Explore topic-wise InterviewSolutions in Current Affairs.

Correct answer is option (a) : Torque and work have the same dimensions.

Conventional flow assumes charges flow from positive to negative.

The number of irrational terms in the expansion is 97.

Explanation:

From binomial theory we know that the (r+1)th term of the expansion

(a+b)ⁿ is (ⁿC_r)*(a^n-r)*(b^r)

In this case a = 5^1/8 ; b = 2^1/6 ; n=100 ;

=> Total number of terms = n+1 = 101

For a term to be rational the powers of ‘a’ and ‘b’ should be integral multiples of 8 and 6 respectively so as to cancel out the fractional exponents.

n-r = 8k and r = 6m where k and m are some non negative integers
n-r should take values 0,8,16,24,32,40,48,56,64,72,80,88,96 (8k)
r should take values among 0,6,12,18,24,30,36,42,48,54,60,66,72,78,84,90,96 (6m)

As n= 100 for n-r = 8k, r should take 100,92,84,76,68,60, 52,44, 36,28,20,12,4

The common values of r which satisfy r = 6m and n-r = 8k simultaneously are r = 12, 36, 60, 84.

There will be 4 rational terms thus it is obvious that remaining 97 terms will be irrational

Hint— When a person is not able to see distant objects clearly but can see nearby objects clearly then he is considered to be myopic. If it is otherwise, he is hypermetropic. Give explanation based on figures.

Limitations of Coulomb’s Law

• The law is applicable only for the point charges at rest.
• Coulomb’s Law can be only applied in those cases where the inverse square law is obeyed.
• It is difficult to implement Coulomb’s law where charges are in arbitrary shape because in such cases we cannot determine the distance’ between the charges.
• The law can’t be used directly to calculate the charge on the big planets.

An implant is a medical device manufactured to replace a missing biological structure, support a damaged biological structure, or enhance an existing biological structure. Medical implants are man-made devices, in contrast to a transplant, which is a transplanted biomedical tissue.

Power will be found by the conjugate method. Using current conjugate, we have

P_VA = (100 − j50) (3 + j4) = 300 + j400 − j150 + 200 = 500 + j 250

Hence, real power is 500 W and reactive power of VAR is 250. Since the second term in the above expression is positive, the reactive volt-amperes of 250 are inductive.

Concept:

The power dissipated through resistor, in a circuit consist of voltage source and a resistor.

P = V² / R = I²R

Where P = Power

V = Source voltage

I = Current through resistance

R = Resistor

Calculation:

Given, V = 105 sin 377t V

R = 10 Ω 

P = V2 / R = (105 sin 377t)² / 15 = 735 sin2 377t W

The instantaneous power in ac circuits can be obtained by taking the product of the instantaneous values of current and voltage.

The current waveform in this case can be represented as

i(t) = I sin (ωt - ϕ)

Now, the expression of the instantaneous power,

p (t) = u(t).i(t)

⇒ p (t) = V sin ωt I sin (ωt - ϕ)

⇒ p (t) = VI sin ωt sin (ωt - ϕ)

⇒ p (t) = VI sin ωt (sin ωt cos ϕ – cos ωt sin ϕ)

⇒ p (t) = VI (sin2ωt cos ϕ – sin ωt cos ωt sin ϕ)

\(\Rightarrow p\left( t \right) = \frac{V}{{\sqrt 2 }}.\frac{I}{{\sqrt 2 }}\left( {2{{\sin }^2}\omega t\cos \phi - 2\sin \omega t\cos \omega t\sin \phi } \right)\)

\(\Rightarrow p\left( t \right) = {V_{rms}}{I_{rms}}\cos \phi \left( {1 - \cos 2\omega t} \right) - {V_{rms}}{I_{rms}}\sin \phi \sin 2\omega t]\)

[Where Vrms and Irms is the root mean square value of voltage and current waveform]

= P (1 – cos 2ωt) – Q sin 2ωt

1. Young’s Modulus (Y) 

2. Bulk Modulus (B) 

3. Rigidity modulus (R)

SI unit of stress: Nm^-2

1. To check the correctness of an equation. 

2. To derive the relation between various physical quantities.

It is the motion in which velocity of all the particles of fluid is same at a given point.

It is the maximum displacement of oscillating particles on either side of its mean position.

Newton’s formula for speed of sound in air:

V = √(P/ρ)

15 x 500 x D/d = 10 x λ₂ x D/d

λ₂ = 15 x 50nm

λ₂ = 750nm

6 degrees of freedom.

i = (12 - 8)/(200 + 200) A = 4/400 = 10^-2A

Power loss in each diode = (4)(10^-2)W = 40mW

Let m be the mass of steam required.
By principle of calorimetry
100×1×(90−24)=m540+m×1(100−90)
i.e.,550m=100×66⇒m=12g

Determine the amplitude, frequency, period and initial phase of the motion. 

x(t) = 5.4 cos (6πt + π/4) 

A = 5.4 m 

n = 3 Hz 

T = 0.333 s 

Initial phase = Φ = π/4

(1) Zero 

A perfect diamagnetic substance will completely expel the magnetic field. Therefore, there will be no magnetic field inside the cavity of sphere. Hence the paramagnetic substance kept inside the cavity will experience no force.

In series circuit the resultant capacitance 1/C_T = 1/c₁+1/c₂ + 1c_n farad. In parallel circuit the resultant capacitance C_T = c₁ + c₂ + c_n farad. 

It is the voltage available at the terminal of the source of supply. It’s symbol is V_T. V_T = emf – P.D  

Compression in spring is x,

Total energy E = constant

\(\frac{1}{2}kx^2\) + \(\frac{1}{2}mv^2\) + 4(\(\frac{1}{2}mv^2\) + \(\frac{1}{2}Iw^2\)) =  constant

Rolling on ground v = Rω

ω = \(\frac{V}{R}\) 

∵ I = \(\frac{MR^2}{2}\) 

\(\frac{1}{2}kx^2\) + \(\frac{1}{2}mv^2\) + 4(\(\frac{1}{2}mv^2\) + \(\frac{1}{4}MR^2\) x \(\frac{V^2}{R^2}\)) = constant

kx² + mv² + 4(Mv² + \(\frac{1}{2}\)Mv²) = constant

kx² + mv² + 4(\(\frac{3}{2}\)Mv²) = constant

kx² + mv² + 6Mv²

kx² = - (m + 6M)v² .........(1)

Different equation (1)

k 2x(\(\frac{dx}{dt}\)) = - (m + 6M) 2v\(\frac{dv}{dt}\)

kx = -(m + 6M)a

x = \(\frac{-(m+6M)a}{k}\)

a = \(\frac{-k}{(m+6M)}x\) .........(2)

a = - ω²x ..........(3)

Compression this equation (2) and (3)

Then,

ω² = \(\frac{k}{m+6M}\) 

⇒ ω = \(\sqrt{\frac{k}{m+6M}}\) 

Frequency formula,

f = \(\frac{\omega}{2\pi}\)

∴ f = \(\frac{1}{2\pi}\)\(\sqrt{\frac{k}{m+6M}}\)

(a) Long distance radio broadcasts use shortwave bands because only these bands can be refracted by the ionosphere.
(b) It is necessary to use satellites for long distance TV transmissions because television signals are of high frequencies and high energies. Thus, these signals are not reflected by the ionosphere. Hence, satellites are helpful in reflecting TV signals. Also, they help in long distance TV transmissions.
(c) With reference to X-ray astronomy, X-rays are absorbed by the atmosphere. However, visible and radio waves can penetrate it. Hence, optical and radio telescopes are built on the ground, while X-ray astronomy is possible only with the help of satellites orbiting the Earth.
(d) The small ozone layer on the top of the atmosphere is crucial for human survival because it absorbs harmful ultraviolet radiations present in sunlight and prevents it from reaching the Earth’s surface.
(e) In the absence of an atmosphere, there would be no greenhouse effect on the surface of the Earth. As a result, the temperature of the Earth would decrease rapidly, making it chilly and difficult for human survival.
(f) A global nuclear war on the surface of the Earth would have disastrous consequences. Post-nuclear war, the Earth will experience severe winter as the war will produce clouds of smoke that would cover maximum parts of the sky, thereby preventing solar light form reaching the atmosphere. Also, it will lead to the depletion of the ozone layer.

(a) Long distance radio broadcasts use shortwave bands because only these bands can be refracted by the ionosphere.
(b) It is necessary to use satellites for long distance TV transmissions because television signals are of high frequencies and high energies. Thus, these signals are not reflected by the ionosphere. Hence, satellites are helpful in reflecting TV signals. Also, they help in long distance TV transmissions.
(c) With reference to X-ray astronomy, X-rays are absorbed by the atmosphere. However, visible and radio waves can penetrate it. Hence, optical and radio telescopes are built on the ground, while X-ray astronomy is possible only with the help of satellites orbiting the Earth.
(d) The small ozone layer on the top of the atmosphere is crucial for human survival because it absorbs harmful ultraviolet radiations present in sunlight and prevents it from reaching the Earth’s surface.
(e) In the absence of an atmosphere, there would be no greenhouse effect on the surface of the Earth. As a result, the temperature of the Earth would decrease rapidly, making it chilly and difficult for human survival.
(f) A global nuclear war on the surface of the Earth would have disastrous consequences. Post-nuclear war, the Earth will experience severe winter as the war will produce clouds of smoke that would cover maximum parts of the sky, thereby preventing solar light from reaching the atmosphere. Also, it will lead to the depletion of the ozone layer.

The force experienced by current carrying conductor placed in a magnetic field is largest when the current carrying conductor is placed in a direction perpendicular to that of magnetic field.

(B) β-D-2-deoxyribose, β -D-ribose

DNA contains ⇒ β – D – 2 – deoxyribose 

RNA contains ⇒ β – D – ribose

Correct option is (C) Buna-N 

Buna – N is synthetic rubber which can be stretched and retains its original status on releasing the force.

As an impurities Buttermilk more than Butter is mixed with butter.

2450 calories are needed for adolescence boys.

Mobile creche moves with workers.