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Curved surface area(C.S.A)= 60 π cm² 

Slant height of the cone(l) = 8 cm 

We know, 

Curved surface area(C.S.A) = πrl 

⇒ πrl = 60 π 

⇒ r x 8 = 60 

or r = 60/8 

= 7.5 

Therefore, radius of the base of the cone is 7.5 cm.

We have, 

Let radius of cone = r =4x, 

Let slant height = l =7x 

Area of curved surface = πrl 

= 792 cm² 

Now, 

πrl = 792

= \(\frac{22}{7}\) \(\times\)4x \(\times\) 7x

= 792

= x² =9

= x = 3 cm

∴r = 4 × 3 = 12 cm 

l = 7×3 = 21 cm

Answer is (C) arg (z₁) = arg (z₂)

Radius of cone(r) = 7 cm 

Curved surface area(C.S.A)= 176 cm² 

We know, C.S.A. = πrl 

⇒ πrl = 176

 ⇒ 22/7 x 7 x l = 176 

or l = 8 

Therefore, slant height of the cone is 8 cm.

Answer is (B) |z²| = |z|²

Volume of a sphere = \(\frac{4}{3}\)πr³
Volume of a cylinder = πr²h

Given, 

A cylindrical tub of radius 16 cm contains water to a depth of 30 cm. A spherical iron ball is dropped into the cylinder and thus the level of water is raised by 9 cm.

Volume of water displaced = Volume of the iron ball

⇒ \(\frac{4}{3}\)πr³ = π x 16² x 9

⇒ r³ = 1728

⇒ r = 12 cm

Volume of a hemisphere = (2/3)πr³ 

Given, 

hemispherical bowl is made of steel 0.25 cm thick. 

The inside radius of the bowl is 5 cm. 

Outer radius = 5 + 0.25 cm = 5.25 cm 

Volume of steel used in the making of the bowl = \(\frac{2}{3}\times\)\(\frac{22}{7}\)\(\times\)(5.25³ - 5³) cm³

⇒ Volume of steel used in the making of the bowl = 41.28 cm³

Volume of a hemisphere =  (\(\frac{2}{3}\))πr³

Volume of a cylinder = πr²h

Given,

Internal radii of the bowl and cylinder are respectively 6 cm and 4 cm.

⇒ \(\frac{2}{3}\) x π x 6³ = π x 4² x h

⇒ h = 9 cm

Radius of base = 21cm and its volume = \(\frac{2}3\) × volume of the hemisphere 

⇒ \(\frac{1}3\)πr²h = \(\frac{2}3\times\frac{2}3\)πr³

⇒ h = \(\frac{4}3r\) = \(\frac{4}3\times21\) = 28 cm

Hence 

surface area of the toy = Surface area of the cone + Surface area of hemisphere 

= π r (\(\sqrt{21^2 + 28^2}\)) + 2π r² 

= π (21)[35 + 2 × 21] 

= 5082 cm²

Surface area of a sphere = 4πr², where r is radius 

(i) r is 10.5 cm 

⇒ surface area = 4 × (22/7) × (10.5)² 

= 1386 cm² 

(ii) r is 5.6 cm 

⇒ surface area = 4 × (22/7) × 5.6² 

= 394.24 cm² 

(iii) r is 14 cm 

⇒ surface area = 4 × (22/7) × 14² 

= 2464 cm²

Surface area of a sphere of diameter ‘d’ = πd² 

(i) d is 14 cm 

⇒ surface area = (22/7) × (14)² = 616 cm² 

(ii) d is 21 cm 

⇒ surface area = (22/7) × (21)² = 1386 cm² 

(iii) d is 3.5 cm 

⇒ surface area = (22/7) × (3.5)² = 38.5cm²

Surface area of a sphere = 4πr² 

Surface area of a sphere of radius 14 cm = 4 × (22/7) × 14² 

= 2464 cm²

Volume of sphere = (1/6)πd³ 

(i) Diameter is 14 cm 

⇒ Volume of the sphere = (1/6) × (22/7) × 14³ 

= 1437.33 cm³ 

(ii) Diameter is 3.5 dm = 35 cm 

⇒ Volume of the sphere = (1/6) × (22/7) × 35³ 

= 22.46 dm³

(iii) Diameter is 2.1 m 

⇒ Volume of the sphere = (1/6) × (22/7) × 2.1³ 

= 4.851 m³

Radius of hemispherical dome = 63 dm or 6.3 m 

Inner surface area of dome = 2πr² 

= 2 x 3.14 x (6.3)² 

= 249.48 

So, Inner surface area of dome is 249.48 m² 

Cost of painting 1m² = Rs.2

Therefore, cost of painting 249.48 m² = Rs. (249.48 x 2) 

= Rs.498.96.

Circumference of a circle = 2πr 

Surface area of a hemisphere = 2πr² 

Given, 

base of the dome is 17.16 m 

⇒ 2 × (22/7) × r = 17.6 

⇒ r = 2.8 m 

Surface area of the hemisphere = 2 × (22/7) × 2.8² 

= 49.28 m² 

= 492800 cm² 

Cost of painting is Rs. 5 per 100 cm², 

Cost of painting the dome = (492800/100) × 5 

= Rs. 24640

No, it will not show any dispersion but show deviation.

During noon hours, the distance to be travelled by the sun rays in atmosphere is less than when compared to morning and evening hours. Therefore all colours reach our eye without scattering. Hence the sunlight appears white in noon hours.

Since magnetic moment is zero, it has no unpaired electrons.

Option : (b) +3

• The colour in lanthanoid ions is due to the presence of unpaired electrons in partially filled 4f sub-shells.
• Due to the absorption of radiations in the visible region there arises the excitations of the unpaired electrons from f-orbital of lower energy to the forbital of higher energy-giving f → f transitions.
• The observed colour is complementary to the colour of the light absorbed.
• The colour of try positive ions (M³⁺) depends upon the number of unpaired electrons in f-orbitals. Hence the lanthanoid ions having equal number of unpaired electrons have similar colour.
• The colours of M³⁺ions of the first seven lanthanoids, La³⁺ to Eu³⁺ are similar to those of seven elements Lu to Tb³⁺ in the reverse order. 

• The least distance a person can see an object comfortably and distinctly is known as least distance of distinct vision.
• Hold the textbook at certain distance with your hands. 
• Try to read the contents on the page. 
• Gradually move the book towards eye, till it reaches very close to your eyes. 
• You may see that printed letters on the page appear blurred or you feel strain to read. 
• Now move the book backwards to a position where you can see clear printed letters without strain. 
• Ask your friend to measure distance between your eye and textbook at this position. 
• Note down its value. 
• Repeat this activity with other friends and note down the distances for distinct vision in each case. 
• Find the average of all these distances of clear vision.
• You will notice that to see an object comfortably and distinctly, you must hold it at a distance about 25 cm from your eyes. 
• This 25 cm distance is called least distance of distinct vision. 
• This value varies from person to person and with age

1. Lanthanoid compounds are used inside the colour television tubes and computer monitor. For example mixed oxide (Eu, Y)₂ O₃ releases an intense red colour when bombarded with high energy electrons.

2. Lanthanoid ions are used as active ions in luminescent materials. (Optoelectronic application)

3. Nd : YAG laser is the most notable application. (Nd : YAG = neodymium doped ytterium aluminium garnet)

4. Erbium doped fibre amplifiers are used in optical fibre communication systems. 

5. Lanthanoids are used in cars, superconductors and permanent magnets.

• Actinoids are 5f-series elements in which electrons progressively enter into 5f-orbitals, which are inner orbitals.
• They have electronic configuration [Rn]⁸⁶ 5f^1-14 6d^0-1 7s².
• They show intermediate properties as compared to electropositive 5-block elements and electronegative p-block elements. Hence they are called second inner transition elements.  

• Position of actinoids in the periodic table : Group-3; Period-7.
• They interrupt the fourth transition series (6d series) in the seventh period in the periodic table. 
• After Actinium, 89Ac which has electronic configuration [Rn] 6d¹ 7s², the electrons enter progressively 5f orbital and they have general electronic configuration, [Rn] 5f^1-14 6d^0-1 7s². 
• They are fourteen elements from ₉₀Th to ₁₀₃Lr and since they follow actinium, they are called actinoids. 
• They are called 5f series or second inner transition series elements and for the convenience they are placed separately below the periodic table. 

• Actinoids : The series of fourteen elements from ₉₀Th to ₁₀₃Lr which follow actinium (₈₉Ac) and in which differentiating electrons are progressively filled in 5f-orbitals in prepenultimate shell are called actinoids.
• Their general electronic configuration is, [Rn]⁸⁶ 5f^1-14 6d^0-1 7s².

Most of the transition metals have simple hexagonal closed packed (hep), cubic closed packed (ccp) or body centred cubic (bcc) lattices.

1. In 3d-series elements as atomic number increases from scandium (Sc [Ar]¹⁸ 3d¹ 4s²) the number of unpaired electrons increases up to 3d⁵ in chromium.

2. As the number of unpaired electrons increases, the metallic character increases, hence the melting points and boiling points increase from ₂₁Sc(3d¹) to ₂₄Cr(3d⁵).

3. After chromium the number of unpaired electrons goes on decreasing due to the pairing of electrons, hence metallic character, melting points and boiling points decrease from ₂₅Mn to ₂₉Cu.

4. Zinc has all electrons paired, hence it is soft, has a low melting and boiling points.

The transition metal alloys are classified into :

• Ferrous alloys 
• Nonferrous alloys. 

In 3d transition elements, 

(i) Scandium (Sc) has lowest density and 

(ii) Zinc (Zn) has the highest density.

• The transition metals form a large number of alloys among themselves, which are hard with high melting points. 
• During alloy formation atoms of one metal are distributed randomly in the lattice of another metal. 
• The metals with similar atomic radii and similar properties readily form alloys. 
• These alloys have industrial importance. 
• The alloys can be ferrous alloys or nonferrous alloys. 

Option : (c) Mn

This is because radon is radioactive element with a short half life (t_1/2) of 3.82 days. This makes the study of chemistry of radon difficult.

Balance reaction is as follow :
XeF₆ + 2H₂O₂ → XeO₂F₂ + 4HF

Because of its low solubility (as compared to nitrogen) in blood, a mixture of oxygen and helium is used in diving kit used by deep sea divers.

(i) It is not soluble in blood even under high pressure. 

(ii) Being monoatomic they have weak dispersion forces. 

(ii) I‐Cl bond is weaker than l‐l bond

Due present of triple bond it has very high bond dissociation energy.

O₂ ⁺ PtF^-₆  of compounds of noble gases

Because fluorine and oxygen are strong oxidizing agents (most electronegative elements) 

Because fluorine and oxygen are strong oxidizing agents (most electronegative elements)

Nitrogen does not expand its covalence beyond four due to absence of d- orbital

Helium is a non-inflammable and light gas

To record sex of student radio buttons are used. Usually radio buttons are provided as a group from which exactly one can be selected at a time by giving the same name for both radio buttons. But here Ram did not give same narfie for both buttons. Therefore the mistake. 

The correct code is as follows,

<HTML>

<HEAD>

<TITLE>RADIO BUTTON</TITLE>

</HEAD>

<BODY>

sex<input type=”radio” name=’’sex” value= “Male”> Male

<input type=”radio” name=”sex” value= “Female”> Female

</BODY>

</HTML

1. Method: 

It determines the method of submission of form data to the server. Get and Post are the two form submission methods. Post method is used to pass large volume of data. Also post method is more secure as data entered is not visible during submission. The get method is faster and is used to send lesser volume of data and it is not secure.

2. Action: 

The URL of the server side program to process the form data is specified by the Action attribute.

Eg: <Form Action=”http://www.scert.com/asp/ process.asp”>

<Input>It is used to create input controls. Its type attribute determines the control type.

Main values of type attribute is given below.

1. Text – To create a text box.

2. Password – To create a password text box.

3. Checkbox – To create a check box.

4. Radio - To create a radio button.

5. Reset – To create a Reset button.

6. Submit - To create a submit button.

7. Button – To create a button

Correct answer is (a) MR²

Correct answer is (b) M \((\frac{R^2}{2} + \frac{l^2}{12})\)

For same external torque, angular acceleration of raw egg will be small than that of Hard boiled egg.

(b) Angular momentum 

L = mV x r 

= 9 x 10^-31 x 4 x 10⁵ x 2 x 10^-10 

= 72 x 10^-36 kg m² s^-1