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70 cm + 40 cm

= 110 cm

= 1 m 10 cm

 cloth for Kurta

cloth for Shirt

∴ 5 m 10 cm cloth in all

William Bedwell

Diameter of the ground d = 350 m 

Distance covered in 1 revolution = Circumference of the circle 

= πd units 

= \(\frac{22}{7}\) × 350 m 

= 22 x 50 

= 1100 m 

Distance covered in 1 rotation = 1100 m 

Distance covered in 4 revolutions = 1100 x 4 

= 4400 m

Given radius of the bangle = 2 inches 

Diameter = 2 x radius 

= 2 x 2 

= 4 inches

(i) Radius of the circle r = 4.2 cm

Area of the circle A = πr² sq. units

= \(\frac{22}{7}\) x 4.2 x 4.2 cm² 

= 5.44 cm²

(ii) Diameter of the circle d = 28 cm

Radius r = \(\frac{d}{2}\) 

= \(\frac{28}{2}\) 

= 14 cm

Area of the circle A = πr² sq.units

= \(\frac{22}{7}\) x 14 x 14 cm² 

= 616 cm²

(i) Radius r = 49 cm 

Circumference C = 2 πr units 

= 2 x \(\frac{22}{7}\) x 49 

= 2 x 22 x 7 

= 44 x 7 

= 308 cm

(ii) Radius r = 91 mm 

Circumference C = 2 πr units 

= 2 x \(\frac{22}{7}\) x 91

= 2 x 22 x 13 

= 44 x 13 

= 572 mm

(ii) more than three times of its diameter

Given the diameter d = 4.2 m 

Circumference C = πd units 

= \(\frac{22}{7}\) x 4.2 m 

= 22 x 0.6 

= 13.2 m

Given radius of a circle r = 3 cm 

Diameter of the circle = 2r 

= 2 x 3 

= 6 cm 

Breadth of the rectangle = Diameter of the circle 

B = 6 cm 

Length of the rectangle L = 4 x diameter of a circle 

L = 4 x 6 

L = 24 cm

(i) Area of the rectangle = L x B sq. units

= 24 x 6 cm²

Area of the rectangle = 144 cm²

(ii) Area of the circle = πr² sq. units

= \(\frac{22}{7}\) x 3 x 3 cm²

= \(\frac{198}{7}\) cm²

= 28.28 cm²

(iii) Area of the shaded area = Area of the rectangle – Area of the 4 circles

= 144 – (4 x \(\frac{198}{7}\)) cm² 

= (144 – \(\frac{792}{7}\)) cm²

= (144 – 113.14) cm² 

= 30.85 cm²

(i) Given radius r = 15 cm 

∴ diameter d = 2 x 15 

= 30 cm 

Circumference C = πd units 

= \(\frac{22}{7}\) x 30 

= \(\frac{660}{7}\)

= 94.28 cm

(ii) Given circumference C = 1760 cm 

2πr = 1760 

2 x \(\frac{22}{7}\) x r = 1760 

r = \(\frac{1760\times7}{2\times22}\)

= \(\frac{160\times7}{2\times2}\)

= 40 x 7 

= 280 cm 

Diameter = 2 x r 

= 2 x 280 

= 560 cm

(iii) diameter d = 24m 

radius r = \(\frac{d}{2}\)

= \(\frac{24}{2}\)

= 12 m 

Circumference C = 2 πr units 

= 2 x \(\frac{22}{7}\) x 12 

= \(\frac{528}{7}\)

= 75.4 m

Tabulating the results

Radius of the portion removed r = 2 cm 

Area of the circular sheet = πr² sq.units 

= \(\frac{22}{7}\) x 2 x 2 cm² 

= \(\frac{88}{7}\) cm² 

= 12.57 cm² 

Length of the rectangular sheet L = 5 cm 

Breadth of the rectangular sheet B = 3 cm 

Area of the rectangle = L x B sq. units

= 5 x 3 cm² 

= 15 cm² 

Area of the sheet left over = Area of the rectangle – Area of the circle 

= 15 cm² – 12.57 cm² 

= 2.43 cm²

Radius of the outer circle R = 5 cm 

Radius of the inner circle r = 3 cm 

Area of the remaining sheet = Area of the outer circle – Area of the inner circle 

= πR² – πr² sq. units 

= π(R² – r²) sq. units 

= \(\frac{22}{7}\)(5² – 3²) cm² 

= \(\frac{22}{7}\) x (5 + 3) (5 – 3) 

= \(\frac{22}{7}\) x (8) (2) 

= \(\frac{352}{7}\)

= 50.28 cm²

Length of the outer rectangle L = 8 cm 

Breadth of the outer rectangle B = 5 cm 

Area of the outer rectangle = L x B sq. units 

= 8 x 5 cm² 

= 40 cm² 

Length of the inner rectangle l = L – 2W 

= 8 – 2(1.5)cm 

= 8 – 3cm 

l = 5cm 

Breadth of thqnner rectangle b = B – 2W 

= 5 – 2(1.5) cm 

= 5 – 3cm 

= 2cm 

∴ Area of the margin = Area of the outer rectangle – Area of the inner rectangle

= (40 – 10) cm² 

= 30cm²

False 

One Celsius degree is an interval 1K is true, but zero degree Celsius is equal to -273.15K.

Rules for rounding off :

1. Decide which is the last digit to keep.

2. Leave it the same, if the next digit is less than 5.

3. Increase it by one, if the next digit is 5 or greater than 5.

86 °F, 303.15 K

In this case, the normal melting point of pure ice is taken as 32 °F and the normal boiling point of pure water is taken as 212 °F. The interval between these two points is divided into 180 equal parts. Each part corresponds to a temperature difference of 1°F. If C is the temperature of a body on the Celsius scale and F is the corresponding temperature on the Fahrenheit scale, the two are related by C = \(\cfrac{5}9\) (F - 32) or F = \(\cfrac{5}9\) C + 32

In this case, the temperature at which pure ice melts at normal atmospheric pressure is taken as zero degree (0°C) and the temperature at which pure water boils at normal atmospheric pressure is taken as hundred degree (100 °C). The interval between them is divided into loo equal parts. Each part corresponds to a temperature difference of 1 °C.

Data: F = 2 C ∴ C = F/2, F = ?

\(\cfrac{F-32}9\) = \(\cfrac{C}5\)

∴ \(\cfrac{F-32}9\) = \(\cfrac{F/2}5\)

∴ F - 32 = \(\cfrac{9}5\) \((\cfrac{F}2)\) = 1.8 \(\cfrac{F}2\) = 0.9 F

∴ F – 0.9 F = 32 

∴ 0.1 F = 32

∴ F = \(\cfrac{32}{0.1}\) = 320 \(^\circ\)F

Mass of water = \(\cfrac{500}5\) = 100 g

Fuels like wood, coal and petrol are sources of chemical energy.

Heat is transferred by conduction, convection and radiation.

The bore of a thermometer is made very small so that even a slight variation in the temperature may cause noticeable variation in the mercury level in the tube of the thermometer. As a result, the sensitivity of the thermometer increases and even small changes in the temperature can be recorded.

1. Sun 

2. earth 

3. fuels like wood, coal, petrol 

4. electricity

5. atomic energy 

6. air.

1. When the bulb of the thermometer is brought in contact with the substance whose temperature is to be measured, there is an exchange of heat between the substance and the mercury in the bulb. 

2. The mercury expands or contracts depending upon whether it gains heat or loses heat. Accordingly there is a rise or fall of the level of mercury in the tube of the thermometer indicating the temperature of the substance when the mercury and the substance are in thermal equilibrium.

Data: m = 3 kg = 3000 g, 

Q = 600 cal, ΔT = 10 °C, c = ? 

Q = mcΔT

∴ C = \(\cfrac{Q}{mΔT}\) = \(\cfrac{600\,cal}{3000\,g\times10^\circ C}\)

= 0.02 cal/(g.°C) 

This is the specific heat of the substance.

Specific heat is expressed in cal/(g.°C).

1. Find the mass (m_i) of the iron ball.

2. Find the total mass (m_c) of the calorimeter (metal container) and the stirrer.

3. Fill the calorimeter to two-thirds of its capacity with water and find its mass (m’_c) along with the stirrer. Hence, find the mass (m_w) of the water in the calorimeter (m_w = m’_c – m_c).

4. Keep the calorimeter in the wooden box and note the temperature (T₁) of the water in the calorimeter with the thermometer.

5. Suspend the iron ball in water in a beaker. Heat the beaker so that the water starts boiling. Note the temperature (T₂) of the boiling water. 

6. Transfer the iron ball quickly to the calorimeter and cover the calorimeter with the lid immediately.

7. Stir the water in the calorimeter gently and continuously for uniformity of temperature and note the maximum temperature (T₃) attained by the mixture.

8. Find the specific heat capacity of iron using the following formula: heat lost by the iron ball = heat gained by the calorimeter, stirrer and water [assuming that there is no exchange of heat between the system (calorimeter, stirrer, water and iron ball) and the surroundings].

∴ m_iC_i (T₂ - T₃) = (m_cC_c + m_wC_w) (T₃ - T₁)

∴ C_i = \(\cfrac{(m_cC_c + m_wC_w) (T_3 - T_1)}{m_i(T_2-T_3)}\)

where cc = specific heat of the material of the calorimeter and stirrer and c_w = specific heat of water. 

Hence, the specific heat of iron (c_i) can be determined when other quantities are known.

Gold, iron, aluminium, water.

0.1 cal/(g.°C)

The specific heat of an object (substance) is the amount of heat required to increase the temperature of unit mass of that substance through one degree.

Density, = \(\rho\) = \(\cfrac{mass}{volume}\)

∴ \(\rho_1\) = \(\cfrac{mass}{V_1}\) and \(\rho_2\) = \(\cfrac{mass}{V_2}\)

∴ \(\cfrac{​​\rho_1}{​​​​\rho_2}\) = \(\cfrac{V_2}{V_1}\) = 1 + \(\beta\) ΔT

∴ \(\rho_2\) = \(\cfrac{​​\rho_1}{​​​​1+\betaΔT}\)

As the temperature increases, density decreases.

Specific heat of a substance,

C = \(\cfrac{Q}{m(T_f-T_i)}\), where m is the mass of the substance and Q is the amount of heat supplied to the substance to increase its temperature from T_i to T_f.

∴ Unit of specific heat = \(\cfrac{Unit\,of\,heat\,(energy)}{Unit\,of\,mass\times Unit\,of\,temperature}\)

The SI unit of heat is the joule (J), that of mass is kg and that of temperature is kelvin (K).

∴ The SI unit of specific heat = \(\cfrac{J}{(kg.K)}\)

[Note: Specific heat is also expressed in J/(kg°C) and cal/(g°C).]

Coefficient of linear expansion of a solid is defined as the increase in the length of a rod of the solid per unit initial length per unit rise in its temperature. Coefficient of linear expansion of a solid, λ = \(\cfrac{l_2-l_1}{l_1ΔT}\), where l₁ and l₂ are respectively the initial and final length of the rod of the solid and ΔT is the rise in its temperature.

Unit of λ = \(\cfrac{unit \,of \,length}{unit \,of \,length\times unit \,of \,temperature}\)

λ = (l₂ – l₁) / (l₁ ΔT).

σ = (A₂ – A₁)/ (A₁ ΔT).

Suppose a rod of length l₁ at temperature T₁ is heated to temperature T₂ such that ΔT = T₂ – T₁ is very small. Let l₂ be the length of the rod at temperature T₂ .

Experimentally, it is found that the increase in the ength of the rod (linear expension), l₂ – l₁ , is proportional to l and ΔT. Therefore, (l₂ – l₁ ) α, l₁ ΔT

∴ l₂ – l₁ = λl₁ ΔT, where X is the constant of proportionality, called the coefficient of linear expansion of the solid.

λ = \(\cfrac{l_2-l_1}{l_1ΔT}\) It is expressed in per °C.

We have l₂ – l₁ + λΔT = l₁(1 + λΔT).

Suppose a sheet of a solid with surface area A₁ at temperature T₁ is heated to temperature T₂ such that ΔT = T₂ – T₁ is very small. Let A₂ be the surface area of the sheet at temperature T₂ . Experimentally, it is found that the increase in the surface area of the sheet (areal expansion), A₂ – A₁ , is proportional to A₁ and ΔT. Therefore,

(A₂ – A₁ )α A₁  ΔT

∴ A₂ – A₁ = σ Al₁ΔT, where a is the constant of proportionality, called the coefficient of areal expansion of the solid.

σ = \(\cfrac{A_2-A_1}{A_1ΔT}\), It is expressed in per °C.

We have A₂ = A₁ + σA₁ ΔT = A₁ (1 + σΔT).

σ is the increase in the area of a solid per unit original area per unit rise in its temperature. [Note: Consider a thin square metal plate of length l. Area of one face of the plate = A = l₂ . Suppose the plate is heated so that the rise in its temperature is ΔT (assumed to be very small). Then in the usual notation, Δl = l λΔT and ΔA = AσΔT = l² σΔT. Also, ΔA = (l + Δl)² – l² = l² + 2l.Δl + Δl² – l2 = 2l.Δl + Δl² . As Δl << 2l.Δl^2, we can write

ΔA = 2l.Δl(approximately)

∴ ΔA = 2l(l λΔT) = 2l² λΔT but ΔA = l² σΔT

∴ σ = 2.λ]

If the temperature of a silver rod of length lm is increased by 1 °C, the length of the rod increases by 18 × 10^-6 m.

β = (V₂ – V₁ ) / (V₁ ΔT).

Data: m₁ = 8 kg, T₁ = 20 °C, m₂ = 4 kg, T = 40 °C,

T₂ = ?

Heat lost by the hot water = heat gained by the cold water (ignoring the heat absorbed by the bucket)

∴ m₂ c (T₂ – T) = m₁ c (T – T₁ ) 

∴ 4 kg × c × (T₂ – 40°C) 

= 8 kg × c × (40°C – 20°C) 

∴ T₂ – 40°C = 2 × 20°C = 40°C 

∴ T₂ = 40 °C + 40 °C = 80 °C 

Temperature of the hot water = 80 °C.

8 × 10^-5 °C^-1

Correct answer is 37 °C

Data: mx = 2 kg, 

c₁ = 0.11 kcal/(kg.°C), T₂ = 400 °C, m₂ = 1 kg, 

c₂ = 1 kcal/(kg.°C), T₂ = 20 °C, T = ? 

Heat lost by the horseshoe = heat gained by the water

∴ m₁ c₁ (T₁ – T) = m₂ c₂(T – T₂ )

∴ 2 kg × 0.11 kcal/(kg.°C) × (400 °C – T) 

= 1 kg × 1 kcal/(kg.°C) × (T – 20 °C) 

∴ 0.22 × (400 °C – T) = T – 20 °C 

∴ 1.22 T= 108 °C

∴ T = \(\cfrac{108}{1.22}\) °C = 88.52 °C 

Maximum temperature of the water = 88.52 °C.

50 paise + 60 paise

= 110 paise

= 1 ₹ 10 paise

∴ ₹ 24, 10 paise

When we heat the neck of the glass bottle, it expands. Due to this, the stopper becomes loose and can be removed with ease.

(i) The density of the gas will remain the same as there is no change on the mass and volume of the gas.

(ii) The piston will move upwards as the gas expands. Therefore, the expansion of a gas is measured by keeping its pressure constant.

The bulb of a thermometer is made of a thin glass so that it can easily conduct the heat from the substance in contact with the mercury in the bulb.

Rise in temperature/boiling, melting, burning, expansion.

The energy stored in a body because of its specific state or position is called its potential energy. The energy possessed by a body because of its motion is called it’s kinetic energy.