucleus of the atom of mass number A = A a.m.u.

Mass of nucleus, m = A × 1.66 × 10^{−27} kg

Volume of nucleus V = 4/3 πR^{3} = 4/3 π(1.1 × 10^{−15})^{3 }× A m^{3}

Density of nucleus, ρ is equal to mass of nucleus divides by volume of nucleus. Therefore

ρ = A × 1.66 × 10^{−27} × A / 4/3 π(1.1 × 10^{−15})^{3 }× A

ρ = 2.97 × 10^{17} kg/m^{3}

Therefore, density of nucleus is independent of mass number.

Question 90 :

Two infinitely long straight wire A_{1} and A_{2} carrying currents I and 2I following in the same direction are kept 'd' distance apart. Where should a third straight wire A_{3} carrying current 1.5 I be placed between A_{1} and A_{2} so that it experiences no net force due to A_{1} and A_{2}? Does the net force acting on A_{3} depend on the current following through it?

Answer / Solution :

Let the wire A_{3} placed at a distance x from the wire A_{1} and the distance of A_{3} from A_{2} be (d − x). The current flowing through them is I, 1.5 I and 2 I in A_{1}, A_{3} and A_{2} respectively. Let the current flowing in the wire A_{3 }be the same direction as A_{1} and A_{2}.

Therefore, force between A_{1} and A_{3} is F_{1} = µ_{0}/2π × (I × 1.5 I)/x .....(i)

Force between A_{3} and A_{2} is F_{2} = µ_{0}/2π × (2 I × 1.5 I)/ (d − x) .....(ii)

Since, the net force on A_{3} is zero, therefore F_{1} = F_{2}

⇒ µ_{0}/2π × (I × 1.5 I)/x = µ_{0}/2π × (2 I × 1.5 I)/ (d − x)

⇒ 1.5/x = 3/(d − x)

⇒ 1.5(d − x) = 3x

⇒ 1.5d − 1.5 x = 3x

⇒ 4.5 x = 1.5d

⇒ x = d/3

The wire A_{3} is placed at a distance of d/3 from A_{1} and 2d/3 from A_{2}.

No, at a same distance, the force of the wire A_{3} is independent of the direction of the current. As if current is in opposite direction, then F_{1} and F_{2} will be in opposite direction, but will be in equilibrium position.

Question 91 :

How is the drift velocity in a conductor affected with the raise in temperature.

(1 Marks)
Answer / Solution :

With the raise in temperature the collision of electrons occurs more frequently, so relaxation time decreases and hence drift velocity increases.

v_{d} = eE/m × τ ⇒ v_{d} ∝ τ

Question 92 :

A Charged particle q is moving in the presence of a magnetic field B, which is inclined to an angle 30 degree with the direction of the motion of the particle. Draw the trajectory followed by the particle in the presence of the field and explain how the particle described this path.

(2 Marks)
Answer / Solution :

When a charged particle enter in a magnetic field making an angle 30 degree then velocity component is resolved into 2 components one is v cos θ along the magnetic field and other one other is v sin θ normal to the magnetic field.

As the charged particle moves along xy plane due to velocity component v sine θ, it also advances linearly due to the velocity component v cos θ . As a result, the charged particle will move in a helical path as shown in the figure.

Question 93 :

A deuteron and an alpha particle having some momentum are in turn allowed to pass through a magnetic field B, acting normal to the direction of motion of the particles. Calculate the ratio of the radii of a circular path described by them.

(2 Marks)
Answer / Solution :

Radius of circular path r = mv/bB ⇒ r = p/qB [∵ p = mv]

Momentum of deuteron and alpha particle are same and in same magnetic field. Therefore,

r_{d} = p/q_{d }B = p/eB [∵ q_{d} = q_{α} = e]

r_{α} = p/q_{α }B = p/2eB [∵ q_{α} = 2e ]

Therefore, **r _{d} : r_{α} = 2 : 1**

Question 94 :

When unpolarized light is incident on the interface separating the rarer medium and the denser medium. Brewester angle is found to be 60 degree. Determine the refractive index of the denser medium.

(1 Marks)
Answer / Solution :

According to the Brewester's law

**tan i _{p} = n**, and

Therefore, n = tan 60 ⇒ n = 1.732

Question 95 :

Draw the exponential surface due to an isolated point charge.

(1 Marks)
Answer / Solution :

Question 96 :

a) State the underlying principle of a moving coil galvanometer.

b) Give two reasons to explain why a galvanometer cannot as such be used to measure the value of the current in a given circuit.

c) Define the terms voltage sensitivity and current sensitivity of a galvanometer.

(3 Marks)
Answer / Solution :

a) When a current flows through the conductor coil, a torque acts on it due to the external radical magnetic field. Counter torque due to suspension balances coil after appropriate deflection due to current in the circuit.

b) A galvanometer can be used as such to measure current due to the following two reasons.

- A galvanometer has a finite large resistance and is connected in series in the circuit, so it will increase the resistance of circuit and hence change the value of current in the circuit.
- A galvanometer is a very sensitive device. It gives a full scale deflection for the current of the order of micro ampere, hence if connected as such it will not measure current of the order of ampere.

c) **Voltage sensitivity :** It is defined as the deflection produced in the galvanometer when a unit voltage is applied across it.

**Current sensitivity : **The ratio of deflection produced by the coil φ to the current in the coil is called the current sensitivity. It is the deflection of the meter per unit current.

Question 97 :

Explain with the help of Einstein's photoelectric equation and to observed features in photoelectric effect which cannot be explained by wave theory.

(2 Marks)
Answer / Solution :

Futures of photoelectric equation which can not be explained be wave theory:

- Maximum kinetic energy of the emitted photoelectrons is independent of intensity of incident light.
- A the wave theory could not explain the instantaneous process of photoelectric effect.

Question 98 :

a) Draw equipotential surfaces corresponding to the electric field that uniformly increase in magnitude along with the z-directions.

b) Two charge +q and −q are located at a point (0, 0, −a) and (0, 0, a). What is the electrostatic potential at the points (0, 0, ±z) and (x, y, 0)

(3 Marks)
Answer / Solution :

a)

b) Since, two charges are on Z-axis. Therefore, the electric potential on an arrival position at (0, 0, ±2) is given by

V_{1} = 1 / 4πε_{0} × 2lq / (r^{2} − l^{2})

= 1 / 4πε_{0} × 2aq / (r^{2} − a^{2}) [because l = a and r = ±z ]

Now electric potential at the position (, y, 0) is given by

V = 1 / 4πε_{0} × 2aq cos θ / r^{2} but θ = 90 degree. Therefore,

V_{2} = 0 at the position of (x, y, 0) due to an electric dipole, placed on Z-axis.

Question 99 :

a) Write the relation between Half life and average life of a radioactive nucleus.

b) In a given sample two isotopes A and B is initially present in the ratio of 1 : 2. Their half lives are 60 years and 30 years respectively. How long will it take so that the sample has these isotopes in the ratio 2 : 1.

(3 Marks)
Answer / Solution :

a) Half life period = 0.693/λ = 0.693 τ = 69.3% of average life.

b) We have, N = N_{0}e^{−λt}

For the time be t after which, N_{A}/N_{B} = 2/1 i.e.t_{A} = t_{B} = t. Therefore,

N_{0}e^{−λA t }= N_{0}e^{−λB t}

⇒ (λ_{B} t −^{ }λ_{A} t) = log_{e }2

⇒ (λ_{B} −^{ }λ_{A}) t = log_{e }2

⇒ [log_{e }2/T_{B} −^{ }log_{e }2/T_{A}) t = log_{e }2 [Because λ = log_{e}2 / T]

⇒ [1/30 − 1/60] t = 1

⇒ t = 30 x 60 / 30 = 60 years.

Question 100 :

a) Define the term **self inductanc** of a coil. Write its S.I. unit.

b) A triangular loop of sides A and B carrying current I_{2} is kept at a distance a from and infinitely long straight wire carrying current I_{1} as shown in the figure. Obtain an expression for the resultant force acting on the loop.

Answer / Solution :

**a) Self-Inductance **: Self inductance of a coil is numerically equal to the amount of magnetic flux linked with the coil when and current flows through the coil. The S.I. unit of self inductance is henery (H) or weber per ampere 1 H = 1wb/A.

**b) **The force on the side AB of a rectangle is attractive as the current is flowing in the same direction and on the side CD will be repulsive the current is flowing in the opposite direction with respect to the straight conductor. The resultant magnetic force on sides AD and BC is zero. The side AB is the straight wire. So the net force will be attractive and rectangular loop will move towards the straight wire.

Now, the force between AB and straigjht wire is

F_{2} = µ_{0 }/ 2π × I_{1 }I_{2 }/ (a + b)

Therefore,

F_{net} = F_{1} − F_{2} = µ_{0}/2π × I_{1 }I_{2 }/ a − µ_{0}/2π × I_{1 }I_{2 }/ (a + b)

F_{net} = µ_{0}/2π × I_{1 }I_{2 }[1/a − 1/(a + b)]

F_{net} = µ_{0}/2π × I_{1 }I_{2 }[b/a(a + b)]

**F _{net} = µ_{0}I_{1 }I_{2 }b/2πa(a + b)**

Question 101 :

What is the speeed of light in a denser medium of a polarising angle 30 degree?

(1 Marks)
Answer / Solution :

Given i_{p} = 30°,

By Brewster's law, µ = tan i_{p}. Therefore

⇒ µ = tan 30° = 1/√3

Since, µ = C/v

v = velocity of light in medium = C/µ

= 3 × 10^{8} / 1/√3

= 3√3 × 10^{8} m/s

= 5.196 × 10^{8} m/s

Question 102 :

Why is wave theoty of electromagnetic radiation not able to explain photoelectric effect? How does photon picture resolve this problem?

(2 Marks)
Answer / Solution :

Wave theory cannot explain the following laws of photoelectric effect.

- The instantaneous emission of photo electrons
- Existence of threshold frequency for metal surface
- Kinetic energy of emitted electrons is independent of intensity of light and depends on frequency.

The concept of photon explain that energy is not only emitted and absorbed a in depends energy quanta, but also it propagates through space in definite quanta with the speed of light.

It can explain all the above photoelectric effect, which wave theory cannot explain.

Question 103 :

Define the term Decay constant of a radioactive sample. The rate of disintegration of a given radioactive nucleus is 10,000 disintegrations and 5,000 disintegrations after 20 hour and 30 hour respectively from start. Calculate the half-life and initial number of nuclei at t equal to zero.

(3 Marks)
Answer / Solution :

Decay constant of a radioactive sample is defined as the ratio of its instantaneous rate of disintegration to the number of atoms present at a time.

dN/dt = −λn ⇒ λ = −dN/dt/N

Let the initial number of nuclei be N_{0 }at t = 0, dN/dt = 10000 at t = 20 Hr and dN'/dt = 5000 at t = 30 Hr.

We have, dN/dt = λn

For first case

⇒ 10000 = λN,

⇒ N = 10000/λ and N = N_{0} e^{−λt}

⇒ 10000/λ = N_{0} e^{−20λ }...(i)

For second case

⇒ 5000 = λN',

⇒ N' = 5000/λ and N' = N_{0} e^{−λt}

⇒ 5000/λ = N_{0} e^{−30λ }...(ii)

On dividing equation first by equation second, we get

⇒ e^{−20λ}/e^{−30λ }= 2

Taking log both the sides

⇒ −20λ + 30λ = log 2

⇒ 10λ = 2.302 × 0.3010

⇒ λ = 0.0693

Therefore, Half life is τ = 0.693/0.0693 = 10 Hr

Question 104 :

a) State Gauss law for magnetism. Explain its significance.

b) Write the four important properties of the magnetic field lines due to a bar magnet.

(3 Marks)
Answer / Solution :

**Solution a) Gauss law of magnetism : **

If a closed surface is imagined in a magnetic field the number of lines of force emerging from the surface must be equal to the number entering it. That is, the net magnetic flux out of any closed surface is zero. Gauss law signifies that magnetic monopoles does not exist.

**Solution b)**

- In a bar magnet, each lines of force, starts from a north pole and reaches the south pole externally and then goes from south pole to our north pole internally. Thus magnetic line of force forms a closed loop.
- No two lines of force will never intersect each other.
- In a uniform field, the lines are parallel and equidistant from each other.
- The lines of force are crowded near the poles.

Question 105 :

Write three points of differences between para-magnetic materials, dia-magnetic materials, and ferro-magnetic materials. Give one examples of each.

(3 Marks)
Answer / Solution :

1. **Properties : **State

**Ferro-magnetic Materials : **They are solid.

**Para-magnetic Materials : **They can be solid, liquid or gas**.**

**Dia-magnetic Materials : **They can be solid, liquid or gas**. **

2. **Properties :** Effect of Magnet

**Ferro-magnetic Materials : **Strongly attracted by a magnet ** **

**Para-magnetic Materials : **Weakly attracted by a magnet** **

**Dia-magnetic Materials : **Weakly repelled by a magnet** **

3. **Properties :** Effect of temperature

**Ferro-magnetic Materials : **Above curie point, it becomes a paramagnetic** **

**Para-magnetic Materials : **With the rise of temperature, it becomes a diamagnetic. ** **

**Dia-magnetic Materials : **No effect.

4. **Properties : **Examples

**Ferro-magnetic Materials : **Iron, Nickel, Cobalt** **

**Para-magnetic Materials : **Lithium, Molybdenum, and Magnesium** **

**Dia-magnetic Materials : ** Copper, Silver and Gold

Question 106 :

a) Three photo diodes D_{1} D_{2} and D_{3} are made of semi conductors having band gaps of 2.5 eV, 2 eV and 3 eV respectively. Which of them will not be able to deduct light of wavelength 600 nm.

b) Why photodiodes are required to operate in reverse bias? Explain.

(3 Marks)
Answer / Solution :

**Solution a)** Energy corresponding to the wavelength 600 nm is −

E = hc/λ

= (6 × 10^{−34} × 3 × 10^{8}) / 600 × 10^{−9 }J = 3.3 × 10^{−19}

= 3.3 × 10^{−19} / 1.6 × 10^{−19} eV = 2.06 eV

The photon energy E = 2.06 eV is greater than the band gap for diode D_{2} only. Hence, diode D_{1} and D_{3} will not be able to detect the given wavelength.

**Solution b) **

A photodiode is operated reverse bias because in reverse bias it is easier to observe change in current with change in light intensity.

Question 107 :

a) In the series LCR circuit connected across an AC source of variable frequency. Obtain the expression for its inpedence and draw a plot showing its variation with frequency of the AC source.

b) What is the phase difference between the voltages across inductor and the capacitor at resonance in the LCR circuit.

c) When an inductor is connected to 200 V DC voltage, a current of 1 A flows through it. When the same inductor is connected to a 200 V, 50 Hz, AC source, only 0.5 A current flows. Explain, Why? Also, calculate the self inductance of the inductor.

(5 Marks)
Answer / Solution :

**Solution a) **Consider an alternating emf is connected in series with an inductor, resistance R and capacitance C. Let, E and I be the instantaneous values of e.m.f. and current in the LCR circuit V_{L}, V_{C} and V_{R} be the instantaneous values of voltage across inductor, capacitor and resistor respectively. Then,

V_{L }= I X_{L },_{ }V_{C }= IX_{C } and V_{R }= IR Here, X_{L } = ωL and X_{C } = 1/ωC

In an AC circuit V_{R} and I are in same phase, V_{L} leads I by π/2 and V_{C} lags I by π/2.

From the graph, in right angled ΔOEA.

OE = √(OA^{2} + AE^{2}) = √(OA^{2} + OD^{2})

OE = √[ V_{R}^{2} + (V_{L} − V_{C})^{2 }]

Therefore,

E = √[ (IR)^{2} + (IX_{L} − IX_{C})^{2 }]

E = I √ [ R^{2} + (X_{L} − X_{C})^{2 }]

I = E / √ [ R^{2} + (X_{L} − X_{C})^{2 }]

Therefore, effective resistance or impedance of LCR circuit is

Z = √ [ R^{2} + (X_{L} − X_{C})^{2 }] = √ [ R^{2} + (ωL − 1/ωC)^{2 }]

b) At resonance X_{L} = X_{C} i.e. i X_{L} = i X_{C} or V_{L} = V_{C}

The voltages across inductance and capacitance are equal and have a phase difference of 180 degree at resonance

**Solution c) **Since the reactance of an inductor is zero for DC circuit, but the inductor offers resistance to an AC circuit. Therefore, the current decreases for the same inductor then it is connected from the AC source.

**When inductor is connected in the AC circuit : **

V = 200 V, f = 50 Hz, i = 0.5 A

⇒ i = V/R = V/XL = V/ωL

⇒ 0.5 = 200 / (2π × 50) [ Because ω = 2πf ]

L = 200 / (100 × 0.5 × 3.14) = 3/3.14 = 1.27H

Question 108 :

Draw the diagram of a device which is used to decrease high ac voltage into low ac voltage and state its working principle. Write your sources of energy loss in the device.

b) A small town with a demand of 1200 kV of electric power of 220 V is situated 20 kilometer away from an electric plant generating power at 440 V. The resistance of the two wire line carrying power of 0.5 Ohm per kilometer. The town gets the power from the line through a 4000-220 C step-down transformer at a sub station in the town. Estimate the line power loss in the form of heat.

(5 Marks)
Answer / Solution :

**Solution a) Step Down Transformer **

It is a device used for converting high alternating voltage at low current into low alternating voltage at high current and vice versa. The device works on the principle of mutual induction. That means if the current of magnetic flux linked with a coil changes than an EMF is induced in the other coil. In step-down transformer N_{P} >N_{S }and transformation ratio is less than 1.

Energy losses are :

**Copper losses : **

Due to resistance of winding in primary and secondary coils, some electrical energy is converted into heat energy.

**Flux losses : **

Some of the flux produced in primary coil is not linked up with the secondary coils.

**Hysteresis losses : **

When the iron core is subjected to a cycle of magnetisation the core gets heated up dure to hysteresis known as is known as hysteresis loss.

**Iron Losses : **

The varying magnetic flux produces eddy current in the iron core which leads to the wastage of energy in the form of heat.

**Solution b) **

Length of wire line = 20 × 2 = 40 km

Resistance of wire line r = 40 × 0.5 = 50 Ω

Power to be supplied = 1200 kW = 1200 × 10^{3} W

Voltage at which power supplied = 4000 V

Since, P = VI ⇒ I = P/V

⇒ I = 1200 x 10^{3 }/ 4000 = 300 A

Therefore, line power loss = I^{2} × R

⇒ (300)^{2} × 200 = 36 × 10^{5 }W = 360 kW

Therefore, Line power loss in the form of heat is 360 kW.

Question 109 :

Draw the circuit diagram of a full wave rectifier and explain its working. Also, give the input and output waveforms.

(3 Marks)
Answer / Solution :

For a full wave rectifier, we used two junctions diodes as shown in the figure

During first half cycle of input AC signal the terminal S_{1} is positive relative to S and S_{2} is negative. Then diode D_{1} is in forward biased and diode D_{2} is reverse biased. Therefore, current flow in D_{1 }not in D_{2}. In the next half cycle S_{1 }is negative and S_{2} is positive relative to S. Then D_{1} is in reverse biased and D_{2} is in forward biased. Therefore, current flows in D_{2} not in D_{1}. Thus for input AC signal the output current is a continuous series of unidirectional pulse. The input and output waveforms are shown in the figure below.

Question 110 :

a) Describe any two characteristics features which distinguish between interference and diffraction phenomena. Derive the expression for the intensity at a point of the interference pattern in Young's double slit experiment.

b) In the diffraction due to single slit experiment, the aperture of the slit is 3 mm. If monochromatic light of wavelength 620 nm is incident normally on the slit. Calculate the separation between the first order minima and the third order maxima on one side of the screen. The distance between the slit and the screen is 1.5 m.

(5 Marks)
Answer / Solution :

**Interference**

- It is the result of interaction of light coming from two different wavefronts originating from two coherent sources.
- All the bright fringes are the same intensity.

**Diffraction**

- It is the result of interaction of light come from different parts of same wavefronts.
- The bright fringes are of varying intensity i.e. intensity of bright fringes decreases from central bright fringe on either sides).

A sources of monochromatic light illuminates two narrow slits S_{1} and S_{2}. The two illuminated slits act as the two coherent sources. The two slits is very close to each other and at equal distance from the source. The wavefront S_{1} and S_{2} is spread in all the direction and superpose and produces dark and bright fringe on a screen. Let the displacement of waves from S_{1} and S_{2} at point P on screen at time t is

y_{1} = a_{1} sin ωt

y_{2} = a_{2} sin (ωt + φ)

The resultant displacement at point P is given by y = y_{1} + y_{2}

= a_{1} sin ωt + a_{2} sin (ωt + φ)

= a_{1} sin ωt + a_{2} sin ωt + a_{2} cos φ + a_{2} cos ωt + a_{2} sin φ

= (a_{1} + a_{2 }cos φ) sin ωt + a_{2} sin φ cos ωt .....(i)

Let

a_{1} + a_{2 }cos φ = A cos φ .......(ii)

a_{2 }sin φ = A sin φ .......(ii)

Therefore, equation (i) becomes

y = A cos θ sin ωt + A sin θ cos ωt = A cos (ωt + θ)

**y = A cos (ωt + θ)**, this is the resultant displacement.

Now, squaring and adding both the equation second and third, we get

A^{2} cos^{2} θ + A^{2} sin^{2} θ = (a_{1 }+ a_{2 }cos φ)^{2} + a_{2}^{2}_{ }sin^{2} φ

⇒ A^{2} (cos^{2} θ + sin^{2} θ) = a_{1}^{2}+ a_{2}^{2}_{ }cos^{2} φ + 2 a_{1 }a_{2 }cos φ + a_{2}^{2}_{ }sin^{2} φ

⇒ A^{2} = a_{1}^{2}+ a_{2}^{2}_{ }(cos^{2} φ + sin^{2} φ) + 2 a_{1 }a_{2 }cos φ

⇒ A^{2} = a_{1}^{2}+ a_{2}^{2}_{ }+ 2 a_{1 }a_{2 }cos φ

This is intensity of light is directly proportional to the square of the amplitude. i.e.

I = a_{1}^{2}+ a_{2}^{2}_{ }+ 2 a_{1 }a_{2 }cos φ

This is the expression for intensity at a point of interference pattern.

I = I_{1}+ I_{2}_{ }+ 2√(I_{1 }I_{2}) cos φ

**Solution b)**

Here, λ = 620 nm = 620 × 10^{−9}, a = 3 × 10^{−3} m, D = 1.5 m

Distance of first order minimum from the center.

y_{1} = Dλ/a = 1.5 × 620 × 10^{−9} / 3 × 10^{−3}

y_{1} = 3.1 × 10^{−4} m

Distance of third order Maxima on the same side is given by

y_{2} = 7Dλ/2a = 7× 1.5 × 620 × 10^{−9} / 2 × 3 × 10^{−3}

y_{2} = 10.85 × 10^{−4} m

Separation between them y = y_{2} − y_{1}

= 10.85 × 10^{−4} − 3.1 × 10^{−4} = (10.85 − 3.1) × 10^{−4 }

= 7.75 × 10^{−4 }m

Question 111 :

a) Under what conditions is the phenomena of total internal reflection of light observed? Obtain the relation between the critical angle of incidence and the refractive index of the medium.

b) Three lenses of focal length +10 cm, −10 and +30 cm are arranged coaxially as in the figure given below. Find the position of the final image formed by the combination.

(6 Marks)
Answer / Solution :

Condition for total internal reflection is

- The ray must travel from a denser medium into a rarer medium.
- The angle of incidence in the denser medium must be greater than the critical angle for the pair of media.

**Relation between critical and angle and refractive index : **

When a ray of light travel from denser to rarer medium, the ray bends away from the normal. When the angle of incidence is equal to the critical angle then the refracted ray grazes the surface of separation represented by ray and .

By Snell's law, **sin i/sin r = 1/n**, where i = C, r = 90 degree, and n = refractive index of a denser medium. Therefore,

sin C / sin 90 = 1/n ⇒ n = 1/sin C

**Solution b)**

For first lens, u_{1} = −30, f_{1} = 10 cm. Therefore, using from lens formula, we get

1/f_{1} = 1/v_{1} − 1/u_{1}

⇒ 1/v_{1} = 1/f_{1} + 1/u_{1} = 1/10 − 1/30

⇒ v_{1} = 15 cm

This means that image formed by first lens is at a distance of 15 cm to the right of the first lens. This images serves as the vertical object for second lens. Therefore, for second lens we have

f_{2} = −10 cm, u = 15 − 5 = 10 cm

Therefore, 1/v_{2 }= 1/f_{2} + 1/u_{2} = −1/10 + 1/10 = 0

⇒ v_{2} = ∞

This means that the real image is formed by second lens his infinite distance. This acts as an object for third lens. For third lens f_{3} = 30 cm and u_{3} = ∞

1/v_{3 }= 1/f_{3} + 1/u_{3} = 1/30 + 1/∞ = 1/30

⇒ v_{3} = 30 cm

Final image is formed at a distance of 30 cm to the right of third lens.

Question 112 :

a) Derive an expression for the electric field at any point on the equatorial line of an electric dipole.

b) The Identical point charges, q each are kept 2 m apart in air. A third point charge Q of unknown magnitude and sign is placed on the line joining the charges such that the system remains in equilibrium. Find a position and nature of Q.

(5 Marks)
Answer / Solution :

**Solution a)**

Consider an electric dipole of −q and +q separated by a distance 2a and placed in a free space. Let P be a point on equitorial line of dipole at a distance r from the center of the dipole.

Let E_{A} and E_{B} be the electric field at point P due to charges −q and +q. Then resultant electric field at point P is E = E_{A} and E_{B}, Now,

| E_{A} | = 1/4πε_{0} • q/AP^{2} = 1/4πε_{0} • q/(r^{2} + a^{2})^{2} (Along PA) and

| E_{A} | = 1/4πε_{0} • q/BP^{2} = 1/4πε_{0} • q/(r^{2} + a^{2})^{2} (Along BP)

THus, the resultant intensity is the vector sum of E_{A} and E_{B}. E_{A} and E_{B }can be resolved into two components. The y-components cancel out each other and x-component will add up to give the resultant field. Therefore,

E = E_{A} cos θ + E_{B} cos θ. Now the right triangle ORB

cos θ = OP/BP = a / √(r^{2} + a^{2})

E = 2 × 1/4πε_{0 }× q / (r^{2} + a^{2}) × a / √(r^{2} + a^{2})

E = 2qa / 4πε_{0 }× (r^{2} + a^{2})^{3/2}

Threfore the required expression is **E = 1/4πε _{0 }× p/(r^{2} + a^{2})^{3/2 }[because 2qa = p]**

**Solution b) **

Let the two charges of +q each placed at point A and B at a distance 2 m apart in air.

Suppose the third charge Q which is unknown magnitude and charge is placed at a point O, on the line joining the other two charges such as OA equals x and OB equals 2 − x. For a system to be in equilibrium, net force on each three charges must be equal to zero.

If we assume that charge Q placed at point O is positive, then the force on it at O may be zero. But the force on charge q at point A or B will not be zero. It is because, the forces on a charge q due to the other chrges will act in the same direction.

If the charge Q is negative, then the force is on q due to other two charges will act in opposite direction.

Hence, Q will be negative in nature. For charge −Q to be in equilibrium. Force at charge −q due to the charge +q at point A should be equal and opposite to charge +Q at B

1/4πε_{0 }× Qq/x^{2} = 1/4πε_{0 }• Qq/(2 − x)^{2}

or (2 − x)^{2} = x^{2}

⇒ x = 2 − x

⇒ 2x = 2

⇒ x = 1 m

Therefore for the system to be in equilibrium a charge −Q is placed at a midpoint between the two charges of +q each.

Question 113 :

Describe briefly the process of transferring the charge between the two plates of a parallel plate capacitor when connected to a battery. Derive an expression for the energy stored in a capacitor.

(5 Marks)

Answer / Solution :

When the plates of the parallel plate capacitor is connected to a battery. Then the first insulated metal plates gets the positive charge till its potential become maximum. Then the charge will be leak to surroundings. So the negative charge will be induced on the nearer face of the second plate and the positive charge will be induced on its further plate.

Consider a capacitor of capacitance C. Initial charge and potential difference be zero. Let, a chart Q be given in small steps. Let any instant when charge on capacitor be q, the potential difference between its plates V = q/C.

Now work done in giving an additional charge dq is, dW = V dq = q/C dq

Total work done in giving charge from 0 to Q is,

W = ∫_{0}^{Q} Vdq = ∫_{0}^{Q} q/C dq

= 1/C [q^{2}/2]_{0}^{Q} = 1/C [q^{2}/2 − 0/2] = Q^{2}/2C

W =(CV)^{2}/2C = 1/2 × CV^{2} = 1/2 QV [Q = CV]

Threfore, Electrostatic potential energy U = Q^{2}/2C = 1/2 CV^{2} = 1/2 QV

Question 114 :

A parallel plate capacitor is charged by a battery to a potential difference V. It is disconnected from battery and then connected to another uncharged capacitor of f the same capacitance. Calculate the ratio of the energy is stored in the combination to the initial energy on the single capacitor.

(5 Marks)
Answer / Solution :

Here C_{1} = C, C_{2} = C, V_{1} = V, V_{2} = 0. Now initially, energy stored in first capacitor as second capacitor is uncharged.

U_{1} = 1/2 CV_{1}^{2 }= 1/2 CV^{2} ....(i)

Now, when C_{1} and C_{2} are connected the two capacitor form a parallel combination. Equivalent capacitance,

C' = C_{1} + C_{2 }= C + C = 2C.

Final potential = Total charge / Total capacitance

V ' = −q/2C [Total charge will be of first capacitor, which is distrubuted]

V ' = CV/2C = V/2

Final energy is stored in a combination.

U_{2} = 1/2 C' V'^{2} = 1/2 × 2C × (V/2)^{2} = CV^{2}/4 ...(ii)

On dividing equation first with second, we get

U_{2}/U_{1} = (1/4 CV^{2}) / (1/2 CV^{2}) = 1/2

**U _{2 }:_{ }U_{1} = 1 : 2**

Question 115 :

Out of Chlorobenzene and Cyclohexyl chloride, which one is more reactive towards nucelophilic substitution reaction and Why?

(1 Marks)
Answer / Solution :

Cyclohexyl chloride his more reactive towards nucelophilic substitution reaction, because the carbon bearing the chlorine atom is deficient in electron and seeks a nucleophile. In chlorobenzene the carbon bearing the halogen is a part of aromatic ring and is electron rich due to the electron density in the ring.

Question 116 :
Answer / Solution :

Out of Chlorobenzene and Cyclohexyl chloride, which one is more reactive towards nucelophilic substitution reaction and Why?

(1 Marks)Cyclohexyl chloride his more reactive towards nucelophilic substitution reaction, because the carbon bearing the chlorine atom is deficient in electron and seeks a nucleophile. In chlorobenzene the carbon bearing the halogen is a part of aromatic ring and is electron rich due to the electron density in the ring.

Question 117 :

What is the basic structural difference between starch and sellulose? Write the products obtained after hydrolysis of DNA.

(1 Marks)
Answer / Solution :

Startch consists of two components amylose and amylopectin. Amylose is a long linear chain of α−D−glucose units joined by C_{1}−C_{4} glycosidic linkage (α link). Amylopectin is a branded chain polymer of α−D−glucose units, in which the chain is formed by C_{1}−C_{4 }glycosidic linkage and the branching occurs by C_{1}−C_{6 }glycosidic linkage. On the other hand, cellulose is a straight chain polysaccharide of β−D−glucose units joining by C_{1}−C_{4} glycosidic linkage (β− link).

Hydrolysis of DNA yields a pentose sugar (i.e. β−D−2deoxyribose), phosphoric acid and nitrogen containing heaterocyclic compounds called bases (Adenine, Guanine, Cytosine and Thymine).

Question 118 :

Give reasons: Cooking is faster in pressure cooker than in cooking pan and Red Blood Cells (RBC) shrink when placed in saline water, but swell in distilled water.

(2 Marks)
Answer / Solution :

Boiling point increase on increasing the pressure in case of liquids. Water used for cooking attends higher temperature than usual boiling temperature inside the pressure cooker due to the existing high pressure inside the pressure cooker vessel. This leads so faster flow of water inside the vegetables or grains etc. resulting in faster cooking of food in a pressure cooker then in the cooking pan.

Red Blood Cells shrink when placed in salline water because of exosmosie that is water comes out from the cell of surrounding more concentrated to equate the concentration. Whereas, when placed his distilled water concentration within the cell becomes more than the surroundings, as water comes inside and endosmosis takes place to equate the concentrations.

Question 119 :

a) Although both [NiCl_{4}]^{2–} and ^{[Ni(CO)4] }has sp^{3} hybridisation yet [NiCl_{4}]^{2– }is paramagnetic and [Ni(CO)_{4}]is damagnetic. Give reasons (Use Atomic number of Ni is 28).

b) Write the electronic configuration of d^{5} on the basics of crystal field theory when Δ_{0} < P and Δ_{0} > P.

Answer / Solution :

**Solution a) **

[NiCl_{4}]^{2–} is a high spin complex and there are two unpaired electrons with 3d^{8} electronic configuration of central metal atom, hence it is paramagnetic. Whereas in [Ni(CO)_{4}] Ni is in zero oxidation state and contains no unpaired electrons, hence, it is diamagnetic in nature.

**Ansertr b)**

Electromagnetic electronic configuration of d^{5} Δ_{0} < P is given by t_{2g}^{3} e_{g}^{2 }and Electromagnetic electronic configuration of d^{5} Δ_{0} > P is given by t_{2g}^{5} e_{g}^{2}.

Question 120 :

The following data were obtained for the reaction A + 2B → C

Experiment | [A]/M | [B]/M | Initial rate of formation of C/M min^{–1} |

1 | 0.2 | 0.3 | 4.2 × 10^{–2} |

2 | 0.1 | 0.1 | 6.0 × 10^{–3} |

3 | 0.4 | 0.3 | 1.68 × 10^{–1} |

4 | 0.1 | 0.4 | 2.40 × 10^{–2} |

a) Find the order of reaction with respect to A and B.

b) Write the rate law and overall order of reaction

c) Calculate the rate of constant k.

(3 Marks)
Answer / Solution :

The reaction is a A + 2B → C

**Solution a) **It can be seen that when concentration of A is doubled keeping B is a constant then the rate increases by a factor of 4 (4.2 × 10^{–2 }to 1.68 × 10^{–1}). This indicates that the rate depends on the square of the concentration of the reactant A. Also when concentration of reactant B is made four times, keeping the concentration of reactant A is constant, the reaction rate also becomes four times (2.40 × 10^{–2} to 6.0 × 10^{–3}). This indicates that the rate depends on concentration of reactant B to the first power

**Solution b) **So, the rate equation will be : Rate = k [A]^{2} [B]. Overall order of reaction will be 2 + 1 = 3.

**Solution c) **Rate constant can be calculated by putting the given values in formula, we get

4.2 × 10^{–2 }M min^{–1 }= k (0.2)^{2} (0.3)^{ }M

k = 0.042/0.012 = 3.5 min^{–1}

Question 121 :

a) Write this dispersed and dispersion medium of dust.

b) Why is physisorption reversible whereas chemisorption is irreversible?

c) A colloidal solution is prepared by the method given in the figure. What is the change of AGI colloidal particles formed in the test tube? How is this Solution represented?

(3 Marks)
Answer / Solution :

**Solution a) **In dust the dispersed phase is solid particles and dispersion medium is air means gas.

**Solution b)** Physisorption occurs only because of physical attractive forces like van der Walls forces between molecules of absorbate and absorbent. Therefore, that can be reversed on application of bigger forces, but chemisorption occurs due to chemical reaction between molecules of odsorbate and absorbent, hence it can't be reversed.

**Solution c) **When KI solution is added to AgNO_{3} a positively charged solution results due to absorption of Ag^{+} ions from dispersion medium AGI / Ag^{+ }(positively charged).

Question 122 :

A solution containing 1.9 gram per 100 mL of KCl M = 74.5 g mol^{–1 }is isotonic with a solution containing 3 gram per 100 mL of urea M = 60 mol^{–1}. Calculate the degree of dissocination of KCL solution. Assume that both the solutions have the same temperature.

Answer / Solution :

Two solutions having same as osmotic pressure at a given temperature are called isotonic solution.

Now in the given problem, the KCl and urea solutions are given to be isotonic. Osmotic pressure π is given by the equation

π = (N_{2}/V)RT, where, n_{2} = molecules of solute, V = volume of a solution in litre.

Also n_{2} = w_{2 }/ M_{2, }where w_{2 }= grams of solute and M_{2} = 74.5 g mol^{–1} Weight of KCL , w_{2} = 1.9 g, V = 100 mL.

So, for KCl

π = (w_{2}/M_{2} × V)RT

πRT_{KCL}= 1.9/(74.5 × 100) = 2.55 × 10^{–4}

Now as the solutions are isotonic at the same temperature : π RT_{KCL}= π RT_{UREA}

Hence, substituting the values of urea : 2.55 × 10^{–4 }= 3/M_{2} × 100 ⇒ M_{2} = 117.6

So, the experimentally determine molecular weight of urea is found to be as 117.6, so the degree of this consecration can be given as :

Osmotic pressure (π) = experimentally determined. Molecular weight/ Actual molecular weight = 117.6/60 = 1.96 ≈ 2.

α = i – 1/ n – 1 = 1.96 – 1 / 2 – 1 = 96 %.

So, Urea diamerised in the given experimental solution.

Question 123 :

Write the name and principal of the method used for refining of Zinc, Germanium and Titanium.

(3 Marks)
Answer / Solution :

**Zine : **Distillation is used for refining zinc. As a zinc is a low boiling matel, the impure metal is evaporated and the pure metal is obtained as a distillate.

**Germanium : **Zone referring is used for refining Germanium. This method is based on principle that the impurities are more soluble in the melt than in the solid state of the matter.

**Titanium : **Titanium is refined by van Arkel method. This method is used for removal of oxygen and nitrogen present as impurity. The crude metal is heated in as an evacuated vessel with iodine to obtain metal iodide, which volatilizes being covalent. Later this metal iodide is decomposed through electrical heating to obtain the pure metal.

Question 124 :

Give reasons for the following :

a) Transition metals from complex compounds

b) E^{o }values of (Zn^{2+}/Zn) and (Mn^{2+}/Mn) are more negative then expected.

c) Actinoids show wide range of oxidation states.

(3 Marks)
Answer / Solution :

**Solution a) **Transaction elements have partly filled d-orbitals due to which they have variable oxidation states which enables them to bind with a variety of ligands and hence from complex compounds.

**Solution b) **Oxidation of Zn and Zn^{2+ }leads to a completely filled d^{10} configuration in in Zn^{2+}, making it more stable. Also Mn/Mn^{2+} conversion leads to a half-filled stable d^{5} configuration of Mn^{2+ }ion. Hence, E^{o }value of Zn/Zn^{2} and Zn/Zn^{2+} conversion have negative values.

**Solution c) **Antinodes show wide range of oxidation states due to their partially filled f-orbitals and they have comparable energies as well.

Question 125 :

Define the following terms with suitable example of each : Antibiotics, Antiseptics and Anionic detergent.

(3 Marks)
Answer / Solution :

**Antibiotics:** These are the compounds produced by a microorganism and synthetically which either inhabit the growth of bacteria or kill a bacteria. Example is Penicillin.

**Antiseptics : **These are the chemical used to kill or prevent the growth of microorganism when applied to the living tissues. Example is Soframicun.

**Anionic detergents : **These are sodium salts of sulphonated long-chain alcohols or hydrocarbons. In these the anionic part of the molecule is involved in the cleansing action. Example sodium lauryl sulphate.

Question 126 :

Write balanced chemical equations for the following processes.

- Cl
_{2}is passed through slaked lime. - SO
_{2}gas is passed through an aqueous solution of Fe(III) salt.

Answer / Solution :

a) Cl_{2} is passed through slaked lime to give bleaching powder Ca(OCl)_{2}.

2Ca(OH)_{2} + 2Cl_{2} → Ca(OCl)_{2} + CaCl_{2} + 2H_{2}0

b) When SO_{2} gas is passed through an Fe(III) aqueous solution, Fe(III) is reduced to Fe(II) ion :

2Fe^{3+} + SO_{2} +2H_{2}O → 2Fe_{2}+ +SO_{4}^{2–} +4H^{+}

Question 127 :

Write two poisonous gases prepared from chlorine gas. Why does Cu^{2+} solution gives blue color on reaction with ammonia?

Answer / Solution :

Two poisonous gas prepared from chlorine is phosgene (COCl_{2}) and tear gas (CCl_{3}NO_{2}).

Nitrogen is ammonia has a lone pair of electrons, which makes it a Lewis base. It donates the electron pair and forms linkage with the metal ions.

Cu^{2+} (aq) (Blue) + 4NH_{3}(aq) ⇋ [Cu(NH_{3})_{4}]^{2+} (aq) (Deep Blue)

Question 128 :

a) Pick out the odd one from the following on the basis of their medicinal properties : Equanil, Seconal, Bithional and Luminal

b) What types of detergents are used in dishwashing liquids?

c) Why the use of aspartame limited to cold foods?

(2 Marks)
Answer / Solution :

a) Bithionol is the odd one here, as it is an antiseptic whereas others are tranquilizers.

b) Liquid dishwashing detergent are non-ionic type.

c) Aspartame is an artificial sweetener which is unstable at cooking temperature hence its use is limited to cold foods.

Question 129 :

Give reasons

a) Benzoic acid is stronger acid than acetic acid.

b) Methanal is more reactive toward.s nucleophilic addition reaction than ethanol.

c) Give a simple chemical test to distinguish between propanal and propanone.

(3 Marks)
Answer / Solution :

a) Benzoic acid is a stronger acid than acetic acid because the benzoate anion which is a conjugate base of benzoic acid formed after loss of H+ is stabilized by resonance, whereas acetate ion (CH3COO-) has no much extra stability. Hence, benzoic acid has more tendency of losing portion compared to acetic acid hence more acidic.

b) Methanal is more reactive towards nucleophilic addition reaction than ethanol because in ethanol there is a methyl group attached to the carbonyl carbon (centre for nucleophile attack) and +I effect of the methyl group decreases the nucleophilicity of carbonyl carbon by increasing the electron density at carbonyl carbon.

c) Propanal and propanone can be distinguished using Tollen;s reagent by silver mirror test. Propanal being an aldehyde reacts with Tollen’s reagent to give silver deposition whereas propanone being a ketone does not give the reaction.

Question 130 :

Account for the following :

(ii) Acidic character increases from H_{2}O to H_{2}Te.

(iii) F_{2} is more reactive than CIF_{3} whereas CIF3 is more reactive than cl_{2}.

(iii) Draw the structure of XeF_{2} and H_{4}P_{2}O_{7}

Answer / Solution :

(ii) Acidic character increases from H_{2}O to H_{2}Te due to decrease in E-H bond dissociation enthalpy down the group. THus it becomes easy to lose proton going down the group.

(iii) F_{2} is more reactive than CIF_{3} because of small size of fluorine atom F-F bond, bond dissociation enthalpy is low thus is reactive whereas CIF_{3} is more reactive than Cl_{2} because CIF_{3} is an interhalogen halogen compound with weak CI-F bond means compared to cl-cl bond due to the difference in atomic sizes (hence in ineffective overlap of orbitals).

Structure of XeF_{2} is a linear:

Structure of h_{4}P_{2}O_{7}

Question 131 :

a) Give an example to show the anomalous reaction of fluorine.

b) What is a structural difference between white phosphorus and red phosphorus.

c) what happens when XeF_{6} reacts with NaF?

d) Why his H_{2}S is better reducing agent than H_{2}O?

e) Arrange the following acids in the increasing order of their acidic character : HF, HCL, HBr, and HI

(5 Marks)
Answer / Solution :

a) Fluorine reacts with cold sodium hydroxide solution to give OF_{2}.

2F_{2} (g) + 2 NaOH (aq) → 2NaF (aq) + OF_{2} (g) + H_{2}O (I)

b) The key difference between red and white phosphorus is that the red phosphorus appears as dark red colored crystals whereas the white phosphorus exists as a translucent waxy solid that quickly becomes yellow when exposed to light.

Phosphorus is a chemical element that occurs in several different allotropes. The most common allotropes are red and white forms, and these are solid compounds. Furthermore, when exposed to light, the white form converts into the red form. However, there are several differences between these two allotropes. Let us discuss more details on the difference between red and white phosphorus.

c) XeF_{6} reacts with NaF as follows : XeF_{6} + NaF → Na^{+ }[XeF_{6}]^{–}

d) Ability to reduce is judged by ease with which an atom can donate its electrons to the species which is getting reduced. Now, the size of oxygen atom in H_{2}O is smaller than that of sulphur atom in H_{2}S, due to which the lone pair of electrons on oxygen are more attracted by the oxygen nucleus making it difficult to denote the electrons i.e. by oxygen compound to sulphur, while in H_{2}S the influence of nucleus is less than lone pair of electrons of sulphur and hence, it can give away its electrons easily compared to oxygen and thus acts as a better reducing agent.

e) The increasing order of acidic character can be written as HF < HCL < HBr < HI

Question 132 :

a) The conductivity of 0.001 molecule per liter and acetic acid is 4.95 × 10^{5 }cm^{–1}. Calculator the dissociation constant if ⋀^{0}_{m} for acetic acid is 390.5 S cm2 mol^{–1}.

b) Write Nernst equation for the reaction at 25°C.

Al (s) + 3Cu^{2+} (aq) → 2 Al^{3+} (aq) + 3 Cu (s)

c) What are secondary batteries give an example?

(5 Marks)
Answer / Solution :

a) Conductivity ⋀_{m} of the solution is given by the following equation ⋀_{m} = k/c, where k is dissociation constant and c is the concentration of solution. Here, given

conductivity k = 4.95 × 10^{–5} S cm^{–1}

Limiting molar conductivity ⋀^{0}_{m }= 390.5 S cm^{2 }mol^{–1 }

Concentration c = 0.001 mol L^{–1 }= 1 × 10^{–3 }mol L^{–1}

Substituting the given values in the above equation

Molar conductivity ⋀_{m} = (4.95 × 10^{–5} S cm^{–1} / 1 × 10^{–3 }mol L^{–1}) 10^{3 }cm^{3} L^{–1}

= 49.5 S cm^{2} mol^{–1}

Now, degree of dissociation α is given by α = ⋀_{m} / ⋀^{0}_{m }. Putting the values

α = 49.5 / 390.5 S cm^{2} mol^{–1} = 0.1267

Dissociation constant, k, for acetic acid, can be given as

k = cα^{2} / (1 – α)

= [1 × 10^{–3} mol L^{–1} × (0.1267)^{2} ] / (1 – 0.1267)

= 0.016 × 10^{–3} mol L^{–1} / 0.8733

= 1.8 × 10^{–5} mol L^{–1}

b) Nernst equation for the given reaction can be written as

E_{cell} = E°_{cell} – RT / 2F ln( [Al^{3+}]^{2} / [Al^{2+}]^{3 })

c) A secondary battery can be recharged after use, bypassing current through it in the opposite direction so that it can be used again. For example, the most important secondary cell is the lead storage cell. It consists of a lead anode and the grid of lead packed with lead dioxide as a cathode. A total 38 percent solution of sulfuric acid is used as an electrolyte.

Question 133 :

Define the following terms

- Invert Sugar
- Native Protein
- Nucleotide

Answer / Solution :

**Invert sugar: **It is a mixture of glucose and fructose obtained after hydrolysis of sucrose. Sucrose is dextrorotatory, but after hydrolysis gives a mixture of dextrorotatory glucose and levorotatory fructose which outweighs magnitude and hence the whole mixture becomes levorotatory. Hence, the mixture obtained is called invert sugar.

**Native protein: **Protein found in a biological system with a unique three-dimensional structure and biological activity is called a native protein.

**Nucleotides: **They are the building blocks of **DNA RNA. **These consist of a pentose sugar moiety attached to a nitrogenous base at one position and a phosphoric acid molecule at five position

Example:

Question 134 :

a) What are the products of hydrolysis of maltose?

b) What type of bonding provides stability to the α-helix structure of a protein?

c) Name the vitamin whose deficiency causes pernicious anemia.

(3 Marks)
Answer / Solution :

a) On hydrolysis maltose gives two molecules of α-D-glucose.

b) α-helix structure of proteins is stabilized by hydrogen bonds between -NH group of each amino acid and -COOH group of the amino acid at the adjacent turn.

c) Deficiency of vitamin B12 causes pernicious anemia.

Question 135 :

a) Represent the cell in the following reaction take place:

**2Al (s) + 3 Ni ^{2+} (0.1 M) → 2Al^{3+} (0.01 M) + 3 Ni(s)**.

Calculate its emf if E^{0}_{cell} = 1.41 V.

b) How does molar conductivity vary with an increase in concentration for a strong electrolyte and weak electrolyte? How can you obtain limiting molar conductivity ⋀^{0}_{m }for weak electrolytes?

Answer / Solution :

a) The cell can be represented as Al | Al^{3+} (0.01 M) | Ni^{2+} (0.1) | Ni

The cell potential is given by the following equation in this case

E_{cell} = E^{o}_{cell} – RT / 2F ln [ (Al^{3+})^{2} / (Ni^{2+})^{3}]

Given that concentration of Al^{3+} ions case 0.01 M and Ni^{2+ }ions is 0.1 M, E^{o}_{cell} = 1.41, putting the values in the equation, we get

E_{cell} = 1.41 V – 0.059 / 2log [ (0.01)^{2} / (0.1)^{3}]

E_{cell} = 1.38 V log [ 1 × 10^{-4} / 1 × 10^{-3}]

E_{cell} = 1.38 – (–1) = 1.38 + 1 = 2.38 V

So EMF of the cell is 2.38 volt.

b) For strong electrolytes the molar conductivity is increased only slightly on dilution. A strong electrolyte is completely dissociated in solution and thus furnishes all irons for conductance. However, at higher concentrations, the dissociated ions are close to each other and thus, the interionic attractions are greater. These forces retard the motion of the iron and thus conductivity is low.

With a decrease in concentration dilution, the ions move away from each other thereby feeling less attractive forces from the counterions. This results in an increase in molar conductivity with dilation. The molar conductivity approaches a maximum limiting value at infinite dilution designated as ⋀^{0}_{m}

In the case of weak electrolytes as the solution of a weak electrolyte is diluted, its ionization is increased. This results in more number of ions in the solution and thus, there is an increase in molar conductivity. Also, there is a large increase in the value of molar conductivity with dilution, especially near-infinite dilution. However, the conductance of a weak electrolyte never approaches a limiting value or in other words, we can say that it is not possible to find conductance as infinite dilution. Means zero concentration.

So, limiting molar conductivity for weak electrolytes is obtained by using KOHLRAUSCH LAW from the limiting molar conductivities of individuals ions (λ°).

KOHLRAUSCH LAW of independent migration of ions in states that limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and the cation of the electrolyte.

⋀^{0}_{m }= λ°_{+} + λ°

Question 136 :

a) Give the equation of the following reactions :

- Phenol his treated with concentrate HNO
_{3}. - Propane is treated with B
_{2}H_{6}followed by H_{2}O_{2}/OH^{−}. - Sodium t-butoxide is treated with CH
_{3}Cl.

b) How will you distinguish between butan−1−ol and butan−2−ol?

c) Arrange the following in increasing order of acidity: Phenol, Ethanol, and water.

(5 Marks)
Answer / Solution :

a) i) Phenol is treated with concentrated HNO_{3} to obtain 2,4,6-trinitrophenol picric acid.

ii) Propene undergoes Hydroboration-oxidation when treated with B_{2}H_{6} followed by hydrogen peroxide in basic medium to give propan-1-ol.

iii) Methyl tert-butyl ether is produced when sodium tert-butoxide is treated with methyl chloride.

b) Butan−1−ol and butan−2−ol can be distinguished using Lucas reagent (ZnCl_{2} + HCL), where butan−2−ol would react with Lucas reagent in arround 5 minutes to give a white precipitate of 2-chlorobutene, whereas butan−1−ol would not give any reaction at room temperature.

c) Increasing order of acidity can be given as

**Ethanol < Water < Phenol**

Question 137 :

If

(6 Marks)
Answer / Solution :

So powders.

Question 138 :

What is HTML Language

(3 Marks)
Answer / Solution :

HTML is a web development language in which we prepare the html pages with website desigining

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