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Data : R = 10 cm = 0.1 m, 

N = 1, I = 0.1 A 

The magnetic moment,

μ = NIA = NI(πR²) 

= (1) (0.1 A) (3.142) (0.1 m)² 

= 3.142 × 10^-3 A∙m²

Correct option is: (D) 20 × 10^-4 N, upward.

Correct option is: (A) remain uncharged.

Correct option is (A) tan^-1 2π 

(A) Period T of its circular motion

(D) the p.d. across the gap between the dees.

 The gain of an op-amp without feedback is called the open-loop gain whereas the gain of an op-amp with a feedback circuit is called the closed-loop gain. 

Input offset voltage is the small voltage that should be applied between two input terminals to  make output voltage zero.

A = 100

A_f = 40

V_in = 1V

A_f = \(\frac{A}{1+A\beta}\)

40 = \(\frac{100}{1+100\beta}\)

1 + 100β = \(\frac{100}{40}\) = 2.5

A = \(\frac{V_o}{V_{in}}\)

100β = 2.5 - 1 = 1.5

100 = \(\frac{V_o}{1}\)

β = \(\frac{1.5}{100}\) = 0.015

V_o = 100V

β = \(\frac{V_f}{V_o}\)

V_f = βV_o

= 0.015 x 100

= 1.5 V

V_in = V_s - V_f

1 = V_s - 1.5

V_s = 2.5 Volt

An amplifier should have high input impedance so that it will not load the preceding stage or input voltage source. High input impedance for an amplifier can be achieved with Voltage series negative feedback.

Let I_i and V_i be input current and input voltage to the basic amplifier. V_s is input to feedback amplifier, V_f is feedback voltage and V_o is the output voltage.

Input impedance of basic amplifier is

\(z_i=\frac{V_i}{I_i}\)   ....(1)

Input impedance of feedback amplifier is

\(z_v= \frac{V_s}{I_i}\)    ....(2)

Input voltage to basic amplifier with negative feedback is

V_i = V_s - V_f

= V_s - βV_o  {∴ β = \(\frac{V_f}{V_o}\), V_f = BV_o}

= V_s - β.AV_i {∴ A = \(\frac{V_o}{V_{in}}\), V_o = AV_in}

V_i(1 + β) = V_s  ....(3)

Put eqn (3) in eqn (2)

\(z_{if}\) = \(\frac{V_s}{I_i}\)

= \(\frac{V_i(1+A\beta)}{I_i}\)

= \((\frac{V_i}{I_i})(1+A\beta)\)

Z_if = Z_i (1 + Aβ) from eq …… (1)
As (1 + Aβ) > 1 for negative feedback, Z_if > Z_i
Hence with negative feedback, input impedance of an amplifier increases.

BW_f  = BW(1 + Aβ)
The gain-bandwidth product of an amplifier is always a constant i.e.,

BW x A = BW_f x A_f

(i) Feedback factor is the ratio of feedback voltage to the output voltage.

(ii) Loop gain of an amplifier is the product of gain and feedback fraction.

(iii) Closed loop gain is the gain of an amplifier with negative feedback.

(i) Positive feedback is the process of applying a portion of output in phase with the input.

(ii) Negative feedback is the process of applying a portion of output out of phase with the input.

Correct Answer is: (a) a^3/2 , (d) m^-1/2

Correct Answer is: (a, b)

Correct Answer is: (d) √(4H/3g)

As the masses of the body and the earth are comparable, they will both move towards their centre of mass, which remains stationary. When they meet, the body will have moved through a distance H/3 and 2H/3 In both cases, the body moves with acceleration g, as H << R, i.e., the body is close to the earth’s surface.

The substance that gets used up in any reaction is called limiting reagent.

Correct option: (A) 20 mL

N₂ + 3H₂ \(\rightleftharpoons\)  2NH₃

10 mL 30 mL 20 mL

Thus, ammonia obtained in the reversible reaction of NH₃ is 20 mL.

1u = 1/12 Mass of the carbon atom (¹²c) 

6.022X10²³ atoms of carbon = 12g 

I atom of carbon = 12/6.022x10²³ g 

= 1.99x10^-23g 

1u = 1.99 x 10⁻²³ /12 = 1.66x10^-24g 

(i) Number of particles in a mole = 6.022 × 10²³

1 attogram = 10^–18g

Number of carbon atom = 1

Atomic mass of carbon = 12

\(\because\) 12 g of carbon contain

= 1 × 6.022 × 10²³ carbon atoms

\(\therefore\) 10^–18 g of carbon contains

= \(\frac{1\times6.022\times10^{23}\times10^{-18}}{12}\)

= 5.02 x 10⁴ carbon atoms

(ii)

4.68 mg of Silicon = 4.68 x 10^-3 g of silicon 

moles of Silicon = Mass of Silicon / Atomic Mass (g) = 4.68x10⁻³/ 28.1 = 1.66x10^-4 mol.

Atomic mass of elements A 

= \(\frac{\%\,of\,elements\,A}{Relative\,number\,of\,moles}\)

= \(\frac{70}{1.25}\)

Atomic mass of element B 

= \(\frac{\%\,of\,elements\,B}{Relative\,number\,of\,moles}\)

=\(\frac{30}{1.88}\)

= 15.957 ≈ 16

(ii)

\(\therefore\) Empirical formula of compound = \(A_2B_3\)

Molecular mass = 160

Empirical formula mass = 2(56) + 3(16)

= 112 + 48 = 160

n = \(\frac{Molecular\,mass}{Empirical\,formula\,mass}\) = \(\frac{160}{160}\) = 1

\(\therefore\) Molecular formula = n x Empirical formula

= 1 x A₂B₃

= A₂B₃

\(\therefore\) Empirical formula of compound = CH₂Cl

n = \(\frac{Molecular\,mass}{Empirical\,formula\,mass}\)

Molecular mass = 98.96 g

Empirical formula mass = 12 + 2(1) + 35.5

= 49.5

n = \(\frac{98.96}{49.5}\) = 2

\(\therefore\) Molecular formula = n × Empirical formula

= 2 x CH₂Cl

= C₂H₄Cl₂

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

The empty space in body centred cubic lattice is 32%

Precision refers to the closeness of various measurements for the same quantity. However, accuracy is the agreement of a particular value to the true value of the result.

Chemical substance which reduce anxiety, stress by acting on nerves e.g. Equanil, luminal, seconal. 

Correct option: (C) Cd²⁺ 

Clemmensen reduction and Wolff-Kishner reduction.

Correct option is d. +4.41 x 10^-17 J/atom

Correct option is a. 20565 cm^-1

In FCH₂ - COOH, tluorine is more electron withdrawing and has stronger ^-l effect than chlorine in CICH₂ - COOH. So, FCH₂COOH is more acidic than CICH₂COOH hence its pK_a value is lesser than CICH₂COOH.

The vertical columns in a Periodic Table are called groups and the horizontal rows are called periods.

Correct option is  B) only ii

Number of lines produced when an electron from nth shell drops to ground state =  n(n-1)/ 2

Now according to question lines produced when an electron drops from 6th shell to ground state

= 6(6-1) / 2 = 15

Benzene reacts with H₂ in the presence of Pt to give Cyclohexane.

FCH₂COOH is stronger acid than CH₃COOH. Due to -I effect of Fluorine electron density is low in the O —H bond and greater stability of FCH₂COO^– ion over CH₃COO^– ion.

R — CHO + 2 Cu²⁺ + 5 OH^– → RCOO^– + Cu₂O + 3 H₂O (Reddish brown ppt.)

Benzaldehyde does not give Fehling soln. test (Aromatic aldehydes do not give this test.)

From graph (a)

As, v_x = \(\frac{dx}{dt}\)= \(\frac{2}{2}\) = 1 m/s

a_x = \(\frac{dv_x}{dt}\) = 0

from (b), y = t²

v_y = \(\frac{dy}{dt}\) = 2t

a_y = \(\frac{dv_y}{dt}\) = 2

F_y = ma_y

= 0.5 × 2 = 1N (toward y-axis)

F_x = 0.5 × 0 = 0 N

F = \(\sqrt{F^2_s+F^2_y}=\sqrt{0^2+1^2}\)

F = 1 N (towards Y-axis)

CH₃CH₂CH₂COCH₃ will give iodoform test as it has a terminal methyl ketone group.

Electronic configuration of Kr (atomic no.=36) = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 

Total no. of s-electrons = 8, total no. of p-electrons = 18. 

Difference = 10 

No. of d- electrons = 10

Correct answer is C.

Let x₁ < x₂ and both are real number

a^x₁ < a^x₂

\(\Rightarrow\) f(x₁) < f(x₂)

\(\Rightarrow\) x₁ < x₂ ∈

only possible on a > 1

Formula:-

(i) The necessary and sufficient condition for differentiable function defined on (a, b) to be strictly increasing on (a, b) is that f’(x) > 0 for all x ∈ (a, b)

If f(x) is strictly increasing function on interval [a, b], then f^-1 exist and it is also a strictly increasing function

Correct answer is D.

Formula:-

(i) The necessary and sufficient condition for differentiable function defined on (a, b) to be strictly increasing on (a, b) is that f’(x) > 0 for all x ∈ (a, b)

Given:-

For x < 0

f(x) = -1

for 0 < x < 1

f(x) = 2x - 1

for x > 1

f(x) = 1

Hence f(x) will increasing in 0 < x < 1

Given as f(x) = 5x³ – 15x² – 120x + 3

Differentiate the above equation with respect x, we get

f'(x) = (d/dx)(5x³ - 15x² - 120x + 3)

⇒ f’(x) = 15x² – 30x – 120

For the function f(x) we have to find critical point, we must have

⇒ f’(x) = 0

⇒ 15x² – 30x – 120 = 0

⇒ 15(x² – 2x – 8) = 0

⇒ 15(x² – 4x + 2x – 8) = 0

⇒ x² – 4x + 2x – 8 = 0

⇒ (x – 4) (x + 2) = 0

⇒ x = 4, – 2

It is clear, f’(x) > 0 if x < –2 and x > 4 and f’(x) < 0 if –2 < x < 4

Hence, f(x) increases on (–∞,–2) ∪ (4, ∞) and f(x) is decreasing on interval x ∈ (–2, 4)

Given as f(x) = 2x³ + 9x² + 12x + 20

Differentiate the above equation we get

f'(x) = (d/dx)(2x³ + 9x² + 12x + 20)

⇒ f’(x) = 6x² + 18x + 12

For the function f(x) we have to find critical point, we must have

⇒ f’(x) = 0

⇒ 6x² + 18x + 12 = 0

⇒ 6(x² + 3x + 2) = 0

⇒ 6(x² + 2x + x + 2) = 0

⇒ x² + 2x + x + 2 = 0

⇒ (x + 2) (x + 1) = 0

⇒ x = –1, –2

It is clear, f’(x) > 0 if –2 < x < –1 and f’(x) < 0 if x < –1 and x > –2

Hence, f(x) increases on x ∈ (–2,–1) and f(x) is decreasing on interval (–∞, –2) ∪ (–2, ∞)

Given as f(x) = 2x³ – 15x² + 36x + 1

Differentiate the above equation with respect x, we get

f'(x) = (d/dx)(2x³ – 15x² + 36x + 1)

⇒ f’(x) = 6x² – 30x + 36

For the function f(x) we have to find critical point, we must have

⇒ f’(x) = 0

⇒ 6x² – 30x + 36 = 0

⇒ 6 (x² – 5x + 6) = 0

⇒ 3(x² – 3x – 2x + 6) = 0

⇒ x² – 3x – 2x + 6 = 0

⇒ (x – 3) (x – 2) = 0

⇒ x = 3, 2

It is clear, f’(x) > 0 if x < 2 and x > 3 and f’(x) < 0 if 2 < x < 3

Hence, f(x) increases on (–∞, 2) ∪ (3, ∞) and f(x) is decreasing on interval x ∈ (2, 3)

Given as f(x) = 8 + 36x + 3x² – 2x³

Differentiate with respect to x

f'(x) = (d/dx)(8 + 36x + 3x² - 2x³)

⇒ f’(x) = 36 + 6x – 6x²

For the f(x) we have to find critical point, we must have

⇒ f’(x) = 0

⇒ 36 + 6x – 6x² = 0

⇒ 6(–x² + x + 6) = 0

⇒ 6(–x² + 3x – 2x + 6) = 0

⇒ –x² + 3x – 2x + 6 = 0

⇒ x² – 3x + 2x – 6 = 0

⇒ (x – 3) (x + 2) = 0

⇒ x = 3, – 2

It is clear, f’(x) > 0 if –2 < x < 3 and f’(x) < 0 if x < –2 and x > 3

Hence, f(x) increases on x ∈ (–2, 3) and f(x) is decreasing on interval (–∞, 2) ∪ (3, ∞)

Given as f(x) = x³ – 6x² – 36x + 2

f'(x) = (d/dx)(x³ – 6x² – 36x + 2)

⇒ f’(x) = 3x² – 12x – 36

For the function f(x) we have to find critical point, we must have

⇒ f’(x) = 0

⇒ 3x² – 12x – 36 = 0

⇒ 3(x² – 4x – 12) = 0

⇒ 3(x² – 6x + 2x – 12) = 0

⇒ x² – 6x + 2x – 12 = 0

⇒ (x – 6) (x + 2) = 0

⇒ x = 6, – 2

It is clear, f’(x) > 0 if x < –2 and x > 6 and f’(x) < 0 if –2< x < 6

Hence, f(x) increases on (–∞,–2) ∪ (6, ∞) and f(x) is decreasing on interval x ∈ (–2, 6)

Given as f(x) = 5 + 36x + 3x² – 2x³

f'(x) = (d/dx)(5 + 36x + 3x² - 2x³)

⇒ f’(x) = 36 + 6x – 6x²

For the function f(x) now we have to find critical point, we must have

⇒ f’(x) = 0

⇒ 36 + 6x – 6x² = 0

⇒ 6(–x² + x + 6) = 0

⇒ 6(–x² + 3x – 2x + 6) = 0

⇒ –x² + 3x – 2x + 6 = 0

⇒ x² – 3x + 2x – 6 = 0

⇒ (x – 3) (x + 2) = 0

⇒ x = 3, – 2

It is clear, f’(x) > 0 if –2< x < 3 and f’(x) < 0 if x < –2 and x > 3

Hence, f(x) increases on x ∈ (–2, 3) and f(x) is decreasing on interval (–∞, –2) ∪ (3, ∞)