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Calcium-Ca,

Carbon-C, 

Sodium- Na, 

Potassium- K, 

Iron- Fe, 

Silver- Ag

Ca = 1, H = 2, S = 2, O = 8

(i) If σ is positive, \(\vec E\) points normally outwards/away from the sheet.

(ii) If σ is negative, \(\vec E\) points normally inwards/towards the sheet.

Time requires in travelling 130 km = 2 hours 

⇒ Time requires in travelling 1 km = (2/130) hours 

⇒ Time requires in travelling 520 km 

= (2 x 520/130) hours = 8 hours 

Thus, 8 hours required to cover 520 km with the same speed.

Cost of a pen = Rs. 10 

Cost of a pencil = Rs. 2 

The ratio of pen’s cost to pencil’s cost 

= 10 : 2 = 10/2 = 5/1 = 5 : 1 

∴ The ratio of pen’s and pencil’s cost = 5 : 1

One score contains 20 pencils

And cost per score = 16

Therefore pencil cost = 16/20

= Rs. 0.80

Cost of one dozen ball pen = 8.40

1 dozen = 12

Therefore cost of pen = 8.40/12

= Rs 0.70

Ratio of the price of pencil to that of ball pen = 0.80/0.70

= 8/7

= 8: 7

(a) 100 litres

Explanation: 

Distance covered by truck using 54 litres diesel = 297 km

Distance covered by truck using 1 litre diesel = 297/54 = 5.5 km

Hence, for 550 km, diesel required = 550/5.5 = 100 litres

Cost of dozen pens = Rs 180

Cost of 1 per = 180/12 = Rs 15

Cost of 8 ball pens = Rs 56

Cost of ball pens= 56/8 = Rs 7

Required ratio of the cost of a pen to the cost of a ball pen = 15/7= 15 : 7

Raju purchase 10 pens for Rs 150

Price of 1 pen = 150/10 = Rs 15

Manish purchase 7 pens for Rs 84

Price of 1 pen = 84/7 = Rs 12

Therefore, manish got the pens cheaper

Let the number of machines required to produce articles in 54 days be x. The following table is obtained.

More the number of machines, lesser will be the number of days that it will take to produce the given number of articles. Thus, this is a case of inverse proportion. Therefore,

42 × 63 = 54× x

x = (42 x 63)/24 = 49

Hence, the required number of machines to produce the given number of articles in 54 days is 49.

Distance can be travelled in 15 litres of petrol = 240 km

∴ Distance can be travelled in 1 litre of petrol = 240/15 km = 16 km

∴ Distance can be travelled in 25 litres of petrol = (25 × 16) km = 400 km

If there are more number of machines, less number of days required to produce the given articles. Therefore it’s inversely proportional.

Let us consider the required of machines be x

42 × 56 = 48 × x

x = (42 × 56) /48 = 49

∴ 49 machines are required.

If the speed of the car is more, time required is less to travel. Therefore it’s inversely proportional.

Let us consider the required time be x hours

5 × 60 = 75 × x

x = (5 × 60) /75 = 4

∴ 4 hours are required.

Correct option is (B) 13 1/2

Work done to complete the wall = Work done by 30 labourers in 9 days

= Work done by 20 labourers in x days

\(\therefore 20\times x=30\times9\)

\(\Rightarrow x=\frac{30\times9}{20}=\frac{27}2\,days\) \(=13\frac12\,days\)

Correct option is  B) 13 1/2

More the speed of car, lesser the time required. So, it is an inverse proportion. 

Let the required time be x hours.

⇒ 5 × 60 = 75 × x

⇒ x = \(\frac{5\,\times\,60}{75}\) = 4 hours

If more number of cows, less number of days required to graze the field. Therefore it’s inversely proportional.

Let us consider the required number of days be x

6 × 28 = 14 × x

x = (6 × 28) /14 = 12

∴ 12 days are required.

Correct option is (A) 27

Given that 36 men can built a wall in 12 days.

\(\therefore\) Total work = 36 \(\times\) 12

Let 16 men can built the wall in x days.

\(\therefore\) Total work = 16 \(\times\) x

\(\because\) Both work are same (to build a same wall)

\(\therefore\) 16 \(\times\) x = 36 \(\times\) 12

\(\Rightarrow\) \(x=\frac{36\times12}{16}=9\times3\) = 27

\(\therefore\) 16 men can built the same wall in 27 days.

Correct option is  A) 27

More the no. of men, lesser the days for which food last. So, it is an inverse proportion. 

200 men had provisions for 45 days. After 15 days,200 men had provisions for 30days.

Now,

(i) Number of men (x₁) = 200 Provisions finished (y₁) = 30 days

(ii) after 15 days number of men joined are 40. Therefore,

Number of men after 15 days (x₂) = 240 

Let food last in number of days = y₂

y₂ = \(\frac{200\,\times\,30}{240}\)

= 25 days

Number of workers and number of days are in inverse proportion.

⇒ x₁y₁ = x₂y₂

⇒ (x + 1) (x + 1) = (x + 2) x k

⇒ k = ((x + 1)(x + 1)) / (x + 2)

∴ k = (x + 1)²/(x + 2)

Time and distance are in direct proportion

⇒ \(\frac{x_1}{y_1}= \frac{x_2}{y_2}\)

x₁ = 14 km , x₂ = ?

y₁ = 25min = 25/60hr = 5/12hr = y₂ = 5hrs

⇒ x₂ = (x₁×y₂) / y₁ = (14 × 5)/5 = (14 × 5 × 12)/5

= 168 km

∴ Lorry travelled in 5 hrs = 168km

More the number of cows, lesser the days required to graze a field. So, it is an inverse proportion. 

Let the required no. of days be x.

⇒ 6 × 28 = 14 × x

⇒ x = \(\frac{6\,\times\,28}{14}\) = 12 days

Correct option is (A) 24

Number of required workers \(=\frac{15\times48}{30}\)

\(=\frac{48}2\) = 24

Correct option is  A) 24

More the no. of persons, lesser the days to build. So, it is an inverse proportion. 

Let the required duration be x days.

⇒ 3 × 4 = 4 × x

⇒ x = \(\frac{3\,\times\,4}{4}\) = 3 days

Number of workers and number of days are in inverse proportion 

∴ x₁ y₁ = x₂ y₂  let y₂ = x (say)

= 36 × 12 = 9 × x 

x = (36 x 12)/9 = 48 

∴ x = 48 days

More the no. of men, lesser the days. So, it is an inverse proportion. 

Let the required time be x days. 

⇒ 30 × 28 = 21 × x

⇒ x = \(\frac{30\,\times\,28}{21}\) = 40 days

It is an inverse proportion. 

If 14 workers can do it in 42 days. 

Then time taken by 1 worker = 14 × 42 = 588 days

Average Speed = \(\frac{Total\,Distance}{Total\,Time}\)

Let distance be x km, time = \((1+\frac{12}{60})\) hr = \((1+\frac{1}{5})\) hr = \(\frac{6}{5}\) hr

⇒ 60 km/hr = \(\frac{\frac{x}{6}}{5}\)

⇒ x = 60 × \(\frac{6}{5}\) = 72 km

Correct option is (C) 24

Let 28 men will take x days to dig the same trench.

Work done by 28 men in x days = Work done to dig the trench

= Work done by 48 men in 14 days.

\(\therefore\) \(28\times x=48\times14\)

\(\Rightarrow\) \(x=\frac{48\times14}{28}=\frac{48}2\) = 24

\(\therefore\) 28 men will dig the same trench in 24 days.

Correct option is  C) 24

∵30 persons can reap a field in 17 days.

1 person can reap the same field in 30×17 i.e. 510 days.

In 10 days, the number of persons required = 510/10 = 51 persons.

M₁D₁H₁ = M₂D₂H₂

∴ M₁ = 24

D₁ = 15 days

H₁ = 8 hrs

M₂ = 20

D₂ = ?

H₂ = 9 hrs

⇒ 24 × 15 × 8 = 20 × x × 9

⇒ x = (24 x 15 x 8)/(20 x 9)

= 16

∴ No. of days are required = 16 

[ ∵ No. of men and working hours are in inverse]

More the no. of men, lesser the days required. So, it is an inverse proportion. 

Let the required no. be x days. 

⇒ 35 × 8 = 20 × x

⇒ x = \(\frac{35\,\times\,8}{20}\) = 14 days

Correct option is (A) ₹ 3375

The cost of food per person per day \(=Rs\;\frac{6300}{35\times24}\)

= Rs 7.5

\(\therefore\) Expenditure of 25 members for 18 days \(=Rs\;(7.5\times18\times25)\)

= Rs 3375

Correct option is  A) ₹ 3375

More the number of men, lesser the days required. So, it is an inverse proportion. 

Let the required no. of days be x.

⇒ 35 × 8 = 20 × x

⇒ x = \(\frac{35\,\times\,8}{20}\) = 14 days

Students: teacher ratio = 35:1

It shows every 35 students-one teachers should available in the school.

In the school, number of teachers required for 280 students = 280/35 = 8 teachers.

Number of days A required do a piece of work : 20

Number of days B required do a piece of work : 12

Work done by A in one day: \(\frac{1}{20}\)

Work done by B in one day: \(\frac{1}{12}\)

B works alone for 9 days, so work completed by B in 9 days : \(9\times\frac{1}{12}=\frac{3}{4}\)

Work left = \(1-\frac{3}{4}=\frac{1}{4}\)

But \(\frac{3}{4}\)th of the work is already done by B

∴ Time required to complete the remaining \(\frac{1}{4}\)th of the work by A : \(\frac{1}{4}\times20\) = 5 days

∴ Time taken to finish the work 9 + 5 = 14 days

∴ A can finish the remaining work in : 5 days

Given, 

A can do a piece of work in = 6 days

 B can do same piece of work in = 4 days

∵ Work done by A in 1 day \(=\frac{1}{6}\)

∴ In 2 Days he finished \(={2}\times\frac{1}{6}=\frac{1}{3}\) part of work 

Remaining work \(={1}-\frac{1}{3}=\frac{2}{3}\)

So, Remaining work will be finished by A and B in \(\frac{\frac{2}{3}}{\left\{ \frac{1}{6}+\frac{1}{4}\right\}}=\)\(\frac{\frac{2}{3}}{\frac{5}{12}}=\frac{2\times12}{5\times3}=\frac{8}{5}={1}\frac{3}{5}\) days

∴ Total time taken to complete the work \(={2}+{1}\frac{3}{5}={3}\frac{3}{5}\) days

Let the number of doctors = x 

Number of lawyers = y 

The average age of doctors = 35 

The total age of doctors = 35 × x = 35 x years 

The average age of lawyers = 50 

∴ The total age of lawyers = 50 x y = 50y 

According to the sum

\(\frac{35x\,\times\,50y}{x\,+\,y}\)  = 40

⇒ 35x + 50y = 40x + 40y 

⇒ 40x – 35x = 50y – 40y 

⇒ 5x = 10y 

⇒ x/y = 10/5 

(or) 

x : y = 2 : 1 

∴ The ratio of number of doctors to lawyers = 2:1

Number of days A required do a piece of work : 14 days 

Number of days B required do a piece of work : 21 days

Work done by A in one day: \(\frac{1}{14}\)

Work done by B in one day: \(\frac{1}{21}\)

Work done by A and B together in one day: \(\frac{1}{14}+\frac{1}{21}=\frac{5}{42}\)

They can do the work together in \(\frac{42}{5}\) days .

A and B worked together for 6 days, so work completed by A and B in 6 days : 6 x \(\frac{5}{42}=\frac{5}{7}\)

Work left = \(1-\frac{5}{7}=\frac{2}{7}\)

Number of days taken by B to complete the left over work : \(\frac{2}{7}\times21\) = 6

6 ( here 21 is days required by B to complete a piece of work).

∴ Time taken to finish the work: 6 + 6 = 12 days.

∴ Total time taken to finish the work: 12 days

In 20 g of sauted fish, protein is 75 g

∴ In 1 g of souted fish, Protein is 20/75 g

In 225 g of sauted fish, protein = 20/75 x 225 = 20 x 3 = 60 g

Number of days A required do a piece of work : 9 days 

Let number of days B required do a piece of work : X days 

Number of hours required by A and B together to do a piece of work : 6 days

Work done by A in one day: \(\frac{1}{9}\)

Work done by B in one day: \(\frac{1}{x}\)

Work done by A and B together in a day : \(\frac{1}{6}\)

Work done by A and B together in one day : \(\frac{1}{9}+\frac{1}{x}=\frac{x\,+\,9}{9x}=\frac{1}{6}\)

\(\therefore \frac{x\,+\,9}{9x}=\frac{1}{6}\)

⇒ 6X + 54 = 9X 

⇒ 3X = 54 

⇒ X = 54/3 = 18 

∴ B can do the work 18 days

If there are more number of sheets then, the weight will also be more. Therefore it’s directly proportional.

Let us consider the sheets be x, 12/0.04 = x/1

0.04 × x = 12 × 1

0.04x = 12

x = 12/0.04

x = 300

∴ 300 sheets are required.

Number of days Rajan required do a piece of work : 24

Number of days Amit required do a piece of work : 30

Work done by Rajan in one day: \(\frac{1}{24}\)

Work done by Amit in one day: \(\frac{1}{30}\)

Work done by Rajan and Amit together in one day: \(\frac{1}{24}+\frac{1}{30}=\frac{54}{720}=\frac{3}{40}\)

∴ They can do the work together in \(\frac{40}{3}\) days = \(13\frac{1}{3}\) days

(d) β-particle

As λ = h/mv, of the given particles β – particle is the lightest, so it will have maximum deBroglie wavelength.

(a) diffraction and photoelectric effect

Diffraction exhibits wave nature while photoelectric effect exhibits particle nature. Hence these two phenomena exhibit dual nature of light.

(a) β – particle

λ = \(\frac{h}{mv}\)

λ ∝ \(\frac{1}{m}\) 

As β – particle (an electron) has the smallest mass, so it has the longest de-Broglie wavelength.

(C) 1.326 × 10^-27 kg∙m/s 

The de Broglie wavelength associated with the particle,

λ = \(\cfrac hp\) = \(\cfrac{6.63\times10^{-34}}{10^{-26}}\) = 6.63 × 10^-8 m

(d) become half

As λ = \(\frac{h}{P}\) when momentum p is doubled, wavelength will become half the initial value.

(b) frequency

For photoelectric emission, the incident light must have a certain minimum frequency, called threshold frequency.

Consider an electron accelerated from rest through a potential difference V. Let v be the final speed of the electron. We consider the nonrelativistic case, v << c, where c is the speed of light in free space. The kinetic energy acquired by the electron is

\(\cfrac12\) mv² = \(\cfrac1{2m}\)(mv)² = eV......(1)

where e and m are the electronic charge and mass (nonrelativistic).

Therefore, the electron momentum,

p = mv = \(\sqrt{2meV}\).....(2)

The de Broglie wavelength associated with the electron is

λ = \(\cfrac hp\)....(3)

where h is Planck’s constant. 

From Eqs. (2) and (3),

λ = \(\cfrac h{\sqrt{2meV}}\)

Equation (4) gives the required expression. 

[Note : Substituting the values of h = 6.63 × 10^-34 J-s, 

m = 9.1 × 10^-31 kg, 

e = 1.6 × 10^-19 C in 

Eq. (4), we obtain 

λ = \(\sqrt{150/V}\) + 10^-10 m = \(\sqrt{150/V}\) Å 

= 12.25 / √V Å, where V is in volt. 

Therefore, electrons accelerated from rest through 150 volts have a de Broglie wavelength of 1 Å. This corresponds to the Xray region of the electromagnetic spectrum.]