100525 + Interview Questions in GENERAL QA IN GENERAL AWARENESS Page 499 InterviewSolution

24901.	A 100 watt bulb working on 200 volt and a 200 watt bulb working on 100 volt haveA. Resistances in the ratio of 4 : 1B. Maximum current ratings in the ratio of 1 : 4C. Resistances in the ratio of 2 : 1D. Maximum current ratings in the ratio of 1 : 2
Answer» Correct Answer - B By definition

Discussion

24902.	In the following question, two statements are numbered as I and II. On solving these statements, we get quantities I and II respectively. Solve both quantities and choose the correct option.Statement I . The price of sugar gets increased by 40% and as a result Ram gets 10 kg less sugar. What was the previous consumption?Statement II. The price of rice increased by 50% and thereby 20% increase in the Ram’s expenditure by consuming 12 kg less rice than previous consumption. What was the previous consumption?1.Quantity I ≥ Quantity II2.Quantity I ≤ Quantity II3.Quantity I < Quantity II4.Quantity I > Quantity II5.Quantity I = Quantity II or No relation
Answer» Correct Answer - Option 3 : Quantity I < Quantity II Quantity I. Let the price of sugar be Rs. 100 per Kg Let the amount of sugar consumed previously be x kg Total expenditure on sugar previously = 100 × x = Rs. 100x Price increases by 40% New price = Rs 140% of 100 = Rs 140 Consumption of sugar decreases by 10 Kg New consumption = (x – 10) Kg Total expenditure = 140 × (x – 10) = Rs. (140x – 1400) Expenditure remains same ⇒ 140x – 1400 = 100x ⇒ 140x – 100x = 1400 ⇒ 40x = 1400 ⇒ x = 35 Previous consumption = 35 Kg Quantity II. Let the price of sugar be Rs 100 per Kg Let the amount of sugar consumed previously be x Kg Total expenditure = 100 × x = Rs. 100x kg Price increases by 50% New price = 150% of 100 = Rs. 150 Consumption decreases by 12 Kg New consumption = (x – 12) Kg Total expenditure on sugar = 150 × (x – 12) = Rs. (150x – 1800) Expenditure increases by 20% New expenditure on sugar = 120% of 100x = Rs. 120x 120x = 150x – 1800 ⇒ 150x – 120x = 1800 ⇒ 30x = 1800 ⇒ x = 60 Previous consumption = 60 Kg ∴ Quantity II > Quantity I

24902.

In the following question, two statements are numbered as I and II. On solving these statements, we get quantities I and II respectively. Solve both quantities and choose the correct option.Statement I . The price of sugar gets increased by 40% and as a result Ram gets 10 kg less sugar. What was the previous consumption?Statement II. The price of rice increased by 50% and thereby 20% increase in the Ram’s expenditure by consuming 12 kg less rice than previous consumption. What was the previous consumption?1.Quantity I ≥ Quantity II2.Quantity I ≤ Quantity II3.Quantity I < Quantity II4.Quantity I > Quantity II5.Quantity I = Quantity II or No relation

Answer» Correct Answer - Option 3 :

Quantity I < Quantity II

Quantity I.

Let the price of sugar be Rs. 100 per Kg

Let the amount of sugar consumed previously be x kg

Total expenditure on sugar previously = 100 × x = Rs. 100x

Price increases by 40%

New price = Rs 140% of 100 = Rs 140

Consumption of sugar decreases by 10 Kg

New consumption = (x – 10) Kg

Total expenditure = 140 × (x – 10) = Rs. (140x – 1400)

Expenditure remains same

⇒ 140x – 1400 = 100x

⇒ 140x – 100x = 1400

⇒ 40x = 1400

⇒ x = 35

Previous consumption = 35 Kg

Quantity II.

Let the price of sugar be Rs 100 per Kg

Let the amount of sugar consumed previously be x Kg

Total expenditure = 100 × x = Rs. 100x kg

Price increases by 50%

New price = 150% of 100 = Rs. 150

Consumption decreases by 12 Kg

New consumption = (x – 12) Kg

Total expenditure on sugar = 150 × (x – 12) = Rs. (150x – 1800)

Expenditure increases by 20%

New expenditure on sugar = 120% of 100x = Rs. 120x

120x = 150x – 1800

⇒ 150x – 120x = 1800

⇒ 30x = 1800

⇒ x = 60

Previous consumption = 60 Kg

∴ Quantity II > Quantity I

Discussion

24903.	The moment of inertia of a squaer lamin about the perpendicluar axis though its centre of mass is ` 20 kg- m^(2)` . Then its moment of inertia about an axis toching its side and in the plane of the lamina will be : -A. `10 kg - m^(2)`B. `30 kg- m^(2)`C. `40 k g- m^(2)`D. `25 kg m^(2)`
Answer» Correct Answer - C `(Ml^2)/(6)=20 rArr Ml^2 = 120 ` `I= (Ml^2)/(12)+(Ml^2)/(4)=(Ml^2)/(3)=40 kg m^2 `

Discussion

24904.	Two pipes can fill a tank in 15 hours and 4 hours, respectively, while a third pipe can empty it in 12 hours. How long (in hours) will it take to fill the empty tank if all the three pipes are opened simultaneously?1. 30 / 72. 20 / 73. 50 / 74. 15 / 7
Answer» Correct Answer - Option 1 : 30 / 7 Given: Time to fill the tank by two pipes = 15 hours and 4 hours Time to empty the tank by third pipe = 12 hours Concept used: If Pipe A take ‘x’ hours and Pipe B takes ‘y’ hours to fill a tank then assume total capacity of the tank equal to LCM of ‘x’ and ‘y’ or a multiple of them. Formula used: Efficiency = Total work/total time Calculations: Let total capacity of tank be LCM of 15, 4 and 12 i.e. 60 units. Efficiency of first pipe = 60/15 ⇒ 4 units/hour Efficiency of second pipe = 60/4 ⇒ 15 units/hour Efficiency of third pipe = 60/12 ⇒ 5 units/hour Combined efficiency of three pipes = 4 + 15 - 5 ⇒ 14 units/hour Time to empty the tank by all the pipes together = 60/14 ⇒ 30/7 hours ∴ It takes 30/7 hours to fill the empty tank if all the three pipes are opened simultaneously.

24904.

Two pipes can fill a tank in 15 hours and 4 hours, respectively, while a third pipe can empty it in 12 hours. How long (in hours) will it take to fill the empty tank if all the three pipes are opened simultaneously?1. 30 / 72. 20 / 73. 50 / 74. 15 / 7

Answer» Correct Answer - Option 1 : 30 / 7

Given:

Time to fill the tank by two pipes = 15 hours and 4 hours

Time to empty the tank by third pipe = 12 hours

Concept used:

If Pipe A take ‘x’ hours and Pipe B takes ‘y’ hours to fill a tank then assume total capacity of the tank equal to LCM of ‘x’ and ‘y’ or a multiple of them.

Formula used:

Efficiency = Total work/total time

Calculations:

Let total capacity of tank be LCM of 15, 4 and 12 i.e. 60 units.

Efficiency of first pipe = 60/15

⇒ 4 units/hour

Efficiency of second pipe = 60/4

⇒ 15 units/hour

Efficiency of third pipe = 60/12

⇒ 5 units/hour

Combined efficiency of three pipes = 4 + 15 - 5

⇒ 14 units/hour

Time to empty the tank by all the pipes together = 60/14

⇒ 30/7 hours

∴ It takes 30/7 hours to fill the empty tank if all the three pipes are opened simultaneously.

Discussion

24905.	Each question below is followed by two statements I and II. You have to determine whether the data given in the statements are sufficient for answering the question. You should use the data and your knowledge of Mathematics to choose the best possible answer.What is the ratio of coconut oil and milk in the final beaker? If contents from four vessels poured in it.I. Vessel B has 10 ml more capacity than vessel A and the ratio of coconut oil and milk in vessel B is 2 ∶ 7. Vessel C has coconut oil and milk in the ratio 2 ∶ 3 and contains 38 ml more capacity than Vessel DII. Vessel A has milk and coconut oil in the ratio 3 ∶ 5. Vessel C has 12 ml more coconut oil than vessel D.1. If the data in statement I alone is sufficient to answer the question, while the data in statement II alone are not sufficient to answer the question.2.If the data in statement II alone is sufficient to answer the question, while the data in statement I alone are not sufficient to answer the question.3.If the data either in statement I alone or in statement II alone is sufficient to answer the question.4. If the data even in both the statements I and II together are not sufficient to answer the question.5. If the data in both statements I and II together are needed to answer the question.
Answer» Correct Answer - Option 4 : If the data even in both the statements I and II together are not sufficient to answer the question. Using statement I Vessel B Ratio of coconut oil and milk = 2 ∶ 7 ⇒ Coconut oil in B = 2x ⇒ Milk in B = 7x Total quantity = 2x + 7x = 9x ∴ Capacity of vessel A = 9x – 10 Vessel C Ratio of coconut oil and milk = 2 ∶ 3 ⇒ Coconut oil in C = 2y ⇒ Milk in C = 3y Total capacity = 5y Capacity of vessel D = 5y – 38 Statement I alone is not sufficient to answer the question Using statement II Vessel A has milk and coconut oil in the ratio 3 ∶ 5 ∴ Coconut oil in vessel A = 5z Milk in vessel A = 3z Total capacity = 8z Vessel C has 12 ml more coconut oil than vessel D Statement II alone is not sufficient to answer the question Using statement I and statement II together, Capacity of vessel A = 9x – 10 ⇒ 8z = 9x – 10 ⇒ 9x – 8z = 10 Coconut oil in vessel C = 2y + 12 No data can be determined ∴ The data even in both the statements I and II together are not sufficient to answer the question.

24905.

Each question below is followed by two statements I and II. You have to determine whether the data given in the statements are sufficient for answering the question. You should use the data and your knowledge of Mathematics to choose the best possible answer.What is the ratio of coconut oil and milk in the final beaker? If contents from four vessels poured in it.I. Vessel B has 10 ml more capacity than vessel A and the ratio of coconut oil and milk in vessel B is 2 ∶ 7. Vessel C has coconut oil and milk in the ratio 2 ∶ 3 and contains 38 ml more capacity than Vessel DII. Vessel A has milk and coconut oil in the ratio 3 ∶ 5. Vessel C has 12 ml more coconut oil than vessel D.1. If the data in statement I alone is sufficient to answer the question, while the data in statement II alone are not sufficient to answer the question.2.If the data in statement II alone is sufficient to answer the question, while the data in statement I alone are not sufficient to answer the question.3.If the data either in statement I alone or in statement II alone is sufficient to answer the question.4. If the data even in both the statements I and II together are not sufficient to answer the question.5. If the data in both statements I and II together are needed to answer the question.

Answer» Correct Answer - Option 4 : If the data even in both the statements I and II together are not sufficient to answer the question.

Using statement I

Vessel B

Ratio of coconut oil and milk = 2 ∶ 7

⇒ Coconut oil in B = 2x

⇒ Milk in B = 7x

Total quantity = 2x + 7x = 9x

∴ Capacity of vessel A = 9x – 10

Vessel C

Ratio of coconut oil and milk = 2 ∶ 3

⇒ Coconut oil in C = 2y

⇒ Milk in C = 3y

Total capacity = 5y

Capacity of vessel D = 5y – 38

Statement I alone is not sufficient to answer the question

Using statement II

Vessel A has milk and coconut oil in the ratio 3 ∶ 5

∴ Coconut oil in vessel A = 5z

Milk in vessel A = 3z

Total capacity = 8z

Vessel C has 12 ml more coconut oil than vessel D

Statement II alone is not sufficient to answer the question

Using statement I and statement II together,

Capacity of vessel A = 9x – 10

⇒ 8z = 9x – 10

⇒ 9x – 8z = 10

Coconut oil in vessel C = 2y + 12

No data can be determined

∴ The data even in both the statements I and II together are not sufficient to answer the question.

Discussion

24906.	If three pipes P, Q, and R together can empty full tank in 25 minutes, if P is twice as efficient as Q and Q is twice as efficient as R then in how much time R will empty 60% of the tank?1. 90 minutes2. 105 minutes3. 95 minutes4. 114 minutes
Answer» Correct Answer - Option 2 : 105 minutes Given: P, Q, and R can empty a tank in 25 minutes P is twice as efficient as Q Q is twice as efficient as R Concept used The efficiency of people/taps is equal to what they can do/destroy or fill/empty the given task per hour/minute/day. Calculation: Let tank emptied by R in a minute be x units, Q be 2x units and P be 4x units Tank emptied by R in 25 minutes = 25x units Tank emptied by Q in 25 minutes = 2x × 25 = 50x units Tank emptied by P in 25 minutes = 4x × 25 = 100x units Total tank emptied by them together in 25 minutes = 25x + 50x + 100x = 175x Time taken to empty the tank by R = (Total units)/(Tank emptied in 1 minute) R will empty 60% of 175x = 105x in 105x/x ⇒ 105 mins ∴ R will empty 60% of the tank in 105 min

24906.

If three pipes P, Q, and R together can empty full tank in 25 minutes, if P is twice as efficient as Q and Q is twice as efficient as R then in how much time R will empty 60% of the tank?1. 90 minutes2. 105 minutes3. 95 minutes4. 114 minutes

Answer» Correct Answer - Option 2 : 105 minutes

Given:

P, Q, and R can empty a tank in 25 minutes

P is twice as efficient as Q

Q is twice as efficient as R

Concept used

The efficiency of people/taps is equal to what they can do/destroy or fill/empty the given task per hour/minute/day.

Calculation:

Let tank emptied by R in a minute be x units, Q be 2x units and P be 4x units

Tank emptied by R in 25 minutes = 25x units

Tank emptied by Q in 25 minutes = 2x × 25 = 50x units

Tank emptied by P in 25 minutes = 4x × 25 = 100x units

Total tank emptied by them together in 25 minutes = 25x + 50x + 100x = 175x

Time taken to empty the tank by R = (Total units)/(Tank emptied in 1 minute)

R will empty 60% of 175x = 105x in 105x/x

⇒ 105 mins

∴ R will empty 60% of the tank in 105 min

Discussion

24907.	The dimensional formula for magnetic flux isA. `ML^(2)T^(-1)C^(-1)`B. `ML^(3)T^(-2)C^(-1)`C. `ML^(2)T^(2)C^(-1)`D. `ML^(2)T^(2)C^(-1)`
Answer» [Magnetic flux `=[ML^(2)T^(-2)A^(-1)]` `=[ML^(2)^(-1)C^(-1)]`

Discussion

24908.	Using mass `(M)` , length `(L)` , time `(T)` , and electric current `(A)` as fundamental quantities , the dimensions of permitivity will beA. `[M^(-1)LT^(-2)A]`B. `[ML^(-2)T^(-2)A^(-1)]`C. `[MLT^(-2)A^(-2)]`D. `[MLT^(-1)A^(-1)]`
Answer» Units of permeability are equivalent to `N/(Amp)^(2)` thus dimensions are `=(M^(1)L^(1)T^(-2))/(A^(2))=M^(1)L^(1)T^(-2)A^(-2)`

Discussion

24909.	If the force is given by `F=at+bt^(2)` with `t` is time. The dimensions of `a` and `b` areA. `[M L T^(-3)]` and `[M L T^(-4)]`B. `[M L T^(-4)]` and `[M L T^(-3)]`C. `[ M L T^(-1)]` and `[M L T^(-2)]`D. `[ M L T^(-2)]` and `[M L T^(0)]`
Answer» `F=at+bt^(2)` dimensions of at and `bt^(2)` are same as that of F `at=FrArra[T]=M^(1)L^(1)T^(-2)` `a=M^(1)L^(1)T^(-3)` `bt^(3)=FrArrb[T]^(2)=M^(1)L^(1)T^(-2)` `b=M^(1)L^(1)T^(-4)`

Discussion

24910.	Find the dimensions of a in the formula `(p+a/V^2)(V-b)=RT`
Answer» `because [(a)/(V^(2))]=[P] therefore [a]=[P][V^(2)]=ML^(-1)T^(-2)xxL^(6)=M^(1)L^(5)T^(-2)`

Discussion

24911.	Are there any physical quantities out of the following which have the same dimension? If yes, identify them. Impulse, Torque, Angular momentum, Energy, Force Moment of inertia
Answer» Yes, torque and energy have same diemensions

Discussion

24912.	Directions: Given below are two quantities named I and II. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose among the possible answers.Quantity I: Rs. 7, 000 is divided between X and Y in the ratio 2 : 3.What is the difference between thrice the share of X and the share of Y?Quantity II: A sum of money Rs. 1, 25, 000 is divided among 4 persons P, Q, R and S. It is divided in such a way that P’s share is (1 / 3)rd of Q's share, R's share is Rs. 25000 and S’s share is equals to P’s share. Find the share of Q.1. Quantity I ≤ Quantity II2. Quantity I ˃ Quantity II3. Quantity I ≥ Quantity II4. Quantity I ˂ Quantity II5. Quantity I = Quantity II
Answer» Correct Answer - Option 4 : Quantity I ˂ Quantity II Quantity I: X’s share = Rs. [7000 × (2/5)] ⇒ Rs. 2800 Y’s share = Rs. [7000 × (3/5)] ⇒ Rs. 4200 Thrice the share of X = Rs. 8400 ⇒ Difference between thrice the share of X and the share of Y = Rs. 8400 - Rs. 4200 = Rs. 4200 Quantity II: Let the Q's share be Rs. x P's share + Q's share + R's share + S's share = 1, 25, 000 ⇒ (x/3) + x + 25, 000 + [x/3] = 1, 25, 000 ⇒ (5x/3) + 25000 = 1, 25, 000 ⇒ 5x/3 = 1, 25, 000 - 25000 ⇒ 5x = 3, 00, 000 ⇒ x = 60, 000 ∴ Quantity I < Quantity II

24912.

Directions: Given below are two quantities named I and II. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose among the possible answers.Quantity I: Rs. 7, 000 is divided between X and Y in the ratio 2 : 3.What is the difference between thrice the share of X and the share of Y?Quantity II: A sum of money Rs. 1, 25, 000 is divided among 4 persons P, Q, R and S. It is divided in such a way that P’s share is (1 / 3)rd of Q's share, R's share is Rs. 25000 and S’s share is equals to P’s share. Find the share of Q.1. Quantity I ≤ Quantity II2. Quantity I ˃ Quantity II3. Quantity I ≥ Quantity II4. Quantity I ˂ Quantity II5. Quantity I = Quantity II

Answer» Correct Answer - Option 4 : Quantity I ˂ Quantity II

Quantity I:

X’s share = Rs. [7000 × (2/5)] ⇒ Rs. 2800

Y’s share = Rs. [7000 × (3/5)] ⇒ Rs. 4200

Thrice the share of X = Rs. 8400

⇒ Difference between thrice the share of X and the share of Y = Rs. 8400 - Rs. 4200

= Rs. 4200

Quantity II:

Let the Q's share be Rs. x

P's share + Q's share + R's share + S's share = 1, 25, 000

⇒ (x/3) + x + 25, 000 + [x/3] = 1, 25, 000

⇒ (5x/3) + 25000 = 1, 25, 000

⇒ 5x/3 = 1, 25, 000 - 25000

⇒ 5x = 3, 00, 000

⇒ x = 60, 000

∴ Quantity I < Quantity II

Discussion

24913.	Directions: Given below are two quantities named I and II. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose among the possible answers.Quantity I: Cost price of an article is 55% of the selling price and the selling price is Rs. 75000. Find his profit percent.Quantity II: A chair is sold at a profit of Rs. 600, which is 50% of its cost price. If its C.P. is increased by 25% and the selling price is Rs. 1800. Then, find the new profit percent.1. Quantity I = Quantity II2. Quantity I ≥ Quantity II3. Quantity I ≤ Quantity II4. Quantity I ˂ Quantity II5. Quantity I ˃ Quantity II
Answer» Correct Answer - Option 5 : Quantity I ˃ Quantity II Quantity I: C.P. = 55% of selling price = 55% × 75000 = Rs. 41250 ⇒ Profit = 75000 - 41250 = Rs. 33750 Profit percent = (33750/41250) × 100 = 81.81% Quantity II: Let the C.P. be Rs. x ⇒ 50% of x = 600 ⇒ x = (600 × 100)/50 ⇒ x = Rs. 1200 Revised cost price = 1200 + 25% of 1200 ⇒ 1200 + 300 = Rs. 1500 ⇒ Profit = Rs. 1800 - Rs. 1500 = Rs. 300 Profit percentage = (300/1500) × 100 ⇒ 20% ∴ Quantity I > Quantity II

24913.

Directions: Given below are two quantities named I and II. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose among the possible answers.Quantity I: Cost price of an article is 55% of the selling price and the selling price is Rs. 75000. Find his profit percent.Quantity II: A chair is sold at a profit of Rs. 600, which is 50% of its cost price. If its C.P. is increased by 25% and the selling price is Rs. 1800. Then, find the new profit percent.1. Quantity I = Quantity II2. Quantity I ≥ Quantity II3. Quantity I ≤ Quantity II4. Quantity I ˂ Quantity II5. Quantity I ˃ Quantity II

Answer» Correct Answer - Option 5 : Quantity I ˃ Quantity II

Quantity I:

C.P. = 55% of selling price

= 55% × 75000 = Rs. 41250

⇒ Profit = 75000 - 41250 = Rs. 33750

Profit percent = (33750/41250) × 100 = 81.81%

Quantity II:

Let the C.P. be Rs. x

⇒ 50% of x = 600

⇒ x = (600 × 100)/50

⇒ x = Rs. 1200

Revised cost price = 1200 + 25% of 1200

⇒ 1200 + 300 = Rs. 1500

⇒ Profit = Rs. 1800 - Rs. 1500 = Rs. 300

Profit percentage = (300/1500) × 100

⇒ 20%

∴ Quantity I > Quantity II

Discussion

24914.	Directions: Given below are two quantities named I and II. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose among the possible answers.Quantity I: A bag contains 7 red, 4 black, 4 yellow, and 5 pink pens. If 4 pens are drawn at random, find the probability that 2 pens are red and 2 pens are black.Quantity II: Find the probability such that 2 jacks are drawn when 5 cards are drawn at random from a pack of 52 cards.1. Quantity I ≤ Quantity II2. Quantity I ≥ Quantity II3. Quantity I ˃ Quantity II4. Quantity I ˂ Quantity II5. Quantity I = Quantity II
Answer» Correct Answer - Option 4 : Quantity I ˂ Quantity II Quantity I: Total number of outcomes = ²⁰C₄ = 20 ! / (4! × 16 !) = (20 × 19 × 18 × 17 × 16!) / (4 × 3 × 2 × 1 × 16!) = 4845 Number of favourable outcomes = ⁷C₂ × ⁴C₂ = 126 ⇒ Required probability = 126 / 4845 ⇒ 0.02600 (approx) Quantity II: Total number of possible ways to drawn a card = ⁵²C₅ = 52! / (5! × 47!) = (52 × 51 × 50 × 49 × 48 × 47!) / (5 × 4 × 3 × 2 × 1 × 47!) = 2598960 Number of sets of 5 with 2 are jack = ⁴C₂× ⁴⁸C₃ = 6 × (48 × 47 × 46 × 45!) / (3! × 45!) = 103776 ⇒ P (E) = 103776 / 2598960 = 0.03992 (approx) ∴ Quantity I ˂ Quantity II

24914.

Directions: Given below are two quantities named I and II. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose among the possible answers.Quantity I: A bag contains 7 red, 4 black, 4 yellow, and 5 pink pens. If 4 pens are drawn at random, find the probability that 2 pens are red and 2 pens are black.Quantity II: Find the probability such that 2 jacks are drawn when 5 cards are drawn at random from a pack of 52 cards.1. Quantity I ≤ Quantity II2. Quantity I ≥ Quantity II3. Quantity I ˃ Quantity II4. Quantity I ˂ Quantity II5. Quantity I = Quantity II

Answer» Correct Answer - Option 4 : Quantity I ˂ Quantity II

Quantity I:

Total number of outcomes = ²⁰C₄

= 20 ! / (4! × 16 !)

= (20 × 19 × 18 × 17 × 16!) / (4 × 3 × 2 × 1 × 16!)

= 4845

Number of favourable outcomes = ⁷C₂ × ⁴C₂

= 126

⇒ Required probability = 126 / 4845

⇒ 0.02600 (approx)

Quantity II:

Total number of possible ways to drawn a card = ⁵²C₅

= 52! / (5! × 47!)

= (52 × 51 × 50 × 49 × 48 × 47!) / (5 × 4 × 3 × 2 × 1 × 47!)

= 2598960

Number of sets of 5 with 2 are jack = ⁴C₂× ⁴⁸C₃

= 6 × (48 × 47 × 46 × 45!) / (3! × 45!) = 103776 ⇒ P (E) = 103776 / 2598960

= 0.03992 (approx)

∴ Quantity I ˂ Quantity II

Discussion

24915.	The dimension of torque is:A. `[ML^(3)L^(-3)]`B. `[ML^(-1)T^(-1)]`C. `[ML(2)T^(-2)]`D. `[ML^(-2)]`
Answer» Torque `vectau=vecrxxvecF`

Discussion

24916.	Directions: Given below are two quantities named I and II. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose among the possible answers.Quantity I: Average of p, q, r, and s is 24.q is twice of r, p is 14 more than r and s is 18. Find the sum of p and rQuantity II: Sum of b, c, and d is 57.b is twice of c and d is 21. Find the sum of c and b.1. Quantity I ≥ Quantity II2. Quantity I = Quantity II3. Quantity I ≤ Quantity II4. Quantity I ˃ Quantity II5. Quantity I ˂ Quantity II
Answer» Correct Answer - Option 4 : Quantity I ˃ Quantity II Quantity I: S = 18, r = r, p = r + 14 and q = 2r (p + q + r + s)/4 = 24 ⇒ p + q + r + s = 96 ⇒ [14 + r + 2r + r + 18] = 96 ⇒ 4r + 32 = 96 ⇒ 4r = 96 - 32 ⇒ 4r = 64 ⇒ r = 16 ⇒ p = 16 + 14 ⇒ p = 30 Sum of p and r = 30 + 16 ⇒ 46 Quantity II: Sum of b, c and d = 57 b = 2c, d = 21 ⇒ b + c + d = 57 ⇒ 2c + c + 21 = 57 ⇒ 3c = 36 ⇒ c = 12 ⇒ b = 24 Sum of b and c is 36 (i.e. 24 + 12) ∴ Quantity I ˃ Quantity II

24916.

Directions: Given below are two quantities named I and II. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose among the possible answers.Quantity I: Average of p, q, r, and s is 24.q is twice of r, p is 14 more than r and s is 18. Find the sum of p and rQuantity II: Sum of b, c, and d is 57.b is twice of c and d is 21. Find the sum of c and b.1. Quantity I ≥ Quantity II2. Quantity I = Quantity II3. Quantity I ≤ Quantity II4. Quantity I ˃ Quantity II5. Quantity I ˂ Quantity II

Answer» Correct Answer - Option 4 : Quantity I ˃ Quantity II

Quantity I:

S = 18, r = r, p = r + 14 and q = 2r

(p + q + r + s)/4 = 24

⇒ p + q + r + s = 96

⇒ [14 + r + 2r + r + 18] = 96

⇒ 4r + 32 = 96

⇒ 4r = 96 - 32

⇒ 4r = 64

⇒ r = 16

⇒ p = 16 + 14

⇒ p = 30

Sum of p and r = 30 + 16

⇒ 46

Quantity II:

Sum of b, c and d = 57

b = 2c, d = 21

⇒ b + c + d = 57

⇒ 2c + c + 21 = 57

⇒ 3c = 36

⇒ c = 12

⇒ b = 24

Sum of b and c is 36 (i.e. 24 + 12)

∴ Quantity I ˃ Quantity II

Discussion

24917.	Which of the following represents the dimensions of FaradA. `M^(-1)L^(-2)T^(4)A^(2)`B. `ML^(2)T^(-2)A^(-2)`C. `ML^(2)T^(-2)A^(-1)`D. `MT^(-2)A^(-1)`
Answer» Frad is the unit of capacity. [Farad]=[capacity]=`[M^(-1)L^(-2)T^(4)A^(2)]`

Discussion

24918.	The amount of time that B takes to fill the tank, pipe A fills it in its one fourth time. Pipe C takes 3 times the time taken by pipe A. If all three pipes are opened simultaneously, it takes 33 hours to fill the tank. If pipe C is not switched on, how many hours will it take to fill the tank?1. 42.42. 41.53. 42.14. 41.8
Answer» Correct Answer - Option 4 : 41.8 Calculations : The amount of time that B takes to fill the tank, pipe A fills it in its one fourth time Pipe C takes 3 times as much as time taken by A Time for completing the work is 33 hours Concept used : Time taken by a person to complete a work is inversely proportional to his efficiency Person's per day work = Total work/Number of days taken by him to complete the work Calculations : Ratio of time for A to B is 1 : 4 Ratio of time taken for A to C is 1 : 3 Ratio of time for A, B and C will be 1 : 4 : 3 So ratio of the efficiencies for A, B and C will be A : B : C = 1 : (1/4) : (1/3) ⇒ 12 : 3 : 4 Let the work done by A, B and C per hour be 12x, 3x and 4x units respectively A, B and C per hour work = 12x + 3x + 4x = 19x units Total work = 19x × 33 Total work = 627x units Time taken by A and B to complete the work = 627x/15x ⇒ 41.8 hours ∴ Total time taken by a and B will be 41.8 hours

24918.

The amount of time that B takes to fill the tank, pipe A fills it in its one fourth time. Pipe C takes 3 times the time taken by pipe A. If all three pipes are opened simultaneously, it takes 33 hours to fill the tank. If pipe C is not switched on, how many hours will it take to fill the tank?1. 42.42. 41.53. 42.14. 41.8

Answer» Correct Answer - Option 4 : 41.8

Calculations :

The amount of time that B takes to fill the tank, pipe A fills it in its one fourth time

Pipe C takes 3 times as much as time taken by A

Time for completing the work is 33 hours

Concept used :

Time taken by a person to complete a work is inversely proportional to his efficiency

Person's per day work = Total work/Number of days taken by him to complete the work

Calculations :

Ratio of time for A to B is 1 : 4

Ratio of time taken for A to C is 1 : 3

Ratio of time for A, B and C will be 1 : 4 : 3

So ratio of the efficiencies for A, B and C will be

A : B : C = 1 : (1/4) : (1/3)

⇒ 12 : 3 : 4

Let the work done by A, B and C per hour be 12x, 3x and 4x units respectively

A, B and C per hour work = 12x + 3x + 4x = 19x units

Total work = 19x × 33

Total work = 627x units

Time taken by A and B to complete the work = 627x/15x

⇒ 41.8 hours

∴ Total time taken by a and B will be 41.8 hours

Discussion

24919.	Directions: Given below are two quantities named I and II. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose among the possible answers.There are 35 cards in a box numbered from 1 to 35. One number is written on each card. A card is drawn from the box.Quantity I: Find the probability that the drawn card is a multiple of 2, 3 and 4.Quantity II: Find the probability that the drawn card is an odd prime number.1. Quantity I ˂ Quantity II2. Quantity I ≥ Quantity II3. Quantity I ≤ Quantity II4. Quantity I = Quantity II5. Quantity I ˃ Quantity II
Answer» Correct Answer - Option 1 : Quantity I ˂ Quantity II Quantity I: Total number of cards = 35 Number of favourable cards = 2 (i.e. 12, 24) ⇒ P (E) = 2/35 = 0.057 Quantity II: Total number of cards: 35 Number of favourable cards = 10 (i.e. 3, 5, 7, 11, 13, 17, 19, 23, 29, 31) ⇒ P (E) = 10/35 = 2/7 = 0.285 ∴ Quantity I ˂ Quantity II

24919.

Directions: Given below are two quantities named I and II. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose among the possible answers.There are 35 cards in a box numbered from 1 to 35. One number is written on each card. A card is drawn from the box.Quantity I: Find the probability that the drawn card is a multiple of 2, 3 and 4.Quantity II: Find the probability that the drawn card is an odd prime number.1. Quantity I ˂ Quantity II2. Quantity I ≥ Quantity II3. Quantity I ≤ Quantity II4. Quantity I = Quantity II5. Quantity I ˃ Quantity II

Answer» Correct Answer - Option 1 : Quantity I ˂ Quantity II

Quantity I:

Total number of cards = 35

Number of favourable cards = 2 (i.e. 12, 24)

⇒ P (E) = 2/35 = 0.057

Quantity II:

Total number of cards: 35

Number of favourable cards = 10 (i.e. 3, 5, 7, 11, 13, 17, 19, 23, 29, 31)

⇒ P (E) = 10/35 = 2/7 = 0.285

∴ Quantity I ˂ Quantity II

Discussion

24920.	Directions: Given below are two quantities named I and II. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose among the possible answers.a, b, c, and d are four positive numbers. The average of a, b and c is 15, and the average of b, c, and d is 12. If a is 13 and c is 18.Quantity I: Find the sum of b and d.Quantity II: Find the sum of a and c.1. Quantity I = Quantity II2. Quantity I ≥ Quantity II3. Quantity I ≤ Quantity II4. Quantity I ˃ Quantity II5. Quantity I ˂ Quantity II
Answer» Correct Answer - Option 5 : Quantity I ˂ Quantity II (a + b + c)/3 = 15 ⇒ 13 + b + 18 = 45 ⇒ b = 45 - 31 = 14 (b + c + d)/3 = 12 ⇒ 14 + 18 + d = 36 ⇒ d = 36 - 32 ⇒ d = 4 Quantity I: (b + d) = 14 + 4 ⇒ 18 Quantity II: (a + c) = 13 + 18 ⇒ 31 ∴ Quantity I ˂ Quantity II.

24920.

Directions: Given below are two quantities named I and II. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose among the possible answers.a, b, c, and d are four positive numbers. The average of a, b and c is 15, and the average of b, c, and d is 12. If a is 13 and c is 18.Quantity I: Find the sum of b and d.Quantity II: Find the sum of a and c.1. Quantity I = Quantity II2. Quantity I ≥ Quantity II3. Quantity I ≤ Quantity II4. Quantity I ˃ Quantity II5. Quantity I ˂ Quantity II

Answer» Correct Answer - Option 5 : Quantity I ˂ Quantity II

(a + b + c)/3 = 15

⇒ 13 + b + 18 = 45

⇒ b = 45 - 31 = 14

(b + c + d)/3 = 12

⇒ 14 + 18 + d = 36 ⇒ d = 36 - 32

⇒ d = 4

Quantity I: (b + d) = 14 + 4

⇒ 18

Quantity II: (a + c) = 13 + 18

⇒ 31

∴ Quantity I ˂ Quantity II.

Discussion

24921.	Directions: Given below are two quantities named I and II. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose among the possible answers.a = -3, b = -1 and c = 6Quantity I: Find the value of (a3 + b2 + c2) / 4Quantity II: Find the value of [a / (a3 + b2)]1. Quantity I = Quantity II2. Quantity I ≥ Quantity II3. Quantity I ≤ Quantity II4. Quantity I ˂ Quantity II5. Quantity I ˃ Quantity II
Answer» Correct Answer - Option 5 : Quantity I ˃ Quantity II Quantity I: (a³ + b² + c²) = [(-3)³ + (-1)² + (6)²] / 4 ⇒ 2.5 Quantity II: [a / (a³ + b²)] = [-3 / (-3³ + (-1)²)] ⇒ 0.115 ∴ Quantity I ˃ Quantity II

24921.

Directions: Given below are two quantities named I and II. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose among the possible answers.a = -3, b = -1 and c = 6Quantity I: Find the value of (a3 + b2 + c2) / 4Quantity II: Find the value of [a / (a3 + b2)]1. Quantity I = Quantity II2. Quantity I ≥ Quantity II3. Quantity I ≤ Quantity II4. Quantity I ˂ Quantity II5. Quantity I ˃ Quantity II

Answer» Correct Answer - Option 5 : Quantity I ˃ Quantity II

Quantity I:

(a³ + b² + c²) = [(-3)³ + (-1)² + (6)²] / 4

⇒ 2.5

Quantity II:

[a / (a³ + b²)] = [-3 / (-3³ + (-1)²)]

⇒ 0.115

∴ Quantity I ˃ Quantity II

Discussion

24922.	When x is added each of 10, 23, 15 and 33, the sums obtained in this order are in proportion. The value of (x + 5) is:1. 12. 83. 34. 5
Answer» Correct Answer - Option 2 : 8 Given: The numbers are 10, 23, 15, and 33 x is added to each number and the sum is in proportion. Concept Used: If a, b, c, and d are in proportion, then, a/b = c/d Calculation: When x is added to each number the numbers will be, (10 + x), (23 + x), (15 + x), and (33 + x) ⇒ (10 + x)/(23 + x) = (15 + x)/(33 + x) ⇒ (10 + x)(33 + x) = (15 + x)(23 + x) ⇒ 330 + 10x + 33x + x² = 345 + 15x + 23x + x² ⇒ 48x – 43x = 345 – 330 ⇒ 5x = 15 ⇒ x = 15/5 ⇒ x = 3 The value of (x + 5) ⇒ 3 + 5 ⇒ 8 ∴ The value of (x + 5) is 8.

24922.

When x is added each of 10, 23, 15 and 33, the sums obtained in this order are in proportion. The value of (x + 5) is:1. 12. 83. 34. 5

Answer» Correct Answer - Option 2 : 8

Given:

The numbers are 10, 23, 15, and 33

x is added to each number and the sum is in proportion.

Concept Used:

If a, b, c, and d are in proportion,

then, a/b = c/d

Calculation:

When x is added to each number the numbers will be,

(10 + x), (23 + x), (15 + x), and (33 + x)

⇒ (10 + x)/(23 + x) = (15 + x)/(33 + x)

⇒ (10 + x)(33 + x) = (15 + x)(23 + x)

⇒ 330 + 10x + 33x + x² = 345 + 15x + 23x + x²

⇒ 48x – 43x = 345 – 330

⇒ 5x = 15

⇒ x = 15/5

⇒ x = 3

The value of (x + 5)

⇒ 3 + 5

⇒ 8

∴ The value of (x + 5) is 8.

Discussion

24923.	Directions: Given below are two quantities named I and II. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose among the possible answers.Quantity I: Three numbers are in the ratio 7 : 21 : 25.Their average is 53. Find the largest number.Quantity II: 341. Quantity I ˃ Quantity II2. Quantity I ˂ Quantity II3. Quantity I ≤ Quantity II4. Quantity I ≥ Quantity II5. Quantity I = Quantity II
Answer» Correct Answer - Option 1 : Quantity I ˃ Quantity II Quantity I: Let the numbers be 7x, 21x and 25x ⇒ (7x + 21x + 25x) / 3 = 53 ⇒ (53x / 3) = 53 ⇒ x = (53 × 3) / 53 ⇒ x = 3 Largest number is 75 (i.e. 25 × 3) Quantity II: 34 ∴ Quantity I ˃ Quantity II

24923.

Directions: Given below are two quantities named I and II. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose among the possible answers.Quantity I: Three numbers are in the ratio 7 : 21 : 25.Their average is 53. Find the largest number.Quantity II: 341. Quantity I ˃ Quantity II2. Quantity I ˂ Quantity II3. Quantity I ≤ Quantity II4. Quantity I ≥ Quantity II5. Quantity I = Quantity II

Answer» Correct Answer - Option 1 : Quantity I ˃ Quantity II

Quantity I:

Let the numbers be 7x, 21x and 25x

⇒ (7x + 21x + 25x) / 3 = 53

⇒ (53x / 3) = 53

⇒ x = (53 × 3) / 53

⇒ x = 3

Largest number is 75 (i.e. 25 × 3)

Quantity II: 34

∴ Quantity I ˃ Quantity II

Discussion

24924.	Directions: Given below are two quantities named I and II. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose among the possible answers.Quantity I: 6 kmphQuantity II: A boat sails in still water at 13 km/h, but takes five times as long to sail upstream than downstream. Find the speed of the stream.1. Quantity I ˃ Quantity II2. Quantity I ˂ Quantity II3. Quantity I ≥ Quantity II4. Quantity II ≥ Quantity I5. Quantity II = Quantity I
Answer» Correct Answer - Option 2 : Quantity I ˂ Quantity II Quantity I: 3 kmph Quantity II: Speed of boat = 13 km/h Let the speed of the stream be x km/h ⇒ Downstream speed = (13 + x) ⇒ Upstream speed = (13 - x) According to the question, (13 + x) = 5 × (13 - x) ⇒ 13 + x = 65 - 5x ⇒ 13 - 65 = -5x - x ⇒ 52 = 6x ⇒ x = 8.66 Speed of stream = 8.66 kmph ∴ Quantity I ˂ Quantity II

24924.

Directions: Given below are two quantities named I and II. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose among the possible answers.Quantity I: 6 kmphQuantity II: A boat sails in still water at 13 km/h, but takes five times as long to sail upstream than downstream. Find the speed of the stream.1. Quantity I ˃ Quantity II2. Quantity I ˂ Quantity II3. Quantity I ≥ Quantity II4. Quantity II ≥ Quantity I5. Quantity II = Quantity I

Answer» Correct Answer - Option 2 : Quantity I ˂ Quantity II

Quantity I: 3 kmph

Quantity II:

Speed of boat = 13 km/h

Let the speed of the stream be x km/h

⇒ Downstream speed = (13 + x)

⇒ Upstream speed = (13 - x)

According to the question,

(13 + x) = 5 × (13 - x)

⇒ 13 + x = 65 - 5x

⇒ 13 - 65 = -5x - x

⇒ 52 = 6x

⇒ x = 8.66

Speed of stream = 8.66 kmph

∴ Quantity I ˂ Quantity II

Discussion

24925.	Directions: Given below are two quantities named I and II. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose among the possible answers.Quantity I: 8 monthsQuantity II: Bhavik and Kunal are partners in a business. Bhavik received (2 / 3) of the profit and Kunal invested (1 / 6) of the capital for 5 months. Find the period of Bhavik’s investment.1. Quantity I = Quantity II2. Quantity I ≥ Quantity II3. Quantity I ≤ Quantity II4. Quantity I ˂ Quantity II5. Quantity I ˃ Quantity II
Answer» Correct Answer - Option 5 : Quantity I ˃ Quantity II Quantity I: 8 months Quantity II: Let the total profit be Rs. x ⇒ Bhavik’s share = 2x / 3 ⇒ Kunal’s share = (x) - (2x / 3) = x / 3 Profit ratio of bhavik and Kunal = 2 : 1 Let total capital be Rs. a and period of investment of Bhavik be ‘t’ months ⇒ [(5a / 6) × t] : [(a / 6) × 5] = 2 : 1 ⇒ [5at / 6] : [5a / 6] = 2 : 1 ⇒ [(5at / 6) × (6 / 5a)] = 2 : 1 ⇒ t = 2 months ∴ Quantity I ˃ Quantity II.

24925.

Directions: Given below are two quantities named I and II. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose among the possible answers.Quantity I: 8 monthsQuantity II: Bhavik and Kunal are partners in a business. Bhavik received (2 / 3) of the profit and Kunal invested (1 / 6) of the capital for 5 months. Find the period of Bhavik’s investment.1. Quantity I = Quantity II2. Quantity I ≥ Quantity II3. Quantity I ≤ Quantity II4. Quantity I ˂ Quantity II5. Quantity I ˃ Quantity II

Answer» Correct Answer - Option 5 : Quantity I ˃ Quantity II

Quantity I: 8 months

Quantity II:

Let the total profit be Rs. x

⇒ Bhavik’s share = 2x / 3

⇒ Kunal’s share = (x) - (2x / 3) = x / 3

Profit ratio of bhavik and Kunal = 2 : 1

Let total capital be Rs. a and period of investment of Bhavik be ‘t’ months

⇒ [(5a / 6) × t] : [(a / 6) × 5] = 2 : 1

⇒ [5at / 6] : [5a / 6] = 2 : 1

⇒ [(5at / 6) × (6 / 5a)] = 2 : 1

⇒ t = 2 months

∴ Quantity I ˃ Quantity II.

Discussion

24926.	If `P` represents radiation pressure , `C` represents the speed of light , and `Q` represents radiation energy striking a unit area per second , then non - zero integers `x, y, z` such that `P^(x) Q^(y) C^(z)` is dimensionless , find the values of `x, y , and z`.A. x =1 , y =1 , z = -1B. x = 1 , y = -1 , z = 1C. x = -1 , y = 1 , z=1D. x = 1 , y = 1 , z =1
Answer» Correct Answer - B `P = (F)/(A) = (M^(1)L^(1)T^(-2))/(L^(2)) M^(1)L^(-1)T^(2)` `C = L^(1) T^(-1)` `Q = (E)/(At) = (ML^(2)T^(-2))/(L^(2)T) = MT^(3)` `P^(x)Q^(y)C^(z) = (M^(1) L^(-1)T^(-2))^(x) (MT^(-3))^(y) (L^(1)T^(-1))^(x) = M^(0)L^(0)T^(0)` `M^(x + y) L^(-x + z) T^(-2x - 3y -z) = M^(0)L^(0)T^(0)` x + y = 0 `-x + 2 = 0 ` `-2x - 3y - z = 0` All these conditions are statisfied by option (2)

Discussion

24927.	A physical quantity `rho` is calculated by using the formula `rho =(1)/(10)(xy^(2))/(z^(1//3))`, where x, y and z are experimentally measured quantities. If the fractional error in the measurement of x, y and z are `2%, 1%` and `3%`, respectively, then the maximum fractional error in the calculation of `rho` isA. 0.005B. 0.05C. 0.06D. 0.07
Answer» `epsilon=(xy^(2))/(10z^(1//3))` Take log and differentiate, `(Deltaepsilon)/(epsilon)=(Deltaex)/(x)+(2Deltay)/(y)-(1)/(3)(Deltaz)/(z)` for maximum error, `(Deltaepsilon)/(epsilon)=(Deltax)/(x)+2(Deltay)/(y)+(1)/(3)(Deltaz)/(z)` `=2%+(2xx1%)+((1)/(3)xx3%)` `=2%+2%+1%=5%`

Discussion

24928.	Change in energy for adiabatic process is
Answer» During an adiabatic process, heat energy is not exchanged between system and its surroundings.

Discussion

24929.	The pressure on a square plate is measured by measureing the force on the plate and the percentage error of measurement of force and length are respectively 4% and 2%, the maximum error in the measurement of pressure is-A. 0.01B. 0.02C. 0.06D. 0.08
Answer» `P=(F)/(A)=(F)/(L^(2))` For max. error `(DeltaP)/(P)=(DeltaF)/(F)+(2DeltaL)/(L)=4%+2(2%)=8%`

Discussion

24930.	The error in measuring the side of a cube is `+- 1%`. The error in the calculation of the volume of the cube will be aboutA. `+-0.001%`B. `+-1%`C. `+-6%`D. `+-3%`
Answer» Volume of cube V- (side `a)^(3)` `therefore V=a^(3)`. Take log and differentiate. `(DeltaV)/(V)=3(Deltaa/(a)=3(+-+-1%)=+-3%`

Discussion

24931.	Define oscillation.
Answer» Oscillation is the process of moving back and forth regularly, like the oscillation of a fan that cools off the whole room, or the oscillation of a movie plot that makes you laugh and cry. Oscillation is from the Latin word oscillare for "to swing," so oscillation is when something is swinging back and forth.

Discussion

24932.	For a perfectly black body a1=1?
Answer» Answer: One which absorbs all radiant energy at all wavelengths

Discussion

24933.	A circuit is shown below. If A is an ideal ammeter, B an ideal Battery of voltage V, and C an ideal volmeter, what will be the reading of C/reading of A ?A. RB. 2RC. R/2D. 0
Answer» Correct Answer - A Reading of C=V {I in that branch=0} Reading of `A=V/R` Rightarrow R

Discussion

24934.	Liposcelidae is the family of ------ a. Bird louse b. Book louse c. Mammal louse d. None of all
Answer» Liposcelidae is the family of Book louse.

Discussion

24935.	Psocus lineatus is a. Bird louse b. Book louse c. Mammal louse d. None of all
Answer» Psocus lineatus is Book louse.

Discussion

24936.	A circuit is shown below. If A is a capacitor, B is an ideal ammeter and C is an ideal battery of voltage V, what is the voltage across the capacitor?A. VB. `V/2`C. 2VD. 0
Answer» Correct Answer - D `V_(B)=iR_(ammeter)=0` `Rightarrow V_("cap")=0`

Discussion

24937.	A current of `(2.5+-0.05)` A flows through a wire and develops a potential difference of `(10+-0.1)` volt. Resistance of the wire in ohm, isA. `4+-0.12`B. `4+-0.04`C. `4+-0.08`D. `4+-0.02`
Answer» `R=(10)/(2.5)=4+-DeltaR` `DeltaR=0.03xx4=0.12`

Discussion

24938.	Describe any two parts of business correspondence.
Answer» Two parts of business correspondence can be: 1. The Heading: In business letter, the name of the firm along with its address is given. It also contains telephone numbers, e-mail address and logo of the firm , if any. 2. Reference Number: The word reference no. is generally printed on the letter head on the left hand side. It is the file no., name of the department, year, dispatch number.

24938.

Describe any two parts of business correspondence.

Answer»

Two parts of business correspondence can be:

1. The Heading: In business letter, the name of the firm along with its address is given. It also contains telephone numbers, e-mail address and logo of the firm , if any.

2. Reference Number: The word reference no. is generally printed on the letter head on the left hand side. It is the file no., name of the department, year, dispatch number.

Discussion

24939.	The skills of the hand embroider lies in ____________ which would result in a product of beauty and grace.
Answer» right selection of the design, embroidery stitches, threads, colours

Discussion

24940.	An article made of metal ‘X’ acquires a blackish tinge on exposure to air after a few days. The metal ‘X’ is1. Gold2. Silver3. Aluminium4. Iron
Answer» Correct Answer - Option 2 : Silver The correct answer is Silver Silver articles when exposed to air become black after some time. This is because the silver metal reacts with the sulphur present in the atmosphere and forms silver sulphide (black colour). The silver article turns black when kept in the air because the silver article reacts with sulphur compounds such as hydrogen sulphide (H₂S) present in the air. Ag+H₂S→Ag₂S+H₂ The phenomenon is called corrosion. For silver particularly, it is called tarnishing of silver. Silver sulphide (Ag₂S) is the black substance formed.

24940.

An article made of metal ‘X’ acquires a blackish tinge on exposure to air after a few days. The metal ‘X’ is1. Gold2. Silver3. Aluminium4. Iron

Answer» Correct Answer - Option 2 : Silver

The correct answer is Silver

Silver articles when exposed to air become black after some time.
This is because the silver metal reacts with the sulphur present in the atmosphere and forms silver sulphide (black colour).

The silver article turns black when kept in the air because the silver article reacts with sulphur compounds such as hydrogen sulphide (H₂S) present in the air. Ag+H₂S→Ag₂S+H₂
The phenomenon is called corrosion. For silver particularly, it is called tarnishing of silver.
Silver sulphide (Ag₂S) is the black substance formed.

Discussion

24941.	An Oligotrophic Lake has1. High levels of nutrients in water2. High aquatic productivity3. Algel Blooms4. Low nutrients and low productivity
Answer» Correct Answer - Option 4 : Low nutrients and low productivity The correct answer is Low nutrients and low productivity. An Oligotrophic Lake has low nutrients and low productivity. These lakes have low algal production, and consequently, often have very clear waters, with high drinking-water quality. An oligotrophic lake is a lake with low primary productivity, as a result of low nutrient content. These lakes have low algal production, and consequently, often have very clear waters, with high drinking-water quality. Oligotrophic lakes are most common in cold, sparsely developed regions that are underlain by crystalline igneous, granitic bedrock. Due to their low algal production, these lakes consequently have very clear waters, with high drinking-water quality. Lakes that have to intermix their layers are classified into the category of holomictic, whereas lakes that do not have interlayer mixing are permanently stratified and thus are termed meromictic.

24941.

An Oligotrophic Lake has1. High levels of nutrients in water2. High aquatic productivity3. Algel Blooms4. Low nutrients and low productivity

Answer» Correct Answer - Option 4 : Low nutrients and low productivity

The correct answer is Low nutrients and low productivity.

An Oligotrophic Lake has low nutrients and low productivity.
These lakes have low algal production, and consequently, often have very clear waters, with high drinking-water quality.

An oligotrophic lake is a lake with low primary productivity, as a result of low nutrient content.
These lakes have low algal production, and consequently, often have very clear waters, with high drinking-water quality.
Oligotrophic lakes are most common in cold, sparsely developed regions that are underlain by crystalline igneous, granitic bedrock.
Due to their low algal production, these lakes consequently have very clear waters, with high drinking-water quality.
Lakes that have to intermix their layers are classified into the category of holomictic, whereas lakes that do not have interlayer mixing are permanently stratified and thus are termed meromictic.

Discussion

24942.	Which of the following has the highest water potential?1. Pure water 2. 5% glucose solution3. 10% glucose solution4. 15% glucose solution
Answer» Correct Answer - Option 1 : Pure water The correct answer is Pure water Water potential depends on the number or concentration of water molecules. The more the number of water molecules in a system, the higher will be the water potential. Hence, pure water has the highest water potential. By convention water potential of pure water has been fixed as zero. This means that the water potential of any solution will always be negative.

24942.

Which of the following has the highest water potential?1. Pure water 2. 5% glucose solution3. 10% glucose solution4. 15% glucose solution

Answer» Correct Answer - Option 1 : Pure water

The correct answer is Pure water

Water potential depends on the number or concentration of water molecules.
The more the number of water molecules in a system, the higher will be the water potential.
Hence, pure water has the highest water potential.
By convention water potential of pure water has been fixed as zero.
This means that the water potential of any solution will always be negative.

Discussion

24943.	When the two input terminals of a diff amp are grounded, a. The base currents are equal b. The collector currents are equal c. An output error voltage usually exists d. The ac output voltage is zero
Answer» (c) An output error voltage usually exists

Discussion

24944.	Machines are used in some form of the other in every office. What should be kept in mind while choosing the machine for the office?
Answer» The factors which should be considered or taken into account at the time of choosing machines are: Ease of Operation Durability Flexibility Portability Adaptability Service Operating cost Reputation of the supplier

24944.

Machines are used in some form of the other in every office. What should be kept in mind while choosing the machine for the office?

Answer»

The factors which should be considered or taken into account at the time of choosing machines are:

Ease of Operation
Durability
Flexibility
Portability
Adaptability
Service
Operating cost
Reputation of the supplier

Discussion

24945.	Mota Ram runs a small business specialising in delivering organic fruits and Vegetables to the local area. He buys from local farms and packages these in boxes and delivers them locally. Total fixed cost incurred in the entire operation is Rs. 1,00,000. What will be the Total Break Even Point for Mota Ram? Organic Fruits (per Kg)Organic Vegetables (per Kg)Selling price per box (in Rs.)350250Variable cost per box (in Rs.)250150Sales Mix %5545a. 100 boxes b. 1000 boxes c. 10,000 boxes d. 1,00,000 boxes
Answer» Answer is b. 1000 boxes

Discussion

24946.	Bhawna, Maya, and Advik’s interior design business was taking off in a big way. Their talent was in high demand. Now, the trio needed to hire more employees. Since the needs of their company were changing, they reviewed their personal circumstances, finances, and goals. They decided to legally organize their organization into one with a limited liability. Identify the type of organization the trio will form. a. Partnership b. Public Companyc. Private Company d. Company
Answer» c. Private Company

Discussion

24947.	Four objects W,X,Y and Z,each with charge +q are held fixed at four points of a square of side d as shown in the figure. Objects X and Z are on object. W on object X is F. Then the magnitude of the force exerted by object W to Z is :(a) \(\frac{F}{7}\)(b) \(\frac{F}{5}\) (c) \(\frac{F}{3}\) (d) \(\frac{F}{2}\)
Answer» Option : (b) \(\frac{F}{5}\)

Discussion

24948.	Two sources of equal emf are connected in series. This combination is, in turn connected to an external resistance R. The internal resistance of two sources are r1 and r2 (r2 > r1). If the potential difference across the source of internal resistance r2 is zero,then R equals to -(a) \(\frac{r_1+r_2}{r_2-r_1}\) (b) r2 - r1(c) \(\frac{r_1r_2}{r_2-r_1}\) (d) \(\frac{r_1+r_2}{r_1r_2}\)
Answer» Option : (b) r₂ - r₁

Discussion

24949.	Which of the following statements is correct ?(a) Magnetic field lines do not form closed loops.(b) Magnetic field lines start from north pole and end at south pole of a magnet.(c) The tangent at a point on a magnetic field line represents the direction of the magnetic field at that point.(d) Two magnetic field lines may intersect each other.
Answer» Option : (c) The tangent at a point on a magnetic field line represents the direction of the magnetic field at that point.

Discussion

24950.	The equivalent resistance between A and B of the network shown in figure is :(a) 3R Ω(b) \((\frac{3}{2})\)R Ω(c) 2R Ω(d) \((\frac{2}{3})\)R Ω
Answer» Option : (c) 2R Ω

Discussion

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A 100 watt bulb working on 200 volt and a 200 watt bulb working on 100 volt haveA. Resistances in the ratio of 4 : 1B. Maximum current ratings in the ratio of 1 : 4C. Resistances in the ratio of 2 : 1D. Maximum current ratings in the ratio of 1 : 2

The moment of inertia of a squaer lamin about the perpendicluar axis though its centre of mass is ` 20 kg- m^(2)` . Then its moment of inertia about an axis toching its side and in the plane of the lamina will be : -A. `10 kg - m^(2)`B. `30 kg- m^(2)`C. `40 k g- m^(2)`D. `25 kg m^(2)`

Two pipes can fill a tank in 15 hours and 4 hours, respectively, while a third pipe can empty it in 12 hours. How long (in hours) will it take to fill the empty tank if all the three pipes are opened simultaneously?1. 30 / 72. 20 / 73. 50 / 74. 15 / 7

If three pipes P, Q, and R together can empty full tank in 25 minutes, if P is twice as efficient as Q and Q is twice as efficient as R then in how much time R will empty 60% of the tank?1. 90 minutes2. 105 minutes3. 95 minutes4. 114 minutes

The dimensional formula for magnetic flux isA. `ML^(2)T^(-1)C^(-1)`B. `ML^(3)T^(-2)C^(-1)`C. `ML^(2)T^(2)C^(-1)`D. `ML^(2)T^(2)C^(-1)`

Using mass `(M)` , length `(L)` , time `(T)` , and electric current `(A)` as fundamental quantities , the dimensions of permitivity will beA. `[M^(-1)LT^(-2)A]`B. `[ML^(-2)T^(-2)A^(-1)]`C. `[MLT^(-2)A^(-2)]`D. `[MLT^(-1)A^(-1)]`

If the force is given by `F=at+bt^(2)` with `t` is time. The dimensions of `a` and `b` areA. `[M L T^(-3)]` and `[M L T^(-4)]`B. `[M L T^(-4)]` and `[M L T^(-3)]`C. `[ M L T^(-1)]` and `[M L T^(-2)]`D. `[ M L T^(-2)]` and `[M L T^(0)]`

Find the dimensions of a in the formula `(p+a/V^2)(V-b)=RT`

Are there any physical quantities out of the following which have the same dimension? If yes, identify them. Impulse, Torque, Angular momentum, Energy, Force Moment of inertia

The dimension of torque is:A. `[ML^(3)L^(-3)]`B. `[ML^(-1)T^(-1)]`C. `[ML(2)T^(-2)]`D. `[ML^(-2)]`

Which of the following represents the dimensions of FaradA. `M^(-1)L^(-2)T^(4)A^(2)`B. `ML^(2)T^(-2)A^(-2)`C. `ML^(2)T^(-2)A^(-1)`D. `MT^(-2)A^(-1)`

When x is added each of 10, 23, 15 and 33, the sums obtained in this order are in proportion. The value of (x + 5) is:1. 12. 83. 34. 5

Change in energy for adiabatic process is

The pressure on a square plate is measured by measureing the force on the plate and the percentage error of measurement of force and length are respectively 4% and 2%, the maximum error in the measurement of pressure is-A. 0.01B. 0.02C. 0.06D. 0.08

The error in measuring the side of a cube is `+- 1%`. The error in the calculation of the volume of the cube will be aboutA. `+-0.001%`B. `+-1%`C. `+-6%`D. `+-3%`

Define oscillation.

For a perfectly black body a1=1?

A circuit is shown below. If A is an ideal ammeter, B an ideal Battery of voltage V, and C an ideal volmeter, what will be the reading of C/reading of A ?A. RB. 2RC. R/2D. 0

Liposcelidae is the family of ------ a. Bird louse b. Book louse c. Mammal louse d. None of all

Psocus lineatus is a. Bird louse b. Book louse c. Mammal louse d. None of all

A circuit is shown below. If A is a capacitor, B is an ideal ammeter and C is an ideal battery of voltage V, what is the voltage across the capacitor?A. VB. `V/2`C. 2VD. 0

A current of `(2.5+-0.05)` A flows through a wire and develops a potential difference of `(10+-0.1)` volt. Resistance of the wire in ohm, isA. `4+-0.12`B. `4+-0.04`C. `4+-0.08`D. `4+-0.02`

Describe any two parts of business correspondence.

The skills of the hand embroider lies in ____________ which would result in a product of beauty and grace.

An article made of metal ‘X’ acquires a blackish tinge on exposure to air after a few days. The metal ‘X’ is1. Gold2. Silver3. Aluminium4. Iron

An Oligotrophic Lake has1. High levels of nutrients in water2. High aquatic productivity3. Algel Blooms4. Low nutrients and low productivity

Which of the following has the highest water potential?1. Pure water 2. 5% glucose solution3. 10% glucose solution4. 15% glucose solution

When the two input terminals of a diff amp are grounded, a. The base currents are equal b. The collector currents are equal c. An output error voltage usually exists d. The ac output voltage is zero

Machines are used in some form of the other in every office. What should be kept in mind while choosing the machine for the office?

The equivalent resistance between A and B of the network shown in figure is :(a) 3R Ω(b) \((\frac{3}{2})\)R Ω(c) 2R Ω(d) \((\frac{2}{3})\)R Ω