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No, e.g., the C.M. of a ring is at the centre where there is no mass.

(i) The centre of mass of sphere is located at its centre.

(ii) The centre of mass of cylinder is at the centre of its axis of symmetry.

(iii) The centre of mass of a ring is at centre of the ring.

(iv) The centre of mass of a cube is at its geometrical centre.

The centre of mass may lie outside the body also, e.g., in case a ring, a hollow sphere, a hollow cylinder, a hollow cube etc.

Geometric centre; No

The centre of mass (C.M.) is a point where the mass of a body is supposed to be concentrated. For the given geometric shapes having a uniform mass density, the C.M. lies at their respective geometric centres.

The centre of mass of a body need not necessarily lie within it. For example, the C.M. of bodies such as a ring, a hollow sphere, etc., lies outside the body.

Given as f(x) = e^2x

f'(x) = (d/dx)(e^2x)

⇒ f’(x) = 2e^2x

For the function f(x) to be increasing, we must have

⇒ f’(x) > 0

⇒ 2e^2x > 0

⇒ e^2x > 0

Here, the value of e lies between 2 and 3

Therefore, whatever be the power of e (that is x in domain R) will be greater than zero.

Hence, f(x) is increasing on interval R

D) Silver sulphide

CH₃- CH = N-NH-CO-NH₂

Correct option is c. -1.5 eV

λ = h/√2 x m x K.E.

= 6.626 x 10^-34Js / √2 x 9.1 x 10^-31kg x 150 x 1.6 x 10-19 J

=6.626 x 10^-34 / √4368 x 10^-50 = 10^–10 m = 1 Å

The percentage of empty space in a body centred cubic arrangement is 32.

(ii) hcp and ccp

(i) Number of moles of Na₂CO₃ 

= Molarity × Volume of solution in litre

= 0.75 × 5

= 3.75 mol

(ii) Number of moles of Fe 

= \(\frac{Mass}{Molecular\,mass}\) 

=\(\frac{7.85}{56}\) 

= 0.14

(iii) Number of moles of sucrose

= \(\frac{Mass}{Molecular\,mass}\)

= \(\frac{34.2}{34.2}\)

= 0.1

1 mole of S₈ molecules = 6. 022x10²³ molecules 

0.1 mole of S₈ molecules = 6. 022x10²³ x 0.1 = 6.022 x 10²² molecules 

One molecule of S₈ contains 8 atoms of Sulphur 

Therefore 6.022x10²² molecules of S₈ contain =6.022 x 10²² x 8 = 4.816 x 10²³ atoms. 

Correct option: (C) 1.33

N₂ + 3H₂ → 2NH₃

(1 vol.) (3 vol.) (2 vol.)

3 volumes of H₂ gives 2 volumes of ammonia

∴ 2 L of H₂ will give = \(\cfrac{2\,{\times\,2}}{3}\) L of ammonia

= 1.33 L of ammonia

Correct option: (D) 4.48 dm³

H₂O → H₂ + \(\frac{1}{2}\) O₂

H₂ : O₂ = 2 : 1

∴ for 2.24 dm³ of O₂, H₂ liberated will be 4.48 dm³

Correct option: (A) 1.568 x 10⁴ 

Since, 0.5 g Se ≡ 100 gm peroxidase anhydrous enzyme

∴ 78.4 g Se = \(\cfrac{100\, \times \, 78.4}{0.5}\) = 1.568 x 10⁴

Minimum molecular mass of peroxidase anhydrous enzyme means molecule atleast contains one selenium atom.

Correct Answer is: (d) d will decrease, v will increase.

mv²/d = G MG/d²  or GM = v²d or V∝ 1/√d

Total energy = 1/2 mv² - G GM/d = - 1/2 G Mm/d.

Correct Answer is: (a, b)

Correct Answer is: (c) √1.5 v

Taking the potential at a large distance from the planet as zero, the potential at the centre of the planet = - 3GM/2R

∴ 1/2 mv² = m [ 0 - (- 3GM/2R)]

or v² = 3GM/2R = 3Rg = 3/2 (2RG) = 3/2 v_e²

or v = √1.5 v_e .

Correct Answer is: (b) v_e/√2

v_e = √2Rg

At the surface of the planet, V_s = - GM/R

At the centre of the planet, V_c = - 3GM/2R

1/2mv² = m (V_s - V_c)

or v² = 2 GM/R (3/2 - 1) = GM/R =Rg = v_e²/2.

Correct Answer is: (c) V/√2 

Let M, R be the mass and radius of the planet, and g be the acceleration due to gravity on its surface.

Then, V = √(2Rg) and GM = R²g.

Gravitational potential at the surface is - GM/R and at the centre is - 3GM/2R. In going from the surface to the centre, loss in gravitational PE

= m [ - GM/R - ( - 3/2 GM/R)] = 1/2 GMm/R = 1/2 mv²

or v² = GM/R = Rg =V²/2 or V/√2.

Correct Answer is: (b) v_e/ √2 < v < v_e

For a satellite orbiting very close to the earth’s surface, the orbital velocity = √Rg (∵ mg = mv²/R). This is equal to the velocity of projection and is the minimum velocity required to go into orbit. Also, the satellite would escape completely and not go into orbit for v ≥ v_e.

∴ v_e/√2 < v < v_e.

The open loop gain of an amplifier is

β = \(\frac{V_f}{V_o}\), V_f = βV_o

V_i = V_s - V_f

V_i = V_s - βV_o

V_i = V_s - β.AV_i

V_i + β AV_i = V_s

V_i(1 + Aβ) = V_s

A_f = \(\frac{V_o}{V_s}\)

= \(\frac{V_o}{V_i(1+A\beta)}\)

A_f = \(\frac{A}{1+A\beta}\)

In F₂, large repulsion occur between non bonding electrons on small sized fluorine atom in the fluorine molecule.

(d) all the above

Electroplating is a process where there is continuous flow of ions for the deposistion of copper, which is not possible in Alternating current. Therefore, electroplating is possible with DC only, for the sake of perfectness and homogeneity of the electroplating.

1. (c) 

2. (e) 

3. (d) 

4. (a) 

5. (b)

Current is the rate at which charges flow past a point on a circuit. Current (I) is represented as, I = \(\frac{q}{t}\)

The standard SI unit for current is ampere with the symbol A.

The electric lines of force are straight or curved paths along which a unit positive charge tends to move in the electric field.

Ohm’s law states that electric potential difference across two points in an electrical circuit is directly proportional to the current passing through it. That is,

V~I 

The proportionality constant is tne resistance (R) offered between the two points. Hence, Ohm’s law is written as, 

V = RI (or) V = IR 

Where, V is the potential difference in volt (V), I is the current flow in ampere (A), R is the resistance in ohm (Ω)

The home appliances are connected in parallel. This is because, when the appliances are connected in parallel, each of them can be switched on and off independently. This is a feature that is essential in a house wiring. Also, if the appliances were wired in series, the potential difference across each appliance would vary depending on the resistance of the appliance.

The direction of the electric field is the direction of the force that would act on a small positive charge. Therefore the lines representing the electric field are called ‘electric lines of force’.

Safety features to be followed are: 

• Ground connection 
• Trip switch
• Fuse.

(c) either (a) or (b)

(c) absence of electron

False  

An ammeter is connected in series with a device to measure its current.

(a) The material is comb which gained electrons and the hair gained it.

Charge on 1 electron, e = 1.6 × 10^-19 C

q = ne or n =\(\frac{q}{e}\)

n = \(\frac{0.4}{1.6}\) x 10^-19 = 0.25 x 10¹⁹ = 2.5 x 10¹⁸

(b) start; end

The electrostatic force between two charges depend on the following factors; 

• value of charges on them, 
• distance between them, and 
• nature of medium between them.

False  

An ammeter is connected in series with a device to measure its current.

The principle of all electric heating appliances like iron box, water heater, toaster etc. work under the principle of heating effect of current.

Given: 

Time ‘t’ = 2 hours = 2 × 60 × 60s 

t = 7200s 

I = 2.5A 

Amount of charge, 

Q = I × t 

= 2.5 × 7200 

Q = 18000C

Electrically neutral means it is either zero or equal positive and negative charges is True.

(b) chemical effect

(a) electrons

i. Acidified permanganate solution oxidizes iron (II) salt to iron (III) salts.

MnO₄^- + 8H⁺ + 5Fe²⁺ → Mn²+ + 4H₂O + 5Fe³⁺

ii. It oxidizes sulphur dioxide to sulphuric acid.

2MnO₄^- 5SO₂ + 2H₂O → 5SO₄^2- + 2Mn²⁺ + 4H⁺

iii. It oxidizes oxalic acid to CO₂ and H₂O.

2MnO₄^- + 16H⁺ + 5C₂O₄^2- → 2Mn²⁺ + 8H₂O + 10CO₂.

2,5 dimethylhexane-1,3-diol 

Correct Option  (c) Monoatomic

Explanation: 

By comparing with relation C_v = R/γ - 1 we get γ - 1 = 0.67 or γ = 1.67 i.e. the gas is monoatomic.

(b) sulphanilamide

(c) The ratio between the number of molecules in A to number of molecules in B is 2 : 1