Explore topic-wise InterviewSolutions in .

1. Newton’s laws of motion are not applicable in a non-inertial (accelerated) frame of reference.

2. Newton’s laws are only applicable to point objects.

3. Newton’s laws are only applicable to rigid bodies.

4. Results obtained by applying Newton’s laws of motion for objects moving with speeds comparable to that of light do not match with the experimental results and Einstein special theory of relativity has to be used.

5. Newton’s laws of motion fail to explain the behaviour and interaction of objects having atomic or molecular sizes, and quantum mechanics has to be used.

Correct option is: (D) non-inertial frame of reference

1. Inertial frame of reference:

• A frame of reference in which Newton ‘s first law of motion is applicable is called inertia/frame of reference.
• A body moves with a constant velocity (which can be zero) in the absence of a net force. The body does not accelerate.
• Example: A rocket in inter-galactic space (gravity free space between galaxies) with all its engine shut.

2. Non-inertial frame of reference:

• A frame of reference in which an object suffers acceleration in absence of net force is called noninertial frame of reference.
• The body undergoes acceleration.
• Example: If a car just start its motion from rest, then during the time of acceleration the car will be in a noninertial frame of reference.

When we move the cycle with its brakes on, then, its wheels can only skid, i.e., slide along the road as they can’t rotate. So, the friction is sliding in nature. As the sliding friction is greater than the rolling friction therefore, it is difficult to move with its brakes on.

The charge on up quark (u) = \(+\frac{2e}{3}\)
The charge on down quark = \(-\frac{e}{3}\)
The charge on proton is +e and it is made of three quarks. Therefore the possible quark composition of a proton is ‘uud'

On the other hand, neutron is a neutral particle but it is also made of three quarks. For this, the possible composition of the neutron is ‘udd’.

The electric fields bind the atoms to neutral entity. Fields are caused by excess charges. There can be no excess charge on the inter surface of an isolated conductor.

A firm is compulsorily dissolved when all partners or when all except one partner become insolvent – True.

Correct option is (B) Credited to Realisation account

In a partnership firm new partners can be admited and few partners can retire or die, it would change the form of partnership and partners, but the firm does not dissolve and its existence continues.

Partnership is compulsorily dissolved when the partners of the firm become insolvent.

Explanation: When the partners of a firm become insolvent, it implies that the assets of the firm have decreased in comparison to the liabilities. Moreover, the partners do not have enough funds to make payment to the creditors; hence, the partnership is compulsorily dissolved as per the order of a court of justice.

As per Indian Partnership Act 1932 the dissolution in two ways:

1. Normal Dissolution
2. Dissolution by the Court.

When a ongoing partnership firm is closed and its business is stopped i.e. in the eyes of law the existence of the firm does not remain any more it is known as Dissolution of Partnership.

All activities of the partnership firm cease (stop) on dissolution of firm.

Explanation: Dissolution of a partnership firm involves discontinuance of the firm’s business, besides the termination of the existing partnership deed. Thus, at the time of dissolution of a partnership firm, all the activities of the firm tend to cease.

C) Law of inertia

Correct option is C) inertia

Correct option is: (C) 5.00

Given, F = 5N

W = 9.80

a = ?

W = mg

\(9.80 = m \times 9.8\)

m = 1 kg

F = ma

\(5 = 1 \times a\)

\(a = 5 m/s^2\)

Correct option is: (C) 5.00

A) isolated system

D) All of these

The force is said to be 1 newton if it produces 1 ms^-2 acceleration in 1 kg mass of a body.

Correct option is (b) Economic

The frequencies of all the harmonics present in an open pipe are

n = (p + 1) \(\frac v{4L}\) = (p + 1) \(\frac v{4(I+2e)'}\)

where p = 0,1,2,3,......

The frequencies of the odd harmonics present in a closed pipe are

n = (2p + 1) \(\frac v{4L}\) = (p + 1) \(\frac v{4(I+e)'}\)

where p = 0,1, 2, 3, … .

m = 10,000kg, v = 72 km/h

= \(\frac {72 \times 1000}{60 \times 60} = 20 m/s\)

r = 200 m.

Acceleration a = v ω

= \(v\frac{v}{\tau}\)

= \(\frac {v^2}{\tau} = \frac {(20)}{200}\)

= 2m/s²

Centripetal force, F = \(\frac {mv^2}{\tau}\)

\(\frac {10000 \times(20)^2}{200}\)

= 20000 N.

Maximum height attained by the bullet, H = 50m,

Horizontal range R = 200m,

Angle of projection 0 =?

From the equation H = \(\frac {u^2 sin^2\theta}{2g}\)

50 = \(\frac {u^2 sin^2\theta}{2g}....(1)\)

From the equation R = \(\frac{u^22sin \theta cos \theta}{g}\)

200 =  \(\frac{u^22sin \theta cos \theta}{g}.....{2}\)

Divide equ (1) by equ (2), we have

\(\frac{50}{200} =\frac {sin\theta}{4cos\theta}\)

i.e \(\frac{1}{4}=\frac {1}{4}\) tanθ Thus tanθ = 1

θ = tan^-1 (1.0000) = 45°.

(i) ∠A+∠D = 180°, क्योंकि AB || DC तथा ये कोण अंतःकोण हैं।

(ii) ∠ B +∠D = 180°, क्योंकि ये कोण चक्रीय चतुर्भुज के सम्मुख कोण हैं।

(iii) ∠A= ∠ B, क्योंकि ∠A+ ∠ D = 180°।

∠ACB +∠APB = 90° + 90° = 180°

( अर्द्धवृत्त पर बना कोण समकोण होता है) ( दिया है)

जीवा AC = जीवा BC

∴ Δ ABC के कोणों का योग = 180°

∴  ∠C = 90°, ∠A + ∠B = 90°

Δ ABC = 45

 (i) AC = CB (सत्य)

(ii) AB = \(\frac12\)AC (असत्य)

(iii) AB = 2AC (सत्य)

(iv) OC= \(\sqrt{OA^2-AC^2}\) (सत्य)

(v) AC= \(\sqrt{OA^2+OC^2}\) (असत्य)

Given: L = 40 cm = 0.4 m,

F = 2000 N = 2 × 10³N

l = 0.5 cm = 0.005 m, A = L² = 0.16 m²

To find: Modulus of rigidity (n)

Formulae:

i. θ = \(\frac{l}{L}\)

ii. n = \(\frac{F}{A\theta}\)

Calculation:

From formula (i),

θ = \(\frac{0.005}{0.4}\) = 0.0125

From formula (ii),

n = \(\frac{2\times10^3}{0.16\times0.0125}\) = 1 × 10⁶ N/m²

The modulus of rigidity of the metal cube is 1 × 10⁶ N/m².

Here, surface tension,

T = 2.50 x 10^-2 Nm^-1

Radius, r = 5.00 mm = 5 x 10^-3 m

Excess pressure,

p = 4T/r (For a bubble) = {4 x 2.50 x 10^-2}/{5 x 10^-3} = 20 Pa

Now, p₁ - p₀ = p

⇒ p₁ = p + p₀ = 1.01 x 10⁵ + 20 = 1.0102 x 10⁵ Pa

If the bubble of the same radius as above is at depth of 40 cm inside the soap solution, then excess of pressure is given by,

p = 2T/r = {2 x 2.50 x 10^-2}/{5 x 10^-3} = 10 Pa

Also, outside pressure is,

p₀ = Pressure of water above the bubble + atmosphere pressure

= ρgh + 1.01 x 10⁵

Here, ρ = 1.2 x 10³ kg m^-3

g = 9.8 ms^-2

and h = 40 x 10^-2 m

p₀ = 1.2 x 10³ x 9.8 x 40 x 10^-2 + 1.0 x 10⁵

= 4.704 x 10³ + 1.01 x 10⁵

= 1.057 x 10⁵ Pa

If p₁ is the inside pressure, then

p₁ - p₀ = p

⇒ p₁ = p + p₀

= 1.057 x 10⁵ + 10 = 1.057 x 10⁵

The terminal velocity is given by v, where v α r² (r = radius of the sphere). The mass of the sphere can be given by

\( m = \frac{4}{2} \pi r^3\rho, thus\,m \propto r^3\)

\(\frac{m_1}{m_2} = \frac{1}{8}=( \frac{r_1}{r_2})^3 \,\,\Rightarrow\frac{r_1}{r_2} = \frac{1}{2}\)

\( and \, since, \frac{v_1}{v_2} = (\frac{r_1}{r_2})^2 = \frac{1}{4}\)

\( \Rightarrow\frac{v}{nv} = \frac{1}{4}\)

\( \Rightarrow n=4\)

The terminal velocity of a body v₁ in a freely falling system is zero, because g = 0 for the systems and v_t ∝ 0

Data : R = 10^-3 m, T = 0.072 N/m, ρ = 10³ kg/m³ , h = 0.1 m

Let the atmospheric pressure be p0. Then, the absolute pressure within the liquid at a depth h is

p = p₀ + ρgh

Hence, the pressure inside the bubble is

p_in = p₀ + \(\frac{2T}R\) = p₀ + pgh + \(\frac{2T}R\)

The excess pressure inside the bubble over the atmospheric pressure is

p_in = p_{0 = pgh +} \(\frac{2T}R\)

= (10³) (9.8) (0.1) + \(\frac{2(0.072)}{10^{-3}}\)

= 980 + 144 = 1124 Pa

Liquids: 

1. Viscosity increases with increase in density of the liquid. Viscosity increases with increase in pressure (except water). 

2. Gases: 

Viscosity decreases with increase in density. Viscosity decreases with increase in pressure.

Viscosity of liquid decreases with increase in temperature and viscosity of gases increases with increase in temperature.

(c) gases increases. 

(d) liquids decreases

The terminal velocity of a body v_c in a freely falling system is zero, because g = 0 for this system and v_c ∝ g.

The smearing of glycerine increases the angle of contact from acute to obtuse.

High viscosity liquids are used as buffers in trains so that the shocks may be absorbed easily.

The surface tension of a liquid at critical temperature is zero

The exact Reynold number at which the flow becomes turbulent is called Critical Reynold number.

Correct option is B) Electroplating

When the velocity of flow of the liquid becomes greater than the critical velocity of the flow.

Given,

T = 27°C, L_T1 = 5.231 m

L_T2 = 5.243 m

So, L_T2 = L_T1 [1 + α₁ (T₂ - T₁)]

⇒ 5.243 = 5.231 [1 + 1.20 x 10^-5°C^-1 (T₂ - 27°C)]

or, T₂ = 218°C

One lines can be drawn through both the given points.

(i) unique 

(ii) lines

(iii) perpendicular, perpendicular

(iv) three, two half planes, line.

One points two distinct lines can intersect.

(i) False

(ii) False

(iii) False

(iv) True

(v) False

(vi) True

(vii) False

(viii) False

(ix) False

(x) False

1. False. A line segment has definite length.
2. False. A ray has one end point.
3. False. A line has no definite length.
4. True.
5. False. BA and AB have different end points.
6. True.
7. True.
8. True.
9. True.
10. True.
11. False. Two lines intersect at only one point.
12. True.

(i) Unique

(ii) Lines

(iii) Perpendicular; Perpendicular

(iv) Three, two half planes, line

(i) Five line segments are: \(\overline{PQ}\), \(\overline{PN}\), \(\overline{RS}\), \(\overline{ND}\), \(\overline{TL}\)

(ii) Five rays are: ray QC, ray PM, ray RB, ray DF, ray LH

(iii) Four Collinear points are: PRQ, PTN, QLD, NSD

(iv) Two pairs of non-intersecting line segments are: PN, RS and PQ, TL

Line segments are PQ, QR, PR.