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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. |
Conisder a system of three charges `q//3`, `q//3` and `-2q//3` placed at point A, B and C, respectively, as shown in the figure. Take O to be centre of the circle of radius R and angle `CAB=60^@` A. the electric field at point O is `q//8piepsilon_0R^2` directed along the negative x-axis.B. the potential energy of the system is zeroC. the magnitude of the force between the charges at C and B is `q^2// 54 piepsilon_0R^2`.D. the potential at point O is `q //12piepsilon_0R` |
| Answer» `F_(BC)=(1)/(4piepsilon_(0))(((q)/(3))((2q)/(3)))/((R//sqrt(3))^(2))=(q^(2))/(54piepsilon_(0)R^(2))` | |
| 2. |
If a ∶ b = 2 ∶ 3, b ∶ c = 5 ∶ 7 and c ∶ d = 4 ∶ 5. Then find a ∶ b ∶ c ∶ d. 1. 40 ∶ 60 ∶ 84 ∶ 1052. 44 ∶ 60 ∶ 84 ∶ 1053. 40 ∶ 60 ∶ 84 ∶ 1074. 40 ∶ 64 ∶ 84 ∶ 105 |
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Answer» Correct Answer - Option 1 : 40 ∶ 60 ∶ 84 ∶ 105 Given: a ∶ b = 2 ∶ 3 b ∶ c = 5 ∶ 7 c ∶ d = 4 ∶ 5 Calculation: a ∶ b ∶ c ∶ d 2 ∶ 3 ∶ 3 ∶ 3 5 ∶ 5 ∶ 7 ∶ 7 4 ∶ 4 ∶ 4 ∶ 5 ⇒ (2 × 5 × 4) ∶ (3 × 5 × 4) ∶ (3 × 7 × 4) ∶(3 × 7 × 5) ⇒ 40 ∶ 60 ∶ 84 ∶ 105 ∴ a ∶ b ∶ c ∶ d = 40 ∶ 60 ∶ 84 ∶ 105 The correct option is 1 i.e. 40 ∶ 60 ∶ 84 ∶ 105 |
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| 3. |
For a dipole, the value of each charge is 10–10 stat coulomb and their separation is 1Å, then its dipole moment is :–A. one debyeB. 2 debyeC. `10^(–3)` debyeD. `3xx 10^(–20)` debye |
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Answer» Correct Answer - A P = q d `=10-10xx1xx10^(-8)` = 10–18 = 1 debye |
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| 4. |
Taps P and Q can fill a tank in 6 hours and R can empty it in 4 hours. If all the taps are opened together. In how much time will the tank be empty? 1. 8 hours2. 11 hours3. 10 hours4. 12 hours |
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Answer» Correct Answer - Option 4 : 12 hours Given: P and Q can fill a tank = 6 hours R can empty fill a tank = 4 hours Formula Used: Work done = Total work/Total work done in 1 hour Calculation:
Work done = Total work/Total work done 1 hour by all three ⇒ Work done by (P + Q + R) = 12/(-1) = 12 hours ∴ Work done by P, Q and R in 12 hours. The correct option is 4 i.e. 12 hours |
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| 5. |
if (1-p) is root of quadratic equation x^2+px+(1-p)=0 then its roots are |
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Answer» As 1 – p is root of x2 + px + 1 – p = 0 => (1 – p)2 + p(1 – p) + (1 – p) = 0 (1 – p) [1 – p + p + 1] = 0 => p = 1 ∴ Given equation becomes x2 + x = 0 So, Roots are x = 0, – 1 |
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| 6. |
sin500-sin700+sin100= |
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Answer» 1) Sin(A+B) = Sin(A)Cos(B)+Cos(A)Sin(B) 2) Sin(A-B) = Sin(A)Cos(B)-Cos(A)Sin(B) Now we write our expression as Sin(60-10) - Sin(60+10) +Sin(10). On applying the above formulas we get: Sin(60)Cos(10) - Cos(60)Sin(10) -{Sin(60)Cos(10) + Cos(60)Sin(10)} + Sin(10) On simplifying we get: -2*Cos(60)Sin(10) + Sin(10) Using Cos(60) = 1/2 in above expression we get: -2*(1/2)*Sin(10) + Sin(10) = 0 |
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| 7. |
the ratio of the number of side of 2 regular polygon is 1:2 and the ratio of there interior angle is 3:4 . find the number of side of polygon. |
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Answer» \(\Rightarrow\) The ratio of the sides of two polygon is 1 : 2. \(\Rightarrow\) Let the polygon A have n sides and polygon B have 2n sides. \(\Rightarrow\) The sum of the interior angles of A is (n - 2) x 180º = 180ºn - 360º \(\Rightarrow\)So each interior angle, \(\Rightarrow\)\(\frac{180^on-360^o}{n}\) ............(1) \(\Rightarrow\) Sum of interior angles of B is (2n - 2)×180º = 360ºn - 360º. \(\Rightarrow\) \(\frac{360^on-360^o}{2n}\) ....................(2) \(\Rightarrow\) Now the ratio of the interior angles of A and B. \(\Rightarrow\) \(\frac{180^on-360^o}{n}\): \(\frac{360^on-360^o}{2n}\): : \(\frac{3}{4}\) .................[from (1) and (2)] \(\Rightarrow\) \(\frac{360^on-720^o}{360^on-360^o}\) = \(\frac{3}{4}\) \(\Rightarrow\) \(\frac{360^o(n-2)}{360^o(n-1)}\) = \(\frac{3}{4}\) \(\Rightarrow\) 4n - 8 = 3n - 3 \(\therefore\) n = 5 and 2n = 10 Thus, the number of sides of each polygon is 5 and 10. |
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| 8. |
The shopping centre is now a pedestrian ______. A) arrival B) palace C) pavement D) precinct |
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Answer» Correct option is D) precinct |
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| 9. |
In olden days, the drawee bank used to send cheques to the drawer’s bank to get it cleared. Now the drawee bank truncates the cheques and send it to the drawer’s bank. What is ‘truncated cheque’ ? a) a cheque cut into 2 pieces b) a cheque in a trunk c) scan of the physical cheque d) Cheque which is tranquilized |
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Answer» Correct answer is c) scan of the physical cheque |
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| 10. |
You can hang your jacket in the ______. A) bedspread B) chest of drawers C) hanger D) wardrobe |
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Answer» Correct option is D) wardrobe |
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| 11. |
A sum of Rs. 13,725 is divided between A, B, C and D such that the ratio of the shares of A and B is 3 : 4 and B and C is 8 : 5 and that of C and D is 7 : 10. What is the share of C?1. Rs. 2,6252. Rs. 2,4753. Rs. 3,1504. Rs. 3,750 |
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Answer» Correct Answer - Option 1 : Rs. 2,625 Given: A : B = 3 : 4 B : C = 8 : 5 C : D = 7 : 10 Calculation: A : B : C = 6 : 8 : 5 A : B : C : D = 42 : 56 : 35 : 50 ∴ Share of C = 35/(42 + 56 + 35 + 50) × 13725 = Rs.2625 |
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| 12. |
A pipe can fill a pool in 3 hours, due to leakage in the bottom, it takes \(3\frac{1}{2}\) hours to fill it. In what time the leakage will empty the pool?1. 12 hours2. 21 hours3. \(6\frac{1}{2}\) hours4. \(10\frac{1}{2}\) hours |
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Answer» Correct Answer - Option 2 : 21 hours Given: A pipe can fill a pool in 3 hours. With leakage pool can be filled in \(3\frac{1}{2}\) hours Concept: If a tap can fill a tank in x hours, then the tank filled by the tap in 1 hour = 1/x of the total tank. Calculation: Work done in 1 hour by the pipe = 1/3 Work done in 1 hour by leakage and pipe = 2/7 Work done by the leakage in 1 hour = (1/3) – (2/7) ⇒ (7 – 6)/21 ⇒ 1/21 ∴ The leakage can empty the pool in 21 hours. |
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| 13. |
Two pipes can fill a tank in 8 hours and 12 hours respectively. Both the pipes are opened together but due to a leakage in the tank, it takes 24 hours to fill the tank. If the tank is full, how long will the leakage take to empty the tank?1. 5 hours2. 6 hours3. 8 hours4. 12 hours5. 16 hours |
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Answer» Correct Answer - Option 2 : 6 hours Given: Two pipes can fill a tank in 8 hours and 12 hours respectively. With the leakage, it takes 24 hours to fill the tank. Concept: If a work can be completed in x hours, work done in 1 hour is 1/x If two pipes are opened together summation of their 1-hour work is the total work for 1 hour If leakage is there, its 1-hour work is subtracted from the total 1-hour work to get the total work done in an hour. Calculation: Total 1 hour work of the pipes = (1/8) + (1/12) = (10/48) Let the leakage can empty the tank in x hours Leakage 1 hour work = 1/x (10/48) – (1/x) = 1/24 ⇒ (1/x) = (10/48) – (1/24) ⇒ (1/x) = (10 – 2)/48 ⇒ (1/x) = (8/48) ⇒ (1/x) = (1/6) ⇒ x = 6 ∴ The leakage will take 6 hours to empty the full tank |
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| 14. |
The following bar graph represents information about the migration of people from rural areas to cities during 2000-2015. Using the data prepare a report in about 120 words. |
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Answer» From Villages to Cities – In Search of Greener Pastures Ever since the world became industrialized, there has been a steady exodus of people from villages to cities. The graph is proof of this disturbing phenomenon and shows an increase in the number of people coming to urban areas from rural areas. In 1970,11 only 20% migrated from the countryside to the cities, in 1980 the percentage rose to 30. Even in the next two decades, the same increase continued and hence in 2000, the increase went up to 50%. This trend is owing to the misconception that people can make easy money in cities. It is also owing to the false notion that city life is easy and full of pleasures. Very often people go through pathetic hardships in cities. But they don’t go back to villages because either they feel ashamed to do that or they have nothing left in the village to go back to. The government should study the problem and take appropriate steps to curb the inflow of people into the cities as it is against the development of both cities and villages. |
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| 15. |
What do the underlined words in the following extract refer to?Rabindranath Tagore, in a poem who prays to the Lord not to remove all obstacles, asks for strength to bear them. Before the start of the Mahabharata war, Arjuna was seized with emotional weakness. Therefore he refused to fight the war. But Lord Krishna rescued him by giving emotional strength. i. Who: .....ii. Them: ..... iii. He: .....iv. Him: ..... |
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Answer» i. Rabindranath Tagore ii. obstacles iii. Arjuna iv. Arjuna. |
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| 16. |
A and B can finish a job in 6 and 8 days respectively. After 2 days of working together, they started to work the alternating way in which A starts first. Find the time in which total work is completed.1. 4 days2. \(4\frac{3}{4}\) days3. 5 days4. \(5\frac{3}{4}\) days5. 3 days |
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Answer» Correct Answer - Option 2 : \(4\frac{3}{4}\) days Given: A and B can finish a job in 6 and 8 days respectively They worked together for 2 days Calculation A can complete the work in 6 days B can complete the work in 8 days Let the total work be LCM of (6, 8) = 24 units Efficiency of A = 24/6 = 4 units Efficiency of B = 24/8 = 3 units Work done by A and B in 2 days = 2 × (4 + 3) = 14 units Remaining work = 24 – 14 = 10 units Now, they started working together starting with A ⇒ Work done by A on day 1 = 4 units ⇒ Work done by B on day 2 = 3 units Remaining work done by B on day = 3/4 units Total work will be completed in = 2 + 1 + 1 + (3/4) ⇒ \(4\frac{3}{4}\) days ∴ Time taken to complete total work is \(4\frac{3}{4}\) days |
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| 17. |
A, B and C contract to do a work for Rs. 6500. A can do the work in 10 days. B can do the same work in 15 days and C can do it in 20 days. Find share of B if they work together?1. Rs. 10002. Rs. 15003. Rs. 20004. Rs. 21005. Rs. 1900 |
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Answer» Correct Answer - Option 3 : Rs. 2000 Given: Contract amount for the work by A, B and C = 6500 A alone can do the work in 10 days B alone can do the work in 15 days C alone can do the work in 20 days Calculation: Work = LCM of 10, 15, 20 = 60 units Work done by A in 1 day = 60/10 = 6 units Work done by B in 1 day = 60/15 = 4 units Work done by C in 1 day = 60/20 = 3 units Share of B = (4/13) × 6500 = Rs. 2000 ∴ The share of B if they work together is Rs. 2000. |
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| 18. |
A and B can complete a piece of work in 12 days and 18 days respectively. They start working but 3 days before the completion of work, A left. In how many days will the total piece of work be completed?1. 9 days2. 6 days3. 7 days4. 5 days |
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Answer» Correct Answer - Option 1 : 9 days Given: A and B can complete a piece of work in 12 days and 18 days respectively. A left before 3 days from the completion of work. Concept used: Time = Work/Efficiency Calculation: A's 1 day work = 1/12 B's 1 day work = 1/18 (A and B)'s 1 day work = [(1/12) + (1/18)] ⇒ 5/36 As B alone worked for last 3 days, ⇒ B's 3 days work = 3/18 = 1/6 Remaining work = 1 – (1/6) = 5/6 Remaining work is done by A and B together. Time = Work/Efficiency ⇒ Time = (5/6) ÷ [(1/18) + (1/12)] ⇒ (5/6) ÷ (5/36) ⇒ 6 days. Total days taken by to complete aa piece of work = 6 days + 3 days ⇒ 9 days. ∴ A and B took 9 days to complete a piece of work. |
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| 19. |
Hitesh can complete a work alone in 24 days and Jinesh can complete the same work in 30 days alone. With the help of Ishaan, they can complete this work in 10 days. In how many days will Ishaan complete this work alone?1. 32 days2. 36 days3. 48 days4. 40 days |
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Answer» Correct Answer - Option 4 : 40 days Given: Time taken by Hitesh to complete the work = 24 days Time taken by Jinesh to complete the work = 30 days Time taken with the help of Ishaan to complete the work = 10 days Formula used: Efficiency = Total work/Time taken Calculation: LCM of (24, 30 and 10) = 120 = Total work Efficiency of Hitesh = 120/24 units/day ⇒ 5 units/day Efficiency of Jinesh = 120/30 units/day ⇒ 4 units/day Efficiency of all of them together = 120/10 units/day ⇒ 12 units/day Efficiency of Ishaan = 12 – (5 + 4) units/day ⇒ (12 – 9) units/day ⇒ 3 units/day Time taken by Ishaan = 120/3 days ⇒ 40 days ∴ Time taken by ishaan to complete the work is 40 days |
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| 20. |
Work done by A in one day is 2 times of the work done by B in one day. Work done by B is 3 times of the work done by C. If C can alone complete the work in 35 days, In how many days can A, B, and C together complete the work?1. 3 days2. 2.5 days3. 1.5 days4. 5 days5. 3.5 days |
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Answer» Correct Answer - Option 5 : 3.5 days Given: Work done by A in one day is 2 times of the work done by B in one day Work done by B is 3 times of the work done by C C can alone complete the work in 35 days Calculation: According to the question: Let work done by C be x days Then, work done by B be 3x days Work done by A be 6x days Now, The ratio of A : B : C = 6x : 3x : x Time taken by together (A + B + C) = (35 × x)/10x ⇒ 35/10 = 3.5 days. ∴ A, B, and C together complete the work in 3.5 days. |
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| 21. |
5 men and 3 boys can do work in 40 days while 6 men and 2 boys can do this work in 36 days. Then, how many much time will be taken by 4 men and 4 boys to complete work?1. 45 days2. 40 days3. 35 days4. 55 days5. 36 days |
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Answer» Correct Answer - Option 1 : 45 days Given: 5 men and 3 boys can do work in 40 days 6 men and 2 boys can do this work in 36 days Calculation: According to the question: (5M + 3B) × 40 = (6M + 2B) × 36 ⇒ 50M + 30B = 54M + 18B ⇒ 50M – 54M = 18B – 30B ⇒ 12B = 4M ⇒ 1M = 3B Now, (5M + 3B) = 5 × 3B + 3B ⇒ 15B + 3B = 18B Again, (4M + 4B) = 4 × 3B + 4B ⇒ 12B + 4B = 16B 18 boys take time 40 days. 1 boys take time = (40 × 1) days 16 boys take time = (40 × 18)/16 = 45 days ∴ 4 men and 4 boys to complete work in 45 days. |
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| 22. |
If the readings `V_(1)` and `V_(3)` are 10. volt each, then reading of `V_(2)` is: A. 0 voltB. 100 voltC. 200 voltD. cannot be determined by given information. |
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Answer» Correct Answer - C `because` L,R,C are in series So amount of current will remains same through each. Here `V_(1)=V_(3)` or `iX_(L)=iX_(C)` so `[X_(L)=X_(C)]` So circuit will behave as pure resistive ckt so `V_(2)=200V` |
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| 23. |
Statement 1: A pendulum made of an insulated rigid massless rod of length l is attached to a small sphere of mass m and charge q. The pendulum is undergoinng oscillations of small amplitude having time period T. Now a uniform horizontal magnetic field `vec B` out of plane of page is switched on. As a result of this change, the time period of oscillations does not change. Statement 2: A force acting along the string on the bob of a simple pendulum (such that tension in string is never zero) does not produce any restoring torque on the bob about the hinge.A. Statement-1 is true, Statement-2: is true, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is true, Statement-2: is true, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is true but statement-2 is falseD. Statement-1 is false, Statement-2 is true |
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Answer» Correct Answer - A The magnetic force on bob does not produced any restoring torque on bob about the hinge. Hence this force has no effect on time period of oscillation. Therefore both statements are correct and statement-2 is the correct explanation. |
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| 24. |
If A + D > C + E, C + D = 2B nd B + E > C + D, it necessarily follows that(A) A + B > 2D (B) B + D > C + E (C) A + D > B + E (D) A + D > B + C |
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Answer» (D) A + B > C + E => A + D > (2B - D) + E (∴ C + D = 2B) => A + D > (B + E) + (B - D) => A + D > (C + D) + (B - D) => A + D > B + C. |
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| 25. |
When the optical power incident on a photodiode is 10 Wµ and the responsivity is 0.8A/W, the photocurrent generated (in Aµ ) is ________. |
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Answer» Explanation: Responsivity(R) = IP/P0 0.8 = IP/(10 x 10-6) ⇒ I8 = 8μA |
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| 26. |
A real (4 × 4) matrix A satisfies the equation A2 = I, where I is the (4 × 4) identity matrix. The positive eigen value of A is __________. |
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Answer» A2 = I ⇒ A = A-1 ⇒ λ is on eigen value of A then 1/λ is also its eigen value. Since, we require positive eigen value. ∴ λ = 1 is the only possibility as no other positive number is self inversed. |
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| 27. |
A train that is 280 metres long, travelling at a uniform speed, crosses a platform in 60 seconds and passes a man standing on the platform in 20 seconds. What is the length of the platform in metres? |
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Answer» For a train to cross a person, it takes 20 seconds for its 280m. So, for second 60 seconds. Total distance travelled should be 840. Including 280 train length so length of plates = 840 - 280 = 560 |
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| 28. |
the sum of three consecutive terms in an AP is 6 & their product is -120 find the three terms |
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Answer» Let the consecutive terms in AP be a-d, a and. a+d So by 1st condition a-d+a+a+d=6 =>a=2 By 2nd condition (a--d)×a×(a+d)=-120 =>2(4-d^2)=-120 =>d^2=64 =>d=+8or-8 So three terms are -6,2,10 Or 10,2,-6 |
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| 29. |
If a : b = 3 : √5, then the value of (2a + b) : (3a - 2b) is:1. \(\frac{1}{{61}}(64 + 21\sqrt 5 )\)2. \(\frac{1}{{64}}(64 + 21\sqrt 5 )\)3. \(\frac{1}{{63}}(64 + 21\sqrt 5 )\)4. \(\frac{1}{{62}}(64 + 21\sqrt 5 )\) |
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Answer» Correct Answer - Option 1 : \(\frac{1}{{61}}(64 + 21\sqrt 5 )\) Given: a ∶ b = 3 ∶ √5 Calculation: Let a, b be 3k, k√5 ⇒ 2a + b = 6k + k√5 ⇒ 3a – 2b = 9k – 2k√5 (2a + b) ∶ (3a – 2b) ⇒ (6k + k√5) ∶ (9k – 2k√5) ⇒ (6 + √5)(9 + 2√5) ∶ (92 – (2√5)2) ⇒ (64 + 21√5)/61 ∴ (2a + b) ∶ (3a – 2b) is (64 + 21√5)/61 |
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| 30. |
∆ABC~∆PQR. If AM and PN are altitudes of ∆ABC and ∆PQR respectively and AB2 : PQ2 = 4 : 9, then AM:PN = (a) 16:81(b) 4:9(c) 3:2(d) 2:3 |
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Answer» Correct answer is: (d) 2:3 Ratio of altitudes = Ratio of sides for similar triangles So AM:PN = AB:PQ = 2:3 |
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| 31. |
Rahul can row 25 km upstream in 5 hour. If the speed of Rahul in still water is 6km/hr. Find the distance he can row downstream in 20 hour.1. 120 km2. 140 km3. 160 km4. 158 km |
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Answer» Correct Answer - Option 2 : 140 km Given Speed of Rahul in still water (u) = 6 km/hr Rahul rows 25 km upstream in 5 hours Formula used The speed of a boat in still water is u km/hr The speed of the stream is v km/hr, Speed downstream = (u + v) km/hr. Speed upstream = (u - v) km/hr. Calculation Speed of water (v) = km/hr ⇒ Speed in still water (u) = 6 km/hr ⇒ Upstream speed = 25/5 km/hr ⇒ 5 km/hr Speed of Stream = (6 – 5) km/hr ⇒ 1 km/hr Downstream speed = (6 + 1) km/hr ⇒ 7 km/hr Distance travelled in 20 hour downstream ⇒ 7 × 20 = 140 km ∴ The required distance covered is 140 km. |
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| 32. |
By selling an article for Rs. 15000 a man allows 8% discount and earns 20.48 % profit. If the article is sold without a discount. What should be the profit percentage?1. 30%2. 25%3. 20%4. 35% |
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Answer» Correct Answer - Option 1 : 30% Given: Selling price = Rs. 15600 Formula used: Profit or Loss% = [(Profit or Loss)/Cost Price] × 100 Calculation: Selling price without discount ⇒ 15000 × (108/100) ⇒ Rs. 16200 Let the CP be ‘X’ ⇒ X × (120.48/100) = 15000 ⇒ Rs. 12450 Profit without discount ⇒ 16200 – 12450 ⇒ Rs. 3750 Profit percentage without discount ⇒ (3750/12450) × 100 ⇒ 30% ∴ The profit% without discount is 30%. |
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| 33. |
A pair of dice are rolled. The probability of obtaining an even prime number on each die is(A) 1/36(B) 1/12(C) 1/6(D) 0 |
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Answer» Correct option: (A) 1/36 |
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| 34. |
A person running at a speed of 4 m/s in a circular track of radius 420 m. If he decrease his speed by 1/4 after completing every one round of track, then what is the time take by him to complete two rounds of the track?1. 1.54 hours2. 1540 seconds3. 1540 minutes4. 1.5 hours 40 minutes5. None of the above/More than one of the above |
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Answer» Correct Answer - Option 2 : 1540 seconds Given: Initial speed of person = 4 m/s Radius of track = 420 m Decrease in speed = 1/4 Calculations: Distance covered in on round = 2 × π × r ⇒ 2 × (22/7) × 420 ⇒ 2,640 m Time taken to complete first round = 2640/4 ⇒ 660 sec After 1 round speed = 4 - {4 × (1/4)} ⇒ 4 - 1 ⇒ 3 m/s Time taken to complete second round = 2640/3 ⇒ 880 sec Total time taken = 660 + 880 ⇒ 1540 sec ∴ The time take by him to complete two rounds of the track 1540 seconds |
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| 35. |
Two friends A and B, with speed in the ratio 4:1, are running on track PQ. A starts from P towards Q and when he reaches point exactly in the middle of the track, B starts running from P towards Q. A reaches Q turns back and continues towards P and meets B at a distance of 175 m from Q. What is the total length of the track? 1. 250 m2. 275 m3. 225 m4. 200 m |
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Answer» Correct Answer - Option 1 : 250 m Let 4x and x be the speeds(m/s) of A and B respectively and L be the length(m) of track. Let d be the distance travelled by B towards Q from P, when A reaches Q from the mid-point. As we know, distance covered is directly proportional to the speed if the time taken is constant. Thus, (L/2):d = 4:1 ⇒ d = L/8 Similarly, 175:(L – d – 175) = 4:1 ⇒ 175 = 4 × (7L/8 – 175) ⇒ 7L/2 = 5 × 175 ⇒ L = 10 × 25 = 250 m |
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| 36. |
CMOS devices use a. Bipolar transistors b. Complementary E-MOSFETs c. Class A operation d. DMOS devices |
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Answer» (b) Complementary E-MOSFETs |
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| 37. |
The main advantage of CMOS is its a. High power rating b. Small-signal operation c. Switching capability d. Low power consumption |
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Answer» (d) Low power consumption |
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| 38. |
The sum of either pair of the opposite angles of a cyclic quadrilateral will be(a) 360° (b) 180° (c) 90° (d) 0° |
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Answer» The sum of either pair of the opposite angles of a cyclic quadrilateral will be 180° |
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| 39. |
The VGS(on) of an n-channel E-MOSFET is a. Less than the threshold voltage b. Equal to the gate-source cutoff voltage c. Greater than VDS(on) d. Greater than VGS(th) |
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Answer» (d) Greater than VGS(th) |
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| 40. |
Which of these may appear on the data sheet of an enhancement-mode MOSFET? a. VGS(th) b. ID(on) c. VGS(on) d. All of the above |
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Answer» (d) All of the above |
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| 41. |
An n-channel E-MOSFET conducts when it has a. VGS > VP b. An n-type inversion layer c. VDS > 0 d. Depletion layers |
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Answer» (b) An n-type inversion layer |
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| 42. |
Fill in the blanks. The value of sin2 25° +cos2 25° is ........... |
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Answer» The value of sin2 25° +cos2 25° is 1. |
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| 43. |
Fill in the blanks. The mean of 3,5,6,7,9 is ............ |
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Answer» The mean of 3,5,6,7,9 is 6. |
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| 44. |
The doping concentrations on the p-side and n-side of a silicon diode are 1 x 1016cm-3 and 1 x 1017cm−3, respectively. A forward bias of 0.3 V is applied to the diode. At T = 300K, the intrinsic carrier concentration of silicon ni = 1.5 x 1010cm-3 and kT/q = 26mV. The electron concentration at the edge of the depletion region on the p-side is (A) 2.3 x 109cm-3(B) 1 x 1016cm-3(C) 1 x 1017cm-3(D) 2.25 x 106cm-3 |
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Answer» Electron concentration, n ≅ ni2/NA(eVhi/VT) = (1.5 x 1010)2/(1 x 1016)e0.3/26mv = 2.3 x 109/cm3 |
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| 45. |
find the value of \( a \) if \( a x^{2}+9 y^{2}-3 x+2 y-1=0 \) Reprents on a circle and find its radius. |
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Answer» Given equation is : ax2 + 9y2 - 3x + 2y -1 = 0 .....(1) Equation (1) is represent an equation of circle if coefficient of x2 = coefficient of y2. ∴ a = 9 then equation (1) represents an equation of circle. Then equation (1) becomes 9x2 + 9y2 - 3x + 2y -1 = 0 ⇒ x2 + y2 - \(\frac{3}{9}x\) + \(\frac{2}{9}y\) - \(\frac{1}{9}\) = 0 ⇒ x2 - \(\frac{3}{9}x\) + (\(\frac{3}{18}\))2 + y2 + \(\frac{2}{9}y\) + \((\frac{1}{9})^2\)- \(\frac{1}{9}\) - (\(\frac{3}{18}\))2 - \((\frac{1}{9})^2\) = 0 ⇒ (x - \(\frac{3}{18}\))2 + (y + \(\frac{1}{9}\))2 = \(\frac{36+9+4}{18^2}\) ⇒ (x - \(\frac{3}{18}\))2 + (y + \(\frac{1}{9}\))2 = \(\frac{49}{18^2}\) = \((\frac{7}{18})^2\) ∴ Radius of the formed circle is \(\frac{7}{18}\). |
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| 46. |
To make someone feel upset or angry is to ______. A) jump them B) get to them C) do them in |
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Answer» Correct option is B) get to them |
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| 47. |
How to encourage yourself after bad marks |
Answer»
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| 48. |
With trembling oars I turned. The figure of speech used here is(A) Synecdoche (B) Metaphor (C) Transferred Epithet (D) Simile |
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Answer» (B) Metaphor |
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| 49. |
Why KMnO4 is used in cleaning surgical instruments in hospitals ? |
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Answer» This is because KMnO4 has a germicidal action. |
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| 50. |
For which among the following reactions, change in entropy is less than zero?A. Sublimation of IodineB. Dissociation of HydrogenC. Formation of HydrogenD. Thermal decomposition of Calcium Carbonate |
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Answer» Correct Answer - C The formation of water. `H_(2)(g) + O_(2)(g) to H_(2)O(l)` `DeltaS = sumS_("Products") -sum S_("reactants")` An Entropies of reactants are more than products Hence `DeltaS` is negtive. (a) For sublimation of Iodine `I_(2)(g) to 2I_(g) DeltaS = +ve` (b) Dissociation of hydrogen `H_(2)(g) to 2H(g)` `DeltaS=+ve` (d) Thermal decomposition of `CaCO_(3)` `CaCO_(3)(s) to CaO(s) + CO_(2)(g), DeltaS= +ve` |
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