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1 | Let f(x) = ∫ 3x7+6x5 (x3+2x8+3)2 dx, with f(0) = 0 and f(1) = 1 K . Find the value of K. | {'A': '1 6', 'B': '6', 'C': '1', 'D': '5 6'} | B | Sol: Given: f(x) = ∫ 3x7+6x5 (x3+2x3+3)2 dx with f(0) = 0 and f(1) = 1 K . Solution: 1. Step 1: Differentiate the integrand to find the derivative of the function: The function we are integrating is: f ′(x) = 3x7+6x5 (x3+2x8+3)2 We will attempt a substitution method to simplify this integral. Let's consider the denomin... | Mathematics |
2 | Let f(x) = x (1+kxm)1/m for m ≥2 and g(x) = (f ∘f ∘⋯∘f) f occurs m times (x). Then ∫xm−2g(x)dx equals: | {'A': '1 km(m−1) (1 + kmxm)1−1 m + C', 'B': '1 k(m−1) (1 + kmxm)1−1 m + C', 'C': '1 km(m+1) (1 + kmxm)1+ 1 m + C', 'D': '1 k(m+1) (1 + kmxm)1+ 1 m + C'} | A | Sol: The function f(x) = x (1+kxm)1/m maps x iteratively. The composition g(x) involves applying f, m times, but because of the structure of f, it simplifies as follows: Simplification of g(x) : When f(x) is applied m times, the effect is to "balance out" the power of 1 + kxm in the denominator. Therefore, g(x) = x (1+... | Mathematics |
3 | Let X = {(x, y) ∈Z × Z : x2 12 + y2 18 < 1 and y2 < 4x}. Three distinct points P, Q, and R are randomly chosen from X. What is the probability that P, Q, and R form a triangle whose area is a positive integer? | {'A': '47 132', 'B': '49 132', 'C': '53 132', 'D': '55 132'} | B | Sol: Characterizing the set X The points in X must satisfy two conditions: 1. x2 12 + y2 18 < 1, which defines an ellipse. 2. y2 < 4x, which is a parabolic inequality. Integer lattice points in X Rewriting the first inequality, x2 12 + y2 18 < 1 ⟹x2 3 + y2 2 < 12 This represents an ellipse centered at (0, 0), with semi... | Mathematics |
4 | Let p and q be the coefficients of the eighth and fourteenth terms, respectively, in the expansion of ( 1 4 x 1 4 + 1 3x 3 4 ) 20 . Then calculate ( q p ) 1 4 . | {'A': '7√3 9', 'B': '8√2 9', 'C': '8√7 9', 'D': '8√3 9'} | D | Sol: The general term in the binomial expansion of ( 1 4 x 1 4 + 1 3x 3 4 ) 20 is given by: Tk = (20 k )( 1 4 x 1 4 ) 20−k ( 1 3x 3 4 ) k Simplify: Tk = (20 k ) 1 420−k 1 3k x 1 4 (20−k)−3 4 k Tk = (20 k ) 1 420−k ⋅3k x 20 4 −k 4 −3k 4 Tk = (20 k ) 1 420−k ⋅3k x5−k For the r-th term in the expansion, the term index r r... | Mathematics |
5 | Consider the equation of a hyperbola: 4x2 −9y2 −12x + 36y −15 = 0 Let the vertex of the hyperbola be denoted by point A. Let B be one of the endpoints of the latus rectum, and C be the focus of the hyperbola closest to A. What is the area of triangle ABC ? | {'A': '1 + √7 3', 'B': '1 + √5 3', 'C': '1 + √1 3', 'D': '1 + √2 3'} | D | Sol: The given equation of the hyperbola is: 4x2 −9y2 −12x + 36y −15 = 0 First, let's rewrite this in the standard form by completing the square. - Group the x-terms and y-terms: (4x2 −12x) −(9y2 −36y) = 15 - Complete the square for the x-terms: 4 (x2 −3x) = 4 ((x −3 2 ) 2 −9 4 ) = 4(x −3 2 ) 2 −9 - Complete the square... | Mathematics |
6 | Let a and b be real numbers such that a ≠0, a3 ≠b, and a3 ≠−b. If α and β are nonzero complex numbers satisfying α + β = −a and α3 + β3 = b, then a quadratic equation having α β and β α as its roots is: | {'A': '(a3 + b)x2 −(a3 −2b)x + (a3 + b) = 0', 'B': '(a3 + b)x2 −(a3 −2b)x + (a3 + b) = 0', 'C': '(a3 −b)x2 −(5a3 −2b)x + (a3 −b) = 0', 'D': '(a3 −b)x2 −(5a3 + 2b)x + (a3 −b) = 0'} | A | Sol: Using the identity for the cube of sums: α3 + β3 = (α + β) ((α + β)2 −3αβ) Substitute α + β = −a : α3 + β3 = (−a) ((−a)2 −3αβ). Simplify: b = −a (a2 −3αβ) Rearranging, we find: αβ = a3+b 3a . Thus, the symmetric sums of α and β are: α + β = −a, αβ = a3+b 3a The roots of the quadratic equation are α β and β α . Den... | Mathematics |
7 | Let y1, y2, … , y12 be twelve observations such that: ∑12 i=1 (yi −3) = 36, ∑12 i=1 (yi −γ)2 = 144, γ > 3 and their variance is 3 4 . If μ and σ2 are the mean and variance of the transformed data: 3 (y1 −2) + 5γ, 3 (y2 −2) + 5γ, … , 3 (y12 −2) + 5γ then find the value of γμ σ2 . | {'A': '137 3 + 16√5', 'B': '80 −4√5', 'C': '40 + 8√5', 'D': '23 + 3√5'} | A | Sol: 1. Calculate the Mean of yi: ∑12 i=1 (yi −3) = 36 ∑12 i=1 yi −12 × 3 = 36 ⟹∑12 i=1 yi = 36 + 36 = 72 Mean M = ∑12 i=1 yi 12 = 72 12 = 6 2. Calculate γ : The variance is given by: σ2 = 1 12 ∑12 i=1 (yi −6)2 = 3 4 ∑12 i=1 (yi −6)2 = 12 × 3 4 = 9 Since: ∑12 i=1 (yi −γ)2 = 144 Using the relation: ∑12 i=1 (yi −γ)2 = ∑1... | Mathematics |
8 | The normal at a point P on the ellipse x2 + 4y2 = 16 meets the x-axis at Q. If M is the mid point of the line segment PQ, then the locus of M intersects the latus rectums of the given ellipse at the points: | {'A': '(±2√3 , ±1/7)', 'B': '(±3√5/2 , ±√19/4)', 'C': '(±3√5/2 , ± 2/7)', 'D': '(±2√3 , ±4√3/7)'} | A | Sol: Given: Ellipse= x2 16 + y2 4 = 1 e = √1 −b2 a2 = √3 2 ∵P is a point on the ellipse So, P = (4 cos θ, 2 sin θ) Equation of normal to the ellipse x2 16 + y2 4 = 1 at point (x1 , y1) = (4 cos θ, 2 sin θ) is given by a2y1 (x −x1) = b2x1 (y −y1) ⟹16 × 2 sin θ (x −4 cos θ) = 4 × 4 cos θ (y −2 sin θ) ⟹2x sin θ −8 sin θ c... | Mathematics |
9 | If two random variables X and Y are connected by relation 2X−3Y 5X+4Y = 4 and X follows Binomial distribution with parameters n = 10 and p = 1 2 , then what is the variance of Y ? | {'A': '810 361', 'B': '9 19', 'C': '21 361', 'D': '121 361'} | A | Sol: Given: Given: If two random variables X and Y are connected by the relation 2X−3Y 5X+4Y = 4 and X follows a Binomial distribution with parameters n = 10 and p = 1 2 Concept Used: Concept Used: For Binomial distribution X ∼ B(n, p) , E(X) = np and Var(X) = np(1-p) Explanation: Explanation: Given relation: 2X−3Y 5X+... | Mathematics |
10 | If →a and → b are vectors in space given by →a = ^i−2^j √5 and → b = 2^i+^j+3^k √14 , then the value of (2→a + → b). [(→a× → b)×(→a −2 → b)] is | {'A': '4', 'B': '5', 'C': '9', 'D': '7'} | B | Sol: Concept Concept - Use the dot product and cross product formula →a. → b = a1b1 + a2b2 + a3b3 (→a −2 → b) × (→a × → b) = [{(→a −2 → b). → b}→a −{(→a −2 → b).→a} → b] Solution Solution - →a. → b = 1 √5 × 2 √14 + ( −2 √5 ) × 1 √14 = 2 √70 − 2 √70 = 0 |→a| = √( 1 √5 ) 2 + ( −2 √5 ) 2 = √1+4 5 = 1 | → b| = √( 2 √14 ) 2... | Mathematics |
11 | If the eqation of the plane containing the line x +2y + 3z – 4 = 0 = 2x + y –z + 5 and perpendicularto the plane→r = (^i −^j) + λ(^i + ^j + ^k) + μ(^i −2^j + 3^k)is ax + by + cz = 4, then (a – b + c) is equal to | {'A': '22', 'B': '24', 'C': '20', 'D': '18'} | A | Sol: Concept - Concept - When two planes are perpendicular, then dot product of their normals is zero. Explanation - Explanation - Equation of plane P containing the given lines is (x + 2y + 3z – 4) + λ(2x + y – z + 5) = 0 ⇒ (1 + 2λ) x + (2 + λ)y + (3 – λ) z + (–4 + 5λ) = 0 Now, plane P is perpendicular to plane P' →r ... | Mathematics |
12 | Let a, b ∈R. If the mirror image of the point P(a, 6, 9) with respect to the line x −3 7 = y −2 5 = z −1 −9 is (20, b, −a −9), then|a + b| is equal to : | {'A': '84', 'B': '86', 'C': '88', 'D': '90'} | C | Sol: Explanation: Explanation: As per the question, Mid - Point ofP(a, 6, 9) and(20, b, −a −9)will lie on the line x −3 7 = y −2 5 = z −1 −9 . ∴ ( a + 20 2 , 6 + b 2 , −a 2 )lies on the line x −3 7 = y −2 5 = z −1 −9 So, a + 20 2 −3 7 = 6 + b 2 −2 5 = −a 2 −1 −9 ⇒a + 14 14 = 2 + b 10 = a + 2 18 ⇒18a + 252 = 14a + 28 ⇒4... | Mathematics |
13 | If log x, log y and log z are in arithmetic progression and logx −log 3y1 log 2y −log 3z and log 3z −log x are also in arithmetic progression, then | {'A': 'x, y and z are in AP', 'B': 'x, y and z form a triangle', 'C': 'x, y and z are in HP', 'D': 'None of these'} | B | Sol: Here, logx, logy andlogz are in AP, Again, log x −log 3y1 log 2y −log 3z and log 3z −log x are also in AP then Now, from above results, ⇒2 log y = log x + log z ⇒log y2 = log xz ⇒y2 = xz 2(log 2y −log 3z) = (log x −log 2y) + (log 3z −log x) ⇒3 log 2y = 3 log 3z ⇒2y = 3z Thus, Since the sum of any two sides is grea... | Mathematics |
14 | The integral I = ∫ sin2 x cos2 x (sin5 x+cos3 x sin2 x+sin3 x cos2 x+cos5 x) 2 dx is equal to: | {'A': '1 1+ cot3x + C', 'B': '−1 1 + cot3x + C', 'C': '1 3(1+ tan3 x) + C', 'D': '− 1 3(1+tan3 x) + C'} | D | Sol: The given equation is, I = ∫ sin2 x cos2 x (sin5 x+cos3 x sin2 x+sin3 x cos2 x+cos5 x) 2 dx Dividing the numerator and denominator by (cos x)10, we get, ⇒∫ sin 2x cos 2x cos 10x [ sin5 x cos5 x + cos3 x sin2 x cos5 x + sin3 x cos2 x cos5 x + cos5 x cos5 x ] 2 dx ⇒∫ tan2 x⋅sec6 x [tan5 + tan2 x+tan3 x+1] 2 dx ----(... | Mathematics |
15 | Solve the equation (sin 10o)tan x + tan 3x = sin2(2π −x) −cos(π −x) sin ( π 2 −x). | {'A': 'x = nπ 4 , n ∈Z', 'B': 'x = nπ 4 , n ≠2k, n ∈Z', 'C': 'x = π', 'D': 'None of these'} | B | Sol: We have to solve the equation, (sin 10o)tan x + tan 3x = sin2(2π −x) −cos(π −x) sin ( π 2 −x) Consider, (sin 10o)tan x + tan 3x = sin2(2π −x) −cos(π −x) sin ( π 2 −x) ⇒ (sin 10o)tan x + tan 3x = sin2x - (-cosx) (cosx) ⇒ (sin 10o)tan x + tan 3x = sin2x + cos2x ⇒ (sin 10o)tan x + tan 3x = 1 = (sin 10o)0 [Using trigo... | Mathematics |
16 | If (x2 −5y + 3) (y2 + y + 1) < 2y For everyone y ∈R, Then the interval in which x lies: | {'A': '( 5−√5 2 , 5+√5 2 )', 'B': '( √5 2 , √7 2 )', 'C': '( 7−√7 2 , 7+√7 2 )', 'D': "Can't determine"} | A | Sol: Here (x2 −5y + 3) (y2 + y + 1) < 2y for all y ∈R ⇒x2 −5y + 3 < 2y y2+y+1 Let 2y y2+y+1 = t ⇒ty2 + ty + t = 2y ⇒ty2 + (t −2)y + t = 0 since y is real, D ≥0 ⇒(t −2)2 −4t2 ≥0 ⇒(t −2 + 2t)(t −2 −2t) ≥0 ⇒(3t −2)(−t −2) ≥0 ⇒(3t −2)(t + 2) ≤0 Apply wavy curve method to solve this inequality ⇒−2 ≤t ≤2/3 Minimum value of 2... | Mathematics |
17 | Let →c = c1^ı + c2^ȷ + c3^k is a unit vector which makes an angle of 45∘with vector 2^ı + 2^ȷ −^k and an angle of 60∘with the vector^ȷ −^k. Find possible values of (c1, c2, c3) | {'A': '( 1 √2 , 0, −1 √2 ) and ( 1 3√2 , 4 3√2 , 1 3√2 )', 'B': '( 1 √2 , −1 √2 , 0) and ( 1 3√2 , 4 3√2 , 1 3√2 )', 'C': '( 1 √2 , 0, −1 √2 ) and ( 1 3√2 , 4 3 , 1 3√2 )', 'D': '( 1 √2 , 0, −1 √2 ) and ( 1 3√2 , 4 3√2 , 3 3√2 )'} | A | Sol: Similarly, (c1^ı + c2^ȷ + c3^k) ⋅(^ȷ −^k) = (√c2 1 + c2 2 + c2 3) (√(1)2 + (1)2) ⋅cos 60∘ ⇒c2−c3 √2 = 1 2 ⇒c2 −c3 = 1 √2 ⇒c2 = c3 + 1 √2 Using (1), equation (2) will become ⇒c1 = −c3 2 + 1 2√2 Also c2 1 + c2 2 + c2 3 = 1 ⇒c2 3 4 + 1 2 + 2c3 √2 + 1 8 − c3 2√2 + c2 3 = 1 ⇒c3 = −1 √2 , 1 3√2 Now put the value of c3 i... | Mathematics |
18 | For the inequality (log (|x + 6|)2 ) log2 (x2 −x −2) ≥1, x lies between? | {'A': '(−5, −2) ∪(4, ∞)', 'B': '(−∞, −7) ∪(−5, −2) ∪(4, ∞)', 'C': '(4, ∞)', 'D': '(−∞, −6) ∪(4, ∞)'} | B | Sol: = (log2 |x+6|) log2 (x2 −x −2) ≥1 →(log2 |x+6|) log|x+6|(x2−x−2) log2 |x+6| →log|x+6| (x2 −x −2) ≥1, |x + 6| ≠1 →x ≠−5, −7 when |x + 6| > 1 →x ∈(−∞, −7) ∪(−5, ∞) −x2 −x −2 > |x + 6 when x ∈(−5, −)x2 −x −2 > x + 6 →x2 −2x −8 > 0 →x ∈(−∞, −2) ∪(4, ∞) x ∈(−5, −2) ∪(4, −∞) when x ∈(−∞, −7) →x2 −x −2 > −x −6 →x2 + 4 > ... | Mathematics |
19 | If a and b are real numbers such that (2 + α)4 = a + bα, where a = −1+i√3 2 , then a + b is equal to | {'A': '24', 'B': '33', 'C': '9', 'D': '57'} | C | Sol: It is given that (2 + α)4 = a + bα, where a, b ∈R and α = −1+i√3 2 ∴α = −1+i√3 2 is non-real complex root of unity and we know that α3 = 1 and 1 + α + α2 = 0, So, (2 + α)4 = 16 + 32α + 24α2 + 8α3 + α4 = 16 + 8α + 24(α + α2) + 8 + α [∵α3 = 1] = 24 + 9α −24 [∵1 + α + α2 = 0] 9α = a + bα [given] ∴a = 0 and b = 9 ∴a +... | Mathematics |
20 | {'A': 'π 12', 'B': 'π 6', 'C': 'π 4', 'D': 'π 3'} | B | Sol: We have, logcos x ( √3 2 sin x) = 2 −logsec x (tan x) ⇒ logcos x ( √3 2 sin x) −logcos x (tan x) = 2 ⇒logcos x ( √3 2 cos x) = 2 (∵log m −log n = log m n ) ⇒√3 2 cos x = cos2 x (∵loga x = b ⇒x = ab) ⇒cos x ≠0 and cos x = √3 2 Therefore, x = π 6 | Mathematics | |
21 | Let [X]denote greatest less than or equal to x, If for n ∈ N, (1 −x + x3) n = ∑3n j=0 ajxj, then ∑[ 3n 2 ] j=0 a2j + 4 ∑[ 3n−1 2 ] j=0 a2j + 1 is equal to : The smallest positive value of x (in radians) satisfying the equation logcosx ( √3 2 sinx) = 2 −logsecx (tanx), is | {'A': '2', 'B': '2n−1', 'C': '1', 'D': 'n'} | C | Sol: (1 −x + x3) n = ∑3n j=0 ajxj (1 −x + x3) n = a0 + a1x + a2x2 . . . . . + a3nx3n ∑[ 3n 2 ] j=0 a2j = Sum of a0 + a2 + a4 . . . . . . . . . . . ∑[ 3n−1 2 ] j=0 a2j + 1 = Sum of a1 + a3 + a5 . . . . . . . . . . . put x = 1 1 = a0 + a1 + a2 + a3 . . . . . . . . . . . . + a3n . . . . . . .(1) put x = –1 1 = a0 −a1 + a2... | Mathematics |
22 | The value of 15 ∑ r=1 (−1)r−1 (1 + 1 2 + 1 3 + . . . . . + 1 r ) 15Cr is equal to | {'A': '−1', 'B': '−1 15', 'C': '− 1 225', 'D': '1 15'} | D | Sol: = n ∑ r=1 (−1)r−1.n Cr ( 1 1 + 1 2 + 1 3 +. . . . . + 1 r ) = n ∑ r=1 ((−1)r−1.n Cr 1 ∫ 0 (1 + x + x2 +. . . + xr−1) dx) = n ∑ r=1 (−1)r−1.n Cr 1 ∫ 0 ( 1−xr 1−x )dx = 1 ∫ 0 n ∑ r=1 (−1)r−1.nCr−(−1)r−1.nCrxr 1−x dx = 1 ∫ 0 −C0 + 1 + (1−x)n 1−x dx = 1 ∫ 0 (1 −x)n−1 dx = 1 n ⇒for n = 15; 15 ∑ r=1 (−1)r−1 15Cr (1 + 1 ... | Mathematics |
23 | {'A': '42n+1', 'B': '42n', 'C': '42n–1', 'D': 'None of these If [x] denotes the greatest integer less than or equal to x, and F = R −[R] where R = (5√5 + 11) 2n+1 , then R F is equal to'} | A | Sol: We have, R = [R] + F Let G = (5√5 −11) 2n+1 . Then 0 < G < 1 as 0 < 5√5 −11 < 1 Now, R −G = (5√5 + 11) 2n+1 −(5√5 −11) 2n+1 ⇒R −G = 2 {2n+1C1(5√5) 2n (11)1 +2n+1 C3(5√5) 2n−2 (11)3+. . . +2 n+1C2 n+1(11)2n+1} ⇒R −G = an even integer ⇒[R] + F −G = an even integer ⇒F −G is an integer ⇒F −G = 0 ⇒F = G ⇒RF = RG = (5√5... | Mathematics | |
24 | {'A': '3', 'B': '5', 'C': '7', 'D': '9'} | D | Sol: If sum of the series a1 + a2 + … … to ∞ is (1 + logeλ) (9 logeλ) then λ2 is equal to, where an = (loge3)n n ∑ r=1 r2 r! (n − r)! , n ∈ N We have an = (loge3)n n ∑ r=1 r2 r!(n − r)! = (loge3)n n! n ∑ r=1 n! (n − r)!r! r2 = (loge3)n n! n ∑ r=1 nCr. r2 = (loge3)n n! [n(n − 1)2n−2 + n.2n−1 [∵ n ∑ r=1 r2.n Cr = n(n − 1... | Mathematics | |
25 | Let S = {z ∈C : |z −1| = 1 and (√2 −1)(z + ¯z) −i(z −¯z) = 2√2}. Let z1, z2 ∈S be such that |z1| = maxz∈S |z| and |z2| = minz∈S |z|. Then √2z1 −z2 2 equals: | {'A': '2', 'B': '4', 'C': '3', 'D': '1'} | A | Sol: Let z = x + iy (x −1)2 + y2 = 1 . . . (1) (√2 −1)x + y = √2 . . . (2) Solving (1) and (2), we get |z −1| = 1 ⇒|x + iy −1| = 1 (√2 −1)(z + ¯z) −i(z −¯z) = 2√2 (Given) (√2 −1)(2x) −i(2iy) = 2√2 (x −1)2 + (√2 −(√2 −1)x)2 = 1 (x2 −2x + 1) + 2 + (√2 −1)2x2 −2√2(√2 −1)x = 0 x2 (1 + (√2 −1)2) + x(−2 −2√2(√2 −1)) + 2 = 0 ... | Mathematics |
26 | Let C1 and C2 be two concentric circles of radius 1/5 and 3 respectively having centres at (5, 0) on the Argand plane. If the complex number z satisfies the inequality, log1/5 ( |z−5|2+1 16|z−5|+2 ) ≥ 1, then : - | {'A': 'z lies outside C1 but inside C2∣ √2z1 −z2 ∣ 2 = ∣ ( 1 √2 + 1)√2 + i −(1 + i) ∣ 2 = (√2)2', 'B': 'z lies inside of both C1 and C2', 'C': 'z lies outside both of C1 and C2', 'D': 'z lies inside C1 but outside C2'} | A | Sol: log 1 5 ( |z−5|2+1 16|z−5|+2 ) ≥ 1 Note that 16|z – 5| + 2 > 0 ⇒ |z−5|2+1 16|z−5|+2 ≤1 5 Let |z – 5| = λ ⇒ 5λ2 + 5 < 16λ + 2 ⇒ 5λ2 – 16λ + 3 < 0 ⇒ 5λ2 – 15λ – λ + 3 < 0 ⇒ (5λ – 1) (λ – 3) < 0 ⇒ 1/5 < λ < 3 ⇒ 1/5 < |z – 5| < 3 ⇒ z lies outside the circle with centre (5, 0) and radius 1/5 and inside of the circle wi... | Mathematics |
27 | The value of x such that the equations log10 x + log10 x1/2 + log10 x1/4 + … … to ∞= y and 1+3+5+……+(2y−1) 4+7+10+……+(3y+1) = 20 7 log10 x simultaneously hold, is | {'A': '105', 'B': '10', 'C': '100', 'D': '1000'} | A | Sol: log10 x + log10 x 1 2 + log10 x 1 4 + ⋯= y ⇒log10 x + 1 2 log10 x + 1 4 log10 x + ⋯= y ⇒log10 x (1 + 1 2 + 1 4 + ⋯) = y ⇒2 log10 x = y ......(1) From the second equation, y 2 (1+2y−1) y 2 (4+3y+1) = 20 7 log10 x From (1), ⇒ 2 3y+5 = 20 7( y 2 ) ⇒7y2 −60y −100 = 0 ⇒(y −10) (7y + 10) = 0 ⇒y = 10 ⇒2 log10 x = 10 ⇒log... | Mathematics |
28 | The value of cos ( 2π 13 ) + cos ( 6π 13 ) + cos ( 8π 13 ) is | {'A': '√13−1 4', 'B': '√13+1 4', 'C': '√17−1 4', 'D': '√17+1 4'} | A | Sol: Let θ = 2π 13 and x = cos ( 2π 13 ) + cos ( 6π 13 ) + cos ( 8π 13 ) Squaring x, we get, x2 = cos2(θ) + cos2(3θ) + cos2(4θ) + 2(cosθ cos3θ + cos3θ cos4θ + cos4θ cosθ) ⇒ x2 = cos2(θ) + cos2(3θ) + cos2(4θ) + [cos4θ + cos2θ + cos7θ + cosθ + cos5θ + cos3θ] But ⇒cos (7θ) = cos (6θ) {∵cos ( 14π 13 ) = −cos ( π 13 ) = cos... | Mathematics |
29 | [4+√(15)] 3 2 +[4−√(15)] 3 2 [6+√(35)] 3 2 −[6−√(35)] 3 2 = | {'A': '1', 'B': '7 13', 'C': '13 7', 'D': 'None of these'} | B | Sol: Let 4 + √15 = x ⇒4 −√15 = 1 x 6 + √35 = y ⇒6 −√35 = 1 y ⇒ x3/2+ 1 x3/2 y3/2− 1 y3/2 = x3+1 y3−1 ( y x ) 3 2 = (4+√15) 3 +1 (6+√35) 3 −1 ( 6+√35 4+√15 ) 3 2 = (4+√15+1){(4+√15) 2 −(4+√15)+1} (6+√35−1){(6+√35) 2 +(6+√35)+1} ( 6+√35 4+√15 ) 3 2 = 5+√15 5+√35 {31+8√5−4−√15+1} {71+12√35+6+√35+1} ( 6+√35 4+√15 ) 3 2 = √... | Mathematics |
30 | If the shortest distance between the lines x−λ −2 = y−2 1 = z−1 1 and x−√3 1 = y−1 −2 = z−2 1 is 1, then the sum of all possible values of λ is : | {'A': '0', 'B': '2√3', 'C': '3√3', 'D': '−2√3'} | B | Sol: Shortest distance = ∣ →V1× →V2 ∣ →V1× →V2 ∣ ⋅ − → AB ∣ = 1 where, V2 = ^i −2^j + ^k |(λ −√3) + 1 −1| = √3 V1 = −2^i + ^j + ^k, − → AB = (λ −√3, 2 −1, 1 −2) →V1 × →V2 = ∣ i j k −2 1 1 1 −2 1 ∣ = 3^i + 3^j + 3^k ⇒→V1 × →V2 = 3(^i + ^j + ^k) ⇒ →V1 × →V2 = 3√3 ⇒ (^i + ^j + ^k) ⋅((λ −√3)^i + ^j −^k √3 = 1 ⇒Sum of λ = 2... | Mathematics |
31 | The plane containing the line x−3 2 = y+2 −1 = z−1 3 and also containing its projection on the plane 2x + 3y −z = 5, contains which one of the following points ? | {'A': '(0, –2, 2)', 'B': '(2, 0, –2)', 'C': '(2, 2, 2)', 'D': '(–2, 2,2)'} | B | Sol: Given line x−3 2 = y+2 −1 = z−1 3 And plane 2x + 3y −z = 5 Normal of required plane (2^i −^j + 3^k) × (2^i + 3^j −^k) = −8^i + 8^j + 8^k Required plane whose direction ratio of normal (–1, 1, 1) ⇒|λ −√3| = √3 ⇒λ −√3 = √3 or λ −√3 = −√3 ⇒λ = 2√3 or 0 −(x −3) + (y + 2) + (z −1) = 0 −x + y + z + 4 = 0 Satisfied by (2... | Mathematics |
32 | If α, β are the roots of quadratic equation x2 −(3 + 2√log2 3 −3√log3 2)x −2 (3log3 2 −2log2 3)= 0, then quadratic equation whole roots are α5, β5 is: | {'A': 'x2 – 33x + 32 = 0', 'B': 'x2 – 32x + 33 = 0', 'C': 'x2 – 33x – 32 = 0', 'D': 'x2 + 32x – 33 = 0'} | A | Sol: let log23 = x ⇒3 = 2x ⇒2 = 31/x then 2√log23 = 2√x = 3 1 x √x = 3 1 √x so 2√log23 = 3 1 √log23 2√log23 = 3√log32 Now x2 −(3 + 2√log23 −3√log32)x −2 (3log32 −2log23)=0 ⇒(x −2) (x −1) = 0 ⇒x = 2, 1 i. e. α = 2, β = 1 So quadratic equation whole roots are = α5, β5 = 32, 1 is: x2 −(32 + 1)x + 32 × 1 = 0 x2 −33x + 32 =... | Mathematics |
33 | If α, β are the roots of the equation x2 −(5 + 3√log3 5 −5√log5 3)x + 3 (3(log3 5) 1 3 −5(log5 3) 2 3 −1) = 0 , then the equation whose roots are α + 1 β and β + 1 α is | {'A': '3x2 – 20x – 12 = 0', 'B': '3x2 – 10x – 4 = 0', 'C': '3x2 – 10x + 2 = 0', 'D': '3x2 – 20x + 16 = 0'} | B | Sol: Given, x2 −(5 + 3√log35 −5√log53)x + 3 (3(log35) 1 3 −5(log53) 2 3 −1) = 0 3√log35 = 3√log35 . √log35 . √log53 = 3log35. √log53 = (3log35) √log53 = 5√log53 3 3√log35 = 3log35. 3√(log53)2 = (3log35) (log53)2/3 = 5(log53)2/3 So, equation is x2 – 5x – 3 = 0 and roots are α & β {α + β = 5 ; αβ = −3} New roots are α + ... | Mathematics |
34 | {'A': 'p(y) = y2, ∀y ∈R', 'B': 'p(y) = y2 + 2 , ∀y ∈R', 'C': 'p(y) = y2 −2, ∀y ∈R', 'D': 'None of these'} | A | Sol: Note that P(a)=a2 ,P(b)=b2 and P(C)=c2 Consider the polynomial Q(x) = P(x) - x2, Q(x) has degree atmost 2. Also, Q(a) = Q(b) = Q(c) = 0 ⇒Q(x) has distinct roots. It follows that Q(x) is identically zero i.e. Q(y) = 0 ∀ y ∈ R Let p(x) = (x−a)(x−b) (c−a)(c−b) . c2 + (x−b)(x−c) (a−b)(a−c) . a2 + (x−c)(x−a) (b−c)(b−a)... | Mathematics | |
35 | {'A': '11', 'B': '12', 'C': '13', 'D': '14'} | C | Sol: If 1 √4x + 1 {(1 + √4x + 1 2 ) n − (1 − √4x + 1 2 ) n } = a0 + a1x + … … . + a5x5 + a6x6, then n equals We have (1 + √4x+1 2 ) n = nC0(1)n( √4x+1 2 ) 0 + nC1(1)n−1 ( √4x+1 2 ) +n C2(1)n−2( √4x+1 2 ) 2 . . . . . . +nCn(1)n−1( √4x+1 2 ) n . . . . . . . . . (i) and (1 −√4x+1 2 ) n = nC0(1)n( 4x+1 2 ) 0 −nC1(1)n−1 ( √... | Mathematics | |
36 | If the coefficient of the term independent of x in the expansion of ( x+1 x 2 3 −x 1 3 +1 − x−1 x−x 1 2 ) 10 is C then the value of C 10√7 is | {'A': '10√2', 'B': '7√3', 'C': '3√7', 'D': '21'} | C | Sol: ( x+1 x2/3−x1/3+1 − x−1 x−x1/2 ) 10 = [ (x1/3) 3 +(1)3 x2/3−x1/3+1 − ((x1/2) 2−(1)2) x1/2(x1/2−1) ] 10 = [ (x1/3+1)(x2/3+1−x1/3) x2/3−x1/3+1 − (x1/2−1)(x1/2+1) x1/2(x1/2−1) ] 10 = [(x1/3 + 1) −(1 + x−1/2)] 10 = (x1/3 −x−1/2) 10 = Tr+1=nCrxn−ryr Tr+1 =10Cr(x1/3) 10−r(−x−1/2) r =10Crx 10−r 3 (−1)rx−r/2 =10Crx 10−r 3... | Mathematics |
37 | If the fourth term in the binomial expansion of ( 2 x + xlog8x) 6(x > 0) is 20 × 87 , then the value of x is | {'A': '8−2', 'B': '83', 'C': '8', 'D': '82'} | D | Sol: Given binomial is ( 2 x + xlog8x) 6 Since, general term in the expansion of (x + a)n is Tr+1 = nCrxn−rar ∴ T4 = T3+1 = 6C3( 2 x ) 6−3(xlog8x)3 = 20 × 87 (Given) ⇒20( 2 x ) 3x3log8x = 20 × 87 [ ∵ 6C3 = 20] ⇒23. x[3(log8x)−3] = (23)7 ⇒x( 3 3 log2x−3) = (23)6 [∵logan(x) = 1 n loga x for a ≠1] ⇒x(log2x−3) = 218 On tak... | Mathematics |
38 | Let K be the sum of the coefficients of the odd powers of x in the expansion of (1 + x) 99. Let a be the middle term in the expansion of (2 + 1 √2 ) 200 . If 200C99 K a = 2l m n where m and n are odd numbers, then the ordered pair (l, n) is equal to | {'A': '(50, 51)', 'B': '(50, 101)', 'C': '(51, 99)', 'D': '(51, 101)'} | B | Sol: In the expansion of (1 + x)99 = C0 + C1x + C2x2 + … . +C99x99 K = C1 + C3 + … . +C99 = 298 a ⟹ Middle in the expansion of (2 + 1 √2 ) 200 T 200 2 +1 = 200C100(2)100( 1 √2 ) 100 = 200C100 ⋅250 So, 200C99×298 200C100×250 = 100 101 × 248 So, 25 101 × 250 = m n 2l ∴m, n are odd so (l, n) become (50,101) | Mathematics |
39 | Let Tn denotes number of triangles that can be formed using the vertices of a regular polygon of side n. If Tn+1 –Tn = 10, then, the number of different selections of n- digit number that can be formed using the digits, one 1’s, two2’s, three 4’s, and four 3’s is equal to: | {'A': '20', 'B': '15', 'C': '22', 'D': '81'} | C | Sol: Tn = nC3 and Tn+1 – Tn = 10 ⇒n+1C3 – nC3 = 10 ⇒ n = 5 Given 1, 2, 2, 4, 4, 4, 3, 3, 3, 3 Out of these digits, we need to select 5-digit number ∴We need to find coefficient of x5 in (x0 + x1) (x0 + x1 + x2) (x0 + x1 + x2 + x3) (x0 + x1 + x2 + x3 + x4) i.e. coefficient of x5 in (1 + x) (1+ x + x2) (1 + x +x2 + x3) (... | Mathematics |
40 | For three positive integers p, q, r, xpq2 = yqr = zp2r and r = pq + 1 such that 3, 3logyx, 3logzy, 7logxz are in A.P. with common difference 1 2 . Then r – p – q is equal to | {'A': '6', 'B': '–6', 'C': '12', 'D': '2'} | D | Sol: pq2 = logx λ qr = logyλ p2r = logzλ logyx = qr pq2 = r pq … (i) logxz = pq2 p2r = q2 pr … (ii) logzy = p2r qr = p2 q … (iii) 3, 3r pq , 3p2 q , 7q2 pr in A. P 3r pq −3 = 1 2 r = 7 6 pq … (iv) r = pq + 1 pq = 6 … (v) r = 7 … (vi) 3p2 q = 4 After solving p = 2 and q = 3 r −p −q = 7 −5 = 2 | Mathematics |
41 | The roots of equation x2 + 2(a −3)x + 9 = 0 lie between -6 and 1 and 1, h1, h2, … h20, [a] are in HP, where [a] denotes the integral part of a and 1, a1, a2, … … , a20, [a] are in AP, then a2h17 is equal to | {'A': '12', 'B': '41 186', 'C': '186 41', 'D': '6'} | C | Sol: Since roots of equation f(x) = x2 + 2(a −3)x + 9 = 0 lie between –6 and 1 ∴ (i) D ≥ 0 (ii) f(–6) > 0 (iii) f(1) > 0 (iv) −6 < α+β 2 (v) 1 > α+β 2 Hence 6 ≤a ≤27 4 [a] = 6 a2 = a + 2d = 1 + 2 ( b−a n+1 ) = 1 + 2 × ( 6−1 21 ) = 1 + 2×5 21 = 21+10 21 = 31 21 1 h17 = 1 a + 17 ( 1 b −1 a ) n+1 = 1 1 + 17 ( 1 6 −1 1 ) 2... | Mathematics |
42 | {'A': '1 2', 'B': '1 3', 'C': '1 4', 'D': '3 4 Ans: b In a ΔABC, sides a, b, c are in AP and 2 1!9! + 2 3!7! + 1 5!5! = 8a (2b)! , then the maximum value of tan A tan B is equal to Sol: Sol: ∵ 2 1!9! + 2 3!7! + 1 5!5! = 8a (2b)! ⇒ 1 1!9! + 1 3!7! + 1 5!5! + 1 3!7! + 1 9!1! = 8a (2b)! ⇒ 1 10! ( 10! 1!9! + 10! 3!7! + 10!... | D | Sol: logcos x cot x + 4logsin x tan x = 1 ⇒ln cos x−ln sin x ln cos x + 4 ln sin x−ln cos x ln sin x = 1 ⇒(ln sin x)2 −4(ln sin x)(ln cos x) + 4(ln cos x)2 = 0 ⇒ln sin x = 2 ln cos x ⇒sin2x + sin x −1 = 0 ⇒sin x = −1+√5 2 ∴ α + β = 4 | Mathematics | |
44 | Let f : N →R be a function such that f (x + y) = 2f (x) f (y) for natural numbers x and y. If f (1) = 2, then the value of α for which 10 ∑ k=1 f (α + k) = 512 3 (220 −1) holds, is | {'A': '2', 'B': '3', 'C': '4', 'D': '6'} | C | Sol: f : N→R, f (x + y) = 2 f (x) f (y) ....(1) f(1) = 2, ∑10 k=1 f(α + k) = 2f(α) ∑10 k=1 f(k) = 2f(α) ....(2) From (1) f (2) = 2 f 2(1) = 23 f (3) = 2 f(2) f(1) = 25 ⁝ ⁝ f (10) = 29f 10(1) = 219 f (α) = 22α - 1 ; α ϵ N from (2) 10 ∑ k=1 f (α + k) = 2 (22α−1) (2 + 23 + 25+. . . . . +219) 512 3 (220 −1) = 22α (2 (220−1... | Mathematics |
45 | A line 'L' is drawn from P(4,3) to meet the lines L1 and L2 given by 3x + 4y + 5 = 0 and 3x + 4y + 15 = 0 at points A and B, respcetively. From 'A', a line perpendicular to L is drawn meeting the line L2 at A1. Similarly from point 'B', a line perpendicular to L is drawn meeting the line L1 at B1. Thus a parallelogram ... | {'A': 'x - 7y + 17 = 0', 'B': '7x + y + 31 = 0', 'C': 'x - 7y - 17 = 0', 'D': 'x - 7y - 31 = 0'} | A | Sol: The given lines (L1 and L2) are parallel and distance between them (BC or AD)is (15 −5)/5 = 2 units. Let ∠BCA = θ ⇒ AB = BC cos ecθ and AA1 = AD sec θ = 2 sec θ. now area of parallelogram AA1BB1 is Δ = AB × AA1 = 4 sec θ cos ec θ = 8 sin 2θ Clearly, Δ is least for θ = π 4 . Let slope AB be m Then, 1 = m+ 3 4 1− 3m... | Mathematics |
46 | {'A': '2x + 11y + 13 = 0', 'B': '11x −2y + 13 = 0', 'C': '2x −11y + 2 = 0', 'D': '11x + 2y −11 = 0'} | B | Sol: Writing the given equation as a quadratic in x, we have 6x2 + (5y + 7)x −(4y2 −13y + 3) = 0 ⇒x = −(5y+7)±√(5y+7)2+24(4y2−13y+3) 12 ⇒−(5y+7)±√121y2+242y+121 12 ⇒−(5y+7)±11(y−1) 12 = 6y−18 12 , −16y+4 12 ⇒2x −y + 3 = 0 and 3x + 4y −1 = 0 are two lines represented by the given equation. Now point of intersection of t... | Mathematics | |
47 | The equation of a circle which passes through the point (2, 0) and whose centre is the limit of the point of intersection of the lines 3x + 5y = 1 and (2 + c) x + 5c2y = 1 as c tends to 1, is If θ is the acute angle between the lines given by 6x2 + 5xy −4y2 + 7x + 13y −3 = 0, then the equation of the line passing throu... | {'A': '25 (x2+y2) + 20x + 2y - 60 = 0', 'B': '25 (x2+y2) - 20x + 2y + 60 = 0', 'C': '25 (x2+y2) -20x + 2y - 60 = 0', 'D': '25 (x2 + y2) + 20x - 2y + 60 = 0'} | C | Sol: LetA = (2, 0) Given lines are 3x + 5y = 1.. . (1) and (2 + c)x + 5c2y = 1.. . (2) Multiplying equation (1)byc2 and subtracting (2) form it, we get (3c2 −c −2)x = c2 −1or x = c2−1 3c2−c−2 Now, lim c→1x = lim c→1 (c−1)(c+1) (c−1)(3c+2) = lim c→1 c+1 3c+2 = 2 5 ; x coordinate of centre = 2 5 From(1), whenx = 2 5 , y ... | Mathematics |
48 | If the point (−5 + λ √2 , −3 + λ 2 ) is an interior point of the larger segment of the circle x2 + y2 = 16 cut off by line x + y = 2, then | {'A': 'λ ∈ (− ∞, 5√2)', 'B': 'λ ∈ (4√2 − √14, 5√2)', 'C': 'λ ∈ (4√2 − √14, 4√2 + √14)', 'D': 'None of these'} | B | Sol: The point (−5 + λ √2 , −3 + λ 2 ) will be an interior point of the larger segment of the circle x2 + y2 = 16 cut off by the line x + y = 2, if (i) it is an interior point of the circle (ii) the centre of the circle and the point lies on the same side of x + y = 2 ∴(−5 + λ √2 ) 2 + (−3 + λ √2 ) 2 −16 < 0 and, (0 + ... | Mathematics |
49 | Consider a hyperbola H : x2 – 2y2 = 4. Let the tangent at a point P (4, √6) meet the x – axis at Q and latus rectum at R (x1, y1), x1 > 0. If F is a focus of H which is nearer to the point P, then the area of ΔQFR is equal to | {'A': '4√6', 'B': '√6 −1', 'C': '7 √6 −2', 'D': '4√6 −1'} | C | Sol: x2 4 −y2 2 = 1 e = √1 + b2 a2 = √3 2 ⸫ Focus F(ae, 0) ⇒ F (√6, 0) Equation of tangent at P to the hyperbola is 2x −y√6 = 2 Tangent meet x-axis at Q(1, 0) & latus rectum x = √6 at R (√6, 2 √6 (√6 −1)) ⸫ Area of ΔQFR = 1 2 (√6 −1). 2 √6 (√6 −1) = 7 √6 −2 | Mathematics |
50 | {'A': '−2 tan ϕ', 'B': '−2 tan(π −ϕ)', 'C': '0', 'D': '2 cot ϕ'} | A | Sol: Slope of tangent at~P~is~ 1 t1 ~and at~Q~is~ 1 t2 ⇒cot θ1 = t1~and~ cot θ2 = t2 Slope of~PQ = 2 t1+t2 ⇒~Slope of OR is~ −t1+t2 2 = tan ϕ (angle in a semicircle is~90∘) ⇒tan ϕ = −1 2 (cot θ1 + cot θ2) ⇒cot θ1 + cot θ2 = −2 tan ϕ Through the vertex~O~of the parabola~y2 = 4ax, ~two chords~OP~and~ OQ~are drawn and the... | Mathematics | |
1 | With the two forces acting at a point, the maximum effect is obtained when their resultant is 4N. If they act at right angles, then their resultant is 3N. Then the forces are: | {'A': '(2 + √2)N and (2 −√2)N', 'B': '(2 + √3)N and (2 −√3)N', 'C': '(2 + 1 2 √2)N and (2 −1 2 √2)N', 'D': '(2 + 1 2 √3)N and (2 −1 2 √3)N'} | C | Sol: Let P and Q are forces. We know that R = √p2 + Q2 + 2PQ cos θ When θ = 0∘, R = 4N R = 4N = √P 2 + Q2 + 2PQ P + Q = 4.....(i) when θ = 90∘, R = 3N P 2 + Q2 = 9......(ii) From equation (i) (P + Q)2 = 16 P 2 + Q2 + 2PQ = 16 9 + 2PQ = 16 2PQ = 16 −9 [using (ii)] 2PQ = 7 Now, (P −Q)2 = P 2 + Q2 −2PQ (P −Q)2 = 9 −7 P −Q... | Physics |
2 | An ideal gas is in thermodynamic equilibrium. The number of degrees of freedom of a molecule of the gas is n. The internal energy of one mole of the gas is Un and the speed of sound in the gas is vn. At a fixed temperature and pressure, which of the following is the correct option? | {'A': 'v3 < v6 and U3 > U6', 'B': 'v5 > v3 and U3 > U5', 'C': 'v5 > v7 and U5 < U7', 'D': 'v6 < v7 and U6 < U7'} | C | Sol: Un = n 2 RT Un ∝n, where n is degree of freedom As n →increases, hence; Un →increases Vn = √(1+ 2 n )RT M where n is degree of freedom As n →increases, hence; Vn →decreases | Physics |
3 | At t = 0. a particle executing SHM with a time period 3 s is in phase with another particle executing SHM. The time period of the second particle is T (less than 3 s ). If they are again in the same phase for the third time after 45 s . then the value of T is ____ | {'A': '1 s', 'B': '1.5 s', 'C': '2 s', 'D': '2.5 s'} | D | Sol: Let initially, both the oscillating particles are at mean position i.e., , (x = 0 at t = 0) The first particle is in same phase at given time instances. So, t = 0 s, 3 s, 6 s, 9 s, 12 s, 15 s, …. . etc The time period of second particle will be selected from the given options which give same phase of the both part... | Physics |
4 | A long horizontal rod has a bead, which can slide along its length and is initially placed at a distance L from one end A of the rod. The rod is set in angular motion about A with constant angular acceleration α. If the coefficient of friction between the rod and the bead is and gravity is neglected, then the time afte... | {'A': '√μ α', 'B': 'μ √α', 'C': '1 √μα', 'D': 'infinitesimal'} | A | Sol: The radius of the circular motion of the bead is r = L. The linear acceleration of the bead is a = αr = αL. If m is the mass of the bead, then Force acting on the bead = ma = mαL Reaction force acting on the bead is R = mαL This bead starts slipping when the frictional force between the bead and the rod becomes eq... | Physics |
5 | The pressure and density of a diatomic gas (Y = 7 5 ) change adiabatically from (P1, ρ1)to (P2, ρ2), if ρ2 ρ1 = 32, then what should be the value of P2 P1 ? | {'A': '16', 'B': '32', 'C': '64', 'D': '128'} | D | Sol: In an adiabatic process. or or PV γ = constant P1V γ 1 = P2V ′ 2 P2 P1 = ( V1 V2 ) Y Volume of gas = Mass Density L.e., Thus, from Eq. (i), we have | Physics |
6 | An engine has an efficiency of 1 6 . When the temperature of the sink is reduced by 62oC, its efficiency is doubled. The temperature of the source is | {'A': '124∘C', 'B': '37∘C', 'C': '62∘C', 'D': '99∘C'} | D | Sol: Maximum efficiency of an engine working between temperatures T2 and T1 is given by the fraction of the heat absorbed by the engine, which can be converted into work. Mathematically, Efficiency, η = (T2 −T1)/T2 = 1/6 V = M ρ or V ∝1 ρ ∴ V1 V2 = ρ2 ρ1 = 32 P2 P1 = (32)γ = (32)7/5 = 27 = 128 6T2 −6T1 = T2 Therefore, ... | Physics |
7 | Atomic masses of two oxygen isotopes 16 8 O and 18 8 O are 15.99491 u and 17.99916 u , respectively, where u is the atomic mass unit. Masses of proton and neutron are given by 1.00727 u and 1.00866 u , respectively. The speed of light is c. What is the difference between the binding energies of 18 8 O and 16 8 O nuclei... | {'A': '0.01307', 'B': '2.00425', 'C': '0.99559', 'D': '3.01291'} | A | Sol: To find the difference in binding energies between the 18 8 O and 16 8 O nuclei in units of u ⋅c2, we will use the mass defect formula. The binding energy is the energy equivalent of the mass defect, which is the difference between the mass of the constituent protons and neutrons and the actual mass of the nucleus... | Physics |
8 | Consider a Carnot heat engine operating between a heat reservoir at temperature 600 K , and an external atmosphere at temperature 300 K . In one cycle, 1000 kJ of heat is extracted from the heat reservoir, and the associated work is input to a reversible refrigerator that operates between 200 K and the same external at... | {'A': '2500 Kj', 'B': '2000 Kj', 'C': '1000 Kj', 'D': '500 Kj'} | B | Sol: The Carnot efficiency η is given by: η = 1 −Tc Th where Th is the temperature of the heat reservoir and Tc is the temperature of the atmosphere. η = 1 −300 600 = 0.5 The work done by the heat engine: WHE = ηQh where Qh is the heat extracted from the heat reservoir. Given, Qh = 1000 kJ : WHE = 0.5 × 1000 = 500 kJ T... | Physics |
9 | Consider the Bohr model of an atom where an electron of charge −e revolves around a nucleus of charge +e in an orbit of radius r. The electron has an orbital angular momentum 2h. If the nucleus had charge +2e, what would have been the radius of the orbit of the electron with the same principal quantum number? | {'A': '2r', 'B': 'r/2', 'C': 'r', 'D': '√2r'} | B | Sol: In the Bohr model of the atom, the radius of the electron's orbit is given by the formula: rn = 4πϵ0n2ℏ2 Ze2me Where, n is the principal quantum number, ℏis the reduced Planck's constant, Z is the atomic number (number of protons in the nucleus), e is the elementary charge, me is the electron mass, ϵ0 is the permi... | Physics |
10 | A ball suspended by a thread swings in a vertical plane so that its magnitude of acceleration in the extreme position and lowest position are equal. The angle (θ) of thread deflection in the extreme position will be : | {'A': 'tan−1(√2)', 'B': '2 tan−1 ( 1 2 )', 'C': 'tan−1 ( 1 2 )', 'D': '2 tan−1 ( 1 √5 )'} | B | Sol: Loss in kinetic energy = Gain in potential energy ⇒1 2 mv2 = mgℓ(1 −cos θ) ⇒v2 ℓ= 2 g(1 −cos θ) Acceleration at lowest point = v2 ℓ Acceleration at extreme point = g sin θ Hence, v2 ℓ= g sin θ ∴sin θ = 2(1 −cos θ) ⇒tan θ 2 = 1 2 ⇒θ = 2 tan−1 ( 1 2 ) | Physics |
11 | The pressure P of a system of N particles contained in a volume V at a temperature T is given by P = nkBT −1 2 an2 + 1 6 bn3, where n is the number density and a and b are temperature independent constants. If the system exhibits a gas-liquid transition, the critical temperature is: | {'A': 'a bkB', 'B': 'a 2b2kB', 'C': 'a2 2bkB', 'D': 'a2 b2kB'} | C | Sol: P = nkBT −1 2 an2 + 1 6 bn3 where, n = N V . For critical condition ∂P ∂V = 0 and ∂2P ∂V 2 = 0 Putting the expressions for n we have: P = N V kBT −1 2 a N 2 V 2 + 1 6 b N 3 V 3 ∂P ∂V = 0 ⇒NkBT = aN 2 V −bN 3 2V 2 ....(i) ∂2P ∂V 2 = 0 ⇒2NkBT = 3aN 2 V −2bN 3 V 2 ...(ii) From equation (i) and (ii) VC = bN a put the ... | Physics |
12 | A piece of wire is bent in the shape of a parabola y = kx2 ( y-axis vertical) with a bead of mass m on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distan... | {'A': 'a gk', 'B': 'a 2gk', 'C': '2a gk', 'D': 'a 4gk'} | B | Sol: While the parabola is moving right side, the bead experiences a force with a magnitude ma to the left. That is, at equilibrium, N cos θ = mg N sin θ = ma Now, tan θ = a g Also, tan θ = dy dx = 2kx ⇒x = a 2gk | Physics |
13 | A man who is running has half the kinetic energy of a boy of half his mass. The man speeds up by 1 ms−1 and then has the same kinetic energy as the boy. The initial speed of the boy is | {'A': '√2 + 1 ms−1', 'B': '2(√2 + 1)ms−1', 'C': '√2 ms−1', 'D': '2 ms−1'} | B | Sol: Let the mass of the man is m and speed is v1, thus the mass of the boy is m 2 and let his initial velocity is v2 According to given condition, After the increase in velocity of man by 1 m/s, the new velocity of the man is v′ 1 = ( v2 2 + 1) According to given condition, Substituting the values, we get (KE)man = 1 ... | Physics |
14 | A rain drop of radius r is falling through air. starting from rest. The work done by all the forces 011 the drop, when it attains terminal velocity, is proportional to | {'A': '3', 'B': 'r7', 'C': 'r5', 'D': 'r4 1'} | B | Sol: The initial velocity of rain drop will be zero, so u = 0. The expression of the final velocity of rain drop is, By work energy theorem, Substitute the value of v in the above expression. W ∝r7 | Physics |
15 | When the air column of a resonance tube is vibrated together with a tuning fork. 3 beats are heard per second, either the temperature of the air column is 51∘C or 16∘C. Hie frequency of the tuning fork is v = 2gr2(ρ −σ) 9η W = 1 2 mv2 −1 2 mu2 W = 1 2 mv2 −1 2 m(0)2 W = 1 2 (ρ V)v2 W = 1 2 ρ ( 4 3 πr3)v2 W = 1 2 ρ ( 4 ... | {'A': '128 Hz', 'B': '98 Hz', 'C': '105 Hz', 'D': '256 Hz'} | C | Sol: The frequency of tuning fork is directly proportional to square root of temperature, therefore n ∝√T From equation (I) | Physics |
16 | All electrons ejected from a metallic surface by incident light of wavelength 400 mn travelled 1 m in the direction of uniform electric field of 2NC−1 and came to rest. The work function of the surface is | {'A': '1.1 eV', 'B': '2.2 eV', 'C': '3.1 eV', 'D': '5.1 eV'} | A | Sol: The energy of the incident light is calculated as, n1 n2 = √T1 T2 … . (I) AtT1 = 51∘C = 324 K, n1 = n + 3 At T2 = 16∘C = 289 K, n2 = n −3 n1 n2 = √324 289 n2+3 n−3 = 18 17 n = 105 Hz The acceleration (retardation) of the emitted electrons in the electric field is, The initial speed of the electron is, The maximum ... | Physics |
17 | A body is projected vertically upwards from the surface of the Earth with a velocity sufficient to carry it to infinity. The time taken by it to reach a height of three times the radius of the Earth is (Acceleration due to gravity = 9.8 ms−2 and radius of the Earth = 6400 km ) E = hc λ E = 6.6×10−34×3×108 4×10−7 J E = ... | {'A': '44.44 minutes', 'B': '22.22 minutes', 'C': '18.76 minutes', 'D': '37.52 minutes'} | A | Sol: From the conservation of mechanical energy, Solve further, Integrate above equation, Substitute values in above expression, 1 2 mv2 + (−GMm r ) = 1 2 mv2 e + (−GMm R ) v2 = v2 e −2GM R + 2GM r v2 = 2gR + 2gR ( R r −1) (As, ve = √2gR, g = GM R2 ) v = dr dt = √2gR2 r dr dt = R√2g r ∫t 0 dt = 1 R√2g ∫R+h R r 1 2 dr t... | Physics |
18 | Two uniform strings A and B made of steel are made to vibrate under the same tension. If the first overtone of A is equal to the second overtone of B and if the radius of A is twice that of B, the ratio of the lengths of the strings is : | {'A': '1:2', 'B': '1 : 3', 'C': '1:4', 'D': '1 : 5'} | B | Sol: Fundamental frequency of string, where, m = mass per unit length of the string For string A first overtone, For string B second overtone, According to question, n = 1 2l √ T m = πr2ρ ∴n = 1 2l √ T πr2ρ nA = 1 l1 √ T1 m1 = 1 l1 √ T1 πr12ρ nB = 3 2l2 √ T2 πr22ρ2 | Physics |
19 | A reversible Carnot heat engine converts 1 4 th its input heat into work. When the temperature of the sink is reduced by 50 K . its efficiency becomes 33 1 3 %. The initial temperatures of the source and the sink respectively are | {'A': '600 K, 550 K', 'B': '600 K, 450 K', 'C': '300 K, 150 K', 'D': '450 K, 350 K'} | B | Sol: Given, W = Q 4 The expression of the efficiency is, nA = nB 1 l1 √ T1 πr12l1 = 3 2l2 √ T2 πr22l2 but T1 = T2 ∴2l2 3l1 = √πr12ρ πr22ρ 4l22 9l12 = r12 r22 ( l1 l2 ) 2 = 4r22 9r12 2l2 3l1 = r1 r2 l2 l1 = 3r1 2r2 l1 l2 = 2r2 3r1 but r1 = 2r2 ∴l1 l2 = 2r2 3(2r2) l1 : l2 = 1 : 3 η = W Q 1 −T2 T1 = W Q Substitute values,... | Physics |
20 | An electric dipole consists of charges −q and +q at points A and B, respectively, with AB = 4a. The point O is the midpoint of the dipole. A charge Q is placed at point P, where OP = 3y, and y ≫2a. The force on Q at point P is F. Point P and P ′ lies on Axis of dipole. If Q is moved to point P ′, where OP ′ = 3y 4 , th... | {'A': '60 F', 'B': '20 F', 'C': '16 F', 'D': '64 F'} | D | Sol: Step 1: Initial Force F at OP = 3y 1 −T2 T1 = Q/4 Q 1 −T2 T1 = 1 4 … … (I) η′ = 100 3 × 1 100 1 −T2−50 T1 = 1 3 1 −T2 T1 + 50 T1 = 1 3 . 50 T1 = 1 3 −1 4 T1 = 600 K T2 = 3 4 T1 T2 = 3 4 (600) T2 = 450 K At distance r1 = 3y, the electric field is: E1 = 2kp (3y)3 = 2kp 27y3 The force on Q at P is: F = Q ⋅E1 = Q ⋅ 2k... | Physics |
21 | When photons of energy 5.00 eV strike the surface of metal A the ejected photoelectrons have maximum kinetic energy TAeV and de Broglie wavelength λA. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 5.50 eV is TB = (TA −2.00)eV. If the de Broglie wavelength of these phot... | {'A': '0.50 eV', 'B': '1.25 eV', 'C': '2.50 eV', 'D': '1.60 eV'} | D | Sol: We will solve this step by step using the photoelectric equation and the de Broglie wavelength formula. Step 1: Photoelectric Equation The photoelectric equation states that: KEmax = hν −ϕ where: KEmax is the maximum kinetic energy of ejected electrons. hν is the energy of the incident photon. ϕ is the work functi... | Physics |
22 | In an AC circuit, a resistor R and an inductor where XL = 7R are connected in series. A capacitor is added such that XC = 6R. Now power factor becomes √50 : x, find x. | {'A': '1', 'B': '3', 'C': '7', 'D': '10'} | D | Sol: Given: Resistor (R): R Inductive Reactance (XL) : XL = 7R Capacitive Reactance ( XC ): XC = 6R Step 1: Calculate the Impedance (Z) In a series RLC circuit, the total impedance Z is given by: Z = √R2 + (XL −XC)2 Substituting the given values: Z = √R2 + (7R −6R)2 = √R2 + R2 = √2R2 = R√2 Step 2: Determine the Power F... | Physics |
23 | A bob of mass 2m is tied to a string wound around a Disc of mass 3m and radius r. The bob falls a vertical distance h. What is the angular velocity of the Disc after the bob has fallen a distance h ? | {'A': '1 r √5gh 6', 'B': '1 r √7gh 8', 'C': '3√2gh √7r', 'D': '2 r √2gh 7'} | D | Sol: Given: Mass of the bob: 2m Mass of the Disc: 3m Radius of the Disc: r Distance fallen by the bob: h Assumptions: The Disc is a solid disk, so its moment of inertia I is given by: I = 1 2 Mr2 where M = 3m is the mass of the Disc. Thus, I = 1 2 (3m)r2 = 3 2 mr2 The bob falls vertically, and the string unwinds withou... | Physics |
24 | A uniformly charged ring of radius 4a and total charge Q is placed in the xy-plane, centered at the origin. A point charge q moves towards the ring along the z-axis with speed u at z = 5a. What is the minimum speed u such that the point charge crosses the origin? | {'A': '√ 3qQ 7πϵ0ma', 'B': '√0.188qQ 4πϵ0ma', 'C': '√ 5qQ 3πϵ0ma', 'D': '√ 3qQ 4πϵ0ma'} | B | Sol: 1. System Setup A ring of radius 4a lies in the xy-plane, centered at the origin, with total charge Q. A point charge q is located at z = 5a on the z-axis, moving towards the ring with speed u . We need to find the minimum speed u such that the point charge q can reach the origin (z = 0). 2. Electric Potential of ... | Physics |
25 | If a light wave is characterized by E = 1000 sin ω (t −x c ), and an electron moves with a velocity of 2.5 × 107 m/s perpendicular to the wave, calculate the maximum magnetic force acting on the electron. | {'A': '1.33 × 10−18 N', 'B': '1.33 × 10−17 N', 'C': '2.67 × 10−18 N', 'D': '2.67 × 10−17 N'} | B | Sol: Step 1: Expression for Magnetic Force The magnetic force on a moving charged particle is given by: F = qvB sin θ Where: q is the charge of the electron (q = e = 1.6 × 10−19C). v is the speed of the electron (v = 2.5 × 107 m/s), B is the magnetic field strength, θ is the angle between the velocity and the magnetic ... | Physics |
26 | A particle of mass m is dropped from a height H above the ground. At the same time, another particle of the same mass is thrown vertically upwards from the ground with a speed of v0 = √4gH. If they collide head-on completely inelastically, the time taken for the combined mass to reach the ground, in units of √H g , is: | {'A': '2', 'B': '3', 'C': '4', 'D': '5 B = E c B = 1000 3.0 × 108 = 3.33 × 10−6 T F = qvB F = (1.6 × 10−19) (2.5 × 107) (3.33 × 10−6) F = 1.6 × 2.5 × 3.33 × 10−19+7−6 F = 13.32 × 10−18 N F = 1.33 × 10−17 N'} | C | Sol: Step 1: Equation of motion for the two particles For the particle dropped from height H : Let the particle's position at time t be y1(t). Starting from rest at y = H, the the equation of motion is: y1(t) = H −1 2 gt2 For the particle thrown upward with speed v0 = √4gH : Let the particle's position at time t be y2(... | Physics |
27 | The electric fields of two plane electromagnetic waves in vacuum are given by: →E1 = E0^k cos(ωt −kz), →E2 = E0^i cos(ωt −ky) At t = 0, a particle of charge q = 2μC is at the origin with a velocity →v = 0.5c^j, where c is the speed of light in vacuum.Find the instantaneous force on the particle at t = 0. | {'A': '→F = 2 × 10−6E0^k + 1 × 10−6E0^i', 'B': '→F = 1 × 10−6E0^k + 2 × 10−6E0^i', 'C': '→F = 3 × 10−6E0^k + 0.5 × 10−6E0^i', 'D': '→F = 1.5 × 10−6E0^k + 2.5 × 10−6E0^i'} | A | Sol: To calculate the force on the charged particle, we use the Lorentz force equation: →F = q( →E + →v × →B) where →E is the total electric field, and →B is the total magnetic field. Step 1: Electric fields The total electric field is: →E = →E1 + →E2 At t = 0 and the origin (z = 0, y = 0) : →E1 = E0^k cos(0) = E0^k, →... | Physics |
28 | A small spherical droplet of density d is floating exactly one-third immersed in a liquid of density ρ and surface tension T. The radius of the droplet is (take note that the surface tension applies an upward force on the droplet): | {'A': 'r = √ 9T g(3d+ρ)', 'B': 'r = √ 9T g(3d−ρ)', 'C': 'r = √ 9T 2g(3d−p)', 'D': 'r = √ T 2g(3d−p)'} | C | Sol: We are tasked with finding the radius r of the spherical droplet, given the following conditions: 1. The droplet floats exactly one-third immersed in a liquid. 2. The densities are: - Density of the droplet: d - Density of the liquid: ρ 3. The liquid has a surface tension T, which applies an upward force on the dr... | Physics |
29 | For an ideal gas, the rate of change of pressure with volume is dp dv = −9p, and the boundary condition is p = 3p0 at v = 0. What is the maximum temperature the gas can attain? | {'A': 'p0 3nRe', 'B': 'p0 2nRe', 'C': 'Zero', 'D': 'p0 nRe'} | A | Sol: To determine the maximum temperature an ideal gas can attain under the given conditions, we'll follow a systematic approach. Here's a detailed breakdown of the solution: 1. Understanding the Given Information We are provided with: The rate of change of pressure with respect to volume: dp dv = −9p A boundary condit... | Physics |
30 | When photons of energy 4.50 eV strike metal A , the ejected photoelectrons have maximum kinetic energy TAeV and de Broglie wavelength λA. The maximum kinetic energy of photoelectrons from metal B when struck by photons of energy 5.00 eV is TB = TA −0.75eV. If the de Broglie wavelength of the photoelectrons from metal B... | {'A': '0.25 eV', 'B': '1.25 eV', 'C': '1.75 eV', 'D': '2.25 eV'} | A | Sol: To determine the maximum kinetic energy TB of photoelectrons ejected from metal B, we'll systematically analyze the given information and apply the relevant physics principles. Here's a detailed breakdown of the solution: 1. Understanding the Given Information We are provided with: Photons striking metal A have en... | Physics |
31 | A 52 kg mass is fastened to the end of a steel wire of unstretched length 2.7 m . The object moves in a vertical circle with an angular velocity of 2.4rev/s at the lowest point. The wire's cross-sectional area is 0.095 cm2. What is the approximate elongation of the wire at the lowest point?( Take Y=2 × 1011 N m2 for st... | {'A': '4.61 cm', 'B': '5.61 cm', 'C': '4.95 cm', 'D': '5.63 cm'} | A | Sol: To calculate the elongation (ΔL) of the steel wire at the lowest point, we consider the forces acting on the mass and use Hooke's law for the wire. At the lowest point, the tension in the wire (T) must provide the centripetal force (Fc) required for circular motion and balance the gravitational force (Fg) acting o... | Physics |
32 | A vessel of depth 3h is half-filled with a liquid of refractive index √3 and the other half with another liquid of refractive index 2 . The liquids are immiscible. The apparent depth of the inner surface of the bottom of the vessel will be: | {'A': 'h(4+3√3) 4√3', 'B': 'h(2+3√3) 4√3', 'C': 'h(6+3√3) 4√3', 'D': 'h(8+3√3) 4√3'} | B | Sol: Total Apparent Depth Calculation We have calculated the apparent depths for both sections of the vessel: 1. The apparent depth for the first liquid (with refractive index √3 ) is: Apparent depth 1 = h 2√3 2. The apparent depth for the second liquid (with refractive index 2 ) is: Apparent depth 2 = 3h 4 Now, the to... | Physics |
33 | A rectangular wooden block with dimensions 40 cm × 20 cm × 10 cm is floating in water at 0∘C. The block has a mass of 4.8 kg , and at 0∘C, the block is submerged to a depth of 8 cm . When the water temperature is increased to 4∘C, the block submerges to a depth of 7.8 cm . Find the ratio of the density of water at 4∘C ... | {'A': '1.234', 'B': '1.026', 'C': '1.058', 'D': '2.026'} | B | Sol: Analyze the Block at 0∘C The volume of the block is: Vblock = 40 cm × 20 cm × 10 cm = 8000 cm3 = 0.008 m3 The volume submerged at 0∘C is: Vsub,0 = Base area × Submerged depth Vsub,0 = (40 cm × 20 cm) × 8 cm = 6400 cm3 = 0.0064 m3 The buoyant force equals the weight of the block: Buoyant Force = ρ0 ⋅g ⋅Vsub,0 = m ⋅... | Physics |
34 | A 1000 Hz, 25 V(rms) source is connected to a resistance of 75Ω, an inductance of 0.15 H , and a capacitance of 5μ F, all in series. The resistance has a heat capacity of 3 J/∘C. Calculate the time it takes for the resistor to heat up by 8∘C, assuming no heat loss to the surroundings. | {'A': '426', 'B': '564', 'C': '223', 'D': '457'} | A | Sol: Impedance of the circuit The total impedance is: Z = √R2 + (XL −XC)2 Given: R = 75Ω, L = 0.15H, C = 5μ F = 5 × 10−6 F, f = 1000 Hz. Calculate XL and XC : XL = 2πfL = 2π(1000)(0.15) = 942.48Ω XC = 1 2πfC = 1 2π(1000)(5×10−6) ≈31.83Ω The net reactance is: XL −XC = 942.48 −31.83 = 910.65Ω Now calculate Z : Z = √R2 + ... | Physics |
35 | The vertical cross-section of an inclined plane is given by y = x2 10 . If the coefficient of friction is μ = 0.25, determine the maximum height (in cm ) where a stationary block will not slip. (x and y are in m) | {'A': '15.6 cm', 'B': '16.6 cm', 'C': '18.5 cm', 'D': '18.6 cm'} | A | Sol: Step 1: Understand the problem At any point on the surface, the block experiences both: The gravitational force ( mg ), acting vertically downward. The normal force (N) from the inclined plane, which is perpendicular to the surface. The frictional force f resists the motion of the block, and the maximum frictional... | Physics |
36 | A wire vibrates at a fundamental frequency of 500 Hz . A second identical wire produces 5 beats per second with it when the tension in the first wire is slightly decreased. The ratio of the tension in the second wire to the tension in the first wire is approximately equal to: | {'A': '1.04', 'B': '1.01', 'C': '1.05', 'D': '1.02'} | D | Sol: Given that, fundamental frequency of first wire, f1 = 500 Hz Let frequency of 2 nd wire be f2. Beat frequency, fb = 5 Using, expression for fundamental frequency of wire, fb = f2 −f1 f2 = fb + f1 = 500 + 5 = 505Hz f = 1 2 L √ T u ⇒f ∝√ T ⇒f2 f1 = √T2 T1 ⇒505 500 = √T2 T1 ⇒ T2 T1 = ( 505 500 ) 2 = 1.02 | Physics |
37 | A square loop made of a metal wire with a diameter of 2 mm and a total length of 20 cm is placed in a uniform magnetic field that is perpendicular to the plane of the loop. The magnetic field is changing at a rate of dB dt = 0.05 T/s. The resistivity of the metal wire is 1.2 × 10−8Ω m. Calculate the induced current in ... | {'A': '', 'B': '', 'C': '', 'D': ''} | B | Sol: 1. Given values: Diameter of the wire: 2 mm = 2 × 10−3 m Length of the wire: 20 cm = 0.2 m Rate of change of magnetic field: dB dt = 0.05 T/s Resistivity of the metal wire: ρ = 1.2 × 10−8Ω m 2. Calculate the resistance of the wire: The wire is made of a metal, and we can treat it as a cylindrical wire. The resista... | Physics |
38 | An object is weighed at a height h above the Earth's surface at the equator using a spring balance. The object's weight at this height is found to be the same as at the Earth's surface at the poles. If the acceleration due to gravity at the Earth's surface is g and the angular velocity of the Earth is ω, find the value... | {'A': 'ω2R2 g', 'B': 'ω2R2 2g', 'C': 'gR2 ω2R', 'D': 'gR2 2ω2'} | B | Sol: Step 1: Weight at the Poles At the poles, the weight Wpoles of an object is given by: Wpoles = mg where m is the mass of the object and g is the acceleration due to gravity at the Earth's surface. Step 2: Weight at Height h Above the Equator At height h above the Earth's surface at the equator, the apparent weight... | Physics |
39 | A ball is thrown vertically upward with an initial velocity u. It experiences a resistive force proportional to the square of the velocity, Fres = −mkv2, where v is the velocity and k is a constant. If the maximum height attained by the ball is H, which of the following expressions for H is correct? | {'A': 'H = 1 2k ln (1 + ku2 g )', 'B': 'H = 1 2k ln (1 −ku2 g )', 'C': 'H = 1 k ln (1 + 2ku2 g )', 'D': 'H = 1 2k ln (1 + 2ku2 g )'} | A | Sol: Given: A ball is thrown vertically upward with initial velocity u. The resistive force on the ball is proportional to the square of its velocity, i.e., Fres = −mkv2, where m is the mass, k is the constant of proportionality, and v is the velocity. We are to find the maximum height attained by the ball. Understandi... | Physics |
40 | A circular coil of radius r = 2 cm and moment of inertia I = 0.5 kg ⋅ m2 carries a current that produces a magnetic moment of μ = 10Am2. The coil is initially in a vertical position and can rotate freely around a horizontal diameter. A uniform magnetic field of strength B = 3 T is applied along the vertical direction. ... | {'A': '5.00rad/s', 'B': '5.25rad/s', 'C': '5.50rad/s', 'D': '5.94rad/s'} | D | Sol: Step 1: Magnetic Potential Energy The magnetic potential energy Umag of the coil in the magnetic field is given by: Umag = −μB cos θ Where: μ = 10Am2 is the magnetic moment, B = 3 T is the magnetic field strength, θ is the angle between the magnetic moment and the magnetic field. At the initial position, the coil ... | Physics |
41 | The effective capacitance of a parallel combination of two capacitors C1 and C2 is 12μ F. When these capacitors are individually connected to a voltage source of 1 V , the energy stored in C2 is 9 times that of C1. If these capacitors are connected in series, what will be their effective capacitance? | {'A': '1.00μ F', 'B': '1.08μ F', 'C': '1.20μ F', 'D': '1.50μ F'} | B | Sol: Given: 1. Effective capacitance in parallel: Cparallel = C1 + C2 = 12μ F 2. Energy stored in C2 is 9 times that of C1 : U2 = 9U1 The energy stored in a capacitor is given by: U = 1 2 CV 2 Since V = 1 V, the energy ratio depends only on their capacitances: U2 U1 = C2 C1 Substituting U2 = 9U1: C2 C1 = 9 ⟹C2 = 9C1 3.... | Physics |
42 | A resistor R, an inductor XL = 6R, and a capacitor XC = 5R are connected in series in an AC circuit. The ratio of the new power factor to the old power factor is √37 : x. Determine x. | {'A': '√2', 'B': '√3', 'C': '√6', 'D': '√9'} | A | Sol: First Case (Before Adding Capacitor): The total impedance is: Z1 = √R2 + X 2 L = √R2 + (6R)2 = √R2 + 36R2 = √37R2 = R√37 The power factor is: cos ϕ1 = R Z1 = R R√37 = 1 √37 Second Case (After Adding Capacitor): The effective reactance is: Xnet = XL −XC = 6R −5R = R The total impedance is: Z2 = √R2 + X 2 net = √R2 ... | Physics |
43 | One mole of an ideal gas follows the law p2 V = constant during expansion. If the initial and final temperatures of the gas are T1 and T2, respectively, find the work done by the gas : | {'A': 'R 3 (T2 −T1)', 'B': 'R (T2 −T1)', 'C': 'R 2 (T2 −T1)', 'D': '2R 3 (T2 −T1)'} | D | Sol: The gas follows p2 V = constant, which implies p2 = CV , where C is a constant. Ideal gas law: pV = RT. From the ideal gas law: p = RT V Substitute p = RT V into p2 = CV : ( RT V ) 2 = CV Simplify: R2T 2 V 2 = CV ⟹V 3 = R2T 2 C Thus: V = ( R2T 2 C ) 1/3 The work done by the gas is: W = ∫pdV Using p = RT V , this b... | Physics |
44 | A spherical gaseous cloud of mass density ρ(r) in free space consists of particles, each of mass m, moving in circular orbits about the center. The force on the particles is due to their mutual gravitational attraction. If the kinetic energy K of each particle is proportional to its radial distance r, then the particle... | {'A': 'r−1', 'B': 'r−2', 'C': 'r−3', 'D': 'r0 (constant)'} | A | Sol: The gravitational force acting on a particle at radius r is given by: Fgrav = GmM(r) r2 where M(r) is the total mass enclosed within radius r, given by: M(r) = ∫r 0 4πr′2ρ (r′)dr′ For circular motion, the centripetal force is provided by the gravitational force: mv2 r = GmM(r) r2 Simplify to: v2 = GM(r) r The kine... | Physics |
45 | A current-carrying wire heats a metal block. The wire delivers constant power P to the block. The block is enclosed in an insulated chamber, and its temperature T(t) changes with time t asT(t) = T1 (1 + αt 1 3 )where α is a constant with appropriate dimensions, and T1 is a constant with the dimension of temperature. De... | {'A': '3P(T(t)−T1)4 α3T 5 1', 'B': '3P(T(t)−T1)3 α3T 4 1', 'C': '3P(T(t)−T1)2 α3T 3 1', 'D': '3P(T(t)−T1) α3T 2 1'} | C | Sol: Differentiating T(t) with respect to time t : dT dt = T1 ⋅d dt (1 + αt 1 3 ) = T1 ⋅(0 + α 3 t−2 3 ) dT dt = αT1 3 t−2 3 The power supplied P is used to increase the temperature, so P = C(T) dT dt where C(T) is the heat capacity. Substituting dT dt : P = C(T) ⋅αT1 3 t−2 3 C(T) = 3P αT1 ⋅t 2 3 From T(t) = T1 (1 + αt... | Physics |
46 | A circular coil of radius R, with n turns, resistance r, and total mass m, lies in the horizontal plane. A uniform magnetic field B(t) is applied perpendicular to the plane of the coil, where B(t) increases linearly from 0 to B0 in time t0, then decreases linearly back to 0 in the same time t0, and this process repeats... | {'A': 'n2A2B2 0 rt0', 'B': '2n2A2B2 0 rt0', 'C': 'n2A2B2 0 2rt0', 'D': '4n2A2B2 0 rt0'} | B | Sol: The magnetic flux through the coil is: The rate of change of the flux induces an emf: Since B(t) changes linearly, dB dt = B0 t0 , so: The current in the coil, using Ohm's law, is: The instantaneous power dissipated due to the coil's resistance is: Substituting E = nA B0 t0 : The energy dissipated during one phase... | Physics |
47 | A wheel with 8 metallic spokes, each 0.7 m long, is rotating at 180rev/min in a plane normal to the Earth's magnetic field of strength HE = 0.5G. However, the Earth's magnetic field is not uniform, and it has a gradient that increases linearly with radial distance from the axle. The magnetic field strength varies as: B... | {'A': '4.70 mV', 'B': '5.62 mV', 'C': '3.62 mV', 'D': "0.38 mV Ans: d Ephase = P ⋅Δt Ephase = (nA B0 t0 ) 2 r ⋅t0 = n2A2B2 0 rt0 Etotal = 2 ⋅Ephase = 2 ⋅n2A2B2 0 rt0 Sol: Sol: Given Data: Number of spokes, n = 8 Length of each spoke, l = 0.7m Rotational speed, 180rev/min Magnetic field strength at the rim, B0 = 0.5G, w... | A | Sol: The equation of a sinusoidal transverse wave traveling in the positive x-direction is: y(x, t) = A sin(kx −ωt) where: - A = 15 cm = 0.15 m (amplitude), - k = 2π λ is the wave number, with λ = 0.4 m, - ω = v ⋅k is the angular frequency, with v = 20 cm/s = 0.2 m/s. Calculate Wave Number k and Angular Frequency ω 1. ... | Physics |
49 | A particle is projected from the ground with an initial velocity u = 20m/s at an angle θ = 60∘with the horizontal. The particle clears two walls of equal height h = 10m at distances x1 = 10m and x2 = 20m from the point of projection. Find the time taken by the particle to travel from the first wall to the second wall. ... | {'A': '3 s', 'B': '2 s', 'C': '1 s', 'D': '5 s'} | C | Sol: Given: - Initial velocity, u = 20m/s - Angle of projection, θ = 60∘ - Height of walls, h = 10m - Distances of walls from the point of projection, x1 = 10m and x2 = 20m - Acceleration due to gravity, g = 10m/s2 Horizontal Motion The horizontal component of velocity is constant: ux = u cos θ = 20 cos 60∘= 20 × 0.5 =... | Physics |
50 | A spherical oil drop carrying a charge q is held stationary in air using a vertical uniform electric field of strength E = 50π 7 × 105 V/m. When the electric field is switched off, the oil drop falls with a terminal velocity of 1.5 × 10−3 m/s. Given: Acceleration due to gravity, g = 9.8 m/s2, Viscosity of air, η = 1.8 ... | {'A': '1.2 × 10−19C', 'B': '2.4 × 10−19C', 'C': '3.6 × 10−19C', 'D': '9.14 × 10−19C'} | D | Sol: When the electric field holds the drop stationary, the forces acting are: 1. Weight of the drop: Fg = mg, 2. Electric force due to the field: FE = qE. At equilibrium: Fg = FE ⇒ qE = mg When the electric field is switched off, the drop falls under gravity with terminal velocity vt. At terminal velocity, the net for... | Physics |
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