The normal glucose tolerance curve reaches peak is (A) 15 min (B) 1 hr (C) 2 hrs (D) 2 ½ hrs

The normal glucose tolerance curve reaches peak is (A) 15 min (B) 1 hr (C) 2 hrs (D) 2 ½ hrs
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Adam and Smith are working on a project. Adam takes 18 hrs to type 108 pages on a computer, while Smith takes 15 hrs to type 120 pages. How much time will they take, working together on two different computers to type a project of 360 pages? a) 23 hrs 24 min b) 45 hrs 42 min c) 25 hrs 42 min d) 42 hrs 45 min

Description : Adam and Smith are working on a project. Adam takes 18 hrs to type 108 pages on a computer, while Smith takes 15 hrs to type 120 pages. How much time will they take, working together on two different computers to type a ... ? a) 23 hrs 24 min b) 45 hrs 42 min c) 25 hrs 42 min d) 42 hrs 45 min

Last Answer : C Number of pages typed by Adam=108 Number of pages typed by Adam in 1 hour = 108/18=6 Number of pages typed by Smith =120 Number of pages typed by Smith in 1 hour = 120/15 =8 Number of pages typed by both in ... (6 + 8) = 14 Time taken by both to type 360 pages, = (360 * 1/14) =25 hrs 42 min

A flight is planned from L to M, distance 850 nm. Wind component out is 35 kt(TWC), TAS 450kt. Mean fuel flow out is 2500kg/hr, mean fuel flow inbound is 1900kg/hr and the fuel available is 6000kg. The time and distance to PSR is: a. 1hr 30 min, 660nm b. 1hr 30 min, 616nm c. 1hr 15 min, 606nm d. 1hr 16 min, 616nm

Description : A flight is planned from L to M, distance 850 nm. Wind component out is 35 kt(TWC), TAS 450kt. Mean fuel flow out is 2500kg/hr, mean fuel flow inbound is 1900kg/hr and the fuel available is 6000kg. The time ... 1hr 30 min, 660nm b. 1hr 30 min, 616nm c. 1hr 15 min, 606nm d. 1hr 16 min, 616nm

Last Answer : d. 1hr 16 min, 616nm

A flight is planned from L to M, distance 850 nm. Wind component out is 35 kt(TWC), TAS 450kt. Mean fuel flow out is 2500kg/hr, mean fuel flow inbound is 1900kg/hr and the fuel available is 6000kg. The time and distance to PSR is: a. 1hr 30 min, 660nm b. 1hr 30 min, 616nm c. 1hr 15 min, 606nm d. 1hr 16 min, 616nm

Description : A flight is planned from L to M, distance 850 nm. Wind component out is 35 kt(TWC), TAS 450kt. Mean fuel flow out is 2500kg/hr, mean fuel flow inbound is 1900kg/hr and the fuel available is 6000kg. The time ... 30 min, 660nm b. 1hr 30 min, 616nm c. 1hr 15 min, 606nm d. 1hr 16 min, 616nm

Last Answer : d. 1hr 16 min, 616nm

A train M speeding with 60km/hr crooses another train N, running in the same direction in 1 min. If the lengths of the trains M and N be 50m and 100m respectively. What is the speed of train N? A) 15 km/hr B) 61km/hr C) 51km/hr D) 50k/hr

Description : A train M speeding with 60km/hr crooses another train N, running in the same direction in 1 min. If the lengths of the trains M and N be 50m and 100m respectively. What is the speed of train N? A) 15 km/hr B) 61km/hr C) 51km/hr D) 50k/hr

Last Answer : ANSWER: C Explanation:  Let the speed of be train N be ‘x’km/hr  Speed of M relative to N = (60 – x)km/hr  = (60-x) * 5/18]m/s = (300 – 5x / 18)m/s  150 / (300 – 5x / 18) =60  2700/300-5x=60  2700=18000-300x 300x=15300  x= 153/3  x=51km/hr  Therefore the speed of the train is 51 km/hr

Dhanvanth can swim at 15 km/hr in stagnant water. In a river with 3 km/hr current, he swims to a certain distance and comes back within 100 min. What is the distance between the two points? A) 11km B) 13km C) 19km D) 12km

Description : Dhanvanth can swim at 15 km/hr in stagnant water. In a river with 3 km/hr current, he swims to a certain distance and comes back within 100 min. What is the distance between the two points? A) 11km B) 13km C) 19km D) 12km

Last Answer : ANSWER: D Explanation: Speed in still water (a) = 15 km/hr Speed in current( b) = 3 km/hr Upstream speed = a - b = 15 - 3 = 12 km/hr Downstream speed = a + b = 15 + 3 = 18 km/hr Let the ... 36 = 5 / 3 15x = 36 * 5 x = 36 * 5 / 15 = 12 km Distance between the two points is 12km

A 250 m goods train takes 50 s to cross a person who is going in the same direction with the speed of 8 km/h. After crossing that person, the train can reach next station in 2 hr. How long that person takes to reach that station after being crossed by them? A) 2 ½ hr B) 5hr C) 3 1/2hr D) 1hr

Description : A 250 m goods train takes 50 s to cross a person who is going in the same direction with the speed of 8 km/h. After crossing that person, the train can reach next station in 2 hr. How long that person takes to reach that station after being crossed by them? A) 2 ½ hr B) 5hr C) 3 1/2hr D) 1hr

Last Answer : ANSWER:A Explanation:  Speed of the person = 8 kmph  = (8*5/18)m/s  = 20/9 m/s  Then relative speed of train = (x-20/9)m/s  As train takes 50 sec to cross the person  Therefore 50 =250/(x-20/ ... 20,000 m  = 20 km Thus time taken by the person to cover the distance of 20km= 20/8 hr=2 ½ hr

The earth completes one rotation on its axis in : (1) 23 hrs. 56 min. 4.9 sec (2) 23 hrs. 10 min. 2 sec (3) 23 hrs. 30 min. (4) 24 hrs.

Description : The earth completes one rotation on its axis in : (1) 23 hrs. 56 min. 4.9 sec (2) 23 hrs. 10 min. 2 sec (3) 23 hrs. 30 min. (4) 24 hrs.

Last Answer : (1) 23 hrs. 56 min. 4.9 sec Explanation: The Earth rotates around its axis once in about 24 hours with respect to the sun and once every 23 hours 56 minutes and 4 seconds with respect to the stars. The Earth rotates from the west towards east. Seasons are caused by Earth's rotation around its axis.

The earth completes one rotation on its axis in : (1) 23 hrs. 56 min. 4.9 sec (2) 23 hrs. 10 min. 2 sec (3) 23 hrs. 30 min. (4) 24 hrs.

Description : The earth completes one rotation on its axis in : (1) 23 hrs. 56 min. 4.9 sec (2) 23 hrs. 10 min. 2 sec (3) 23 hrs. 30 min. (4) 24 hrs.

Last Answer : 23 hrs. 56 min. 4.9 sec

A leak in the bottom of a bunker can empty the full bunker in 4 hrs. An inlet pump fills water at the rate of 3 litres a min. when the bunker is full, the inlet is opened and due to the leak, the bunker is empty in 6 hrs. How many litres does the bunker field? A) 1160 litrs B) 1610 litres C) 2160 litres D) 2610 litres

Description : A leak in the bottom of a bunker can empty the full bunker in 4 hrs. An inlet pump fills water at the rate of 3 litres a min. when the bunker is full, the inlet is opened and due to the leak, ... hrs. How many litres does the bunker field? A) 1160 litrs B) 1610 litres C) 2160 litres D) 2610 litres

Last Answer :  C Work done by the inlet in 1 hr = (1/4 – 1/6) = (3-2/12) = 1/12 Work done by the inlet in 1 min = 1/12 * 1/60 = 1/720 Volume of 1/720 part = 3 litres Volume of whole = (720 * 3 )=2160 litres.

Two filling tap can fill a dumper in 14 and 28 min. respectively and when the waste tap is open, they can together fill it in 35 min. The waste pipe can empty the full dumper in – A) 152/9 hrs. B) 137/7 hrs. C) 60/9 hrs. D) 140/11 hrs.

Description : Two filling tap can fill a dumper in 14 and 28 min. respectively and when the waste tap is open, they can together fill it in 35 min. The waste pipe can empty the full dumper in – A) 152/9 hrs. B) 137/7 hrs. C) 60/9 hrs. D) 140/11 hrs.

Last Answer :  D 1/14 + 1/28 – 1/35 =>10 + 5 – 4/140 =>11/140 =>140/11 => hence answer is 140/11 hours.

Acromegaly results in all the following except (A) Overgrowth of the bones of face, hands and feet (B) Increased stature (C) Enlargements of viscera (D) Impaired glucose tolerance

Description : Acromegaly results in all the following except (A) Overgrowth of the bones of face, hands and feet (B) Increased stature (C) Enlargements of viscera (D) Impaired glucose tolerance

Last Answer : Answer : B

Metallic constituent of “Glucose tolerance factor” is (A) Sulphur (B) Cobalt (C) Chromium (D) Selenium

Description : Metallic constituent of “Glucose tolerance factor” is (A) Sulphur (B) Cobalt (C) Chromium (D) Selenium

Last Answer : Answer : C

Impaired galactose tolerance test suggests (A) Defect in glucose utilisation (B) Liver cell injury (C) Renal defect (D) Muscle injury

Description : Impaired galactose tolerance test suggests (A) Defect in glucose utilisation (B) Liver cell injury (C) Renal defect (D) Muscle injury

Last Answer : Answer : B

Glucose tolerance is decreased in (A) Diabetes mellitus (B) Hypopituitarisme (C) Addison’s disease (D) Hypothyroidism

Description : Glucose tolerance is decreased in (A) Diabetes mellitus (B) Hypopituitarisme (C) Addison’s disease (D) Hypothyroidism

Last Answer : A

Glucose tolerance is increased in (A) Diabetes mellitus (B) Adrenalectomy (C) Acromegaly (D) Thyrotoxicosis

Description : Glucose tolerance is increased in (A) Diabetes mellitus (B) Adrenalectomy (C) Acromegaly (D) Thyrotoxicosis

Last Answer : B

An 16-year-old man presents with polyuria and polydipsia. Which of the following may confirm the diagnosis of diabetes mellitus? 1) A random plasma glucose of >7.5 mmol/L 2) A finding of 3+ ketonuria 3) An HbA1c of 7.0% 4) A fasting plasma glucose of 7.5 mmol/L 5) An abnormal glucose tolerance test

Description : An 16-year-old man presents with polyuria and polydipsia. Which of the following may confirm the diagnosis of diabetes mellitus? 1) A random plasma glucose of >7.5 mmol/L 2) A finding of 3+ ketonuria 3) An HbA1c of 7.0% 4) A fasting plasma glucose of 7.5 mmol/L 5) An abnormal glucose tolerance test

Last Answer : Answers-4 The diagnosis is usually relatively easy to confirm in a symptomatic subject. A random glucose of >11.1 mmol/L or a fasting glucose of >7.0 mmol/L would be regarded as ... haemoglobin (HbA1c) is also highly suggestive but not diagnostic. A glucose tolerance test is rarely needed.

A 35 year old male presents with weakness and tiredness. He is noted to be hyertensive. Electrolytes show a hypokalaemia and a hypomagnesaemia. What investigation would you select for this patient? 1) Colonoscopy 2) Plasma renin toaldosterone ratio 3) Serum amylase 4) Serum calcium 5) Oral glucose tolerance test

Description : A 35 year old male presents with weakness and tiredness. He is noted to be hyertensive. Electrolytes show a hypokalaemia and a hypomagnesaemia. What investigation would you select for this patient? ... Plasma renin toaldosterone ratio 3) Serum amylase 4) Serum calcium 5) Oral glucose tolerance test

Last Answer : Answers-2 The hypokalaemic hypertension with hypomagnesaemia suggests primary hyperaldosteronism. The most reliable assessment for this would be renin to aldosterone ratio.

What are other conditions which may cause impaired glucose tolerance?

Description : What are other conditions which may cause impaired glucose tolerance?

Last Answer : Alimentary glucosuria, renal glucosuria.

What is impaired glucose tolerance (IGT)?

Description : What is impaired glucose tolerance (IGT)? 

Last Answer : When fasting plasma glucose level is between 110 and 126 mg/dl and 2-hour post-glucose value is between 140 and 200 mg/dl.

What is the major indication for doing an oral glucose tolerance test (OGTT)?

Description : What is the major indication for doing an oral glucose tolerance test (OGTT)? 

Last Answer : Patient has symptoms suggestive of diabetes mellitus, but fasting blood sugar value is inconclusive (between 100 and 126 mg/dl).

A motor car starts with the speed of 70 km/hr with its speed increasing every two hours by 10 kmph. In how many hours will it cover 345 kms? A.4 hrs B.4 hrs 5 mins C.4 1/2 hrs D.2 hrs E.None of these

Description : A motor car starts with the speed of 70 km/hr with its speed increasing every two hours by 10 kmph. In how many hours will it cover 345 kms? A.4 hrs B.4 hrs 5 mins C.4 1/2 hrs D.2 hrs E.None of these

Last Answer : Answer - C (4 1/2 hrs) Explanation - Distance covered in first 2 hours = (70 x 2) km = 140 km Distance covered in next 2 hours = (80 x 2) km = 160 km Remaining distance = 345 - (140 + 160) = 45 km. Speed in the ... ..1 hr 45km .? ?=45/90hr=1/2hr Total time taken = 2 + 2 +1/2=4 (1/2)hrs

How many hours in advance of departure time should a flight plan be filed in the case of flights into areas subject to air traffic flow management (ATFM)? a. 3.00 hrs b. 0.30 hrs c. 1.00 hr d. 0.10 hr

Description : How many hours in advance of departure time should a flight plan be filed in the case of flights into areas subject to air traffic flow management (ATFM)? a. 3.00 hrs b. 0.30 hrs c. 1.00 hr d. 0.10 hr

Last Answer : a. 3.00 hrs

How many hours in advance of departure time should a flight plan be filed in the case of flights into areas subject to air traffic flow management (ATFM)? a. 3.00 hrs b. 0.30 hrs c. 1.00 hr d. 0.10 hr

Description : How many hours in advance of departure time should a flight plan be filed in the case of flights into areas subject to air traffic flow management (ATFM)? a. 3.00 hrs b. 0.30 hrs c. 1.00 hr d. 0.10 hr

Last Answer : a. 3.00 hrs

Two bike A & B start at the same time from salem to Rasipuram which are 60km apart. If the 2 bikes travel in opposite directions, they meet after 1 hr and if they travel in same direction(from salem towards Rasipuram), then A meets B after 3 hrs. what is the speed of bike A? a) 50 b) 70 c) 90 d) 30

Description : Two bike A & B start at the same time from salem to Rasipuram which are 60km apart. If the 2 bikes travel in opposite directions, they meet after 1 hr and if they travel in same direction(from salem towards Rasipuram), then A meets B after 3 hrs. what is the speed of bike A? a) 50 b) 70 c) 90 d) 30

Last Answer : B Let their speed be x km/hr and y km/hr respectively Then 60 / (x+y) = ½ X+y = 120 ------------->1 Now when they move in same direction (distance travelled by A in 3 hrs) - (Distance travelled by B in ... = 140 X=70 Put x value in 1 we get Y=120 -70=50 Y=50 Therefore A's speed = 70 km/hr

Karthiga jogs a speed of 18 km/hr at a distance of 27 km. at what speed would she need to jog during the next 4.5 hrs to have an average of 27km/hr for the entire jogging session. a) 40km/hr b) 30 km/hr c) 20 km/hr d) 35 km/hr

Description : Karthiga jogs a speed of 18 km/hr at a distance of 27 km. at what speed would she need to jog during the next 4.5 hrs to have an average of 27km/hr for the entire jogging session. a) 40km/hr b) 30 km/hr c) 20 km/hr d) 35 km/hr

Last Answer : B Let the speed of jagging be x km/hr Total time taken = (27/18 hrs + 4.5 hrs) =(1.5 hrs + 4.5 hrs) =6 hrs Total distance covered = (27+4.5 x) km Therefore (27 +4.5x) / 6= 27 27+4.5x = 162 4.5x = 135 X=135/4.5 =30 So jagging speed is 30 km/hr

P,Q & R are on a journey by scorpio. P drives during the 1st 2 hrs at an average speed of 60 km/hr. Q drives during the next 3 hrs at an average speed of 58 km/hr R drives for the next 4 hrs at an average speed of 62 km/hr. They reached their destination after exactly 5 hrs. Their main speed is. a) 60 2/9 b) 62 2/9 c) 60 4/8 d) 62 4 /8

Description : P,Q & R are on a journey by scorpio. P drives during the 1st 2 hrs at an average speed of 60 km/hr. Q drives during the next 3 hrs at an average speed of 58 km/hr R drives for the next 4 hrs at an average speed of 62 ... exactly 5 hrs. Their main speed is. a) 60 2/9 b) 62 2/9 c) 60 4/8 d) 62 4 /8 

Last Answer : A Distance covered by P = 2*60 = 120 km Distance covered by Q = 3* 58 = 174 km Distance covered by R = 4*62 = 248 km Total distance = (120+174+248)km =542 km Total time taken = 2+3+4 = 9 hrs Mean speed = diatance/ time =(542 / 9)km/hr =60 2/9 km/hr

Jackson travels a distance of 100km in 5 hrs. How much faster in kilometer per hr, on an average, must he travel to make a journey in 5/3 hr less time? a) 5 km/hr b) 10km/hr c) 6 km/hr d) 8 km/hr

Description : Jackson travels a distance of 100km in 5 hrs. How much faster in kilometer per hr, on an average, must he travel to make a journey in 5/3 hr less time? a) 5 km/hr b) 10km/hr c) 6 km/hr d) 8 km/hr

Last Answer : B Time required = (5 hrs – 5/3 hrs) =(5 hrs – 100 mints) = (5 hrs – 1hr 40 mints) = 3 hrs 20 mints = 3 1/3 hrs Required speed = (100*3/10) km/hr =30 km/hr Original speed = (100/5)km/hr = 20 km/hr Difference speed = required speed – original speed =(30-20) =10km/hr

A royal enfield bike starts with the speed of 210 km/hr with its speed increasing every 2 hrs by 30km/hr, in how many hrs will it cover 1035km? a) 2 1/2 hr b) 4 1 /2 hr c) 6 1 /2 hr d) 7 1 /2 hr

Description : A royal enfield bike starts with the speed of 210 km/hr with its speed increasing every 2 hrs by 30km/hr, in how many hrs will it cover 1035km? a) 2 1/2 hr b) 4 1 /2 hr c) 6 1 /2 hr d) 7 1 /2 hr

Last Answer : B Distance covered in 1st 2 hrs = (210*2)km =420km Distance covered in next 2 hrs = (240*2)km =480km Remaining distance = 1035 – (420+480) =1035 – 900 =135km Speed in 5th hr = 270 km/hr Time taken to cover 135 km = 135/270 = ½ hr Total time taken = (2+2+1/2) =4 ½ hrs

Three taps P,Q,R can fill a bunker in 3 hrs. After working at it together for 1 hr, R is closed and P,Q can fill the remaining part in 3 hrs. The number of hrs taken by R alone to fill the bunker. A) 9hrs B) 7hrs C) 11hrs D) 5hrs

Description : Three taps P,Q,R can fill a bunker in 3 hrs. After working at it together for 1 hr, R is closed and P,Q can fill the remaining part in 3 hrs. The number of hrs taken by R alone to fill the bunker. A) 9hrs B) 7hrs C) 11hrs D) 5hrs

Last Answer : A  Part filled in 1 hr = 1/3  Remaining part = 1 – 1/3 = 2/3  (P + Q)’s 3 hours work = 2/3  (P+Q)’s 1 hour work = 2/9  R’s 1 hr work = [(P+Q+R)’s 1 hr work – (P+Q)’s 1 hr work]  = 1/3 – 2/9 = 3-2/9 = 1/9  R alone can fill the tank in 9 hours.

A boat man lakes 5hrs 30 mins to row a boat 30km downstream of a river and 4 hrs 15mints to cover a distance of 10km upstream. Find the speed of the river current in km/hr. A) 1.5 km/hr B) 3km/hr C) 5km/hr D) 7 km/hr

Description : A boat man lakes 5hrs 30 mins to row a boat 30km downstream of a river and 4 hrs 15mints to cover a distance of 10km upstream. Find the speed of the river current in km/hr. A) 1.5 km/hr B) 3km/hr C) 5km/hr D) 7 km/hr

Last Answer : ANSWER: A  Explanation:  Rate downstream = (30 / 5 ½ ) km/hr  = 30 / (11/2) km/hr  = 30 * 2 / 11 km/hr = 60/11 km/hr  Rate upstream = (10 / 4 ¼)km/hr  = 10 / (17/4)km/hr  = 10 * 4 / 17 km/hr = ...  Speed of current = 1 / 2 (a - b) km/hr  =1 / 2 (60 / 11 - 40 / 17)  = 1.5 km/hr (approx.)

A boat takes 27 hrs to travel a distance upstream and takes 9hrs to travel the same distance downstream. If the speed of the boat in still water is 12km/hr, then what is the velocity of the stream? A) 8km/hr B) 6km/hr C) 4km/hr D) None of these

Description : A boat takes 27 hrs to travel a distance upstream and takes 9hrs to travel the same distance downstream. If the speed of the boat in still water is 12km/hr, then what is the velocity of the stream? A) 8km/hr B) 6km/hr C) 4km/hr D) None of these

Last Answer : ANSWER : B  Explanation: Let the velocity of the stream be ‘ y’ km/hr Then the speed of the downstream = (12 + y)km/hr The speed of the upstream = (12 – y)km/hr 9 (12 + y) = 27 (12 – y) 108 + 9y = 324 – 27y 27y + 9y = 324 - 108 36y = 216 y = 6 km/hr

A boat takes 38 hrs for travelling downstream from point p to point q and coming back to a point r midway between p and q. If the velocity of the stream is 8 km/hr and the speed of the boat in still water is 28 km/hr. What is the distance between p and q? A) 720 B) 640 C) 510 D) 450

Description : A boat takes 38 hrs for travelling downstream from point p to point q and coming back to a point r midway between p and q. If the velocity of the stream is 8 km/hr and the speed of the boat in still water is 28 km/hr. What is the distance between p and q? A) 720 B) 640 C) 510 D) 450

Last Answer : ANSWER: A Explanation: Speed downstream = (28 + 8)km/hr = 36 km/hr Speed upstream = (28 – 8)km/hr = 20 km/hr Let the distance between p and q be ‘x’km Then, x/36 + (x/2)/20 = 38 x/36 +x/40 = 38 19x = 13680 X = 720 km Therefore the distance between p and q is 720km

Light from the Sun reaches us in nearly (1) 8 min (2) 2 min (3) 6 min (4) 4 min

Description : Light from the Sun reaches us in nearly (1) 8 min (2) 2 min (3) 6 min (4) 4 min

Last Answer : (1) 8 min Explanation: The sun's light takes about 8 minutes to reach the Earth after it has been emitted from the sun's surface. The time it takes for light to reach planets in our Solar System varies from about 3 minutes for Mercury, to about 5.3 hours for Pluto.

Light from the Sun reaches us in nearly (1) 8 min (2) 2 min (3) 6 min (4) 4 min

Description : Light from the Sun reaches us in nearly (1) 8 min (2) 2 min (3) 6 min (4) 4 min

Last Answer : 8 min

In man, the amount of calcium in gms filtered in 24 hrs period by the renal glomeruli is (A) 5 (B) 10 (C) 15 (D) 20

Description : In man, the amount of calcium in gms filtered in 24 hrs period by the renal glomeruli is (A) 5 (B) 10 (C) 15 (D) 20

Last Answer : Answer : B

A bus starts at bus stop and reaches a destination in 4 hours. If it travels first and second half at the speed of 40 km/hr and 50 km/hr respectively then the distance between bus stop and destination is a) 177.7 km b) 154.2 km c) 163.5 km d) 147.3 km

Description : A bus starts at bus stop and reaches a destination in 4 hours. If it travels first and second half at the speed of 40 km/hr and 50 km/hr respectively then the distance between bus stop and destination is a) 177.7 km b) 154.2 km c) 163.5 km d) 147.3 km

Last Answer : A Let the distance between bus stop and destination be X. The total time taken by the bus to cover X = 4 hours Since X/2 by 40km/hr and remaining X/2 by 50km/hr Then by Time = distance/speed, we have ... = 1600 9X = 1600 X = 1600/9 = 177.7 Hence 177.7 km is the required answer.

One tonrefrigeration corresponds to (a) 50 kcal/min (b) 50 kcal/kr (c) 80 kcal/min (d) 80 kcal/hr

Description : One tonrefrigeration corresponds to (a) 50 kcal/min (b) 50 kcal/kr (c) 80 kcal/min (d) 80 kcal/hr

Last Answer : Ans: a

What is the Time Scale for Meso Climate? [a] 1-10 hr [b] 1-10 min [c] 1-10 seconds [d] 1-10 days [e] None of the above

Description : What is the Time Scale for Meso Climate? [a] 1-10 hr [b] 1-10 min [c] 1-10 seconds [d] 1-10 days [e] None of the above

Last Answer : [d] 1-10 days

We have defined EER by a) BTU/hr/W c) BTU/min/W b) BTU/sec/W d) BTU/hr/V

Description : We have defined EER by a) BTU/hr/W c) BTU/min/W b) BTU/sec/W d) BTU/hr/V

Last Answer : BTU/hr/W

Which hormone level reaches peak during luteal phase of menstrual cycle?

Description : Which hormone level reaches peak during luteal phase of menstrual cycle?

Last Answer : Which hormone level reaches peak during luteal phase of menstrual cycle? A. Luteinising hormone B ... C. Pollicle stimulating hormone D. Estrogen.

The net head under which the turbine reaches its peak efficiency at synchronous speed is called (A) Design head (B) Rated head (C) Gross head (D) Operating head

Description : The net head under which the turbine reaches its peak efficiency at synchronous speed is called (A) Design head (B) Rated head (C) Gross head (D) Operating head

Last Answer : Answer: Option A

The net speed under which the turbine reaches its peak efficiency is called (A) Design speed (B) Rated speed (C) Gross speed (D) Operating speed

Description : The net speed under which the turbine reaches its peak efficiency is called (A) Design speed (B) Rated speed (C) Gross speed (D) Operating speed

Last Answer : Answer: Option A

If worker takes 15 min as a standard time for as job in which total allowance is 20% of normal time. If the rating of the worker is 100 %. Find the actual time required by the worker

Description : If worker takes 15 min as a standard time for as job in which total allowance is 20% of normal time. If the rating of the worker is 100 %. Find the actual time required by the worker

Last Answer : Standard Time (ST) = 15 Minutes Rating Factor (RF) = 100 % Allowance = 20 % of Normal Time (NT) Standard Time (ST) = Normal Time (NT) + Allowance 15 = NT + (20/100) NT 15 = ... (Actual) Time Actual (Observed) Time (AT) required by the worker to complete the job is 12.5 Minutes.

An intense rain is falling at a uniform rate of 7.5 cm/hour for a period of 60 minutes on a basin whose area is 500 hectares. If the average infiltration capacity during the entire rain period is assumed to be 1.5 cm/hr, the maximum run-off rate based on 10 minute peak percentage of 16% from distributing graph of the basin, is (A) 40 cumecs (B) 60 cumecs (C) 80 cumecs (D) 100 cumecs

Description : An intense rain is falling at a uniform rate of 7.5 cm/hour for a period of 60 minutes on a basin whose area is 500 hectares. If the average infiltration capacity during the entire rain period is assumed to be 1.5 ... of the basin, is (A) 40 cumecs (B) 60 cumecs (C) 80 cumecs (D) 100 cumecs

Last Answer : Answer: Option C

Glucose tolerance test?

Description : Glucose tolerance test?

Last Answer : DefinitionThe glucose tolerance test is a laboratory method to check how the body breaks down (metabolizes) sugar.Alternative NamesOral glucose tolerance testHow the test is performedThe most common glucose tolerance ... the injection, and again at 1 and 3 minutes after the injection, although th

Diazoxide is an effective hypotensive, but is not used in the long-term treatment of hypertension because: A. It impairs glucose tolerance by inhibiting insulin release B. It inhibits uric acid excretion C. It causes marked Na+ and water retention leading to development of tolerance D. All of the above

Description : Diazoxide is an effective hypotensive, but is not used in the long-term treatment of hypertension because: A. It impairs glucose tolerance by inhibiting insulin release B. It inhibits uric acid ... causes marked Na+ and water retention leading to development of tolerance D. All of the above

Last Answer : D. All of the above

Metabolic actions of estrogens tend to cause the following except: A. Anabolism B. Impaired glucose tolerance C. Rise in plasma LDL-cholesterol D. Salt and water retention

Description : Metabolic actions of estrogens tend to cause the following except: A. Anabolism B. Impaired glucose tolerance C. Rise in plasma LDL-cholesterol D. Salt and water retention

Last Answer : C. Rise in plasma LDL-cholesterol