This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
The Nobel Prize 2018 in Physiology or Medicine was won byA. Yoshinori OhsumiB. Kazuo IshiguroC. James P. Allision and Tasuku HonjoD. William D.Northaus and Paul M. Romer |
| Answer» Correct Answer - C | |
| 2. |
The paris agreements of the United Nations Framework Convention on Climate Change is to limit global warming well belowA. `1.0^(@)C`B. `1.5^(@)C`C. `2.0^(@)C`D. `2.5^(@)C` |
| Answer» Correct Answer - C | |
| 3. |
Define Anaphylaxis. |
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Answer» Anaphylaxis is the classical immediate hypersensitivity reaction. It is a sudden, systematic, severe and immediate hypersensitivity reaction occurring as a result of rapid generalized mast-cell degranulation |
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| 4. |
Tiya said, ”Shut the door.” – to open the door.According to you, did he give the correct command?(a) True(b) False |
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Answer» Correct answer is (b) False |
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| 5. |
James P. Allison and Tasuku Honjo won the Nobel Prize-2018 in the filed of Physiology or Medicine for their discovery ofA. molecular mechanisms controlling the circadian rhythmB. cancer therapy by inhibition of negative immune regulationC. mechanisms for autophagyD. novel therapy against malaria |
| Answer» Correct Answer - B | |
| 6. |
State any two unique features of ELISA test. |
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Answer» ELISA is highly sensitive and can detect antigen even in nanograms. ELISA test does not require radioisotopes or radiation counting apparatus. |
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| 7. |
What is the child careful about according to the poem? |
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Answer» The child is careful about not tripping others. |
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| 8. |
Read the poem below and choose the word that is not an action word:-a) hopb) spinc) floord) roll |
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Answer» Correct answer is (c) floor |
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| 9. |
Choose the correct statement:-1. The child in the poem is careless2. The child in the poem is careful. |
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Answer» Correct statement is 2. The child in the poem is careful |
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| 10. |
Spread of cancerous cells to distant sites is termed as _________. (a) Metastasis (b) Oncogenes (c) Proto-oncogenes (d) Malignant neoplasm |
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Answer» Spread of cancerous cells to distant sites is termed as Metastasis. |
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| 11. |
Name the active chemical found in the medicinal plan Rauwolfia vomitoria. What type of diversity does it belongs to? |
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Answer» Rauwolfia vomitoria can be cited as an example for genetic diversity. Reserpine is an active chemical extracted from Rauwolfia vomitoria. |
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| 12. |
Read and recite the poem with action. Here in the poem, legs are for a. clappingb. snapc. jumpingd. see |
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Answer» Correct option is C. jumping |
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| 13. |
Assertion (A): Streptomycin is an antibiotic. Reason (R): Antibiotic are microbial chemicals inhibits the growth of pathogenic microbe.(a) A is right R is wrong(b) R explains A (c) A and R are wrong (d) A is wrong but R is right |
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Answer» (b) R explains A |
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| 14. |
Match the body parts names with their action.List AList B1. handsa) jumping2. fingersb) see3. legsc) snap4. eyesd) clapping |
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Answer» 1. d 2. c 3. a 4. b |
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| 15. |
Write whether the given statement is True or False .My eyes help me to see.(a) True(b) False |
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Answer» Correct answer is (a) True |
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| 16. |
Write whether the given sentence is True or False. A pencil is used to write. (a) True(b) False |
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Answer» Correct answer is True |
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| 17. |
Match the picture with its name List AList Ba) pencilb) flowerc) cupd) book |
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Answer» 1.d 2.c 3.a 4.b |
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| 18. |
Look at the pictures carefully and fill in the blanks with the correct preposition- in, on, under-1. The dog is ......... the kennel.2. The bird is ............. the roof of the kennel. |
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Answer» 1. In 2. on |
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| 19. |
\( \sin ^{-1} x+\sin ^{-1} 2 x=\frac{\pi}{3} \) |
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Answer» sin-1 x + sin-1 2x = \(\frac{π}{3}\) sin-1 ( x\((x { \sqrt{1-4x^2} \ } + 2x { \sqrt{1-x^2} \ }) =\frac{π}{3}\) ( ∵ sin-1 x + sin-1 y = sin-1 \((2x { \ \sqrt{1-y2} \ }+{ 4 \sqrt{1-x^2} \ }))\)\(\) ⇒ \(x { \sqrt{1-4x^2} \ } +2x { \sqrt{1-x^2} \ } =sin\frac{π}{3}=\frac{√3}{2}\) ⇒ x2 (1- 4x2) + 4 x2 (1-x2) + 4x2 \( { \ \sqrt{(1-4x^2)(1-x^2)} \ }=\frac{3}{4}\) (by squaring on both sides) ⇒ 16x2 \( { \ \sqrt{(1-x^2)(1-4x^2)} }=3-(x^2 -4x^4+4x^2-4x^4)\) ⇒ \(16x^2 = { \sqrt{(1-x^2)(1-4x^2)} \ } =3-5x^2+8x^4\) ⇒ 256x4 (1 - x2) (1- 4x2) = 9 + 25x4+ 64x8 - 30 x2 + 48 x4 - 80x6 ⇒ 256x4 (1 - 5x2 +4x4 ) = 64x8 - 80x6 + 73x4 - 30 x2 + 9 ⇒ 256x4 - 1280x6 + 1024 x8 = 64 x8 - 80 x6 + 73x4 - 30x2 +9 ⇒ 960x8 - 1200x6 + 183x4 +30x2 - 9 = 0 |
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| 20. |
Which of the following statement is/are correct?1. The directorate of mines and minerals is located in Jaipur.2. Rajasthan is the largest producer of copper in India.3. Rajasthan is endowed with the largest lead-zinc ore deposit.1. 1 and 2 only2. 2 and 3 only3. 3 only4. All the above |
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Answer» Correct Answer - Option 3 : 3 only The correct answer is 3 only.
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| 21. |
In which state are the Khetri mines located?1. Orisha2. Rajasthan3. Bihar4. Uttar Pradesh |
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Answer» Correct Answer - Option 2 : Rajasthan The correct answer is Rajasthan.
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| 22. |
Which of the following, a major film actor, has been selected for the prestigious award in France called 'Knight of the Order of Arts and Letters'.1. Kamal Hassan2. Shabana Azmi3. Rajinikanth4. Anupam Kher |
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Answer» Correct Answer - Option 1 : Kamal Hassan The correct answer is Kamal Hassan.
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| 23. |
In which state is the Singareni Mines located?1. Odisha2. Telangana3. Maharashtra4. Chhattisgarh |
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Answer» Correct Answer - Option 2 : Telangana The correct answer is Telangana.
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| 24. |
Apart from Virat Kohli, who is the only other member of the 2008 Under 19 World Cup winning squad to have played a test match for India? |
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Answer» Abhinav Mukund |
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| 25. |
प्रत्येक रविवार का जिन 3 मील जॉगिंग करता है। यदि सोमवार को वह 1 मील जॉगिंग करता है और सप्ताह के शेष दिनों में वह प्रत्येक दिन पिछले दिन की अपेक्षा 1 मील अधिक जॉगिंग करता है। बताइए 2 सप्ताह में जिन कितने मील की जॉगिंग करता है?A. 42B. 63C. 48D. 98 |
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Answer» Correct Answer - C `((S, M, T, W, T, F, S),(3,1 , 2, 3, 4, 5, 6))` Jogs in one weak `=24` Total Jogs in two week `=24xx2` `=48` |
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| 26. |
एक व्यक्ति जिसकी लम्बाई 169 सेमी है, एक खम्भे के सहारे खड़ा हुआ है। उसकी छाया की लम्बाई 130 सेमी है। तो उस खम्भे की लम्बाई ज्ञात करें, जिसकी छाया की लम्बाई 420 सेमी है?A. 550 cmB. 589 cmC. 323 cmD. 546 cm |
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Answer» Correct Answer - D यदि छाया 130 cm है तो वास्तविक लम्बाई 169 cm है। यदि छाया `1cm` है तो वास्तविक लम्बाई `169/130` यदि छाया 420 cm है वास्तविक लम्बाई है। `=169/130xx420` `=546` cm |
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| 27. |
3 बजे मिनट व घण्टे की सुई द्वारा बनाये गए कोण का मान बताओ?A. `pi//4`B. `pi//3`C. `5pi//12`D. `pi//2` |
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Answer» Correct Answer - D hour `=3` minute `=00` angle `=|30H-11/2M|` `=|30xx3-11/2xx0|` `theta=90` `theta=90xx(pi)/180=(pi)/2` |
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| 28. |
if `omega` is a complex fifth roots of unity and `(1+omega+omega^(2)+omega^(3))^(2005)=A+Bomega+Comega^(2)+Domega^(3)` (Where A,B,C,D are real), the `A^(2)+B^(3)+C^(4)+D^(5)=`A. 2B. 1C. 3D. None of these |
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Answer» Correct Answer - B `because omega` is a complex fifth roots of unity. `therefore1+omega+omega^(2)+omega^(3)+omega^(4)=0` `implies(-omega^(4))^(2005)=A+Bomega+Comega^(2)+Domega^(3)` `implies-omega^(8020)=A+Bomega+Comega^(2)+Domega^(3)` `implies-(omega^(5))^(1604)=A+Bomega+Comega^(2)+Domega^(3)` `implies-1=A+Bomega+Comega^(2)+Domega^(3)` `impliesA=-1,B=0,C=0,D=0` |
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| 29. |
If a complex number having absolute value of √2 is making 45° angle with x-axis in third quadrant, then it can be written as –1. [r, θ] = \(\left[ {\sqrt 2 ,\frac{{3\pi }}{4}} \right]\)2. [r, θ] = \(\left[ {\sqrt 2 ,\frac{{\pi }}{4}} \right]\)3. [r, θ] = \(\left[ {\sqrt 2 ,-\frac{{3\pi }}{4}} \right]\)4. [r, θ] = \(\left[ {\sqrt 2 ,-\frac{{\pi }}{4}} \right]\) |
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Answer» Correct Answer - Option 3 : [r, θ] = \(\left[ {\sqrt 2 ,-\frac{{3\pi }}{4}} \right]\) CONCEPT: Point P is uniquely determined by the ordered pair of real numbers (r, θ), called the polar coordinates of the point P. If P represent the nonzero complex number z = x + iy. Here \(r = \sqrt {{x^2} + {y^2}} = \left| z \right|\) is called modulus of the given complex number. The argument of Z is measured from the positive x-axis only. Let z = r (cos θ + i sin θ) is a polar form of any complex number then following ways are used while writing θ for different quadrants – For the first quadrant, \({\rm{\theta }} = {\tan ^{ - 1}}\frac{{\rm{y}}}{{\rm{x}}}\) For the second quadrant \({\rm{\theta }} = {\rm{\pi }} - {\tan ^{ - 1}}\frac{{\rm{y}}}{{\rm{x}}}\) For the third quadrant \({\rm{\theta }} = - {\rm{\pi }} + {\tan ^{ - 1}}\frac{{\rm{y}}}{{\rm{x}}}\) For the fourth quadrant \({\rm{\theta }} = - {\rm{\;ta}}{{\rm{n}}^{ - 1}}\frac{{\rm{y}}}{{\rm{x}}}\) CALCULATION: Given that \(\left| z \right| = \sqrt 2 \) and angle with x-axis is 45°. Since its in the third quadrant - ∴ \(\theta = - \pi + \frac{\pi }{4} = - \frac{{3\pi }}{4}\) ⇒ [r, θ] = \(\left[ {\sqrt 2 , - \frac{{3\pi }}{4}} \right]\) |
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| 30. |
Two series X (x1, x2, x3, x4, ….xn) and Y (y1, y2, y3, y4, ….., yn) are in A.P., such that xn – yn = n − 2. It is also known that x3= b5. Find the value of x99 – y197. (a) 47 (b) 48 (c) 49 (d) 50 |
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Answer» Correct option (b) 48 Explanation: Let the common difference of the series X be d1 and that of Y be d2 Since xn – yn = n − 2, x1 − y1 = −1 or y1 = x1 + 1 x3 = y5 x1 + 2d1 = y1 + 4d2 x1 + 2d1= x1 + 1 + 4d2 2d1 − 4d2 = 1 x99 – y197 = x1+ 98d1 − y1- 196d2 = −1 + 49(2d1− 4d2) = −1 + 49 = 48. |
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| 31. |
If `A, B, C, D` be the angles of a quadrilateral, prove that : `(tanA+tanB+tanC+tanD)/(cotA+cotB+cotC+cotD) = tan A tan B tan C tan D`A. tan A tan B tan C tan DB. cot A cot B cot C cot DC. `tan^(2)A + tan^(2)B + tan^(2)C + tan^(2)D`D. `sum tan A tan B tanC` |
| Answer» Correct Answer - A | |
| 32. |
If AB and AC are two tangents to a circle with centre O such that ∠BOC=110° then ∠CAO = (A) 60° (B) 45° (C) 35° (D) 150° |
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Answer» Correct answer is (C) 35° |
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| 33. |
If `(tan(A+B+C))/(tan(A-B+C))=(tanC)/(tanB)` then `sin2A+sin2B+sin2C` is equal to_______ |
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Answer» `(sin(A+B+C).cos(A-B+C))/(sin(A-B+C)cos(A+B-C))=(sinCcosB)/(sinBcosC)` `implies(sin2A+sin(2B-2C))/(sin2A-sin(2B-2C))=(sinCcosB)/(sinBcosC)` Use `C` and `D` rule will get `sin2A+sin2B+sin2C=0` |
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| 34. |
In ∆ABC, tan\((\frac{A-B}{2})\).tan\(\frac{c}{2}\) =(A) a+b/a-b(B) a2+b2/a2-b2(C) a2-b2/a2+b2(D) a-b/a+b |
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Answer» Correct option is: (D) \(\frac{a-b}{a+b}\) |
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| 35. |
In the given figure QOR is a diameter of a circle and PQ=PR, then ∠PQR = (A) 90° (B) 60° (C) 45° (D) 30° |
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Answer» Correct answer is (C) 45° |
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| 36. |
In the given figure, O is the centre of the circle and ∠QPR = 60° then ∠OQR =(A) 60° (B) 30° (C) 120° (D) 40° |
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Answer» Correct answer is (B) 30° |
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| 37. |
If A is a 3 × 3 non-singular matrix such that AAT = AT A and B = A-1 AT , then BBT = ________. (a) A (b) B (c) I3 (d) BT |
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Answer» The correct answer is : (c) l3 |
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| 38. |
Sin α x cot α =(A) 1 (B) cos α(C) cos2α(D) sin2α |
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Answer» Correct answer is (B) cos α |
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| 39. |
If A=[aij]3x3 is any orthogonal matrix then which of the following cannot be true. |A| = 1|A| = -1|A|=0None of these |
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Answer» \(\because\) A is orthogonal matrix. \(\therefore \) ATA = A.AT = I, where I is identify matrix. ⇒ |ATA| = | I | ⇒ |ATA| |A| = 1 (\(\because\) | I | = 1) ⇒ |A| |A| = 1 (\(\because\) |ATA| = |A|) ⇒ |A|2 = 1 ⇒ |A| = -1 or |A| = 1 Hence, |A| = 0 can not be true for an orthogonal matrix A. |
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| 40. |
Velocity and acceleration vector of a charged particle moving in a magnetic field at some instant are `vecv=3hati+4hatj and veca=2hati+xhatj`. Select the correct options.A. `x=-1.5`B. `x=3`C. Magnetic field is along `z` -directionD. Kinetic energy of the particle is constant |
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Answer» Correct Answer - A::C::D acd |
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| 41. |
Assertion: The lightening conductor at the top of high building has sharp pointed ends. Reason: The surface density of charge at sharp points is very high resulting in setting up of electric wind.A. If both assertion and reason are true and reason correct explanation of assertion.B. If both assertion and reason are true but reason the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - A We know that surface density of charge is very large on the sharp ends of a conductor. This charge from pointed ends sets up a charged electric wind. This charged electric wind comes in contact with the charged cloubs and then source of its building neturalised and so potential drops between building is reduced. Even if lightning strikes the building, charge is conducted by the lightning conductor to the earth and there is no harm to building. |
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| 42. |
Assertion: The lightening conductor at the top of high building has sharp pointed ends. Reason: The surface density of charge at sharp points is very high resulting in setting up of electric wind.A. Statement I is true, Statement II is True, Statement II is a correct explanation for statement I.B. Statement I is true, Statement II is True, Statement II is Not a correct explanation for statement I.C. Statement I is True, Statement II is False.D. Statement I is False, Statement II is True. |
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Answer» Correct Answer - A We know that surface density of a charge is very large on the sharp ends of a conductor. Hence, the lighing conductor conducts chares of lighting to earth and there is no harm to the building. |
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| 43. |
Two small drops of mercury, each of radius `R`, coalesce to form a single large drop. The ratio of the total surface energies before and after the change isA. `2^(2//3) : 1`B. `sqrt2 : 1`C. `sqrt(1//3) : 1`D. `2 : 1` |
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Answer» Correct Answer - C Given that, the two small drops of mercury and radius R of each drop coalesce to form a large drop of radius r, so the net volume remains constant i.e, initial volume = final volume `V_("initial")=V_("final")` `2xx4/3piR^(3)=1dot4/3pir^(3)rArrr=2^(1//3)R` ...(i) As, surface tension is constant for both the drops. So, the surface energy of two small drops, `E_(1)=2xxTA_(1)=2xx4piR^(2)xxT` Surface energy of one big drop, `E_(2)=TA_(2)=4pir^(2)xxT=2^(1//3).4piR^(2)T` (using Eq. (i)) Ratio the total surface energy before and after the change is given as, `E_(1)/E_(2)=(8piR^(2)T)/(2^(2//3)4piR^(2)T)=(2^(1-2/3))/1=2^(1/3):1` Hence, the ratio of suface energy of the each drop is `2^(1/3) : 1` |
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| 44. |
The figure shows an isosceles triangle plate of mass `M` and base `L`. The angle at the apex is `90^@`. The apex lies at the origin and the base is parallel to `X-`axis. The moment of inertia of the plate about the `y-`axis isA. `(ML^(2))/(6)`B. `(ML^(2))/(8)`C. `(ML^(2))/(24)`D. None of these |
| Answer» Correct Answer - A | |
| 45. |
The figure shows an isosceles triangle plate of mass `M` and base `L`. The angle at the apex is `90^@`. The apex lies at the origin and the base is parallel to `X-`axis. The moment of inertia of the plate about its base parallel to the `x-`axis isA. `(ML^(2))/(18)`B. `(ML^(2))/(36)`C. `(ML^(2))/(24)`D. None of these |
| Answer» Correct Answer - A | |
| 46. |
Two infinitely long conducting parallel rails are connected through a capacitor C as shown in Fig. A conductor of length l is moved with constant speed `v_(0)`. Which of the following graph truly depicts the variation of current through the conductor with time? A. B. C. D. |
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Answer» Correct Answer - C emf across capacitor is constant and equal to Blv. In steady state, no current flows through the circuit. |
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| 47. |
A body is dropped from rest from a height h. It covers a distance `(9h)/(25)` in the last seconds of fall. The height h is : (use `g = 9.8 m//sec^(2)` )A. `102.5 m`B. `112.5 m`C. `122.5 m`D. `132.5 m` |
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Answer» Correct Answer - C t is the time to reach ground. `h= (1)/(2) at^(2) , (1 - (9)/(25)) h = (1)/(2) a (t - 1)^(2)` `(1 - (9)/(25)) = ((t - 1)^(2))/(t^(2)) , (16)/(25) = ((t - 1)^(2))/(t^(2))` or `(4)/(5) = (t - 1)/(t) :. T = 5 sec` `h = (1)/(2) xx 9.8 xx 5^(2) = 122.5 m` |
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| 48. |
Two particles are projected simultaneously from the top of the towers as shown. Find the distance between the towers if they collide in mid air during their flight. A. 15mB. 20mC. `15sqrt(3)m`D. `5sqrt(3)m` |
| Answer» `vecV+(A//B)` is in directed along AB. | |
| 49. |
If `vec(A)=hat(i)+2hat(j)+3 hat(k), vec(B)=-hat(i)+hat(j)+4hat(k)` and `vec(C)=3hat(i)-3hat(j)-12 hat(k)`, then find the angle between the vectors `(vec(A)+vec(B)+vec(C))` and `(vec(A)xxvec(B))` in degrees. |
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Answer» Correct Answer - 90 `vecP=vecA+vecB+vecC=3hati-5hatk` and `vecQ=vecAxxvecB=|(hati,hatj,hatk),(1,2,3),(-1,1,4)|=5hati-7hatj+3hatk` Angle between `vecP` & `vecQ` is given by `costheta=(vecP.vecQ)/(PQ)=(15-15)/(PQ)=0rArrtheta=90^(@)` |
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| 50. |
What is the percentage change in the tension necessary in a somometer of fixed length to produce a note one octave lower (half of original frequency) than before?A. `25%`B. `50%`C. `67%`D. `75%` |
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Answer» Correct Answer - D In Sonometer `Vprop sqrt(T)` `rArr T_(2)=(T_(1))/(4)` `(T_(1)-T_(2))/(T_(1))xx100=(T_(1)-(T_(1))/(4))/(T_(1))xx100=75%` |
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