TextBook Solutions
- A point charge causes an electric flux of -1.0 x 103 Nm2/C to pass through a spherical Gaussian surface of 10.0 cm (a) If the radius of the gaussian surface is doubled, the electric flux through the new surface will still remain the same i.e. -1 x 10³Nm²C... 1
- An infinite line charge produces a field of 9 x 104 N/C at a distance of 2cm. Calculate the linear charge density. E = 9 x 10⁴ NC⁻¹, r = 2 cm = 2 x 10⁻²m E = 9 x 10⁹ x 2λ / 2 λ = fracEr9 × 10⁹ × 2 = frac9 × 10⁴ × 2 × 10⁻²9 × 10⁹ × 2 = 0.1µC/m 1
- Although ordinary rubber is an insulator, the special rubber tyres of aircraft are made slightly conducting The aircraft while in motion gets charged due to atmospheric air friction. The continuous charging may raise the sharp conducting metal part... 1
- Is coulomb a very big unit of charge? Yes. The force of repulsion between two one coulomb charges placed one meter apart in air or vacuum is 9 x 10⁹ N, which is enormous. 1
- What is the effect of keeping the dipole in the field? If the dipole is placed parallel to the field, it is in stable equilibrium, and if placed antiparallel, it is in the unstable equilibrium. I... 1
- How does the electric field (E) vary with distance (r) in the following cases? (i) Outside ∝ frac 1 r ² , on surface = frac 1 r ² , inside = zero (ii) On the surface = frac 1 r ² , inside ∝ r (iii) Constant, E = fracσ2... 2
- An electric flux of – 6 x 103 Nm2/C passes normally through a spherical Gaussian surface of a radius of 10 cm (i) φ = – 6 x 103Nm2C-1 By Gauss’s theorem, φ = fracqε₀ q= ε₀ x φ = 8.85 x 1o-12 x – 6 x 103 = – 5.31 x 10-8C (ii) Same (- 6 x 103 Nm2C-1) 1
- In the matrix A = 2 5 19 -7 35 -2 5/2 12 (i) The order of the matrix = 3 x 4. (ii) Hie number of elements = 3 x 4 = 12. (iii) a13 = 19. a21 = 35, a33 = – 5, a24 = 12, a23 = 5 / 2 . 1
- If a matrix has 24 elements, what are possible orders it can have? What, if it has 13 elements? (i) 24 = 1 x 24 = 2 x 12 = 3 x 8 = 4 x 6 Thus, there are 8 matrices having 24 elements. Their orders are 1 x 24, 24 x 1, 2 x 12, 12 x 2, 3 x... 1
- If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements? (i) 18 = 1 x 18 = 2 x 9 = 3 x 6 There are 6 matrices of orders 1 x 18, 18 x 1, 2 x 9, 9 x 2, 3 x 6, 6 x 3. (ii) 5 = 1 x 5 There are two matr... 1
- Find the values of x, y and z from the following equations: (i) [beginarrayll 4 & 3 \\ x & 5 endarray] = [beginarrayll y & 3 \\ 1 & 5 endarray] Equating corresponding elements of the two matrices, we... 1
- Find the values of a, b, c and d from the equation \(\left[\begin{array}{cc} a-b & 3 a+c \end{array}\right]\) Comparing the corresponding elements of [beginarraycc a-b & 3 a+c \\ 2 a-b & 3 c+d endarray] = [beginarraycc -1 & 5 \\ 0 & 13 endarray] , we... 1
- A = [aij]mxn is a square matrix, if (A) m < n (B) m > n (C) m-n (D) None of these. For a square matrix m = n. 2
- The number of all possible matrices of order 3 x 3 with each entry 0 or 1 is (A) 27 (B) 18 (C) 81 (D) 512 There are 9 entries in a 3 x 3 matrix. Each place can be filled with 0 or 1. ∴ 9 places can be filled in 2⁹ = 512 ways. Number of such matri... 1
- cos-1(cos\(\frac { 13π }{ 6 }\)) cos⁻¹(cos 13π / 6 ) = cos⁻¹cos(2π + π / 6 = cos⁻¹cos π / 6 ) = π / 6 1
- tan-1(tan\(\frac { 7π }{ 6 }\)) tan⁻¹(tan 7π / 6 ) = tan⁻¹tan(π + π / 6 = tan⁻¹tan π / 6 ) = π / 6 . 1
- tan-1\(\sqrt{x}\) = \(\frac { 1 }{ 2 }\)cos-1\(\frac { 1-x }{ 1+x }\), x ∈ [0, 1] Put x = tan²θ. ∴ θ = tan⁻¹ √(x) R.H.S. = 1 / 2 cos⁻¹ 1-x / 1+x = 1 / 2 cos⁻¹ (frac1-tan ² θ1+tan ² θ) = 1 / 2 cos⁻¹(cos2θ) = 1 / 2 x 2θ = θ.... 1
- 2tan-1(cos x) = tan-1(cosec x) Now L.H.S = 2tan⁻¹(cos x) = tan⁻¹ (frac2 cos x1-cos ² x) = tan⁻¹ (frac2 cos xsin ² x) Putting this value in the given equation, we get tan⁻¹... 1
- sin tan-1x, |x| < 1 is equal to (A) \(\frac{x}{\sqrt{1-x^{2}}}\) (B) \(\frac{1}{\sqrt{1-x^{2}}}\) Let tan⁻¹x = α ∴ tan α = x. So, sin α = fracxsqrt1+x² or α = sin⁻¹ fracxsqrt1+x² Now sin tan⁻¹x = sin α = sin(sin⁻¹ fracxsqrt1+x² ) = fracxs... 1
- If sin-1(1-x)-2 sin-1x = \(\frac { π }{ 2 }\), then x is equal to (A) 0, \(\frac { 1 }{ 2 }\) sin⁻¹(1-x)-2 sin⁻¹x = π / 2 Putting π / 2 = sin⁻¹(1 – x) + cos⁻¹(1 – x), sin⁻¹(1 – x) – 2 sin⁻¹ = sin⁻¹(1 – x) + cos⁻¹(1 – x) ⇒ – 2 sin⁻¹x =... 1
- 3sin-1x = sin-1(3x – 4x³), x ∈ [- \(\frac { 1 }{ 2 }\), \(\frac { 1 }{ 2 }\) ] Let sin⁻¹ = θ ∴ sin θ = x Also, ∴ sin 3θ = 2 sinθ – 4sin³θ = 3x – 4x³ ∴ 3θ = sin⁻¹(3x – 4x³) or 3 sin⁻¹x = sin⁻¹(3x – 4x³), x ∈ [- 1 / 2 , 1... 1
- 3cos-1x = cos-1(4x³ – 3x), x ∈ [\(\frac { 1 }{ 2 }\), 1] Let cos⁻¹x = θ. cos θ = x. Now, cos 3θ = 4cos³θ – 3 cos θ = 4x³ – 3x. or 3θ = cos⁻¹(4x³ – 3x) or 3 cos⁻¹x = cos⁻¹(4x³ – 3x), x ∈ [- 1 / 2 ,... 1
- If sin(sin-1\(\frac { 1 }{ 5 }\) + cos-1x) = 1, then find the value of x. sin(sin⁻¹ 1 / 5 + cos⁻¹x) = sin π / 2 ⇒ sin⁻¹ 1 / 5 + cos⁻¹x = sin π / 2 or (sin⁻¹ 1 / 5 + cos⁻¹ 1 / 5 ) + (- cos⁻¹ 1 / 5 x) = π / 2 or π /... 1
- If tan-1\(\frac { x-1 }{ x+2 }\) + tan-1\(\frac { x+1 }{ x+2 }\) = \(\frac { π }{ 4 }\), then find the value of x. L.H.S. = tan⁻¹ x-1 / x+2 + tan⁻¹ x+1 / x+2 = tan⁻¹[ x-1 / x+2 + x+1 / x+2 /1-( x-1 / x+2 )( x+1 / x+2 )] = tan⁻¹[2x²-4/-3] = πover4=tan⁻¹(1)... 1