Explore topic-wise InterviewSolutions in Current Affairs.

Correct answer is

d. Ce^3⊕, Eu^3⊕, Yb^3⊕

x’x = ff’, जहाँ x’ तथा x क्रमश: प्रथम एवं द्वितीय फोकस से वस्तु की दूरियाँ एवं f’ तथा f क्रमशः लेन्स की प्रथम तथा द्वितीय फोकस दूरियाँ हैं।

(i), (iv)

(i) As₂O₃ < SiO₂ < P₂O₃ < SO₂   Acid strength.

(iv) H₂O > H₂S > H₂Se > H₂Te    Thermal stability.

जब उभयोत्तल लेन्स की वक्रता त्रिज्याएँ समान होती हैं तब उसकी फोकस दूरी वक्रता त्रिज्या के समान होती है। अत: f = R = 20 सेमी।

(i), (ii)

(i) S–S bond is present in H₂S₂O₆.

(ii) In peroxosulphuric acid (H₂SO₅) sulphur is in +6 oxidation state.

(ii), (iii)

(ii) 2HI + H₂SO₄ → I₂ + SO₂ + 2H₂O

(iii) Cu + 2H₂SO₄ → CuSO₄ + SO₂ + 2H₂O

(i) Cu + 2H₂SO₄ → CuSO₄+ SO₂ + H₂O

(ii) C + 2H₂SO₄ → CO₂ + 2SO₂ + 2H₂O

When Chlorine reacts with cold sodium hydroxide, it forms sodium chloride.

(i) 2NaOH + Cl₂ = NaCl + NaOCl + H₂O 

When Chlorine reacts withhot sodium hydroxide, it forms sodium chloride and other products.

(ii) 6NaOH + 3Cl₂ = 5NaCl + NaClO₃ + 3H₂O

It is hydrolysed in moisture and form fumes of HCl.

PCl₃ + 3H₂O  →  3HCl (Fumes) + H₃PO₄

(i) 0°C, (ii) 4°C.

Correct answer is (A) I and II are both correct

Correct answer is (C) 0.46 °C

What comes to your mind first when you think of a ‘pilot’ or a ‘traffic policeman? Discuss in pairs and share your thoughts with the class: A pilot controls and steers an airplane. He operates the directional flight controls. He wears milk white uniform and golden stripes on his shoulders. He wears a shiny black cap. A traffic policeman wears white stripes on his shoulders in a khaki uniform. In some states, a traffic policeman wears a white and white uniform also. He regulates traffic, fines people who violate traffic rules. He prevents accidents by monitoring over speeding vehicles and by discouraging drunken driving.

We know that

(i) f(x) = x² is a polynomial which is continuous for all x ϵ R

Hence, f(x) = x² is continuous on [- 1, 1]

(ii) f’(x) = 2x exist in [- 1, 1]

Hence, f(x) = x² is differentiable on (- 1, 1)

(iii) We know that

f(- 1) = (- 1)² = 1

Similarly f(1) = 1¹ = 1

Here f(-1) = f(1)

The conditions of Rolle’s Theorem are satisfied.

There exist at least one c ϵ (-1, 1) where f’(c) = 0

2c = 0 which gives c = 0

We know that value of c = 0 ϵ (-1, 1)

Therefore, Rolle’s Theorem is satisfied.

We know that

(i) f(x) = x² – x – 12 is a polynomial which is continuous for all x ϵ R

Hence, f(x) = x² – x – 12 is continuous on [-3, 4]

(ii) f’(x) = 2x – 1 exist in [-3, 4]

Hence, f(x) = x² – x – 12 is differentiable on (-3, 4)

(iii) We know that

f(- 3) = (- 3)² – 3 – 12= 0

Similarly f(4) = 4² – 4 – 12 = 0

Here f(-3) = f(4)

The conditions of Rolle’s Theorem are satisfied.

There exist at least one c ϵ (-3, 4) where f’(c) = 0

2c – 1 = 0 which gives c = 1/2

We know that value of c = 1/2 ϵ (-3, 4)

Therefore, Rolle’s Theorem is satisfied.

f(x)= x^{​​​​​​2} - X -12

f(x) being a polynomial function is continuous for all x €[-3,4]

f'(x) = 2x-1

Where f(x) is clearly differentiable for all x€[-3,4].

f(-3)= (-3)² -(-3)-12

f(-3)= 0

f(4)= (4)²-4-12

f(4)= 0

f(-3)=f(4)

There must exist a real number c such that 

f'(c)= 0

Or. 2c-1=0

c=½ €[-3,4]

Hence Rolle's theorem is verified

Hope it helps...

Volume of a hemisphere = 2425 1/2 = 4851/2 cm³

We know, Volume of a hemisphere = 2/3πr³

4851/2 = 2/3 πr³

r³ = (4851 x 21)/88

r = 10.5

So, radius of hemisphere is 10.5 cm.

Curved Surface Area of hemisphere = 2πr²

= 2 × 22/7 × (10.5)²

= 693

Curved surface area of hemisphere is 693 cm².

Radius of circle = 14 cm 

Angle = 60° 

Area of sector = \(\frac{θ}{360}πr^2\)

= \(\frac{60}{360}π\times14\times14\) = \(\frac{98}3π\) = 102.57 cm²

Area of triangle OAB = \(\frac{1}2r^2\) sin θ

= \(\frac{1}{2}\times14\times14\times{sin}θ\)

= \(\frac{1}{2}\times14\times14\times\frac{\sqrt3}2\) 

= \(49\sqrt3\) = 84.77 cm²

So, 

Area of minor segment = 102.57 – 84.77 = 17.80 cm²

Given, 

Radius of circle = 6 cm 

Length of chord = 10 cm 

Angle subtend by chord = 110° 

(I). Circumference of circle = 2πr 

= 2 × 3.14 × 6 = 37.68 cm 

(II). Are of circle = πr² 

= 3.14 × 6× = 113.1 cm² 

(III). Length of arc = radius × angle subtend 

= 6 x \(\frac{120π}{180}\)

= 6 x \(\frac{2}3\times\frac{22}7\) = 11.51 cm

(IV). Area of sector = \(\frac{θ}{360}\timesπr^2\)

= \(\frac{110}{360}\times\frac{22}7\times6\times6\) = \(\frac{274}7\)

= 34.5 cm²

Volume of hemi-sphere = \(\frac{2}3πr^3\)

⇒ \(\frac{2}3πr^3\) = 4851

⇒ r³ = \(\frac{{4851}\times{3}\times{7}}{{22}\times{2}\times{2}}\)

⇒ r = 10.5 cm 

Curved surface area = 2πr²

^{= \(2\times\frac{22}7\times10.5^2\)}

= 693 cm²

K.E. because friction does its work against motion.

External force, because in many-particle system, the internal forces in system cancel out.

Dimensional formula of wavelength = [M⁰L¹T⁰] 

Dimensional formula of frequency = [M⁰L⁰T^-1 ]

The dimensional formula of torque is [ML²T^-2]

The dimensions of torque [ML²T^-2].

(i)1, (ii) 3, (iii) 4, (iv) 4, (v) 3, (vi) 5 since it comes from a measurement the last two zeros become significant.

1 l/y = 9.46 x 10¹⁵m

1m = \(\frac{1}{9.46 \times 10^{15}}\) = 1.057 x 10^-16 ly

(i) 1, 

(ii) 3, 

(iii) 4, 

(iv) 4, 

(v) 3, 

(vi) 5 since it comes from a measurement the last two zeros become significant.

(a) SONAR → Sound Navigation and Ranging. 

(b) RADAR → Radio Detection and Ranging.

(a) SONAR ➝ Sound Navigation and Ranging. 

(b) RADAR ➝ Radio Detection and Ranging.

Correct answer is (a) [ML² T^-1] 

(c) Relative error

Correct answer is (a) 0

Correct answer is (d) Faraday

Correct answer is (c) ML²T⁰

Correct answer is (c) A /B

An atomic mass unit (symbolized AMU or amu) is defined as precisely 1/12 the mass of an atom of carbon-12. The carbon-12 (C-12) atom has six protons and six neutrons in its nucleus. 

In imprecise terms, one AMU is the average of the proton rest mass and the neutron rest mass. This is approximately 1.67377 × 10^-27 kilogram (kg), or 1.67377 × 10^-24 gram (g). The mass of an atom in AMU is roughly equal to the sum of the number of protons and neutrons in the nucleus. 

The AMU is used to express the relative masses of, and thereby differentiate between, various isotopes of elements. Thus, for example, uranium235 (U-235) has an AMU of approximately 235, while uranium-238 (U-238) is slightly more massive. The difference results from the fact that U-238, the most abundant naturally occurring isotope of uranium, has three more neutrons than U-235, an isotope that has been used in nuclear reactors and atomic bombs.

Correct answer is (c) 160 m

Correct answer is (a) Kg²

Correct answer is (d) 7%

Correct answer is (a) Nm^-1

Correct answer is (a) 10° s

Correct answer is (d) 10³ s

Work, energy and torque.

Correct answer is (d) 6%

Correct answer is (b) NS

(b) Boltzmann constant

Physical quantities are classified into two types.

There are fundamental and derived quantities. Fundamental or base quantities are quantities which cannot be expressed in terms of any other physical quantities. These are length, mass, time, electric current, temperature, luminous intensity and amount of substance. 

Quantities that can be expressed in terms of fundamental quantities are called derived quantities. 

For example, area, volume, velocity, acceleration, force.

Given 

Let length of parallel sides be 84cm and y cm 

Area of trapezium = 1586 cm² 

Let Height (h) = 26 cm

We know that area of trapezium is \(\frac{1}{2}\)× (sum of parallel sides) × height

Therefore Area of trapezium is \(\frac{1}{2}\)× (84 + y) × 26 =1586 cm².

\(\therefore \frac{1}{2}\)× (84 +y) × 26 = 1586

⇒ (84 + y) × 13 =1586

⇒ 84 + y = \(\frac{1586}{13}\)

⇒ y = 122 -  84 = 38

∴ Length of the other parallel side is 38 cm.

Correct answer is (b) error

Correct answer is (d) 4%